# Thermodynamics An Engineering Approach 7th Edition Solution Manual

##### 1-1 Solutions Manual for Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McG

32,986 26,023 19MB

Pages 2070 Page size 612 x 792 pts (letter) Year 2011

##### Citation preview

1-1

Solutions Manual for

Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011

Chapter 1 INTRODUCTION AND BASIC CONCEPTS

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-2 Thermodynamics 1-1C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of energy principle.

1-2C A car going uphill without the engine running would increase the energy of the car, and thus it would be a violation of the first law of thermodynamics. Therefore, this cannot happen. Using a level meter (a device with an air bubble between two marks of a horizontal water tube) it can shown that the road that looks uphill to the eye is actually downhill.

1-3C There is no truth to his claim. It violates the second law of thermodynamics.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-3 Mass, Force, and Units 1-4C The “pound” mentioned here must be “lbf” since thrust is a force, and the lbf is the force unit in the English system. You should get into the habit of never writing the unit “lb”, but always use either “lbm” or “lbf” as appropriate since the two units have different dimensions.

1-5C In this unit, the word light refers to the speed of light. The light-year unit is then the product of a velocity and time. Hence, this product forms a distance dimension and unit.

1-6C There is no acceleration, thus the net force is zero in both cases.

1-7E The weight of a man on earth is given. His weight on the moon is to be determined. Analysis Applying Newton's second law to the weight force gives

W = mg ⎯ ⎯→ m =

210 lbf W = g 32.10 ft/s 2

⎛ 32.174 lbm ⋅ ft/s 2 ⎜ ⎜ 1 lbf ⎝

⎞ ⎟ = 210.5 lbm ⎟ ⎠

Mass is invariant and the man will have the same mass on the moon. Then, his weight on the moon will be 1 lbf ⎞ ⎛ W = mg = (210.5 lbm)(5.47 ft/s 2 )⎜ ⎟ = 35.8 lbf 2 ⎝ 32.174 lbm ⋅ ft/s ⎠

1-8 The interior dimensions of a room are given. The mass and weight of the air in the room are to be determined. Assumptions The density of air is constant throughout the room. Properties The density of air is given to be ρ = 1.16 kg/m3. Analysis The mass of the air in the room is 3

3

m = ρV = (1.16 kg/m )(6 × 6 × 8 m ) = 334.1 kg

ROOM AIR 6X6X8 m3

Thus, ⎛ 1N W = mg = (334.1 kg)(9.81 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 3277 N ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-4 1-9 The variation of gravitational acceleration above the sea level is given as a function of altitude. The height at which the weight of a body will decrease by 0.5% is to be determined. z Analysis The weight of a body at the elevation z can be expressed as W = mg = m(9.807 − 3.32 × 10−6 z )

In our case, W = 0.995W s = 0.995mg s = 0.995(m)(9.81)

Substituting, 0.995(9.81) = (9.81 − 3.32 × 10

0 −6

z) ⎯ ⎯→ z = 14,774 m ≅ 14,770 m

Sea level

1-10 The mass of an object is given. Its weight is to be determined. Analysis Applying Newton's second law, the weight is determined to be

W = mg = (200 kg)(9.6 m/s 2 ) = 1920 N

1-11E The constant-pressure specific heat of air given in a specified unit is to be expressed in various units. Analysis Applying Newton's second law, the weight is determined in various units to be ⎛ 1 kJ/kg ⋅ K ⎞ ⎟⎟ = 1.005 kJ/kg ⋅ K c p = (1.005 kJ/kg ⋅ °C)⎜⎜ ⎝ 1 kJ/kg ⋅ °C ⎠ ⎛ 1000 J ⎞⎛ 1 kg ⎞ ⎟⎟ = 1.005 J/g ⋅ °C c p = (1.005 kJ/kg ⋅ °C)⎜ ⎟⎜⎜ ⎝ 1 kJ ⎠⎝ 1000 g ⎠ ⎛ 1 kcal ⎞ c p = (1.005 kJ/kg ⋅ °C)⎜ ⎟ = 0.240 kcal/kg ⋅ °C ⎝ 4.1868 kJ ⎠ ⎛ 1 Btu/lbm ⋅ °F ⎞ ⎟⎟ = 0.240 Btu/lbm ⋅ °F c p = (1.005 kJ/kg ⋅ °C)⎜⎜ ⎝ 4.1868 kJ/kg ⋅ °C ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-5 1-12

A rock is thrown upward with a specified force. The acceleration of the rock is to be determined.

Analysis The weight of the rock is ⎛ 1N W = mg = (3 kg)(9.79 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 29.37 N ⎟ ⎠

Then the net force that acts on the rock is

Fnet = Fup − Fdown = 200 − 29.37 = 170.6 N

Stone

From the Newton's second law, the acceleration of the rock becomes

a=

F 170.6 N ⎛⎜ 1 kg ⋅ m/s 2 = m 3 kg ⎜⎝ 1 N

⎞ ⎟ = 56.9 m/s 2 ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-6 1-13 Problem 1-12 is reconsidered. The entire EES solution is to be printed out, including the numerical results with proper units. Analysis The problem is solved using EES, and the solution is given below. "The weight of the rock is" W=m*g m=3 [kg] g=9.79 [m/s2] "The force balance on the rock yields the net force acting on the rock as" F_up=200 [N] F_net = F_up - F_down F_down=W "The acceleration of the rock is determined from Newton's second law." F_net=m*a "To Run the program, press F2 or select Solve from the Calculate menu." SOLUTION a=56.88 [m/s^2] F_down=29.37 [N] F_net=170.6 [N] F_up=200 [N] g=9.79 [m/s2] m=3 [kg] W=29.37 [N] 200

160

2

a [m/s2] 190.2 90.21 56.88 40.21 30.21 23.54 18.78 15.21 12.43 10.21

a [m/s ]

m [kg] 1 2 3 4 5 6 7 8 9 10

120

80

40

0 1

2

3

4

5

6

m [kg]

7

8

9

10

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-7 1-14 During an analysis, a relation with inconsistent units is obtained. A correction is to be found, and the probable cause of the error is to be determined. Analysis The two terms on the right-hand side of the equation

E = 25 kJ + 7 kJ/kg do not have the same units, and therefore they cannot be added to obtain the total energy. Multiplying the last term by mass will eliminate the kilograms in the denominator, and the whole equation will become dimensionally homogeneous; that is, every term in the equation will have the same unit. Discussion Obviously this error was caused by forgetting to multiply the last term by mass at an earlier stage.

1-15 A resistance heater is used to heat water to desired temperature. The amount of electric energy used in kWh and kJ are to be determined. Analysis The resistance heater consumes electric energy at a rate of 4 kW or 4 kJ/s. Then the total amount of electric energy used in 2 hours becomes

Total energy = (Energy per unit time)(Time interval) = (4 kW)(2 h) = 8 kWh Noting that 1 kWh = (1 kJ/s)(3600 s) = 3600 kJ, Total energy = (8 kWh)(3600 kJ/kWh) = 28,800 kJ Discussion Note kW is a unit for power whereas kWh is a unit for energy.

1-16 A gas tank is being filled with gasoline at a specified flow rate. Based on unit considerations alone, a relation is to be obtained for the filling time. Assumptions Gasoline is an incompressible substance and the flow rate is constant. Analysis The filling time depends on the volume of the tank and the discharge rate of gasoline. Also, we know that the unit of time is ‘seconds’. Therefore, the independent quantities should be arranged such that we end up with the unit of seconds. Putting the given information into perspective, we have

t [s] ↔

V [L],

and V& [L/s}

It is obvious that the only way to end up with the unit “s” for time is to divide the tank volume by the discharge rate. Therefore, the desired relation is

t=

V V&

Discussion Note that this approach may not work for cases that involve dimensionless (and thus unitless) quantities.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-8 1-17 A pool is to be filled with water using a hose. Based on unit considerations, a relation is to be obtained for the volume of the pool. Assumptions Water is an incompressible substance and the average flow velocity is constant. Analysis The pool volume depends on the filling time, the cross-sectional area which depends on hose diameter, and flow velocity. Also, we know that the unit of volume is m3. Therefore, the independent quantities should be arranged such that we end up with the unit of seconds. Putting the given information into perspective, we have

V [m3]

is a function of t [s], D [m], and V [m/s}

It is obvious that the only way to end up with the unit “m3” for volume is to multiply the quantities t and V with the square of D. Therefore, the desired relation is

V = CD2Vt where the constant of proportionality is obtained for a round hose, namely, C =π/4 so that V = (πD2/4)Vt. Discussion Note that the values of dimensionless constants of proportionality cannot be determined with this approach.

1-18 It is to be shown that the power needed to accelerate a car is proportional to the mass and the square of the velocity of the car, and inversely proportional to the time interval. Assumptions The car is initially at rest. Analysis The power needed for acceleration depends on the mass, velocity change, and time interval. Also, the unit of power W& is watt, W, which is equivalent to

W = J/s = N⋅m/s = (kg⋅m/s2)m/s = kg⋅m2/s3 Therefore, the independent quantities should be arranged such that we end up with the unit kg⋅m2/s3 for power. Putting the given information into perspective, we have W& [ kg⋅m2/s3] is a function of m [kg], V [m/s], and t [s]

It is obvious that the only way to end up with the unit “kg⋅m2/s3” for power is to multiply mass with the square of the velocity and divide by time. Therefore, the desired relation is

W& is proportional to mV 2 / t or, W& = CmV 2 / t

where C is the dimensionless constant of proportionality (whose value is ½ in this case). Discussion Note that this approach cannot determine the numerical value of the dimensionless numbers involved.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-9 Systems, Properties, State, and Processes 1-19C This system is a region of space or open system in that mass such as air and food can cross its control boundary. The system can also interact with the surroundings by exchanging heat and work across its control boundary. By tracking these interactions, we can determine the energy conversion characteristics of this system.

1-20C The system is taken as the air contained in the piston-cylinder device. This system is a closed or fixed mass system since no mass enters or leaves it.

1-21C Any portion of the atmosphere which contains the ozone layer will work as an open system to study this problem. Once a portion of the atmosphere is selected, we must solve the practical problem of determining the interactions that occur at the control surfaces which surround the system's control volume.

1-22C Intensive properties do not depend on the size (extent) of the system but extensive properties do.

1-23C If we were to divide the system into smaller portions, the weight of each portion would also be smaller. Hence, the weight is an extensive property.

1-24C If we were to divide this system in half, both the volume and the number of moles contained in each half would be one-half that of the original system. The molar specific volume of the original system is

v =

V N

and the molar specific volume of one of the smaller systems is

v =

V/ 2 V = N /2 N

which is the same as that of the original system. The molar specific volume is then an intensive property.

1-25C For a system to be in thermodynamic equilibrium, the temperature has to be the same throughout but the pressure does not. However, there should be no unbalanced pressure forces present. The increasing pressure with depth in a fluid, for example, should be balanced by increasing weight.

1-26C A process during which a system remains almost in equilibrium at all times is called a quasi-equilibrium process. Many engineering processes can be approximated as being quasi-equilibrium. The work output of a device is maximum and the work input to a device is minimum when quasi-equilibrium processes are used instead of nonquasi-equilibrium processes.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-10 1-27C A process during which the temperature remains constant is called isothermal; a process during which the pressure remains constant is called isobaric; and a process during which the volume remains constant is called isochoric.

1-28C The state of a simple compressible system is completely specified by two independent, intensive properties.

1-29C The pressure and temperature of the water are normally used to describe the state. Chemical composition, surface tension coefficient, and other properties may be required in some cases.

As the water cools, its pressure remains fixed. This cooling process is then an isobaric process.

1- 30C When analyzing the acceleration of gases as they flow through a nozzle, the proper choice for the system is the volume within the nozzle, bounded by the entire inner surface of the nozzle and the inlet and outlet cross-sections. This is a control volume since mass crosses the boundary.

1-31C A process is said to be steady-flow if it involves no changes with time anywhere within the system or at the system boundaries.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-11 1-32 The variation of density of atmospheric air with elevation is given in tabular form. A relation for the variation of density with elevation is to be obtained, the density at 7 km elevation is to be calculated, and the mass of the atmosphere using the correlation is to be estimated. Assumptions 1 Atmospheric air behaves as an ideal gas. 2 The earth is perfectly sphere with a radius of 6377 km, and the thickness of the atmosphere is 25 km. Properties The density data are given in tabular form as 1.4 1.2 1 3

ρ, kg/m3 1.225 1.112 1.007 0.9093 0.8194 0.7364 0.6601 0.5258 0.4135 0.1948 0.08891 0.04008

0.8

ρ, kg/m

z, km 0 1 2 3 4 5 6 8 10 15 20 25

r, km 6377 6378 6379 6380 6381 6382 6383 6385 6387 6392 6397 6402

0.6 0.4 0.2 0 0

5

10

15

20

25

z, km

Analysis Using EES, (1) Define a trivial function rho= a+z in equation window, (2) select new parametric table from Tables, and type the data in a two-column table, (3) select Plot and plot the data, and (4) select plot and click on “curve fit” to get curve fit window. Then specify 2nd order polynomial and enter/edit equation. The results are: ρ(z) = a + bz + cz2 = 1.20252 – 0.101674z + 0.0022375z2

for the unit of kg/m3,

(or, ρ(z) = (1.20252 – 0.101674z + 0.0022375z2)×109 for the unit of kg/km3) where z is the vertical distance from the earth surface at sea level. At z = 7 km, the equation would give ρ = 0.60 kg/m3. (b) The mass of atmosphere can be evaluated by integration to be m=

V

ρdV =

h

z =0

(a + bz + cz 2 )4π (r0 + z ) 2 dz = 4π

[

h

z =0

(a + bz + cz 2 )(r02 + 2r0 z + z 2 )dz

= 4π ar02 h + r0 (2a + br0 )h 2 / 2 + (a + 2br0 + cr02 )h 3 / 3 + (b + 2cr0 )h 4 / 4 + ch 5 / 5

]

where r0 = 6377 km is the radius of the earth, h = 25 km is the thickness of the atmosphere, and a = 1.20252, b = 0.101674, and c = 0.0022375 are the constants in the density function. Substituting and multiplying by the factor 109 for the density unity kg/km3, the mass of the atmosphere is determined to be m = 5.092×1018 kg Discussion Performing the analysis with excel would yield exactly the same results.

EES Solution for final result: a=1.2025166;

b=-0.10167

c=0.0022375;

r=6377;

h=25

m=4*pi*(a*r^2*h+r*(2*a+b*r)*h^2/2+(a+2*b*r+c*r^2)*h^3/3+(b+2*c*r)*h^4/4+c*h^5/5)*1E+9

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-12 Temperature 1-33C The zeroth law of thermodynamics states that two bodies are in thermal equilibrium if both have the same temperature reading, even if they are not in contact.

1-34C They are Celsius (°C) and kelvin (K) in the SI, and fahrenheit (°F) and rankine (R) in the English system.

1-35C Probably, but not necessarily. The operation of these two thermometers is based on the thermal expansion of a fluid. If the thermal expansion coefficients of both fluids vary linearly with temperature, then both fluids will expand at the same rate with temperature, and both thermometers will always give identical readings. Otherwise, the two readings may deviate.

1-36 A temperature is given in °C. It is to be expressed in K. Analysis The Kelvin scale is related to Celsius scale by

T(K] = T(°C) + 273 Thus, T(K] = 37°C + 273 = 310 K

1-37E The temperature of air given in °C unit is to be converted to °F and R unit. Analysis Using the conversion relations between the various temperature scales,

T (°F) = 1.8T (°C) + 32 = (1.8)(150) + 32 = 302°F T (R ) = T (°F) + 460 = 302 + 460 = 762 R

1-38 A temperature change is given in °C. It is to be expressed in K. Analysis This problem deals with temperature changes, which are identical in Kelvin and Celsius scales. Thus,

∆T(K] = ∆T(°C) = 45 K

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-13 1-39E The flash point temperature of engine oil given in °F unit is to be converted to K and R units. Analysis Using the conversion relations between the various temperature scales,

T (R ) = T (°F) + 460 = 363 + 460 = 823 R T (K ) =

T (R ) 823 = = 457 K 1.8 1.8

1-40E The temperature of ambient air given in °C unit is to be converted to °F, K and R units. Analysis Using the conversion relations between the various temperature scales, T = −40°C = (−40)(1.8) + 32 = −40°C T = −40 + 273.15 = 233.15 K T = −40 + 459.67 = 419.67 R

1-41E The change in water temperature given in °F unit is to be converted to °C, K and R units. Analysis Using the conversion relations between the various temperature scales, ∆T = 10 / 1.8 = 5.6°C ∆T = 10 / 1.8 = 5.6 K ∆T = 10°F = 10 R

1-42E A temperature range given in °F unit is to be converted to °C unit and the temperature difference in °F is to be expressed in K, °C, and R. Analysis The lower and upper limits of comfort range in °C are T (°C) =

T (°F) − 32 65 − 32 = = 18.3 °C 1. 8 1. 8

T (°C) =

T (°F) − 32 75 − 32 = = 23.9 °C 1. 8 1. 8

A temperature change of 10°F in various units are ∆T (R ) = ∆T (°F) = 10 R ∆T (°F) 10 ∆T (°C) = = = 5.6°C 1.8 1 .8 ∆T (K ) = ∆T (°C) = 5.6 K

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-14 Pressure, Manometer, and Barometer 1-43C The pressure relative to the atmospheric pressure is called the gage pressure, and the pressure relative to an absolute vacuum is called absolute pressure.

1-44C The blood vessels are more restricted when the arm is parallel to the body than when the arm is perpendicular to the body. For a constant volume of blood to be discharged by the heart, the blood pressure must increase to overcome the increased resistance to flow.

1-45C No, the absolute pressure in a liquid of constant density does not double when the depth is doubled. It is the gage pressure that doubles when the depth is doubled.

1-46C If the lengths of the sides of the tiny cube suspended in water by a string are very small, the magnitudes of the pressures on all sides of the cube will be the same.

1-47C Pascal’s principle states that the pressure applied to a confined fluid increases the pressure throughout by the same amount. This is a consequence of the pressure in a fluid remaining constant in the horizontal direction. An example of Pascal’s principle is the operation of the hydraulic car jack.

1-48E The pressure given in psia unit is to be converted to kPa. Analysis Using the psia to kPa units conversion factor, ⎛ 6.895 kPa ⎞ ⎟⎟ = 1034 kPa P = (150 psia )⎜⎜ ⎝ 1 psia ⎠

1-49 The pressure in a tank is given. The tank's pressure in various units are to be determined. Analysis Using appropriate conversion factors, we obtain

(a)

⎛ 1 kN/m 2 P = (1500 kPa )⎜ ⎜ 1 kPa ⎝

⎞ ⎟ = 1500 kN/m 2 ⎟ ⎠

(b)

⎛ 1 kN/m 2 P = (1500 kPa )⎜ ⎜ 1 kPa ⎝

⎞⎛ 1000 kg ⋅ m/s 2 ⎟⎜ ⎟⎜ 1 kN ⎠⎝

⎞ ⎟ = 1,500,000 kg/m ⋅ s 2 ⎟ ⎠

(c)

⎛ 1 kN/m 2 P = (1500 kPa )⎜ ⎜ 1 kPa ⎝

⎞⎛ 1000 kg ⋅ m/s 2 ⎟⎜ ⎟⎜ 1 kN ⎠⎝

⎞⎛ 1000 m ⎞ ⎟⎜ = 1,500,000,000 kg/km ⋅ s 2 ⎟⎝ 1 km ⎟⎠ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-15 1-50E The pressure in a tank in SI unit is given. The tank's pressure in various English units are to be determined. Analysis Using appropriate conversion factors, we obtain

(a)

⎛ 20.886 lbf/ft 2 P = (1500 kPa )⎜ ⎜ 1 kPa ⎝

⎞ ⎟ = 31,330 lbf/ft 2 ⎟ ⎠

(b)

⎛ 20.886 lbf/ft 2 P = (1500 kPa )⎜ ⎜ 1 kPa ⎝

⎞⎛ 1 ft 2 ⎟⎜ ⎟⎜ 144 in 2 ⎠⎝

⎞⎛ 1 psia ⎞ ⎟⎜ = 217.6 psia ⎟⎝ 1 lbf/in 2 ⎟⎠ ⎠

1-51E The pressure given in mm Hg unit is to be converted to psia. Analysis Using the mm Hg to kPa and kPa to psia units conversion factors, ⎛ 0.1333 kPa ⎞⎛ 1 psia ⎞ ⎟⎟⎜ P = (1500 mm Hg)⎜⎜ ⎟ = 29.0 psia ⎝ 1 mm Hg ⎠⎝ 6.895 kPa ⎠

1-52 The pressure given in mm Hg unit is to be converted to kPa. Analysis Using the mm Hg to kPa units conversion factor, ⎛ 0.1333 kPa ⎞ ⎟⎟ = 166.6 kPa P = (1250 mm Hg)⎜⎜ ⎝ 1 mm Hg ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-16 1-53 The pressure in a pressurized water tank is measured by a multi-fluid manometer. The gage pressure of air in the tank is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus we can determine the pressure at the air-water interface. Properties The densities of mercury, water, and oil are given to be 13,600, 1000, and 850 kg/m3, respectively. Analysis Starting with the pressure at point 1 at the air-water interface, and moving along the tube by adding (as we go down) or subtracting (as we go up) th e ρgh terms until we reach point 2, and setting the result equal to Patm since the tube is open to the atmosphere gives

P1 + ρ water gh1 + ρ oil gh2 − ρ mercury gh3 = Patm Solving for P1,

P1 = Patm − ρ water gh1 − ρ oil gh2 + ρ mercury gh3 or,

P1 − Patm = g ( ρ mercury h3 − ρ water h1 − ρ oil h2 ) Noting that P1,gage = P1 - Patm and substituting, P1,gage = (9.81 m/s 2 )[(13,600 kg/m 3 )(0.46 m) − (1000 kg/m 3 )(0.2 m) ⎛ 1N − (850 kg/m 3 )(0.3 m)]⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ = 56.9 kPa

⎞⎛ 1 kPa ⎞ ⎟⎜ ⎟⎝ 1000 N/m 2 ⎟⎠ ⎠

Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly.

1-54 The barometric reading at a location is given in height of mercury column. The atmospheric pressure is to be determined. Properties The density of mercury is given to be 13,600 kg/m3. Analysis The atmospheric pressure is determined directly from Patm = ρgh ⎛ 1N = (13,600 kg/m 3 )(9.81 m/s 2 )(0.750 m)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ = 100.1 kPa

⎞⎛ 1 kPa ⎞ ⎟⎜ ⎟⎝ 1000 N/m 2 ⎟⎠ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-17 1-55 The gage pressure in a liquid at a certain depth is given. The gage pressure in the same liquid at a different depth is to be determined. Assumptions The variation of the density of the liquid with depth is negligible. Analysis The gage pressure at two different depths of a liquid can be expressed as P1 = ρgh1

and

P2 = ρgh2

Taking their ratio,

h1

P2 ρgh2 h2 = = P1 ρgh1 h1

1

h2

Solving for P2 and substituting gives

2

h 9m P2 = 2 P1 = (42 kPa) = 126 kPa h1 3m

Discussion Note that the gage pressure in a given fluid is proportional to depth.

1-56 The absolute pressure in water at a specified depth is given. The local atmospheric pressure and the absolute pressure at the same depth in a different liquid are to be determined. Assumptions The liquid and water are incompressible. Properties The specific gravity of the fluid is given to be SG = 0.85. We take the density of water to be 1000 kg/m3. Then density of the liquid is obtained by multiplying its specific gravity by the density of water,

ρ = SG × ρ H 2O = (0.85)(100 0 kg/m 3 ) = 850 kg/m 3 Analysis (a) Knowing the absolute pressure, the atmospheric pressure can be determined from Patm = P − ρgh

⎛ 1 kPa = (145 kPa) − (1000 kg/m 3 )(9.81 m/s 2 )(5 m)⎜ ⎜ 1000 N/m 2 ⎝ = 96.0 kPa

⎞ ⎟ ⎟ ⎠

Patm h P

(b) The absolute pressure at a depth of 5 m in the other liquid is P = Patm + ρgh

⎛ 1 kPa = (96.0 kPa) + (850 kg/m 3 )(9.81 m/s 2 )(5 m)⎜ ⎜ 1000 N/m 2 ⎝ = 137.7 kPa

⎞ ⎟ ⎟ ⎠

Discussion Note that at a given depth, the pressure in the lighter fluid is lower, as expected.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-18 2

1-57E It is to be shown that 1 kgf/cm = 14.223 psi . Analysis Noting that 1 kgf = 9.80665 N, 1 N = 0.22481 lbf, and 1 in = 2.54 cm, we have ⎛ 0.22481 lbf 1 kgf = 9.80665 N = (9.80665 N )⎜⎜ 1N ⎝

⎞ ⎟⎟ = 2.20463 lbf ⎠

and 2

⎛ 2.54 cm ⎞ ⎟⎟ = 14.223 lbf/in 2 = 14.223 psi 1 kgf/cm 2 = 2.20463 lbf/cm 2 = (2.20463 lbf/cm 2 )⎜⎜ 1 in ⎝ ⎠

1-58E The pressure in chamber 3 of the two-piston cylinder shown in the figure is to be determined. Analysis The area upon which pressure 1 acts is

A1 = π

D12 (3 in) 2 =π = 7.069 in 2 4 4

F2

and the area upon which pressure 2 acts is

F3

D2 (1.5 in) 2 A2 = π 2 = π = 1.767 in 2 4 4 The area upon which pressure 3 acts is given by

A3 = A1 − A2 = 7.069 − 1.767 = 5.302 in 2 The force produced by pressure 1 on the piston is then ⎛ 1 lbf/in F1 = P1 A1 = (150 psia )⎜ ⎜ 1 psia ⎝

2

F1

⎞ ⎟(7.069 in 2 ) = 1060 lbf ⎟ ⎠

while that produced by pressure 2 is

F1 = P2 A2 = (250 psia )(1.767 in 2 ) = 441.8 lbf According to the vertical force balance on the piston free body diagram F3 = F1 − F2 = 1060 − 441.8 = 618.3 lbf

Pressure 3 is then P3 =

F3 618.3 lbf = = 117 psia A3 5.302 in 2

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-19 1-59 The pressure in chamber 1 of the two-piston cylinder shown in the figure is to be determined. Analysis Summing the forces acting on the piston in the vertical direction gives F2 + F3 = F1 P2 A2 + P3 ( A1 − A2 ) = P1 A1

F2

which when solved for P1 gives P1 = P2

F3

⎛ A ⎞ A2 + P3 ⎜⎜1 − 2 ⎟⎟ A1 ⎠ A1 ⎝

since the areas of the piston faces are given by A = πD 2 / 4 the above equation becomes ⎛D P1 = P2 ⎜⎜ 2 ⎝ D1

2 ⎡ ⎛D ⎞ ⎟⎟ + P3 ⎢1 − ⎜⎜ 2 ⎢ ⎝ D1 ⎠ ⎣

⎞ ⎟⎟ ⎠

2⎤

⎥ ⎥ ⎦

F1

2 ⎡ ⎛ 4 ⎞2 ⎤ ⎛ 4⎞ = (2000 kPa)⎜ ⎟ + (700 kPa) ⎢1 − ⎜ ⎟ ⎥ ⎝ 10 ⎠ ⎢⎣ ⎝ 10 ⎠ ⎥⎦

= 908 kPa

1-60 The mass of a woman is given. The minimum imprint area per shoe needed to enable her to walk on the snow without sinking is to be determined. Assumptions 1 The weight of the person is distributed uniformly on the imprint area of the shoes. 2 One foot carries the entire weight of a person during walking, and the shoe is sized for walking conditions (rather than standing). 3 The weight of the shoes is negligible. Analysis The mass of the woman is given to be 70 kg. For a pressure of 0.5 kPa on the snow, the imprint area of one shoe must be A= =

W mg = P P (70 kg)(9.81 m/s 2 ) ⎛⎜ 1N ⎜ 1 kg ⋅ m/s 2 0.5 kPa ⎝

⎞⎛ 1 kPa ⎞ ⎟⎜ = 1.37 m 2 ⎟⎝ 1000 N/m 2 ⎟⎠ ⎠

Discussion This is a very large area for a shoe, and such shoes would be impractical to use. Therefore, some sinking of the snow should be allowed to have shoes of reasonable size.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-20 1-61 The vacuum pressure reading of a tank is given. The absolute pressure in the tank is to be determined. Properties The density of mercury is given to be ρ = 13,590 kg/m3. Analysis The atmospheric (or barometric) pressure can be expressed as Patm = ρ g h

⎛ 1N = (13,590 kg/m )(9.807 m/s )(0.750 m)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ = 100.0 kPa 3

2

⎞⎛ 1 kPa ⎟⎜ ⎟⎜ 1000 N/m 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

Then the absolute pressure in the tank becomes

Pabs

30 kPa

Patm = 750 mmHg

Pabs = Patm − Pvac = 100.0 − 30 = 70.0 kPa

1-62E The vacuum pressure given in kPa unit is to be converted to various units. Analysis Using the definition of vacuum pressure, Pgage = not applicable for pressures below atmospheric pressure Pabs = Patm − Pvac = 98 − 80 = 18 kPa

Then using the conversion factors,

⎛ 1 kN/m 2 Pabs = (18 kPa)⎜⎜ ⎝ 1 kPa

⎞ ⎟ = 18 kN/m2 ⎟ ⎠

⎛ 1 lbf/in 2 ⎞ ⎟ = 2.61 lbf/in2 Pabs = (18 kPa)⎜ ⎜ 6.895 kPa ⎟ ⎠ ⎝ ⎛ 1 psi ⎞ Pabs = (18 kPa)⎜ ⎟ = 2.61 psi ⎝ 6.895 kPa ⎠ ⎛ 1 mm Hg ⎞ Pabs = (18 kPa)⎜ ⎟ = 135 mm Hg ⎝ 0.1333 kPa ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-21 1-63 A mountain hiker records the barometric reading before and after a hiking trip. The vertical distance climbed is to be determined.

630 mbar

Assumptions The variation of air density and the gravitational acceleration with altitude is negligible. Properties The density of air is given to be ρ = 1.20 kg/m3.

h=?

Analysis Taking an air column between the top and the bottom of the mountain and writing a force balance per unit base area, we obtain Wair / A = Pbottom − Ptop

740 mbar

( ρgh) air = Pbottom − Ptop ⎛ 1N (1.20 kg/m 3 )(9.81 m/s 2 )(h)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞⎛ 1 bar ⎟⎜ ⎟⎜ 100,000 N/m 2 ⎠⎝

⎞ ⎟ = (0.740 − 0.630) bar ⎟ ⎠

It yields h = 934 m which is also the distance climbed.

1-64 A barometer is used to measure the height of a building by recording reading at the bottom and at the top of the building. The height of the building is to be determined. Assumptions The variation of air density with altitude is negligible. Properties The density of air is given to be ρ = 1.18 kg/m3. The density of mercury is 13,600 kg/m3.

675 mmHg

Analysis Atmospheric pressures at the top and at the bottom of the building are Ptop = ( ρ g h) top

⎛ 1N = (13,600 kg/m 3 )(9.81 m/s 2 )(0.675 m)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ = 90.06 kPa

Pbottom = ( ρ g h) bottom

⎛ 1N = (13,600 kg/m 3 )(9.81 m/s 2 )(0.695 m)⎜ ⎜ 1kg ⋅ m/s 2 ⎝ = 92.72 kPa

⎞⎛ 1 kPa ⎟⎜ ⎟⎜ 1000 N/m 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

⎞⎛ 1 kPa ⎟⎜ ⎟⎜ 1000 N/m 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

h

695 mmHg

Taking an air column between the top and the bottom of the building and writing a force balance per unit base area, we obtain

Wair / A = Pbottom − Ptop ( ρgh) air = Pbottom − Ptop ⎛ 1N (1.18 kg/m 3 )(9.81 m/s 2 )(h)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞⎛ 1 kPa ⎟⎜ ⎟⎜ 1000 N/m 2 ⎠⎝

⎞ ⎟ = (92.72 − 90.06) kPa ⎟ ⎠

It yields h = 231 m which is also the height of the building.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-22 1-65 Problem 1-64 is reconsidered. The entire EES solution is to be printed out, including the numerical results with proper units. Analysis The problem is solved using EES, and the solution is given below. P_bottom=695 [mmHg] P_top=675 [mmHg] g=9.81 [m/s^2] "local acceleration of gravity at sea level" rho=1.18 [kg/m^3] DELTAP_abs=(P_bottom-P_top)*CONVERT(mmHg, kPa) "[kPa]" "Delta P reading from the barometers, converted from mmHg to kPa." DELTAP_h =rho*g*h*Convert(Pa, kPa) "Delta P due to the air fluid column height, h, between the top and bottom of the building." DELTAP_abs=DELTAP_h SOLUTION DELTAP_abs=2.666 [kPa] DELTAP_h=2.666 [kPa] g=9.81 [m/s^2] h=230.3 [m] P_bottom=695 [mmHg] P_top=675 [mmHg] rho=1.18 [kg/m^3]

1-66 A man is standing in water vertically while being completely submerged. The difference between the pressures acting on the head and on the toes is to be determined. Assumptions Water is an incompressible substance, and thus the density does not change with depth. Properties We take the density of water to be ρ =1000 kg/m3.

Analysis The pressures at the head and toes of the person can be expressed as Phead = Patm + ρghhead

and

Ptoe = Patm + ρghtoe

where h is the vertical distance of the location in water from the free surface. The pressure difference between the toes and the head is determined by subtracting the first relation above from the second,

htoe

Substituting, ⎛ 1N Ptoe − Phead = (1000 kg/m 3 )(9.81 m/s 2 )(1.75 m - 0)⎜ ⎜ 1kg ⋅ m/s 2 ⎝

⎞⎛ 1kPa ⎟⎜ ⎟⎜ 1000 N/m 2 ⎠⎝

⎞ ⎟ = 17.2 kPa ⎟ ⎠

Discussion This problem can also be solved by noting that the atmospheric pressure (1 atm = 101.325 kPa) is equivalent to 10.3-m of water height, and finding the pressure that corresponds to a water height of 1.75 m.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-23 1-67 A gas contained in a vertical piston-cylinder device is pressurized by a spring and by the weight of the piston. The pressure of the gas is to be determined. Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield

Fspring

PA = Patm A + W + Fspring

Patm

Thus, P = Patm +

mg + Fspring

A (3.2 kg)(9.81 m/s 2 ) + 150 N ⎛⎜ 1 kPa = (95 kPa) + ⎜ 1000 N/m 2 35 × 10 − 4 m 2 ⎝ = 147 kPa

P

⎞ ⎟ ⎟ ⎠

W = mg

Problem 1-67 is reconsidered. The effect of the spring force in the range of 0 to 500 N on the pressure inside 1-68 the cylinder is to be investigated. The pressure against the spring force is to be plotted, and results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. g=9.81 [m/s^2] P_atm= 95 [kPa] m_piston=3.2 [kg] {F_spring=150 [N]} A=35*CONVERT(cm^2, m^2) W_piston=m_piston*g F_atm=P_atm*A*CONVERT(kPa, N/m^2) "From the free body diagram of the piston, the balancing vertical forces yield:" F_gas= F_atm+F_spring+W_piston P_gas=F_gas/A*CONVERT(N/m^2, kPa)

Pgas [kPa] 104 118.3 132.5 146.8 161.1 175.4 189.7 204 218.3 232.5 246.8

260 240 220

Pgas [kPa]

Fspring [N] 0 50 100 150 200 250 300 350 400 450 500

200 180 160 140 120 100 0

100

200

300

400

500

Fspring [N]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-24 1-69 Both a gage and a manometer are attached to a gas to measure its pressure. For a specified reading of gage pressure, the difference between the fluid levels of the two arms of the manometer is to be determined for mercury and water. Properties The densities of water and mercury are given to be ρwater = 1000 kg/m3 and be ρHg = 13,600 kg/m3. Analysis The gage pressure is related to the vertical distance h between the two fluid levels by

⎯→ h = Pgage = ρ g h ⎯

Pgage

ρg

(a) For mercury, h= =

Pgage

ρ Hg g ⎛ 1 kN/m 2 ⎜ (13,600 kg/m 3 )(9.81 m/s 2 ) ⎜⎝ 1 kPa 80 kPa

⎞⎛ 1000 kg/m ⋅ s 2 ⎟⎜ ⎟⎜ 1 kN ⎠⎝

⎞ ⎟ = 0.60 m ⎟ ⎠

(b) For water, h=

Pgage

ρ H 2O g

=

⎛ 1 kN/m 2 ⎜ (1000 kg/m 3 )(9.81 m/s 2 ) ⎜⎝ 1 kPa 80 kPa

⎞⎛ 1000 kg/m ⋅ s 2 ⎟⎜ ⎟⎜ 1 kN ⎠⎝

⎞ ⎟ = 8.16 m ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-25

1-70 Problem 1-69 is reconsidered. The effect of the manometer fluid density in the range of 800 to 13,000 kg/m3 on the differential fluid height of the manometer is to be investigated. Differential fluid height against the density is to be plotted, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Let's modify this problem to also calculate the absolute pressure in the tank by supplying the atmospheric pressure. Use the relationship between the pressure gage reading and the manometer fluid column height. " Function fluid_density(Fluid\$) "This function is needed since if-then-else logic can only be used in functions or procedures. The underscore displays whatever follows as subscripts in the Formatted Equations Window." If fluid\$='Mercury' then fluid_density=13600 else fluid_density=1000 end {Input from the diagram window. If the diagram window is hidden, then all of the input must come from the equations window. Also note that brackets can also denote comments - but these comments do not appear in the formatted equations window.} {Fluid\$='Mercury' P_atm = 101.325 [kPa] DELTAP=80 [kPa] "Note how DELTAP is displayed on the Formatted Equations Window."} g=9.807 [m/s^2] "local acceleration of gravity at sea level" rho=Fluid_density(Fluid\$) "Get the fluid density, either Hg or H2O, from the function" "To plot fluid height against density place {} around the above equation. Then set up the parametric table and solve." DELTAP = RHO*g*h/1000 "Instead of dividiing by 1000 Pa/kPa we could have multiplied by the EES function, CONVERT(Pa,kPa)" h_mm=h*convert(m, mm) "The fluid height in mm is found using the built-in CONVERT function." P_abs= P_atm + DELTAP "To make the graph, hide the diagram window and remove the {}brackets from Fluid\$ and from P_atm. Select New Parametric Table from the Tables menu. Choose P_abs, DELTAP and h to be in the table. Choose Alter Values from the Tables menu. Set values of h to range from 0 to 1 in steps of 0.2. Choose Solve Table (or press F3) from the Calculate menu. Choose New Plot Window from the Plot menu. Choose to plot P_abs vs h and then choose Overlay Plot from the Plot menu and plot DELTAP on the same scale."

Manometer Fluid Height vs Manometer Fluid Density hmm [mm] 10197 3784 2323 1676 1311 1076 913.1 792.8 700.5 627.5

11000 8800

hmm [mm]

ρ [kg/m3] 800 2156 3511 4867 6222 7578 8933 10289 11644 13000

6600 4400 2200 0 0

2000

4000

6000

8000

10000 12000 14000

ρ [kg/m^3] PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-26

1-71 The air pressure in a tank is measured by an oil manometer. For a given oil-level difference between the two columns, the absolute pressure in the tank is to be determined. Properties The density of oil is given to be ρ = 850 kg/m3. Analysis The absolute pressure in the tank is determined from P = Patm + ρgh

⎛ 1kPa = (98 kPa) + (850 kg/m 3 )(9.81m/s 2 )(0.36 m)⎜ ⎜ 1000 N/m 2 ⎝ = 101.0 kPa

⎞ ⎟ ⎟ ⎠

0.36 m

AIR

Patm = 98 kPa

1-72 The air pressure in a duct is measured by a mercury manometer. For a given mercury-level difference between the two columns, the absolute pressure in the duct is to be determined. Properties The density of mercury is given to be ρ = 13,600 kg/m3. Analysis (a) The pressure in the duct is above atmospheric pressure since the fluid column on the duct side is at a lower level.

AIR

(b) The absolute pressure in the duct is determined from P = Patm + ρgh

⎛ 1N = (100 kPa) + (13,600 kg/m 3 )(9.81 m/s 2 )(0.015 m)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ = 102 kPa

15 mm

P ⎞⎛ 1 kPa ⎟⎜ ⎟⎜ 1000 N/m 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

1-73 The air pressure in a duct is measured by a mercury manometer. For a given mercury-level difference between the two columns, the absolute pressure in the duct is to be determined. Properties The density of mercury is given to be ρ = 13,600 kg/m3. Analysis (a) The pressure in the duct is above atmospheric pressure since the fluid column on the duct side is at a lower level.

AIR

(b) The absolute pressure in the duct is determined from P = Patm + ρgh

⎛ 1N = (100 kPa) + (13,600 kg/m 3 )(9.81 m/s 2 )(0.045 m)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ = 106 kPa

45 mm

P ⎞⎛ 1 kPa ⎟⎜ ⎟⎜ 1000 N/m 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-27

1-74E The systolic and diastolic pressures of a healthy person are given in mmHg. These pressures are to be expressed in kPa, psi, and meter water column. Assumptions Both mercury and water are incompressible substances. Properties We take the densities of water and mercury to be 1000 kg/m3 and 13,600 kg/m3, respectively. Analysis Using the relation P = ρgh for gage pressure, the high and low pressures are expressed as ⎞⎛ 1 kPa ⎞ ⎛ 1N ⎟ = 16.0 kPa ⎟⎜ Phigh = ρghhigh = (13,600 kg/m3 )(9.81 m/s 2 )(0.12 m)⎜⎜ 2 ⎟⎜ 2⎟ ⎝ 1 kg ⋅ m/s ⎠⎝ 1000 N/m ⎠ ⎞⎛ 1 kPa ⎞ ⎛ 1N ⎟ = 10.7 kPa ⎟⎜ Plow = ρghlow = (13,600 kg/m 3 )(9.81 m/s 2 )(0.08 m)⎜⎜ 2 ⎟⎜ 2⎟ 1 kg m/s 1000 N/m ⋅ ⎠ ⎠⎝ ⎝

Noting that 1 psi = 6.895 kPa, ⎛ 1 psi ⎞ ⎟⎟ = 2.32 psi Phigh = (16.0 Pa)⎜⎜ ⎝ 6.895 kPa ⎠

and

⎛ 1 psi ⎞ ⎟⎟ = 1.55 psi Plow = (10.7 Pa)⎜⎜ ⎝ 6.895 kPa ⎠

For a given pressure, the relation P = ρgh can be expressed for mercury and water as P = ρ water ghwater and P = ρ mercury ghmercury . Setting these two relations equal to each other and solving for water height gives

P = ρ water ghwater = ρ mercury ghmercury → hwater =

ρ mercury ρ water

hmercury h

Therefore, hwater, high = hwater, low =

ρ mercury ρ water ρ mercury ρ water

hmercury, high = hmercury, low =

13,600 kg/m 3 1000 kg/m 3

13,600 kg/m 3 1000 kg/m 3

(0.12 m) = 1.63 m

(0.08 m) = 1.09 m

Discussion Note that measuring blood pressure with a “water” monometer would involve differential fluid heights higher than the person, and thus it is impractical. This problem shows why mercury is a suitable fluid for blood pressure measurement devices.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-28 1-75 A vertical tube open to the atmosphere is connected to the vein in the arm of a person. The height that the blood will rise in the tube is to be determined. Assumptions 1 The density of blood is constant. 2 The gage pressure of blood is 120 mmHg. Properties The density of blood is given to be ρ = 1050 kg/m3. Analysis For a given gage pressure, the relation P = ρgh can be expressed

for mercury and blood as P = ρ blood ghblood and P = ρ mercury ghmercury .

Blood

Setting these two relations equal to each other we get

h

P = ρ blood ghblood = ρ mercury ghmercury Solving for blood height and substituting gives hblood =

ρ mercury ρ blood

hmercury =

13,600 kg/m 3 1050 kg/m 3

(0.12 m) = 1.55 m

Discussion Note that the blood can rise about one and a half meters in a tube connected to the vein. This explains why IV tubes must be placed high to force a fluid into the vein of a patient.

1-76 A diver is moving at a specified depth from the water surface. The pressure exerted on the surface of the diver by water is to be determined. Assumptions The variation of the density of water with depth is negligible. Properties The specific gravity of seawater is given to be SG = 1.03. We take the density of water to be 1000 kg/m3. Analysis The density of the seawater is obtained by multiplying its specific gravity by the density of water which is taken to be 1000 kg/m3:

Patm Sea

ρ = SG × ρ H 2 O = (1.03)(100 0 kg/m 3 ) = 1030 kg/m 3

h

The pressure exerted on a diver at 30 m below the free surface of the sea is the absolute pressure at that location: P = Patm + ρgh

⎛ 1 kPa = (101 kPa) + (1030 kg/m 3 )(9.807 m/s 2 )(30 m)⎜ ⎜ 1000 N/m 2 ⎝ = 404 kPa

P ⎞ ⎟ ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-29 1-77 Water is poured into the U-tube from one arm and oil from the other arm. The water column height in one arm and the ratio of the heights of the two fluids in the other arm are given. The height of each fluid in that arm is to be determined. Assumptions Both water and oil are incompressible substances.

Water

Properties The density of oil is given to be ρ = 790 kg/m3. We take the density of water to be ρ =1000 kg/m3.

ha

Analysis The height of water column in the left arm of the monometer is given to be hw1 = 0.70 m. We let the height of water and oil in the right arm to be hw2 and ha, respectively. Then, ha = 4hw2. Noting that both arms are open to the atmosphere, the pressure at the bottom of the U-tube can be expressed as Pbottom = Patm + ρ w gh w1

oil

hw1

hw2

Pbottom = Patm + ρ w ghw2 + ρ a gha

and

Setting them equal to each other and simplifying,

ρ w ghw1 = ρ w ghw2 + ρ a gha

ρ w h w1 = ρ w h w2 + ρ a ha

h w1 = h w2 + ( ρ a / ρ w )ha

Noting that ha = 4hw2, the water and oil column heights in the second arm are determined to be 0.7 m = h w2 + (790/1000) 4 hw 2 →

h w2 = 0.168 m

0.7 m = 0.168 m + (790/1000)ha →

ha = 0.673 m

Discussion Note that the fluid height in the arm that contains oil is higher. This is expected since oil is lighter than water.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-30 1-78 Fresh and seawater flowing in parallel horizontal pipelines are connected to each other by a double U-tube manometer. The pressure difference between the two pipelines is to be determined. Assumptions 1 All the liquids are incompressible. 2 The effect of air column on pressure is negligible.

Air

Properties The densities of seawater and mercury are given to be ρsea = 1035 kg/m3 and ρHg = 13,600 kg/m3. We take the density of water to be ρ w =1000 kg/m3. Analysis Starting with the pressure in the fresh water pipe (point 1) and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the sea water pipe (point 2), and setting the result equal to P2 gives

hsea

Fresh Water

Sea Water

hair hw

P1 + ρ w ghw − ρ Hg ghHg − ρ air ghair + ρ sea ghsea = P2

hHg

Rearranging and neglecting the effect of air column on pressure,

Mercury

P1 − P2 = − ρ w ghw + ρ Hg ghHg − ρ sea ghsea = g ( ρ Hg hHg − ρ w hw − ρ sea hsea ) Substituting, P1 − P2 = (9.81 m/s 2 )[(13600 kg/m 3 )(0.1 m) ⎛ 1 kN − (1000 kg/m 3 )(0.6 m) − (1035 kg/m 3 )(0.4 m)]⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝

⎞ ⎟ ⎟ ⎠

= 3.39 kN/m 2 = 3.39 kPa

Therefore, the pressure in the fresh water pipe is 3.39 kPa higher than the pressure in the sea water pipe. Discussion A 0.70-m high air column with a density of 1.2 kg/m3 corresponds to a pressure difference of 0.008 kPa. Therefore, its effect on the pressure difference between the two pipes is negligible.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-31 1-79 Fresh and seawater flowing in parallel horizontal pipelines are connected to each other by a double U-tube manometer. The pressure difference between the two pipelines is to be determined. Assumptions All the liquids are incompressible.

Oil

Properties The densities of seawater and mercury are given to be ρsea = 1035 kg/m3 and ρHg = 13,600 kg/m3. We take the density of water to be ρ w =1000 kg/m3. The specific gravity of oil is given to be 0.72, and thus its density is 720 kg/m3.

hsea

Analysis Starting with the pressure in the fresh water pipe (point 1) and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the sea water pipe (point 2), and setting the result equal to P2 gives

Fresh Water

Sea Water

hoil hw

P1 + ρ w ghw − ρ Hg ghHg − ρ oil ghoil + ρ sea ghsea = P2

hHg

Rearranging,

Mercury

P1 − P2 = − ρ w ghw + ρ Hg ghHg + ρ oil ghoil − ρ sea ghsea = g ( ρ Hg hHg + ρ oil hoil − ρ w hw − ρ sea hsea )

Substituting, P1 − P2 = (9.81 m/s 2 )[(13600 kg/m 3 )(0.1 m) + (720 kg/m 3 )(0.7 m) − (1000 kg/m 3 )(0.6 m) ⎛ 1 kN − (1035 kg/m 3 )(0.4 m)]⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝

⎞ ⎟ ⎟ ⎠

= 8.34 kN/m 2 = 8.34 kPa

Therefore, the pressure in the fresh water pipe is 8.34 kPa higher than the pressure in the sea water pipe.

1-80 The pressure indicated by a manometer is to be determined. Properties The specific weights of fluid A and fluid B are given to be 10 kN/m3 and 8 kN/m3, respectively. Analysis The absolute pressure P1 is determined from P1 = Patm + ( ρgh) A + ( ρgh) B

= hB

= Patm + γ A h A + γ B h B ⎛ 0.1333 kPa ⎞ ⎟⎟ = (758 mm Hg)⎜⎜ ⎝ 1 mm Hg ⎠

hA =

+ (10 kN/m 3 )(0.05 m) + (8 kN/m 3 )(0.15 m) = 102.7 kPa

Note that 1 kPa = 1 kN/m2.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-32 1-81 The pressure indicated by a manometer is to be determined. Properties The specific weights of fluid A and fluid B are given to be 100 kN/m3 and 8 kN/m3, respectively. Analysis The absolute pressure P1 is determined from P1 = Patm + ( ρgh) A + ( ρgh) B = Patm + γ A h A + γ B h B

= hB

= 90 kPa + (100 kN/m 3 )(0.05 m) + (8 kN/m 3 )(0.15 m) = 96.2 kPa

hA = 2

Note that 1 kPa = 1 kN/m .

100 kN/m3

1-82 The pressure indicated by a manometer is to be determined. Properties The specific weights of fluid A and fluid B are given to be 10 kN/m3 and 12 kN/m3, respectively. Analysis The absolute pressure P1 is determined from P1 = Patm + ( ρgh) A + ( ρgh) B

= hB

= Patm + γ A h A + γ B h B ⎛ 0.1333 kPa ⎞ ⎟⎟ = (720 mm Hg)⎜⎜ ⎝ 1 mm Hg ⎠ + (10 kN/m 3 )(0.05 m) + (12 kN/m 3 )(0.15 m) = 98.3 kPa

hA =

12 kN/m3

Note that 1 kPa = 1 kN/m2.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-33 1-83 The gage pressure of air in a pressurized water tank is measured simultaneously by both a pressure gage and a manometer. The differential height h of the mercury column is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air-water interface is the same as the indicated gage pressure. Properties We take the density of water to be ρw =1000 kg/m3. The specific gravities of oil and mercury are given to be 0.72 and 13.6, respectively. Analysis Starting with the pressure of air in the tank (point 1), and moving along the tube by adding (as we go down) or subtracting (as we go u p) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm gives

P1 + ρ w ghw − ρ Hg ghHg − ρ oil ghoil = Patm Rearranging,

P1 − Patm = ρ oil ghoil + ρ Hg ghHg − ρ w ghw or,

P1,gage

ρw g

= SG oil hoil + SG Hg hHg − hw

Substituting,

⎛ ⎞⎛ 1000 kg ⋅ m/s 2 80 kPa ⎜ ⎟⎜ ⎜ (1000 kg/m 3 )(9.81 m/s 2 ) ⎟⎜ 1 kPa. ⋅ m 2 ⎝ ⎠⎝

⎞ ⎟ = 0.72 × (0.75 m) + 13.6 × hHg − 0.3 m ⎟ ⎠

Solving for hHg gives hHg = 0.582 m. Therefore, the differential height of the mercury column must be 58.2 cm. Discussion Double instrumentation like this allows one to verify the measurement of one of the instruments by the measurement of another instrument.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-34 1-84 The gage pressure of air in a pressurized water tank is measured simultaneously by both a pressure gage and a manometer. The differential height h of the mercury column is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air-water interface is the same as the indicated gage pressure. Properties We take the density of water to be ρ w =1000 kg/m3. The specific gravities of oil and mercury are given to be 0.72 and 13.6, respectively. Analysis Starting with the pressure of air in the tank (point 1), and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm gives

P1 + ρ w ghw − ρ Hg ghHg − ρ oil ghoil = Patm 40 kPa

Rearranging,

P1 − Patm = ρ oil ghoil + ρ Hg ghHg − ρ w ghw P1,gage

or,

ρw g

AIR hoil

= SG oil hoil + SG

Hg hHg

− hw

Water hw

Substituting,

⎡ ⎤⎛ 1000 kg ⋅ m/s 2 40 kPa ⎜ ⎢ 3 2 ⎥⎜ 2 ⎣⎢ (1000 kg/m )(9.81 m/s ) ⎦⎥⎝ 1 kPa. ⋅ m

hHg

⎞ ⎟ = 0.72 × (0.75 m) + 13.6 × hHg − 0.3 m ⎟ ⎠

Solving for hHg gives hHg = 0.282 m. Therefore, the differential height of the mercury column must be 28.2 cm. Discussion Double instrumentation like this allows one to verify the measurement of one of the instruments by the measurement of another instrument.

1-85 The top part of a water tank is divided into two compartments, and a fluid with an unknown density is poured into one side. The levels of the water and the liquid are measured. The density of the fluid is to be determined. Assumptions 1 Both water and the added liquid are incompressible substances. 2 The added liquid does not mix with water.

Fluid

Properties We take the density of water to be ρ =1000 kg/m3. Analysis Both fluids are open to the atmosphere. Noting that the pressure of both water and the added fluid is the same at the contact surface, the pressure at this surface can be expressed as

Water hf

hw

Pcontact = Patm + ρ f ghf = Patm + ρ w ghw

Simplifying and solving for ρf gives

ρ f ghf = ρ w gh w →

ρf =

hw 55 cm ρw = (1000 kg/m 3 ) = 846 kg/m 3 hf 65 cm

Discussion Note that the added fluid is lighter than water as expected (a heavier fluid would sink in water).

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-35 1-86 The fluid levels in a multi-fluid U-tube manometer change as a result of a pressure drop in the trapped air space. For a given pressure drop and brine level change, the area ratio is to be determined. Assumptions 1 All the liquids are incompressible. 2 Pressure in the brine pipe remains constant. 3 The variation of pressure in the trapped air space is negligible.

A Air

Properties The specific gravities are given to be 13.56 for mercury and 1.1 for brine. We take the standard density of water to be ρw =1000 kg/m3.

Area, A1

Analysis It is clear from the problem statement and the figure that the brine pressure is much higher than the air pressure, and when the air pressure drops by 0.7 kPa, the pressure difference between the brine and the air space increases also by the same amount.

B Brine pipe

Water

SG=1.1 ∆hb = 5 mm

Mercury SG=13.56

Starting with the air pressure (point A) and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the brine pipe (point B), and setting the result equal to PB before and after the pressure change of air give Before:

PA1 + ρ w ghw + ρ Hg ghHg, 1 − ρ br ghbr,1 = PB

After:

PA2 + ρ w ghw + ρ Hg ghHg, 2 − ρ br ghbr,2 = PB

Area, A2

Subtracting,

PA2 − PA1 + ρ Hg g∆hHg − ρ br g∆hbr = 0 →

PA1 − PA2 = SG Hg ∆hHg − SG br ∆h br = 0 ρwg

(1)

where ∆hHg and ∆hbr are the changes in the differential mercury and brine column heights, respectively, due to the drop in air pressure. Both of these are positive quantities since as the mercury-brine interface drops, the differential fluid heights for both mercury and brine increase. Noting also that the volume of mercury is constant, we have A1∆hHg,left = A2 ∆hHg,right and

PA2 − PA1 = −0.7 kPa = −700 N/m 2 = −700 kg/m ⋅ s 2 ∆h br = 0.005 m

∆hHg = ∆hHg,right + ∆hHg,left = ∆hbr + ∆hbr A2 /A 1 = ∆hbr (1 + A2 /A 1 ) Substituting, 700 kg/m ⋅ s 2 (1000 kg/m 3 )(9.81 m/s 2 )

= [13.56 × 0.005(1 + A2 /A1 ) − (1.1× 0.005)] m

It gives A2/A1 = 0.134

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-36 1-87 A multi-fluid container is connected to a U-tube. For the given specific gravities and fluid column heights, the gage pressure at A and the height of a mercury column that would create the same pressure at A are to be determined. Assumptions 1 All the liquids are incompressible. 2 The multifluid container is open to the atmosphere. Properties The specific gravities are given to be 1.26 for glycerin and 0.90 for oil. We take the standard density of water to be ρw =1000 kg/m3, and the specific gravity of mercury to be 13.6.

70 cm

Oil SG=0.90

Analysis Starting with the atmospheric pressure on the top surface of the container and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach point A, and setting the result equal to PA give

30 cm

Water

Patm + ρ oil ghoil + ρ w ghw − ρ gly ghgly = PA

20 cm

Glycerin SG=1.26

A

90 cm

Rearranging and using the definition of specific gravity,

15 cm

PA − Patm = SG oil ρ w ghoil + SG w ρ w ghw − SG gly ρ w ghgly or

PA,gage = gρ w (SG oil hoil + SG w hw − SG gly hgly ) Substituting, ⎛ 1 kN PA,gage = (9.81 m/s 2 )(1000 kg/m 3 )[0.90(0.70 m) + 1(0.3 m) − 1.26(0.70 m)]⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝

⎞ ⎟ ⎟ ⎠

= 0.471 kN/m 2 = 0.471 kPa

The equivalent mercury column height is

hHg =

PA,gage

ρ Hg g

=

⎛ 1000 kg ⋅ m/s 2 ⎜ 1 kN (13,600 kg/m 3 )(1000 kg/m 3 )(9.81 m/s 2 ) ⎜⎝ 0.471 kN/m 2

⎞ ⎟ = 0.00353 m = 0.353 cm ⎟ ⎠

Discussion Note that the high density of mercury makes it a very suitable fluid for measuring high pressures in manometers.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-37

Solving Engineering Problems and EES 1-88C Despite the convenience and capability the engineering software packages offer, they are still just tools, and they will not replace the traditional engineering courses. They will simply cause a shift in emphasis in the course material from mathematics to physics. They are of great value in engineering practice, however, as engineers today rely on software packages for solving large and complex problems in a short time, and perform optimization studies efficiently.

Determine a positive real root of the following equation using EES:

1-89 3

2x – 10x0.5 – 3x = -3 Solution by EES Software (Copy the following line and paste on a blank EES screen to verify solution): 2*x^3-10*x^0.5-3*x = -3

Answer: x = 2.063 (using an initial guess of x=2)

1-90

Solve the following system of 2 equations with 2 unknowns using EES: x3 – y2 = 7.75 3xy + y = 3.5

Solution by EES Software (Copy the following lines and paste on a blank EES screen to verify solution): x^3-y^2=7.75 3*x*y+y=3.5

1-91

Solve the following system of 3 equations with 3 unknowns using EES: 2x – y + z = 7 3x2 + 2y = z + 3 xy + 2z = 4

Solution by EES Software (Copy the following lines and paste on a blank EES screen to verify solution): 2*x-y+z=7 3*x^2+2*y=z+3 x*y+2*z=4

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-38 1-92

Solve the following system of 3 equations with 3 unknowns using EES: x2y – z = 1 x – 3y0.5 + xz = - 2 x+y–z=2

Solution by EES Software (Copy the following lines and paste on a blank EES screen to verify solution): x^2*y-z=1 x-3*y^0.5+x*z=-2 x+y-z=2

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-39 Specific heat of water is to be expressed at various units using unit conversion capability of EES.

1-93E

Analysis The problem is solved using EES, and the solution is given below.

EQUATION WINDOW "GIVEN" C_p=4.18 [kJ/kg-C] "ANALYSIS" C_p_1=C_p*Convert(kJ/kg-C, kJ/kg-K) C_p_2=C_p*Convert(kJ/kg-C, Btu/lbm-F) C_p_3=C_p*Convert(kJ/kg-C, Btu/lbm-R) C_p_4=C_p*Convert(kJ/kg-C, kCal/kg-C)

FORMATTED EQUATIONS WINDOW GIVEN C p = 4.18

[kJ/kg-C]

ANALYSIS kJ/kg–K

C p,1 =

Cp ·

1 ·

C p,2 =

Cp ·

0.238846 ·

C p,3 =

Cp ·

0.238846 ·

C p,4 =

Cp ·

0.238846 ·

kJ/kg–C Btu/lbm–F kJ/kg–C Btu/lbm–R kJ/kg–C kCal/kg–C kJ/kg–C

SOLUTION C_p=4.18 [kJ/kg-C] C_p_1=4.18 [kJ/kg-K] C_p_2=0.9984 [Btu/lbm-F] C_p_3=0.9984 [Btu/lbm-R] C_p_4=0.9984 [kCal/kg-C]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-40

Review Problems 1-94 The weight of a lunar exploration module on the moon is to be determined. Analysis Applying Newton's second law, the weight of the module on the moon can be determined from W moon = mg moon =

Wearth 2800 N g moon = (1.64 m/s 2 ) = 469 N g earth 9.8 m/s 2

1-95 The deflection of the spring of the two-piston cylinder with a spring shown in the figure is to be determined. Analysis Summing the forces acting on the piston in the vertical direction gives

F2

Fs + F2 + F3 = F1 kx + P2 A2 + P3 ( A1 − A2 ) = P1 A1

which when solved for the deflection of the spring and substituting A = πD 2 / 4 gives

[P D − P D − P (D − D )] π [5000 × 0.08 − 10,000 × 0.03 = 4 × 800

x=

π

4k

1

2 1

2

2 2

3

2

2 1

F3

Fs

2 2

2

− 1000(0.08 2 − 0.03 2 )

] F1

= 0.0172 m = 1.72 cm

We expressed the spring constant k in kN/m, the pressures in kPa (i.e., kN/m2) and the diameters in m units.

1-96 An airplane is flying over a city. The local atmospheric pressure in that city is to be determined. Assumptions The gravitational acceleration does not change with altitude. Properties The densities of air and mercury are given to be 1.15 kg/m3 and 13,600 kg/m3. Analysis The local atmospheric pressure is determined from Patm = Pplane + ρgh ⎛ 1 kN = 25 kPa + (1.15 kg/m 3 )(9.81 m/s 2 )(9000 m)⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 126.5 kN/m 2 ≅ 127 kPa ⎟ ⎠

The atmospheric pressure may be expressed in mmHg as

hHg =

Patm 126.5 kPa ⎛ 1000 Pa ⎞⎛ 1000 mm ⎞ = ⎟ = 948 mmHg ⎟⎜ ⎜ 3 2 ρg (13,600 kg/m )(9.81 m/s ) ⎝ 1 kPa ⎠⎝ 1 m ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-41 1-97 The gravitational acceleration changes with altitude. Accounting for this variation, the weights of a body at different locations are to be determined. Analysis The weight of an 80-kg man at various locations is obtained by substituting the altitude z (values in m) into the relation

⎛ 1N W = mg = (80kg)(9.807 − 3.32 × 10 −6 z m/s 2 )⎜ ⎜ 1kg ⋅ m/s 2 ⎝

⎞ ⎟ ⎟ ⎠

Sea level:

(z = 0 m): W = 80×(9.807-3.32x10-6×0) = 80×9.807 = 784.6 N

Denver:

(z = 1610 m): W = 80×(9.807-3.32x10-6×1610) = 80×9.802 = 784.2 N

Mt. Ev.:

(z = 8848 m): W = 80×(9.807-3.32x10-6×8848) = 80×9.778 = 782.2 N

1-98 A man is considering buying a 12-oz steak for \$3.15, or a 300-g steak for \$2.95. The steak that is a better buy is to be determined. Assumptions The steaks are of identical quality. Analysis To make a comparison possible, we need to express the cost of each steak on a common basis. Let us choose 1 kg as the basis for comparison. Using proper conversion factors, the unit cost of each steak is determined to be

12 ounce steak: ⎛ \$3.15 ⎞ ⎛ 16 oz ⎞ ⎛ 1 lbm ⎞ ⎟⎟ = \$9.26/kg Unit Cost = ⎜ ⎟⎜ ⎟ ⎜⎜ ⎝ 12 oz ⎠ ⎝ 1 lbm ⎠ ⎝ 0.45359 kg ⎠

300 gram steak: ⎛ \$2.95 ⎞ ⎛ 1000 g ⎞ ⎟⎟ ⎜⎜ ⎟⎟ = \$9.83/kg Unit Cost = ⎜⎜ ⎝ 300 g ⎠ ⎝ 1 kg ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-42 1.99E The mass of a substance is given. Its weight is to be determined in various units. Analysis Applying Newton's second law, the weight is determined in various units to be

⎛ 1N W = mg = (1 kg)(9.81 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 9.81 N ⎟ ⎠

⎛ 1 kN W = mg = (1 kg)(9.81 m/s 2 )⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 0.00981 kN ⎟ ⎠

W = mg = (1 kg)(9.81 m/s 2 ) = 1 kg ⋅ m/s 2 ⎛ 1N W = mg = (1 kg)(9.81 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞⎛ 1 kgf ⎞ ⎟⎜ ⎟ = 1 kgf ⎟⎜ 9.81 N ⎟ ⎠ ⎠⎝

⎛ 2.205 lbm ⎞ ⎟⎟(32.2 ft/s 2 ) = 71 lbm ⋅ ft/s 2 W = mg = (1 kg)⎜⎜ 1 kg ⎠ ⎝

⎛ ⎛ 2.205 lbm ⎞ 1 lbf ⎟⎟(32.2 ft/s 2 )⎜ W = mg = (1 kg)⎜⎜ ⎜ 32.2 lbm ⋅ ft/s 2 1 kg ⎝ ⎠ ⎝

⎞ ⎟ = 2.21 lbf ⎟ ⎠

1-100E The efficiency of a refrigerator increases by 3% per °C rise in the minimum temperature. This increase is to be expressed per °F, K, and R rise in the minimum temperature. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the increase in efficiency is

(a) 3% for each K rise in temperature, and (b), (c) 3/1.8 = 1.67% for each R or °F rise in temperature.

1-101E The boiling temperature of water decreases by 3°C for each 1000 m rise in altitude. This decrease in temperature is to be expressed in °F, K, and R. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the decrease in the boiling temperature is

(a) 3 K for each 1000 m rise in altitude, and (b), (c) 3×1.8 = 5.4°F = 5.4 R for each 1000 m rise in altitude.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-43 1-102E Hyperthermia of 5°C is considered fatal. This fatal level temperature change of body temperature is to be expressed in °F, K, and R. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the fatal level of hypothermia is

(a) 5 K (b) 5×1.8 = 9°F (c) 5×1.8 = 9 R

1-103E A house is losing heat at a rate of 2700 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. The rate of heat loss is to be expressed per °F, K, and R of temperature difference between the indoor and the outdoor temperatures. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the rate of heat loss from the house is

(a) 2700 kJ/h per K difference in temperature, and (b), (c) 2700/1.8 = 1500 kJ/h per R or °F rise in temperature.

1-104 The average temperature of the atmosphere is expressed as Tatm = 288.15 – 6.5z where z is altitude in km. The temperature outside an airplane cruising at 12,000 m is to be determined. Analysis Using the relation given, the average temperature of the atmosphere at an altitude of 12,000 m is determined to be

Tatm = 288.15 - 6.5z = 288.15 - 6.5×12 = 210.15 K = - 63°C Discussion This is the “average” temperature. The actual temperature at different times can be different.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-44 1-105 A new “Smith” absolute temperature scale is proposed, and a value of 1000 S is assigned to the boiling point of water. The ice point on this scale, and its relation to the Kelvin scale are to be determined. Analysis All linear absolute temperature scales read zero at absolute zero pressure, and are constant multiples of each other. For example, T(R) = 1.8 T(K). That is, multiplying a temperature value in K by 1.8 will give the same temperature in R.

S

K

1000

373.15

The proposed temperature scale is an acceptable absolute temperature scale since it differs from the other absolute temperature scales by a constant only. The boiling temperature of water in the Kelvin and the Smith scales are 315.15 K and 1000 K, respectively. Therefore, these two temperature scales are related to each other by T (S ) =

1000 T ( K ) = 2.6799T(K ) 373.15

The ice point of water on the Smith scale is T(S)ice = 2.6799 T(K)ice = 2.6799×273.15 = 732.0 S

0

1-106E An expression for the equivalent wind chill temperature is given in English units. It is to be converted to SI units. Analysis The required conversion relations are 1 mph = 1.609 km/h and T(°F) = 1.8T(°C) + 32. The first thought that comes to mind is to replace T(°F) in the equation by its equivalent 1.8T(°C) + 32, and V in mph by 1.609 km/h, which is the “regular” way of converting units. However, the equation we have is not a regular dimensionally homogeneous equation, and thus the regular rules do not apply. The V in the equation is a constant whose value is equal to the numerical value of the velocity in mph. Therefore, if V is given in km/h, we should divide it by 1.609 to convert it to the desired unit of mph. That is, Tequiv (° F) = 91.4 − [ 91.4 − Tambient (° F)][ 0.475 − 0.0203(V / 1.609) + 0.304 V / 1.609 ]

or Tequiv (° F) = 91.4 − [ 91.4 − Tambient (° F)][ 0.475 − 0.0126V + 0.240 V ]

where V is in km/h. Now the problem reduces to converting a temperature in °F to a temperature in °C, using the proper convection relation: 1.8Tequiv (° C ) + 32 = 914 . − [914 . − (18 . Tambient (° C ) + 32 )][0.475 − 0.0126V + 0.240 V ]

which simplifies to Tequiv (° C) = 33.0 − ( 33.0 − Tambient )(0.475 − 0.0126V + 0.240 V )

where the ambient air temperature is in °C.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-45 1-107E Problem 1-106E is reconsidered. The equivalent wind-chill temperatures in °F as a function of wind velocity in the range of 4 mph to 40 mph for the ambient temperatures of 20, 40, and 60°F are to be plotted, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. T_ambient=20 "V=20" T_equiv=91.4-(91.4-T_ambient)*(0.475 - 0.0203*V + 0.304*sqrt(V))

Tequiv [F] 59.94 54.59 51.07 48.5 46.54 45.02 43.82 42.88 42.16 41.61

The table is for Tambient=60°F

60

T amb = 60°F

50 40 30

Tequiv [F]

V [mph] 4 8 12 16 20 24 28 32 36 40

T amb = 40°F

20 10 0

T amb = 20°F

-10 -20 0

5

10

15

20

25

30

35

40

V [mph]

1-108 One section of the duct of an air-conditioning system is laid underwater. The upward force the water will exert on the duct is to be determined. Assumptions 1 The diameter given is the outer diameter of the duct (or, the thickness of the duct material is negligible). 2 The weight of the duct and the air in is negligible. Properties The density of air is given to be ρ = 1.30 kg/m3. We take the density of water to be 1000 kg/m3. Analysis Noting that the weight of the duct and the air in it is negligible, the net upward force acting on the duct is the buoyancy force exerted by water. The volume of the underground section of the duct is

D =15 cm L = 20 m FB

V = AL = (πD 2 / 4) L = [π (0.15 m) 2 /4](20 m) = 0.353 m 3 Then the buoyancy force becomes ⎛ 1 kN FB = ρgV = (1000 kg/m 3 )(9.81 m/s 2 )(0.353 m 3 )⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 3.46 kN ⎟ ⎠

Discussion The upward force exerted by water on the duct is 3.46 kN, which is equivalent to the weight of a mass of 353 kg. Therefore, this force must be treated seriously.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-46 1-109 A helium balloon tied to the ground carries 2 people. The acceleration of the balloon when it is first released is to be determined. Assumptions The weight of the cage and the ropes of the balloon is negligible. Properties The density of air is given to be ρ = 1.16 kg/m3. The density of helium gas is 1/7th of this. Analysis The buoyancy force acting on the balloon is

V balloon = 4π r 3 /3 = 4 π(6 m) 3 /3 = 904.8 m 3 FB = ρ air gV balloon

⎛ 1N = (1.16 kg/m )(9.81m/s )(904.8 m )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ 3

2

3

⎞ ⎟ = 10,296 N ⎟ ⎠

D =12 m

The total mass is

⎞ ⎛ 1.16 kg/m 3 ⎟(904.8 m 3 ) = 149.9 kg m He = ρ HeV = ⎜ ⎠ ⎝ 7 m total = m He + m people = 149.9 + 2 × 85 = 319.9 kg The total weight is

⎛ 1N W = m total g = (319.9 kg)(9.81 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 3138 N ⎟ ⎠ m = 170 kg

Thus the net force acting on the balloon is Fnet = FB − W = 10,296 − 3138 = 7157 N

Then the acceleration becomes a=

Fnet 7157 N ⎛⎜ 1kg ⋅ m/s 2 = m total 319.9 kg ⎜⎝ 1 N

⎞ ⎟ = 22.4 m/s 2 ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-47 1-110 Problem 1-109 is reconsidered. The effect of the number of people carried in the balloon on acceleration is to be investigated. Acceleration is to be plotted against the number of people, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Given" D=12 [m] N_person=2 m_person=85 [kg] rho_air=1.16 [kg/m^3] rho_He=rho_air/7 "Analysis" g=9.81 [m/s^2] V_ballon=pi*D^3/6 F_B=rho_air*g*V_ballon m_He=rho_He*V_ballon m_people=N_person*m_person m_total=m_He+m_people W=m_total*g F_net=F_B-W a=F_net/m_total

35 30 25 2

1 2 3 4 5 6 7 8 9 10

a [m/s2] 34 22.36 15.61 11.2 8.096 5.79 4.01 2.595 1.443 0.4865

a [m/s ]

Nperson

20 15 10 5 0 1

2

3

4

5

6

7

8

9

10

N pe rson

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-48 1-111 A balloon is filled with helium gas. The maximum amount of load the balloon can carry is to be determined. Assumptions The weight of the cage and the ropes of the balloon is negligible. Properties The density of air is given to be ρ = 1.16 kg/m3. The density of helium gas is 1/7th of this.

D =12 m

Analysis

The buoyancy force acting on the balloon is

V balloon = 4π r 3 /3 = 4 π(6 m) 3 /3 = 904.8 m 3 FB = ρ air gV balloon

⎛ 1N = (1.16 kg/m 3 )(9.81m/s 2 )(904.8 m 3 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 10,296 N ⎟ ⎠

The mass of helium is ⎛ 1.16 ⎞ m He = ρ HeV = ⎜ kg/m 3 ⎟(904.8 m 3 ) = 149.9 kg 7 ⎝ ⎠

In the limiting case, the net force acting on the balloon will be zero. That is, the buoyancy force and the weight will balance each other: W = mg = FB m total =

FB 10,296 N = = 1050 kg g 9.81 m/s 2

Thus,

m people = m total − m He = 1050 − 149.9 = 900 kg

1-112 A 10-m high cylindrical container is filled with equal volumes of water and oil. The pressure difference between the top and the bottom of the container is to be determined. Properties The density of water is given to be ρ = 1000 kg/m3. The specific gravity of oil is given to be 0.85. Oil SG = 0.85

Analysis The density of the oil is obtained by multiplying its specific gravity by the density of water,

h = 10 m

ρ = SG × ρ H 2 O = (0.85)(100 0 kg/m 3 ) = 850 kg/m 3 Water

The pressure difference between the top and the bottom of the cylinder is the sum of the pressure differences across the two fluids, ∆Ptotal = ∆Poil + ∆Pwater = ( ρgh) oil + ( ρgh) water

[

]

⎛ 1 kPa = (850 kg/m 3 )(9.81 m/s 2 )(5 m) + (1000 kg/m 3 )(9.81 m/s 2 )(5 m) ⎜ ⎜ 1000 N/m 2 ⎝ = 90.7 kPa

⎞ ⎟ ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-49 1-113 The pressure of a gas contained in a vertical piston-cylinder device is measured to be 180 kPa. The mass of the piston is to be determined. Assumptions There is no friction between the piston and the cylinder.

Patm

Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield W = PA − Patm A mg = ( P − Patm ) A

⎛ 1000 kg/m ⋅ s 2 (m)(9.81 m/s 2 ) = (180 − 100 kPa)(25 × 10 − 4 m 2 )⎜ ⎜ 1kPa ⎝

It yields

P

⎞ ⎟ ⎟ ⎠

W = mg

m = 20.4 kg

1-114 The gage pressure in a pressure cooker is maintained constant at 100 kPa by a petcock. The mass of the petcock is to be determined. Assumptions There is no blockage of the pressure release valve. Analysis Atmospheric pressure is acting on all surfaces of the petcock, which balances itself out. Therefore, it can be disregarded in calculations if we use the gage pressure as the cooker pressure. A force balance on the petcock (ΣFy = 0) yields W = Pgage A (100 kPa)(4 × 10 − 6 m 2 ) ⎛⎜ 1000 kg/m ⋅ s 2 ⎜ g 1 kPa 9.81 m/s 2 ⎝ = 0.0408 kg

m=

Pgage A

=

Patm

P

W = mg

⎞ ⎟ ⎟ ⎠

1-115 A glass tube open to the atmosphere is attached to a water pipe, and the pressure at the bottom of the tube is measured. It is to be determined how high the water will rise in the tube. Properties The density of water is given to be ρ = 1000 kg/m3. Analysis The pressure at the bottom of the tube can be expressed as P = Patm + ( ρ g h) tube

Solving for h, h=

Patm= 99 kPa

P − Patm ρg

⎛ 1 kg ⋅ m/s 2 ⎜ = 1N (1000 kg/m 3 )(9.81 m/s 2 ) ⎜⎝ = 2.14 m (120 − 99) kPa

h

Water ⎞⎛ 1000 N/m 2 ⎟⎜ ⎟⎜ 1 kPa ⎠⎝

⎞ ⎟ ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-50

1-116 The air pressure in a duct is measured by an inclined manometer. For a given vertical level difference, the gage pressure in the duct and the length of the differential fluid column are to be determined. Assumptions The manometer fluid is an incompressible substance. Properties The density of the liquid is given to be ρ = 0.81 kg/L = 810 kg/m3. Analysis The gage pressure in the duct is determined from

Pgage = Pabs − Patm = ρgh

⎛ 1N = (810 kg/m 3 )(9.81 m/s 2 )(0.12 m)⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ = 954 Pa

⎞⎛ 1 Pa ⎟⎜ ⎟⎜ 1 N/m 2 ⎠⎝

12 cm ⎞ ⎟ ⎟ ⎠

45°

The length of the differential fluid column is L = h / sin θ = (12 cm) / sin 45° = 17.0 cm

Discussion Note that the length of the differential fluid column is extended considerably by inclining the manometer arm for better readability.

1-117E Equal volumes of water and oil are poured into a U-tube from different arms, and the oil side is pressurized until the contact surface of the two fluids moves to the bottom and the liquid levels in both arms become the same. The excess pressure applied on the oil side is to be determined. Assumptions 1 Both water and oil are incompressible substances. 2 Oil does not mix with water. 3 The cross-sectional area of the U-tube is constant. Properties The density of oil is given to be ρoil = 49.3 lbm/ft3. We take the density of water to be ρw = 62.4 lbm/ft3. Analysis Noting that the pressure of both the water and the oil is the same at the contact surface, the pressure at this surface can be expressed as Pcontact = Pblow + ρ a gha = Patm + ρ w gh w

Noting that ha = hw and rearranging, Pgage,blow = Pblow − Patm = ( ρ w − ρ oil ) gh

⎛ 1 lbf = (62.4 - 49.3 lbm/ft 3 )(32.2 ft/s 2 )(30/12 ft)⎜ ⎜ 32.2 lbm ⋅ ft/s 2 ⎝ = 0.227 psi

⎞⎛ 1 ft 2 ⎟⎜ ⎟⎜ 144 in 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

Discussion When the person stops blowing, the oil will rise and some water will flow into the right arm. It can be shown that when the curvature effects of the tube are disregarded, the differential height of water will be 23.7 in to balance 30-in of oil.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-51 1-118 It is given that an IV fluid and the blood pressures balance each other when the bottle is at a certain height, and a certain gage pressure at the arm level is needed for sufficient flow rate. The gage pressure of the blood and elevation of the bottle required to maintain flow at the desired rate are to be determined. Assumptions 1 The IV fluid is incompressible. 2 The IV bottle is open to the atmosphere. Properties The density of the IV fluid is given to be ρ = 1020 kg/m3. Analysis (a) Noting that the IV fluid and the blood pressures balance each other when the bottle is 0.8 m above the arm level, the gage pressure of the blood in the arm is simply equal to the gage pressure of the IV fluid at a depth of 0.8 m,

Pgage, arm = Pabs − Patm = ρgharm-bottle

⎛ 1 kN = (1020 kg/m 3 )(9.81 m/s 2 )(0.8 m)⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝ = 8.0 k Pa

⎞⎛ 1 kPa ⎟⎜ ⎟⎜ 1 kN/m 2 ⎠⎝

80 cm

⎞ ⎟ ⎟ ⎠

(b) To provide a gage pressure of 15 kPa at the arm level, the height of the bottle from the arm level is again determined from Pgage, arm = ρgharm-bottle to be harm- bottle = =

Pgage, arm

ρg ⎛ 1000 kg ⋅ m/s 2 ⎜ 1 kN (1020 kg/m 3 )(9.81 m/s 2 ) ⎜⎝ 15 kPa

⎞⎛ 1 kN/m 2 ⎟⎜ ⎟⎜ 1 kPa ⎠⎝

⎞ ⎟ = 1.5 m ⎟ ⎠

Discussion Note that the height of the reservoir can be used to control flow rates in gravity driven flows. When there is flow, the pressure drop in the tube due to friction should also be considered. This will result in raising the bottle a little higher to overcome pressure drop.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-52 1-119E A water pipe is connected to a double-U manometer whose free arm is open to the atmosphere. The absolute pressure at the center of the pipe is to be determined. Assumptions 1 All the liquids are incompressible. 2 The solubility of the liquids in each other is negligible. Properties The specific gravities of mercury and oil are given to be 13.6 and 0.80, respectively. We take the density of water to be ρw = 62.4 lbm/ft3. Analysis Starting with the pressure at the center of the water pipe, and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm gives

Pwater pipe − ρ water ghwater + ρ oil ghoil − ρ Hg ghHg − ρ oil ghoil = Patm Solving for Pwater

pipe,

Pwater pipe = Patm + ρ water g (hwater − SGoil hoil + SG Hg hHg + SGoil hoil ) Substituting, Pwater pipe = 14.2 psia + (62.4 lbm/ft 3 )(32.2 ft/s 2 )[(35/12 ft) − 0.8(60/12 ft) + 13.6(15/12 ft) ⎛ 1 lbf + 0.8(40/12 ft)] × ⎜ ⎜ 32.2 lbm ⋅ ft/s 2 ⎝ = 22.3 psia

⎞⎛ 1 ft 2 ⎟⎜ ⎟⎜ 144 in 2 ⎠⎝

⎞ ⎟ ⎟ ⎠

Therefore, the absolute pressure in the water pipe is 22.3 psia. Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly.

1-120 The average atmospheric pressure is given as Patm = 101.325(1 − 0.02256z )5.256 where z is the altitude in km. The atmospheric pressures at various locations are to be determined. Analysis The atmospheric pressures at various locations are obtained by substituting the altitude z values in km into the relation Patm = 101325 . (1 − 0.02256z )5.256

Atlanta:

(z = 0.306 km): Patm = 101.325(1 - 0.02256×0.306)5.256 = 97.7 kPa

Denver:

(z = 1.610 km): Patm = 101.325(1 - 0.02256×1.610)5.256 = 83.4 kPa

M. City:

(z = 2.309 km): Patm = 101.325(1 - 0.02256×2.309)5.256 = 76.5 kPa

Mt. Ev.:

(z = 8.848 km): Patm = 101.325(1 - 0.02256×8.848)5.256 = 31.4 kPa

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-53 1-121 The temperature of the atmosphere varies with altitude z as T = T0 − βz , while the gravitational acceleration varies by g ( z ) = g 0 /(1 + z / 6,370,320) 2 . Relations for the variation of pressure in atmosphere are to be obtained (a) by ignoring and (b) by considering the variation of g with altitude. Assumptions The air in the troposphere behaves as an ideal gas. Analysis (a) Pressure change across a differential fluid layer of thickness dz in the vertical z direction is dP = − ρgdz

From the ideal gas relation, the air density can be expressed as ρ = dP = −

P P = . Then, RT R (T0 − βz )

P gdz R(T0 − β z )

Separating variables and integrating from z = 0 where P = P0 to z = z where P = P,

dP =− P

P

P0

z

0

gdz R(T0 − βz )

Performing the integrations.

T − βz g P = ln 0 P0 Rβ T0

ln

Rearranging, the desired relation for atmospheric pressure for the case of constant g becomes g

⎛ β z ⎞ βR P = P0 ⎜⎜1 − ⎟⎟ ⎝ T0 ⎠

(b) When the variation of g with altitude is considered, the procedure remains the same but the expressions become more complicated,

dP = −

g0 P dz R(T0 − βz ) (1 + z / 6,370,320) 2

Separating variables and integrating from z = 0 where P = P0 to z = z where P = P,

P

P0

dP =− P

z

g 0 dz

0

R(T0 − β z )(1 + z / 6,370,320) 2

Performing the integrations, P

ln P

P0

g 1 + kz 1 1 = 0 − ln Rβ (1 + kT0 / β )(1 + kz ) (1 + kT0 / β ) 2 T0 − β z

z

0

where R = 287 J/kg⋅K = 287 m2/s2⋅K is the gas constant of air. After some manipulations, we obtain

⎡ ⎛ 1 g0 1 1 + kz ⎜ P = P0 exp ⎢− ⎜ 1 + 1 / kz + 1 + kT / β ln 1 − βz / T R ( kT ) β + 0 ⎝ 0 0 ⎣⎢

⎞⎤ ⎟⎥ ⎟ ⎠⎦⎥

where T0 = 288.15 K, β = 0.0065 K/m, g0 = 9.807 m/s2, k = 1/6,370,320 m-1, and z is the elevation in m. Discussion When performing the integration in part (b), the following expression from integral tables is used, together with a transformation of variable x = T0 − βz ,

dx

∫ x(a + bx)

2

=

a + bx 1 1 − ln a(a + bx ) a 2 x

Also, for z = 11,000 m, for example, the relations in (a) and (b) give 22.62 and 22.69 kPa, respectively.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-54 1-122 The variation of pressure with density in a thick gas layer is given. A relation is to be obtained for pressure as a function of elevation z. Assumptions The property relation P = Cρ n is valid over the entire region considered. Analysis The pressure change across a differential fluid layer of thickness dz in the vertical z direction is given as, dP = − ρgdz

Also, the relation P = Cρ n can be expressed as C = P / ρ n = P0 / ρ 0n , and thus

ρ = ρ 0 ( P / P0 ) 1 / n Substituting,

dP = − gρ 0 ( P / P0 ) 1 / n dz Separating variables and integrating from z = 0 where P = P0 = Cρ 0n to z = z where P = P,

P

P0

z

( P / P0 ) −1 / n dP = − ρ 0 g dz 0

Performing the integrations. P0

( P / P0 ) −1 / n +1 −1/ n + 1

P

= − ρ 0 gz P0

⎛ P ⎜ ⎜P ⎝ 0

⎞ ⎟ ⎟ ⎠

( n −1) / n

−1 = −

n − 1 ρ 0 gz n P0

Solving for P, ⎛ n − 1 ρ 0 gz ⎞ ⎟ P = P0 ⎜⎜1 − n P0 ⎟⎠ ⎝

n /( n −1)

which is the desired relation. Discussion The final result could be expressed in various forms. The form given is very convenient for calculations as it facilitates unit cancellations and reduces the chance of error.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-55 1-123 A pressure transducers is used to measure pressure by generating analogue signals, and it is to be calibrated by measuring both the pressure and the electric current simultaneously for various settings, and the results are tabulated. A calibration curve in the form of P = aI + b is to be obtained, and the pressure corresponding to a signal of 10 mA is to be calculated. Assumptions Mercury is an incompressible liquid. Properties The specific gravity of mercury is given to be 13.56, and thus its density is 13,560 kg/m3. Analysis For a given differential height, the pressure can be calculated from

P = ρg∆h For ∆h = 28.0 mm = 0.0280 m, for example, ⎛ 1 kN P = 13.56(1000 kg/m 3 )(9.81 m/s 2 )(0.0280 m)⎜⎜ 2 ⎝ 1000 kg ⋅ m/s

⎞⎛ 1 kPa ⎞ ⎟⎜ ⎟⎝ 1 kN/m 2 ⎟⎠ = 3.75 kPa ⎠

Repeating the calculations and tabulating, we have ∆h(mm)

28.0

181.5

297.8

413.1

765.9

1027

1149

1362

1458

1536

P(kPa)

3.73

24.14

39.61

54.95

101.9

136.6

152.8

181.2

193.9

204.3

I (mA)

4.21

5.78

6.97

8.15

11.76

14.43

15.68

17.86

18.84

19.64

A plot of P versus I is given below. It is clear that the pressure varies linearly with the current, and using EES, the best curve fit is obtained to be P = 13.00I - 51.00

(kPa)

for 4.21 ≤ I ≤ 19.64 .

For I = 10 mA, for example, we would get P = 79.0 kPa

225

180

P, kPa

135

90

45

0 4

6

8

10

12

14

16

18

20

I, mA Discussion Note that the calibration relation is valid in the specified range of currents or pressures.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-56 1-124 The flow of air through a wind turbine is considered. Based on unit considerations, a proportionality relation is to be obtained for the mass flow rate of air through the blades. Assumptions Wind approaches the turbine blades with a uniform velocity. Analysis The mass flow rate depends on the air density, average wind velocity, and the cross-sectional area which depends & is kg/s. Therefore, the independent quantities should be arranged such on hose diameter. Also, the unit of mass flow rate m that we end up with the proper unit. Putting the given information into perspective, we have

m& [kg/s] is a function of ρ [kg/m3], D [m], and V [m/s} It is obvious that the only way to end up with the unit “kg/s” for mass flow rate is to multiply the quantities ρ and V with the square of D. Therefore, the desired proportionality relation is

m& is proportional to ρ D 2V or,

m& = CρD 2V

& = where the constant of proportionality is C =π/4 so that m

ρ (πD 2 / 4 )V

Discussion Note that the dimensionless constants of proportionality cannot be determined with this approach.

1-125 A relation for the air drag exerted on a car is to be obtained in terms of on the drag coefficient, the air density, the car velocity, and the frontal area of the car. Analysis The drag force depends on a dimensionless drag coefficient, the air density, the car velocity, and the frontal area. Also, the unit of force F is newton N, which is equivalent to kg⋅m/s2. Therefore, the independent quantities should be arranged such that we end up with the unit kg⋅m/s2 for the drag force. Putting the given information into perspective, we have

FD [ kg⋅m/s2] ↔ CDrag [], Afront [m2], ρ [kg/m3], and V [m/s] It is obvious that the only way to end up with the unit “kg⋅m/s2” for drag force is to multiply mass with the square of the velocity and the fontal area, with the drag coefficient serving as the constant of proportionality. Therefore, the desired relation is

FD = C Drag ρAfrontV 2 Discussion Note that this approach is not sensitive to dimensionless quantities, and thus a strong reasoning is required.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-57 Fundamentals of Engineering (FE) Exam Problems 1-126 Consider a fish swimming 5 m below the free surface of water. The increase in the pressure exerted on the fish when it dives to a depth of 25 m below the free surface is

(a) 196 Pa

(b) 5400 Pa

(c) 30,000 Pa

(d) 196,000 Pa

(e) 294,000 Pa

Answer (d) 196,000 Pa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1000 "kg/m3" g=9.81 "m/s2" z1=5 "m" z2=25 "m" DELTAP=rho*g*(z2-z1) "Pa" "Some Wrong Solutions with Common Mistakes:" W1_P=rho*g*(z2-z1)/1000 "dividing by 1000" W2_P=rho*g*(z1+z2) "adding depts instead of subtracting" W3_P=rho*(z1+z2) "not using g" W4_P=rho*g*(0+z2) "ignoring z1"

1-127 The atmospheric pressures at the top and the bottom of a building are read by a barometer to be 96.0 and 98.0 kPa. If the density of air is 1.0 kg/m3, the height of the building is

(a) 17 m

(b) 20 m

(c) 170 m

(d) 204 m

(e) 252 m

Answer (d) 204 m Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1.0 "kg/m3" g=9.81 "m/s2" P1=96 "kPa" P2=98 "kPa" DELTAP=P2-P1 "kPa" DELTAP=rho*g*h/1000 "kPa" "Some Wrong Solutions with Common Mistakes:" DELTAP=rho*W1_h/1000 "not using g" DELTAP=g*W2_h/1000 "not using rho" P2=rho*g*W3_h/1000 "ignoring P1" P1=rho*g*W4_h/1000 "ignoring P2"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-58 1-128 An apple loses 4.5 kJ of heat as it cools per °C drop in its temperature. The amount of heat loss from the apple per °F drop in its temperature is

(a) 1.25 kJ

(b) 2.50 kJ

(c) 5.0 kJ

(d) 8.1 kJ

(e) 4.1 kJ

Answer (b) 2.50 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Q_perC=4.5 "kJ" Q_perF=Q_perC/1.8 "kJ" "Some Wrong Solutions with Common Mistakes:" W1_Q=Q_perC*1.8 "multiplying instead of dividing" W2_Q=Q_perC "setting them equal to each other"

1-129 Consider a 2-m deep swimming pool. The pressure difference between the top and bottom of the pool is

(a) 12.0 kPa

(b) 19.6 kPa

(c) 38.1 kPa

(d) 50.8 kPa

(e) 200 kPa

Answer (b) 19.6 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1000 "kg/m^3" g=9.81 "m/s2" z1=0 "m" z2=2 "m" DELTAP=rho*g*(z2-z1)/1000 "kPa" "Some Wrong Solutions with Common Mistakes:" W1_P=rho*(z1+z2)/1000 "not using g" W2_P=rho*g*(z2-z1)/2000 "taking half of z" W3_P=rho*g*(z2-z1) "not dividing by 1000"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

1-59 1-130 At sea level, the weight of 1 kg mass in SI units is 9.81 N. The weight of 1 lbm mass in English units is

(a) 1 lbf

(b) 9.81 lbf

(c) 32.2 lbf

(d) 0.1 lbf

(e) 0.031 lbf

Answer (a) 1 lbf Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=1 "lbm" g=32.2 "ft/s2" W=m*g/32.2 "lbf" "Some Wrong Solutions with Common Mistakes:" gSI=9.81 "m/s2" W1_W= m*gSI "Using wrong conversion" W2_W= m*g "Using wrong conversion" W3_W= m/gSI "Using wrong conversion" W4_W= m/g "Using wrong conversion"

1-131 During a heating process, the temperature of an object rises by 10°C. This temperature rise is equivalent to a temperature rise of

(a) 10°F

(b) 42°F

(c) 18 K

(d) 18 R

(e) 283 K

Answer (d) 18 R Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T_inC=10 "C" T_inR=T_inC*1.8 "R" "Some Wrong Solutions with Common Mistakes:" W1_TinF=T_inC "F, setting C and F equal to each other" W2_TinF=T_inC*1.8+32 "F, converting to F " W3_TinK=1.8*T_inC "K, wrong conversion from C to K" W4_TinK=T_inC+273 "K, converting to K"

1-132, 1-133 Design and Essay Problems

KJ

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-1

Solutions Manual for

Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011

Chapter 2 ENERGY, ENERGY TRANSFER, AND GENERAL ENERGY ANALYSIS

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-2 Forms of Energy

2-1C The macroscopic forms of energy are those a system possesses as a whole with respect to some outside reference frame. The microscopic forms of energy, on the other hand, are those related to the molecular structure of a system and the degree of the molecular activity, and are independent of outside reference frames.

2-2C The sum of all forms of the energy a system possesses is called total energy. In the absence of magnetic, electrical and surface tension effects, the total energy of a system consists of the kinetic, potential, and internal energies.

2-3C Thermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in daily life.

2-4C The mechanical energy is the form of energy that can be converted to mechanical work completely and directly by a mechanical device such as a propeller. It differs from thermal energy in that thermal energy cannot be converted to work directly and completely. The forms of mechanical energy of a fluid stream are kinetic, potential, and flow energies.

2-5C Hydrogen is also a fuel, since it can be burned, but it is not an energy source since there are no hydrogen reserves in the world. Hydrogen can be obtained from water by using another energy source, such as solar or nuclear energy, and then the hydrogen obtained can used as a fuel to power cars or generators. Therefore, it is more proper to view hydrogen is an energy carrier than an energy source.

2-6E The total kinetic energy of an object is given is to be determined. Analysis The total kinetic energy of the object is given by KE = m

(100 ft/s) 2 V2 = (15 lbm) 2 2

⎛ 1 Btu/lbm ⎜ ⎜ 25,037 ft 2 /s 2 ⎝

⎞ ⎟ = 3.00 Btu ⎟ ⎠

2-7 The total kinetic energy of an object is given is to be determined. Analysis The total kinetic energy of the object is given by

KE = m

(20 m/s) 2 ⎛ 1 kJ/kg ⎞ V2 = (100 kg) ⎜ ⎟ = 20.0 kJ 2 2 ⎝ 1000 m 2 /s 2 ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-3 2-8E The specific potential energy of an object is to be determined. Analysis In the English unit system, the specific potential energy in Btu is given by ⎛ 1 Btu/lbm pe = gz = (32.1 ft/s 2 )(100 ft)⎜⎜ 2 2 ⎝ 25,037 ft /s

⎞ ⎟ = 0.128 Btu/lbm ⎟ ⎠

2-9E The total potential energy of an object is to be determined. Analysis Substituting the given data into the potential energy expression gives ⎛ 1 Btu/lbm PE = mgz = (200 lbm)(32.2 ft/s 2 )(10 ft)⎜⎜ 2 2 ⎝ 25,037 ft /s

⎞ ⎟ = 2.57 Btu ⎟ ⎠

2-10 The total potential energy of an object that is below a reference level is to be determined. Analysis Substituting the given data into the potential energy expression gives ⎛ 1 kJ/kg ⎞ PE = mgz = (20 kg)(9.5 m/s 2 )(−20 m)⎜ ⎟ = −3.8 kJ ⎝ 1000 m 2 /s 2 ⎠

2-11 A person with his suitcase goes up to the 10th floor in an elevator. The part of the energy of the elevator stored in the suitcase is to be determined. Assumptions 1 The vibrational effects in the elevator are negligible. Analysis The energy stored in the suitcase is stored in the form of potential energy, which is mgz. Therefore,

⎛ 1 kJ/kg ⎞ ∆E suitcase = ∆PE = mg∆z = (30 kg )(9.81 m/s 2 )(35 m)⎜ ⎟ = 10.3 kJ ⎝ 1000 m 2 /s 2 ⎠ Therefore, the suitcase on 10th floor has 10.3 kJ more energy compared to an identical suitcase on the lobby level. Discussion Noting that 1 kWh = 3600 kJ, the energy transferred to the suitcase is 10.3/3600 = 0.0029 kWh, which is very small.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-4 2-12 A hydraulic turbine-generator is to generate electricity from the water of a large reservoir. The power generation potential is to be determined. Assumptions 1 The elevation of the reservoir remains constant. 2 The mechanical energy of water at the turbine exit is negligible. Analysis The total mechanical energy water in a reservoir possesses is equivalent to the potential energy of water at the free surface, and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. e mech

160 m

Turbine

Generator

⎛ 1 kJ/kg ⎞ = pe = gz = (9.81 m/s 2 )(160 m)⎜ ⎟ = 1.574 kJ/kg ⎝ 1000 m 2 /s 2 ⎠

Then the power generation potential becomes

⎛ 1 kW ⎞ W& max = E& mech = m& e mech = (3500 kg/s)(1.574 kJ/kg)⎜ ⎟ = 5509 kW ⎝ 1 kJ/s ⎠ Therefore, the reservoir has the potential to generate 1766 kW of power. Discussion This problem can also be solved by considering a point at the turbine inlet, and using flow energy instead of potential energy. It would give the same result since the flow energy at the turbine inlet is equal to the potential energy at the free surface of the reservoir.

2-13 Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass and the power generation potential are to be determined. Assumptions The wind is blowing steadily at a constant uniform velocity. Properties The density of air is given to be ρ = 1.25 kg/m .

Wind

Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate:

10 m/s

3

e mech = ke =

Wind turbine 60 m

V 2 (10 m/s ) 2 ⎛ 1 kJ/kg ⎞ = ⎜ ⎟ = 0.050 kJ/kg 2 2 ⎝ 1000 m 2 /s 2 ⎠

m& = ρVA = ρV

πD 2 4

= (1.25 kg/m3 )(10 m/s)

π (60 m)2 4

= 35,340 kg/s

W& max = E& mech = m& e mech = (35,340 kg/s)(0.050 kJ/kg) = 1770 kW Therefore, 1770 kW of actual power can be generated by this wind turbine at the stated conditions. Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power generation will change strongly with the wind conditions.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-5 2-14 A water jet strikes the buckets located on the perimeter of a wheel at a specified velocity and flow rate. The power generation potential of this system is to be determined. Assumptions Water jet flows steadily at the specified speed and flow rate. Analysis Kinetic energy is the only form of harvestable mechanical energy the water jet possesses, and it can be converted to work entirely. Therefore, the power potential of the water jet is its kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate:

emech = ke =

V 2 (60 m/s)2 ⎛ 1 kJ/kg ⎞ = 1.8 kJ/kg = ⎜ 2 2⎟ 2 2 ⎝ 1000 m /s ⎠

W&max = E& mech = m& emech ⎛ 1 kW ⎞ = (120 kg/s)(1.8 kJ/kg)⎜ ⎟ = 216 kW ⎝ 1 kJ/s ⎠

Shaft Nozzle

Vj

Therefore, 216 kW of power can be generated by this water jet at the stated conditions. Discussion An actual hydroelectric turbine (such as the Pelton wheel) can convert over 90% of this potential to actual electric power.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-6 2-15 Two sites with specified wind data are being considered for wind power generation. The site better suited for wind power generation is to be determined. Assumptions 1The wind is blowing steadily at specified velocity during specified times. 2 The wind power generation is negligible during other times. Properties We take the density of air to be ρ = 1.25 kg/m3 (it does not affect the final answer). Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate. Considering a unit flow area (A = 1 m2), the maximum wind power and power generation becomes

Wind

Wind turbine

V, m/s

e mech, 1 = ke1 =

V12 (7 m/s ) 2 ⎛ 1 kJ/kg ⎞ = ⎜ ⎟ = 0.0245 kJ/kg 2 2 ⎝ 1000 m 2 /s 2 ⎠

e mech, 2 = ke 2 =

V 22 (10 m/s ) 2 ⎛ 1 kJ/kg ⎞ = ⎜ ⎟ = 0.050 kJ/kg 2 2 ⎝ 1000 m 2 /s 2 ⎠

W& max, 1 = E& mech, 1 = m& 1e mech, 1 = ρV1 Ake1 = (1.25 kg/m 3 )(7 m/s)(1 m 2 )(0.0245 kJ/kg) = 0.2144 kW W& max, 2 = E& mech, 2 = m& 2 e mech, 2 = ρV 2 Ake 2 = (1.25 kg/m 3 )(10 m/s)(1 m 2 )(0.050 kJ/kg) = 0.625 kW

since 1 kW = 1 kJ/s. Then the maximum electric power generations per year become E max, 1 = W& max, 1 ∆t1 = (0.2144 kW)(3000 h/yr) = 643 kWh/yr (per m 2 flow area) E max, 2 = W& max, 2 ∆t 2 = (0.625 kW)(2000 h/yr) = 1250 kWh/yr (per m 2 flow area)

Therefore, second site is a better one for wind generation. Discussion Note the power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the average wind velocity is the primary consideration in wind power generation decisions.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-7 2-16 A river flowing steadily at a specified flow rate is considered for hydroelectric power generation by collecting the water in a dam. For a specified water height, the power generation potential is to be determined. Assumptions 1 The elevation given is the elevation of the free surface of the river. 2 The mechanical energy of water at the turbine exit is negligible. Properties We take the density of water to be ρ = 1000 kg/m3.

River

Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam (relative to free surface of discharge water), and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. ⎛ 1 kJ/kg e mech = pe = gz = (9.81 m/s 2 )(80 m)⎜ ⎝ 1000 m 2 /s 2

80 m

⎞ ⎟ = 0.7848 kJ/kg ⎠

The mass flow rate is

m& = ρV& = (1000 kg/m 3 )(175 m 3 /s) = 175,000 kg/s Then the power generation potential becomes

⎛ 1 MW ⎞ W& max = E& mech = m& emech = (175,000 kg/s)(0.7848 kJ/kg)⎜ ⎟ = 137 MW ⎝ 1000 kJ/s ⎠ Therefore, 137 MW of power can be generated from this river if its power potential can be recovered completely. Discussion Note that the power output of an actual turbine will be less than 137 MW because of losses and inefficiencies.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-8 2-17 A river is flowing at a specified velocity, flow rate, and elevation. The total mechanical energy of the river water per unit mass, and the power generation potential of the entire river are to be determined. Assumptions 1 The elevation given is the elevation of the free surface of the river. 2 The velocity given is the average velocity. 3 The mechanical energy of water at the turbine exit is negligible. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis Noting that the sum of the flow energy and the potential energy is constant for a given fluid body, we can take the elevation of the entire river water to be the elevation of the free surface, and ignore the flow energy. Then the total mechanical energy of the river water per unit mass becomes

emech = pe + ke = gh +

River

3 m/s

90 m

V 2 ⎛⎜ (3 m/s) 2 ⎞⎟⎛ 1 kJ/kg ⎞ = 0.887 kJ/kg = (9.81 m/s 2 )(90 m) + ⎜ ⎟⎜⎝ 1000 m 2 /s 2 ⎟⎠ 2 2 ⎝ ⎠

The power generation potential of the river water is obtained by multiplying the total mechanical energy by the mass flow rate,

m& = ρV& = (1000 kg/m 3 )(500 m 3 /s) = 500,000 kg/s W&max = E& mech = m& emech = (500,000 kg/s)(0.887 kJ/kg) = 444,000 kW = 444 MW Therefore, 444 MW of power can be generated from this river as it discharges into the lake if its power potential can be recovered completely. Discussion Note that the kinetic energy of water is negligible compared to the potential energy, and it can be ignored in the analysis. Also, the power output of an actual turbine will be less than 444 MW because of losses and inefficiencies.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-9 Energy Transfer by Heat and Work

2-18C Energy can cross the boundaries of a closed system in two forms: heat and work.

2-19C The form of energy that crosses the boundary of a closed system because of a temperature difference is heat; all other forms are work.

2-20C An adiabatic process is a process during which there is no heat transfer. A system that does not exchange any heat with its surroundings is an adiabatic system.

2-21C Point functions depend on the state only whereas the path functions depend on the path followed during a process. Properties of substances are point functions, heat and work are path functions.

2-22C (a) The car's radiator transfers heat from the hot engine cooling fluid to the cooler air. No work interaction occurs in the radiator.

(b) The hot engine transfers heat to cooling fluid and ambient air while delivering work to the transmission. (c) The warm tires transfer heat to the cooler air and to some degree to the cooler road while no work is produced. No work is produced since there is no motion of the forces acting at the interface between the tire and road. (d) There is minor amount of heat transfer between the tires and road. Presuming that the tires are hotter than the road, the heat transfer is from the tires to the road. There is no work exchange associated with the road since it cannot move. (e) Heat is being added to the atmospheric air by the hotter components of the car. Work is being done on the air as it passes over and through the car.

2-23C When the length of the spring is changed by applying a force to it, the interaction is a work interaction since it involves a force acting through a displacement. A heat interaction is required to change the temperature (and, hence, length) of the spring.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-10 2-24C (a) From the perspective of the contents, heat must be removed in order to reduce and maintain the content's temperature. Heat is also being added to the contents from the room air since the room air is hotter than the contents.

(b) Considering the system formed by the refrigerator box when the doors are closed, there are three interactions, electrical work and two heat transfers. There is a transfer of heat from the room air to the refrigerator through its walls. There is also a transfer of heat from the hot portions of the refrigerator (i.e., back of the compressor where condenser is placed) system to the room air. Finally, electrical work is being added to the refrigerator through the refrigeration system. (c) Heat is transferred through the walls of the room from the warm room air to the cold winter air. Electrical work is being done on the room through the electrical wiring leading into the room.

2-25C (a) As one types on the keyboard, electrical signals are produced and transmitted to the processing unit. Simultaneously, the temperature of the electrical parts is increased slightly. The work done on the keys when they are depressed is work done on the system (i.e., keyboard). The flow of electrical current (with its voltage drop) does work on the keyboard. Since the temperature of the electrical parts of the keyboard is somewhat higher than that of the surrounding air, there is a transfer of heat from the keyboard to the surrounding air.

(b) The monitor is powered by the electrical current supplied to it. This current (and voltage drop) is work done on the system (i.e., monitor). The temperatures of the electrical parts of the monitor are higher than that of the surrounding air. Hence there is a heat transfer to the surroundings. (c) The processing unit is like the monitor in that electrical work is done on it while it transfers heat to the surroundings. (d) The entire unit then has electrical work done on it, and mechanical work done on it to depress the keys. It also transfers heat from all its electrical parts to the surroundings.

2-26 The power produced by an electrical motor is to be expressed in different units. Analysis Using appropriate conversion factors, we obtain

(a)

⎛ 1 J/s ⎞⎛ 1 N ⋅ m ⎞ W& = (5 W )⎜ ⎟⎜ ⎟ = 5 N ⋅ m/s ⎝ 1 W ⎠⎝ 1 J ⎠

(b)

⎛ 1 J/s ⎞⎛ 1 N ⋅ m ⎞⎛⎜ 1 kg ⋅ m/s W& = (5 W)⎜ ⎟⎜ ⎟ 1N ⎝ 1 W ⎠⎝ 1 J ⎠⎜⎝

2

⎞ ⎟ = 5 kg ⋅ m 2 /s 3 ⎟ ⎠

2-27E The power produced by a model aircraft engine is to be expressed in different units. Analysis Using appropriate conversion factors, we obtain

(a)

⎛ 1 Btu/s ⎞⎛ 778.169 lbf ⋅ ft/s ⎞ W& = (10 W )⎜ ⎟⎜ ⎟ = 7.38 lbf ⋅ ft/s 1 Btu/s ⎝ 1055.056 W ⎠⎝ ⎠

(b)

⎛ 1 hp ⎞ W& = (10 W )⎜ ⎟ = 0.0134 hp ⎝ 745.7 W ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-11 Mechanical Forms of Work

2-28C The work done is the same, but the power is different.

2-29 A car is accelerated from rest to 100 km/h. The work needed to achieve this is to be determined. Analysis The work needed to accelerate a body the change in kinetic energy of the body, Wa =

⎛ ⎛ 100,000 m ⎞ 2 ⎞⎛ ⎞ 1 1 1 kJ ⎟ = 309 kJ ⎟⎟ − 0 ⎟⎜ m(V22 − V12 ) = (800 kg)⎜ ⎜⎜ 2 2⎟ ⎜ ⎜ ⎟ 2 2 3600 s ⎠ 1000 kg ⋅ m /s ⎠ ⎝⎝ ⎠⎝

2-30E A construction crane lifting a concrete beam is considered. The amount of work is to be determined considering (a) the beam and (b) the crane as the system. Analysis (a) The work is done on the beam and it is determined from 1 lbf ⎛ ⎞ W = mg∆z = (2 × 3000 lbm)(32.174 ft/s 2 )⎜ ⎟(24 ft ) 2 ⎝ 32.174 lbm ⋅ ft/s ⎠ = 144,000 lbf ⋅ ft

24 ft

1 Btu ⎛ ⎞ = (144,000 lbf ⋅ ft)⎜ ⎟ = 185 Btu 778 . 169 lbf ft ⋅ ⎝ ⎠

(b) Since the crane must produce the same amount of work as is required to lift the beam, the work done by the crane is

W = 144,000 lbf ⋅ ft = 185 Btu

2-31E A man is pushing a cart with its contents up a ramp that is inclined at an angle of 10° from the horizontal. The work needed to move along this ramp is to be determined considering (a) the man and (b) the cart and its contents as the system. Analysis (a) Considering the man as the system, letting l be the displacement along the ramp, and letting θ be the inclination angle of the ramp, W = Fl sin θ = (100 + 180 lbf )(100 ft)sin(10) = 4862 lbf ⋅ ft 1 Btu ⎛ ⎞ = (4862 lbf ⋅ ft)⎜ ⎟ = 6.248 Btu ⎝ 778.169 lbf ⋅ ft ⎠

This is work that the man must do to raise the weight of the cart and contents, plus his own weight, a distance of lsinθ. (b) Applying the same logic to the cart and its contents gives W = Fl sin θ = (100 lbf )(100 ft)sin(10) = 1736 lbf ⋅ ft 1 Btu ⎛ ⎞ = (1736 lbf ⋅ ft)⎜ ⎟ = 2.231 Btu ⎝ 778.169 lbf ⋅ ft ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-12 2-32E The work required to compress a spring is to be determined. Analysis Since there is no preload, F = kx. Substituting this into the work expression gives 2

2

2

W = Fds = kxdx = k xdx = 1

1

1

[

k 2 ( x 2 − x12 ) 2

F

]

200 lbf/in ⎛ 1 ft ⎞ (1 in ) 2 − 0 2 ⎜ = ⎟ = 8.33 lbf ⋅ ft 2 ⎝ 12 in ⎠ 1 Btu ⎞ ⎛ = (8.33 lbf ⋅ ft)⎜ ⎟ = 0.0107 Btu ⎝ 778.169 lbf ⋅ ft ⎠

x

2-33E The work required to expand a soap bubble is to be determined. Analysis The surface tension work is determined from

[

2

W = σ s dA = σ ( A1 − A2 ) = (0.005 lbf/ft )4π (2 / 12 ft ) 2 − (0.5 / 12 ft ) 2

]

1

1 Btu ⎛ ⎞ -6 = 0.00164 lbf ⋅ ft = (0.00164 lbf ⋅ ft)⎜ ⎟ = 2.11 × 10 Btu ⎝ 778.2 lbf ⋅ ft ⎠

2-34E The work required to stretch a steel rod in a specieid length is to be determined. Assumptions The Young’s modulus does not change as the rod is stretched. Analysis The original volume of the rod is

V0 =

πD 2 4

L=

π (0.5 in) 2 4

(12 in) = 2.356 in 3

The work required to stretch the rod 0.125 in is

V0 E

(ε 22 − ε 12 ) 2 (2.356 in 3 )(30,000 lbf/in 2 ) = (0.125 / 12 in) 2 2 1 Btu ⎛ = 2.835 lbf ⋅ in = (2.835 lbf ⋅ in)⎜ ⎝ 9338 lbf ⋅ in

W=

[

− 02

]

⎞ -4 ⎟ = 4.11 × 10 Btu ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-13 2-35E The work required to compress a spring is to be determined. Analysis The force at any point during the deflection of the spring is given by F = F0 + kx, where F0 is the initial force and x is the deflection as measured from the point where the initial force occurred. From the perspective of the spring, this force acts in the direction opposite to that in which the spring is deflected. Then, 2

2

W = Fds = ( F0 + kx)dx 1

F

1

k 2 ( x 2 − x12 ) 2 200 lbf/in 2 (1 − 0 2 )in 2 = (100 lbf)[(1 − 0)in ] + 2 = 200 lbf ⋅ in

x

= F0 ( x 2 − x1 ) +

1 Btu ⎛ ⎞⎛ 1 ft ⎞ = (200 lbf ⋅ in)⎜ ⎟⎜ ⎟ = 0.0214 Btu ⎝ 778.169 lbf ⋅ ft ⎠⎝ 12 in ⎠

2-36 The work required to compress a spring is to be determined. Analysis Since there is no preload, F = kx. Substituting this into the work expression gives 2

2

2

W = Fds = kxdx = k xdx = 1

1

[

1

300 kN/m (0.03 m) 2 − 0 2 2 = 0.135 kN ⋅ m =

k 2 ( x 2 − x12 ) 2

]

F x

⎛ 1 kJ ⎞ = (0.135 kN ⋅ m)⎜ ⎟ = 0.135 kJ ⎝ 1 kN ⋅ m ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-14 2-37 A ski lift is operating steadily at 10 km/h. The power required to operate and also to accelerate this ski lift from rest to the operating speed are to be determined. Assumptions 1 Air drag and friction are negligible. 2 The average mass of each loaded chair is 250 kg. 3 The mass of chairs is small relative to the mass of people, and thus the contribution of returning empty chairs to the motion is disregarded (this provides a safety factor). Analysis The lift is 1000 m long and the chairs are spaced 20 m apart. Thus at any given time there are 1000/20 = 50 chairs being lifted. Considering that the mass of each chair is 250 kg, the load of the lift at any given time is

Load = (50 chairs)(250 kg/chair) = 12,500 kg Neglecting the work done on the system by the returning empty chairs, the work needed to raise this mass by 200 m is

⎛ 1 kJ W g = mg (z 2 − z1 ) = (12,500 kg)(9.81 m/s 2 )(200 m)⎜ ⎜ 1000 kg ⋅ m 2 /s 2 ⎝

⎞ ⎟ = 24,525 kJ ⎟ ⎠

At 10 km/h, it will take ∆t =

distance 1 km = = 0.1 h = 360 s velocity 10 km / h

to do this work. Thus the power needed is W& g =

Wg ∆t

=

24,525 kJ 360 s

= 68.1 kW

The velocity of the lift during steady operation, and the acceleration during start up are ⎛ 1 m/s ⎞ V = (10 km/h)⎜ ⎟ = 2.778 m/s ⎝ 3.6 km/h ⎠ a=

∆V 2.778 m/s - 0 = = 0.556 m/s 2 ∆t 5s

During acceleration, the power needed is ⎛ 1 kJ/kg 1 1 W& a = m(V 22 − V12 ) / ∆t = (12,500 kg) (2.778 m/s) 2 − 0 ⎜ ⎜ 1000 m 2 /s 2 2 2 ⎝

(

)

⎞ ⎟/(5 s) = 9.6 kW ⎟ ⎠

Assuming the power applied is constant, the acceleration will also be constant and the vertical distance traveled during acceleration will be h=

1 2 1 200 m 1 at sin α = at 2 = (0.556 m/s 2 )(5 s) 2 (0.2) = 1.39 m 2 2 1000 m 2

and

⎛ 1 kJ/kg W& g = mg (z 2 − z1 ) / ∆t = (12,500 kg)(9.81 m/s 2 )(1.39 m)⎜ ⎜ 1000 kg ⋅ m 2 /s 2 ⎝

⎞ ⎟ /(5 s) = 34.1 kW ⎟ ⎠

Thus, W& total = W& a + W& g = 9.6 + 34.1 = 43.7 kW

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-15 2-38 A car is to climb a hill in 12 s. The power needed is to be determined for three different cases. Assumptions Air drag, friction, and rolling resistance are negligible. Analysis The total power required for each case is the sum of the rates of changes in potential and kinetic energies. That is, W& total = W& a + W& g

(a) W& a = 0 since the velocity is constant. Also, the vertical rise is h = (100 m)(sin 30°) = 50 m. Thus, ⎛ 1 kJ W& g = mg ( z 2 − z1 ) / ∆t = (1150 kg)(9.81 m/s 2 )(50 m)⎜ ⎜ 1000 kg ⋅ m 2 /s 2 ⎝

and

⎞ ⎟ /(12 s) = 47.0 kW ⎟ ⎠

W& total = W& a + W& g = 0 + 47.0 = 47.0 kW

(b) The power needed to accelerate is

[

1 1 W& a = m(V22 − V12 ) / ∆t = (1150 kg) (30 m/s )2 − 0 2 2 and

]⎛⎜⎜ 1000 kg1 kJ⋅ m /s 2

2

⎞ ⎟ /(12 s) = 43.1 kW ⎟ ⎠

W& total = W& a + W& g = 47.0 + 43.1 = 90.1 kW

(c) The power needed to decelerate is

[

1 1 W& a = m(V22 − V12 ) / ∆t = (1150 kg) (5 m/s)2 − (35 m/s)2 2 2 and

]⎛⎜⎜ 1000 kg1 kJ⋅ m /s ⎝

2

2

⎞ ⎟ /(12 s) = −57.5 kW ⎟ ⎠

W& total = W& a + W& g = −57.5 + 47.1 = −10.5 kW (breaking power)

2-39 A damaged car is being towed by a truck. The extra power needed is to be determined for three different cases. Assumptions Air drag, friction, and rolling resistance are negligible. Analysis The total power required for each case is the sum of the rates of changes in potential and kinetic energies. That is, W& total = W& a + W& g

(a) Zero. (b) W& a = 0 . Thus, ∆z = mgV z = mgV sin 30 o W& total = W& g = mg ( z 2 − z1 ) / ∆t = mg ∆t ⎛ 50,000 m ⎞⎛ 1 kJ/kg ⎞ ⎟ ⎟⎜ = (1200 kg)(9.81m/s 2 )⎜⎜ ⎟⎜ 1000 m 2 /s 2 ⎟(0.5) = 81.7 kW 3600 s ⎠⎝ ⎝ ⎠

(c) W& g = 0 . Thus, ⎛ ⎛ 90,000 m ⎞ 2 ⎞⎛ 1 kJ/kg ⎞ 1 1 2 2 & & ⎟ /(12 s) = 31.3 kW ⎟⎟ − 0 ⎟⎜ Wtotal = Wa = m(V2 − V1 ) / ∆t = (1200 kg)⎜ ⎜⎜ ⎜ 3600 s ⎟⎜ 1000 m 2 /s 2 ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-16 The First Law of Thermodynamics

2-40C No. This is the case for adiabatic systems only.

2-41C Energy can be transferred to or from a control volume as heat, various forms of work, and by mass transport.

2-42C Warmer. Because energy is added to the room air in the form of electrical work.

2-43E The high rolling resistance tires of a car are replaced by low rolling resistance ones. For a specified unit fuel cost, the money saved by switching to low resistance tires is to be determined. Assumptions 1The low rolling resistance tires deliver 2 mpg over all velocities. 2 The car is driven 15,000 miles per year. Analysis The annual amount of fuel consumed by this car on high- and low-rolling resistance tires are

Annual Fuel Consumption High =

Miles driven per year 15,000 miles/year = = 428.6 gal/year Miles per gallon 35 miles/gal

Annual Fuel Consumption Low =

Miles driven per year 15,000 miles/year = = 405.4 gal/year Miles per gallon 37 miles/gal

Then the fuel and money saved per year become Fuel Savings = Annual Fuel Consumptio n High − Annual Fuel Consumptio n Low = 428.6 gal/year − 405.4 gal/year = 23.2 gal/year

Cost savings = (Fuel savings)( Unit cost of fuel) = (23.2 gal/year)(\$2.20/gal) = \$51.0/year

Discussion A typical tire lasts about 3 years, and thus the low rolling resistance tires have the potential to save about \$150 to the car owner over the life of the tires, which is comparable to the installation cost of the tires.

2-44 The specific energy change of a system which is accelerated is to be determined. Analysis Since the only property that changes for this system is the velocity, only the kinetic energy will change. The change in the specific energy is

∆ke =

V 22 − V12 (30 m/s) 2 − (0 m/s) 2 ⎛ 1 kJ/kg ⎞ = ⎜ ⎟ = 0.45 kJ/kg 2 2 ⎝ 1000 m 2 /s 2 ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-17 2-45 The specific energy change of a system which is raised is to be determined. Analysis Since the only property that changes for this system is the elevation, only the potential energy will change. The change in the specific energy is then ⎛ 1 kJ/kg ⎞ ∆pe = g ( z 2 − z1 ) = (9.8 m/s 2 )(100 − 0) m⎜ ⎟ = 0.98 kJ/kg ⎝ 1000 m 2 /s 2 ⎠

2-46E A water pump increases water pressure. The power input is to be determined. Analysis The power input is determined from W& = V& ( P2 − P1 )

50 psia

⎛ 1 Btu = (1.2 ft 3 /s)(50 − 10)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 12.6 hp

⎞⎛ 1 hp ⎞ ⎟⎜ ⎟⎝ 0.7068 Btu/s ⎟⎠ ⎠

Water 10 psia

The water temperature at the inlet does not have any significant effect on the required power.

2-47 A classroom is to be air-conditioned using window air-conditioning units. The cooling load is due to people, lights, and heat transfer through the walls and the windows. The number of 5-kW window air conditioning units required is to be determined. Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. Analysis The total cooling load of the room is determined from Q& cooling = Q& lights + Q& people + Q& heat gain

where Q& lights = 10 × 100 W = 1 kW Q& people = 40 × 360 kJ / h = 4 kW Q& heat gain = 15,000 kJ / h = 4.17 kW

Room 15,000 kJ/h

40 people 10 bulbs

·

Qcool

Substituting, Q& cooling = 1 + 4 + 4.17 = 9.17 kW

Thus the number of air-conditioning units required is 9.17 kW 5 kW/unit

= 1.83 ⎯ ⎯→ 2 units

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-18 2-48 The lighting energy consumption of a storage room is to be reduced by installing motion sensors. The amount of energy and money that will be saved as well as the simple payback period are to be determined. Assumptions The electrical energy consumed by the ballasts is negligible. Analysis The plant operates 12 hours a day, and thus currently the lights are on for the entire 12 hour period. The motion sensors installed will keep the lights on for 3 hours, and off for the remaining 9 hours every day. This corresponds to a total of 9×365 = 3285 off hours per year. Disregarding the ballast factor, the annual energy and cost savings become

Energy Savings = (Number of lamps)(Lamp wattage)(Reduction of annual operating hours) = (24 lamps)(60 W/lamp )(3285 hours/year) = 4730 kWh/year Cost Savings = (Energy Savings)(Unit cost of energy) = (4730 kWh/year)(\$0.08/kWh) = \$378/year The implementation cost of this measure is the sum of the purchase price of the sensor plus the labor, Implementation Cost = Material + Labor = \$32 + \$40 = \$72 This gives a simple payback period of Simple payback period =

Implementation cost \$72 = = 0.19 year (2.3 months) Annual cost savings \$378 / year

Therefore, the motion sensor will pay for itself in about 2 months.

2-49 The classrooms and faculty offices of a university campus are not occupied an average of 4 hours a day, but the lights are kept on. The amounts of electricity and money the campus will save per year if the lights are turned off during unoccupied periods are to be determined. Analysis The total electric power consumed by the lights in the classrooms and faculty offices is E& lighting, classroom = (Power consumed per lamp) × (No. of lamps) = (200 × 12 × 110 W) = 264,000 = 264 kW E& lighting, offices = (Power consumed per lamp) × (No. of lamps) = (400 × 6 × 110 W) = 264,000 = 264 kW E& lighting, total = E& lighting, classroom + E& lighting, offices = 264 + 264 = 528 kW

Noting that the campus is open 240 days a year, the total number of unoccupied work hours per year is Unoccupied hours = (4 hours/day)(240 days/year) = 960 h/yr Then the amount of electrical energy consumed per year during unoccupied work period and its cost are Energy savings = ( E& lighting, total )( Unoccupied hours) = (528 kW)(960 h/yr) = 506,880 kWh Cost savings = (Energy savings)(Unit cost of energy) = (506,880 kWh/yr)(\$0.082/kWh) = \$41,564/yr

Discussion Note that simple conservation measures can result in significant energy and cost savings.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-19 2-50 A room contains a light bulb, a TV set, a refrigerator, and an iron. The rate of increase of the energy content of the room when all of these electric devices are on is to be determined. Assumptions 1 The room is well sealed, and heat loss from the room is negligible. 2 All the appliances are kept on. Analysis Taking the room as the system, the rate form of the energy balance can be written as

E& − E& 1in424out 3

Rate of net energy transfer by heat, work, and mass

=

dE system / dt 14243

dE room / dt = E& in

Rate of change in internal, kinetic, potential, etc. energies

ROOM

since no energy is leaving the room in any form, and thus E& out = 0 . Also, E& in = E& lights + E& TV + E& refrig + E& iron = 100 + 110 + 200 + 1000 W

Electricity

= 1410 W

- Lights - TV - Refrig - Iron

Substituting, the rate of increase in the energy content of the room becomes

dE room / dt = E& in = 1410 W Discussion Note that some appliances such as refrigerators and irons operate intermittently, switching on and off as controlled by a thermostat. Therefore, the rate of energy transfer to the room, in general, will be less.

2-51 A fan is to accelerate quiescent air to a specified velocity at a specified flow rate. The minimum power that must be supplied to the fan is to be determined. Assumptions The fan operates steadily. Properties The density of air is given to be ρ = 1.18 kg/m3. Analysis A fan transmits the mechanical energy of the shaft (shaft power) to mechanical energy of air (kinetic energy). For a control volume that encloses the fan, the energy balance can be written as

E& − E& 1in424out 3

Rate of net energy transfer by heat, work, and mass

E& in = E& out

Rate of change in internal, kinetic, potential, etc. energies

V2 W& sh, in = m& air ke out = m& air out 2 where

m& air = ρV& = (1.18 kg/m 3 )(9 m 3 /s) = 10.62 kg/s Substituting, the minimum power input required is determined to be

V2 (8 m/s) 2 ⎛ 1 J/kg ⎞ W& sh, in = m& air out = (10.62 kg/s) ⎜ ⎟ = 340 J/s = 340 W 2 2 ⎝ 1 m 2 /s 2 ⎠ Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the power required will be considerably higher because of the losses associated with the conversion of mechanical shaft energy to kinetic energy of air.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-20 2-52E A fan accelerates air to a specified velocity in a square duct. The minimum electric power that must be supplied to the fan motor is to be determined. Assumptions 1 The fan operates steadily. 2 There are no conversion losses. Properties The density of air is given to be ρ = 0.075 lbm/ft3. Analysis A fan motor converts electrical energy to mechanical shaft energy, and the fan transmits the mechanical energy of the shaft (shaft power) to mechanical energy of air (kinetic energy). For a control volume that encloses the fan-motor unit, the energy balance can be written as

E& − E& 1in424out 3

Rate of net energy transfer by heat, work, and mass

E& in = E& out

Rate of change in internal, kinetic, potential, etc. energies

V2 W& elect, in = m& air ke out = m& air out 2 where

m& air = ρVA = (0.075 lbm/ft3 )(3 × 3 ft 2 )(22 ft/s) = 14.85 lbm/s Substituting, the minimum power input required is determined to be V2 (22 ft/s) 2 W& in = m& air out = (14.85 lbm/s) 2 2

⎛ 1 Btu/lbm ⎜⎜ 2 2 ⎝ 25,037 ft /s

⎞ ⎟⎟ = 0.1435 Btu/s = 151 W ⎠

since 1 Btu = 1.055 kJ and 1 kJ/s = 1000 W. Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the power required will be considerably higher because of the losses associated with the conversion of electrical-to-mechanical shaft and mechanical shaft-to-kinetic energy of air.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-21 2-53 A gasoline pump raises the pressure to a specified value while consuming electric power at a specified rate. The maximum volume flow rate of gasoline is to be determined. Assumptions 1 The gasoline pump operates steadily. 2 The changes in kinetic and potential energies across the pump are negligible. Analysis For a control volume that encloses the pump-motor unit, the energy balance can be written as

E& − E& out 1in 424 3

Rate of net energy transfer by heat, work, and mass

=0

E&in = E& out 3.8 kW

Rate of change in internal, kinetic, potential, etc. energies

W&in + m& ( Pv )1 = m& ( Pv ) 2 → W&in = m& ( P2 − P1 )v = V& ∆P since m& = V&/v and the changes in kinetic and potential energies of gasoline are negligible, Solving for volume flow rate and substituting, the maximum flow rate is determined to be

V&max =

W& in 3.8 kJ/s ⎛ 1 kPa ⋅ m 3 ⎞ ⎜ ⎟ = 0.543 m 3 /s = ∆P 7 kPa ⎜⎝ 1 kJ ⎟⎠

PUMP

Motor

Pump inlet

Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the volume flow rate will be less because of the losses associated with the conversion of electrical-to-mechanical shaft and mechanical shaft-to-flow energy.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-22 2-54 An inclined escalator is to move a certain number of people upstairs at a constant velocity. The minimum power required to drive this escalator is to be determined. Assumptions 1 Air drag and friction are negligible. 2 The average mass of each person is 75 kg. 3 The escalator operates steadily, with no acceleration or breaking. 4 The mass of escalator itself is negligible. Analysis At design conditions, the total mass moved by the escalator at any given time is

Mass = (30 persons)(75 kg/person) = 2250 kg The vertical component of escalator velocity is Vvert = V sin 45° = (0.8 m/s)sin45°

Under stated assumptions, the power supplied is used to increase the potential energy of people. Taking the people on elevator as the closed system, the energy balance in the rate form can be written as E& − E& out 1in 424 3

Rate of net energy transfer by heat, work, and mass

=

dEsystem / dt 14243

=0

∆Esys E& in = dEsys / dt ≅ ∆t

Rate of change in internal, kinetic, potential, etc. energies

∆PE mg∆z = W&in = = mgVvert ∆t ∆t

That is, under stated assumptions, the power input to the escalator must be equal to the rate of increase of the potential energy of people. Substituting, the required power input becomes ⎛ 1 kJ/kg ⎞ ⎟ = 12.5 kJ/s = 12.5 kW W&in = mgVvert = (2250 kg)(9.81 m/s 2 )(0.8 m/s)sin45°⎜⎜ 2 2⎟ ⎝ 1000 m /s ⎠

When the escalator velocity is doubled to V = 1.6 m/s, the power needed to drive the escalator becomes ⎛ 1 kJ/kg ⎞ ⎟ = 25.0 kJ/s = 25.0 kW W&in = mgVvert = (2250 kg)(9.81 m/s 2 )(1.6 m/s)sin45°⎜⎜ 2 2⎟ ⎝ 1000 m /s ⎠

Discussion Note that the power needed to drive an escalator is proportional to the escalator velocity.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-23 2-55 An automobile moving at a given velocity is considered. The power required to move the car and the area of the effective flow channel behind the car are to be determined. Analysis The absolute pressure of the air is ⎛ 0.1333 kPa ⎞ ⎟⎟ = 93.31 kPa P = (700 mm Hg)⎜⎜ ⎝ 1 mm Hg ⎠

and the specific volume of the air is

v=

RT (0.287 kPa ⋅ m 3 /kg ⋅ K)(293 K) = = 0.9012 m 3 /kg P 93.31 kPa

The mass flow rate through the control volume is m& =

A1V1

v

=

(3 m 2 )(90/3.6 m/s) 0.9012 m 3 /kg

= 83.22 kg/s

The power requirement is

V 2 − V22 (90 / 3.6 m/s) 2 − (82 / 3.6 m/s) 2 ⎛ 1 kJ/kg ⎞ = (83.22 kg/s) W& = m& 1 ⎜ ⎟ = 4.42 kW 2 2 ⎝ 1000 m 2 /s 2 ⎠ The outlet area is

m& =

A2V2

v

⎯ ⎯→ A2 =

m& v (83.22 kg/s)(0.9012 m 3 /kg) = = 3.29 m 2 (82/3.6) m/s V2

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-24 Energy Conversion Efficiencies

2-56C Mechanical efficiency is defined as the ratio of the mechanical energy output to the mechanical energy input. A mechanical efficiency of 100% for a hydraulic turbine means that the entire mechanical energy of the fluid is converted to mechanical (shaft) work.

2-57C The combined pump-motor efficiency of a pump/motor system is defined as the ratio of the increase in the mechanical energy of the fluid to the electrical power consumption of the motor,

η pump-motor = η pumpη motor =

W& pump E& mech,out − E& mech,in ∆E& mech,fluid = = W& elect,in W& elect,in W& elect,in

The combined pump-motor efficiency cannot be greater than either of the pump or motor efficiency since both pump and motor efficiencies are less than 1, and the product of two numbers that are less than one is less than either of the numbers.

2-58C The turbine efficiency, generator efficiency, and combined turbine-generator efficiency are defined as follows:

η turbine =

W& shaft,out Mechanical energy output = Mechanical energy extracted from the fluid | ∆E& mech,fluid |

η generator =

Electrical power output W& elect,out = Mechanical power input W& shaft,in

η turbine -gen = η turbineηgenerator = & E

W&elect,out − E&

mech,in

mech,out

=

W&elect,out | ∆E& mech,fluid |

2-59C No, the combined pump-motor efficiency cannot be greater that either of the pump efficiency of the motor efficiency. This is because η pump - motor = η pumpη motor , and both η pump and η motor are less than one, and a number gets

smaller when multiplied by a number smaller than one.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-25 2-60 A hooded electric open burner and a gas burner are considered. The amount of the electrical energy used directly for cooking and the cost of energy per “utilized” kWh are to be determined. Analysis The efficiency of the electric heater is given to be 73 percent. Therefore, a burner that consumes 3-kW of electrical energy will supply

η gas = 38% η electric = 73% Q& utilized = (Energy input) × (Efficiency) = (2.4 kW)(0.73) = 1.75 kW of useful energy. The unit cost of utilized energy is inversely proportional to the efficiency, and is determined from Cost of utilized energy =

Cost of energy input \$0.10 / kWh = = \$0.137/kWh Efficiency 0.73

Noting that the efficiency of a gas burner is 38 percent, the energy input to a gas burner that supplies utilized energy at the same rate (1.75 kW) is

Q& utilized 1.75 kW Q& input, gas = = = 4.61 kW (= 15,700 Btu/h) Efficiency 0.38 since 1 kW = 3412 Btu/h. Therefore, a gas burner should have a rating of at least 15,700 Btu/h to perform as well as the electric unit. Noting that 1 therm = 29.3 kWh, the unit cost of utilized energy in the case of gas burner is determined the same way to be Cost of utilized energy =

Cost of energy input \$1.20 /( 29.3 kWh) = = \$0.108/kWh Efficiency 0.38

2-61 A worn out standard motor is replaced by a high efficiency one. The reduction in the internal heat gain due to the higher efficiency under full load conditions is to be determined. Assumptions 1 The motor and the equipment driven by the motor are in the same room. 2 The motor operates at full load so that fload = 1. Analysis The heat generated by a motor is due to its inefficiency, and the difference between the heat generated by two motors that deliver the same shaft power is simply the difference between the electric power drawn by the motors,

W& in, electric, standard = W& shaft / η motor = (75 × 746 W)/0.91 = 61,484 W W& in, electric, efficient = W& shaft / η motor = (75 × 746 W)/0.954 = 58,648 W Then the reduction in heat generation becomes Q& reduction = W& in, electric, standard − W& in, electric, efficient = 61,484 − 58,648 = 2836 W

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-26 2-62 An electric car is powered by an electric motor mounted in the engine compartment. The rate of heat supply by the motor to the engine compartment at full load conditions is to be determined. Assumptions The motor operates at full load so that the load factor is 1. Analysis The heat generated by a motor is due to its inefficiency, and is equal to the difference between the electrical energy it consumes and the shaft power it delivers, W& in, electric = W& shaft / η motor = (90 hp)/0.91 = 98.90 hp Q& generation = W& in, electric − W& shaft out = 98.90 − 90 = 8.90 hp = 6.64 kW

since 1 hp = 0.746 kW. Discussion Note that the electrical energy not converted to mechanical power is converted to heat.

2-63 A worn out standard motor is to be replaced by a high efficiency one. The amount of electrical energy and money savings as a result of installing the high efficiency motor instead of the standard one as well as the simple payback period are to be determined. Assumptions The load factor of the motor remains constant at 0.75. Analysis The electric power drawn by each motor and their difference can be expressed as W& electric in, standard = W& shaft / η standard = (Power rating)(Load factor) / η standard W& electric in, efficient = W& shaft / η efficient = (Power rating)(Load factor) / η efficient Power savings = W& electric in, standard − W& electric in, efficient = (Power rating)(Load factor)[1 / η standard − 1 / η efficient ]

where ηstandard is the efficiency of the standard motor, and ηefficient is the efficiency of the comparable high efficiency motor. Then the annual energy and cost savings associated with the installation of the high efficiency motor are determined to be Energy Savings = (Power savings)(Operating Hours) = (Power Rating)(Operating Hours)(Load Factor)(1/ηstandard- 1/ηefficient) = (75 hp)(0.746 kW/hp)(4,368 hours/year)(0.75)(1/0.91 - 1/0.954)

η old = 91.0% η new = 95.4%

= 9,290 kWh/year Cost Savings = (Energy savings)(Unit cost of energy) = (9,290 kWh/year)(\$0.08/kWh) = \$743/year The implementation cost of this measure consists of the excess cost the high efficiency motor over the standard one. That is, Implementation Cost = Cost differential = \$5,520 - \$5,449 = \$71 This gives a simple payback period of Simple payback period =

Implementation cost \$71 = = 0.096 year (or 1.1 months) Annual cost savings \$743 / year

Therefore, the high-efficiency motor will pay for its cost differential in about one month.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-27

2-64E The combustion efficiency of a furnace is raised from 0.7 to 0.8 by tuning it up. The annual energy and cost savings as a result of tuning up the boiler are to be determined. Assumptions The boiler operates at full load while operating. Analysis The heat output of boiler is related to the fuel energy input to the boiler by

Boiler output = (Boiler input)(Combustion efficiency) or

Q& out = Q& inη furnace

Boiler 70% 5.5×106

The current rate of heat input to the boiler is given to be Q& in, current = 5.5 × 10 6 Btu/h . Then the rate of useful heat output of the boiler becomes

Q& out = (Q& inη furnace ) current = (5.5 × 10 6 Btu/h)(0.7) = 3.85 × 10 6 Btu/h The boiler must supply useful heat at the same rate after the tune up. Therefore, the rate of heat input to the boiler after the tune up and the rate of energy savings become Q& in, new = Q& out / η furnace, new = (3.85 × 10 6 Btu/h)/0.8 = 4.81 × 10 6 Btu/h Q& in, saved = Q& in, current − Q& in, new = 5.5 × 10 6 − 4.81 × 10 6 = 0.69 × 10 6 Btu/h

Then the annual energy and cost savings associated with tuning up the boiler become Energy Savings = Q& in, saved (Operation hours) = (0.69×106 Btu/h)(4200 h/year) = 2.89×109 Btu/yr Cost Savings = (Energy Savings)(Unit cost of energy) = (2.89×109 Btu/yr)(\$4.35/106 Btu) = \$12,600/year Discussion Notice that tuning up the boiler will save \$12,600 a year, which is a significant amount. The implementation cost of this measure is negligible if the adjustment can be made by in-house personnel. Otherwise it is worthwhile to have an authorized representative of the boiler manufacturer to service the boiler twice a year.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-28

2-65E Problem 2-64E is reconsidered. The effects of the unit cost of energy and combustion efficiency on the annual energy used and the cost savings as the efficiency varies from 0.7 to 0.9 and the unit cost varies from \$4 to \$6 per million Btu are the investigated. The annual energy saved and the cost savings are to be plotted against the efficiency for unit costs of \$4, \$5, and \$6 per million Btu. Analysis The problem is solved using EES, and the solution is given below. "Given" Q_dot_in_current=5.5E6 [Btu/h] eta_furnace_current=0.7 eta_furnace_new=0.8 Hours=4200 [h/year] UnitCost=4.35E-6 [\$/Btu] "Analysis" Q_dot_out=Q_dot_in_current*eta_furnace_current Q_dot_in_new=Q_dot_out/eta_furnace_new Q_dot_in_saved=Q_dot_in_current-Q_dot_in_new Energysavings=Q_dot_in_saved*Hours CostSavings=EnergySavings*UnitCost 6x 109

0.7 0.72 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9

EnergySaving s [Btu/year] 0.00E+00 6.42E+08 1.25E+09 1.82E+09 2.37E+09 2.89E+09 3.38E+09 3.85E+09 4.30E+09 4.73E+09 5.13E+09

CostSaving s [\$/year] 0 3208 6243 9118 11846 14437 16902 19250 21488 23625 25667

Energysavings [Btu/year]

ηfurnace,new

5x 109 4x 109 3x 109 2x 109 109 0x 100 0.68

0.72

0.76

0.8

0.84

η furnace,new

0.88

0.92

30000

CostSavings [\$/year]

25000 20000 15000

6x10-6 \$/Btu 5x10-6 \$/Btu

10000

4x10-6 \$/Btu 5000 0 0.68

0.72

0.76

0.8

0.84

η furnace ,ne w

0.88

0.92

Table values are for UnitCost = 5E-5 [\$/Btu]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-29

2-66 Several people are working out in an exercise room. The rate of heat gain from people and the equipment is to be determined. Assumptions The average rate of heat dissipated by people in an exercise room is 525 W. Analysis The 8 weight lifting machines do not have any motors, and thus they do not contribute to the internal heat gain directly. The usage factors of the motors of the treadmills are taken to be unity since they are used constantly during peak periods. Noting that 1 hp = 746 W, the total heat generated by the motors is Q& motors = ( No. of motors) × W& motor × f load × f usage / η motor = 4 × (2.5 × 746 W) × 0.70 × 1.0/0.77 = 6782 W

The heat gain from 14 people is Q& people = ( No. of people) × Q& person = 14 × (525 W) = 7350 W

Then the total rate of heat gain of the exercise room during peak period becomes Q& total = Q& motors + Q& people = 6782 + 7350 = 14,132 W

2-67 A room is cooled by circulating chilled water through a heat exchanger, and the air is circulated through the heat exchanger by a fan. The contribution of the fan-motor assembly to the cooling load of the room is to be determined. Assumptions The fan motor operates at full load so that fload = 1. Analysis The entire electrical energy consumed by the motor, including the shaft power delivered to the fan, is eventually dissipated as heat. Therefore, the contribution of the fan-motor assembly to the cooling load of the room is equal to the electrical energy it consumes, Q& internal generation = W& in, electric = W& shaft / η motor = (0.25 hp)/0.54 = 0.463 hp = 345 W

since 1 hp = 746 W.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-30

2-68 A hydraulic turbine-generator is to generate electricity from the water of a lake. The overall efficiency, the turbine efficiency, and the shaft power are to be determined. Assumptions 1 The elevation of the lake and that of the discharge site remains constant. 2 Irreversible losses in the pipes are negligible. Properties The density of water can be taken to be ρ = 1000 kg/m3. The gravitational acceleration is g = 9.81 m/s2. Analysis (a) We take the bottom of the lake as the reference level for convenience. Then kinetic and potential energies of water are zero, and the mechanical energy of water consists of pressure energy only which is e mech,in − e mech,out =

P

ρ

= gh

⎛ 1 kJ/kg ⎞ = (9.81 m/s 2 )(50 m)⎜ ⎟ ⎝ 1000 m 2 /s 2 ⎠ = 0.491 kJ/kg

Then the rate at which mechanical energy of fluid supplied to the turbine and the overall efficiency become | ∆E& mech,fluid |= m& (e mech,in − e mech,in ) = (5000 kg/s)(0.491 kJ/kg) = 2455 kW

η overall = η turbine-gen =

W& elect,out 1862 kW = = 0.760 & | ∆E mech,fluid | 2455 kW

(b) Knowing the overall and generator efficiencies, the mechanical efficiency of the turbine is determined from

η turbine-gen = η turbineη generator → η turbine =

η turbine-gen η generator

=

0.76 = 0.800 0.95

(c) The shaft power output is determined from the definition of mechanical efficiency, W& shaft,out = η turbine | ∆E& mech,fluid |= (0.800)(2455 kW) = 1964 kW ≈ 1960 kW

Therefore, the lake supplies 2455 kW of mechanical energy to the turbine, which converts 1964 kW of it to shaft work that drives the generator, which generates 1862 kW of electric power.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-31

2-69 Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass, the power generation potential, and the actual electric power generation are to be determined. Assumptions 1 The wind is blowing steadily at a constant uniform velocity. 2 The efficiency of the wind turbine is independent of the wind speed. Properties The density of air is given to be ρ = 1.25 kg/m3. Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate:

e mech = ke =

Wind

Wind turbine

7 m/s

80 m

V 2 (7 m/s) 2 ⎛ 1 kJ/kg ⎞ = ⎜ ⎟ = 0.0245 kJ/kg 2 2 ⎝ 1000 m 2 /s 2 ⎠

m& = ρVA = ρV

πD 2 4

= (1.25 kg/m 3 )(7 m/s)

π (80 m) 2 4

= 43,982 kg/s

W& max = E& mech = m& emech = (43,982 kg/s)(0.0245 kJ/kg) = 1078 kW The actual electric power generation is determined by multiplying the power generation potential by the efficiency,

W& elect = η wind turbineW& max = (0.30)(1078 kW) = 323 kW Therefore, 323 kW of actual power can be generated by this wind turbine at the stated conditions. Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power generation will change strongly with the wind conditions.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-32

2-70 Problem 2-69 is reconsidered. The effect of wind velocity and the blade span diameter on wind power generation as the velocity varies from 5 m/s to 20 m/s in increments of 5 m/s, and the diameter varies from 20 m to 120 m in increments of 20 m is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Given" V=7 [m/s] D=80 [m] eta_overall=0.30 rho=1.25 [kg/m^3] "Analysis" g=9.81 [m/s^2] A=pi*D^2/4 m_dot=rho*A*V W_dot_max=m_dot*V^2/2*Convert(m^2/s^2, kJ/kg) W_dot_elect=eta_overall*W_dot_max

16000 14000

Welect [kW]

12000 10000

D=120 m

8000

D=100 m D=80 m

6000 4000

D=60 m D=40 m

2000

D=20 m 0 4

6

8

10

12

14

16

18

20

V [m/s]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-33

2-71 Water is pumped from a lake to a storage tank at a specified rate. The overall efficiency of the pump-motor unit and the pressure difference between the inlet and the exit of the pump are to be determined. Assumptions 1 The elevations of the tank and the lake remain constant. 2 Frictional losses in the pipes are negligible. 3 The changes in kinetic energy are negligible. 4 The elevation difference across the pump is negligible. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis (a) We take the free surface of the lake to be point 1 and the free surfaces of the storage tank to be point 2. We also take the lake surface as the reference level (z1 = 0), and thus the potential energy at points 1 and 2 are pe1 = 0 and pe2 = gz2. The flow energy at both points is zero since both 1 and 2 are open to the atmosphere (P1 = P2 = Patm). Further, the kinetic energy at both points is zero (ke1 = ke2 = 0) since the water at both locations is essentially stationary. The mass flow rate of water and its potential energy at point 2 are

2 Storage tank

20 m

Pump

1

m& = ρV& = (1000 kg/m 3 )(0.070 m 3/s) = 70 kg/s

⎛ 1 kJ/kg ⎞ pe 2 = gz 2 = (9.81 m/s 2 )(20 m)⎜ ⎟ = 0.196 kJ/kg ⎝ 1000 m 2 /s 2 ⎠ Then the rate of increase of the mechanical energy of water becomes ∆E& mech,fluid = m& (e mech,out − e mech,in ) = m& ( pe 2 − 0) = m& pe 2 = (70 kg/s)(0.196 kJ/kg) = 13.7 kW

The overall efficiency of the combined pump-motor unit is determined from its definition,

η pump-motor =

∆E& mech,fluid 13.7 kW = = 0.672 or 67.2% 20.4 kW W& elect,in

(b) Now we consider the pump. The change in the mechanical energy of water as it flows through the pump consists of the change in the flow energy only since the elevation difference across the pump and the change in the kinetic energy are negligible. Also, this change must be equal to the useful mechanical energy supplied by the pump, which is 13.7 kW: ∆E& mech,fluid = m& (e mech,out − e mech,in ) = m&

P2 − P1

ρ

= V&∆P

Solving for ∆P and substituting, ∆P =

∆E& mech,fluid 13.7 kJ/s ⎛ 1 kPa ⋅ m 3 ⎜ = 0.070 m 3 /s ⎜⎝ 1 kJ V&

⎞ ⎟ = 196 kPa ⎟ ⎠

Therefore, the pump must boost the pressure of water by 196 kPa in order to raise its elevation by 20 m. Discussion Note that only two-thirds of the electric energy consumed by the pump-motor is converted to the mechanical energy of water; the remaining one-third is wasted because of the inefficiencies of the pump and the motor.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-34

2-72 A large wind turbine is installed at a location where the wind is blowing steadily at a certain velocity. The electric power generation, the daily electricity production, and the monetary value of this electricity are to be determined. Assumptions 1 The wind is blowing steadily at a constant uniform velocity. 2 The efficiency of the wind turbine is independent of the wind speed. Properties The density of air is given to be ρ = 1.25 kg/m3.

Wind

Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate: e mech = ke =

Wind turbine

8 m/s

100 m

V 2 (8 m/s ) 2 ⎛ 1 kJ/kg ⎞ = ⎜ ⎟ = 0.032 kJ/kg 2 2 ⎝ 1000 m 2 /s 2 ⎠

m& = ρVA = ρV

πD 2 4

= (1.25 kg/m 3 )(8 m/s)

π (100 m) 2 4

= 78,540 kg/s

W& max = E& mech = m& e mech = (78,540 kg/s)(0.032 kJ/kg) = 2513 kW The actual electric power generation is determined from

W& elect = η wind turbineW& max = (0.32)(2513 kW) = 804.2 kW Then the amount of electricity generated per day and its monetary value become Amount of electricity = (Wind power)(Operating hours)=(804.2 kW)(24 h) =19,300 kWh Revenues = (Amount of electricity)(Unit price) = (19,300 kWh)(\$0.06/kWh) = \$1158 (per day) Discussion Note that a single wind turbine can generate several thousand dollars worth of electricity every day at a reasonable cost, which explains the overwhelming popularity of wind turbines in recent years.

2-73E A water pump raises the pressure of water by a specified amount at a specified flow rate while consuming a known amount of electric power. The mechanical efficiency of the pump is to be determined. ∆P = 1.2 psi Assumptions 1 The pump operates steadily. 2 The changes in velocity and elevation across the pump are negligible. 3 Water is incompressible. Analysis To determine the mechanical efficiency of the pump, we need to know the increase in the mechanical energy of the fluid as it flows through the pump, which is

6 hp PUMP

∆E& mech,fluid = m& (emech,out − e mech,in ) = m& [( Pv ) 2 − ( Pv )1 ] = m& ( P2 − P1 )v ⎛ 1 Btu = V& ( P2 − P1 ) = (15 ft 3 /s)(1.2 psi)⎜ ⎜ 5.404 psi ⋅ ft 3 ⎝

⎞ ⎟ = 3.33 Btu/s = 4.71 hp ⎟ ⎠

Pump inlet

since 1 hp = 0.7068 Btu/s, m& = ρV& = V& / v , and there is no change in kinetic and potential energies of the fluid. Then the mechanical efficiency of the pump becomes

η pump =

∆E& mech,fluid 4.71 hp = = 0.786 or 78.6% 6 hp W& pump, shaft

Discussion The overall efficiency of this pump will be lower than 83.8% because of the inefficiency of the electric motor that drives the pump.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-35

2-74 Water is pumped from a lower reservoir to a higher reservoir at a specified rate. For a specified shaft power input, the power that is converted to thermal energy is to be determined. Assumptions 1 The pump operates steadily. 2 The elevations of the reservoirs remain constant. 3 The changes in kinetic energy are negligible.

2

Properties We take the density of water to be ρ = 1000 kg/m3.

Reservoir

Analysis The elevation of water and thus its potential energy changes during pumping, but it experiences no changes in its velocity and pressure. Therefore, the change in the total mechanical energy of water is equal to the change in its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. That is,

45 m

Pump

1 Reservoir

∆E& mech = m& ∆e mech = m& ∆pe = m& g∆z = ρV&g∆z ⎛ 1N = (1000 kg/m 3 )(0.03 m 3 /s)(9.81 m/s 2 )(45 m)⎜⎜ ⋅ m/s 2 1 kg ⎝

⎞⎛ 1 kW ⎞ ⎟⎜ ⎟⎝ 1000 N ⋅ m/s ⎟⎠ = 13.2 kW ⎠

Then the mechanical power lost because of frictional effects becomes W& frict = W& pump, in − ∆E& mech = 20 − 13.2 kW = 6.8 kW

Discussion The 6.8 kW of power is used to overcome the friction in the piping system. The effect of frictional losses in a pump is always to convert mechanical energy to an equivalent amount of thermal energy, which results in a slight rise in fluid temperature. Note that this pumping process could be accomplished by a 13.2 kW pump (rather than 20 kW) if there were no frictional losses in the system. In this ideal case, the pump would function as a turbine when the water is allowed to flow from the upper reservoir to the lower reservoir and extract 13.2 kW of power from the water.

2-75 The mass flow rate of water through the hydraulic turbines of a dam is to be determined. Analysis The mass flow rate is determined from W& = m& g ( z 2 − z1 ) ⎯ ⎯→ m& =

W& = g ( z 2 − z1 )

100,000 kJ/s ⎛ 1 kJ/kg ⎞ (9.8 m/s 2 )(206 − 0) m⎜ ⎟ ⎝ 1000 m 2 /s 2 ⎠

= 49,500 kg/s

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-36

2-76 A pump is pumping oil at a specified rate. The pressure rise of oil in the pump is measured, and the motor efficiency is specified. The mechanical efficiency of the pump is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference across the pump is negligible. Properties The density of oil is given to be ρ = 860 kg/m3. Analysis Then the total mechanical energy of a fluid is the sum of the potential, flow, and kinetic energies, and is expressed per unit mass as emech = gh + Pv + V 2 / 2 . To determine the mechanical efficiency of the pump, we need to know the increase in the mechanical energy of the fluid as it flows through the pump, which is

⎛ V2 V2 ∆E& mech,fluid = m& (e mech,out − e mech,in ) = m& ⎜ ( Pv ) 2 + 2 − ( Pv ) 1 − 1 ⎜ 2 2 ⎝

⎞ &⎛ V 2 − V12 ⎟ = V ⎜ ( P2 − P1 ) + ρ 2 ⎜ ⎟ 2 ⎝ ⎠

since m& = ρV& = V& / v , and there is no change in the potential energy of the fluid. Also,

V1 = V2 =

V& A1

V& A2

V&

=

πD12

=

/4

V& πD 22 / 4

=

0.1 m 3 /s

π (0.08 m) 2 / 4

=

3

0.1 m /s

π (0.12 m) 2 / 4

2

⎞ ⎟ ⎟ ⎠ 35 kW

= 19.9 m/s PUMP

= 8.84 m/s

Motor

Pump inlet

Substituting, the useful pumping power is determined to be 1

W& pump,u = ∆E& mech,fluid ⎛ (8.84 m/s) 2 − (19.9 m/s) 2 = (0.1 m 3 /s)⎜ 400 kN/m 2 + (860 kg/m 3 ) ⎜ 2 ⎝ = 26.3 kW

⎛ 1 kN ⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝

⎞ ⎞⎛ 1 kW ⎞ ⎟ ⎟⎜ ⎟ ⎟⎝ 1 kN ⋅ m/s ⎟⎠ ⎠⎠

Then the shaft power and the mechanical efficiency of the pump become W& pump, shaft = η motor W& electric = (0.90)(35 kW) = 31.5 kW

η pump =

W& pump, u W&

pump, shaft

=

26.3 kW = 0.836 = 83.6% 31.5 kW

Discussion The overall efficiency of this pump/motor unit is the product of the mechanical and motor efficiencies, which is 0.9×0.836 = 0.75.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-37

2-77E Water is pumped from a lake to a nearby pool by a pump with specified power and efficiency. The mechanical power used to overcome frictional effects is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference between the lake and the free surface of the pool is constant. 3 The average flow velocity is constant since pipe diameter is constant. Properties We take the density of water to be ρ = 62.4 lbm/ft3. Analysis The useful mechanical pumping power delivered to water is W& pump,u = η pumpW& pump = (0.80)(20 hp) = 16 hp

The elevation of water and thus its potential energy changes during pumping, but it experiences no changes in its velocity and pressure. Therefore, the change in the total mechanical energy of water is equal to the change in its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. That is,

Pool

2

Pump

80 ft Lake

1

∆E& mech = m& ∆e mech = m& ∆pe = m& g∆z = ρV&g∆z Substituting, the rate of change of mechanical energy of water becomes 1 hp 1 lbf ⎛ ⎞⎛ ⎞ ∆E& mech = (62.4 lbm/ft 3 )(1.5 ft 3 /s)(32.2 ft/s 2 )(80 ft )⎜ ⎟⎜ ⎟ = 13.63 hp 2 550 lbf ⋅ ft/s ⎠ ⎝ 32.2 lbm ⋅ ft/s ⎠⎝

Then the mechanical power lost in piping because of frictional effects becomes W& frict = W& pump, u − ∆E& mech = 16 − 13.63 hp = 2.37 hp

Discussion Note that the pump must supply to the water an additional useful mechanical power of 2.37 hp to overcome the frictional losses in pipes.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-38

2-78 A wind turbine produces 180 kW of power. The average velocity of the air and the conversion efficiency of the turbine are to be determined. Assumptions The wind turbine operates steadily. Properties The density of air is given to be 1.31 kg/m3. Analysis (a) The blade diameter and the blade span area are Vtip D= = πn&

A=

πD 2 4

=

⎛ 1 m/s ⎞ (250 km/h)⎜ ⎟ ⎝ 3.6 km/h ⎠ = 88.42 m ⎛ 1 min ⎞ π (15 L/min)⎜ ⎟ ⎝ 60 s ⎠

π (88.42 m) 2 4

= 6140 m 2

Then the average velocity of air through the wind turbine becomes

V=

42,000 kg/s m& = = 5.23 m/s ρA (1.31 kg/m 3 )(6140 m 2 )

(b) The kinetic energy of the air flowing through the turbine is

KE& =

1 1 m& V 2 = (42,000 kg/s)(5.23 m/s) 2 = 574.3 kW 2 2

Then the conversion efficiency of the turbine becomes

η=

W& 180 kW = = 0.313 = 31.3% & KE 574.3 kW

Discussion Note that about one-third of the kinetic energy of the wind is converted to power by the wind turbine, which is typical of actual turbines.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-39 Energy and Environment

2-79C Energy conversion pollutes the soil, the water, and the air, and the environmental pollution is a serious threat to vegetation, wild life, and human health. The emissions emitted during the combustion of fossil fuels are responsible for smog, acid rain, and global warming and climate change. The primary chemicals that pollute the air are hydrocarbons (HC, also referred to as volatile organic compounds, VOC), nitrogen oxides (NOx), and carbon monoxide (CO). The primary source of these pollutants is the motor vehicles.

2-80C Smog is the brown haze that builds up in a large stagnant air mass, and hangs over populated areas on calm hot summer days. Smog is made up mostly of ground-level ozone (O3), but it also contains numerous other chemicals, including carbon monoxide (CO), particulate matter such as soot and dust, volatile organic compounds (VOC) such as benzene, butane, and other hydrocarbons. Ground-level ozone is formed when hydrocarbons and nitrogen oxides react in the presence of sunlight in hot calm days. Ozone irritates eyes and damage the air sacs in the lungs where oxygen and carbon dioxide are exchanged, causing eventual hardening of this soft and spongy tissue. It also causes shortness of breath, wheezing, fatigue, headaches, nausea, and aggravate respiratory problems such as asthma.

2-81C Fossil fuels include small amounts of sulfur. The sulfur in the fuel reacts with oxygen to form sulfur dioxide (SO2), which is an air pollutant. The sulfur oxides and nitric oxides react with water vapor and other chemicals high in the atmosphere in the presence of sunlight to form sulfuric and nitric acids. The acids formed usually dissolve in the suspended water droplets in clouds or fog. These acid-laden droplets are washed from the air on to the soil by rain or snow. This is known as acid rain. It is called “rain” since it comes down with rain droplets. As a result of acid rain, many lakes and rivers in industrial areas have become too acidic for fish to grow. Forests in those areas also experience a slow death due to absorbing the acids through their leaves, needles, and roots. Even marble structures deteriorate due to acid rain.

2-82C Carbon monoxide, which is a colorless, odorless, poisonous gas that deprives the body's organs from getting enough oxygen by binding with the red blood cells that would otherwise carry oxygen. At low levels, carbon monoxide decreases the amount of oxygen supplied to the brain and other organs and muscles, slows body reactions and reflexes, and impairs judgment. It poses a serious threat to people with heart disease because of the fragile condition of the circulatory system and to fetuses because of the oxygen needs of the developing brain. At high levels, it can be fatal, as evidenced by numerous deaths caused by cars that are warmed up in closed garages or by exhaust gases leaking into the cars.

2-83C Carbon dioxide (CO2), water vapor, and trace amounts of some other gases such as methane and nitrogen oxides act like a blanket and keep the earth warm at night by blocking the heat radiated from the earth. This is known as the greenhouse effect. The greenhouse effect makes life on earth possible by keeping the earth warm. But excessive amounts of these gases disturb the delicate balance by trapping too much energy, which causes the average temperature of the earth to rise and the climate at some localities to change. These undesirable consequences of the greenhouse effect are referred to as global warming or global climate change. The greenhouse effect can be reduced by reducing the net production of CO2 by consuming less energy (for example, by buying energy efficient cars and appliances) and planting trees.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-40

2-84E A person trades in his Ford Taurus for a Ford Explorer. The extra amount of CO2 emitted by the Explorer within 5 years is to be determined. Assumptions The Explorer is assumed to use 940 gallons of gasoline a year compared to 715 gallons for Taurus. Analysis The extra amount of gasoline the Explorer will use within 5 years is

Extra Gasoline

= (Extra per year)(No. of years) = (940 – 715 gal/yr)(5 yr) = 1125 gal

Extra CO2 produced

= (Extra gallons of gasoline used)(CO2 emission per gallon) = (1125 gal)(19.7 lbm/gal) = 22,163 lbm CO2

Discussion Note that the car we choose to drive has a significant effect on the amount of greenhouse gases produced.

2-85 A power plant that burns natural gas produces 0.59 kg of carbon dioxide (CO2) per kWh. The amount of CO2 production that is due to the refrigerators in a city is to be determined. Assumptions The city uses electricity produced by a natural gas power plant. Properties 0.59 kg of CO2 is produced per kWh of electricity generated (given). Analysis Noting that there are 300,000 households in the city and each household consumes 700 kWh of electricity for refrigeration, the total amount of CO2 produced is Amount of CO 2 produced = (Amount of electricity consumed)(Amount of CO 2 per kWh) = (300,000 household)(700 kWh/year household)(0.59 kg/kWh) = 1.23 × 10 8 CO 2 kg/year = 123,000 CO 2 ton/year

Therefore, the refrigerators in this city are responsible for the production of 123,000 tons of CO2.

2-86 A power plant that burns coal, produces 1.1 kg of carbon dioxide (CO2) per kWh. The amount of CO2 production that is due to the refrigerators in a city is to be determined. Assumptions The city uses electricity produced by a coal power plant. Properties 1.1 kg of CO2 is produced per kWh of electricity generated (given). Analysis Noting that there are 300,000 households in the city and each household consumes 700 kWh of electricity for refrigeration, the total amount of CO2 produced is Amount of CO 2 produced = (Amount of electricity consumed)(Amount of CO 2 per kWh) = (300,000 household)(700 kWh/household)(1.1 kg/kWh) = 2.31 × 10 8 CO 2 kg/year = 231,000 CO 2 ton/year

Therefore, the refrigerators in this city are responsible for the production of 231,000 tons of CO2.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-41

2-87E A household uses fuel oil for heating, and electricity for other energy needs. Now the household reduces its energy use by 20%. The reduction in the CO2 production this household is responsible for is to be determined. Properties The amount of CO2 produced is 1.54 lbm per kWh and 26.4 lbm per gallon of fuel oil (given). Analysis Noting that this household consumes 11,000 kWh of electricity and 1500 gallons of fuel oil per year, the amount of CO2 production this household is responsible for is Amount of CO 2 produced = (Amount of electricity consumed)(Amount of CO 2 per kWh) + (Amount of fuel oil consumed)(Amount of CO 2 per gallon) = (11,000 kWh/yr)(1.54 lbm/kWh) + (1500 gal/yr)(26.4 lbm/gal) = 56,540 CO 2 lbm/year

Then reducing the electricity and fuel oil usage by 15% will reduce the annual amount of CO2 production by this household by Reduction in CO 2 produced = (0.15)(Current amount of CO 2 production) = (0.15)(56,540 CO 2 kg/year) = 8481 CO 2 lbm/year

Therefore, any measure that saves energy also reduces the amount of pollution emitted to the environment.

2-88 A household has 2 cars, a natural gas furnace for heating, and uses electricity for other energy needs. The annual amount of NOx emission to the atmosphere this household is responsible for is to be determined. Properties The amount of NOx produced is 7.1 g per kWh, 4.3 g per therm of natural gas, and 11 kg per car (given). Analysis Noting that this household has 2 cars, consumes 1200 therms of natural gas, and 9,000 kWh of electricity per year, the amount of NOx production this household is responsible for is

Amount of NO x produced = ( No. of cars)(Amount of NO x produced per car) + ( Amount of electricity consumed)(Amount of NO x per kWh) + ( Amount of gas consumed)(Amount of NO x per gallon) = (2 cars)(11 kg/car) + (9000 kWh/yr)(0.0071 kg/kWh) + (1200 therms/yr)(0.0043 kg/therm) = 91.06 NOx kg/year Discussion Any measure that saves energy will also reduce the amount of pollution emitted to the atmosphere.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-42 Special Topic: Mechanisms of Heat Transfer

2-89C The three mechanisms of heat transfer are conduction, convection, and radiation.

2-90C Diamond has a higher thermal conductivity than silver, and thus diamond is a better conductor of heat.

2-91C No. It is purely by radiation.

2-92C In forced convection, the fluid is forced to move by external means such as a fan, pump, or the wind. The fluid motion in natural convection is due to buoyancy effects only.

2-93C A blackbody is an idealized body that emits the maximum amount of radiation at a given temperature, and that absorbs all the radiation incident on it. Real bodies emit and absorb less radiation than a blackbody at the same temperature.

2-94C Emissivity is the ratio of the radiation emitted by a surface to the radiation emitted by a blackbody at the same temperature. Absorptivity is the fraction of radiation incident on a surface that is absorbed by the surface. The Kirchhoff's law of radiation states that the emissivity and the absorptivity of a surface are equal at the same temperature and wavelength.

2-95 The inner and outer surfaces of a brick wall are maintained at specified temperatures. The rate of heat transfer through the wall is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Thermal properties of the wall are constant. Properties The thermal conductivity of the wall is given to be W/m⋅°C.

k = 0.69

Analysis Under steady conditions, the rate of heat transfer through the wall is ∆T (20 − 5)°C Q& cond = kA = (0.69 W/m ⋅ °C)(5 × 6 m 2 ) = 1035 W L 0.3 m

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-43

2-96 The inner and outer surfaces of a window glass are maintained at specified temperatures. The amount of heat transferred through the glass in 5 h is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. 2 Thermal properties of the glass are constant. Properties The thermal conductivity of the glass is given to be k = 0.78 W/m⋅°C.

Glass

Analysis Under steady conditions, the rate of heat transfer through the glass by conduction is (15 − 6)°C ∆T = (0.78 W/m ⋅ °C)(2 × 2 m 2 ) = 5616 W Q& cond = kA L 0.005 m

Then the amount of heat transferred over a period of 10 h becomes

Q = Q& cond ∆t = (5.616 kJ/s)(10 × 3600s) = 202,200 kJ

15°C

6°C

0.5 cm

If the thickness of the glass is doubled to 1 cm, then the amount of heat transferred will go down by half to 101,100 kJ.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-44

2-97 Reconsider Prob. 2-96. Using EES (or other) software, investigate the effect of glass thickness on heat loss for the specified glass surface temperatures. Let the glass thickness vary from 0.2 cm to 2 cm. Plot the heat loss versus the glass thickness, and discuss the results. Analysis The problem is solved using EES, and the solution is given below. FUNCTION klookup(material\$) If material\$='Glass' then klookup:=0.78 If material\$='Brick' then klookup:=0.72 If material\$='Fiber Glass' then klookup:=0.043 If material\$='Air' then klookup:=0.026 If material\$='Wood(oak)' then klookup:=0.17 END L=2 [m] W=2 [m] material\$='Glass' T_in=15 [C] T_out=6 [C] k=0.78 [W/m-C] t=10 [hr] thickness=0.5 [cm] A=L*W Q_dot_loss=A*k*(T_in-T_out)/(thickness*convert(cm,m)) Q_loss_total=Q_dot_loss*t*convert(hr,s)*convert(J,kJ)

Qloss,total [kJ 505440 252720 168480 126360 101088 84240 72206 63180 56160 50544

600000 500000

Qloss,total [kJ]

Thicknes s [cm] 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

400000 300000 200000 100000 0 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

thickness [cm]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-45

2-98 Heat is transferred steadily to boiling water in the pan through its bottom. The inner surface temperature of the bottom of the pan is given. The temperature of the outer surface is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the pan remain constant at the specified values. 2 Thermal properties of the aluminum pan are constant. Properties The thermal conductivity of the aluminum is given to be k = 237 W/m⋅°C. Analysis The heat transfer surface area is

A = π r² = π(0.1 m)² = 0.0314 m² Under steady conditions, the rate of heat transfer through the bottom of the pan by conduction is

105°C

∆T T −T Q& = kA = kA 2 1 L L

500 W 0.4 cm

Substituting, 500 W = (237 W / m⋅o C)(0.0314 m2 )

T2 − 105o C 0.004 m

which gives

T2 = 105.3°C

2-99 The inner and outer glasses of a double pane window with a 1-cm air space are at specified temperatures. The rate of heat transfer through the window is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. 2 Heat transfer through the window is one-dimensional. 3 Thermal properties of the air are constant. 4 The air trapped between the two glasses is still, and thus heat transfer is by conduction only.

18°C

Air 6°C · Q

Properties The thermal conductivity of air at room temperature is k = 0.026 W/m.°C (Table 2-3). Analysis Under steady conditions, the rate of heat transfer through the window by conduction is

1cm

o

(18 − 6) C ∆T = (0.026 W/m⋅ o C)(2 × 2 m 2 ) = 125 W = 0.125 kW Q& cond = kA 0.01 m L

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-46

2-100 Two surfaces of a flat plate are maintained at specified temperatures, and the rate of heat transfer through the plate is measured. The thermal conductivity of the plate material is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the plate remain constant at the specified values. 2 Heat transfer through the plate is one-dimensional. 3 Thermal properties of the plate are constant. Analysis The thermal conductivity is determined directly from the steady one-dimensional heat conduction relation to be

Plate 2 cm 0°C

100°C

T −T Q& = kA 1 2 L & (Q / A) L (500 W/m 2 )(0.02 m) k= = = 0.1 W/m.°C (100 - 0)°C T1 − T2

2-101 A person is standing in a room at a specified temperature. The rate of heat transfer between a person and the surrounding air by convection is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is not considered. 3 The environment is at a uniform temperature.

500 W/m2

Q& Ts =34°C

Analysis The heat transfer surface area of the person is

A = πDL = π(0.3 m)(1.70 m) = 1.60 m² Under steady conditions, the rate of heat transfer by convection is

Q& conv = hA∆T = (15 W/m2 ⋅ °C)(1.60 m2 )(34 − 20)°C = 336 W

2-102 A spherical ball whose surface is maintained at a temperature of 110°C is suspended in the middle of a room at 20°C. The total rate of heat transfer from the ball is to be determined. Assumptions 1 Steady operating conditions exist since the ball surface and the surrounding air and surfaces remain at constant temperatures. 2 The thermal properties of the ball and the convection heat transfer coefficient are constant and uniform.

Air 20°C 110°C

Properties The emissivity of the ball surface is given to be ε = 0.8. Analysis The heat transfer surface area is

A = πD² = π (0.09 m)2 = 0.02545 m2

D = 9 cm

. Q

Under steady conditions, the rates of convection and radiation heat transfer are Q& conv = hA∆T = (15 W/m 2 ⋅ o C)(0.02545 m 2 )(110 − 20) o C = 34.35 W Q& rad = εσA(Ts4 − To4 ) = 0.8(0.02545 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(383 K) 4 − (293 K) 4 ] = 16.33 W Therefore,

Q& total = Q& conv + Q& rad = 34.35 + 16.33 = 50.7 W

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-47

2-103 Reconsider Prob. 2-102. Using EES (or other) software, investigate the effect of the convection heat transfer coefficient and surface emissivity on the heat transfer rate from the ball. Let the heat transfer coefficient vary from 5 W/m2.°C to 30 W/m2.°C. Plot the rate of heat transfer against the convection heat transfer coefficient for the surface emissivities of 0.1, 0.5, 0.8, and 1, and discuss the results. Analysis The problem is solved using EES, and the solution is given below. "Given" D=0.09 [m] T_s=ConvertTemp(C,K,110) T_f=ConvertTemp(C,K,20) h=15 [W/m^2-C] epsilon=0.8 "Properties" sigma=5.67E-8 [W/m^2-K^4] "Analysis" A=pi*D^2 Q_dot_conv=h*A*(T_s-T_f) Q_dot_rad=epsilon*sigma*A*(T_s^4-T_f^4) Q_dot_total=Q_dot_conv+Q_dot_rad

Qtotal [W] 27.8 33.53 39.25 44.98 50.7 56.43 62.16 67.88 73.61 79.33 85.06

90 80 70

Qtotal [W]

h [W/m2-C] 5 7.5 10 12.5 15 17.5 20 22.5 25 27.5 30

ε=1

ε=0.8 ε=0.5

60 50

ε=0.1

40 30 5

10

15

20

25

30

2

h [W/m -°C]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-48

2-104 Hot air is blown over a flat surface at a specified temperature. The rate of heat transfer from the air to the plate is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is not considered. 3 The convection heat transfer coefficient is constant and uniform over the surface.

80°C Air

Analysis Under steady conditions, the rate of heat transfer by convection is

30°C

Q& conv = hA∆T = (55 W/m 2 ⋅ °C)(2 × 4 m 2 )(80 − 30)o C = 22,000 W = 22 kW

2-105 A 1000-W iron is left on the iron board with its base exposed to the air at 20°C. The temperature of the base of the iron is to be determined in steady operation. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the iron base and the convection heat transfer coefficient are constant and uniform. 3 The temperature of the surrounding surfaces is the same as the temperature of the surrounding air. Properties The emissivity of the base surface is given to be

Iron 1000 W Air 20°C

ε = 0.6.

Analysis At steady conditions, the 1000 W of energy supplied to the iron will be dissipated to the surroundings by convection and radiation heat transfer. Therefore, Q& total = Q& conv + Q& rad = 1000 W

where

Q& conv = hA∆T = (35 W/m 2 ⋅ K)(0.02 m 2 )(Ts − 293 K) = 0.7(Ts − 293 K) W and Q& rad = εσA(Ts4 − To4 ) = 0.6(0.02 m 2 )(5.67 × 10 −8 W / m 2 ⋅ K 4 )[Ts4 − (293 K) 4 ] = 0.06804 × 10 −8 [Ts4 − (293 K) 4 ] W Substituting,

1000 W = 0.7(Ts − 293 K ) + 0.06804 ×10 −8 [Ts4 − (293 K) 4 ] Solving by trial and error gives Ts = 947 K = 674°C

Discussion We note that the iron will dissipate all the energy it receives by convection and radiation when its surface temperature reaches 947 K.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-49

2-106 The backside of the thin metal plate is insulated and the front side is exposed to solar radiation. The surface temperature of the plate is to be determined when it stabilizes. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the insulated side of the plate is negligible. 3 The heat transfer coefficient is constant and uniform over the plate. 4 Heat loss by radiation is negligible. Properties The solar absorptivity of the plate is given to be α = 0.8. Analysis When the heat loss from the plate by convection equals the solar radiation absorbed, the surface temperature of the plate can be determined from

Q& solarabsorbed = Q& conv

αQ& solar = hA(Ts − To ) 0.8 × A × 450 W/m 2 = (50 W/m 2 ⋅ °C) A(Ts − 25)

450 W/m2 α = 0.8 25°C

Canceling the surface area A and solving for Ts gives Ts = 32.2°C

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-50

2-107 Reconsider Prob. 2-106. Using EES (or other) software, investigate the effect of the convection heat transfer coefficient on the surface temperature of the plate. Let the heat transfer coefficient vary from 10 W/m2.°C to 90 W/m2.°C. Plot the surface temperature against the convection heat transfer coefficient, and discuss the results. Analysis The problem is solved using EES, and the solution is given below. "Given" alpha=0.8 q_dot_solar=450 [W/m^2] T_f=25 [C] h=50 [W/m^2-C] "Analysis" q_dot_solarabsorbed=alpha*q_dot_solar q_dot_conv=h*(T_s-T_f) q_dot_solarabsorbed=q_dot_conv

Ts [C] 61 49 43 39.4 37 35.29 34 33 32.2 31.55 31 30.54 30.14 29.8 29.5 29.24 29

65 60 55 50

Ts (°C)

h [W/m2-C] 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90

45 40 35 30 25 10

20

30

40

50

60

70

80

90

2

h (W/m -°C)

2-108 A hot water pipe at 80°C is losing heat to the surrounding air at 5°C by natural convection with a heat transfer coefficient of 25 W/ m2.°C. The rate of heat loss from the pipe by convection is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is not considered. 3 The convection heat transfer coefficient is constant and uniform over the surface.

80°C

D = 5 cm

Analysis The heat transfer surface area is

A = (πD)L = 3.14x(0.05 m)(10 m) = 1.571 m²

L = 10 m

Under steady conditions, the rate of heat transfer by convection is

Q Air, 5°C

Q& conv = hA∆T = (25 W/m2 ⋅ °C)(1.571 m 2 )(80 − 5)°C = 2945 W = 2.95 kW

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-51

2-109 A spacecraft in space absorbs solar radiation while losing heat to deep space by thermal radiation. The surface temperature of the spacecraft is to be determined when steady conditions are reached.. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Thermal properties of the spacecraft are constant. Properties The outer surface of a spacecraft has an emissivity of 0.6 and an absorptivity of 0.2. Analysis When the heat loss from the outer surface of the spacecraft by radiation equals the solar radiation absorbed, the surface temperature can be determined from Q& solar absorbed = Q& rad 4 αQ& solar = εσA(Ts4 − Tspace )

1000 W/m2 α = 0.2 ε = 0.6

0.2 × A × (1000 W/m 2 ) = 0.6 × A × (5.67 × 10 −8 W/m 2 ⋅ K 4 )[Ts4 − (0 K) 4 ]

Canceling the surface area A and solving for Ts gives Ts = 276.9 K

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-52

2-110 Reconsider Prob. 2-109. Using EES (or other) software, investigate the effect of the surface emissivity and absorptivity of the spacecraft on the equilibrium surface temperature. Plot the surface temperature against emissivity for solar absorptivities of 0.1, 0.5, 0.8, and 1, and discuss the results. Analysis The problem is solved using EES, and the solution is given below. "Given" epsilon=0.2 alpha=0.6 q_dot_solar=1000 [W/m^2] T_f=0 [K] "space temperature" "Properties" sigma=5.67E-8 [W/m^2-K^4] "Analysis" q_dot_solarabsorbed=alpha*q_dot_solar q_dot_rad=epsilon*sigma*(T_s^4-T_f^4) q_dot_solarabsorbed=q_dot_rad

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Ts [K] 648 544.9 492.4 458.2 433.4 414.1 398.4 385.3 374.1 364.4

Table for ε = 1

650 600 550 500

Ts (K)

ε

450

α=1

400

α=0.8

350

α=0.5

α=0.1

300 250 0.1

0.2

0.3

0.4

0.5

ε

0.6

0.7

0.8

0.9

1

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-53

2-111 A hollow spherical iron container is filled with iced water at 0°C. The rate of heat loss from the sphere and the rate at which ice melts in the container are to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Heat transfer through the shell is one-dimensional. 3 Thermal properties of the iron shell are constant. 4 The inner surface of the shell is at the same temperature as the iced water, 0°C. Properties The thermal conductivity of iron is k = 80.2 W/m⋅°C (Table 2-3). The heat of fusion of water is at 1 atm is 333.7 kJ/kg. 5°C Analysis This spherical shell can be approximated as a plate of thickness 0.4 cm and surface area

A = πD² = 3.14×(0.2 m)² = 0.126 m² Then the rate of heat transfer through the shell by conduction is ∆T (5 − 0)°C Q& cond = kA = (80.2 W/m⋅o C)(0.126 m 2 ) = 12,632 W L 0.004 m

Iced water 0°C

0.4 cm

Considering that it takes 333.7 kJ of energy to melt 1 kg of ice at 0°C, the rate at which ice melts in the container can be determined from

m& ice =

Q& 12.632 kJ/s = = 0.038 kg/s hif 333.7 kJ/kg

Discussion We should point out that this result is slightly in error for approximating a curved wall as a plain wall. The error in this case is very small because of the large diameter to thickness ratio. For better accuracy, we could use the inner surface area (D = 19.2 cm) or the mean surface area (D = 19.6 cm) in the calculations.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-54

Review Problems

2-112 A classroom has a specified number of students, instructors, and fluorescent light bulbs. The rate of internal heat generation in this classroom is to be determined. Assumptions 1 There is a mix of men, women, and children in the classroom. 2 The amount of light (and thus energy) leaving the room through the windows is negligible. Properties The average rate of heat generation from people seated in a room/office is given to be 100 W. Analysis The amount of heat dissipated by the lamps is equal to the amount of electrical energy consumed by the lamps, including the 10% additional electricity consumed by the ballasts. Therefore, Q& lighting = (Energy consumed per lamp) × (No. of lamps) Q& people

= (40 W)(1.1)(18) = 792 W = ( No. of people) × Q& = 56 × (100 W) = 5600 W person

Then the total rate of heat gain (or the internal heat load) of the classroom from the lights and people become Q& total = Q& lighting + Q& people = 792 + 5600 = 6392 W

2-113 A decision is to be made between a cheaper but inefficient natural gas heater and an expensive but efficient natural gas heater for a house. Assumptions The two heaters are comparable in all aspects other than the initial cost and efficiency. Analysis Other things being equal, the logical choice is the heater that will cost less during its lifetime. The total cost of a system during its lifetime (the initial, operation, maintenance, etc.) can be determined by performing a life cycle cost analysis. A simpler alternative is to determine the simple payback period.

The annual heating cost is given to be \$1200. Noting that the existing heater is 55% efficient, only 55% of that energy (and thus money) is delivered to the house, and the rest is wasted due to the inefficiency of the heater. Therefore, the monetary value of the heating load of the house is

Gas Heater η1 = 82% η2 = 95%

Cost of useful heat = (55%)(Current annual heating cost) = 0.55×(\$1200/yr)=\$660/yr This is how much it would cost to heat this house with a heater that is 100% efficient. For heaters that are less efficient, the annual heating cost is determined by dividing \$660 by the efficiency:

82% heater:

Annual cost of heating = (Cost of useful heat)/Efficiency = (\$660/yr)/0.82 = \$805/yr

95% heater:

Annual cost of heating = (Cost of useful heat)/Efficiency = (\$660/yr)/0.95 = \$695/yr

Annual cost savings with the efficient heater = 805 - 695 = \$110 Excess initial cost of the efficient heater = 2700 - 1600 = \$1100 The simple payback period becomes Simple payback period =

Excess initial cost \$1100 = = 10 years Annaul cost savings \$110 / yr

Therefore, the more efficient heater will pay for the \$1100 cost differential in this case in 10 years, which is more than the 8-year limit. Therefore, the purchase of the cheaper and less efficient heater is a better buy in this case.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-55

2-114 A wind turbine is rotating at 20 rpm under steady winds of 30 km/h. The power produced, the tip speed of the blade, and the revenue generated by the wind turbine per year are to be determined. Assumptions 1 Steady operating conditions exist. 2 The wind turbine operates continuously during the entire year at the specified conditions. Properties The density of air is given to be ρ = 1.20 kg/m3. Analysis (a) The blade span area and the mass flow rate of air through the turbine are

A = πD 2 / 4 = π (80 m) 2 / 4 = 5027 m 2 ⎛ 1000 m ⎞⎛ 1 h ⎞ V = (30 km/h)⎜ ⎟⎜ ⎟ = 8.333 m/s ⎝ 1 km ⎠⎝ 3600 s ⎠ m& = ρAV = (1.2 kg/m 3 )(5027 m 2 )(8.333 m/s) = 50,270 kg/s Noting that the kinetic energy of a unit mass is V2/2 and the wind turbine captures 35% of this energy, the power generated by this wind turbine becomes 1 ⎛ 1 kJ/kg ⎞ ⎛1 ⎞ W& = η ⎜ m& V 2 ⎟ = (0.35) (50,270 kg/s)(8.333 m/s ) 2 ⎜ ⎟ = 610.9 kW 2 ⎝2 ⎠ ⎝ 1000 m 2 /s 2 ⎠ (b) Noting that the tip of blade travels a distance of πD per revolution, the tip velocity of the turbine blade for an rpm of n& becomes

V tip = πDn& = π (80 m)(20 / min) = 5027 m/min = 83.8 m/s = 302 km/h (c) The amount of electricity produced and the revenue generated per year are Electricity produced = W& ∆t = (610.9 kW)(365 × 24 h/year) = 5.351× 10 6 kWh/year Revenue generated = (Electricity produced)(Unit price) = (5.351× 10 6 kWh/year)(\$0.06/kWh) = \$321,100/y ear

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-56

2-115 A wind turbine is rotating at 20 rpm under steady winds of 20 km/h. The power produced, the tip speed of the blade, and the revenue generated by the wind turbine per year are to be determined. Assumptions 1 Steady operating conditions exist. 2 The wind turbine operates continuously during the entire year at the specified conditions. Properties The density of air is given to be ρ = 1.20 kg/m3. Analysis (a) The blade span area and the mass flow rate of air through the turbine are

A = πD 2 / 4 = π (80 m) 2 / 4 = 5027 m 2 ⎛ 1000 m ⎞⎛ 1 h ⎞ V = (20 km/h)⎜ ⎟⎜ ⎟ = 5.556 m/s ⎝ 1 km ⎠⎝ 3600 s ⎠ m& = ρAV = (1.2 kg/m 3 )(5027 m 2 )(5.556 m/s) = 33,510 kg/s Noting that the kinetic energy of a unit mass is V2/2 and the wind turbine captures 35% of this energy, the power generated by this wind turbine becomes 1 ⎛ 1 kJ/kg ⎞ ⎛1 ⎞ W& = η ⎜ m& V 2 ⎟ = (0.35) (33,510 kg/s)(6.944 m/s) 2 ⎜ ⎟ = 181.0 kW 2 2 ⎝ ⎠ ⎝ 1000 m 2 /s 2 ⎠

(b) Noting that the tip of blade travels a distance of πD per revolution, the tip velocity of the turbine blade for an rpm of n& becomes

Vtip = πDn& = π (80 m)(20 / min) = 5027 m/min = 83.8 m/s = 302 km/h (c) The amount of electricity produced and the revenue generated per year are Electricity produced = W& ∆t = (181.0 kW)(365 × 24 h/year) = 1,585,535 kWh/year Revenue generated = (Electricity produced)(Unit price) = (1,585,535 kWh/year)(\$0.06/kWh) = \$95,130/ye ar

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-57

2-116E The energy contents, unit costs, and typical conversion efficiencies of various energy sources for use in water heaters are given. The lowest cost energy source is to be determined. Assumptions The differences in installation costs of different water heaters are not considered. Properties The energy contents, unit costs, and typical conversion efficiencies of different systems are given in the problem statement. Analysis The unit cost of each Btu of useful energy supplied to the water heater by each system can be determined from

Unit cost of useful energy =

Unit cost of energy supplied Conversion efficiency

Substituting,

⎛ 1 ft 3 ⎞ −6 ⎜ ⎟ ⎜ 1025 Btu ⎟ = \$21.3 × 10 / Btu ⎝ ⎠

Natural gas heater:

Unit cost of useful energy =

\$0.012/ft 3 0.55

Heating by oil heater:

Unit cost of useful energy =

\$1.15/gal ⎛ 1 gal ⎞ ⎟ = \$15.1× 10 − 6 / Btu ⎜ 0.55 ⎜⎝ 138,700 Btu ⎟⎠

Electric heater:

Unit cost of useful energy =

\$0.084/kWh) ⎛ 1 kWh ⎞ −6 ⎜ ⎟ = \$27.4 × 10 / Btu 0.90 ⎝ 3412 Btu ⎠

Therefore, the lowest cost energy source for hot water heaters in this case is oil.

2-117 A home owner is considering three different heating systems for heating his house. The system with the lowest energy cost is to be determined. Assumptions The differences in installation costs of different heating systems are not considered. Properties The energy contents, unit costs, and typical conversion efficiencies of different systems are given in the problem statement. Analysis The unit cost of each Btu of useful energy supplied to the house by each system can be determined from

Unit cost of useful energy =

Unit cost of energy supplied Conversion efficiency

Substituting, Natural gas heater:

Unit cost of useful energy =

\$1.24/therm ⎛ 1 therm ⎞ −6 ⎜⎜ 105,500 kJ ⎟⎟ = \$13.5 × 10 / kJ 0.87 ⎠ ⎝

Heating oil heater:

Unit cost of useful energy =

\$1.25/gal ⎛ 1 gal ⎞ ⎟ = \$10.4 × 10 − 6 / kJ ⎜ 0.87 ⎜⎝ 138,500 kJ ⎟⎠

Electric heater:

Unit cost of useful energy =

\$0.09/kWh) ⎛ 1 kWh ⎞ −6 ⎜ ⎟ = \$25.0 × 10 / kJ 1.0 ⎝ 3600 kJ ⎠

Therefore, the system with the lowest energy cost for heating the house is the heating oil heater.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-58

2-118 The heating and cooling costs of a poorly insulated house can be reduced by up to 30 percent by adding adequate insulation. The time it will take for the added insulation to pay for itself from the energy it saves is to be determined. Assumptions It is given that the annual energy usage of a house is \$1200 a year, and 46% of it is used for heating and cooling. The cost of added insulation is given to be \$200.

Heat loss

Analysis The amount of money that would be saved per year is determined directly from Money saved = (\$1200 / year)(0.46)(0.30) = \$166 / yr

Then the simple payback period becomes Payback period =

Cost \$200 = = 1.2 yr Money saved \$166/yr

House

Therefore, the proposed measure will pay for itself in less than one and a half year.

2-119 Caulking and weather-stripping doors and windows to reduce air leaks can reduce the energy use of a house by up to 10 percent. The time it will take for the caulking and weather-stripping to pay for itself from the energy it saves is to be determined. Assumptions It is given that the annual energy usage of a house is \$1100 a year, and the cost of caulking and weatherstripping a house is \$60. Analysis The amount of money that would be saved per year is determined directly from Money saved = (\$1100 / year)(0.10) = \$110 / yr

Then the simple payback period becomes Payback period =

Cost \$60 = = 0.546 yr Money saved \$110/yr

Therefore, the proposed measure will pay for itself in less than half a year.

2-120E The energy stored in the spring of a railroad car is to be expressed in different units. Analysis Using appropriate conversion factors, we obtain

⎞ ⎟ = 160,870 lbm ⋅ ft 2 /s 2 ⎟ ⎠

(a)

⎛ 32.174 lbm ⋅ ft/s 2 W = (5000 lbf ⋅ ft )⎜ ⎜ 1 lbf ⎝

(b)

⎛ 0.33303 yd ⎞ W = (5000 lbf ⋅ ft )⎜ ⎟ = 1665 lbf ⋅ yd 1 ft ⎠ ⎝

(c)

⎛ 32.174 lbm ⋅ ft/s 2 W = (5000 lbf ⋅ ft )⎜ ⎜ 1 lbf ⎝

⎞⎛ 1 mile ⎞ 2 ⎛ 3600 s ⎞ 2 ⎟⎜ = 74,785 lbm ⋅ mile 2 /h 2 ⎟⎝ 5280 ft ⎟⎠ ⎜⎝ 1 h ⎟⎠ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-59

2-121E The work required to compress a gas in a gas spring is to be determined. Assumptions All forces except that generated by the gas spring will be neglected. Analysis When the expression given in the problem statement is substituted into the work integral relation, and advantage is taken of the fact that the force and displacement vectors are collinear, the result is 2

W = Fds = 1

2

∫ 1

Constant xk

dx

Constant 1− k = ( x 2 − x11− k ) 1− k

[

F

x

]

200 lbf ⋅ in 1.4 ⎛ 1 ft ⎞ (4 in) − 0.4 − (1 in) − 0.4 ⎜ ⎟ 1 − 1.4 ⎝ 12 in ⎠ = 17.74 lbf ⋅ ft =

1 Btu ⎛ ⎞ = (17.74 lbf ⋅ ft)⎜ ⎟ = 0.0228 Btu ⎝ 778.169 lbf ⋅ ft ⎠

2-122E A man pushes a block along a horizontal plane. The work required to move the block is to be determined considering (a) the man and (b) the block as the system. Analysis The work applied to the block to overcome the friction is found by using the work integral, 2

W = Fds = 1

2

∫ fW ( x

2

− x1 )

x

1

= (0.2)(100 lbf )(100 ft) = 2000 lbf ⋅ ft

fW

1 Btu ⎞ ⎛ = (2000 lbf ⋅ ft)⎜ ⎟ = 2.57 Btu ⎝ 778.169 lbf ⋅ ft ⎠

The man must then produce the amount of work

fW W

W = 2.57 Btu

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-60

2-123 A diesel engine burning light diesel fuel that contains sulfur is considered. The rate of sulfur that ends up in the exhaust and the rate of sulfurous acid given off to the environment are to be determined. Assumptions 1 All of the sulfur in the fuel ends up in the exhaust. 2 For one kmol of sulfur in the exhaust, one kmol of sulfurous acid is added to the environment. Properties The molar mass of sulfur is 32 kg/kmol. Analysis The mass flow rates of fuel and the sulfur in the exhaust are m& fuel =

(336 kg air/h) m& air = = 18.67 kg fuel/h AF (18 kg air/kg fuel)

m& Sulfur = (750 × 10 -6 )m& fuel = (750 × 10 -6 )(18.67 kg/h) = 0.014 kg/h The rate of sulfurous acid given off to the environment is m& H2SO3 =

M H2SO3 2 × 1 + 32 + 3 × 16 m& Sulfur = (0.014 kg/h) = 0.036 kg/h M Sulfur 32

Discussion This problem shows why the sulfur percentage in diesel fuel must be below certain value to satisfy regulations.

2-124 Lead is a very toxic engine emission. Leaded gasoline contains lead that ends up in the exhaust. The amount of lead put out to the atmosphere per year for a given city is to be determined. Assumptions 35% of lead is exhausted to the environment. Analysis The gasoline consumption and the lead emission are

Gasoline Consumption = (5000 cars)(15,000 km/car - year)(8.5 L/100 km) = 6.375 × 10 6 L/year Lead Emission = (GaolineConsumption)mlead f lead = (6.375 × 10 6 L/year)(0.15 × 10 -3 kg/L)(0.35) = 335 kg/year Discussion Note that a huge amount of lead emission is avoided by the use of unleaded gasoline.

2-125E The power required to pump a specified rate of water to a specified elevation is to be determined. Properties The density of water is taken to be 62.4 lbm/ft3 (Table A-3E). Analysis The required power is determined from W& = m& g ( z 2 − z1 ) = ρV&g ( z 2 − z1 ) ⎛ 35.315 ft 3 /s ⎞ 1 lbf ⎞ ⎟(32.174 ft/s 2 )(300 ft)⎛⎜ = (62.4 lbm/ft 3 )(200 gal/min)⎜⎜ ⎟ 2 ⎟ ⎝ 32.174 lbm ⋅ ft/s ⎠ ⎝ 15,850 gal/min ⎠ 1 kW ⎛ ⎞ = 8342 lbf ⋅ ft/s = (8342 lbf ⋅ ft/s)⎜ ⎟ = 11.3 kW ⎝ 737.56 lbf ⋅ ft/s ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-61

2-126 The power that could be produced by a water wheel is to be determined. Properties The density of water is taken to be 1000 m3/kg (Table A-3). Analysis The power production is determined from

W& = m& g ( z 2 − z1 ) = ρV&g ( z 2 − z1 ) ⎛ 1 kJ/kg ⎞ = (1000 kg/m 3 )(0.320/60 m 3 /s)(9.81 m/s 2 )(14 m)⎜ ⎟ = 0.732 kW ⎝ 1000 m 2 /s 2 ⎠

2-127 The flow of air through a flow channel is considered. The diameter of the wind channel downstream from the rotor and the power produced by the windmill are to be determined. Analysis The specific volume of the air is

v=

RT (0.287 kPa ⋅ m 3 /kg ⋅ K)(293 K) = = 0.8409 m 3 /kg P 100 kPa

The diameter of the wind channel downstream from the rotor is ⎯→(πD12 / 4)V1 = (πD 22 / 4)V 2 A1V1 = A2V 2 ⎯ ⎯ ⎯→ D 2 = D1

V1 10 m/s = (7 m) = 7.38 m 9 m/s V2

The mass flow rate through the wind mill is m& =

A1V1

v

=

π (7 m) 2 (10 m/s) 4(0.8409 m 3 /kg)

= 457.7 kg/s

The power produced is then

V 2 − V22 (10 m/s) 2 − (9 m/s) 2 ⎛ 1 kJ/kg ⎞ W& = m& 1 = (457.7 kg/s) ⎜ ⎟ = 4.35 kW 2 2 ⎝ 1000 m 2 /s 2 ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-62

2-128 The available head, flow rate, and efficiency of a hydroelectric turbine are given. The electric power output is to be determined. Assumptions 1 The flow is steady. 2 Water levels at the reservoir and the discharge site remain constant. 3 Frictional losses in piping are negligible. Properties We take the density of water to be ρ = 1000 kg/m3 = 1 kg/L. Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam (relative to free surface of discharge water), and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. ⎛ 1 kJ/kg e mech = pe = gz = (9.81 m/s 2 )(90 m)⎜ ⎝ 1000 m 2 /s 2

⎞ ⎟ = 0.8829 kJ/kg ⎠

1

90 m

The mass flow rate is

m& = ρV& = (1000 kg/m 3 )(65 m 3 /s) = 65,000 kg/s

ηoverall = 84%

Generator

Turbin 2

Then the maximum and actual electric power generation become

⎛ 1 MW ⎞ W& max = E& mech = m& e mech = (65,000 kg/s)(0.8829 kJ/kg)⎜ ⎟ = 57.39 MW ⎝ 1000 kJ/s ⎠ W& electric = η overallW& max = 0.84(57.39 MW) = 48.2 MW Discussion Note that the power generation would increase by more than 1 MW for each percentage point improvement in the efficiency of the turbine–generator unit.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-63

2-129 An entrepreneur is to build a large reservoir above the lake level, and pump water from the lake to the reservoir at night using cheap power, and let the water flow from the reservoir back to the lake during the day, producing power. The potential revenue this system can generate per year is to be determined. Assumptions 1 The flow in each direction is steady and incompressible. 2 The elevation difference between the lake and the reservoir can be taken to be constant, and the elevation change of reservoir during charging and discharging is disregarded. 3 Frictional losses in piping are negligible. 4 The system operates every day of the year for 10 hours in each mode. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis The total mechanical energy of water in an upper reservoir relative to water in a lower reservoir is equivalent to the potential energy of water at the free surface of this reservoir relative to free surface of the lower reservoir. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. This also represents the minimum power required to pump water from the lower reservoir to the higher reservoir.

2 Reservoir

Pumpturbine

40 m Lake

1

W&max, turbine = W&min, pump = W&ideal = ∆E& mech = m& ∆emech = m& ∆pe = m& g∆z = ρV&g∆z ⎞⎛ ⎛ 1N 1 kW ⎞ ⎟ = (1000 kg/m3 )(2 m3/s)(9.81 m/s2 )(40 m)⎜⎜ ⎟ 2 ⎟⎜ 1000 N m/s ⋅ 1 kg m/s ⋅ ⎠ ⎠⎝ ⎝ = 784.8 kW The actual pump and turbine electric powers are W& pump, elect =

W& ideal

η pump -motor

=

784.8 kW = 1046 kW 0.75

W& turbine = η turbine -gen W& ideal = 0.75(784.8 kW) = 588.6 kW

Then the power consumption cost of the pump, the revenue generated by the turbine, and the net income (revenue minus cost) per year become Cost = W& pump, elect ∆t × Unit price = (1046 kW)(365 × 10 h/year)(\$ 0.03/kWh) = \$114,500/y ear

Reveue = W& turbine ∆t × Unit price = (588.6 kW)(365 × 10 h/year)(\$0.08/kWh) = \$171,900/year Net income = Revenue – Cost = 171,900 –114,500 = \$57,400/year Discussion It appears that this pump-turbine system has a potential to generate net revenues of about \$57,000 per year. A decision on such a system will depend on the initial cost of the system, its life, the operating and maintenance costs, the interest rate, and the length of the contract period, among other things.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-64 Fundamentals of Engineering (FE) Exam Problems

2-130 A 2-kW electric resistance heater in a room is turned on and kept on for 50 min. The amount of energy transferred to the room by the heater is (a) 2 kJ

(b) 100 kJ

(c) 3000 kJ

(d) 6000 kJ

(e) 12,000 kJ

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). We= 2 "kJ/s" time=50*60 "s" We_total=We*time "kJ" "Some Wrong Solutions with Common Mistakes:" W1_Etotal=We*time/60 "using minutes instead of s" W2_Etotal=We "ignoring time"

2-131 In a hot summer day, the air in a well-sealed room is circulated by a 0.50-hp (shaft) fan driven by a 65% efficient motor. (Note that the motor delivers 0.50 hp of net shaft power to the fan). The rate of energy supply from the fan-motor assembly to the room is (a) 0.769 kJ/s

(b) 0.325 kJ/s

(c) 0.574 kJ/s

(d) 0.373 kJ/s

(e) 0.242 kJ/s

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Eff=0.65 W_fan=0.50*0.7457 "kW" E=W_fan/Eff "kJ/s" "Some Wrong Solutions with Common Mistakes:" W1_E=W_fan*Eff "Multiplying by efficiency" W2_E=W_fan "Ignoring efficiency" W3_E=W_fan/Eff/0.7457 "Using hp instead of kW"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-65 3

3

2-132 A fan is to accelerate quiescent air to a velocity to 12 m/s at a rate of 3 m /min. If the density of air is 1.15 kg/m , the minimum power that must be supplied to the fan is (a) 248 W

(b) 72 W

(c) 497 W

(d) 216 W

(e) 162 W

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1.15 V=12 Vdot=3 "m3/s" mdot=rho*Vdot "kg/s" We=mdot*V^2/2 "Some Wrong Solutions with Common Mistakes:" W1_We=Vdot*V^2/2 "Using volume flow rate" W2_We=mdot*V^2 "forgetting the 2" W3_We=V^2/2 "not using mass flow rate"

2-133 A 900-kg car cruising at a constant speed of 60 km/h is to accelerate to 100 km/h in 4 s. The additional power needed to achieve this acceleration is (a) 56 kW

(b) 222 kW

(c) 2.5 kW

(d) 62 kW

(e) 90 kW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=900 "kg" V1=60 "km/h" V2=100 "km/h" Dt=4 "s" Wa=m*((V2/3.6)^2-(V1/3.6)^2)/2000/Dt "kW" "Some Wrong Solutions with Common Mistakes:" W1_Wa=((V2/3.6)^2-(V1/3.6)^2)/2/Dt "Not using mass" W2_Wa=m*((V2)^2-(V1)^2)/2000/Dt "Not using conversion factor" W3_Wa=m*((V2/3.6)^2-(V1/3.6)^2)/2000 "Not using time interval" W4_Wa=m*((V2/3.6)-(V1/3.6))/1000/Dt "Using velocities"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-66

2-134 The elevator of a large building is to raise a net mass of 400 kg at a constant speed of 12 m/s using an electric motor. Minimum power rating of the motor should be (a) 0 kW

(b) 4.8 kW

(c) 47 kW

(d) 12 kW

(e) 36 kW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=400 "kg" V=12 "m/s" g=9.81 "m/s2" Wg=m*g*V/1000 "kW" "Some Wrong Solutions with Common Mistakes:" W1_Wg=m*V "Not using g" W2_Wg=m*g*V^2/2000 "Using kinetic energy" W3_Wg=m*g/V "Using wrong relation"

2-135 Electric power is to be generated in a hydroelectric power plant that receives water at a rate of 70 m3/s from an elevation of 65 m using a turbine–generator with an efficiency of 85 percent. When frictional losses in piping are disregarded, the electric power output of this plant is (a) 3.9 MW

(b) 38 MW

(c) 45 MW

(d) 53 MW

(e) 65 MW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Vdot=70 "m3/s" z=65 "m" g=9.81 "m/s2" Eff=0.85 rho=1000 "kg/m3" We=rho*Vdot*g*z*Eff/10^6 "MW" "Some Wrong Solutions with Common Mistakes:" W1_We=rho*Vdot*z*Eff/10^6 "Not using g" W2_We=rho*Vdot*g*z/Eff/10^6 "Dividing by efficiency" W3_We=rho*Vdot*g*z/10^6 "Not using efficiency"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-67

2-136 A 75 hp (shaft) compressor in a facility that operates at full load for 2500 hours a year is powered by an electric motor that has an efficiency of 93 percent. If the unit cost of electricity is \$0.06/kWh, the annual electricity cost of this compressor is (a) \$7802

(b) \$9021

(c) \$12,100

(d) \$8389

(e) \$10,460

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Wcomp=75 "hp" Hours=2500 “h/year” Eff=0.93 price=0.06 “\$/kWh” We=Wcomp*0.7457*Hours/Eff Cost=We*price "Some Wrong Solutions with Common Mistakes:" W1_cost= Wcomp*0.7457*Hours*price*Eff “multiplying by efficiency” W2_cost= Wcomp*Hours*price/Eff “not using conversion” W3_cost= Wcomp*Hours*price*Eff “multiplying by efficiency and not using conversion” W4_cost= Wcomp*0.7457*Hours*price “Not using efficiency”

2-137 Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarter of the time and the unit cost of electricity is \$0.09/kWh, the electricity cost of this refrigerator per month (30 days) is (a) \$3.56

(b) \$5.18

(c) \$8.54

(d) \$9.28

(e) \$20.74

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). We=0.320 "kW" Hours=0.25*(24*30) "h/year" price=0.09 "\$/kWh" Cost=We*hours*price "Some Wrong Solutions with Common Mistakes:" W1_cost= We*24*30*price "running continuously"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-68

2-138 A 2-kW pump is used to pump kerosene (ρ = 0.820 kg/L) from a tank on the ground to a tank at a higher elevation. Both tanks are open to the atmosphere, and the elevation difference between the free surfaces of the tanks is 30 m. The maximum volume flow rate of kerosene is (a) 8.3 L/s

(b) 7.2 L/s

(c) 6.8 L/s

(d) 12.1 L/s

(e) 17.8 L/s

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). W=2 "kW" rho=0.820 "kg/L" z=30 "m" g=9.81 "m/s2" W=rho*Vdot*g*z/1000 "Some Wrong Solutions with Common Mistakes:" W=W1_Vdot*g*z/1000 "Not using density"

2-139 A glycerin pump is powered by a 5-kW electric motor. The pressure differential between the outlet and the inlet of the pump at full load is measured to be 211 kPa. If the flow rate through the pump is 18 L/s and the changes in elevation and the flow velocity across the pump are negligible, the overall efficiency of the pump is (a) 69%

(b) 72%

(c) 76%

(d) 79%

(e) 82%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). We=5 "kW" Vdot= 0.018 "m3/s" DP=211 "kPa" Emech=Vdot*DP Emech=Eff*We

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-69 The following problems are based on the optional special topic of heat transfer

2-140 A 10-cm high and 20-cm wide circuit board houses on its surface 100 closely spaced chips, each generating heat at a rate of 0.08 W and transferring it by convection to the surrounding air at 25°C. Heat transfer from the back surface of the board is negligible. If the convection heat transfer coefficient on the surface of the board is 10 W/m2.°C and radiation heat transfer is negligible, the average surface temperature of the chips is (a) 26°C

(b) 45°C

(c) 15°C

(d) 80°C

(e) 65°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). A=0.10*0.20 "m^2" Q= 100*0.08 "W" Tair=25 "C" h=10 "W/m^2.C" Q= h*A*(Ts-Tair) "W" "Some Wrong Solutions with Common Mistakes:" Q= h*(W1_Ts-Tair) "Not using area" Q= h*2*A*(W2_Ts-Tair) "Using both sides of surfaces" Q= h*A*(W3_Ts+Tair) "Adding temperatures instead of subtracting" Q/100= h*A*(W4_Ts-Tair) "Considering 1 chip only"

2-141 A 50-cm-long, 0.2-cm-diameter electric resistance wire submerged in water is used to determine the boiling heat transfer coefficient in water at 1 atm experimentally. The surface temperature of the wire is measured to be 130°C when a wattmeter indicates the electric power consumption to be 4.1 kW. Then the heat transfer coefficient is (a) 43,500 W/m2.°C

(b) 137 W/m2.°C

(c) 68,330 W/m2.°C

(d) 10,038 W/m2.°C

(e) 37,540 W/m2.°C Answer (a) 43,500 W/m2.°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). L=0.5 "m" D=0.002 "m" A=pi*D*L "m^2" We=4.1 "kW" Ts=130 "C" Tf=100 "C (Boiling temperature of water at 1 atm)" We= h*A*(Ts-Tf) "W" "Some Wrong Solutions with Common Mistakes:" We= W1_h*(Ts-Tf) "Not using area" We= W2_h*(L*pi*D^2/4)*(Ts-Tf) "Using volume instead of area" We= W3_h*A*Ts "Using Ts instead of temp difference"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-70 2

2-142 A 3-m hot black surface at 80°C is losing heat to the surrounding air at 25°C by convection with a convection heat transfer coefficient of 12 W/m2.°C, and by radiation to the surrounding surfaces at 15°C. The total rate of heat loss from the surface is (a) 1987 W

(b) 2239 W

(c) 2348 W

(d) 3451 W

(e) 3811 W

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). sigma=5.67E-8 "W/m^2.K^4" eps=1 A=3 "m^2" h_conv=12 "W/m^2.C" Ts=80 "C" Tf=25 "C" Tsurr=15 "C" Q_conv=h_conv*A*(Ts-Tf) "W" Q_rad=eps*sigma*A*((Ts+273)^4-(Tsurr+273)^4) "W" Q_total=Q_conv+Q_rad "W" "Some Wrong Solutions with Common Mistakes:" W1_Ql=Q_conv "Ignoring radiation" W2_Q=Q_rad "ignoring convection" W3_Q=Q_conv+eps*sigma*A*(Ts^4-Tsurr^4) "Using C in radiation calculations" W4_Q=Q_total/A "not using area"

2-143 Heat is transferred steadily through a 0.2-m thick 8 m by 4 m wall at a rate of 2.4 kW. The inner and outer surface temperatures of the wall are measured to be 15°C to 5°C. The average thermal conductivity of the wall is (a) 0.002 W/m.°C

(b) 0.75 W/m.°C

(c) 1.0 W/m.°C

(d) 1.5 W/m.°C

(e) 3.0 W/m.°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). A=8*4 "m^2" L=0.2 "m" T1=15 "C" T2=5 "C" Q=2400 "W" Q=k*A*(T1-T2)/L "W" "Some Wrong Solutions with Common Mistakes:" Q=W1_k*(T1-T2)/L "Not using area" Q=W2_k*2*A*(T1-T2)/L "Using areas of both surfaces" Q=W3_k*A*(T1+T2)/L "Adding temperatures instead of subtracting" Q=W4_k*A*L*(T1-T2) "Multiplying by thickness instead of dividing by it"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

2-71

2-144 The roof of an electrically heated house is 7 m long, 10 m wide, and 0.25 m thick. It is made of a flat layer of concrete whose thermal conductivity is 0.92 W/m.°C. During a certain winter night, the temperatures of the inner and outer surfaces of the roof are measured to be 15°C and 4°C, respectively. The average rate of heat loss through the roof that night was (a) 41 W

(b) 177 W

(c) 4894 W

(d) 5567 W

(e) 2834 W

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). A=7*10 "m^2" L=0.25 "m" k=0.92 "W/m.C" T1=15 "C" T2=4 "C" Q_cond=k*A*(T1-T2)/L "W" "Some Wrong Solutions with Common Mistakes:" W1_Q=k*(T1-T2)/L "Not using area" W2_Q=k*2*A*(T1-T2)/L "Using areas of both surfaces" W3_Q=k*A*(T1+T2)/L "Adding temperatures instead of subtracting" W4_Q=k*A*L*(T1-T2) "Multiplying by thickness instead of dividing by it"

2-145 … 2-151 Design and Essay Problems

KJ

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-1

Solutions Manual for

Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011

Chapter 3 PROPERTIES OF PURE SUBSTANCES

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-2 Pure Substances, Phase Change Processes, Property Diagrams 3-1C A liquid that is about to vaporize is saturated liquid; otherwise it is compressed liquid.

3-2C A vapor that is about to condense is saturated vapor; otherwise it is superheated vapor.

3-3C No.

3-4C The temperature will also increase since the boiling or saturation temperature of a pure substance depends on pressure.

3-5C Because one cannot be varied while holding the other constant. In other words, when one changes, so does the other one.

3-6C At critical point the saturated liquid and the saturated vapor states are identical. At triple point the three phases of a pure substance coexist in equilibrium.

3-7C Yes.

3-8C Case (c) when the pan is covered with a heavy lid. Because the heavier the lid, the greater the pressure in the pan, and thus the greater the cooking temperature.

3-9C At supercritical pressures, there is no distinct phase change process. The liquid uniformly and gradually expands into a vapor. At subcritical pressures, there is always a distinct surface between the phases.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-3 Property Tables 3-10C A perfectly fitting pot and its lid often stick after cooking as a result of the vacuum created inside as the temperature and thus the corresponding saturation pressure inside the pan drops. An easy way of removing the lid is to reheat the food. When the temperature rises to boiling level, the pressure rises to atmospheric value and thus the lid will come right off.

3-11C The molar mass of gasoline (C8H18) is 114 kg/kmol, which is much larger than the molar mass of air that is 29 kg/kmol. Therefore, the gasoline vapor will settle down instead of rising even if it is at a much higher temperature than the surrounding air. As a result, the warm mixture of air and gasoline on top of an open gasoline will most likely settle down instead of rising in a cooler environment

3-12C Yes. Otherwise we can create energy by alternately vaporizing and condensing a substance.

3-13C No. Because in the thermodynamic analysis we deal with the changes in properties; and the changes are independent of the selected reference state.

3-14C The term hfg represents the amount of energy needed to vaporize a unit mass of saturated liquid at a specified temperature or pressure. It can be determined from hfg = hg - hf .

3-15C Yes. It decreases with increasing pressure and becomes zero at the critical pressure.

3-16C Yes; the higher the temperature the lower the hfg value.

3-17C Quality is the fraction of vapor in a saturated liquid-vapor mixture. It has no meaning in the superheated vapor region.

3-18C Completely vaporizing 1 kg of saturated liquid at 1 atm pressure since the higher the pressure, the lower the hfg .

3-19C No. Quality is a mass ratio, and it is not identical to the volume ratio.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-4 3-20C The compressed liquid can be approximated as a saturated liquid at the given temperature. Thus v T , P ≅ v f @ T .

3-21C Ice can be made by evacuating the air in a water tank. During evacuation, vapor is also thrown out, and thus the vapor pressure in the tank drops, causing a difference between the vapor pressures at the water surface and in the tank. This pressure difference is the driving force of vaporization, and forces the liquid to evaporate. But the liquid must absorb the heat of vaporization before it can vaporize, and it absorbs it from the liquid and the air in the neighborhood, causing the temperature in the tank to drop. The process continues until water starts freezing. The process can be made more efficient by insulating the tank well so that the entire heat of vaporization comes essentially from the water.

3-22

Complete the following table for H2 O:

T, °C

P, kPa

v, m3 / kg

Phase description

50

12.35

7.72

Saturated mixture

143.6

400

0.4624

Saturated vapor

250

500

0.4744

Superheated vapor

110

350

0.001051

Compressed liquid

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-5 3-23 Problem 3-22 is reconsidered. The missing properties of water are to be determined using EES, and the solution is to be repeated for refrigerant-134a, refrigerant-22, and ammonia. Analysis The problem is solved using EES, and the solution is given below. "Given" T[1]=50 [C] v[1]=7.72 [m^3/kg] P[2]=400 [kPa] x[2]=1 T[3]=250 [C] P[3]=500 [kPa] T[4]=110 [C] P[4]=350 [kPa] "Analysis" Fluid\$='steam_iapws' "Change the Fluid to R134a, R22 and Ammonia and solve" P[1]=pressure(Fluid\$, T=T[1], v=v[1]) x[1]=quality(Fluid\$, T=T[1], v=v[1]) T[2]=temperature(Fluid\$, P=P[2], x=x[2]) v[2]=volume(Fluid\$, P=P[2], x=x[2]) v[3]=volume(Fluid\$, P=P[3], T=T[3]) x[3]=quality(Fluid\$, P=P[3], T=T[3]) v[4]=volume(Fluid\$, P=P[4], T=T[4]) x[4]=quality(Fluid\$, P=P[4], T=T[4]) "x = 100 for superheated vapor and x = -100 for compressed liquid"

SOLUTION for water T [C]

P [kPa]

x

v [kg/m3]

50.00

12.35

0.641 9

7.72

143.6 1

400.00

1

0.4624

250.0 0

500.00

100

0.4744

110.0 0

350.00

-100

0.00105 1

SOLUTION for R-134a T [C]

P [kPa]

x

v [kg/m3]

50.00

3.41

100

7.72

8.91

400.00

1

0.0512

250.0 0

500.00

-

-

110.0 0

350.00

100

0.08666

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-6 SOLUTION for R-22 T [C]

P [kPa]

X

v [kg/m3]

50.00

4.02

100

7.72

-6.56

400.00

1

0.05817

250.0 0

500.00

100

0.09959

110.0 0

350.00

100

0.103

SOLUTION for Ammonia T [C]

P [kPa]

X

v [kg/m3]

50.00

20.40

100

7.72

-1.89

400.00

1

0.3094

250.0 0

500.00

100

0.5076

110.0 0

350.00

100

0.5269

Steam

700 600 500

T [C]

400 300

8600 kPa 2600 kPa

200

500 kPa

100 45 kPa

0 0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

s [kJ/kg-K]

Steam

700 600 500

T [C]

400 300

8600 kPa 2600 kPa

200

500 kPa

100 0 10-4

45 kPa

10-3

10-2

10-1

100

101

102

103

v [m 3/kg] PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-7 Steam

105

104 250 C

P [kPa]

10

3 170 C

102

110 C 75 C

101

100 10-3

10-2

10-1

100

101

102

3

v [m /kg]

Steam

105

104 250 C

P [kPa]

10

3 170 C

102

110 C 75 C

101

100 0

500

1000

1500

2000

2500

3000

h [kJ/kg]

Steam

4000

8600 kPa

h [kJ/kg]

3500

2600 kPa

3000

500 kPa

2500

45 kPa

2000 1500 1000 500 0 0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

s [kJ/kg-K]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-8 3-24E Complete the following table for H2 O: T, °F

P, psia

u, Btu / lbm

Phase description

300

67.03

782

Saturated mixture

267.22

40

236.02

Saturated liquid

500

120

1174.4

Superheated vapor

400

400

373.84

Compressed liquid

Problem 3-24E is reconsidered. The missing properties of water are to be determined using EES, and the 3-25E solution is to be repeated for refrigerant-134a, refrigerant-22, and ammonia. Analysis The problem is solved using EES, and the solution is given below. "Given" T[1]=300 [F] u[1]=782 [Btu/lbm] P[2]=40 [psia] x[2]=0 T[3]=500 [F] P[3]=120 [psia] T[4]=400 [F] P[4]=420 [psia] "Analysis" Fluid\$='steam_iapws' P[1]=pressure(Fluid\$, T=T[1], u=u[1]) x[1]=quality(Fluid\$, T=T[1], u=u[1]) T[2]=temperature(Fluid\$, P=P[2], x=x[2]) u[2]=intenergy(Fluid\$, P=P[2], x=x[2]) u[3]=intenergy(Fluid\$, P=P[3], T=T[3]) x[3]=quality(Fluid\$, P=P[3], T=T[3]) u[4]=intenergy(Fluid\$, P=P[4], T=T[4]) x[4]=quality(Fluid\$, P=P[4], T=T[4]) "x = 100 for superheated vapor and x = -100 for compressed liquid" Solution for steam T, ºF

P, psia

x

u, Btu/lbm

300

67.028

0.6173

782

267.2

40

0

236

500

120

100

1174

400

400

-100

373.8

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-9 3-26 Complete the following table for H2 O: T, °C

P, kPa

h, kJ / kg

x

Phase description

120.21

200

2045.8

0.7

Saturated mixture

140

361.53

1800

0.565

Saturated mixture

177.66

950

752.74

0.0

Saturated liquid

80

500

335.37

---

Compressed liquid

350.0

800

3162.2

---

Superheated vapor

3-27 Complete the following table for Refrigerant-134a:

T, °C

P, kPa

v, m3 / kg

Phase description

-12

320

0.000750

Compressed liquid

30

770.64

0.0065

Saturated mixture

18.73

550

0.03741

Saturated vapor

60

600

0.04139

Superheated vapor

3-28 Complete the following table for water:

*

P, kPa

T, oC

v, m3/kg

h, kJ/kg

200

120.2

0.8858

2706.3

Condition description and quality, if applicable x = 1, Saturated vapor

270.3

130

1959.3

x = 0.650, Two-phase mixture

201.8

400

1.5358

3277.0

Superheated vapor

800

30

0.001004*

125.74*

Compressed liquid

450

147.90

-

-

Insufficient information

Approximated as saturated liquid at the given temperature of 30oC

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-10 3-29E Complete the following table for Refrigerant-134a:

T, °F

P, psia

h, Btu / lbm

x

Phase description

65.89

80

78

0.566

Saturated mixture

15

29.759

69.92

0.6

Saturated mixture

10

70

15.35

---

Compressed liquid

160

180

129.46

---

Superheated vapor

110

161.16

117.23

1.0

Saturated vapor

3-30 A piston-cylinder device contains R-134a at a specified state. Heat is transferred to R-134a. The final pressure, the volume change of the cylinder, and the enthalpy change are to be determined. Analysis (a) The final pressure is equal to the initial pressure, which is determined from

P2 = P1 = Patm +

mpg 2

πD /4

= 88 kPa +

(12 kg)(9.81 m/s 2 ) ⎛⎜ 1 kN 2 ⎜ π (0.25 m) /4 ⎝ 1000 kg.m/s 2

⎞ ⎟ = 90.4 kPa ⎟ ⎠

(b) The specific volume and enthalpy of R-134a at the initial state of 90.4 kPa and -10°C and at the final state of 90.4 kPa and 15°C are (from EES)

v1 = 0.2302 m3/kg 3

v 2 = 0.2544 m /kg

h1 = 247.76 kJ/kg h2 = 268.16 kJ/kg

The initial and the final volumes and the volume change are

V1 = mv 1 = (0.85 kg)(0.2302 m 3 /kg) = 0.1957 m 3 V 2 = mv 2 = (0.85 kg)(0.2544 m 3 /kg) = 0.2162 m 3

R-134a 0.85 kg -10°C

Q

∆V = 0.2162 − 0.1957 = 0.0205 m 3 (c) The total enthalpy change is determined from ∆H = m(h2 − h1 ) = (0.85 kg)(268.16 − 247.76) kJ/kg = 17.4 kJ/kg

3-31E The temperature of R-134a at a specified state is to be determined. Analysis Since the specified specific volume is higher than vg for 120 psia, this is a superheated vapor state. From R-134a tables, P = 120 psia

⎫ ⎬ T = 140°F (Table A - 13E) v = 0.4619 ft /lbm ⎭ 3

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-11 3-32 A rigid container that is filled with water is cooled. The initial temperature and the final pressure are to be determined. Analysis This is a constant volume process. The specific volume is

H2O 2 MPa 1 kg 150 L

0.150 m 3 = 0.150 m 3 /kg v1 = v 2 = = m 1 kg

V

Q

The initial state is superheated vapor. The temperature is determined to be P1 = 2 MPa

⎫ ⎬ T1 = 395°C

v 1 = 0.150 m 3 /kg ⎭

(Table A - 6)

P

This is a constant volume cooling process (v = V /m = constant). The final state is saturated mixture and thus the pressure is the saturation pressure at the final temperature:

1

2

T2 = 40°C ⎫ ⎬ P = Psat @ 40°C = 7.385 kPa (Table A - 4) v 2 = v 1 = 0.150 m 3 /kg ⎭ 2

v

3-33 A rigid container that is filled with R-134a is heated. The final temperature and initial pressure are to be determined. Analysis This is a constant volume process. The specific volume is

v1 = v 2 =

V m

=

R-134a -40°C 10 kg 1.348 m3

1.348 m 3 = 0.1348 m 3 /kg 10 kg

The initial state is determined to be a mixture, and thus the pressure is the saturation pressure at the given temperature

P1 = Psat @ -40°C = 51.25 kPa (Table A - 11) The final state is superheated vapor and the temperature is determined by interpolation to be P2 = 200 kPa

⎫ ⎬ T2 = 66.3°C (Table A - 13)

v 2 = 0.1348 m 3 /kg ⎭

P

2

1

v

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-12 3-34E Left chamber of a partitioned system contains water at a specified state while the right chamber is evacuated. The partition is now ruptured and heat is transferred to the water. The pressure and the total internal energy at the final state are to be determined. Analysis The final specific volume is 3 ft 3 v2 = = = 1.5 ft 3 /lbm m 2 lbm

V2

At this specific volume and the final temperature, the state is a saturated mixture, and the pressure is the saturation pressure

Water 500 psia 2 lbm 1.5 ft3

Evacuated 1.5 ft3

P2 = Psat @ 300°F = 67.03 psia (Table A - 4E) The quality and internal energy at the final state are x2 =

v 2 −v f

(1.5 − 0.01745) ft 3 /lbm

= 0.2299 (6.4663 − 0.01745) ft 3 /lbm u 2 = u f + x 2 u fg = 269.51 + (0.2299)(830.25) = 460.38 Btu/lbm

v fg

=

The total internal energy is then U 2 = mu 2 = (2 lbm)(460.38 Btu/lbm) = 920.8 Btu

3-35 The enthalpy of R-134a at a specified state is to be determined. Analysis The specific volume is

v=

V 9 m3 = = 0.03 m 3 /kg m 300 kg

Inspection of Table A-11 indicates that this is a mixture of liquid and vapor. Using the properties at 10°C line, the quality and the enthalpy are determined to be x=

v −v f v fg

=

(0.03 − 0.0007930) m 3 /kg (0.049403 − 0.0007930) m 3 /kg

= 0.6008

h = h f + xh fg = 65.43 + (0.6008)(190.73) = 180.02 kJ/kg

3-36 The specific volume of R-134a at a specified state is to be determined. Analysis Since the given temperature is higher than the saturation temperature for 200 kPa, this is a superheated vapor state. The specific volume is then P = 200 kPa ⎫ 3 ⎬ v = 0.11646 m /kg (Table A - 13) ⎭

T = 25°C

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-13 3-37E A spring-loaded piston-cylinder device is filled with R-134a. The water now undergoes a process until its volume increases by 40%. The final temperature and the enthalpy are to be determined. Analysis From Table A-11E, the initial specific volume is

v 1 = v f + x1v fg = 0.01143 + (0.80)(4.4300 − 0.01143) = 3.5463 ft 3 /lbm and the initial volume will be

V1 = mv 1 = (0.2 lbm)(3.5463 ft 3 /lbm) = 0.7093 ft 3

P

With a 40% increase in the volume, the final volume will be 3

V 2 = 1.4V1 = 1.4(0.7093 ft ) = 0.9930 ft

2

3

1

The distance that the piston moves between the initial and final conditions is ∆x =

v

4(0.9930 − 0.7093)ft 3 ∆V ∆V = = = 0.3612 ft A p πD 2 / 4 π (1 ft) 2

As a result of the compression of the spring, the pressure difference between the initial and final states is ∆P =

4(37 lbf/in)(0.3612 × 12 in) ∆F k∆x k∆x = = = = 1.42 lbf/in 2 = 1.42 psia Ap A p πD 2 / 4 π (12 in) 2

The initial pressure is

P1 = Psat @ -30°F = 9.87 psia (Table A - 11E) The final pressure is then P2 = P1 + ∆P = 9.87 + 1.42 = 11.29 psia

and the final specific volume is

v2 =

V2 m

=

0.9930 ft 3 = 4.965 ft 3 /lbm 0.2 lbm

At this final state, the temperature and enthalpy are P2 = 11.29 psia

⎫ T1 = 81.5°F ⎬ v 2 = 4.965 ft /lbm⎭ h1 = 119.9 Btu/lbm 3

(from EES)

Note that it is very difficult to get the temperature and enthalpy from Table A-13E accurately.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-14 3-38E A piston-cylinder device that is filled with water is cooled. The final pressure and volume of the water are to be determined. Analysis The initial specific volume is

v1 =

V1 m

=

2.4264 ft 3 = 2.4264 ft 3 /lbm 1 lbm

H2O 600°F 1 lbm 2.4264 ft3

This is a constant-pressure process. The initial state is determined to be superheated vapor and thus the pressure is determined to be T1 = 600°F ⎫ ⎬ P = P2 = 250 psia (Table A - 6E) 3 v 1 = 2.4264 ft /lbm ⎭ 1

P

The saturation temperature at 250 psia is 400.1°F. Since the final temperature is less than this temperature, the final state is compressed liquid. Using the incompressible liquid approximation,

1

2

v 2 = v f @ 200° F = 0.01663 ft 3 /lbm (Table A - 4E)

v

The final volume is then

V 2 = mv 2 = (1 lbm)(0.01663 ft 3 /lbm) = 0.01663 ft 3

3-39 A piston-cylinder device that is filled with R-134a is heated. The final volume is to be determined. Analysis This is a constant pressure process. The initial specific volume is

v1 =

V 1.595 m 3 = = 0.1595 m 3 /kg 10 kg m

R-134a -26.4°C 10 kg 1.595 m3

The initial state is determined to be a mixture, and thus the pressure is the saturation pressure at the given temperature

P1 = Psat @ -26.4°C = 100 kPa (Table A - 12) The final state is superheated vapor and the specific volume is P2 = 100 kPa ⎫ 3 ⎬ v 2 = 0.30138 m /kg (Table A - 13) T2 = 100°C ⎭

P

1

2

The final volume is then

V 2 = mv 2 = (10 kg)(0.30138 m 3 /kg) = 3.0138 m 3

v

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-15 3-40E The total internal energy and enthalpy of water in a container are to be determined. Analysis The specific volume is

v=

V m

=

2 ft 3 = 2 ft 3 /lbm 1 lbm

Water 100 psia 2 ft3

At this specific volume and the given pressure, the state is a saturated mixture. The quality, internal energy, and enthalpy at this state are (Table A-5E) x=

v −v f

=

(2 − 0.01774) ft 3 /lbm

= 0.4490 (4.4327 − 0.01774) ft 3 /lbm u = u f + xu fg = 298.19 + (0.4490)(807.29) = 660.7 Btu/lbm h = h f + xh fg = 298.51 + (0.4490)(888.99) = 697.7 Btu/lbm

v fg

The total internal energy and enthalpy are then

U = mu = (1 lbm)(660.7 Btu/lbm) = 660.7 Btu H = mh = (1 lbm)(697.7 Btu/lbm) = 697.7 Btu

3-41 The volume of a container that contains water at a specified state is to be determined. Analysis The specific volume is determined from steam tables by interpolation to be P = 100 kPa ⎫ 3 ⎬ v = 2.4062 m /kg (Table A - 6) T = 250°C ⎭

The volume of the container is then

V = mv = (3 kg)(2.4062 m 3 /kg) = 7.22 m 3

Water 3 kg 100 kPa 250°C

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-16 3-42 A rigid container that is filled with R-134a is heated. The temperature and total enthalpy are to be determined at the initial and final states. Analysis This is a constant volume process. The specific volume is

R-134a 300 kPa 10 kg 14 L

V 0.014 m 3 v1 = v 2 = = = 0.0014 m 3 /kg 10 kg m The initial state is determined to be a mixture, and thus the temperature is the saturation temperature at the given pressure. From Table A-12 by interpolation P

T1 = Tsat @ 300 kPa = 0.61°C

2

and x1 =

v1 −v f v fg

=

(0.0014 − 0.0007736) m 3 /kg (0.067978 − 0.0007736) m 3 /kg

= 0.009321

h1 = h f + x1 h fg = 52.67 + (0.009321)(198.13) = 54.52 kJ/kg

1

v

The total enthalpy is then H 1 = mh1 = (10 kg )(54.52 kJ/kg ) = 545.2 kJ

The final state is also saturated mixture. Repeating the calculations at this state,

T2 = Tsat @ 600 kPa = 21.55°C x2 =

v 2 −v f v fg

=

(0.0014 − 0.0008199) m 3 /kg (0.034295 − 0.0008199) m 3 /kg

= 0.01733

h2 = h f + x 2 h fg = 81.51 + (0.01733)(180.90) = 84.64 kJ/kg

H 2 = mh2 = (10 kg )(84.64 kJ/kg ) = 846.4 kJ

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-17 3-43 A piston-cylinder device that is filled with R-134a is cooled at constant pressure. The final temperature and the change of total internal energy are to be determined. Analysis The initial specific volume is

v1 =

V 12.322 m 3 = = 0.12322 m 3 /kg 100 kg m

R-134a 200 kPa 100 kg 12.322 m3

The initial state is superheated and the internal energy at this state is P1 = 200 kPa

⎫ ⎬ u1 = 263.08 kJ/kg (Table A - 13)

v 1 = 0.12322 m 3 /kg ⎭

P

The final specific volume is

v2 =

v1 2

=

0.12322 m 3 / kg = 0.06161 m 3 /kg 2

2

This is a constant pressure process. The final state is determined to be saturated mixture whose temperature is

1

v

T2 = Tsat @ 200 kPa = −10.09°C (Table A - 12) The internal energy at the final state is (Table A-12) x2 =

v 2 −v f v fg

=

(0.06161 − 0.0007533) m 3 /kg (0.099867 − 0.0007533) m 3 /kg

= 0.6140

u 2 = u f + x 2 u fg = 38.28 + (0.6140)(186.21) = 152.61 kJ/kg

Hence, the change in the internal energy is ∆u = u 2 − u1 = 152.61 − 263.08 = −110.47 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-18 3-44 A piston-cylinder device fitted with stops contains water at a specified state. Now the water is cooled until a final pressure. The process is to be indicated on the T-v diagram and the change in internal energy is to be determined. Analysis The process is shown on T-v diagram. The internal energy at the initial state is P1 = 300 kPa ⎫ ⎬ u1 = 2728.9 kJ/kg (Table A - 6) T1 = 250°C ⎭

Water 300 kPa 250°C

State 2 is saturated vapor at the initial pressure. Then, P2 = 300 kPa ⎫ 3 ⎬ v 2 = 0.6058 m /kg (Table A - 5) x 2 = 1 (sat. vapor) ⎭

Process 2-3 is a constant-volume process. Thus, P3 = 100 kPa

⎫ ⎬ u = 1163.3 kJ/kg (Table A - 5) 3 v 3 = 0.6058 m /kg ⎭ 3

300 kPa

T

1

250°C

The overall change in internal energy is ∆u = u1 − u 3 = 2728.9 − 1163.3 = 1566 kJ/kg

Q

100 kPa 2 3

v

3-45E The local atmospheric pressure, and thus the boiling temperature, changes with the weather conditions. The change in the boiling temperature corresponding to a change of 0.2 in of mercury in atmospheric pressure is to be determined. Properties The saturation pressures of water at 200 and 212°F are 11.538 and 14.709 psia, respectively (Table A-4E). One in. of mercury is equivalent to 1 inHg = 3.387 kPa = 0.491 psia (inner cover page). Analysis A change of 0.2 in of mercury in atmospheric pressure corresponds to ⎛ 0.491 psia ⎞ ⎟⎟ = 0.0982 psia ∆P = (0.2 inHg)⎜⎜ ⎝ 1 inHg ⎠

P ± 0.2 inHg

At about boiling temperature, the change in boiling temperature per 1 psia change in pressure is determined using data at 200 and 212°F to be (212 − 200)°F ∆T = = 3.783 °F/psia ∆P (14.709 − 11.538) psia

Then the change in saturation (boiling) temperature corresponding to a change of 0.147 psia becomes

∆Tboiling = (3.783 °F/psia)∆P = (3.783 °F/psia)(0.0982 psia) = 0.37°F which is very small. Therefore, the effect of variation of atmospheric pressure on the boiling temperature is negligible.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-19 3-46 A person cooks a meal in a pot that is covered with a well-fitting lid, and leaves the food to cool to the room temperature. It is to be determined if the lid will open or the pan will move up together with the lid when the person attempts to open the pan by lifting the lid up. Assumptions 1 The local atmospheric pressure is 1 atm = 101.325 kPa. 2 The weight of the lid is small and thus its effect on the boiling pressure and temperature is negligible. 3 No air has leaked into the pan during cooling. Properties The saturation pressure of water at 20°C is 2.3392 kPa (Table A-4). Analysis Noting that the weight of the lid is negligible, the reaction force F on the lid after cooling at the pan-lid interface can be determined from a force balance on the lid in the vertical direction to be

PA +F = PatmA or, F = A( Patm − P) = (πD 2 / 4)( Patm − P ) =

π (0.3 m) 2

P

2.3392 kPa

(101,325 − 2339.2) Pa

4 = 6997 m 2 Pa = 6997 N (since 1 Pa = 1 N/m 2 )

Patm = 1 atm

The weight of the pan and its contents is

W = mg = (8 kg)(9.81 m/s2 ) = 78.5 N which is much less than the reaction force of 6997 N at the pan-lid interface. Therefore, the pan will move up together with the lid when the person attempts to open the pan by lifting the lid up. In fact, it looks like the lid will not open even if the mass of the pan and its contents is several hundred kg.

3-47 Water is boiled at 1 atm pressure in a pan placed on an electric burner. The water level drops by 10 cm in 45 min during boiling. The rate of heat transfer to the water is to be determined. Properties The properties of water at 1 atm and thus at a saturation temperature of Tsat = 100°C are hfg = 2256.5 kJ/kg and vf = 0.001043 m3/kg (Table A-4). Analysis The rate of evaporation of water is mevap = m& evap =

Vevap (πD 2 / 4) L [π (0.25 m)2 / 4](0.10 m) = = = 4.704 kg vf vf 0.001043 mevap ∆t

=

H2O 1 atm

4.704 kg = 0.001742 kg/s 45 × 60 s

Then the rate of heat transfer to water becomes Q& = m& evap h fg = (0.001742 kg/s)(2256 .5 kJ/kg) = 3.93 kW

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-20 3-48 Water is boiled at a location where the atmospheric pressure is 79.5 kPa in a pan placed on an electric burner. The water level drops by 10 cm in 45 min during boiling. The rate of heat transfer to the water is to be determined. Properties The properties of water at 79.5 kPa are Tsat = 93.3°C, hfg = 2273.9 kJ/kg and vf = 0.001038 m3/kg (Table A-5). Analysis The rate of evaporation of water is m evap = m& evap =

V evap vf m evap ∆t

= =

(πD 2 / 4) L

vf

=

H2O 79.5 kPa

[π (0.25 m) 2 / 4](0.10 m) = 4.727 kg 0.001038

4.727 kg = 0.001751 kg/s 45× 60 s

Then the rate of heat transfer to water becomes Q& = m& evap h fg = (0.001751 kg/s)(2273 .9 kJ/kg) = 3.98 kW

3-49 Saturated steam at Tsat = 40°C condenses on the outer surface of a cooling tube at a rate of 130 kg/h. The rate of heat transfer from the steam to the cooling water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The condensate leaves the condenser as a saturated liquid at 30°C. Properties The properties of water at the saturation temperature of 40°C are hfg = 2406.0 kJ/kg (Table A-4). Analysis Noting that 2406.0 kJ of heat is released as 1 kg of saturated vapor at 40°C condenses, the rate of heat transfer from the steam to the cooling water in the tube is determined directly from Q& = m& evap h fg

40°C L = 35 m

D = 3 cm

= (130 kg/h)(2406.0 kJ/kg) = 312,780 kJ/h = 86.9 kW

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-21 3-50 The boiling temperature of water in a 5-cm deep pan is given. The boiling temperature in a 40-cm deep pan is to be determined. Assumptions Both pans are full of water. Properties The density of liquid water is approximately ρ = 1000 kg/m3. Analysis The pressure at the bottom of the 5-cm pan is the saturation pressure corresponding to the boiling temperature of 98°C: P = [email protected] C = 94.39 kPa

(Table A-4)

40 cm 5 cm

The pressure difference between the bottoms of two pans is

⎛ 1 kPa ⎜ 1000 kg/m ⋅ s 2 ⎝

∆P = ρ g h = (1000 kg/m 3 )(9.807 m/s 2 )(0.35 m)⎜

⎞ ⎟ = 3.43 kPa ⎟ ⎠

Then the pressure at the bottom of the 40-cm deep pan is P = 94.39 + 3.43 = 97.82 kPa Then the boiling temperature becomes

Tboiling = [email protected]

kPa

= 99.0°C

(Table A-5)

3-51 A vertical piston-cylinder device is filled with water and covered with a 20-kg piston that serves as the lid. The boiling temperature of water is to be determined. Analysis The pressure in the cylinder is determined from a force balance on the piston,

PA = PatmA + W Patm

or, P = Patm +

mg A

= (100 kPa) + = 119.61 kPa

⎞ (20 kg)(9.81 m/s 2 ) ⎛ 1 kPa ⎜ ⎟ 2 2⎟ ⎜ 0.01 m ⎝ 1000 kg/m ⋅ s ⎠

P

W = mg

The boiling temperature is the saturation temperature corresponding to this pressure,

T = Tsat @119.61 kPa = 104.7°C

(Table A-5)

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-22 3-52 A rigid tank that is filled with saturated liquid-vapor mixture is heated. The temperature at which the liquid in the tank is completely vaporized is to be determined, and the T-v diagram is to be drawn. Analysis This is a constant volume process (v = V /m = constant), and the specific volume is determined to be

v=

V m

=

H2O 90°C

1.8 m 3 = 0.12 m 3 /kg 15 kg

When the liquid is completely vaporized the tank will contain saturated vapor only. Thus,

T

v 2 = v g = 0.12 m 3 /kg

2

The temperature at this point is the temperature that corresponds to this vg value, T = Tsat @v

g = 0.12

m 3 /kg

= 202.9 °C

1

v

(Table A-4)

3-53 A piston-cylinder device contains a saturated liquid-vapor mixture of water at 800 kPa pressure. The mixture is heated at constant pressure until the temperature rises to 200°C. The initial temperature, the total mass of water, the final volume are to be determined, and the P-v diagram is to be drawn. Analysis (a) Initially two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. Then the temperature in the tank must be the saturation temperature at the specified pressure,

T = [email protected] kPa = 158.8°C (b) The total mass in this case can easily be determined by adding the mass of each phase, mf = mg =

Vf vf Vg vg

= =

0.005 m 3 0.001101 m 3 /kg 0.9 m 3 0.3156 m 3 /kg

= 4.543 kg P

= 2.852 kg

mt = m f + m g = 4.543 + 2.852 = 7.395 kg

(c) At the final state water is superheated vapor, and its specific volume is

1

2

v

P2 = 600 kPa ⎫ 3 ⎬ v = 0.3521 m /kg (Table A-6) T2 = 200 o C ⎭ 2

Then,

V 2 = mt v 2 = (7.395 kg)(0.3521 m 3 /kg) = 2.604 m3

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-23 3-54 Problem 3-53 is reconsidered. The effect of pressure on the total mass of water in the tank as the pressure varies from 0.1 MPa to 1 MPa is to be investigated. The total mass of water is to be plotted against pressure, and results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. P[1]=600 [kPa] P[2]=P[1] T[2]=200 [C] V_f1 = 0.005 [m^3] V_g1=0.9 [m^3] spvsat_f1=volume(Steam_iapws, P=P[1],x=0) "sat. liq. specific volume, m^3/kg" spvsat_g1=volume(Steam_iapws,P=P[1],x=1) "sat. vap. specific volume, m^3/kg" m_f1=V_f1/spvsat_f1 "sat. liq. mass, kg" m_g1=V_g1/spvsat_g1 "sat. vap. mass, kg" m_tot=m_f1+m_g1 V[1]=V_f1+V_g1 spvol[1]=V[1]/m_tot "specific volume1, m^3" T[1]=temperature(Steam_iapws, P=P[1],v=spvol[1])"C" "The final volume is calculated from the specific volume at the final T and P" spvol[2]=volume(Steam_iapws, P=P[2], T=T[2]) "specific volume2, m^3/kg" V[2]=m_tot*spvol[2] Steam IAPWS

106

P1 [kPa]

mtot [kg]

100 200 300 400 500 600 700 800 900 1000

5.324 5.731 6.145 6.561 6.978 7.395 7.812 8.23 8.648 9.066

105

P [kPa]

104 200°C

103

2

1

P=600 kPa

102 101 100 10-3

10-2

10-1

100

101

102

3

v [m /kg] 9.5 9 8.5

mtot [kg]

8 7.5 7 6.5 6 5.5 5 100

200

300

400

500

600

700

800

900

1000

P1 [kPa]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-24 3-55E Superheated water vapor cools at constant volume until the temperature drops to 250°F. At the final state, the pressure, the quality, and the enthalpy are to be determined. Analysis This is a constant volume process (v = V/m = constant), and the initial specific volume is determined to be P1 = 180 psia ⎫ 3 ⎬ v = 3.0433 ft /lbm T1 = 500 o F ⎭ 1

(Table A-6E) H2O 180 psia 500°F

At 250°F, vf = 0.01700 ft3/lbm and vg = 13.816 ft3/lbm. Thus at the final state, the tank will contain saturated liquid-vapor mixture since vf < v < vg , and the final pressure must be the saturation pressure at the final temperature, P = Psat @ 250 o F = 29.84 psia

(b) The quality at the final state is determined from x2 =

v 2 −v f v fg

=

T

1

3.0433 − 0.01700 = 0.219 13.816 − 0.01700

(c) The enthalpy at the final state is determined from

h = h f + xh fg = 218.63 + 0.219 × 945.41 = 426.0 Btu/lbm

2 v

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-25 3-56E Problem 3-55E is reconsidered. The effect of initial pressure on the quality of water at the final state as the pressure varies from 100 psi to 300 psi is to be investigated. The quality is to be plotted against initial pressure, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. T[1]=500 [F] P[1]=180 [psia] T[2]=250 [F] v[ 1]=volume(steam_iapws,T=T[1],P=P[1]) v[2]=v[1] P[2]=pressure(steam_iapws,T=T[2],v=v[2]) h[2]=enthalpy(steam_iapws,T=T[2],v=v[2]) x[2]=quality(steam_iapws,T=T[2],v=v[2]) Steam

1400

x2

1200

0.4037 0.3283 0.2761 0.2378 0.2084 0.1853 0.1665 0.151 0.1379 0.1268

1.21.31.4 1.5 Btu/lbm -R

1000

T [°F]

P1 [psia] 100 122.2 144.4 166.7 188.9 211.1 233.3 255.6 277.8 300

800 600 400

1600 psia 780 psia

1

180 psia

2

29.82 psia

200

0.050.1 0.2 0.5

0 10 -2

10 -1

10 0

10 1

10 2

10 3

10 4

3

v [ft /lb ] m

0.45 0.4

x[2]

0.35 0.3 0.25 0.2 0.15 0.1 100

140

180

220

260

300

P[1] [psia]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-26 3-57 The properties of compressed liquid water at a specified state are to be determined using the compressed liquid tables, and also by using the saturated liquid approximation, and the results are to be compared. Analysis Compressed liquid can be approximated as saturated liquid at the given temperature. Then from Table A-4,

T = 80°C ⇒

v ≅ v f @ 80°C = 0.001029 m 3 /kg (0.90% error) u ≅ u f @ 80°C = 334.97 kJ/kg h ≅ h f @ 80°C = 335.02 kJ/kg

(1.35% error) (4.53% error)

From compressed liquid table (Table A-7),

v = 0.00102 m 3 /kg

P = 20 MPa ⎫ u = 330.50 kJ/kg T = 80°C ⎬⎭ h = 350.90 kJ/kg

The percent errors involved in the saturated liquid approximation are listed above in parentheses.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-27 3-58 Problem 3-57 is reconsidered. Using EES, the indicated properties of compressed liquid are to be determined, and they are to be compared to those obtained using the saturated liquid approximation. Analysis The problem is solved using EES, and the solution is given below. "Given" T=80 [C] P=20000 [kPa] "Analysis" Fluid\$='steam_iapws' "Saturated liquid assumption" v_f=volume(Fluid\$, T=T, x=0) u_f=intenergy(Fluid\$, T=T, x=0) h_f=enthalpy(Fluid\$, T=T, x=0) "Compressed liquid treatment" v=volume(Fluid\$, T=T, P=P) u=intenergy(Fluid\$, T=T, P=P) h=enthalpy(Fluid\$, T=T, P=P) "Percentage Errors" error_v=100*(v_f-v)/v error_u=100*(u_f-u)/u error_h=100*(h-h_f)/h SOLUTION error_h=4.527 error_u=1.351 error_v=0.8987 Fluid\$='steam_iapws' h=350.90 [kJ/kg] h_f=335.02 [kJ/kg] P=20000 [kPa] T=80 [C] u=330.50 [kJ/kg] u_f=334.97 [kJ/kg] v=0.00102 [m^3/kg] v_f=0.001029 [m^3/kg]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-28 3-59 Superheated steam in a piston-cylinder device is cooled at constant pressure until half of the mass condenses. The final temperature and the volume change are to be determined, and the process should be shown on a T-v diagram. Analysis (b) At the final state the cylinder contains saturated liquidvapor mixture, and thus the final temperature must be the saturation temperature at the final pressure,

T = [email protected] MPa = 179.88°C

(Table A-5)

H2O 300°C 1 MPa

(c) The quality at the final state is specified to be x2 = 0.5. The specific volumes at the initial and the final states are P1 = 1.0 MPa T1 = 300 o C

⎫ 3 ⎬ v 1 = 0.25799 m /kg ⎭

P2 = 1.0 MPa x2 = 0.5

⎫ ⎬ v 2 = v f + x2v fg ⎭ = 0.001127 + 0.5 × (0.19436 − 0.001127 ) = 0.09775 m3/kg

(Table A-6) T 1 2

Thus,

v

∆V = m(v 2 − v 1 ) = (0.8 kg)(0.09775 − 0.25799)m 3 /kg = −0.1282 m 3

3-60 The water in a rigid tank is cooled until the vapor starts condensing. The initial pressure in the tank is to be determined. Analysis This is a constant volume process (v = V /m = constant), and the initial specific volume is equal to the final specific volume that is

25

3

v 1 = v 2 = v g @124°C = 0.79270 m /kg (Table A-4) since the vapor starts condensing at 150°C. Then from Table A-6, T1 = 250°C ⎫ ⎬ P = 0.30 MPa 3 v 1 = 0.79270 m /kg ⎭ 1

T °C 1

H 2O T1= 250°C P1 = ?

15

2

v

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-29 3-61 Heat is supplied to a piston-cylinder device that contains water at a specified state. The volume of the tank, the final temperature and pressure, and the internal energy change of water are to be determined. Properties The saturated liquid properties of water at 200°C are: vf = 0.001157 m3/kg and uf = 850.46 kJ/kg (Table A-4). Analysis (a) The cylinder initially contains saturated liquid water. The volume of the cylinder at the initial state is

V1 = mv 1 = (1.4 kg)(0.001157 m 3 /kg) = 0.001619 m 3 The volume at the final state is

V = 4(0.001619) = 0.006476 m 3

Water 1.4 kg, 200°C sat. liq.

(b) The final state properties are

v2 =

V m

=

0.006476 m3 = 0.004626 m3 / kg 1.4 kg

v 2 = 0.004626 m3 / kg ⎫⎪ x2 = 1

Q

T2 = 371.3 °C

⎬ P2 = 21,367 kPa ⎭⎪ u = 2201.5 kJ/kg 2

(Table A-4 or A-5 or EES)

(c) The total internal energy change is determined from ∆U = m(u 2 − u1 ) = (1.4 kg)(2201.5 - 850.46) kJ/kg = 1892 kJ

3-62E The error involved in using the enthalpy of water by the incompressible liquid approximation is to be determined. Analysis The state of water is compressed liquid. From the steam tables, P = 1500 psia ⎫ ⎬ h = 376.51 Btu/lbm (Table A - 7E) T = 400°F ⎭

Based upon the incompressible liquid approximation, P = 1500 psia ⎫ ⎬ h ≅ h f @ 400° F = 375.04 Btu/lbm (Table A - 4E) T = 400°F ⎭

The error involved is Percent Error =

376.51 − 375.04 × 100 = 0.39% 376.51

which is quite acceptable in most engineering calculations.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-30 3-63 The errors involved in using the specific volume and enthalpy of water by the incompressible liquid approximation are to be determined. Analysis The state of water is compressed liquid. From the steam tables, P = 20 MPa ⎫ v = 0.0010679 m 3 /kg ⎬ T = 140°C ⎭ h = 602.07 kJ/kg

(Table A - 7)

Based upon the incompressible liquid approximation, P = 20 MPa ⎫ v ≅ v f @ 140°C = 0.001080 m 3 /kg (Table A - 4) ⎬ T = 140°C ⎭ h ≅ h f @ 140°C = 589.16 kJ/kg

The errors involved are Percent Error (specific volume) = Percent Error (enthalpy) =

0.0010679 − 0.001080 × 100 = −1.13% 0.0010679 602.07 − 589.16 × 100 = 2.14% 602.07

which are quite acceptable in most engineering calculations.

3-64 A piston-cylinder device that is filled with R-134a is heated. The volume change is to be determined. Analysis The initial specific volume is P1 = 60 kPa ⎫ 3 ⎬ v 1 = 0.33608 m /kg (Table A - 13) T1 = −20°C ⎭

R-134a 60 kPa -20°C 100 g

and the initial volume is

V1 = mv 1 = (0.100 kg)(0.33608 m 3 /kg) = 0.033608 m 3 At the final state, we have P2 = 60 kPa ⎫ 3 ⎬ v 2 = 0.50410 m /kg (Table A - 13) T2 = 100°C ⎭

V 2 = mv 2 = (0.100 kg)(0.50410 m 3 /kg) = 0.050410 m 3

P 1

2

The volume change is then

∆V = V 2 −V1 = 0.050410 − 0.033608 = 0.0168 m 3

v

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-31 3-65 A rigid vessel is filled with refrigerant-134a. The total volume and the total internal energy are to be determined. Properties The properties of R-134a at the given state are (Table A-13).

P = 800 kPa ⎫ u = 327.87 kJ/kg T = 120 o C ⎬⎭ v = 0.037625 m 3 /kg Analysis The total volume and internal energy are determined from

V = mv = (2 kg)(0.037625 m 3 /kg) = 0.0753 m 3

R-134a 2 kg 800 kPa 120°C

U = mu = (2 kg)(327.87 kJ/kg) = 655.7 kJ

3-66 A rigid vessel contains R-134a at specified temperature. The pressure, total internal energy, and the volume of the liquid phase are to be determined. Analysis (a) The specific volume of the refrigerant is

v=

V m

=

0.5 m 3 = 0.05 m 3 /kg 10 kg

At -20°C, vf = 0.0007362 m3/kg and vg = 0.14729 m3/kg (Table A-11). Thus the tank contains saturated liquid-vapor mixture since vf < v < vg , and the pressure must be the saturation pressure at the specified temperature,

R-134a 10 kg -20°C

P = Psat @ − 20o C = 132.82 kPa

(b) The quality of the refrigerant-134a and its total internal energy are determined from x=

v −v f v fg

=

0.05 − 0.0007362 = 0.3361 0.14729 − 0.0007362

u = u f + xu fg = 25.39 + 0.3361× 193.45 = 90.42 kJ/kg U = mu = (10 kg)(90.42 kJ/kg) = 904.2 kJ (c) The mass of the liquid phase and its volume are determined from

m f = (1 − x)mt = (1 − 0.3361) × 10 = 6.639 kg

V f = m f v f = (6.639 kg)(0.0007362 m3/kg) = 0.00489 m 3

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-32 3-67 The Pessure-Enthalpy diagram of R-134a showing some constant-temperature and constant-entropy lines are obtained using Property Plot feature of EES.

P [kPa]

1.2

kJ

/k g

1

0.8

0.5

0.3

0.2

104

-K

R134a

105

70°C

103

40°C 10°C -10°C

102

-30°C

101 -100

0

100

200

300

400

500

h [kJ/kg]

3-68 A rigid vessel that contains a saturated liquid-vapor mixture is heated until it reaches the critical state. The mass of the liquid water and the volume occupied by the liquid at the initial state are to be determined. Analysis This is a constant volume process (v = V /m = constant) to the critical state, and thus the initial specific volume will be equal to the final specific volume, which is equal to the critical specific volume of water,

v 1 = v 2 = v cr = 0.003106 m 3 /kg

(last row of Table A-4)

The total mass is 0.3 m 3 V m= = = 96.60 kg v 0.003106 m 3 /kg

At 150°C, vf = 0.001091 m3/kg and vg = 0.39248 m3/kg (Table A4). Then the quality of water at the initial state is x1 =

v1 −v f v fg

=

T

cp

H 2O 150°C

0.003106 − 0.001091 = 0.005149 0.39248 − 0.001091

vcr

v

Then the mass of the liquid phase and its volume at the initial state are determined from

m f = (1 − x1 )mt = (1 − 0.005149)(96.60) = 96.10 kg

V f = m f v f = (96.10 kg)(0.001091 m3/kg) = 0.105 m 3

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-33

Ideal Gas 3-69C Mass m is simply the amount of matter; molar mass M is the mass of one mole in grams or the mass of one kmol in kilograms. These two are related to each other by m = NM, where N is the number of moles.

3-70C A gas can be treated as an ideal gas when it is at a high temperature or low pressure relative to its critical temperature and pressure.

3-71C Ru is the universal gas constant that is the same for all gases whereas R is the specific gas constant that is different for different gases. These two are related to each other by R = Ru / M, where M is the molar mass of the gas.

3-72C Propane (molar mass = 44.1 kg/kmol) poses a greater fire danger than methane (molar mass = 16 kg/kmol) since propane is heavier than air (molar mass = 29 kg/kmol), and it will settle near the floor. Methane, on the other hand, is lighter than air and thus it will rise and leak out.

3-73 The specific volume of nitrogen at a specified state is to be determined. Assumptions At specified conditions, nitrogen behaves as an ideal gas. Properties The gas constant of nitrogen is R = 0.2968 kJ/kg⋅K (Table A-1). Analysis According to the ideal gas equation of state,

v=

RT (0.2968 kPa ⋅ m 3 /kg ⋅ K)(227 + 273 K) = = 0.495 m 3 /kg 300 kPa P

3-74E The temperature in a container that is filled with oxygen is to be determined. Assumptions At specified conditions, oxygen behaves as an ideal gas. Properties The gas constant of oxygen is R = 0.3353 psia⋅ft3/lbm⋅R (Table A-1E). Analysis The definition of the specific volume gives

v=

V m

=

3 ft 3 = 1.5 ft 3 /lbm 2 lbm

Using the ideal gas equation of state, the temperature is T=

(80 psia)(1.5 ft 3 /lbm) Pv = = 358 R R 0.3353 psia ⋅ ft 3 /lbm ⋅ R

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-34 3-75 The volume of a container that is filled with helium at a specified state is to be determined. Assumptions At specified conditions, helium behaves as an ideal gas. Properties The gas constant of helium is R = 2.0769 kJ/kg⋅K (Table A-1). Analysis According to the ideal gas equation of state,

V =

mRT (2 kg)(2.0769 kPa ⋅ m 3 /kg ⋅ K)(27 + 273 K) = = 4.154 m 3 300 kPa P

3-76 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined. Assumptions At specified conditions, helium behaves as an ideal gas. Properties The universal gas constant is Ru = 8.314 kPa.m3/kmol.K. The molar mass of helium is 4.0 kg/kmol (Table A-1). Analysis The volume of the sphere is

4 3

4 3

V = π r 3 = π (4.5 m) 3 = 381.7 m 3 Assuming ideal gas behavior, the mole numbers of He is determined from N=

(200 kPa)(381.7 m 3 ) PV = = 30.61 kmol Ru T (8.314 kPa ⋅ m 3 /kmol ⋅ K)(300 K)

He D=9m 27°C 200 kPa

Then the mass of He can be determined from

m = NM = (30.61 kmol)(4.0 kg/kmol) = 123 kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-35 3-77 Problem 3-76 is to be reconsidered. The effect of the balloon diameter on the mass of helium contained in the balloon is to be determined for the pressures of (a) 100 kPa and (b) 200 kPa as the diameter varies from 5 m to 15 m. The mass of helium is to be plotted against the diameter for both cases. Analysis The problem is solved using EES, and the solution is given below. "Given Data" {D=9 [m]} T=27 [C] P=200 [kPa] R_u=8.314 [kJ/kmol-K] "Solution" P*V=N*R_u*(T+273) V=4*pi*(D/2)^3/3 m=N*MOLARMASS(Helium)

m [kg] 21.01 38.35 63.31 97.25 141.6 197.6 266.9 350.6 450.2 567.2

600 500 400

m [kg]

D [m] 5 6.111 7.222 8.333 9.444 10.56 11.67 12.78 13.89 15 P=200 kPa

300

P=200 kPa 200

P=100 kPa

100 0 5

7

9

D [m]

11

13

15

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-36 3-78 Two rigid tanks connected by a valve to each other contain air at specified conditions. The volume of the second tank and the final equilibrium pressure when the valve is opened are to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Analysis Let's call the first and the second tanks A and B. Treating air as an ideal gas, the volume of the second tank and the mass of air in the first tank are determined to be ⎛ m1RT1 ⎞ (5 kg)(0.287 kPa ⋅ m3/kg ⋅ K)(308 K) ⎟ = = 2.21 m 3 ⎟ P 200 kPa ⎝ 1 ⎠B

V B = ⎜⎜

⎛ PV ⎞ (500 kPa)(1.0 m3 ) = 5.846 kg m A = ⎜⎜ 1 ⎟⎟ = 3 ⎝ RT1 ⎠ A (0.287 kPa ⋅ m /kg ⋅ K)(298 K)

Thus,

A

B

Air

V = 1 m3 T = 25°C P = 500 kPa

×

Air m = 5 kg T = 35°C P = 200 kPa

V = V A + V B = 1.0 + 2.21 = 3.21 m 3 m = m A + mB = 5.846 + 5.0 = 10.846 kg

Then the final equilibrium pressure becomes

P2 =

mRT2

V

=

(10.846 kg)(0.287 kPa ⋅ m3 /kg ⋅ K)(293 K) 3.21 m3

= 284.1 kPa

3-79E An elastic tank contains air at a specified state. The volume is doubled at the same pressure. The initial volume and the final temperature are to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Analysis According to the ideal gas equation of state,

PV = nRu T (32 psia)V = (2.3 lbmol)(10.73 psia ⋅ ft 3 /lbmol ⋅ R)(65 + 460) R

V = 404.9 ft 3 T2 V 2 T2 = ⎯ ⎯→ 2 = ⎯ ⎯→ T2 = 1050 R = 590°F V 1 T1 (65 + 460) R

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-37

3-80 An ideal gas in a rigid tank is cooled to a final gage pressure. The final temperature is to be determined. Assumptions The gas is specified as an ideal gas so that ideal gas relation can be used. Analysis According to the ideal gas equation of state at constant volume, m1 = m1 P1V1 P2V 2 = T1 T2

Since

Patm = 100 kPa

V1 = V 2

Then,

Ideal gas 1227°C 200 kPa (gage)

Q

P1 P2 = T1 T2 T2 = T1

P2 (50 + 100) kPa = [(1227 + 273) K ] = 750 K = 477°C P2 (200 + 100) kPa

3-81 One side of a two-sided tank contains an ideal gas while the other side is evacuated. The partition is removed and the gas fills the entire tank. The gas is also heated to a final pressure. The final temperature is to be determined. Assumptions The gas is specified as an ideal gas so that ideal gas relation can be used. Analysis According to the ideal gas equation of state, P2 = P1

V 2 = V 1 + 2V 1 = 3V 1 Applying these, m1 = m1

Ideal gas 927°C

V1

Evacuated 2V1

Q

P1V 1 P2V 2 = T1 T2

V1 T1

=

V2 T2

T2 = T1

3V V2 = T1 1 = 3T1 = 3[927 + 273) K ] = 3600 K = 3327 °C V1 V1

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-38

3-82 A piston-cylinder device containing argon undergoes an isothermal process. The final pressure is to be determined. Assumptions At specified conditions, argon behaves as an ideal gas. Properties The gas constant of argon is R = 0.2081 kJ/kg⋅K (Table A-1). Analysis Since the temperature remains constant, the ideal gas equation gives

Argon 0.6 kg 0.05 m3 550 kPa

PV PV m= 1 1 = 2 2 ⎯ ⎯→ P1V1 = P2V 2 RT RT

which when solved for final pressure becomes P2 = P1

V1 V = P1 1 = 0.5P1 = 0.5(550 kPa ) = 275 kPa V2 2V1

3-83 An automobile tire is inflated with air. The pressure rise of air in the tire when the tire is heated and the amount of air that must be bled off to reduce the temperature to the original value are to be determined. Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Analysis Initially, the absolute pressure in the tire is

P1 = Pg + Patm = 210 + 100 = 310kPa Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire can be determined from

Tire 25°C

T 323 K P1V1 P2V 2 = ⎯ ⎯→ P2 = 2 P1 = (310 kPa) = 336 kPa T1 298 K T1 T2

Thus the pressure rise is ∆P = P2 − P1 = 336 − 310 = 26 kPa

The amount of air that needs to be bled off to restore pressure to its original value is m1 =

P1V (310 kPa)(0.025 m3 ) = = 0.0906 kg RT1 (0.287 kPa ⋅ m3/kg ⋅ K)(298 K)

m2 =

P1V (310 kPa)(0.025 m3 ) = = 0.0836 kg RT2 (0.287 kPa ⋅ m3/kg ⋅ K)(323 K) ∆m = m1 − m2 = 0.0906 − 0.0836 = 0.0070 kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-39 Compressibility Factor

3-84C It represent the deviation from ideal gas behavior. The further away it is from 1, the more the gas deviates from ideal gas behavior.

3-85C All gases have the same compressibility factor Z at the same reduced temperature and pressure.

3-86C Reduced pressure is the pressure normalized with respect to the critical pressure; and reduced temperature is the temperature normalized with respect to the critical temperature.

3-87 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,

R = 0.4615 kPa·m3/kg·K,

Tcr = 647.1 K,

Pcr = 22.06 MPa

Analysis (a) From the ideal gas equation of state,

v=

RT (0.4615 kPa ⋅ m 3 /kg ⋅ K)(623.15 K) = = 0.01917 m 3 /kg (67.0% error) 15,000 kPa P

(b) From the compressibility chart (Fig. A-15), 10 MPa P ⎫ = = 0.453 ⎪ Pcr 22.06 MPa ⎪ ⎬ Z = 0.65 673 K T ⎪ = = 1.04 TR = ⎪⎭ Tcr 647.1 K PR =

H2O 15 MPa 350°C

Thus,

v = Zv ideal = (0.65)(0.01917 m 3 /kg) = 0.01246 m 3 /kg (8.5% error) (c) From the superheated steam table (Table A-6),

P = 15 MPa T = 350°C

} v = 0.01148 m /kg 3

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-40

3-88 Problem 3-87 is reconsidered. The problem is to be solved using the general compressibility factor feature of EES (or other) software. The specific volume of water for the three cases at 15 MPa over the temperature range of 350°C to 600°C in 25°C intervals is to be compared, and the % error involved in the ideal gas approximation is to be plotted against temperature. Analysis The problem is solved using EES, and the solution is given below. P=15 [MPa]*Convert(MPa,kPa) {T_Celsius= 350 [C]} T=T_Celsius+273 "[K]" T_critical=T_CRIT(Steam_iapws) P_critical=P_CRIT(Steam_iapws) {v=Vol/m} P_table=P; P_comp=P;P_idealgas=P T_table=T; T_comp=T;T_idealgas=T v_table=volume(Steam_iapws,P=P_table,T=T_table) "EES data for steam as a real gas" {P_table=pressure(Steam_iapws, T=T_table,v=v)} {T_sat=temperature(Steam_iapws,P=P_table,v=v)} MM=MOLARMASS(water) R_u=8.314 [kJ/kmol-K] "Universal gas constant" R=R_u/MM "[kJ/kg-K], Particular gas constant" P_idealgas*v_idealgas=R*T_idealgas "Ideal gas equation" z = COMPRESS(T_comp/T_critical,P_comp/P_critical) P_comp*v_comp=z*R*T_comp "generalized Compressibility factor" Error_idealgas=Abs(v_table-v_idealgas)/v_table*Convert(, %) Error_comp=Abs(v_table-v_comp)/v_table*Convert(, %) Errorcomp [%] 9.447 2.725 0.4344 0.5995 1.101 1.337 1.428 1.437 1.397 1.329 1.245

Errorideal gas [%] 67.22 43.53 32.21 25.23 20.44 16.92 14.22 12.1 10.39 8.976 7.802

TCelcius [C] 350 375 400 425 450 475 500 525 550 575 600

Specific Volume

Percent Error [%]

70 60

Ideal Gas

50

Compressibility Factor

40

Steam at 15 MPa

30 20 10 0 300

350

400

450

500

550

600

TCe lsius [C]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-41

3-89 The specific volume of R-134a is to be determined using the ideal gas relation, the compressibility chart, and the R134a tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are, from Table A-1,

R = 0.08149 kPa·m3/kg·K,

Tcr = 374.2 K,

Pcr = 4.059 MPa

Analysis (a) From the ideal gas equation of state,

v=

RT (0.08149 kPa ⋅ m3/kg ⋅ K)(343 K) = = 0.03105 m3 /kg P 900 kPa

(13.3% error )

(b) From the compressibility chart (Fig. A-15), 0.9 MPa P ⎫ = = 0.222 ⎪ Pcr 4.059 MPa ⎪ ⎬ Z = 0.894 343 K T TR = = = 0.917 ⎪ ⎪⎭ Tcr 374.2 K

R-134a 0.9 MPa

PR =

70°C

Thus,

v = Zv ideal = (0.894)(0.03105 m 3 /kg) = 0.02776 m 3 /kg

(1.3%error)

(c) From the superheated refrigerant table (Table A-13),

}

P = 0.9 MPa v = 0.027413 m3 /kg T = 70°C

3-90 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,

R = 0.4615 kPa·m3/kg·K,

Tcr = 647.1 K,

Pcr = 22.06 MPa

Analysis (a) From the ideal gas equation of state,

v=

RT (0.4615 kPa ⋅ m3/kg ⋅ K)(723 K) = = 0.09533 m3 /kg 3500 kPa P

(3.7% error)

(b) From the compressibility chart (Fig. A-15), 3.5 MPa P ⎫ = = 0.159 ⎪ Pcr 22.06 MPa ⎪ ⎬ Z = 0.961 723 K T ⎪ = = 1.12 TR = ⎪⎭ Tcr 647.1 K PR =

H 2O 3.5 MPa 450°C

Thus,

v = Zv ideal = (0.961)(0.09533 m 3 /kg) = 0.09161 m 3 /kg

(0.4% error)

(c) From the superheated steam table (Table A-6),

}

P = 3.5 MPa v = 0.09196 m3 /kg T = 450°C

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-42

3-91E Ethane in a rigid vessel is heated. The final pressure is to be determined using the compressibility chart. Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A-1E,

R = 0.3574 psia·ft3/lbm·R,

Tcr = 549.8 R,

Pcr = 708 psia

Analysis From the compressibility chart at the initial state (Fig. A-15), ⎫ ⎪ ⎪ ⎬ Z1 = 0.963 P 80 psia PR1 = 1 = = 0.1130 ⎪ ⎪⎭ Pcr 708 psia

TR1 =

T1 560 R = = 1.019 Tcr 549.8 R

Ethane 80 psia

Q

100°F

The specific volume does not change during the process. Then,

v1 = v 2 =

Z1 RT1 (0.963)(0.3574 psia ⋅ ft 3 /lbm ⋅ R)(560 R) = = 2.409 ft 3 /lbm P1 80 psia

At the final state, ⎫ ⎪ ⎪ ⎬ Z 2 = 1 .0 3 v 2,actual 2.4091ft /lbm ⎪ = = = 8 . 68 ⎪ RTcr /Pcr (0.3574 psia ⋅ ft 3 /lbm ⋅ R)(549.8 R)/(708 psia) ⎭

TR 2 =

v R2

T2 1000 R = = 1.819 Tcr 549.8 R

Thus,

P2 =

Z 2 RT2

v2

=

(1.0)(0.3574 psia ⋅ ft 3 /lbm ⋅ R)(1000 R) 2.409 ft 3 /lbm

= 148 psia

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-43

3-92 Ethylene is heated at constant pressure. The specific volume change of ethylene is to be determined using the compressibility chart. Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A-1,

R = 0.2964 kPa·m3/kg·K,

Tcr = 282.4 K,

Pcr = 5.12 MPa

Analysis From the compressibility chart at the initial and final states (Fig. A-15), ⎫ ⎪ ⎪ ⎬ Z1 = 0.56 P 5 MPa = 1 = = 0.977 ⎪ ⎪⎭ Pcr 5.12 MPa

T R1 = PR1

T1 293 K = = 1.038 Tcr 282.4 K

T2 473 K ⎫ = = 1.675 ⎪ Tcr 282.4 KR ⎬ Z1 = 0.961 ⎪ = PR1 = 0.977 ⎭

TR 2 = PR 2

Ethylene 5 MPa 2 0 °C

Q

The specific volume change is R ( Z 2 T2 − Z 1T1 ) P 0.2964 kPa ⋅ m 3 /kg ⋅ K [(0.961)(473 K) − (0.56)(293 K)] = 5000 kPa = 0.0172 m 3 /kg

∆v =

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-44

3-93 Water vapor is heated at constant pressure. The final temperature is to be determined using ideal gas equation, the compressibility charts, and the steam tables. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,

R = 0.4615 kPa·m3/kg·K,

Tcr = 647.1 K,

Pcr = 22.06 MPa

Analysis (a) From the ideal gas equation, T2 = T1

v2 = (350 + 273 K )(2) = 1246 K v1

(b) The pressure of the steam is Water 350°C sat. vapor

P1 = P2 = [email protected]°C = 16,529 kPa From the compressibility chart at the initial state (Fig. A-15), ⎫ ⎪ ⎪ ⎬ Z1 = 0.593, v R1 = 0.75 P1 16.529 MPa = = = 0.749 ⎪ ⎪⎭ Pcr 22.06 MPa

T R1 = PR1

Q

T1 623 K = = 0.963 Tcr 647.1 KR

At the final state, PR 2 = PR1 = 0.749

v R 2 = 2v R1

⎫ ⎬ Z 2 = 0.88 = 2(0.75) = 1.50 ⎭

Thus, T2 =

P2v 2 P v T 16,529 kPa (1.50)(647.1 K) = 2 R 2 cr = = 826 K 0.88 22,060 kPa Z 2 R Z 2 Pcr

(c) From the superheated steam table, T1 = 350°C ⎫ 3 ⎬ v 1 = 0.008806 m /kg x1 = 1 ⎭ P2 = 16,529 kPa

(Table A-4)

⎫ ⎬ T2 = 477°C = 750 K

v 2 = 2v 1 = 0.01761 m 3 /kg ⎭

(from Table A-6 or EES)

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-45

3-94E Water vapor is heated at constant pressure. The final temperature is to be determined using ideal gas equation, the compressibility charts, and the steam tables. Properties The critical pressure and the critical temperature of water are, from Table A-1E,

R = 0.5956 psia·ft3/lbm·R,

Tcr = 1164.8 R,

Pcr = 3200 psia

Analysis (a) From the ideal gas equation, T2 = T1

v2 = (400 + 460 R )(2) = 1720 R v1

(b) The properties of steam are (Table A-4E)

Water 400°F sat. vapor

P1 = P2 = [email protected]° F = 247.26 psia

v 1 = v [email protected]°F = 1.8639 ft 3 /lbm

Q

v 2 = 2v 1 = 3.7278 ft 3 /lbm At the final state, from the compressibility chart (Fig. A-15),

⎫ ⎪ ⎪ ⎬ Z 2 = 0.985 v 2,actual 3.7278 ft 3 /lbm = = = 17.19 ⎪ ⎪ RTcr /Pcr (0.5956 psia ⋅ ft 3 /lbm ⋅ R)(1164.8 R)/(3200 psia) ⎭

PR 2 =

v R2

P2 247.26 psia = = 0.0773 3200 psia Pcr

Thus, T2 =

P2v 2 (247.26 psia)(3.7278 ft 3 /lbm) = = 1571 R Z 2 R (0.985)(0.5956 psia ⋅ ft 3 /lbm ⋅ R)

(c) From the superheated steam table, P2 = 247.26 psia

⎫ ⎬ T2 = 1100°F = 1560 R

v 2 = 3.7278 ft 3 /lbm ⎭

(from Table A-6E or EES)

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-46 3-95 Methane is heated at constant pressure. The final temperature is to be determined using ideal gas equation and the compressibility charts. Properties The gas constant, the critical pressure, and the critical temperature of methane are, from Table A-1,

R = 0.5182 kPa·m3/kg·K,

Tcr = 191.1 K,

Pcr = 4.64 MPa

Analysis From the ideal gas equation, T2 = T1

v2 = (300 K )(1.5) = 450 K v1

From the compressibility chart at the initial state (Fig. A-15),

PR1

Methane 8 MPa 300 K

⎫ ⎪ ⎪ ⎬ Z1 = 0.88, v R1 = 0.80 P1 8 MPa = = = 1.724 ⎪ ⎪⎭ Pcr 4.64 MPa

T R1 =

T1 300 K = = 1.570 Tcr 191.1 K

Q

At the final state, PR 2 = PR1 = 1.724

v R 2 = 1.5v R1

⎫ ⎬ Z 2 = 0.975 = 1.5(0.80) = 1.2 ⎭

Thus, T2 =

P2v 2 P v T 8000 kPa (1.2)(191.1 K) = 2 R 2 cr = = 406 K 0.975 4640 kPa Z 2 R Z 2 Pcr

Of these two results, the accuracy of the second result is limited by the accuracy with which the charts may be read. Accepting the error associated with reading charts, the second temperature is the more accurate.

3-96 The percent error involved in treating CO2 at a specified state as an ideal gas is to be determined. Properties The critical pressure, and the critical temperature of CO2 are, from Table A-1, Tcr = 304.2K and Pcr = 7.39MPa

Analysis From the compressibility chart (Fig. A-15), 5 MPa P ⎫ = = 0.677 ⎪ PR = Pcr 7.39 MPa ⎪ ⎬ Z = 0.69 298 K T = = 0.980 ⎪ TR = ⎪⎭ Tcr 304.2 K

CO2 5 MPa 25°C

Then the error involved in treating CO2 as an ideal gas is Error =

v − v ideal 1 1 = 1− = 1− = −0.44.9 or 44.9% v Z 0.69

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-47 3-97 CO2 gas flows through a pipe. The volume flow rate and the density at the inlet and the volume flow rate at the exit of the pipe are to be determined. Properties The gas constant, the critical pressure, and the critical temperature of CO2 are (Table A-1)

R = 0.1889 kPa·m3/kg·K,

Tcr = 304.2 K,

Pcr = 7.39 MPa

Analysis

3 MPa 500 K 2 kg/s

CO2

450 K

(a) From the ideal gas equation of state,

V&1 =

m& RT1 (2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K) = = 0.06297 m 3 /kg (2.1% error) P1 (3000 kPa)

ρ1 =

P1 (3000 kPa) = = 31.76 kg/m 3 (2.1% error) RT1 (0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K)

V&2 =

m& RT2 (2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(450 K) = = 0.05667 m 3 /kg (3.6% error) P2 (3000 kPa)

(b) From the compressibility chart (EES function for compressibility factor is used) 3 MPa P1 ⎫ = = 0.407 ⎪ Pcr 7.39 MPa ⎪ ⎬ Z1 = 0.9791 500 K T = 1 = = 1.64 ⎪ ⎪⎭ Tcr 304.2 K

PR = TR ,1

3 MPa P2 ⎫ = = 0.407 ⎪ Pcr 7.39 MPa ⎪ ⎬ Z 2 = 0.9656 450 K T2 = = = 1.48 ⎪ ⎪⎭ Tcr 304.2 K

PR = TR , 2

Thus,

V&1 =

Z 1 m& RT1 (0.9791)(2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K) = = 0.06165 m 3 /kg P1 (3000 kPa)

ρ1 =

P1 (3000 kPa) = = 32.44 kg/m 3 Z 1 RT1 (0.9791)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K)

V&2 =

Z 2 m& RT2 (0.9656)(2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(450 K) = = 0.05472 m 3 /kg P2 (3000 kPa)

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-48 3-98 The specific volume of nitrogen gas is to be determined using the ideal gas relation and the compressibility chart. The errors involved in these two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of nitrogen are, from Table A-1,

R = 0.2968 kPa·m3/kg·K,

Tcr = 126.2 K,

Pcr = 3.39 MPa

Analysis (a) From the ideal gas equation of state,

v=

RT (0.2968 kPa ⋅ m 3 /kg ⋅ K)(150 K) = = 0.004452 m 3 /kg P 10,000 kPa

(86.4% error)

(b) From the compressibility chart (Fig. A-15), 10 MPa P ⎫ = = 2.95 ⎪ Pcr 3.39 MPa ⎪ ⎬ Z = 0.54 150 K T = = 1.19 ⎪ TR = ⎪⎭ Tcr 126.2 K

N2 10 MPa 150 K

PR =

Thus,

v = Zv ideal = (0.54)(0.004452 m 3 /kg) = 0.002404 m 3 /kg

(0.7% error)

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-49 Other Equations of State

3-99C The constant a represents the increase in pressure as a result of intermolecular forces; the constant b represents the volume occupied by the molecules. They are determined from the requirement that the critical isotherm has an inflection point at the critical point.

3-100 Carbon monoxide is heated in a piston-cylinder device. The final volume of the carbon monoxide is to be determined using the ideal gas equation and the Benedict-Webb-Rubin equation of state. Properties The gas constant and molar mass of CO are (Table A-1)

R = 0.2968 kPa·m3/kg·K, M = 28.011 kg/kmol Analysis (a) From the ideal gas equation of state, mRT2 (0.100 kg)(0.2968 kPa ⋅ m 3 /kg ⋅ K)(773 K) V2 = = = 0.02294 m 3 P 1000 kPa

CO 1000 kPa 200°C

Q

(b) Using the coefficients of Table 3-4 for carbon dioxide and the given data, the Benedict-Webb-Rubin equation of state for state 2 is ⎛ C ⎞ 1 bR T − a aα c ⎛ γ ⎞ + 6 + 3 2 ⎜1 + 2 ⎟ exp(−γ / v 2 ) + ⎜⎜ B0 RuT2 − A0 − 02 ⎟⎟ 2 + u 23 v2 ⎝ T2 ⎠ v v v v T2 ⎝ v ⎠ (8.314)(773) ⎛⎜ 8.673 ×105 ⎞⎟ 1 0.002632× 8.314 × 773 − 3.71 + 0.05454 × 8.314 × 773 − 135.9 − 1000 = + ⎜ v2 7732 ⎟⎠ v 2 v3 ⎝ P2 =

+

RuT2

3.71× 0.000135 1.054 ×105 ⎛ 0.0060 ⎞ 2 + 3 ⎜1 + ⎟ exp(−0.0060 / v ) v6 v (773)2 ⎝ v2 ⎠

The solution of this equation by an equation solver such as EES gives

v 2 = 6.460 m 3 /kmol Then,

v2 =

v2 M

=

6.460 m 3 /kmol = 0.2306 m 3 /kg 28.011 kg/kmol

V 2 = mv 2 = (0.100 kg)(0.2306 m 3 /kg) = 0.02306 m 3

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-50

3-101 Methane is heated in a rigid container. The final pressure of the methane is to be determined using the ideal gas equation and the Benedict-Webb-Rubin equation of state. Analysis (a) From the ideal gas equation of state, P2 = P1

T2 573 K = (80 kPa) = 156.5 kPa 293 K T1

(b) The specific molar volume of the methane is

v1 = v 2 =

Methane 80 kPa

Q

20°C

Ru T1 (8.314 kPa ⋅ m 3 /kmol ⋅ K)(293 K) = = 30.45 m 3 /kmol P1 80 kPa

Using the coefficients of Table 3-4 for methane and the given data, the Benedict-Webb-Rubin equation of state for state 2 gives

P2 = =

RuT2

v2

⎛ C ⎞ 1 bR T − a aα c ⎛ γ ⎞ + ⎜⎜ B0 RuT2 − A0 − 02 ⎟⎟ 2 + u 23 + 6 + 3 2 ⎜1 + 2 ⎟ exp(−γ / v 2 ) T2 ⎠ v v v v T2 ⎝ v ⎠ ⎝

(8.314)(573) ⎛⎜ 2.286 ×106 ⎞⎟ 1 0.003380× 8.314 × 573 − 5.00 + 0.04260× 8.314 × 573 − 187.91 − + 2 2 ⎜ ⎟ 30.45 573 30.453 ⎝ ⎠ 30.45

5.00 ×1.244 ×10−4 2.578 ×105 ⎛ 0.0060 ⎞ 2 + ⎜1 + ⎟ exp(−0.0060 / 30.45 ) 30.456 30.453 (573)2 ⎝ 30.452 ⎠ = 156.5 kPa +

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-51

3-102E Carbon monoxide is heated in a rigid container. The final pressure of the CO is to be determined using the ideal gas equation and the Benedict-Webb-Rubin equation of state. Properties The gas constant and molar mass of CO are (Table A-1)

R = 0.2968 kPa·m3/kg·K, M = 28.011 kg/kmol Analysis (a) From the ideal gas equation of state, T 1260 R = 34.95 psia P2 = P1 2 = (14.7 psia) 530 R T1

CO 14.7 psia

Q

70°F

(b) The specific molar volume of the CO in SI units is

v1 = v 2 =

Ru T1 (8.314 kPa ⋅ m 3 /kmol ⋅ K)(294 K) = = 24.20 m 3 /kmol P1 101 kPa

Using the coefficients of Table 3-4 for CO and the given data, the Benedict-Webb-Rubin equation of state for state 2 gives P2 = =

RuT2

v2

⎛ C ⎞ 1 bR T − a aα c ⎛ γ ⎞ + ⎜⎜ B0 RuT2 − A0 − 02 ⎟⎟ 2 + u 23 + 6 + 3 2 ⎜1 + 2 ⎟ exp(−γ / v 2 ) T2 ⎠ v v v v T2 ⎝ v ⎠ ⎝

(8.314)(700) ⎛⎜ 8.673×105 ⎞⎟ 1 0.002632× 8.314 × 700 − 3.71 + 0.05454× 8.314 × 700 − 135.87 − + 2 2 ⎜ ⎟ 24.20 700 24.203 ⎝ ⎠ 24.20

3.71×1.350 ×10−4 1.054 ×105 ⎛ 0.0060 ⎞ 2 + ⎜1 + ⎟ exp(−0.0060 / 24.20 ) 24.206 24.203 (700)2 ⎝ 24.202 ⎠ = 240.8 kPa +

The pressure in English unit is ⎛ 1 psia ⎞ P2 = (240.8 kPa)⎜ ⎟ = 34.92 psia ⎝ 6.8948 kPa ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-52

3-103E The temperature of R-134a in a tank at a specified state is to be determined using the ideal gas relation, the van der Waals equation, and the refrigerant tables. Properties The gas constant, critical pressure, and critical temperature of R-134a are (Table A-1E)

R = 0.1052 psia·ft3/lbm·R,

Tcr = 673.6 R,

Pcr = 588. 7 psia

Analysis (a) From the ideal gas equation of state, T=

Pv (160 psia)(0.3479 ft 3 /lbm) = = 529 R R 0.1052 psia ⋅ ft 3 /lbm ⋅ R

(b) The van der Waals constants for the refrigerant are determined from a=

27 R 2Tcr2 (27)(0.1052 psia ⋅ ft 3 /lbm ⋅ R) 2 (673.6 R) 2 = = 3.591 ft 6 ⋅ psia/lbm 2 64 Pcr (64)(588.7 psia)

b=

RTcr (0.1052 psia ⋅ ft 3 /lbm ⋅ R)(673.6 R) = 0.0150 ft 3 /lbm = 8 Pcr 8 × 588.7 psia

T=

1⎛ a ⎜P + 2 R⎝ v

Then, 1 ⎛⎜ 3.591 ⎞ 160 + ⎟(v − b ) = ⎜ 0.1052 (0.3479) 2 ⎠ ⎝

⎞ ⎟(0.3479 − 0.0150) = 600 R ⎟ ⎠

(c) From the superheated refrigerant table (Table A-13E),

P = 160 psia

v = 0.3479 ft 3 /lbm ⎬⎭

T = 160°F (620 R)

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-53

3-104 The pressure of nitrogen in a tank at a specified state is to be determined using the ideal gas relation and the Beattie-Bridgeman equation. The error involved in each case is to be determined. Properties The gas constant and molar mass of nitrogen are (Table A-1)

R = 0.2968 kPa·m3/kg·K and M = 28.013 kg/kmol Analysis (a) From the ideal gas equation of state,

P=

RT

v

=

(0.2968 kPa ⋅ m3 /kg ⋅ K)(150 K) 0.041884 m3/kg

= 1063 kPa (6.3% error)

N2 0.041884 m3/kg 150 K

(b) The constants in the Beattie-Bridgeman equation are ⎛ a⎞ ⎛ 0.02617 ⎞ A = Ao ⎜1 − ⎟ = 136.2315⎜1 − ⎟ = 133.193 1.1733 ⎠ ⎝ v ⎠ ⎝ ⎛ b⎞ ⎛ − 0.00691 ⎞ B = Bo ⎜1 − ⎟ = 0.05046⎜1 − ⎟ = 0.05076 1.1733 ⎠ ⎝ v ⎠ ⎝ c = 4.2 × 10 4 m 3 ⋅ K 3 /kmol

since

v = Mv = (28.013 kg/kmol)(0.041884 m 3 /kg) = 1.1733 m 3 /kmol . Substituting, 4.2 × 10 4 ⎞⎟ RuT ⎛ c ⎞ A 8.314 × 150 ⎛⎜ + − = − − 1 1 ( ) (1.1733 + 0.05076 ) − 133.1932 B v ⎜ ⎟ 2 3 2 2 ⎜ 3⎟ (1.1733) ⎝ 1.1733 × 150 ⎠ v ⎝ vT ⎠ v (1.1733) = 1000.4 kPa (negligible error)

P=

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-54

3-105 Problem 3-104 is reconsidered. Using EES (or other) software, the pressure results of the ideal gas and Beattie-Bridgeman equations with nitrogen data supplied by EES are to be compared. The temperature is to be plotted versus specific volume for a pressure of 1000 kPa with respect to the saturated liquid and saturated vapor lines of nitrogen over the range of 110 K < T < 150 K. Analysis The problem is solved using EES, and the solution is given below. Function BeattBridg(T,v,M,R_u) v_bar=v*M "Conversion from m^3/kg to m^3/kmol" "The constants for the Beattie-Bridgeman equation of state are found in text" Ao=136.2315; aa=0.02617; Bo=0.05046; bb=-0.00691; cc=4.20*1E4 B=Bo*(1-bb/v_bar) A=Ao*(1-aa/v_bar) "The Beattie-Bridgeman equation of state is" BeattBridg:=R_u*T/(v_bar**2)*(1-cc/(v_bar*T**3))*(v_bar+B)-A/v_bar**2 End T=150 [K] v=0.041884 [m^3/kg] P_exper=1000 [kPa] T_table=T; T_BB=T;T_idealgas=T P_table=PRESSURE(Nitrogen,T=T_table,v=v) "EES data for nitrogen as a real gas" {T_table=temperature(Nitrogen, P=P_table,v=v)} M=MOLARMASS(Nitrogen) R_u=8.314 [kJ/kmol-K] "Universal gas constant" R=R_u/M "Particular gas constant" P_idealgas=R*T_idealgas/v "Ideal gas equation" P_BB=BeattBridg(T_BB,v,M,R_u) "Beattie-Bridgeman equation of state Function" PBB [kPa] 1000 1000 1000 1000 1000 1000 1000

160

Ptable [kPa] 1000 1000 1000 1000 1000 1000 1000

Pidealgas [kPa] 1000 1000 1000 1000 1000 1000 1000

v [m3/kg] 0.01 0.02 0.025 0.03 0.035 0.04 0.05

TBB [K] 91.23 95.52 105 116.8 130.1 144.4 174.6

Tideal gas [K] 33.69 67.39 84.23 101.1 117.9 134.8 168.5

Ttable [K] 103.8 103.8 106.1 117.2 130.1 144.3 174.5

Nitrogen, T vs v for P=1000 kPa Ideal Gas

150 140

Beattie-Bridgeman EES Table Value

T [K]

130 120 110 1000 kPa

100 90 80 70 10-3

10-2

10-1

v [m3/kg] PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-55

3-106 Carbon dioxide is compressed in a piston-cylinder device in a polytropic process. The final temperature is to be determined using the ideal gas and van der Waals equations. Properties The gas constant, molar mass, critical pressure, and critical temperature of carbon dioxide are (Table A-1)

R = 0.1889 kPa·m3/kg·K,

M = 44.01 kg/kmol,

Tcr = 304.2 K,

Pcr = 7.39 MPa

Analysis (a) The specific volume at the initial state is

v1 =

RT1 (0.1889 kPa ⋅ m 3 /kg ⋅ K)(473 K) = = 0.08935 m 3 /kg P1 1000 kPa

According to process specification,

⎛ P1 ⎝ P2

v 2 = v 1 ⎜⎜

⎞ ⎟⎟ ⎠

1/ n

⎛ 1000 kPa ⎞ = (0.08935 m 3 /kg)⎜ ⎟ ⎝ 3000 kPa ⎠

1 / 1.2

= 0.03577 m 3 /kg

CO2 1 MPa 200°C

The final temperature is then T2 =

P2v 2 (3000 kPa)(0.03577 m 3 /kg) = = 568 K R 0.1889 kPa ⋅ m 3 /kg ⋅ K

(b) The van der Waals constants for carbon dioxide are determined from a=

27 R 2 Tcr2 (27)(0.1889 kPa ⋅ m 3 /kg ⋅ K) 2 (304.2 K) 2 = = 0.1885 m 6 ⋅ kPa/kg 2 64 Pcr (64)(7390 kPa)

b=

RTcr (0.1889 kPa ⋅ m 3 /kg ⋅ K)(304.2 K) = = 0.0009720 m 3 /kg 8 Pcr 8 × 7390 kPa

Applying the van der Waals equation to the initial state, a ⎞ ⎛ ⎜ P + 2 ⎟(v − b) = RT v ⎠ ⎝ 0.1885 ⎞ ⎛ ⎜1000 + ⎟(v − 0.0009720) = (0.1889)(473) v2 ⎠ ⎝

Solving this equation by trial-error or by EES gives

v 1 = 0.08821 m 3 /kg According to process specification,

⎛P v 2 = v 1 ⎜⎜ 1 ⎝ P2

⎞ ⎟⎟ ⎠

1/ n

⎛ 1000 kPa ⎞ = (0.08821 m 3 /kg)⎜ ⎟ ⎝ 3000 kPa ⎠

1 / 1.2

= 0.03531 m 3 /kg

Applying the van der Waals equation to the final state, a ⎛ ⎜P+ 2 v ⎝

⎞ ⎟(v − b) = RT ⎠

0.1885 ⎞ ⎛ ⎟(0.03531 − 0.0009720) = (0.1889)T ⎜ 3000 + 0.03531 2 ⎠ ⎝

Solving for the final temperature gives T2 = 573 K

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-56 Special Topic: Vapor Pressure and Phase Equilibrium

3-107 A glass of water is left in a room. The vapor pressures at the free surface of the water and in the room far from the glass are to be determined. Assumptions The water in the glass is at a uniform temperature. Properties The saturation pressure of water is 2.339 kPa at 20°C, and 1.706 kPa at 15°C (Table A-4). Analysis The vapor pressure at the water surface is the saturation pressure of water at the water temperature,

Pv, water surface = Psat @ Twater = [email protected]°C = 1.706 kPa

H 2O 15°C

Noting that the air in the room is not saturated, the vapor pressure in the room far from the glass is

Pv, air = φPsat @ Tair = φ[email protected]°C = (0.4)(2.339 kPa) = 0.936 kPa

3-108 The vapor pressure in the air at the beach when the air temperature is 30°C is claimed to be 5.2 kPa. The validity of this claim is to be evaluated. Properties The saturation pressure of water at 30°C is 4.247 kPa (Table A-4). Analysis The maximum vapor pressure in the air is the saturation pressure of water at the given temperature, which is

Pv, max = Psat @ Tair = [email protected]°C = 4.247 kPa

30°C WATER

which is less than the claimed value of 5.2 kPa. Therefore, the claim is false.

3-109 The temperature and relative humidity of air over a swimming pool are given. The water temperature of the swimming pool when phase equilibrium conditions are established is to be determined. Assumptions The temperature and relative humidity of air over the pool remain constant. Properties The saturation pressure of water at 20°C is 2.339 kPa (Table A-4). Analysis The vapor pressure of air over the swimming pool is

Patm, 20°C

Pv, air = φPsat @ Tair = φ[email protected]°C = (0.4)(2.339 kPa) = 0.9357 kPa Phase equilibrium will be established when the vapor pressure at the water surface equals the vapor pressure of air far from the surface. Therefore,

POOL

Pv , water surface = Pv , air = 0.9357 kPa

and

Twater = Tsat @ Pv = Tsat @ 0.9357 kPa = 6.0°C

Discussion Note that the water temperature drops to 6.0°C in an environment at 20°C when phase equilibrium is established.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-57

3-110 A person buys a supposedly cold drink in a hot and humid summer day, yet no condensation occurs on the drink. The claim that the temperature of the drink is below 10°C is to be evaluated. Properties The saturation pressure of water at 35°C is 5.629 kPa (Table A-4). Analysis The vapor pressure of air is

Pv, air = φPsat @ Tair = φ[email protected]°C = (0.7)(5.629 kPa) = 3.940 kPa

35°C 70%

The saturation temperature corresponding to this pressure (called the dew-point temperature) is

Tsat = Tsat @ Pv = [email protected] kPa = 28.7°C That is, the vapor in the air will condense at temperatures below 28.7°C. Noting that no condensation is observed on the can, the claim that the drink is at 10°C is false.

3-111E A thermos bottle half-filled with water is left open to air in a room at a specified temperature and pressure. The temperature of water when phase equilibrium is established is to be determined. Assumptions The temperature and relative humidity of air over the bottle remain constant. Properties The saturation pressure of water at 70°F is 0.3633 psia (Table A-4E). Analysis The vapor pressure of air in the room is

Pv, air = φPsat @ Tair = φ[email protected]°F = (0.35)(0.3633 psia) = 0.1272 psia Phase equilibrium will be established when the vapor pressure at the water surface equals the vapor pressure of air far from the surface. Therefore, Pv , water surface = Pv , air = 0.1272 psia

Thermos bottle 70°F 35%

and

Twater = Tsat @ Pv = Tsat @ 0.1272 psia = 41.1°F Discussion Note that the water temperature drops to 41°F in an environment at 70°F when phase equilibrium is established.

3-112 Two rooms are identical except that they are maintained at different temperatures and relative humidities. The room that contains more moisture is to be determined. Properties The saturation pressure of water is 2.339 kPa at 20°C, and 3.17 kPa at 25°C (Table A-4). Analysis The vapor pressures in the two rooms are

Room 1:

Pv1 = φ1 Psat @ T1 = φ1 [email protected]°C = (0.4)(3.17 kPa) = 1.27 kPa

Room 2:

Pv 2 = φ 2 Psat @ T2 = φ 2 [email protected]°C = (0.55)(2.339 kPa) = 1.29 kPa

Therefore, room 1 at 30°C and 40% relative humidity contains more moisture.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-58 Review Problems

3-113 Nitrogen gas in a rigid tank is heated to a final gage pressure. The final temperature is to be determined. Assumptions At specified conditions, nitrogen behaves as an ideal gas. Analysis According to the ideal gas equation of state at constant volume, m1 = m1 P1V 1 P2V 2 = T1 T2

Patm = 100 kPa

Since V1 = V 2 P1 P2 = T1 T2 T2 = T1

Nitrogen gas 227°C 100 kPa (gage)

Q

P2 (250 + 100) kPa = [(227 + 273) K ] = 875 K = 602°C (100 + 100) kPa P2

3-114 Carbon dioxide flows through a pipe at a given state. The volume and mass flow rates and the density of CO2 at the given state and the volume flow rate at the exit of the pipe are to be determined. Analysis

3 MPa 500 K 0.4 kmol/s

CO2

450 K

(a) The volume and mass flow rates may be determined from ideal gas relation as

V&1 =

N& Ru T1 (0.4 kmol/s)(8.314 kPa.m 3 /kmol.K)(500 K) = = 0.5543 m 3 /s P 3000 kPa

m& 1 =

P1V&1 (3000 kPa)(0.5543 m3 / s) = = 17.60 kg/s RT1 (0.1889 kPa.m3 /kg.K)(500 K)

The density is

ρ1 =

m& 1 (17.60 kg/s) = = 31.76 kg/m 3 & V 1 (0.5543 m 3 /s)

(b) The volume flow rate at the exit is

V&2 =

N& Ru T2 (0.4 kmol/s)(8.314 kPa.m 3 /kmol.K)(450 K) = = 0.4988 m 3 /s P 3000 kPa

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-59

3-115 The cylinder conditions before the heat addition process is specified. The pressure after the heat addition process is to be determined. Assumptions 1 The contents of cylinder are approximated by the air properties. 2 Air is an ideal gas. Analysis The final pressure may be determined from the ideal gas relation P2 =

Combustion chamber 1.4 MPa 450°C

T2 ⎛ 1600 + 273 K ⎞ P1 = ⎜ ⎟(1400 kPa) = 3627 kPa T1 ⎝ 450 + 273 K ⎠

3-116 The cylinder conditions before the heat addition process is specified. The temperature after the heat addition process is to be determined. Assumptions 1 The contents of cylinder is approximated by the air properties. 2 Air is an ideal gas.

Combustion chamber 950 K 75 cm3

Analysis The ratio of the initial to the final mass is m1 AF 22 22 = = = m2 AF + 1 22 + 1 23

The final temperature may be determined from ideal gas relation T2 =

3 m1 V 2 ⎛ 22 ⎞⎛ 150 cm T1 = ⎜ ⎟⎜ m 2 V1 ⎝ 23 ⎠⎜⎝ 75 cm 3

⎞ ⎟(950 K) = 1817 K ⎟ ⎠

3-117 A rigid container that is filled with R-13a is heated. The initial pressure and the final temperature are to be determined. Analysis The initial specific volume is 0.090 m3/kg. Using this with the initial temperature reveals that the initial state is a mixture. The initial pressure is then the saturation pressure,

R-134a -40°C 1 kg 0.090 m3

T1 = −40°C

⎫ ⎬ P = Psat @ - 40°C = 51.25 kPa (Table A - 11) v 1 = 0.090 m /kg ⎭ 1 3

This is a constant volume cooling process (v = V /m = constant). The final state is superheated vapor and the final temperature is then

P

2

P2 = 280 kPa

⎫ ⎬ T = 50°C (Table A - 13) v 2 = v 1 = 0.090 m /kg ⎭ 2 3

1

v

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-60

3-118E A piston-cylinder device that is filled with water is cooled. The final pressure and volume of the water are to be determined. Analysis The initial specific volume is

v1 =

V1 m

=

2.649 ft 3 = 2.649 ft 3 /lbm 1 lbm

H2O 400°F 1 lbm 2.649 ft3

This is a constant-pressure process. The initial state is determined to be superheated vapor and thus the pressure is determined to be T1 = 400°F

⎫ ⎬ P = P2 = 180 psia (Table A - 6E) v 1 = 2.649 ft /lbm ⎭ 1 3

The saturation temperature at 180 psia is 373.1°F. Since the final temperature is less than this temperature, the final state is compressed liquid. Using the incompressible liquid approximation,

P

1

2

v 2 = v f @ 100° F = 0.01613 ft 3 /lbm (Table A - 4E)

v

The final volume is then

V 2 = mv 2 = (1 lbm)(0.01613 ft 3 /lbm) = 0.01613 ft 3

3-119 The volume of chamber 1 of the two-piston cylinder shown in the figure is to be determined. Assumptions At specified conditions, helium behaves as an ideal gas. Properties The gas constant of helium is R = 2.0769 kJ/kg⋅K (Table A-1).

P2A2

Analysis Since the water vapor in chamber 2 is condensing, the pressure in this chamber is the saturation pressure,

P2 = Psat @ 200°C = 1555 kPa

(Table A-4)

Summing the forces acting on the piston in the vertical direction gives

P1 = P2

⎛D A2 = P2 ⎜⎜ 2 A1 ⎝ D1

2

2

⎞ ⎛ 4⎞ ⎟⎟ = (1555 kPa)⎜ ⎟ = 248.8 kPa ⎝ 10 ⎠ ⎠

According to the ideal gas equation of state,

V1 =

P1A1

mRT (1 kg)(2.0769 kPa ⋅ m 3 /kg ⋅ K)(200 + 273 K) = = 3.95 m 3 P1 248.8 kPa

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-61

3-120E The volume of chamber 1 of the two-piston cylinder shown in the figure is to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1E). Analysis Since R-134a in chamber 2 is condensing, the pressure in this chamber is the saturation pressure,

P2 = Psat @ 120°F = 186.0 psia

(Table A-11E)

Summing the forces acting on the piston in the vertical direction gives

F2

F2 + F3 = F1

F3

P2 A2 + P3 ( A1 − A2 ) = P1 A1

which when solved for P1 gives P1 = P2

⎛ A ⎞ A2 + P3 ⎜⎜1 − 2 ⎟⎟ A1 ⎠ A1 ⎝

since the areas of the piston faces are given by A = πD 2 / 4 the above equation becomes

⎛D P1 = P2 ⎜⎜ 2 ⎝ D1

2 ⎡ ⎛D ⎞ ⎟⎟ + P3 ⎢1 − ⎜⎜ 2 ⎢ ⎝ D1 ⎠ ⎣

⎞ ⎟⎟ ⎠

F1

2⎤

⎥ ⎥ ⎦

⎡ ⎛2⎞ ⎛2⎞ = (186.0 psia)⎜ ⎟ + (30 psia) ⎢1 − ⎜ ⎟ ⎝3⎠ ⎢⎣ ⎝ 3 ⎠ 2

2⎤

⎥ ⎥⎦

= 99.33 psia According to the ideal gas equation of state,

V1 =

mRT (0.5 lbm)(0.3704 psia ⋅ ft 3 /lbm ⋅ R)(120 + 460 R) = = 1.08 ft 3 P1 99.33 psia

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-62

3-121E The difference in the volume of chamber 1 for two cases of pressure in chamber 3 is to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1). Analysis Since R-134a in chamber 2 is condensing, the pressure in this chamber is the saturation pressure,

P2 = Psat @ 120°F = 186.0 psia

(Table A-11E)

Summing the forces acting on the piston in the vertical direction gives

F2

F2 + F3 = F1 P2 A2 + P3 ( A1 − A2 ) = P1 A1

F3

which when solved for P1 gives P1 = P2

⎛ A ⎞ A2 + P3 ⎜⎜1 − 2 ⎟⎟ A1 ⎠ A1 ⎝

since the areas of the piston faces are given by A = πD 2 / 4 the above equation becomes ⎛D P1 = P2 ⎜⎜ 2 ⎝ D1

2 ⎡ ⎛D ⎞ ⎟⎟ + P3 ⎢1 − ⎜⎜ 2 ⎢ ⎝ D1 ⎠ ⎣

⎞ ⎟⎟ ⎠

F1

2⎤

⎥ ⎥ ⎦

⎡ ⎛2⎞ ⎛2⎞ = (186.0 psia)⎜ ⎟ + (60 kPa) ⎢1 − ⎜ ⎟ ⎝3⎠ ⎢⎣ ⎝ 3 ⎠ 2

2⎤

⎥ ⎥⎦

= 116 psia According to the ideal gas equation of state,

V1 =

mRT (0.5 lbm)(0.3704 psia ⋅ ft 3 /lbm ⋅ R)(120 + 460 R) = = 0.926 ft 3 P1 116 psia

For a chamber 3 pressure of 30 psia, the volume of chamber 1 was determined to be 1.08 ft3. Then the change in the volume of chamber 1 is

∆V = V 2 − V1 = 1.08 − 0.926 = 0.154 ft 3

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-63

3-122 Ethane is heated at constant pressure. The final temperature is to be determined using ideal gas equation and the compressibility charts. Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A-1,

R = 0.2765 kPa·m3/kg·K,

Tcr = 305.5 K,

Pcr = 4.48 MPa

Analysis From the ideal gas equation, T2 = T1

v2 = (373 K )(1.6) = 596.8 K v1

From the compressibility chart at the initial state (Fig. A-15), ⎫ ⎪ ⎪ ⎬ Z1 = 0.61, v R1 = 0.35 P1 10 MPa = = = 2.232 ⎪ ⎪⎭ Pcr 4.48 MPa

T R1 = PR1

T1 373 K = = 1.221 Tcr 305.5 K

Ethane 10 MPa 100°C

Q

At the final state, PR 2 = PR1 = 2.232

v R 2 = 1.6v R1

⎫ ⎬ Z 2 = 0.83 = 1.6(0.35) = 0.56 ⎭

Thus, T2 =

P2v 2 P v T 10,000 kPa (0.56)(305.5 K) = 2 R 2 cr = = 460 K 0.83 4480 kPa Z 2 R Z 2 Pcr

Of these two results, the accuracy of the second result is limited by the accuracy with which the charts may be read. Accepting the error associated with reading charts, the second temperature is the more accurate.

3-123 A large tank contains nitrogen at a specified temperature and pressure. Now some nitrogen is allowed to escape, and the temperature and pressure of nitrogen drop to new values. The amount of nitrogen that has escaped is to be determined. Properties The gas constant for nitrogen is 0.2968 kPa·m3/kg·K (Table A-1). Analysis Treating N2 as an ideal gas, the initial and the final masses in the tank are determined to be m1 =

P1V (600 kPa)(20 m 3 ) = = 136.6 kg RT1 (0.2968kPa ⋅ m 3 /kg ⋅ K)(296 K)

m2 =

P2V (400 kPa)(20 m 3 ) = = 92.0 kg RT2 (0.2968 kPa ⋅ m 3 /kg ⋅ K)(293 K)

Thus the amount of N2 that escaped is

∆m = m1 − m 2 = 136.6 − 92.0 = 44.6 kg

N2 600 kPa 23°C 20 m3

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-64

3-124 The rigid tank contains saturated liquid-vapor mixture of water. The mixture is heated until it exists in a single phase. For a given tank volume, it is to be determined if the final phase is a liquid or a vapor. Analysis This is a constant volume process (v = V /m = constant), and thus the final specific volume will be equal to the initial specific volume,

v 2 = v1

H2O

The critical specific volume of water is 0.003106 m3/kg. Thus if the final specific volume is smaller than this value, the water will exist as a liquid, otherwise as a vapor.

V = 4 L ⎯⎯→v =

V m

V = 400 L ⎯⎯→v =

=

V m

V=4L m = 2 kg T = 50°C

0.004 m3 = 0.002 m3/kg < v cr Thus, liquid. 2 kg =

0.4 m3 = 0.2 m3/kg > v cr . Thus, vapor. 2 kg

3-125 Two rigid tanks that contain hydrogen at two different states are connected to each other. Now a valve is opened, and the two gases are allowed to mix while achieving thermal equilibrium with the surroundings. The final pressure in the tanks is to be determined. Properties The gas constant for hydrogen is 4.124 kPa·m3/kg·K (Table A-1). Analysis Let's call the first and the second tanks A and B. Treating H2 as an ideal gas, the total volume and the total mass of H2 are

A

V = V A + V B = 0.5 + 0.5 = 1.0 m 3 ⎛ PV ⎞ (400 kPa)(0.5 m 3 ) = 0.1655 kg m A = ⎜⎜ 1 ⎟⎟ = 3 ⎝ RT1 ⎠ A (4.124 kPa ⋅ m /kg ⋅ K)(293 K) ⎛ PV ⎞ (150 kPa)(0.5 m 3 ) m B = ⎜⎜ 1 ⎟⎟ = = 0.0563 kg 3 ⎝ RT1 ⎠ B (4.124 kPa ⋅ m /kg ⋅ K)(323 K) m = m A + m B = 0.1655 + 0.0563 = 0.2218 kg

B

H2

V = 0.5 m3 T=20°C P=400 kPa

×

H2

V = 0.5 m3 T=50°C P=150 kPa

Then the final pressure can be determined from P=

mRT2

V

=

(0.2218 kg)(4.124 kPa ⋅ m 3 /kg ⋅ K)(288 K) 1.0 m 3

= 264 kPa

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-65

3-126 Problem 3-125 is reconsidered. The effect of the surroundings temperature on the final equilibrium pressure in the tanks is to be investigated. The final pressure in the tanks is to be plotted versus the surroundings temperature, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Given Data" V_A=0.5 [m^3] T_A=20 [C] P_A=400 [kPa] V_B=0.5 [m^3] T_B=50 [C] P_B=150 [kPa] {T_2=15 [C]} "Solution" R=R_u/MOLARMASS(H2) R_u=8.314 [kJ/kmol-K] V_total=V_A+V_B m_total=m_A+m_B P_A*V_A=m_A*R*(T_A+273) P_B*V_B=m_B*R*(T_B+273) P_2*V_total=m_total*R*(T_2+273)

T2 [C] -10 -5 0 5 10 15 20 25 30

280

270

P2 [kPa]

P2 [kPa] 240.6 245.2 249.7 254.3 258.9 263.5 268 272.6 277.2

260

250

240 -10

-5

0

5

10

15

20

25

30

T2 [C]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-66

3-127 The pressure in an automobile tire increases during a trip while its volume remains constant. The percent increase in the absolute temperature of the air in the tire is to be determined. Assumptions 1 The volume of the tire remains constant. 2 Air is an ideal gas.

TIRE 200 kPa 0.035 m3

Properties The local atmospheric pressure is 90 kPa. Analysis The absolute pressures in the tire before and after the trip are P1 = Pgage,1 + Patm = 200 + 90 = 290 kPa P2 = Pgage,2 + Patm = 220 + 90 = 310 kPa

Noting that air is an ideal gas and the volume is constant, the ratio of absolute temperatures after and before the trip are P1V1 P2V 2 T P 310 kPa = 1.069 = → 2 = 2 = T1 T2 T1 P1 290 kPa

Therefore, the absolute temperature of air in the tire will increase by 6.9% during this trip.

3-128 The temperature of steam in a tank at a specified state is to be determined using the ideal gas relation, the generalized chart, and the steam tables. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,

R = 0.4615 kPa ⋅ m3/kg ⋅ K,

Tcr = 647.1 K,

Pcr = 22.06 MPa

Analysis (a) From the ideal gas equation of state,

P=

RT

v

=

(0.4615 kPa ⋅ m3/kg ⋅ K)(673 K) = 15,529 kPa 0.02 m3/kg

H 2O 0.02 m3/kg 400°C

(b) From the compressibility chart (Fig. A-15a), ⎫ ⎪ ⎪ ⎬ PR = 0.57 (0.02 m3/kg)(22,060 kPa) v actual = = 1.48 ⎪ vR = ⎪ RTcr / Pcr (0.4615 kPa ⋅ m3 /kg ⋅ K)(647.1 K) ⎭ TR =

673 K T = = 1.040 Tcr 647.1 K

Thus, P = PR Pcr = 0.57 × 22,060 = 12,574 kPa

(c) From the superheated steam table,

T = 400°C

⎫ P = 12,515 kPa

v = 0.02 m 3 /kg ⎬⎭

(from EES)

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-67

3-129 One section of a tank is filled with saturated liquid R-134a while the other side is evacuated. The partition is removed, and the temperature and pressure in the tank are measured. The volume of the tank is to be determined. Analysis The mass of the refrigerant contained in the tank is m=

V1 0.03 m 3 = = 33.58 kg v 1 0.0008934 m 3 /kg

since

v 1 = v f @1.4 MPa = 0.0008934 m 3 /kg

R-134a P=1.2 MPa V =0.03 m3

Evacuated

At the final state (Table A-13), P2 = 400 kPa ⎫ 3 ⎬ v 2 = 0.05680 m /kg T2 = 30°C ⎭

Thus,

V tank = V 2 = mv 2 = (33.58 kg)(0.05680 m 3 /kg) = 1.91 m 3

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-68

3-130 Problem 3-129 is reconsidered. The effect of the initial pressure of refrigerant-134 on the volume of the tank is to be investigated as the initial pressure varies from 0.5 MPa to 1.5 MPa. The volume of the tank is to be plotted versus the initial pressure, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Given Data" x_1=0.0 Vol_1=0.03 [m^3] P_1=1200 [kPa] T_2=30 [C] P_2=400 [kPa] "Solution" v_1=volume(R134a,P=P_1,x=x_1) Vol_1=m*v_1 v_2=volume(R134a,P=P_2,T=T_2) Vol_2=m*v_2

m [kg] 37.23 36.59 36.01 35.47 34.96 34.48 34.02 33.58 33.15 32.73 32.32

2.15 2.1 2.05 3

Vol2 [m3] 2.114 2.078 2.045 2.015 1.986 1.958 1.932 1.907 1.883 1.859 1.836

Vol2 [m ]

P1 [kPa] 500 600 700 800 900 1000 1100 1200 1300 1400 1500

2 1.95 1.9 1.85 1.8 500

700

900

1100

1300

1500

P1 [kPa]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-69

3-131 A propane tank contains 5 L of liquid propane at the ambient temperature. Now a leak develops at the top of the tank and propane starts to leak out. The temperature of propane when the pressure drops to 1 atm and the amount of heat transferred to the tank by the time the entire propane in the tank is vaporized are to be determined. Properties The properties of propane at 1 atm are Tsat = -42.1°C, ρ = 581 kg / m 3 , and hfg = 427.8 kJ/kg (Table A-3). Analysis The temperature of propane when the pressure drops to 1 atm is simply the saturation pressure at that temperature,

Propane 5L

T = Tsat @1 atm = −42.1° C

20°C

The initial mass of liquid propane is

m = ρV = (581 kg/m3 )(0.005 m3 ) = 2.905 kg The amount of heat absorbed is simply the total heat of vaporization,

Leak

Qabsorbed = mh fg = (2.905 kg)(427.8 kJ / kg) = 1243 kJ

3-132 An isobutane tank contains 5 L of liquid isobutane at the ambient temperature. Now a leak develops at the top of the tank and isobutane starts to leak out. The temperature of isobutane when the pressure drops to 1 atm and the amount of heat transferred to the tank by the time the entire isobutane in the tank is vaporized are to be determined. Properties The properties of isobutane at 1 atm are Tsat = -11.7°C, ρ = 593.8 kg / m 3 , and hfg = 367.1 kJ/kg (Table A-3). Analysis The temperature of isobutane when the pressure drops to 1 atm is simply the saturation pressure at that temperature,

T = Tsat @1 atm = −11.7° C

Isobutane 5L

The initial mass of liquid isobutane is

20°C

m = ρV = (593.8 kg/m 3 )(0.005 m 3 ) = 2.969kg The amount of heat absorbed is simply the total heat of vaporization, Leak

Qabsorbed = mh fg = (2.969 kg)(367.1 kJ / kg) = 1090 kJ

3-133 A tank contains helium at a specified state. Heat is transferred to helium until it reaches a specified temperature. The final gage pressure of the helium is to be determined. Assumptions 1 Helium is an ideal gas. Properties The local atmospheric pressure is given to be 100 kPa. Analysis Noting that the specific volume of helium in the tank remains constant, from ideal gas relation, we have P2 = P1

T2 (300 + 273)K = (110 + 100 kPa) = 343.8 kPa T1 (77 + 273)K

Then the gage pressure becomes

Helium 77ºC 110 kPa gage

Q

Pgage,2 = P2 − Patm = 343.8 − 100 = 244 kPa

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-70

3-134 The first eight virial coefficients of a Benedict-Webb-Rubin gas are to be obtained. Analysis The Benedict-Webb-Rubin equation of state is given by P=

γ ⎞ C ⎞ 1 bR T − a aα c ⎛ ⎛ + 6 + 3 2 ⎜1 + 2 ⎟ exp(−γ / v 2 ) + ⎜ B0 RuT − A0 − 02 ⎟ 2 + u 3 v v v v v T T ⎠ ⎝ ⎠ ⎝

RuT

v

Expanding the last term in a series gives exp(−γ / v 2 ) = 1 −

γ 1 γ2 1 γ3 + − + .... v 2 2! v 4 3! v 6

Substituting this into the Benedict-Webb-Rubin equation of state and rearranging the first terms gives P=

Ru T

v

+

R u TB 0 − A0 − C 0 / T 2

v

2

+

bR u T − a

v

3

+

c (1 + γ )

v 5T 2

+

v

6

c γ (1 + γ )

v 7T 2

+

1 c γ 2 (1 + γ ) 2! v 9 T 2

The virial equation of state is P=

Ru T

v

+

a (T )

v

2

+

b (T )

v

3

+

c (T )

v

4

d (T )

+

v

5

+

e (T )

v

6

+

f (T )

v

7

+

g (T )

v

8

+

h (T )

v

9

...

Comparing the Benedict-Webb-Rubin equation of state to the virial equation of state, the virial coefficients are a (T ) = R u TB 0 − A0 − C 0 / T 2 b (T ) = bR u T − a c (T ) = 0 d (T ) = c (1 + γ ) / T 2 e (T ) = a α f (T ) = cγ (1 + γ ) / T 2 g (T ) = 0 h (T ) =

1 cγ 2 (1 + γ ) 2! T2

3-135 The table is completed as follows:

*

P, kPa

T, oC

v, m3/kg

u, kJ/kg

300

250

0.7921

2728.9

300

133.52

0.3058

101.42

100

-

-

Insufficient information

3000

180

0.001127*

761.92*

Compressed liquid

1560.0

Condition description and quality, if applicable Superheated vapor x = 0.504, Two-phase mixture

Approximated as saturated liquid at the given temperature of 180oC

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-71

3-136 The table is completed as follows:

*

P, kPa

T, oC

v, m3/kg

u, kJ/kg

200

120.2

0.8858

2529.1

Condition description and quality, if applicable Saturated vapor

232.23

125

0.5010

1831.0

x = 0.650, Two-phase mixture

7829

400

0.0352

1000

30

0.001004*

125.73*

Compressed liquid

120.90

105

-

-

Insufficient information

2967.2

Superheated vapor

Approximated as saturated liquid at the given temperature of 30oC

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-72

3-137 Water at a specified state is contained in a tank. It is now cooled. The process will be indicated on the P-v and T- v diagrams. Analysis The properties at the initial and final states are P1 = 300 kPa ⎫ 3 ⎬ v 1 = 0.7964 m /kg (Table A - 6) T1 = 250°C ⎭ P2 = 150 kPa

⎫ ⎬ T = 111.35°C (Table A - 5) v 2 = v 1 = 0.7964 m /kg ⎭ 2 3

Using Property Plot feature of EES, and by adding state points we obtain following diagrams.

Steam IAPWS

106 105

P [kPa]

104 103

1

300 kPa

102

250°C 150 kPa

111,4°C

2

101 100 10-4

10-3

10-2

10-1 v [m3/kg]

100

101

102

Steam IAPWS

700 600

T [°C]

500 400 300 200 100 0 10-3

250°C

1

300 kPa 111.35°C

150 kPa

10-2

10-1

2 100

101

102

v [m3/kg]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-73

3-138 Water at a specified state is contained in a piston-cylinder device fitted with stops. Water is now heated until a final pressure. The process will be indicated on the P-v and T- v diagrams. Analysis The properties at the three states are P1 = 300 kPa

⎫ ⎬ T1 = 133.5°C (Table A - 5)

v 1 = 0.5 m 3 /kg ⎭ P2 = 300 kPa

⎫ 3 ⎬ v 2 = 0.6058 m /kg, T2 = 133.5°C (Table A - 5) x 2 = 1 (sat. vap.) ⎭

Q

Water 300 kPa 0.5 m3/kg

P2 = 600 kPa

⎫ ⎬ T = 517.8°C (Table A - 6) v 3 = 0.6058 m /kg ⎭ 2 3

Using Property Plot feature of EES, and by adding state points we obtain following diagrams.

Steam IAPWS

106 105

P [kPa]

104 517.8°C

103

600 kPa

102

300 kPa

3

158.8°C 133.5°C

1

2

101 100 10-4

10-3

10-2

10-1

0.5

100

101

102

v [m3/kg]

Steam IAPWS

700

600 kPa

600

300 kPa

517.8°C

3

T [°C]

500 400 300 200

158.8°C

100

133.5°C

0 10-3

1

10-2

10-1

2

100 3 v [m /kg] 0.5

101

102

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-74

3-139E Argon contained in a piston-cylinder device at a given state undergoes a polytropic process. The final temperature is to be determined using the ideal gas relation and the Beattie-Bridgeman equation. Analysis (a) The polytropic relations for an ideal gas give

⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

n −1 / n

⎛ 2000 psia ⎞ ⎟⎟ = (300 + 460 R)⎜⎜ ⎝ 1000 psia ⎠

0.6 / 1.6

= 986 R

(b) The constants in the Beattie-Bridgeman equation are expressed as a⎞ ⎛ 0.02328 ⎞ ⎛ A = Ao ⎜1 − ⎟ = 130.7802⎜1 − ⎟ v ⎠ ⎝ ⎝ v ⎠ 0⎞ b⎞ ⎛ ⎛ B = Bo ⎜1 − ⎟ = 0.03931⎜1 − ⎟ v v ⎝ ⎠ ⎝ ⎠

Argon 1000 psia 300°F

c = 5.99 × 10 4 m 3 ⋅ K 3 /kmol Substituting these coefficients into the Beattie-Bridgeman equation and using data in SI units (P = 1000 psia = 6895 kPa, T=760 R = 422.2 K, Ru = 8.314 kJ/kmol·K) P=

Ru T ⎛ A c ⎞ ⎟(v + B ) − 2 ⎜1 − 2 3 v ⎝ vT ⎠ v

and solving using an equation solver such as EES gives

v = 0.5120 m 3 /kmol = 8.201 ft 3 /lbmol From the polytropic equation ⎛ P1 ⎝ P2

v 2 = v 1 ⎜⎜

⎞ ⎟⎟ ⎠

1/ n

⎛1⎞ = (0.5120 m 3 /kmol)⎜ ⎟ ⎝2⎠

1 / 1.6

= 0.3319 m 3 /kmol

Substituting this value into the Beattie-Bridgeman equation and using data in SI units (P = 2000 psia = 13790 kPa and Ru = 8.314 kJ/kmol·K), P=

Ru T ⎛ A c ⎞ ⎟(v + B ) − 2 ⎜1 − 2 3 v ⎝ vT ⎠ v

and solving using an equation solver such as EES gives T2 = 532.2 K = 958 R

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-75

3-140E The specific volume of nitrogen at a given state is to be determined using the ideal gas relation, the BenedictWebb-Rubin equation, and the compressibility factor. Properties The properties of nitrogen are (Table A-1E)

R = 0.3830 psia·ft3/lbm·R, M = 28.013 lbm/lbmol,

Tcr = 227.1 R,

Pcr = 492 psia

Analysis (a) From the ideal gas equation of state,

v=

RT (0.3830 psia ⋅ ft 3 /lbm ⋅ R)(360 R) = = 0.3447 ft 3 /lbm P 400 psia

Nitrogen 400 psia, -100°F

(b) Using the coefficients of Table 3-4 for nitrogen and the given data in SI units, the Benedict-Webb-Rubin equation of state is ⎛ C ⎞ 1 bR T − a aα c ⎛ γ ⎞ + ⎜ B0 RuT − A0 − 02 ⎟ + u 3 + 6 + 3 2 ⎜1 + 2 ⎟ exp(−γ / v 2 ) ⎜ ⎟ v T ⎠v v v v T ⎝ v ⎠ ⎝

RuT

P=

(8.314)(200) ⎛⎜ 8.164 ×105 ⎞⎟ 1 0.002328× 8.314 × 200 − 2.54 + + 0.04074× 8.314 × 200 − 106.73 − ⎜ v2 2002 ⎟⎠ v 2 v3 ⎝

2758 = +

2.54 × 1.272 ×10−4

v6

+

7.379 ×104 ⎛ 0.0053 ⎞ 2 ⎟ exp(−0.0053 / v ) ⎜1 + v 3 (200)2 ⎝ v2 ⎠

The solution of this equation by an equation solver such as EES gives

v = 0.5666 m 3 /kmol Then,

v=

v M

=

0.5666 m 3 /kmol ⎛⎜ 16.02 ft 3 /lbm ⎞⎟ = 0.3240 ft 3 /lbm 28.013 kg/kmol ⎜⎝ 1 m 3 /kg ⎟⎠

(c) From the compressibility chart (Fig. A-15), T 360 R ⎫ = = 1.585 ⎪ Tcr 227.1 R ⎪ ⎬ Z = 0.94 400 psia P PR = = = 0.813 ⎪ ⎪⎭ Pcr 492 psia

TR =

Thus,

v = Zv ideal = (0.94)(0.3447 ft 3 /lbm) = 0.3240 ft 3 /lbm

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-76 Fundamentals of Engineering (FE) Exam Problems

3-141 A rigid tank contains 2 kg of an ideal gas at 4 atm and 40°C. Now a valve is opened, and half of mass of the gas is allowed to escape. If the final pressure in the tank is 2.2 atm, the final temperature in the tank is (a) 71°C

(b) 44°C

(c) -100°C

(d) 20°C

(e) 172°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). "When R=constant and V= constant, P1/P2=m1*T1/m2*T2" m1=2 "kg" P1=4 "atm" P2=2.2 "atm" T1=40+273 "K" m2=0.5*m1 "kg" P1/P2=m1*T1/(m2*T2) T2_C=T2-273 "C" "Some Wrong Solutions with Common Mistakes:" P1/P2=m1*(T1-273)/(m2*W1_T2) "Using C instead of K" P1/P2=m1*T1/(m1*(W2_T2+273)) "Disregarding the decrease in mass" P1/P2=m1*T1/(m1*W3_T2) "Disregarding the decrease in mass, and not converting to deg. C" W4_T2=(T1-273)/2 "Taking T2 to be half of T1 since half of the mass is discharged"

3-142 The pressure of an automobile tire is measured to be 190 kPa (gage) before a trip and 215 kPa (gage) after the trip at a location where the atmospheric pressure is 95 kPa. If the temperature of air in the tire before the trip is 25°C, the air temperature after the trip is (a) 51.1°C

(b) 64.2°C

(c) 27.2°C

(d) 28.3°C

(e) 25.0°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). "When R, V, and m are constant, P1/P2=T1/T2" Patm=95 P1=190+Patm "kPa" P2=215+Patm "kPa" T1=25+273 "K" P1/P2=T1/T2 T2_C=T2-273 "C" "Some Wrong Solutions with Common Mistakes:" P1/P2=(T1-273)/W1_T2 "Using C instead of K" (P1-Patm)/(P2-Patm)=T1/(W2_T2+273) "Using gage pressure instead of absolute pressure" (P1-Patm)/(P2-Patm)=(T1-273)/W3_T2 "Making both of the mistakes above" W4_T2=T1-273 "Assuming the temperature to remain constant"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-77 3

3-143 A 300-m rigid tank is filled with saturated liquid-vapor mixture of water at 200 kPa. If 25% of the mass is liquid and the 75% of the mass is vapor, the total mass in the tank is (a) 451 kg

(b) 556 kg

(c) 300 kg

(d) 331 kg

(e) 195 kg

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V_tank=300 "m3" P1=200 "kPa" x=0.75 v_f=VOLUME(Steam_IAPWS, x=0,P=P1) v_g=VOLUME(Steam_IAPWS, x=1,P=P1) v=v_f+x*(v_g-v_f) m=V_tank/v "kg" "Some Wrong Solutions with Common Mistakes:" R=0.4615 "kJ/kg.K" T=TEMPERATURE(Steam_IAPWS,x=0,P=P1) P1*V_tank=W1_m*R*(T+273) "Treating steam as ideal gas" P1*V_tank=W2_m*R*T "Treating steam as ideal gas and using deg.C" W3_m=V_tank "Taking the density to be 1 kg/m^3"

3-144 Water is boiled at 1 atm pressure in a coffee maker equipped with an immersion-type electric heating element. The coffee maker initially contains 1 kg of water. Once boiling started, it is observed that half of the water in the coffee maker evaporated in 10 minutes. If the heat loss from the coffee maker is negligible, the power rating of the heating element is (a) 3.8 kW

(b) 2.2 kW

(c) 1.9 kW

(d) 1.6 kW

(e) 0.8 kW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_1=1 "kg" P=101.325 "kPa" time=10*60 "s" m_evap=0.5*m_1 Power*time=m_evap*h_fg "kJ" h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" W1_Power*time=m_evap*h_g "Using h_g" W2_Power*time/60=m_evap*h_g "Using minutes instead of seconds for time" W3_Power=2*Power "Assuming all the water evaporates"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-78 3

3-145 A 1-m rigid tank contains 10 kg of water (in any phase or phases) at 160°C. The pressure in the tank is (a) 738 kPa

(b) 618 kPa

(c) 370 kPa

(d) 2000 kPa

(e) 1618 kPa

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V_tank=1 "m^3" m=10 "kg" v=V_tank/m T=160 "C" P=PRESSURE(Steam_IAPWS,v=v,T=T) "Some Wrong Solutions with Common Mistakes:" R=0.4615 "kJ/kg.K" W1_P*V_tank=m*R*(T+273) "Treating steam as ideal gas" W2_P*V_tank=m*R*T "Treating steam as ideal gas and using deg.C"

3-146 Water is boiling at 1 atm pressure in a stainless steel pan on an electric range. It is observed that 2 kg of liquid water evaporates in 30 minutes. The rate of heat transfer to the water is (a) 2.51 kW

(b) 2.32 kW

(c) 2.97 kW

(d) 0.47 kW

(e) 3.12 kW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_evap=2 "kg" P=101.325 "kPa" time=30*60 "s" Q*time=m_evap*h_fg "kJ" h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" W1_Q*time=m_evap*h_g "Using h_g" W2_Q*time/60=m_evap*h_g "Using minutes instead of seconds for time" W3_Q*time=m_evap*h_f "Using h_f"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-79

3-147 Water is boiled in a pan on a stove at sea level. During 10 min of boiling, its is observed that 200 g of water has evaporated. Then the rate of heat transfer to the water is (a) 0.84 kJ/min

(b) 45.1 kJ/min

(c) 41.8 kJ/min

(d) 53.5 kJ/min

(e) 225.7 kJ/min

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_evap=0.2 "kg" P=101.325 "kPa" time=10 "min" Q*time=m_evap*h_fg "kJ" h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" W1_Q*time=m_evap*h_g "Using h_g" W2_Q*time*60=m_evap*h_g "Using seconds instead of minutes for time" W3_Q*time=m_evap*h_f "Using h_f"

3-148 A rigid 3-m3 rigid vessel contains steam at 4 MPa and 500°C. The mass of the steam is (a) 3 kg

(b) 9 kg

(c) 26 kg

(d) 35 kg

(e) 52 kg

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V=3 "m^3" m=V/v1 "m^3/kg" P1=4000 "kPa" T1=500 "C" v1=VOLUME(Steam_IAPWS,T=T1,P=P1) "Some Wrong Solutions with Common Mistakes:" R=0.4615 "kJ/kg.K" P1*V=W1_m*R*(T1+273) "Treating steam as ideal gas" P1*V=W2_m*R*T1 "Treating steam as ideal gas and using deg.C"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-80

3-149 Consider a sealed can that is filled with refrigerant-134a. The contents of the can are at the room temperature of 25°C. Now a leak developes, and the pressure in the can drops to the local atmospheric pressure of 90 kPa. The temperature of the refrigerant in the can is expected to drop to (rounded to the nearest integer) (a) 0°C

(b) -29°C

(c) -16°C

(d) 5°C

(e) 25°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 "C" P2=90 "kPa" T2=TEMPERATURE(R134a,x=0,P=P2) "Some Wrong Solutions with Common Mistakes:" W1_T2=T1 "Assuming temperature remains constant"

3-150 … 3-152 Design and Essay Problems

KJ

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-1

Solutions Manual for

Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011

Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-2 Moving Boundary Work 4-1C Yes.

4-2C The area under the process curve, and thus the boundary work done, is greater in the constant pressure case.

4-3 1 kPa ⋅ m 3 = 1 k(N / m 2 ) ⋅ m 3 = 1 kN ⋅ m = 1 kJ

4-4 Helium is compressed in a piston-cylinder device. The initial and final temperatures of helium and the work required to compress it are to be determined. Assumptions The process is quasi-equilibrium. Properties The gas constant of helium is R = 2.0769 kJ/kg⋅K (Table A-1). Analysis The initial specific volume is

v1 =

V1 m

7 m3 = 7 m 3 /kg 1 kg

=

Using the ideal gas equation, T1 =

P1v 1 (150 kPa)(7 m 3 /kg ) = = 505.1 K 2.0769 kJ/kg ⋅ K R

P (kPa) 200

Since the pressure stays constant, T2 =

2

1

3

7

3

V (m3)

V2 3m T1 = (505.1 K ) = 216.5 K V1 7 m3

and the work integral expression gives

Wb,out =

2

1

⎛ 1 kJ P dV = P(V 2 − V1 ) = (150 kPa)(3 − 7) m 3 ⎜ ⎜ 1 kPa ⋅ m 3 ⎝

⎞ ⎟ = −600 kJ ⎟ ⎠

That is, Wb,in = 600 kJ

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-3 4-5E The boundary work done during the process shown in the figure is to be determined. Assumptions The process is quasi-equilibrium. Analysis The work done is equal to the the sumof the areas under the process lines 1-2 and 2-3: P1 + P2 (V 2 −V1 ) + P2 (V 3 −V 2 ) 2 ⎛ (300 + 15)psia 1 Btu (3.3 - 1) ft 3 )⎜ = 3 ⎜ 2 ⎝ 5.404 psia ⋅ ft ⎛ ⎞ 1 Btu ⎟ + (300 psia)(2 − 3.3)ft 3 ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎟ ⎝ ⎠ = -5.14 Btu

P (psia)

Wb,out = Area =

300 ⎞ ⎟ ⎟ ⎠

15

3

2

2

3.3 V (ft3)

1 1

The negative sign shows that the work is done on the system.

4-6 The work done during the isothermal process shown in the figure is to be determined. Assumptions The process is quasi-equilibrium. Analysis From the ideal gas equation, P=

RT

v

For an isothermal process,

v1 = v 2

P2 600 kPa = (0.2 m 3 /kg) = 0.6 m 3 /kg P1 200 kPa

Substituting ideal gas equation and this result into the boundary work integral produces Wb,out =

2

1

P dv = mRT

= mP1v 1 ln

2

1

dv

v

v2 0.2 m 3 = (3 kg)(200 kPa)(0.6 m 3 ) ln v1 0.6 m 3

⎛ 1 kJ ⎜ ⎜ 1 kPa ⋅ m 3 ⎝

⎞ ⎟ ⎟ ⎠

= −395.5 kJ The negative sign shows that the work is done on the system.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-4 4-7 A piston-cylinder device contains nitrogen gas at a specified state. The boundary work is to be determined for the polytropic expansion of nitrogen. Properties The gas constant for nitrogen is 0.2968 kJ/kg.K (Table A-2). Analysis The mass and volume of nitrogen at the initial state are m=

P1V1 (130 kPa)(0.07 m 3 ) = = 0.07802 kg RT1 (0.2968 kJ/kg.K)(120 + 273 K)

V2 =

N2 130 kPa 120°C

mRT2 (0.07802 kg)(0.2968 kPa.m 3 /kg.K)(100 + 273 K) = = 0.08637 m 3 P2 100 kPa

The polytropic index is determined from

P1V1n = P2V 2n ⎯ ⎯→(130 kPa)(0.07 m 3 ) n = (100 kPa)(0.08637 m 3 ) n ⎯ ⎯→ n = 1.249 The boundary work is determined from Wb =

P2V 2 − P1V 1 (100 kPa)(0.08637 m 3 ) − (130 kPa)(0.07 m 3 ) = = 1.86 kJ 1− n 1 − 1.249

4-8 A piston-cylinder device with a set of stops contains steam at a specified state. Now, the steam is cooled. The compression work for two cases and the final temperature are to be determined. Analysis (a) The specific volumes for the initial and final states are (Table A-6) P1 = 1 MPa ⎫ 3 ⎬v1 = 0.30661 m /kg T1 = 400°C ⎭

P2 = 1 MPa ⎫ 3 ⎬v 2 = 0.23275 m /kg T2 = 250°C ⎭

Noting that pressure is constant during the process, the boundary work is determined from

Steam 0.3 kg 1 MPa 400°C

Q

3

Wb = mP (v 1 − v 2 ) = (0.3 kg)(1000 kPa)( 0.30661 − 0.23275) m /kg = 22.16 kJ

(b) The volume of the cylinder at the final state is 60% of initial volume. Then, the boundary work becomes Wb = mP (v 1 − 0.60v 1 ) = (0.3 kg)(1000 kPa)( 0.30661 − 0.60 × 0.30661) m 3 /kg = 36.79 kJ

The temperature at the final state is P2 = 0.5 MPa

⎫⎪ ⎬T2 = 151.8°C (Table A-5) v 2 = (0.60 × 0.30661) m /kg ⎪⎭ 3

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-5 4-9 A piston-cylinder device contains nitrogen gas at a specified state. The final temperature and the boundary work are to be determined for the isentropic expansion of nitrogen. Properties The properties of nitrogen are R = 0.2968 kJ/kg.K , k = 1.395 (Tables A-2a, A-2b) Analysis The mass and the final volume of nitrogen are m=

N2 130 kPa 180°C

P1V1 (130 kPa)(0.07 m 3 ) = = 0.06768 kg RT1 (0.2968 kJ/kg.K)(1 80 + 273 K)

P1V1k = P2V 2k ⎯ ⎯→(130 kPa)(0.07 m 3 )1.395 = (80 kPa)V 21.395 ⎯ ⎯→V 2 = 0.09914 m 3 The final temperature and the boundary work are determined as T2 =

P2V 2 (80 kPa)(0.09914 m 3 ) = = 395 K mR (0.06768 kg)(0.2968 kPa.m 3 /kg.K)

Wb =

P2V 2 − P1V1 (80 kPa)(0.09914 m 3 ) − (130 kPa)(0.07 m 3 ) = = 2.96 kJ 1− k 1 − 1.395

4-10 Saturated water vapor in a cylinder is heated at constant pressure until its temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-4 through A-6)

P1 = 300 kPa ⎫ 3 ⎬ v 1 = v g @ 300 kPa = 0.60582 m /kg Sat. vapor ⎭ P2 = 300 kPa ⎫ 3 ⎬ v 2 = 0.71643 m /kg T2 = 200°C ⎭

P (kPa) 300

1

2

Analysis The boundary work is determined from its definition to be

Wb,out =

2

1

P dV = P(V 2 − V1 ) = mP(v 2 − v1 )

V

⎛ 1 kJ ⎞ ⎟ = (5 kg)(300 kPa)(0.71643 − 0.60582) m3/kg⎜⎜ 3⎟ ⎝ 1 kPa ⋅ m ⎠ = 165.9 kJ Discussion The positive sign indicates that work is done by the system (work output).

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-6 4-11 Refrigerant-134a in a cylinder is heated at constant pressure until its temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-11 through A-13)

P1 = 500 kPa ⎫ 3 ⎬ v 1 = v f @ 500 kPa = 0.0008059 m /kg Sat. liquid ⎭ P2 = 500 kPa ⎫ 3 ⎬ v 2 = 0.052427 m /kg T2 = 70°C ⎭

P (kPa) 1

500

2

Analysis The boundary work is determined from its definition to be m=

V1 0.05 m 3 = = 62.04 kg v 1 0.0008059 m 3 /kg

v

and Wb,out =

2

1

P dV = P(V 2 − V1 ) = mP(v 2 − v 1 )

⎛ 1 kJ = (62.04 kg)(500 kPa)(0.052427 − 0.0008059)m 3 /kg⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 1600 kJ

⎞ ⎟ ⎟ ⎠

Discussion The positive sign indicates that work is done by the system (work output).

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-7 4-12 Problem 4-11 is reconsidered. The effect of pressure on the work done as the pressure varies from 400 kPa to 1200 kPa is to be investigated. The work done is to be plotted versus the pressure.

Analysis The problem is solved using EES, and the solution is given below. "Knowns" Vol_1L=200 [L] x_1=0 "saturated liquid state" P=900 [kPa] T_2=70 [C] "Solution" Vol_1=Vol_1L*convert(L,m^3) "The work is the boundary work done by the R-134a during the constant pressure process." W_boundary=P*(Vol_2-Vol_1) "The mass is:" Vol_1=m*v_1 v_1=volume(R134a,P=P,x=x_1) Vol_2=m*v_2 v_2=volume(R134a,P=P,T=T_2) "Plot information:" v[1]=v_1 v[2]=v_2 P[1]=P P[2]=P T[1]=temperature(R134a,P=P,x=x_1) T[2]=T_2

Wboundary [kJ] 1801 1661 1601 1546 1493 1442 1393 1344 1297 1250

1800

1700

Wboundary [kJ]

P [kPa] 200 400 500 600 700 800 900 1000 1100 1200

1600

1500

1400

1300

1200 200

300

400

500

600

700

800

900

1000 1100 1200

P [kPa]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-8 R134a

104

P [kPa]

103

102

101 10-4

10-3

10-2

10-1

3

v [m /kg]

R134a

250 200 150

T [°C]

100 50

500 kPa

0 -50 -100 10-4

10-3

10-2

10-1

3

v [m /kg]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-9 4-13 Water is expanded isothermally in a closed system. The work produced is to be determined. Assumptions The process is quasi-equilibrium. Analysis From water table

P1 = P2 = Psat @ 200°C = 1554.9 kPa

v 1 = v f @ 200°C = 0.001157 m 3 /kg v 2 = v f + xv fg = 0.001157 + 0.80(0.12721 − 0.001157)

P (kPa) 1

1555

2

= 0.10200 m 3 /kg The definition of specific volume gives

V 2 = V1

1

88.16

V (m3)

3

v2 0.10200 m /kg = (1 m 3 ) = 88.16 m 3 v1 0.001157 m 3 /kg

The work done during the process is determined from Wb,out =

2

1

⎛ 1 kJ ⎞ P dV = P(V 2 −V1 ) = (1554.9 kPa)(88.16 − 1)m 3 ⎜ ⎟ = 1.355 × 10 5 kJ 3 ⋅ 1 kPa m ⎝ ⎠

4-14 Air in a cylinder is compressed at constant temperature until its pressure rises to a specified value. The boundary work done during this process is to be determined. Assumptions 1 The process is quasi-equilibrium. 2 Air is an ideal gas. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). Analysis The boundary work is determined from its definition to be

Wb, out =

2

1

P dV = P1V1 ln

P 2

V2 P = mRT ln 1 V1 P2

150 kPa = (2.4 kg)(0.287 kJ/kg ⋅ K)(285 K)ln 600 kPa

T = 12°C 1

V

= −272 kJ

Discussion The negative sign indicates that work is done on the system (work input).

4-15 Several sets of pressure and volume data are taken as a gas expands. The boundary work done during this process is to be determined using the experimental data. Assumptions The process is quasi-equilibrium. Analysis Plotting the given data on a P-V diagram on a graph paper and evaluating the area under the process curve, the work done is determined to be 0.25 kJ.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-10 4-16 A gas in a cylinder expands polytropically to a specified volume. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis The boundary work for this polytropic process can be determined directly from ⎛V P2 = P1 ⎜⎜ 1 ⎝V 2

n

⎛ 0.03 m 3 ⎞ ⎟⎟ = (350 kPa)⎜ ⎜ 0.2 m 3 ⎠ ⎝

⎞ ⎟ ⎟ ⎠

1.5

P (kPa)

= 20.33 kPa

15

and Wb,out =

2

1

P dV =

P2V 2 − P1V1 1− n

(20.33 × 0.2 − 350 × 0.03) kPa ⋅ m 3 = 1 − 1.5 = 12.9 kJ

1 PV

⎛ 1 kJ ⎜ ⎜ 1 kPa ⋅ m 3 ⎝

⎞ ⎟ ⎟ ⎠

2 0.0

0.2

V

(m3)

Discussion The positive sign indicates that work is done by the system (work output).

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-11 4-17 Problem 4-16 is reconsidered. The process described in the problem is to be plotted on a P-V diagram, and the effect of the polytropic exponent n on the boundary work as the polytropic exponent varies from 1.1 to 1.6 is to be plotted. Analysis The problem is solved using EES, and the solution is given below. Function BoundWork(P[1],V[1],P[2],V[2],n) "This function returns the Boundary Work for the polytropic process. This function is required since the expression for boundary work depens on whether n=1 or n1" If n1 then BoundWork:=(P[2]*V[2]-P[1]*V[1])/(1-n)"Use Equation 3-22 when n=1" else BoundWork:= P[1]*V[1]*ln(V[2]/V[1]) "Use Equation 3-20 when n=1" endif end "Inputs from the diagram window" {n=1.5 P[1] = 350 [kPa] V[1] = 0.03 [m^3] V[2] = 0.2 [m^3] Gas\$='AIR'} "System: The gas enclosed in the piston-cylinder device." "Process: Polytropic expansion or compression, P*V^n = C" P[2]*V[2]^n=P[1]*V[1]^n "n = 1.3" "Polytropic exponent" "Input Data" W_b = BoundWork(P[1],V[1],P[2],V[2],n)"[kJ]" "If we modify this problem and specify the mass, then we can calculate the final temperature of the fluid for compression or expansion" m[1] = m[2] "Conservation of mass for the closed system" "Let's solve the problem for m[1] = 0.05 kg" m[1] = 0.05 [kg] "Find the temperatures from the pressure and specific volume." T[1]=temperature(gas\$,P=P[1],v=V[1]/m[1]) T[2]=temperature(gas\$,P=P[2],v=V[2]/m[2])

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-12 160 140 120

P [kPa]

100 80 60 40 20 0 0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

3

V [m ]

Wb [kJ] 18.14 17.25 16.41 15.63 14.9 14.22 13.58 12.98 12.42 11.89

19 18 17 16

Wb [kJ]

n 1.1 1.156 1.211 1.267 1.322 1.378 1.433 1.489 1.544 1.6

15 14 13 12 11 1.1

1.2

1.3

1.4

1.5

1.6

n

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-13 4-18 Nitrogen gas in a cylinder is compressed polytropically until the temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions 1 The process is quasi-equilibrium. 2 Nitrogen is an ideal gas. Properties The gas constant for nitrogen is R = 0.2968 kJ/kg.K (Table A-2a) Analysis The boundary work for this polytropic process can be determined from

P 2

P V − PV mR(T2 − T1 ) P dV = 2 2 1 1 = = 1 1− n 1− n (2 kg)(0.2968 kJ/kg ⋅ K)(360 − 300)K = 1 − 1.4 = −89.0 kJ

Wb,out

2

PV n =C 1

V

Discussion The negative sign indicates that work is done on the system (work input).

A gas whose equation of state is v ( P + 10 / v 2 ) = Ru T expands in a cylinder isothermally to a specified 4-19 volume. The unit of the quantity 10 and the boundary work done during this process are to be determined. Assumptions The process is quasi-equilibrium. Analysis (a) The term 10 / v is added to P.

2

must have pressure units since it

P

Thus the quantity 10 must have the unit kPa·m6/kmol2.

T = 350 K

(b) The boundary work for this process can be determined from P=

Ru T

v

10

v

2

=

Ru T NR u T 10 N 2 10 − = − V / N (V / N ) 2 V V2

2

and ⎞ V ⎟dV = NRu T ln 2 ⎟ 1 1 V1 ⎠ 3 4m = (0.2 kmol)(8.314 kJ/kmol ⋅ K)(350 K)ln 2 m3 ⎛ 1 1 + (10 kPa ⋅ m 6 /kmol 2 )(0.5kmol) 2 ⎜ − ⎜ 4 m3 2 m3 ⎝ = 403 kJ

Wb,out =

2

P dV =

2⎛

NRu T 10 N 2 ⎜ − ⎜ V V2 ⎝

4

V

⎛ 1 1 ⎞ + 10 N 2 ⎜⎜ − ⎟⎟ ⎝ V 2 V1 ⎠

⎞⎛ 1 kJ ⎟⎜ ⎟⎜ 1 kPa ⋅ m 3 ⎠⎝

⎞ ⎟ ⎟ ⎠

Discussion The positive sign indicates that work is done by the system (work output).

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-14

4-20 Problem 4-19 is reconsidered. Using the integration feature, the work done is to be calculated and compared, and the process is to be plotted on a P-V diagram. Analysis The problem is solved using EES, and the solution is given below. "Input Data" N=0.2 [kmol] v1_bar=2/N "[m^3/kmol]" v2_bar=4/N "[m^3/kmol]" T=350 [K] R_u=8.314 [kJ/kmol-K] "The quation of state is:" v_bar*(P+10/v_bar^2)=R_u*T "P is in kPa" "using the EES integral function, the boundary work, W_bEES, is" W_b_EES=N*integral(P,v_bar, v1_bar, v2_bar,0.01) "We can show that W_bhand= integeral of Pdv_bar is (one should solve for P=F(v_bar) and do the integral 'by hand' for practice)." W_b_hand = N*(R_u*T*ln(v2_bar/v1_bar) +10*(1/v2_bar-1/v1_bar)) "To plot P vs v_bar, define P_plot =f(v_bar_plot, T) as" {v_bar_plot*(P_plot+10/v_bar_plot^2)=R_u*T} " P=P_plot and v_bar=v_bar_plot just to generate the parametric table for plotting purposes. To plot P vs v_bar for a new temperature or v_bar_plot range, remove the '{' and '}' from the above equation, and reset the v_bar_plot values in the Parametric Table. Then press F3 or select Solve Table from the Calculate menu. Next select New Plot Window under the Plot menu to plot the new data."

vplot 10 11.11 12.22 13.33 14.44 15.56 16.67 17.78 18.89 20

320

1

280 240

Pplot (kPa)

Pplot 290.9 261.8 238 218.2 201.4 187 174.6 163.7 154 145.5

T = 350 K

200 160

2

120

Area = Wboundary

80 40 0 9

11

13

15

17

19

21

3

vplot (m /kmol)

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-15

4-21 CO2 gas in a cylinder is compressed until the volume drops to a specified value. The pressure changes during the process with volume as P = aV −2 . The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium.

P

Analysis The boundary work done during this process is determined from Wb,out =

2

∫ PdV = ∫ 1

2

⎛ 1 a ⎞ 1 ⎞ − ⎟⎟ ⎜ 2 ⎟dV = −a⎜⎜ ⎝V ⎠ ⎝ V 2 V1 ⎠

2⎛

1

⎛ 1 1 = −(8 kPa ⋅ m 6 )⎜ − ⎜ 0.1 m 3 0.3 m 3 ⎝ = −53.3 kJ

⎞⎛ 1 kJ ⎟⎜ ⎟⎜ 1 kPa ⋅ m 3 ⎠⎝

P = aV--2 ⎞ ⎟ ⎟ ⎠

1 0.3

0.1

V

(m3)

Discussion The negative sign indicates that work is done on the system (work input).

4-22E A gas in a cylinder is heated and is allowed to expand to a specified pressure in a process during which the pressure changes linearly with volume. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis (a) The pressure of the gas changes linearly with volume, and thus the process curve on a P-V diagram will be a straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus,

At state 1: P1 = aV 1 + b 3

3

15 psia = (5 psia/ft )(7 ft ) + b b = −20 psia

P (psia) 100

At state 2:

P2 = aV 2 + b

P = aV + b 2

15

1

100 psia = (5 psia/ft 3 )V 2 + (−20 psia)

V 2 = 24 ft 3

V

7

(ft3)

and, Wb,out = Area =

⎛ P1 + P2 1 Btu (100 + 15)psia (24 − 7)ft 3 ⎜ (V 2 −V1 ) = 3 ⎜ 2 2 ⎝ 5.4039 psia ⋅ ft

⎞ ⎟ ⎟ ⎠

= 181 Btu

Discussion The positive sign indicates that work is done by the system (work output).

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-16

4-23 A piston-cylinder device contains nitrogen gas at a specified state. The boundary work is to be determined for the isothermal expansion of nitrogen. Properties The properties of nitrogen are R = 0.2968 kJ/kg.K , k = 1.4 (Table A-2a). Analysis We first determine initial and final volumes from ideal gas relation, and find the boundary work using the relation for isothermal expansion of an ideal gas

V1 =

mRT (0.25 kg)(0.2968 kJ/kg.K)(180 + 273 K) = = 0.2586 m 3 P1 (130 kPa)

V2 =

mRT (0.25 kg)(0.2968 kJ/kg.K)(180 + 273 K) = = 0.4202 m 3 P2 80 kPa

⎛V Wb = P1V1 ln⎜⎜ 2 ⎝ V1

⎛ 0.4202 m 3 ⎞ ⎞ ⎟ = 16.3 kJ ⎟⎟ = (130 kPa)(0.2586 m 3 ) ln⎜ ⎜ 0.2586 m 3 ⎟ ⎠ ⎝ ⎠

N2 130 kPa 180°C

4-24 A piston-cylinder device contains air gas at a specified state. The air undergoes a cycle with three processes. The boundary work for each process and the net work of the cycle are to be determined. Properties The properties of air are R = 0.287 kJ/kg.K , k = 1.4 (Table A-2a). Analysis For the isothermal expansion process:

V1 = V2 =

mRT (0.15 kg)(0.287 kJ/kg.K)(350 + 273 K) = = 0.01341 m 3 P1 (2000 kPa) mRT (0.15 kg)(0.287 kJ/kg.K)(350 + 273 K) = = 0.05364 m 3 (500 kPa) P2

Air 2 MPa 350°C

⎛ 0.05364 m3 ⎞ ⎛V ⎞ ⎟ = 37.18 kJ Wb,1− 2 = P1V1 ln⎜⎜ 2 ⎟⎟ = (2000 kPa)(0.01341 m3 ) ln⎜ ⎜ 0.01341 m3 ⎟ ⎝ V1 ⎠ ⎠ ⎝ For the polytropic compression process:

P2V 2n = P3V 3n ⎯ ⎯→(500 kPa)(0.05364 m 3 )1.2 = (2000 kPa)V 31.2 ⎯ ⎯→V 3 = 0.01690 m 3 Wb , 2 − 3 =

P3V 3 − P2V 2 (2000 kPa)(0.01690 m 3 ) − (500 kPa)(0.05364 m 3 ) = = -34.86 kJ 1− n 1 − 1.2

For the constant pressure compression process:

Wb,3−1 = P3 (V1 −V 3 ) = (2000 kPa)(0.01341 − 0.01690)m3 = -6.97 kJ The net work for the cycle is the sum of the works for each process Wnet = Wb,1− 2 + Wb, 2 −3 + Wb,3−1 = 37.18 + (−34.86) + (−6.97) = -4.65 kJ

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-17

4-25 A saturated water mixture contained in a spring-loaded piston-cylinder device is heated until the pressure and temperature rises to specified values. The work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis The initial state is saturated mixture at 90°C. The pressure and the specific volume at this state are (Table A-4), P1 = 70.183 kPa

P 2

800 kPa

v 1 = v f + xv fg = 0.001036 + (0.10)(2.3593 − 0.001036)

1

3

= 0.23686 m /kg

v

The final specific volume at 800 kPa and 250°C is (Table A-6)

v 2 = 0.29321 m 3 /kg Since this is a linear process, the work done is equal to the area under the process line 1-2: P1 + P2 m(v 2 − v 1 ) 2 (70.183 + 800)kPa ⎛ 1 kJ ⎞ = (1 kg)(0.29321 − 0.23686)m 3 ⎜ ⎟ 2 ⎝ 1 kPa ⋅ m 3 ⎠ = 24.52 kJ

Wb,out = Area =

4-26 A saturated water mixture contained in a spring-loaded piston-cylinder device is cooled until it is saturated liquid at a specified temperature. The work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis The initial state is saturated mixture at 1 MPa. The specific volume at this state is (Table A-5),

P

v 1 = v f + xv fg = 0.001127 + (0.30)(0.19436 − 0.001127)

1 MPa

1

3

= 0.059097 m /kg

2

The final state is saturated liquid at 100°C (Table A-4)

P2 = 101.42 kPa

v

v 2 = v f = 0.001043 m 3 /kg Since this is a linear process, the work done is equal to the area under the process line 1-2: P1 + P2 m(v 2 − v 1 ) 2 (1000 + 101.42)kPa ⎛ 1 kJ ⎞ = (1.5 kg)(0.001043 − 0.059097)m 3 ⎜ ⎟ 3 2 ⎝ 1 kPa ⋅ m ⎠ = −48.0 kJ

Wb,out = Area =

The negative sign shows that the work is done on the system in the amount of 48.0 kJ.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-18

4-27 An ideal gas undergoes two processes in a piston-cylinder device. The process is to be sketched on a P-V diagram; an expression for the ratio of the compression to expansion work is to be obtained and this ratio is to be calculated for given values of n and r. Assumptions The process is quasi-equilibrium. Analysis (a) The processes on a P-V diagram is as follows:

P

2

P = const

(b) The ratio of the compression-to-expansion work is called the back-work ratio BWR. Process 1-2:

Wb ,1− 2 =

2

1

The process is PVn = constant, , P = Wb,1-2 =

PVn = const

P dV

constant

V

n

, and the integration results in

3 1

V

P2V 2 − P1V 1 mR (T2 − T1 ) = 1− n 1− n

where the ideal gas law has been used. However, the compression work is W comp = −W b,1- 2 =

Process 2-3:

Wb , 2 −3 =

mR (T2 − T1 ) n −1

3 2

P dV

The process is P = constant and the integration gives Wb, 2 −3 = P(V 3 −V 2 )

where P = P2 = P3. Using the ideal gas law, the expansion work is W exp = W b,2 -3 = mR (T3 − T2 )

The back-work ratio is defined as BWR =

W comp W exp

mR (T2 − T1 ) 1 (T2 − T1 ) 1 T2 (1 − T1 / T2 ) 1 (1 − T1 / T2 ) n −1 = = = = mR (T3 − T2 ) n − 1 (T3 − T2 ) n − 1 T2 (T3 / T2 - 1) n − 1 (T3 / T2 - 1)

Since process 1-2 is polytropic, the temperature-volume relation for the ideal gas is T1 ⎛ V 2 ⎞ =⎜ ⎟ T2 ⎜⎝ V 1 ⎟⎠

n −1

⎛1⎞ =⎜ ⎟ ⎝r⎠

n −1

= r 1− n

where r is the compression ratio V1/ V 2. Since process 2-3 is constant pressure, the combined ideal gas law gives

V P3V 3 P2V 2 T V = and P3 = P2 , then 3 = 3 = 1 = r T3 T2 T2 V 2 V 2 The back-work ratio becomes BWR =

1 (1 − r 1− n ) n − 1 (r − 1)

(c) For n = 1.4 and r = 6, the value of the BWR is BWR =

1 (1 − 61−1.4 ) = 0.256 1.4 − 1 (6 − 1)

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-19 Closed System Energy Analysis

4-28E The table is to be completed using conservation of energy principle for a closed system. Analysis The energy balance for a closed system can be expressed as E −E 1in424out 3

=

Net energy transfer by heat, work, and mass

∆E system 1 424 3

Change in internal, kinetic, potential, etc. energies

Qin − Wout = E 2 − E1 = m(e 2 − e1 )

Application of this equation gives the following completed table: Qin

Wout

E1

E2

m

(Btu)

(Btu)

(Btu)

(Btu)

(lbm)

e2 − e1 (Btu/lbm)

350

510

1020

860

3

-53.3

350

130

550

770

5

44.0

560

260

600

900

2

150

-500

0

1400

900

7

-71.4

-650

-50

1000

400

3

-200

4-29E A piston-cylinder device involves expansion work and work input by a stirring device. The net change of internal energy is to be determined. Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved 3 The thermal energy stored in the cylinder itself is negligible. Analysis This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

E −E 1in424out 3

Net energy transfer by heat, work, and mass

=

∆E system 1 424 3

Change in internal, kinetic, potential, etc. energies

Win − Wout = ∆U

(since KE = PE = 0)

where Win = 10.28 Btu and W out = 15,000 lbf ⋅ ft = (15,000 lbf ⋅ ft)

1 Btu = 19.28 Btu 778.17 lbf ⋅ ft

Substituting, ∆U = Win − Wout = 10.28 − 19.28 = −9.00 Btu

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-20

4-30E The heat transfer during a process that a closed system undergoes without any internal energy change is to be determined. Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero. 2 The compression or expansion process is quasi-equilibrium. Analysis The energy balance for this stationary closed system can be expressed as E −E 1in424out 3

=

Net energy transfer by heat, work, and mass

∆E system 1 424 3

Change in internal, kinetic, potential, etc. energies

Qin − Wout = ∆U = 0

(since KE = PE = 0)

Qin = Wout

Then, 1 Btu ⎛ ⎞ Qin = 1.1× 10 6 lbf ⋅ ft⎜ ⎟ = 1414 Btu 778.17 lbf ⋅ ft ⎝ ⎠

4-31 Motor oil is contained in a rigid container that is equipped with a stirring device. The rate of specific energy increase is to be determined. Analysis This is a closed system since no mass enters or leaves. The energy balance for closed system can be expressed as

E −E 1in424out 3

Net energy transfer by heat, work, and mass

=

∆E system 1 424 3

Change in internal, kinetic, potential, etc. energies

Q& in + W& sh,in = ∆E& Then,

∆E& = Q& in + W& sh,in = 1 + 1.5 = 2.5 = 2.5 W Dividing this by the mass in the system gives

∆e& =

∆E& 2.5 J/s = = 1.67 J/kg ⋅ s m 1.5 kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-21

4-32 An insulated rigid tank is initially filled with a saturated liquid-vapor mixture of water. An electric heater in the tank is turned on, and the entire liquid in the tank is vaporized. The length of time the heater was kept on is to be determined, and the process is to be shown on a P-v diagram. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 The device is wellinsulated and thus heat transfer is negligible. 3 The energy stored in the resistance wires, and the heat transferred to the tank itself is negligible. Analysis We take the contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E − Eout 1in424 3

Net energy transfer by heat, work, and mass

=

∆Esystem 1 424 3

H 2O

Change in internal, kinetic, potential, etc. energies

We,in = ∆U = m(u2 − u1 )

V = const.

(since Q = KE = PE = 0)

VI∆t = m(u2 − u1 )

The properties of water are (Tables A-4 through A-6) P1 = 150 kPa ⎫ v f = 0.001053, v g = 1.1594 m 3 /kg ⎬ x1 = 0.25 ⎭ u f = 466.97, u fg = 2052.3 kJ/kg

We

T

v 1 = v f + x1v fg = 0.001053 + [0.25 × (1.1594 − 0.001053)] = 0.29065 m 3 /kg

2

v 2 = v 1 = 0.29065 m 3 /kg ⎫⎪

1

u1 = u f + x1u fg = 466.97 + (0.25 × 2052.3) = 980.03 kJ/kg

sat.vapor

⎬ u 2 = u g @0.29065 m3 /kg = 2569.7 kJ/kg ⎪⎭

v

Substituting, ⎛ 1000 VA ⎞ ⎟⎟ (110 V)(8 A)∆t = (2 kg)(2569.7 − 980.03)kJ/kg⎜⎜ ⎝ 1 kJ/s ⎠ ∆t = 33613 s = 60.2 min

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-22

4-33 Problem 4-32 is reconsidered. The effect of the initial mass of water on the length of time required to completely vaporize the liquid as the initial mass varies from 1 kg to 10 kg is to be investigated. The vaporization time is to be plotted against the initial mass. Analysis The problem is solved using EES, and the solution is given below. PROCEDURE P2X2(v[1]:P[2],x[2]) Fluid\$='Steam_IAPWS' If v[1] > V_CRIT(Fluid\$) then P[2]=pressure(Fluid\$,v=v[1],x=1) x[2]=1 else P[2]=pressure(Fluid\$,v=v[1],x=0) x[2]=0 EndIf End "Knowns" {m=2 [kg]} P[1]=150 [kPa] y=0.75 "moisture" Volts=110 [V] I=8 [amp] "Solution" "Conservation of Energy for the closed tank:" E_dot_in-E_dot_out=DELTAE_dot E_dot_in=W_dot_ele "[kW]" W_dot_ele=Volts*I*CONVERT(J/s,kW) "[kW]" E_dot_out=0 "[kW]" DELTAE_dot=m*(u[2]-u[1])/DELTAt_s "[kW]" DELTAt_min=DELTAt_s*convert(s,min) "[min]" "The quality at state 1 is:" Fluid\$='Steam_IAPWS' x[1]=1-y u[1]=INTENERGY(Fluid\$,P=P[1], x=x[1]) "[kJ/kg]" v[1]=volume(Fluid\$,P=P[1], x=x[1]) "[m^3/kg]" T[1]=temperature(Fluid\$,P=P[1], x=x[1]) "[C]" "Check to see if state 2 is on the saturated liquid line or saturated vapor line:" Call P2X2(v[1]:P[2],x[2]) u[2]=INTENERGY(Fluid\$,P=P[2], x=x[2]) "[kJ/kg]" v[2]=volume(Fluid\$,P=P[2], x=x[2]) "[m^3/kg]" T[2]=temperature(Fluid\$,P=P[2], x=x[2]) "[C]"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-23 Steam IAPW S

700 600

T [°C]

500 400 300 2

200 655 kPa 150 kPa

100

1

0 10-3

10-2

10 -1

100

10 1

102

3

v [m /kg]

m [kg] 1 2 3 4 5 6 7 8 9 10

350 300 250

∆tmin [min]

∆tmin [min] 30.11 60.21 90.32 120.4 150.5 180.6 210.7 240.9 271 301.1

200 150 100 50 0 1

2

3

4

5

6

7

8

9

10

m [kg]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-24

4-34 Saturated water vapor is isothermally condensed to a saturated liquid in a piston-cylinder device. The heat transfer and the work done are to be determined. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

E −E 1in424out 3

Net energy transfer by heat, work, and mass

=

∆E system 1 424 3

Change in internal, kinetic, potential, etc. energies

Wb,in − Qout = ∆U = m(u 2 − u1 )

Water 200°C sat. vapor

(since KE = PE = 0)

Qout = Wb,in − m(u 2 − u1 )

Heat

The properties at the initial and final states are (Table A-4) T1 = 200°C ⎫ v 1 = v g = 0.12721 m 3 / kg ⎬ x1 = 1 ⎭ u1 = u g = 2594.2 kJ/kg P1 = P2 = 1554.9 kPa T2 = 200°C ⎫ v 2 = v f = 0.001157 m 3 / kg ⎬ x2 = 0 ⎭ u 2 = u f = 850.46 kJ/kg

T

2

1

The work done during this process is wb,out =

2

1

⎛ 1 kJ P dV = P(v 2 − v 1 ) = (1554.9 kPa)(0.001157 − 0.12721) m 3 /kg⎜ ⎜ 1 kPa ⋅ m 3 ⎝

v ⎞ ⎟ = −196.0 kJ/kg ⎟ ⎠

That is, wb ,in = 196.0 kJ/kg

Substituting the energy balance equation, we get q out = wb,in − (u 2 − u1 ) = wb,in + u fg = 196.0 + 1743.7 = 1940 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-25

4-35 Water contained in a rigid vessel is heated. The heat transfer is to be determined. Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved 3 The thermal energy stored in the vessel itself is negligible. Analysis We take water as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E −E 1in424out 3

Net energy transfer by heat, work, and mass

=

∆E system 1 424 3

Change in internal, kinetic, potential, etc. energies

Qin = ∆U = m(u 2 − u1 )

(since KE = PE = 0)

The properties at the initial and final states are (Table A-4) T1 = 100°C ⎫ v 1 = v f + xv fg = 0.001043 + (0.123)(1.6720 − 0.001043) = 0.2066 m 3 / kg ⎬ x1 = 0.123 ⎭ u1 = u f + xu fg = 419.06 + (0.123)(2087.0) = 675.76 kJ/kg v 2 −v f 0.2066-0.001091 = = 0.5250 x2 = T T2 = 150°C v fg 0.39248-0.001091 ⎪⎫ ⎬ v 2 = v 1 = 0.2066 m 3 / kg ⎪⎭ u 2 = u f + x 2 u fg

Water 10 L 100°C x = 0.123

Q

2

= 631.66 + (0.5250)(1927.4) = 1643.5 kJ/kg

The mass in the system is

V 0.100 m 3 m= 1 = = 0.04841 kg v 1 0.2066 m 3 /kg

1

v

Substituting, Qin = m(u 2 − u1 ) = (0.04841 kg)(1643.5 − 675.76) kJ/kg = 46.9 kJ

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-26

4-36 Saturated vapor water is cooled at constant temperature (and pressure) to a saturated liquid. The heat rejected is to be determined. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E −E 1in424out 3

Net energy transfer by heat, work, and mass

=

∆E system 1 424 3

Change in internal, kinetic, potential, etc. energies

− Qout − Wb,out = ∆U = m(u 2 − u1 )

Water 300 kPa sat. vap.

(since KE = PE = 0)

− Qout = Wb,out + m(u 2 − u1 )

Q

− Qout = m(h2 − h1 ) Qout = m(h1 − h2 ) q out = h1 − h2

since ∆U + Wb = ∆H during a constant pressure quasi-equilibrium process. Since water changes from saturated liquid to saturated vapor, we have

T

2

1

q out = h g − h f = h fg @ 300 kPa = 2163.5 kJ/kg (Table A-5)

Note that the temperature also remains constant during the process and it is the saturation temperature at 300 kPa, which is 133.5°C.

v

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-27

4-37 Saturated vapor water is cooled at constant pressure to a saturated liquid. The heat transferred and the work done are to be determined. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E −E 1in424out 3

Net energy transfer by heat, work, and mass

=

∆E system 1 424 3

Change in internal, kinetic, potential, etc. energies

− q out − wb,out = ∆u = u 2 − u1

(since KE = PE = 0)

Water 40 kPa sat. vap.

− q out = wb,out + (u 2 − u1 ) − q out = h2 − h1

Q

q out = h1 − h2

since ∆u + wb = ∆h during a constant pressure quasi-equilibrium process. Since water changes from saturated liquid to saturated vapor, we have q out = h g − h f = h fg @ 40 kPa = 2318.4 kJ/kg (Table A-5)

P

2

1

The specific volumes at the initial and final states are

v 1 = v g @ 40 kPa = 3.993 m 3 / kg

v

3

v 2 = v f @ 40 kPa = 0.001026 m / kg Then the work done is determined from wb,out =

2

1

⎛ 1 kJ ⎞ P dV = P(v 2 − v 1 ) = (40 kPa)(0.001026 − 3.9933)m 3 ⎜ ⎟ = 159.7 kJ/kg ⎝ 1 kPa ⋅ m 3 ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-28

4-38 A cylinder is initially filled with saturated liquid water at a specified pressure. The water is heated electrically as it is stirred by a paddle-wheel at constant pressure. The voltage of the current source is to be determined, and the process is to be shown on a P-v diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The cylinder is wellinsulated and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E −E 1in424out 3

=

Net energy transfer by heat, work, and mass

∆E system 1 424 3

Change in internal, kinetic, potential, etc. energies

We,in + W pw,in − W b,out = ∆U

(since Q = KE = PE = 0)

We,in + W pw,in = m(h2 − h1 )

H2O P = const.

( VI∆t ) + W pw,in = m(h2 − h1 )

since ∆U + Wb = ∆H during a constant pressure quasi-equilibrium process. The properties of water are (Tables A-4 through A-6) P1 = 175 kPa ⎫ h1 = h f @175 kPa = 487.01 kJ/kg ⎬ 3 sat.liquid ⎭ v1 = v f @175 kPa = 0.001057 m /kg P2 = 175 kPa ⎫ ⎬ h2 = h f + x2 h fg = 487.01 + (0.5 × 2213.1) = 1593.6 kJ/kg x2 = 0.5 ⎭ m=

0.005 m3 V1 = = 4.731 kg v1 0.001057 m3/kg

Wpw

We

P

1

2

Substituting, VI∆t + (400kJ) = (4.731 kg)(1593.6 − 487.01)kJ/kg

v

VI∆t = 4835 kJ V=

⎛ 1000 VA ⎞ 4835 kJ ⎜ ⎟ = 223.9 V (8 A)(45 × 60 s) ⎜⎝ 1 kJ/s ⎟⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-29

4-39 A cylinder equipped with an external spring is initially filled with steam at a specified state. Heat is transferred to the steam, and both the temperature and pressure rise. The final temperature, the boundary work done by the steam, and the amount of heat transfer are to be determined, and the process is to be shown on a P-v diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The thermal energy stored in the cylinder itself is negligible. 3 The compression or expansion process is quasi-equilibrium. 4 The spring is a linear spring. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. Noting that the spring is not part of the system (it is external), the energy balance for this stationary closed system can be expressed as

E −E 1in424out 3

=

Net energy transfer by heat, work, and mass

∆E system 1 424 3

Change in internal, kinetic, potential, etc. energies

Qin − Wb,out = ∆U = m(u 2 − u1 )

(since KE = PE = 0)

H2O 200 kPa 200°C

Qin = m(u 2 − u1 ) + Wb,out The properties of steam are (Tables A-4 through A-6) P1 = 200 kPa ⎫ v 1 = 1.08049 m 3 /kg ⎬ T1 = 200°C ⎭ u1 = 2654.6 kJ/kg

Q

P 2

V 0.4 m 3 m= 1 = = 0.3702 kg v 1 1.08049 m 3 /kg

1

0.6 m 3 v2 = = = 1.6207 m 3 /kg m 0.3702 kg

V2

P2 = 250 kPa

⎫⎪ T2 = 606°C ⎬ v 2 = 1.6207 m /kg ⎪⎭ u 2 = 3312.0 kJ/kg

v

3

(b) The pressure of the gas changes linearly with volume, and thus the process curve on a P-V diagram will be a straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus, Wb = Area =

⎛ P1 + P2 (V 2 − V1 ) = (200 + 250)kPa (0.6 − 0.4)m 3 ⎜⎜ 1 kJ 3 2 2 ⎝ 1 kPa ⋅ m

⎞ ⎟ = 45 kJ ⎟ ⎠

(c) From the energy balance we have Qin = (0.3702 kg)(3312.0 - 2654.6)kJ/kg + 45 kJ = 288 kJ

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

4-30

4-40 Problem 4-39 is reconsidered. The effect of the initial temperature of steam on the final temperature, the work done, and the total heat transfer as the initial temperature varies from 150°C to 250°C is to be investigated. The final results are to be plotted against the initial temperature. Analysis The problem is solved using EES, and the solution is given below. "The process is given by:" "P[2]=P[1]+k*x*A/A, and as the spring moves 'x' amount, the volume changes by V[2]-V[1]." P[2]=P[1]+(Spring_const)*(V[2] - V[1]) "P[2] is a linear function of V[2]" "where Spring_const = k/A, the actual spring constant divided by the piston face area" "Conservation of mass for the closed system is:" m[2]=m[1] "The conservation of energy for the closed system is" "E_in - E_out = DeltaE, neglect DeltaKE and DeltaPE for the system" Q_in - W_out = m[1]*(u[2]-u[1]) DELTAU=m[1]*(u[2]-u[1]) "Input Data" P[1]=200 [kPa] V[1]=0.4 [m^3] T[1]=200 [C] P[2]=250 [kPa] V[2]=0.6 [m^3] Fluid\$='Steam_IAPWS' m[1]=V[1]/spvol[1] spvol[1]=volume(Fluid\$,T=T[1], P=P[1]) u[1]=intenergy(Fluid\$, T=T[1], P=P[1]) spvol[2]=V[2]/m[2] "The final temperature is:" T[2]=temperature(Fluid\$,P=P[2],v=spvol[2]) u[2]=intenergy(Fluid\$, P=P[2], T=T[2]) Wnet_other = 0 W_out=Wnet_other + W_b "W_b = integral of P[2]*dV[2] for 0.5 W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature(air,h=h[2]) T[4]=temperature(air,h=h[4]) s[2]=entropy(air,T=T[2],P=P[2]) s[4]=entropy(air,T=T[4],P=P[4])

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-65

Bwr

η

Pratio

0.5229 0.6305 0.7038 0.7611 0.8088 0.85 0.8864 0.9192 0.9491 0.9767

0.1 0.1644 0.1814 0.1806 0.1702 0.1533 0.131 0.1041 0.07272 0.03675

2 4 6 8 10 12 14 16 18 20

Wc [kW] 1818 4033 5543 6723 7705 8553 9304 9980 10596 11165

Wnet [kW] 1659 2364 2333 2110 1822 1510 1192 877.2 567.9 266.1

0.36

Wt [kW] 3477 6396 7876 8833 9527 10063 10496 10857 11164 11431

Qin [kW] 16587 14373 12862 11682 10700 9852 9102 8426 7809 7241

4500

0.32 0.28

η

3600 0.24 3150 0.2

Wnet [kW]

4050

2700

0.16 0.12 2

4

6

8

10

12

14

16

18

2250 20

Pratio

Air

1500

3

T [K]

1000

2 4s

2s

4

500 1000 kPa 100 kPa 0 4.5

5.0

5.5

1

6.0

6.5

7.0

7.5

8.0

8.5

s [kJ/kg-K]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-66

9-91 A simple Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis (a) Using the compressor and turbine efficiency relations, T2 s

⎛P = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

(k −1) / k

(k −1) / k

= (295 K )(10 )

T 0.4/1.4

= 569.6 K

⎛P ⎞ ⎛1⎞ T4 s = T3 ⎜⎜ 4 ⎟⎟ = (1240 K )⎜ ⎟ = 642.3 K ⎝ 10 ⎠ ⎝ P3 ⎠ h − h1 c p (T2 s − T1 ) T − T1 η C = 2s ⎯ ⎯→ T2 = T1 + 2 s = ηC h2 − h1 c p (T2 − T1 )

(b)

qin

0.4/1.4

= 295 +

ηT =

3

1240 K 2s 295 K

569.6 − 295 = 625.8 K 0.83

1

2

qout

4s

4 s

c p (T3 − T4 ) h3 − h4 ⎯ ⎯→ T4 = T3 − η T (T3 − T4 s ) = h3 − h4 s c p (T3 − T4 s ) = 1240 − (0.87 )(1240 − 642.3) = 720 K

q in = h3 − h2 = c p (T3 − T2 ) = (1.005 kJ/kg ⋅ K )(1240 − 625.8)K = 617.3 kJ/kg q out = h4 − h1 = c p (T4 − T1 ) = (1.005 kJ/kg ⋅ K )(720 − 295)K = 427.1 kJ/kg wnet,out = q in − q out = 617.3 − 427.1 = 190.2 kJ/kg

(c)

η th =

wnet,out q in

=

190.2 kJ/kg = 0.3081 = 30.8% 617.3 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-67

9-92E A simple ideal Brayton cycle with helium has a pressure ratio of 14. The power output is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Helium is an ideal gas with constant specific heats. Properties The properties of helium are cp = 1.25 Btu/lbm·R and k = 1.667 (Table A-2Ea). Analysis Using the isentropic relations for an ideal gas,

⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

T

( k −1) / k

= T1 r p( k −1) / k

= (520 R)(14)

0.667/1.667

= 1495 R

3

1760 R

qin 2

Similarly,

⎛P T4 = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛ 1 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1760 R)⎜ ⎟ ⎝ 14 ⎠

0.667/1.667

= 612.2 K

520 R

4 1

qout s

Applying the first law to the constant-pressure heat addition process 2-3 produces

q in = c p (T3 − T2 ) = (1.25 Btu/lbm ⋅ R )(1760 − 1495)R = 331.3 Btu/lbm Similarly,

q out = c p (T4 − T1 ) = (1.25 Btu/lbm ⋅ R )(612.2 − 520)R = 115.3 Btu/lbm The net work production is then wnet = q in − q out = 331.3 − 115.3 = 216.0 Btu/lbm

and 1 hp ⎛ ⎞ W& net = m& w net = (100 lbm/min)(216.0 Btu/lbm)⎜ ⎟ = 509.3 hp 42.41 Btu/min ⎝ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-68

9-93E A simple Brayton cycle with helium has a pressure ratio of 14. The power output is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Helium is an ideal gas with constant specific heats. Properties The properties of helium at room temperature are cp = 1.25 Btu/lbm·R and k = 1.667 (Table A-2Ea). Analysis For the compression process,

⎛P T2 s = T1 ⎜⎜ 2 ⎝ P1

ηC =

⎞ ⎟⎟ ⎠

( k −1) / k

= T1 r p( k −1) / k = (520 R)(14) 0.667/1.667 = 1495 R

T 1760 R

2s 2

h2 s − h1 c p (T2 s − T1 ) T −T = ⎯ ⎯→ T2 = T1 + 2 s 1 ηC h2 − h1 c p (T2 − T1 ) = 520 +

3

qin

1495 − 520 = 1546 R 0.95

520 R

1

qout 4 s

For the isentropic expansion process,

⎛P T4 = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛ 1 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1760 R)⎜ ⎟ ⎝ 14 ⎠

0.667/1.667

= 612.2 R

Applying the first law to the constant-pressure heat addition process 2-3 produces

q in = c p (T3 − T2 ) = (1.25 Btu/lbm ⋅ R )(1760 − 1546)R = 267.5 Btu/lbm Similarly,

q out = c p (T4 − T1 ) = (1.25 Btu/lbm ⋅ R )(612.2 − 520)R = 115.3 Btu/lbm The net work production is then wnet = q in − q out = 267.5 − 115.3 = 152.2 Btu/lbm

and 1 hp ⎛ ⎞ W& net = m& w net = (100 lbm/min)(152.2 Btu/lbm)⎜ ⎟ = 358.9 hp ⎝ 42.41 Btu/min ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-69

9-94 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The effects of non-isentropic compressor and turbine on the back-work ratio is to be compared. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis For the compression process,

T2 s

⎛P = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

T = (288 K)(12) 0.4/1.4 = 585.8 K

c p (T2 s − T1 ) h −h T −T η C = 2s 1 = ⎯ ⎯→ T2 = T1 + 2 s 1 ηC h2 − h1 c p (T2 − T1 ) 585.8 − 288 = 288 + = 618.9 K 0.90

873 K

qin

3

2s 2 288 K

1

4

qout 4s s

For the expansion process, ⎛P T4 s = T3 ⎜⎜ 4 ⎝ P3

ηT =

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (873 K)⎜ ⎟ ⎝ 12 ⎠

0.4/1.4

= 429.2 K

c p (T3 − T4 ) h3 − h 4 = ⎯ ⎯→ T4 = T3 − η T (T3 − T4 s ) h3 − h4 s c p (T3 − T4 s ) = 873 − (0.90)(873 − 429.2) = 473.6 K

The isentropic and actual work of compressor and turbine are

WComp,s = c p (T2 s − T1 ) = (1.005 kJ/kg ⋅ K )(585.8 − 288)K = 299.3 kJ/kg

WComp = c p (T2 − T1 ) = (1.005 kJ/kg ⋅ K )(618.9 − 288)K = 332.6 kJ/kg WTurb,s = c p (T3 − T4 s ) = (1.005 kJ/kg ⋅ K )(873 − 429.2)K = 446.0 kJ/kg WTurb = c p (T3 − T4 ) = (1.005 kJ/kg ⋅ K )(873 − 473.6)K = 401.4 kJ/kg The back work ratio for 90% efficient compressor and isentropic turbine case is rbw =

WComp WTurb, s

=

332.6 kJ/kg = 0.7457 446.0 kJ/kg

The back work ratio for 90% efficient turbine and isentropic compressor case is

rbw =

WComp,s WTurb

=

299.3 kJ/kg = 0.7456 401.4 kJ/kg

The two results are almost identical.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-70

9-95 A gas turbine power plant that operates on the simple Brayton cycle with air as the working fluid has a specified pressure ratio. The required mass flow rate of air is to be determined for two cases. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.

T

Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).

⎛P = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

⎛P T4 s = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

T2 s

(k −1) / k

(k −1) / k

= (300 K )(12 )0.4/1.4 = 610.2 K ⎛1⎞ = (1000 K )⎜ ⎟ ⎝ 12 ⎠

3

1000 K

Analysis (a) Using the isentropic relations,

0.4/1.4

= 491.7 K

2s 300 K

2

1

4s

4 s

ws,C,in = h2 s − h1 = c p (T2 s − T1 ) = (1.005 kJ/kg ⋅ K )(610.2 − 300)K = 311.75 kJ/kg ws,T,out = h3 − h4 s = c p (T3 − T4 s ) = (1.005 kJ/kg ⋅ K )(1000 − 491.7 )K = 510.84 kJ/kg ws, net,out = ws,T,out − ws,C,in = 510.84 − 311.75 = 199.1 kJ/kg m& s =

W& net,out ws, net,out

=

70,000 kJ/s = 352 kg/s 199.1 kJ/kg

(b) The net work output is determined to be wa,net,out = wa,T,out − wa,C,in = η T ws,T,out − ws,C,in / η C = (0.85)(510.84 ) − 311.75 0.85 = 67.5 kJ/kg

m& a =

W& net,out wa,net,out

=

70,000 kJ/s = 1037 kg/s 67.5 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-71

9-96 An actual gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis (a) Using the isentropic relations, T1 = 300 K

⎯ ⎯→

h1 = 300.19 kJ / kg

T2 = 580 K

⎯ ⎯→

h2 = 586.04 kJ / kg

rp =

P2 700 = =7 P1 100

⎯→ h3 = 950 + 586.04 = 1536.04kJ/kg q in = h3 − h2 ⎯ → Pr3 = 474.11 Pr 4 =

T 950 kJ/kg 580 K 300 K

2s 1

3

2 4s

4 s

P4 ⎛1⎞ Pr3 = ⎜ ⎟(474.11) = 67.73 ⎯ ⎯→ h4 s = 905.83 kJ/kg P3 ⎝7⎠

wC,in = h2 − h1 = 586.04 − 300.19 = 285.85 kJ/kg wT,out = η T (h3 − h4 s ) = (0.86 )(1536.04 − 905.83) = 542.0 kJ/kg

Thus, rbw =

(b)

wC,in wT,out

=

285.85 kJ/kg = 52.7% 542.0 kJ/kg

wnet.out = wT,out − wC,in = 542.0 − 285.85 = 256.15 kJ/kg

η th =

wnet,out q in

=

256.15 kJ/kg = 27.0% 950 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-72

9-97 A gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.

T

Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).

950 kJ/kg

Analysis (a) Using constant specific heats,

580 K

2s

P2 700 = =7 P1 100

300 K

1

rp =

3

2

q in = h3 − h2 = c p (T3 − T2 )⎯ ⎯→ T3 = T2 + q in /c p

4s

4 s

= 580 K + (950 kJ/kg )/ (1.005 kJ/kg ⋅ K ) = 1525.3 K

⎛P T4s = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

(k −1)/k

⎛1⎞ = (1525.3 K )⎜ ⎟ ⎝7⎠

0.4/1.4

= 874.8 K

wC,in = h2 − h1 = c p (T2 − T1 ) = (1.005kJ/kg ⋅ K )(580 − 300)K = 281.4 kJ/kg wT,out = η T (h3 − h4 s ) = η T c p (T3 − T4 s ) = (0.86)(1.005 kJ/kg ⋅ K )(1525.3 − 874.8)K = 562.2 kJ/kg

Thus, rbw =

(b)

wC,in wT,out

=

281.4 kJ/kg = 50.1% 562.2 kJ/kg

wnet,out = wT,out − wC,in = 562.2 − 281.4 = 280.8 kJ/kg

η th =

wnet,out q in

=

280.8 kJ/kg = 29.6% 950 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-73

9-98 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis For the isentropic compression process, T2 = T1 r p( k −1) / k = ( 273 K)(10) 0.4/1.4 = 527.1 K

T 3

q in =

qin

Q& in 500 kW = = 500 kJ/kg m& 1 kg/s

2

Applying the first law to the heat addition process,

273 K

q in = c p (T3 − T2 )

4 1

qout s

q 500 kJ/kg T3 = T2 + in = 527.1 K + = 1025 K cp 1.005 kJ/kg ⋅ K

The temperature at the exit of the turbine is ⎛ 1 T4 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1025 K)⎜ ⎟ ⎝ 10 ⎠

0.4/1.4

= 530.9 K

Applying the first law to the adiabatic turbine and the compressor produce

wT = c p (T3 − T4 ) = (1.005 kJ/kg ⋅ K )(1025 − 530.9)K = 496.6 kJ/kg wC = c p (T2 − T1 ) = (1.005 kJ/kg ⋅ K )(527.1 − 273)K = 255.4 kJ/kg The net power produced by the engine is then

W& net = m& ( wT − wC ) = (1 kg/s)(496.6 − 255.4)kJ/kg = 241.2 kW Finally the thermal efficiency is

η th =

W& net 241.2 kW = = 0.482 500 kW Q& in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-74

9-99 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis For the isentropic compression process, T2 = T1 r p( k −1) / k = ( 273 K)(15) 0.4/1.4 = 591.8 K

T 3

q in =

qin

Q& in 500 kW = = 500 kJ/kg m& 1 kg/s

2

Applying the first law to the heat addition process,

273 K

q in = c p (T3 − T2 )

4 1

qout s

q 500 kJ/kg T3 = T2 + in = 591.8 K + = 1089 K cp 1.005 kJ/kg ⋅ K

The temperature at the exit of the turbine is ⎛ 1 T4 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1089 K)⎜ ⎟ ⎝ 15 ⎠

0.4/1.4

= 502.3 K

Applying the first law to the adiabatic turbine and the compressor produce

wT = c p (T3 − T4 ) = (1.005 kJ/kg ⋅ K )(1089 − 502.3)K = 589.6 kJ/kg wC = c p (T2 − T1 ) = (1.005 kJ/kg ⋅ K )(591.8 − 273)K = 320.4 kJ/kg The net power produced by the engine is then

W& net = m& ( wT − wC ) = (1 kg/s)(589.6 − 320.4)kJ/kg = 269.2 kW Finally the thermal efficiency is

η th =

W& net 269.2 kW = = 0.538 500 kW Q& in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-75

9-100 A gas-turbine plant operates on the simple Brayton cycle. The net power output, the back work ratio, and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.

Combustion chamber

Process 1-2: Compression

3

⎯→ h1 = 313.6 kJ/kg T1 = 40°C ⎯

2

T1 = 40°C

⎫ ⎬s1 = 5.749 kJ/kg ⋅ K P1 = 100 kPa ⎭

Compress.

P2 = 2000 kPa ⎫ ⎬h2 s = 736.7 kJ/kg s 2 = s1 = 5.749 kJ/kg.K ⎭

ηC =

2 MPa

1

100 kPa 40°C

Turbine 650°C

4

h2 s − h1 736.7 − 313.6 ⎯ ⎯→ h2 = 811.4 kJ/kg ⎯ ⎯→ 0.85 = h2 − h1 h2 − 313.6

Process 3-4: Expansion T4 = 650°C ⎯ ⎯→ h4 = 959.2 kJ/kg

ηT =

h3 − h4 h − 959.2 ⎯ ⎯→ 0.88 = 3 h3 − h4 s h3 − h4 s

We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with the isentropic efficiency relation, we find h3 = 1873 kJ/kg, T3 = 1421ºC, s3 = 6.736 kJ/kg.K. The solution by hand would require a trialerror approach. h_3=enthalpy(Air, T=T_3) s_3=entropy(Air, T=T_3, P=P_2) h_4s=enthalpy(Air, P=P_1, s=s_3)

The mass flow rate is determined from m& =

P1V&1 (100 kPa)(700/60 m 3 / s) = = 12.99 kg/s RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (40 + 273 K )

(

)

The net power output is W& C,in = m& (h2 − h1 ) = (12.99 kg/s)(811.4 − 313.6)kJ/kg = 6464 kW W& T,out = m& (h3 − h4 ) = (12.99 kg/s)(1873 − 959.2)kJ/kg = 11,868 kW W& net = W& T,out − W& C,in = 11,868 − 6464 = 5404 kW

(b) The back work ratio is rbw =

W& C,in 6464 kW = = 0.545 & WT,out 11,868 kW

(c) The rate of heat input and the thermal efficiency are

Q& in = m& (h3 − h2 ) = (12.99 kg/s)(1873 − 811.4)kJ/kg = 13,788 kW

η th =

W& net 5404 kW = = 0.392 = 39.2% Q& in 13,788 kW

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-76

9-101 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The cycle is to be sketched on the T-s cycle and the isentropic efficiency of the turbine and the cycle thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air are given as cv = 0.718 kJ/kg·K, cp = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K, k = 1.4. Analysis (b) For the compression process,

T

W& Comp = m& c p (T2 − T1 ) = (200 kg/s)(1.005 kJ/kg ⋅ K )(330 − 30)K = 60,300 kW

2s 2

For the turbine during the isentropic process, T4 s

⎛P = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛ 100 kPa ⎞ = (1400 K)⎜ ⎟ ⎝ 800 kPa ⎠

3

873 K

303 K

1

0.4/1.4

= 772.9 K

4

4s s

W& Turb, s = m& c p (T3 − T4 s ) = ( 200 kg/s) (1.005 kJ/kg ⋅ K )(1400 − 772.9) K = 126,050 kW

The actual power output from the turbine is W& net = W& Turb − W& Comp W& Turb = W& net + W& Turb = 60,000 + 60,300 = 120,300 kW The isentropic efficiency of the turbine is then

η Turb =

W& Turb 120,300 kW = = 0.954 = 95.4% & W Turb,s 126,050 kW

(c) The rate of heat input is Q& in = m& c p (T3 − T2 ) = ( 200 kg/s) (1.005 kJ/kg ⋅ K )[(1400 − (330 + 273)]K = 160,200 kW

The thermal efficiency is then

η th =

W& net 60,000 kW = = 0.375 = 37.5% & 160,200 kW Qin

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-77

9-102 A modified Brayton cycle with air as the working fluid operates at a specified pressure ratio. The T-s diagram is to be sketched and the temperature and pressure at the exit of the high-pressure turbine and the mass flow rate of air are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air are given as cv = 0.718 kJ/kg·K, cp = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K, k = 1.4. Analysis (b) For the compression process,

⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

( k −1) / k

T

= (273 K)(8) 0.4/1.4 = 494.5 K

The power input to the compressor is equal to the power output from the high-pressure turbine. Then,

P3 = P2

3

1500 K

P4

2 4

W& Comp,in = W& HP Turb,out

P5

5

m& c p (T2 − T1 ) = m& c p (T3 − T4 )

273 K

T2 − T1 = T3 − T4

1 s

T4 = T3 + T1 − T2 = 1500 + 273 − 494.5 = 1278.5 K The pressure at this state is P4 ⎛ T4 ⎞ =⎜ ⎟ P3 ⎜⎝ T3 ⎟⎠

k /( k −1)

⎛T ⎯ ⎯→ P4 = rP1 ⎜⎜ 4 ⎝ T3

⎞ ⎟ ⎟ ⎠

k /( k −1)

⎛ 1278.5 K ⎞ = 8(100 kPa)⎜ ⎟ ⎝ 1500 K ⎠

1.4 / 0.4

= 457.3 kPa

(c) The temperature at state 5 is determined from ⎛P T5 = T4 ⎜⎜ 5 ⎝ P4

⎞ ⎟⎟ ⎠

( k −1) / k

⎛ 100 kPa ⎞ = (1278.5 K)⎜ ⎟ ⎝ 457.3 kPa ⎠

0.4/1.4

= 828.1 K

The net power is that generated by the low-pressure turbine since the power output from the high-pressure turbine is equal to the power input to the compressor. Then, W& LP Turb = m& c p (T4 − T5 ) m& =

W& LP Turb 200,000 kW = = 441.8 kg/s c p (T4 − T5 ) (1.005 kJ/kg ⋅ K )(1278.5 − 828.1)K

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-78

9-103 A simple Brayton cycle with air as the working fluid operates at a specified pressure ratio and between the specified temperature and pressure limits. The cycle is to be sketched on the T-s cycle and the volume flow rate of the air into the compressor is to be determined. Also, the effect of compressor inlet temperature on the mass flow rate and the net power output are to be investigated. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air are given as cv = 0.718 kJ/kg·K, cp = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K, k = 1.4. Analysis (b) For the compression process,

T2 s

⎛P = T1 ⎜⎜ 2 ⎝ P1

η Comp

⎞ ⎟⎟ ⎠

( k −1) / k

= (273 K)(7 )0.4/1.4 = 476.0 K T

W& Comp,s m& c p (T2 s − T1 ) T2 s − T1 = = = m& c p (T2 − T1 ) T2 − T1 W& Comp

476.0 − 273 0.80 = ⎯ ⎯→ T2 = 526.8 K T2 − 273

2s 2 273 K

For the expansion process, ⎛P T4 s = T3 ⎜⎜ 4 ⎝ P3

η Turb = 0.90 =

⎞ ⎟ ⎟ ⎠

( k −1) / k

3

1500 K

1

4

4s s

⎛1⎞ = (1500 K)⎜ ⎟ ⎝7⎠

0.4/1.4

= 860.3 K

m& c p (T3 − T4 ) T3 − T4 W& Turb = = W& Turb,s m& c p (T3 − T4 s ) T3 − T4 s 1500 − T4 ⎯ ⎯→ T4 = 924.3 K 1500 − 860.3

Given the net power, the mass flow rate is determined from W& net = W& Turb − W& Comp = m& c p (T3 − T4 ) − m& c p (T2 − T1 ) W& net = m& c p [(T3 − T4 ) − (T2 − T1 )] m& =

W& net c p [(T3 − T4 ) − (T2 − T1 )]

150,000 kW (1.005 kJ/kg ⋅ K )[(1500 − 924.3) − (526.8 − 273)] = 463.7 kg/s =

The specific volume and the volume flow rate at the inlet of the compressor are

v1 =

RT1 (0.287 kJ/kg ⋅ K )(273 K) = = 0.7835 m 3 /kg P1 100 kPa

V&1 = m& v 1 = (463.7 kg/s)(0.7835 m 3 /kg) = 363.2 m 3 /s (c) For a fixed compressor inlet velocity and flow area, when the compressor inlet temperature increases, the specific RT V& . When specific volume increases, the mass flow rate decreases since m& = . Note that volume increases since v = P v & volume flow rate is the same since inlet velocity and flow area are fixed (V = AV ). When mass flow rate decreases, the net power decreases since W& net = m& ( wTurb − wComp ) . Therefore, when the inlet temperature increases, both mass flow rate and the net power decrease.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-79

Brayton Cycle with Regeneration

9-104C Regeneration increases the thermal efficiency of a Brayton cycle by capturing some of the waste heat from the exhaust gases and preheating the air before it enters the combustion chamber.

9-105C Yes. At very high compression ratios, the gas temperature at the turbine exit may be lower than the temperature at the compressor exit. Therefore, if these two streams are brought into thermal contact in a regenerator, heat will flow to the exhaust gases instead of from the exhaust gases. As a result, the thermal efficiency will decrease.

9-106C The extent to which a regenerator approaches an ideal regenerator is called the effectiveness ε, and is defined as ε = qregen, act /qregen, max.

9-107C (b) turbine exit.

9-108C The steam injected increases the mass flow rate through the turbine and thus the power output. This, in turn, increases the thermal efficiency since η = W / Qin and W increases while Qin remains constant. Steam can be obtained by utilizing the hot exhaust gases.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-80

9-109 A Brayton cycle with regeneration produces 150 kW power. The rates of heat addition and rejection are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis According to the isentropic process expressions for an ideal gas, T2 = T1 r p( k −1) / k = ( 293 K)(8) 0.4/1.4 = 530.8 K

⎛ 1 T5 = T4 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1073 K)⎜ ⎟ ⎝8⎠

T

0.4/1.4

= 592.3 K

3

When the first law is applied to the heat exchanger, the result is T3 − T2 = T5 − T6

4

qin

1073 K

5

2 293 K

while the regenerator temperature specification gives

1

T3 = T5 − 10 = 592.3 − 10 = 582.3 K

6 qout s

The simultaneous solution of these two results gives T6 = T5 − (T3 − T2 ) = 592.3 − (582.3 − 530.8) = 540.8 K

Application of the first law to the turbine and compressor gives wnet = c p (T4 − T5 ) − c p (T2 − T1 ) = (1.005 kJ/kg ⋅ K )(1073 − 592.3) K − (1.005 kJ/kg ⋅ K )(530.8 − 293) K = 244.1 kJ/kg Then,

m& =

W& net 150 kW = = 0.6145 kg/s wnet 244.1 kJ/kg

Applying the first law to the combustion chamber produces Q& in = m& c p (T4 − T3 ) = (0.6145 kg/s) (1.005 kJ/kg ⋅ K )(1073 − 582.3) K = 303.0 kW

Similarly, Q& out = m& c p (T6 − T1 ) = (0.6145 kg/s) (1.005 kJ/kg ⋅ K )(540.8 − 293) K = 153.0 kW

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-81

9-110 A Brayton cycle with regeneration produces 150 kW power. The rates of heat addition and rejection are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis For the compression and expansion processes we have T2 s = T1 r p( k −1) / k = ( 293 K)(8) 0.4/1.4 = 530.8 K

ηC =

c p (T2 s − T1 ) c p (T2 − T1 )

⎯ ⎯→ T2 = T1 +

T2 s − T1

⎛ 1 T5 s = T4 ⎜ ⎜ rp ⎝

ηT =

⎞ ⎟ ⎟ ⎠

c p (T4 − T5 ) c p (T4 − T5 s )

⎛1⎞ = (1073 K)⎜ ⎟ ⎝8⎠

4

qin

1073 K

ηC

530.8 − 293 = 293 + = 566.3 K 0.87 ( k −1) / k

T

2s 293 K

1

0.4/1.4

= 592.3 K

2

5

3 6

5s

qout s

⎯ ⎯→ T5 = T4 − η T (T4 − T5 s ) = 1073 − (0.93)(1073 − 592.3) = 625.9 K

When the first law is applied to the heat exchanger, the result is T3 − T2 = T5 − T6

while the regenerator temperature specification gives T3 = T5 − 10 = 625.9 − 10 = 615.9 K

The simultaneous solution of these two results gives T6 = T5 − (T3 − T2 ) = 625.9 − (615.9 − 566.3) = 576.3 K

Application of the first law to the turbine and compressor gives wnet = c p (T4 − T5 ) − c p (T2 − T1 ) = (1.005 kJ/kg ⋅ K )(1073 − 625.9) K − (1.005 kJ/kg ⋅ K )(566.3 − 293) K = 174.7 kJ/kg Then,

m& =

W& net 150 kW = = 0.8586 kg/s wnet 174.7 kJ/kg

Applying the first law to the combustion chamber produces Q& in = m& c p (T4 − T3 ) = (0.8586 kg/s) (1.005 kJ/kg ⋅ K )(1073 − 615.9)K = 394.4 kW

Similarly, Q& out = m& c p (T6 − T1 ) = (0.8586 kg/s) (1.005 kJ/kg ⋅ K )(576.3 − 293) K = 244.5 kW

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-82

9-111 A Brayton cycle with regeneration is considered. The thermal efficiencies of the cycle for parallel-flow and counterflow arrangements of the regenerator are to be compared. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis According to the isentropic process expressions for an ideal gas, T2 = T1 r p( k −1) / k = ( 293 K)(7) 0.4/1.4 = 510.9 K

⎛ 1 T5 = T4 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1000 K)⎜ ⎟ ⎝7⎠

T

0.4/1.4

= 573.5 K

When the first law is applied to the heat exchanger as originally arranged, the result is

4

qin

1000 K 3

5

2 293 K

6 qout

1

T3 − T2 = T5 − T6

s

while the regenerator temperature specification gives T3 = T5 − 6 = 573.5 − 6 = 567.5 K

The simultaneous solution of these two results gives T6 = T5 − T3 + T2 = 573.5 − 567.5 + 510.9 = 516.9 K

The thermal efficiency of the cycle is then

η th = 1 −

q out T −T 516.9 − 293 = 0.482 = 1− 6 1 = 1− q in T4 − T3 1000 − 567.5

For the rearranged version of this cycle, T3 = T6 − 6

An energy balance on the heat exchanger gives T3 − T2 = T5 − T6

T

The solution of these two equations is T3 = 539.2 K T6 = 545.2 K

The thermal efficiency of the cycle is then

η th = 1 −

2 293 K

4

qin

1000 K

1

3

6

5

qout s

q out T −T 545.2 − 293 = 0.453 = 1− 6 1 = 1− q in T4 − T3 1000 − 539.2

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-83

9-112E An ideal Brayton cycle with regeneration has a pressure ratio of 11. The thermal efficiency of the cycle is to be determined with and without regenerator cases. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea). Analysis According to the isentropic process expressions for an ideal gas, T2 = T1 r p( k −1) / k = (560 R)(11) 0.4/1.4 = 1111 R

⎛ 1 T5 = T4 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (2400 R)⎜ ⎟ ⎝ 11 ⎠

0.4/1.4

= 1210 R

T 4

qin

2400 R 3

560 R

T3 = T5 = 1210 R

5

2

The regenerator is ideal (i.e., the effectiveness is 100%) and thus,

6 qout

1

T6 = T2 = 1111 R

s

The thermal efficiency of the cycle is then

η th = 1 −

q out T − T1 1111 − 560 = 0.537 = 53.7% =1− =1− 6 q in T4 − T3 2400 − 1210

The solution without a regenerator is as follows: T2 = T1 r p( k −1) / k = (560 R)(11) 0.4/1.4 = 1111 R

⎛ 1 T4 = T3 ⎜ ⎜ rp ⎝

η th

⎞ ⎟ ⎟ ⎠

( k −1) / k

T 3

2400 R

⎛1⎞ = (2400 R)⎜ ⎟ ⎝ 11 ⎠

qin

0.4/1.4

= 1210 R

q T − T1 1210 − 560 = 0.496 = 49.6% = 1 − out = 1 − 4 =1− q in T3 − T2 2400 − 1111

2 560 R

4 1

qout s

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-84

9-113E A car is powered by a gas turbine with a pressure ratio of 4. The thermal efficiency of the car and the mass flow rate of air for a net power output of 95 hp are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 The ambient air is 540 R and 14.5 psia. 4 The effectiveness of the regenerator is 0.9, and the isentropic efficiencies for both the compressor and the turbine are 80%. 5 The combustion gases can be treated as air. Properties The properties of air at the compressor and turbine inlet temperatures can be obtained from Table A-17E. Analysis The gas turbine cycle with regeneration can be analyzed as follows: T1 = 540 R ⎯ ⎯→ Pr2 =

h1 = 129.06 Btu/lbm Pr1 = 1.386

P2 Pr = (4)(1.386 ) = 5.544 ⎯ ⎯→ h2 s = 192.0 Btu/lbm P1 1

T3 = 2160 R ⎯ ⎯→

T qin

2160 R 5

h3 = 549.35 Btu/lbm

2

Pr3 = 230.12

P ⎛1⎞ Pr4 = 4 Pr3 = ⎜ ⎟(230.12 ) = 57.53 ⎯ ⎯→ h4 s = 372.2 Btu/lbm P3 ⎝4⎠

3 4 4s

2s 540 R

1 s

and

η comp =

h2 s − h1 192.0 − 129.06 → 0.80 = → h2 = 207.74 Btu/lbm h2 − h1 h2 − 129.06

η turb =

h3 − h4 549.35 − h4 → 0.80 = → h4 = 407.63 Btu/lbm h3 − h4 s 549.35 − 372.2

Then the thermal efficiency of the gas turbine cycle becomes

q regen = ε (h4 − h2 ) = 0.9(407.63 − 207.74) = 179.9 Btu/lbm q in = (h3 − h2 ) − q regen = (549.35 − 207.74) − 179.9 = 161.7 Btu/lbm w net,out = wT,out − wC,in = (h3 − h4 ) − (h2 − h1 ) = (549.35 − 407.63) − (207.74 − 129.06) = 63.0 Btu/lbm

η th =

w net,out q in

=

63.0 Btu/lbm = 0.39 = 39% 161.7 Btu/lbm

Finally, the mass flow rate of air through the turbine becomes m& air =

W& net ⎛ 0.7068 Btu/s ⎞ 95 hp ⎜ ⎟⎟ = 1.07 lbm/s = 1 hp w net 63.0 Btu/lbm ⎜⎝ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-85

9-114 The thermal efficiency and power output of an actual gas turbine are given. The isentropic efficiency of the turbine and of the compressor, and the thermal efficiency of the gas turbine modified with a regenerator are to be determined. Assumptions 1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible. 3 The mass flow rates of air and of the combustion gases are the same, and the properties of combustion gases are the same as those of air. Properties The properties of air are given in Table A-17. Analysis The properties at various states are T1 = 30°C = 303 K ⎯ ⎯→

h1 = 303.21 kJ/kg Pr1 = 1.4356

P Pr2 = 2 Pr1 = (14.7 )(1.4356 ) = 21.10 ⎯ ⎯→ h2 s = 653.25 kJ/kg P1 T3 = 1288°C = 1561 K ⎯ ⎯→

T

P ⎛ 1 ⎞ Pr4 = 4 Pr3 = ⎜ ⎯→ h4 s = 825.23 kJ/kg ⎟(712.5) = 48.47 ⎯ P3 ⎝ 14.7 ⎠

3

5 2

h3 = 1710.0 kJ/kg Pr3 = 712.5

qin

1561 K

4 4s

2s 303 K

1 s

The net work output and the heat input per unit mass are W& net 159,000 kW ⎛ 3600 s ⎞ = ⎜ ⎟ = 372.66 kJ/kg m& 1,536,000 kg/h ⎝ 1 h ⎠ w 372.66 kJ/kg q in = net = = 1038.0 kJ/kg 0.359 η th wnet =

q in = h3 − h2 → h2 = h3 − q in = 1710 − 1038 = 672.0 kJ/kg q out = q in − wnet = 1038.0 − 372.66 = 665.34 kJ/kg q out = h4 − h1 → h4 = q out + h1 = 665.34 + 303.21 = 968.55 kJ/kg → T4 = 931.7 K = 658.7°C Then the compressor and turbine efficiencies become

ηT =

h3 − h4 1710 − 968.55 = 0.838 = 83.8% = h3 − h4 s 1710 − 825.23

ηC =

h2 s − h1 653.25 − 303.21 = 0.949 = 94.9% = h2 − h1 672 − 303.21

When a regenerator is added, the new heat input and the thermal efficiency become q regen = ε (h4 − h2 ) = (0.65)(968.55 - 672.0) = 192.8 kJ/kg q in,new = q in − q regen = 1038 − 192.8 = 845.2 kJ/kg

η th,new =

wnet 372.66 kJ/kg = = 0.441 = 44.1% q in, new 845.2 kJ/kg

Discussion Note a 65% efficient regenerator would increase the thermal efficiency of this gas turbine from 35.9% to 44.1%.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-86

9-115 Problem 9-114 is reconsidered. A solution that allows different isentropic efficiencies for the compressor and turbine is to be developed and the effect of the isentropic efficiencies on net work done and the heat supplied to the cycle is to be studied. Also, the T-s diagram for the cycle is to be plotted. Analysis Using EES, the problem is solved as follows: "Input data" T[3] = 1288 [C] Pratio = 14.7 T[1] = 30 [C] P[1]= 100 [kPa] {T[4]=659 [C]} {W_dot_net=159 [MW] }"We omit the information about the cycle net work" m_dot = 1536000 [kg/h]*Convert(kg/h,kg/s) {Eta_th_noreg=0.359} "We omit the information about the cycle efficiency." Eta_reg = 0.65 Eta_c = 0.84 "Compressor isentropic efficiency" Eta_t = 0.95 "Turbien isentropic efficiency" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = W_dot_compisen/W_dot_comp "compressor adiabatic efficiency, W_dot_comp > W_dot_compisen" "Conservation of energy for the compressor for the isentropic case: E_dot_in - E_dot_out = DELTAE_dot=0 for steady-flow" m_dot*h[1] + W_dot_compisen = m_dot*h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" m_dot*h[1] + W_dot_comp = m_dot*h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 E_dot_in - E_dot_out =DELTAE_dot_cv =0 for steady flow" m_dot*h[2] + Q_dot_in_noreg = m_dot*h[3] q_in_noreg=Q_dot_in_noreg/m_dot h[3]=ENTHALPY(Air,T=T[3]) P[3]=P[2]"process 2-3 is SSSF constant pressure" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[3] /Pratio s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4])"T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = W_dot_turb /W_dot_turbisen "turbine adiabatic efficiency, W_dot_turbisen > W_dot_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 E_dot_in -E_dot_out = DELTAE_dot_cv = 0 for steady-flow" m_dot*h[3] = W_dot_turbisen + m_dot*h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4]) PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-87

"Actual Turbine analysis:" m_dot*h[3] = W_dot_turb + m_dot*h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Cycle analysis" "Using the definition of the net cycle work and 1 MW = 1000 kW:" W_dot_net*1000=W_dot_turb-W_dot_comp "kJ/s" Eta_th_noreg=W_dot_net*1000/Q_dot_in_noreg"Cycle thermal efficiency" Bwr=W_dot_comp/W_dot_turb"Back work ratio" "With the regenerator the heat added in the external heat exchanger is" m_dot*h[5] + Q_dot_in_withreg = m_dot*h[3] q_in_withreg=Q_dot_in_withreg/m_dot h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[2])/(h[4]-h[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" m_dot*h[2] + m_dot*h[4]=m_dot*h[5] + m_dot*h[6] h[6]=ENTHALPY(Air, T=T[6]) s[6]=ENTROPY(Air,T=T[6], P=P[6]) P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=W_dot_net*1000/Q_dot_in_withreg "The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-88

ηt

ηc

ηth,noreg

ηth,withreg

0.7 0.75 0.8 0.85 0.9 0.95 1

0.84 0.84 0.84 0.84 0.84 0.84 0.84

0.2044 0.2491 0.2939 0.3386 0.3833 0.4281 0.4728

0.27 0.3169 0.3605 0.4011 0.4389 0.4744 0.5076

Qinnoreg [kW] 422152 422152 422152 422152 422152 422152 422152

Qinwithreg [kW] 319582 331856 344129 356403 368676 380950 393223

Wnet [kW] 86.3 105.2 124.1 142.9 161.8 180.7 199.6

T-s Diagram for Gas Turbine with Regeneration 200

1700 1500

1470 kPa

180

3

1300

160

900

100 kPa

700

5 2

500

2s

6

300 100 -100 4.5

Wnet [kW]

T [C]

1100

4 4s

140 120 100

1 5.0

5.5

80 0.7

6.0

6.5

7.0

7.5

8.0

0.75

0.8

0.85

0.9

ηt

8.5

0.95

1

s [kJ/kg-K]

430000

no regeneration

0.5

390000

0.45

370000

0.4

with regeneration η th

Qin [kW]

410000

0.55

with regeneration

0.35

350000

no regeneration

0.3 330000

0.25 310000 0.7

0.75

0.8

0.85

ηt

0.9

0.95

1

0.2 0.7

0.75

0.8

0.85

ηt

0.9

0.95

1

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-89

9-116 A Brayton cycle with regeneration using air as the working fluid is considered. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17.

T qin

1150 K 5

Analysis (a) The properties of air at various states are T1 = 310 K ⎯ ⎯→

h1 = 310.24 kJ/kg Pr1 = 1.5546

P Pr 2 = 2 Pr1 = (7 )(1.5546 ) = 10.88 ⎯ ⎯→ h2 s = 541.26 kJ/kg P1

ηC =

3

2s 310 K

4 4s

2 6

1 s

h2 s − h1 ⎯ ⎯→ h2 = h1 + (h2 s − h1 ) / η C = 310.24 + (541.26 − 310.24 )/ (0.75) = 618.26 kJ/kg h2 − h1

T3 = 1150 K ⎯ ⎯→

h3 = 1219.25 kJ/kg Pr3 = 200.15

Pr 4 =

P4 ⎛1⎞ Pr3 = ⎜ ⎟(200.15) = 28.59 ⎯ ⎯→ h4 s = 711.80 kJ/kg P3 ⎝7⎠

ηT =

h3 − h 4 ⎯ ⎯→ h4 = h3 − η T (h3 − h4 s ) = 1219.25 − (0.82 )(1219.25 − 711.80 ) = 803.14 kJ/kg h3 − h 4 s

Thus, T4 = 782.8 K (b)

wnet = wT,out − wC,in = (h3 − h4 ) − (h2 − h1 )

= (1219.25 − 803.14) − (618.26 − 310.24) = 108.09 kJ/kg

(c)

ε=

h5 − h 2 ⎯ ⎯→ h5 = h2 + ε (h4 − h2 ) h4 − h2 = 618.26 + (0.65)(803.14 − 618.26 ) = 738.43 kJ/kg

Then, q in = h3 − h5 = 1219.25 − 738.43 = 480.82 kJ/kg

η th =

wnet 108.09 kJ/kg = = 22.5% q in 480.82 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-90

9-117 A stationary gas-turbine power plant operating on an ideal regenerative Brayton cycle with air as the working fluid is considered. The power delivered by this plant is to be determined for two cases. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas. 3 Kinetic and potential energy changes are negligible. Properties When assuming constant specific heats, the properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). When assuming variable specific heats, the properties of air are obtained from Table A-17. Analysis (a) Assuming constant specific heats, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

⎛P T4 = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

(k −1) / k

(k −1) / k

= (290 K )(8)0.4/1.4 = 525.3 K ⎛1⎞ = (1100 K )⎜ ⎟ ⎝8⎠

T 1100 K

0.4/1.4

= 607.2 K

ε = 100% ⎯⎯→ T5 = T4 = 607.2 K and T6 = T2 = 525.3 K η th = 1 −

c p (T6 − T1 )

2 290 K

1

75,000 kW

3

5

4 6

qout

s

q out T −T 525.3 − 290 = 1− = 1− 6 1 = 1− = 0.5225 1100 − 607.2 q in c p (T3 − T5 ) T3 − T5

W& net = η T Q& in = (0.5225)(75,000 kW ) = 39,188 kW

(b) Assuming variable specific heats, T1 = 290K ⎯ ⎯→ Pr 2 =

h1 = 290.16 kJ/kg Pr1 = 1.2311

P2 Pr = (8)(1.2311) = 9.8488 ⎯ ⎯→ h2 = 526.12 kJ/kg P1 1

T3 = 1100K ⎯ ⎯→ Pr 4 =

h3 = 1161.07 kJ/kg Pr3 = 167.1

P4 ⎛1⎞ Pr = ⎜ ⎟(167.1) = 20.89 ⎯ ⎯→ h4 = 651.37 kJ/kg P3 3 ⎝ 8 ⎠

ε = 100% ⎯⎯→ h5 = h4 = 651.37 kJ/kg and h6 = h2 = 526.12 kJ/kg q out h −h 526.12 − 290.16 = 1− 6 1 = 1− = 0.5371 q in h3 − h5 1161.07 − 651.37 = η T Q& in = (0.5371)(75,000 kW ) = 40,283 kW

η th = 1 − W& net

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-91

9-118 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The properties at various states are r p = P2 / P1 = 900 / 100 = 9

T qin

1400 K

3

5

T1 = 310 K ⎯ ⎯→ h1 = 310.24 kJ/kg

4

2

T2 = 650 K ⎯ ⎯→ h2 = 659.84 kJ/kg

650 K

T3 = 1400 K ⎯ ⎯→ h3 = 1515.42 kJ/kg Pr3 = 450.5

2s

310 K

1

4s 6 s

P ⎛1⎞ Pr4 = 4 Pr3 = ⎜ ⎟(450.5) = 50.06 ⎯ ⎯→ h4 s = 832.44 kJ/kg P3 ⎝9⎠ h − h4 ηT = 3 ⎯ ⎯→ h4 = h3 − η T (h3 − h4 s ) h3 − h4 s = 1515.42 − (0.90 )(1515.42 − 832.44 ) = 900.74 kJ/kg

q regen = ε (h4 − h2 ) = (0.80 )(900.74 − 659.84 ) = 192.7 kJ/kg

(b)

wnet = wT,out − wC,in = (h3 − h4 ) − (h2 − h1 ) = (1515.42 − 900.74 ) − (659.84 − 310.24 ) = 265.08 kJ/kg q in = (h3 − h2 ) − q regen = (1515.42 − 659.84 ) − 192.7 = 662.88 kJ/kg

η th =

wnet 265.08 kJ/kg = = 0.400 = 40.0% q in 662.88 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-92

9-119 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) Using the isentropic relations and turbine efficiency,

T qin

1400 K

r p = P2 / P1 = 900 / 100 = 9 (k −1) / k

3

5

650 K ⎛P ⎞ ⎛1⎞ T4 s = T3 ⎜⎜ 4 ⎟⎟ = 747.3 K = (1400 K )⎜ ⎟ ⎝9⎠ ⎝ P3 ⎠ 310 K c p (T3 − T4 ) h3 − h4 ηT = = ⎯ ⎯→ T4 = T3 − η T (T3 − T4 s ) h3 − h4 s c p (T3 − T4 s ) = 1400 − (0.90)(1400 − 747.3) = 812.6 K 0.4 / 1.4

4

2 2s 1

4s 6 s

q regen = ε (h4 − h2 ) = ε c p (T4 − T2 ) = (0.80)(1.005 kJ/kg ⋅ K )(812.6 − 650)K = 130.7 kJ/kg (b)

wnet = wT,out − wC,in = c p (T3 − T4 ) − c p (T2 − T1 )

= (1.005 kJ/kg ⋅ K )[(1400 − 812.6 ) − (650 − 310 )]K = 248.7 kJ/kg

q in = (h3 − h2 ) − q regen = c p (T3 − T2 ) − q regen = (1.005 kJ/kg ⋅ K )(1400 − 650 )K − 130.7 = 623.1 kJ/kg

η th =

wnet 248.7 kJ/kg = = 0.399 = 39.9% q in 623.1 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-93

9-120 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17.

T

Analysis (a) The properties at various states are r p = P2 / P1 = 900 / 100 = 9

qin

1400 K

3

5

T1 = 310 K ⎯ ⎯→ h1 = 310.24 kJ/kg

4

2

T2 = 650 K ⎯ ⎯→ h2 = 659.84 kJ/kg

650 K

2s

T3 = 1400 K ⎯ ⎯→ h3 = 1515.42 kJ/kg Pr3 = 450.5

310 K

1

4s 6 s

P4 ⎛1⎞ Pr3 = ⎜ ⎟(450.5) = 50.06 ⎯ ⎯→ h4 s = 832.44 kJ/kg P3 ⎝9⎠ h − h4 ηT = 3 ⎯ ⎯→ h4 = h3 − η T (h3 − h4 s ) h3 − h4 s = 1515.42 − (0.90)(1515.42 − 832.44) = 900.74 kJ/kg q regen = ε (h4 − h2 ) = (0.70 )(900.74 − 659.84) = 168.6 kJ/kg Pr4 =

(b)

wnet = wT,out − wC,in = (h3 − h4 ) − (h2 − h1 ) = (1515.42 − 900.74 ) − (659.84 − 310.24 ) = 265.08 kJ/kg q in = (h3 − h2 ) − q regen = (1515.42 − 659.84 ) − 168.6 = 687.18 kJ/kg

η th =

wnet 265.08 kJ/kg = = 0.386 = 38.6% q in 687.18 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-94

9-121 An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator is to be developed. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Analysis The expressions for the isentropic compression and expansion processes are T2 = T1 r p( k −1) / k

⎛ 1 T4 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

For an ideal regenerator, T5 = T4

T qin 5 4

2 1

T6 = T 2

3

6 qout s

The thermal efficiency of the cycle is

η th = 1 −

q out T −T T (T / T ) − 1 = 1− 6 1 = 1− 1 6 1 q in T3 − T5 T3 1 − (T5 / T3 )

= 1−

T1 (T2 / T1 ) − 1 T3 1 − (T4 / T3 )

= 1−

( k −1) / k −1 T1 r p ( k 1 )/k − − T3 1 − r p

= 1−

T1 ( k −1) / k rp T3

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-95

Brayton Cycle with Intercooling, Reheating, and Regeneration

9-122C As the number of compression and expansion stages are increased and regeneration is employed, the ideal Brayton cycle will approach the Ericsson cycle.

9-123C Because the steady-flow work is proportional to the specific volume of the gas. Intercooling decreases the average specific volume of the gas during compression, and thus the compressor work. Reheating increases the average specific volume of the gas, and thus the turbine work output.

9-124C (a) decrease, (b) decrease, and (c) decrease.

9-125C (a) increase, (b) decrease, and (c) decrease.

9-126C (a) increase, (b) decrease, (c) decrease, and (d) increase.

9-127C (a) increase, (b) decrease, (c) increase, and (d) decrease.

9-128C (c) The Carnot (or Ericsson) cycle efficiency.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-96

9-129 An ideal gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17.

T 1200 K

qin 9

Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine since this is an ideal cycle. Then, T1 = 300 K ⎯ ⎯→

h1 = 300.19 kJ/kg Pr1 = 1.386

300 K

P Pr 2 = 2 Pr1 = (3)(1.386) = 4.158 ⎯ ⎯→ h2 = h4 = 411.26 kJ/kg P1 ⎯→ T5 = 1200 K ⎯

4

2

3

1

5

7

6

8

10 s

h5 = h7 = 1277.79 kJ/kg Pr5 = 238

P6 ⎛1⎞ ⎯→ h6 = h8 = 946.36 kJ/kg Pr5 = ⎜ ⎟(238) = 79.33 ⎯ P5 ⎝ 3⎠ = 2(h2 − h1 ) = 2(411.26 − 300.19 ) = 222.14 kJ/kg

Pr6 = wC,in

wT,out = 2(h5 − h6 ) = 2(1277.79 − 946.36) = 662.86 kJ/kg

Thus, rbw =

wC,in wT,out

=

222.14 kJ/kg = 33.5% 662.86 kJ/kg

q in = (h5 − h4 ) + (h7 − h6 ) = (1277.79 − 411.26 ) + (1277.79 − 946.36 ) = 1197.96 kJ/kg w net = wT,out − wC,in = 662.86 − 222.14 = 440.72 kJ/kg

η th =

wnet 440.72 kJ/kg = = 36.8% q in 1197.96 kJ/kg

(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes q regen = ε (h8 − h4 ) = (0.75)(946.36 − 411.26 ) = 401.33 kJ/kg q in = q in,old − q regen = 1197.96 − 401.33 = 796.63 kJ/kg

η th =

wnet 440.72 kJ/kg = = 55.3% q in 796.63 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-97

9-130 A gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. Then, ⎯→ h1 = 300.19 kJ/kg T1 = 300 K ⎯ Pr1 = 1.386 Pr2 =

ηC =

T

P2 ⎯→ h2 s = h4 s = 411.26 kJ/kg Pr = (3)(1.386) = 4.158 ⎯ P1 1 h2 s − h1 ⎯ ⎯→ h2 = h4 = h1 + (h2 s − h1 ) / η C h2 − h1 = 300.19 + (411.26 − 300.19 ) / (0.84 ) = 432.42 kJ/kg

T5 = 1200 K ⎯ ⎯→ h5 = h7 = 1277.79 kJ/kg

5

qin

6 8 6s 8

9

4

4

2s

3

2

7

1

1 s

Pr5 = 238 P6 ⎛1⎞ ⎯→ h6 s = h8 s = 946.36 kJ/kg Pr = ⎜ ⎟(238) = 79.33 ⎯ P5 5 ⎝ 3 ⎠ h − h6 ηT = 5 ⎯ ⎯→ h6 = h8 = h5 − η T (h5 − h6 s ) h5 − h6 s = 1277.79 − (0.88)(1277.79 − 946.36) = 986.13 kJ/kg Pr6 =

wC,in = 2(h2 − h1 ) = 2(432.42 − 300.19 ) = 264.46 kJ/kg

wT,out = 2(h5 − h6 ) = 2(1277.79 − 986.13) = 583.32 kJ/kg

Thus,

rbw =

wC,in wT,out

=

264.46 kJ/kg = 0.453 = 45.3% 583.32 kJ/kg

q in = (h5 − h4 ) + (h7 − h6 ) = (1277.79 − 432.42) + (1277.79 − 986.13) = 1137.03 kJ/kg wnet = wT,out − wC,in = 583.32 − 264.46 = 318.86 kJ/kg

η th =

wnet 318.86 kJ/kg = = 0.280 = 28.0% q in 1137.03 kJ/kg

(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes q regen = ε (h8 − h4 ) = (0.75)(986.13 − 432.42 ) = 415.28 kJ/kg q in = q in,old − q regen = 1137.03 − 415.28 = 721.75 kJ/kg

η th =

wnet 318.86 kJ/kg = = 0.442 = 44.2% q in 721.75 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-98

9-131E An ideal regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The power produced and consumed by each compression and expansion stage, and the rate of heat rejected are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea). Analysis The pressure ratio for each stage is

r p = 12 = 3.464

T

According to the isentropic process expressions for an ideal gas, 5

T2 = T4 = T1 r p( k −1) / k = (520 R)(3.464) 0.4/1.4 = 741.6 R

Since this is an ideal cycle, T5 = T7 = T9 = T4 + 50 = 741.6 + 50 = 791.6 R

520 R

4

2

3

1

6

8

7

9

1

For the isentropic expansion processes,

s

T6 = T8 = T7 r p( k −1) / k = (791.6 R)(3.464) 0.4/1.4 = 1129 R

The heat input is

q in = 2c p (T6 − T5 ) = 2(0.24 Btu/lbm ⋅ R)(1129 − 791.6) R = 162.0 Btu/lbm The mass flow rate is then

m& =

Q& in 500 Btu/s = = 3.086 lbm/s q in 162.0 Btu/lbm

Application of the first law to the expansion process 6-7 gives W& 6-7,out = m& c p (T6 − T7 ) 1 kW ⎞ ⎛ = (3.086 lbm/s)(0.24 Btu/lbm ⋅ R)(1129 − 791.6) R ⎜ ⎟ ⎝ 0.94782 Btu/s ⎠ = 263.6 kW

The same amount of power is produced in process 8-9. When the first law is adapted to the compression process 1-2 it becomes W&12,in = m& c p (T2 − T1 ) 1 kW ⎛ ⎞ = (3.086 lbm/s)(0.24 Btu/lbm ⋅ R)(741.6 − 520) R ⎜ ⎟ ⎝ 0.94782 Btu/s ⎠ = 173.2 kW

Compression process 3-4 uses the same amount of power. The rate of heat rejection from the cycle is Q& out = 2m& c p (T2 − T3 ) = 2(3.086 lbm/s)(0.24 Btu/lbm ⋅ R)(741.6 − 520) R = 328.3 Btu/s

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-99

9-132E An ideal regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The power produced and consumed by each compression and expansion stage, and the rate of heat rejected are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea). Analysis The pressure ratio for each stage is

r p = 12 = 3.464

T

For the compression processes,

6 5

T2 s = T4 s = T1 r p( k −1) / k = (520 R)(3.464) 0.4/1.4 = 741.6 R

ηC =

c p (T2 s − T1 ) c p (T2 − T1 )

⎯ ⎯→ T2 = T4 = T1 + = 520 +

T2 s − T1

4 2 4s 2s

ηC

741.6 − 520 = 780.7 R 0.85

520 R

3

8

9 7s 7 9s 1

1 s

Since the regenerator is ideal, T5 = T7 = T9 = T4 + 50 = 780.7 + 50 = 830.7 R

For the expansion processes, T6 s = T8 s = T7 r p( k −1) / k = (830.7 R)(3.464) 0.4/1.4 = 1185 R

ηT =

c p (T6 − T7 ) c p (T6 s − T7 )

⎯ ⎯→ T6 = T8 = T7 + η T (T6 s − T7 ) = 830.7 + (0.90)(1185 − 830.7) = 1150 R

The heat input is

q in = 2c p (T6 − T5 ) = 2(0.24 Btu/lbm ⋅ R)(1150 − 830.7) R = 153.3 Btu/lbm The mass flow rate is then

m& =

Q& in 500 Btu/s = = 3.262 lbm/s q in 153.3 Btu/lbm

Application of the first law to the expansion process 6-7 gives W& 6-7,out = m& c p (T6 − T7 ) 1 kW ⎛ ⎞ = (3.262 lbm/s)(0.24 Btu/lbm ⋅ R)(1150 − 830.7) R ⎜ ⎟ = 263.7 kW ⎝ 0.94782 Btu/s ⎠

The same amount of power is produced in process 8-9. When the first law is adapted to the compression process 1-2 it becomes W&12,in = m& c p (T2 − T1 ) 1 kW ⎛ ⎞ = (3.262 lbm/s)(0.24 Btu/lbm ⋅ R)(780.7 − 520) R ⎜ ⎟ = 215.3 kW ⎝ 0.94782 Btu/s ⎠

Compression process 3-4 uses the same amount of power. The rate of heat rejection from the cycle is Q& out = 2m& c p (T2 − T3 ) = 2(3.262 lbm/s)(0.24 Btu/lbm ⋅ R)(780.7 − 520) R = 408.2 Btu/s

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-100

9-133 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis The temperatures at various states are obtained as follows T2 = T4 = T1 r p( k −1) / k = ( 290 K)(4) 0.4/1.4 = 430.9 K

T5 = T4 + 20 = 430.9 + 20 = 450.9 K

T

8 6

q in = c p (T6 − T5 ) T6 = T5 +

⎛ 1 T7 = T6 ⎜ ⎜ rp ⎝ T8 = T7 +

q in cp

⎞ ⎟ ⎟ ⎠

q in cp

⎛ 1 T9 = T8 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

= 450.9 K +

( k −1) / k

⎛1⎞ = (749.4 K)⎜ ⎟ ⎝4⎠

= 504.3 K + ( k −1) / k

7

300 kJ/kg = 749.4 K 1.005 kJ/kg ⋅ K 0.4/1.4

= 504.3 K

5 290 K

4

2

3

1

9

10

s

300 kJ/kg = 802.8 K 1.005 kJ/kg ⋅ K

⎛1⎞ = (802.8 K)⎜ ⎟ ⎝4⎠

0.4/1.4

= 540.2 K

T10 = T9 − 20 = 540.2 − 20 = 520.2 K

The heat input is

q in = 300 + 300 = 600 kJ/kg The heat rejected is q out = c p (T10 − T1 ) + c p (T2 − T3 ) = (1.005 kJ/kg ⋅ K)(520.2 − 290 + 430.9 − 290) R = 373.0 kJ/kg The thermal efficiency of the cycle is then

η th = 1 −

q out 373.0 = 1− = 0.378 600 q in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-101

9-134 A regenerative gas-turbine cycle with three stages of compression and three stages of expansion is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis The temperatures at various states are obtained as follows T2 = T4 = T6 = T1 r p( k −1) / k = ( 290 K)(4) 0.4/1.4 = 430.9 K

T

T7 = T6 + 20 = 430.9 + 20 = 450.9 K q in = c p (T8 − T7 ) q in

T8 = T7 +

cp

⎛ 1 T9 = T8 ⎜ ⎜ rp ⎝ T10 = T9 +

⎞ ⎟ ⎟ ⎠

cp

⎛ 1 T11 = T10 ⎜ ⎜ rp ⎝ T12 = T11 +

T13

⎞ ⎟ ⎟ ⎠

q in cp

⎛ 1 = T12 ⎜ ⎜ rp ⎝

= 450.9 K +

( k −1) / k

q in

⎞ ⎟ ⎟ ⎠

9 7

300 kJ/kg = 749.4 K 1.005 kJ/kg ⋅ K

6 290 K

⎛1⎞ = (749.4 K)⎜ ⎟ ⎝4⎠

= 504.3 K + ( k −1) / k

= 504.3 K

5

2

3

1

13

11 14

s

300 kJ/kg = 802.8 K 1.005 kJ/kg ⋅ K

⎛1⎞ = (802.8 K)⎜ ⎟ ⎝4⎠

= 540.2 K + ( k −1) / k

0.4/1.4

4

12

1

8

0.4/1.4

= 540.2 K

300 kJ/kg = 838.7 K 1.005 kJ/kg ⋅ K

⎛1⎞ = (838.7 K)⎜ ⎟ ⎝4⎠

0.4/1.4

= 564.4 K

T14 = T13 − 20 = 564.4 − 20 = 544.4 K

The heat input is

q in = 300 + 300 + 300 = 900 kJ/kg The heat rejected is q out = c p (T14 − T1 ) + c p (T2 − T3 ) + c p (T4 − T5 ) = (1.005 kJ/kg ⋅ K)(544.4 − 290 + 430.9 − 290 + 430.9 − 290) R = 538.9 kJ/kg The thermal efficiency of the cycle is then

η th = 1 −

q out 538.9 =1− = 0.401 = 40.1% q in 900

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-102

9-135 A regenerative gas-turbine cycle with three stages of compression and three stages of expansion is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis Since all compressors share the same compression ratio and begin at the same temperature, T2 = T4 = T6 = T1 r p( k −1) / k = ( 290 K)(4) 0.4/1.4 = 430.9 K

T

From the problem statement, T7 = T13 − 65

The relations for heat input and expansion processes are q in

q in = c p (T8 − T7 ) ⎯ ⎯→ T8 = T7 +

⎛ 1 T9 = T8 ⎜ ⎜ rp ⎝ T10 = T9 +

T12 = T11 +

⎞ ⎟ ⎟ ⎠

9 7

cp

6 290 K

( k −1) / k

q in cp q in cp

5

4

2

3

1

12

1

8

13

11 14

s

( k −1) / k

,

⎛ 1 T11 = T10 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

,

⎛ 1 = T12 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

T13

( k −1) / k

The simultaneous solution of above equations using EES software gives the following results T7 = 520.7 K,

T8 = 819.2 K,

T9 = 551.3 K

T10 = 849.8 K,

T11 = 571.9 K,

T12 = 870.4 K,

T13 = 585.7 K

From an energy balance on the regenerator, T7 − T6 = T13 − T14 (T13 − 65) − T6 = T13 − T14 ⎯ ⎯→ T14 = T6 + 65 = 430.9 + 65 = 495.9 K

The heat input is

q in = 300 + 300 + 300 = 900 kJ/kg The heat rejected is q out = c p (T14 − T1 ) + c p (T2 − T3 ) + c p (T4 − T5 ) = (1.005 kJ/kg ⋅ K)(495.9 − 290 + 430.9 − 290 + 430.9 − 290) R = 490.1 kJ/kg The thermal efficiency of the cycle is then

η th = 1 −

q out 490.1 =1− = 0.455 = 45.5% q in 900

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-103

Jet-Propulsion Cycles

9-136C The power developed from the thrust of the engine is called the propulsive power. It is equal to thrust times the aircraft velocity.

9-137C The ratio of the propulsive power developed and the rate of heat input is called the propulsive efficiency. It is determined by calculating these two quantities separately, and taking their ratio.

9-138C It reduces the exit velocity, and thus the thrust.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-104

9-139E A turboprop engine operating on an ideal cycle is considered. The thrust force generated is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1E), cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea). Analysis Working across the two isentropic processes of the cycle yields T2 = T1 r p( k −1) / k = ( 450 R)(10) 0.4/1.4 = 868.8 R

⎛ 1 T5 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1400 R)⎜ ⎟ ⎝ 10 ⎠

T

3

qin

4

0.4/1.4

2

= 725.1 R

5 1

Since the work produced by expansion 3-4 equals that used by compression 1-2, an energy balance gives

qout s

T4 = T3 − (T2 − T1 ) = 1400 − (868.8 − 450) = 981.2 R

The excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller, m& e c p (T4 − T5 ) = m& p

2 2 Vexit − Vinlet 2

which when solved for the velocity at which the air leaves the propeller gives Vexit

⎤ ⎡ m& 2 = ⎢2 e c p (T4 − T5 ) + Vinlet ⎥ ⎦⎥ ⎣⎢ m& p

1/ 2

⎡ 1 ⎛ 25,037 ft 2 /s 2 (0.24 Btu/lbm ⋅ R )(981.2 − 725.1)R ⎜ = ⎢2 ⎜ 1 Btu/lbm ⎢⎣ 20 ⎝ = 716.9 ft/s

⎤ ⎞ ⎟ + (600 ft/s) 2 ⎥ ⎟ ⎥⎦ ⎠

1/ 2

The mass flow rate through the propeller is

v1 = m& p =

RT1 (0.3704 psia ⋅ ft 3 )(450 R) = = 20.84 ft 3 /lbm P1 8 psia AV1

v1

=

πD 2 V1 π (10 ft) 2 600 ft/s = = 2261 lbm/s 4 v1 4 20.84 ft 3 /lbm

The thrust force generated by this propeller is then ⎛ 1 lbf F = m& p (Vexit − Vinlet ) = (2261 lbm/s)(716.9 − 600)ft/s⎜ ⎜ 32.174 lbm ⋅ft/s 2 ⎝

⎞ ⎟ = 8215 lbf ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-105

9-140E A turboprop engine operating on an ideal cycle is considered. The thrust force generated is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1E), cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea). Analysis Working across the two isentropic processes of the cycle yields

T

T2 = T1 r p( k −1) / k = ( 450 R)(10) 0.4/1.4 = 868.8 R

⎞ ⎟ ⎟ ⎠

⎛ 1 T5 = T3 ⎜ ⎜ rp ⎝

( k −1) / k

3

qin

4 ⎛1⎞ = (1400 R)⎜ ⎟ ⎝ 10 ⎠

2

0.4/1.4

= 725.1 R

5 qout

1

Since the work produced by expansion 3-4 equals that used by compression 1-2, an energy balance gives

s

T4 = T3 − (T2 − T1 ) = 1400 − (868.8 − 450) = 981.2 R

The mass flow rate through the propeller is

v1 = m& p =

RT (0.3704 psia ⋅ ft 3 )(450 R) = = 20.84 ft 3 /lbm 8 psia P AV1

v1

=

600 ft/s πD 2 V1 π (8 ft) 2 = = 1447 lbm/s 4 v1 4 20.84 ft 3 /lbm

According to the previous problem, m& e =

m& p 20

=

2261 lbm/s = 113.1 lbm/s 20

The excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller, m& e c p (T4 − T5 ) = m& p

2 2 Vexit − Vinlet 2

which when solved for the velocity at which the air leaves the propeller gives Vexit

⎤ ⎡ m& 2 = ⎢2 e c p (T4 − T5 ) + Vinlet ⎥ ⎦⎥ ⎣⎢ m& p

1/ 2

⎡ 113.1 lbm/s ⎛ 25,037 ft 2 /s 2 (0.24 Btu/lbm ⋅ R )(981.2 − 725.1)R ⎜ = ⎢2 ⎜ 1 Btu/lbm ⎢⎣ 1447 lbm/s ⎝ = 775.0 ft/s

⎤ ⎞ ⎟ + (600 ft/s) 2 ⎥ ⎟ ⎥⎦ ⎠

1/ 2

The thrust force generated by this propeller is then ⎛ 1 lbf F = m& p (Vexit − Vinlet ) = (1447 lbm/s)(775 − 600)ft/s⎜ ⎜ 32.174 lbm ⋅ft/s 2 ⎝

⎞ ⎟ = 7870 lbf ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-106

9-141 A turbofan engine operating on an ideal cycle produces 50,000 N of thrust. The air temperature at the fan outlet needed to produce this thrust is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K, cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A2a). Analysis The total mass flow rate is RT (0.287 kPa ⋅ m 3 )(253 K) = = 1.452 m 3 /kg P 50 kPa AV1 πD 2 V1 π (2.5 m) 2 200 m/s m& = = = = 676.1 kg/s v1 4 v1 4 1.452 m 3 /kg

v1 =

T

3

qin

4 2 5

Now, 1

m& 676.1 kg/s m& e = = = 84.51 kg/s 8 8

qout s

The mass flow rate through the fan is

m& f = m& − m& e = 676.1 − 84.51 = 591.6 kg/s In order to produce the specified thrust force, the velocity at the fan exit will be F = m& f (Vexit − Vinlet ) Vexit = Vinlet +

50,000 N ⎛⎜ 1 kg ⋅m/s 2 F = (200 m/s) + 591.6 kg/s ⎜⎝ 1 N m& f

⎞ ⎟ = 284.5 m/s ⎟ ⎠

An energy balance on the stream passing through the fan gives 2 2 − Vinlet Vexit 2 2 V 2 − Vinlet T5 = T4 − exit 2c p

c p (T4 − T5 ) =

= 253 K −

(284.5 m/s) 2 − (200 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎜ ⎟ 2(1.005 kJ/kg ⋅ K ) ⎝ 1000 m 2 /s 2 ⎠

= 232.6 K

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-107

9-142 A pure jet engine operating on an ideal cycle is considered. The velocity at the nozzle exit and the thrust produced are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K, cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A2a). Analysis (a) We assume the aircraft is stationary and the air is moving T towards the aircraft at a velocity of V 1 = 240 m/s. Ideally, the air will 4 leave the diffuser with a negligible velocity (V 2 ≅ 0). q in Diffuser: 5 ⎯→ E& in = E& out E& in − E& out = ∆E& system ©0 (steady) ⎯ 3 h1 + V12

/2=

h2 + V22

/2⎯ ⎯→ 0 = h2 − h1 +

V22

0 = c p (T2 − T1 ) − V12 / 2 T2 = T1 +

− V12 2

2 1

⎞ ⎟⎟ ⎠

k / (k −1)

qout s

V12 (240 m/s )2 ⎛⎜ 1 kJ/kg = 260 K + 2c p (2)(1.005 kJ/kg ⋅ K ) ⎜⎝ 1000 m 2 /s 2

⎛T P2 = P1 ⎜⎜ 2 ⎝ T1

6

⎛ 288.7 K ⎞ ⎟⎟ = (45 kPa )⎜⎜ ⎝ 260 K ⎠

⎞ ⎟ = 288.7 K ⎟ ⎠

1.4/0.4

= 64.88 kPa

Compressor: P3 = P4 = r p (P2 ) = (13)(64.88 kPa ) = 843.5 kPa

( )

⎛P T3 = T2 ⎜⎜ 3 ⎝ P2

⎞ ⎟⎟ ⎠

(k −1) / k

= (288.7 K )(13)0.4/1.4 = 600.7 K

Turbine:

wcomp,in = wturb,out ⎯ ⎯→ h3 − h2 = h4 − h5 ⎯ ⎯→ c p (T3 − T2 ) = c p (T4 − T5 ) or Nozzle:

T5 = T4 − T3 + T2 = 830 − 600.7 + 288.7 = 518.0 K (k −1) / k 0.4/1.4 ⎛ P6 ⎞ ⎛ 45 kPa ⎞ ⎟⎟ = (830 K )⎜⎜ = 359.3 K T6 = T4 ⎜⎜ ⎟⎟ ⎝ 843.5 kPa ⎠ ⎝ P4 ⎠ E& in − E& out = ∆E& system ©0 (steady) ⎯ ⎯→ E& in = E& out

h5 + V52 / 2 = h6 + V62 / 2 V 2 − V52 0 = h6 − h5 + 6 2

or

V6 = Vexit =

⎯ ⎯→ 0 = c p (T6 − T5 ) + V62 / 2

/s 2 1 kJ/kg

(2)(1.005 kJ/kg ⋅ K )(518.0 − 359.3)K⎜⎜ 1000 m ⎝

2

⎞ ⎟ = 564.8 m/s ⎟ ⎠

The mass flow rate through the engine is RT (0.287 kPa ⋅ m 3 )(260 K) = = 1.658 m 3 /kg P 45 kPa AV1 πD 2 V1 π (1.6 m) 2 240 m/s = = = 291.0 kg/s m& = v1 4 v1 4 1.658 m 3 /kg

v1 =

The thrust force generated is then ⎛ 1N F = m& (Vexit − Vinlet ) = (291.0 kg/s)(564.8 − 240)m/s⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 94,520 N ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-108

9-143 A turbojet aircraft flying at an altitude of 9150 m is operating on the ideal jet propulsion cycle. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will T leave the diffuser with a negligible velocity (V 2 ≅ 0). 4 · Qi Diffuser: 5 E& in − E& out = ∆E& system ©0 (steady) ⎯ ⎯→ E& in = E& out 3 ©0 V 22 − V12 2 2 h1 + V1 / 2 = h2 + V 2 / 2 ⎯ ⎯→ 0 = h2 − h1 + 2 6 2 1 0 = c p (T2 − T1 ) − V12 / 2 s 2 2 ⎞ ⎛ V (320 m/s ) ⎜ 1 kJ/kg ⎟ = 291.9 K T2 = T1 + 1 = 241 K + (2)(1.005 kJ/kg ⋅ K ) ⎜⎝ 1000 m 2 /s 2 ⎟⎠ 2c p ⎛T P2 = P1 ⎜⎜ 2 ⎝ T1

⎞ ⎟⎟ ⎠

k / (k −1)

⎛ 291.9 K ⎞ ⎟⎟ = (32 kPa )⎜⎜ ⎝ 241 K ⎠

1.4/0.4

= 62.6 kPa

Compressor: P3 = P4 = r p (P2 ) = (12 )(62.6 kPa ) = 751.2 kPa

( )

⎛P T3 = T2 ⎜⎜ 3 ⎝ P2

⎞ ⎟⎟ ⎠

(k −1) / k

= (291.9 K )(12 )0.4/1.4 = 593.7 K

Turbine:

wcomp,in = wturb,out ⎯ ⎯→ h3 − h2 = h4 − h5 ⎯ ⎯→ c p (T3 − T2 ) = c p (T4 − T5 ) or

T5 = T4 − T3 + T2 = 1400 − 593.7 + 291.9 = 1098.2K

Nozzle: (k −1) / k

⎛P ⎞ ⎛ 32 kPa ⎞ ⎟⎟ T6 = T4 ⎜⎜ 6 ⎟⎟ = (1400 K )⎜⎜ ⎝ 751.2 kPa ⎠ ⎝ P4 ⎠ E& in − E& out = ∆E& system ©0 (steady) ⎯ ⎯→ E& in = E& out

0.4/1.4

= 568.2 K

h5 + V52 / 2 = h6 + V62 / 2 V 2 − V52 0 = h 6 − h5 + 6 2

or

(b) (c)

V6 =

⎯ ⎯→ 0 = c p (T6 − T5 ) + V62 / 2 ⎛

/s 2 1 kJ/kg

(2)(1.005 kJ/kg ⋅ K )(1098.2 − 568.2)K⎜⎜ 1000 m ⎝

2

⎞ ⎟ = 1032 m/s ⎟ ⎠

⎛ 1 kJ/kg ⎞ ⎟ = 13,670 kW W& p = m& (Vexit − Vinlet )Vaircraft = (60 kg/s )(1032 − 320)m/s(320 m/s )⎜ ⎜ 1000 m 2 /s 2 ⎟ ⎝ ⎠ Q& = m& (h − h ) = m& c (T − T ) = (60 kg/s )(1.005 kJ/kg ⋅ K )(1400 − 593.7 )K = 48,620 kJ/s in

m& fuel =

4

3

p

4

3

Q& in 48,620 kJ/s = = 1.14 kg/s HV 42,700 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-109

9-144 A turbojet aircraft is flying at an altitude of 9150 m. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0).

Diffuser:

T

E& in − E& out = ∆E& system ©0 (steady) h1 + V12

/2=

5

h2 + V 22

3

− V12 2 0 = c p (T2 − T1 ) − V12 / 2 0 = h2 − h1 +

T2 = T1 +

V 22

2

⎞ ⎟⎟ ⎠

k / (k −1)

5s 6

1 s

V12 (320 m/s)2 ⎛⎜ 1 kJ/kg = 241 K + (2)(1.005 kJ/kg ⋅ K ) ⎜⎝ 1000 m 2 /s 2 2c p

⎛T P2 = P1 ⎜⎜ 2 ⎝ T1

Compressor:

4

· Qin

E& in = E& out

⎛ 291.9 K ⎞ ⎟⎟ = (32 kPa )⎜⎜ ⎝ 241 K ⎠

⎞ ⎟ = 291.9 K ⎟ ⎠

1.4/0.4

= 62.6 kPa

( )

P3 = P4 = r p (P2 ) = (12 )(62.6 kPa ) = 751.2 kPa ⎛P T3s = T2 ⎜⎜ 3 ⎝ P2

⎞ ⎟⎟ ⎠

(k −1) / k

= (291.9 K )(12 )0.4/1.4 = 593.7 K

h3s − h2 c p (T3s − T2 ) = h3 − h 2 c p (T3 − T2 )

ηC =

T3 = T2 + (T3s − T2 ) / η C = 291.9 + (593.7 − 291.9 )/ (0.80 ) = 669.2 K

Turbine: wcomp,in = w turb,out ⎯ ⎯→ h3 − h2 = h4 − h5 ⎯ ⎯→ c p (T3 − T2 ) = c p (T4 − T5 )

or, T5 = T4 − T3 + T2 = 1400 − 669.2 + 291.9 = 1022.7 K

ηT =

c p (T4 − T5 ) h4 − h5 = h4 − h5 s c p (T4 − T5 s )

T5 s = T4 − (T4 − T5 ) / η T = 1400 − (1400 − 1022.7 ) / 0.85 = 956.1 K ⎛T P5 = P4 ⎜⎜ 5 s ⎝ T4

⎞ ⎟⎟ ⎠

k / (k −1)

⎛ 956.1 K ⎞ ⎟⎟ = (751.2 kPa )⎜⎜ ⎝ 1400 K ⎠

1.4/0.4

= 197.7 kPa

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-110

Nozzle: (k −1) / k

⎛P ⎞ ⎛ 32 kPa ⎞ ⎟⎟ = (1022.7 K )⎜⎜ T6 = T5 ⎜⎜ 6 ⎟⎟ ⎝ 197.7 kPa ⎠ ⎝ P5 ⎠ E& in − E& out = ∆E& system ©0 (steady)

0.4/1.4

= 607.8 K

E& in = E& out h5 + V52 / 2 = h6 + V62 / 2 ©0

V62 − V52 2 0 = c p (T6 − T5 ) + V62 / 2 0 = h 6 − h5 +

or, V6 =

/s 2 1 kJ/kg

(2)(1.005 kJ/kg ⋅ K )(1022.7 − 607.8)K⎜⎜ 1000 m ⎝

2

⎞ ⎟ = 913.2 m/s ⎟ ⎠

(b)

W& p = m& (Vexit − Vinlet )Vaircraft

(c)

Q& in = m& (h4 − h3 ) = m& c p (T4 − T3 ) = (60 kg/s )(1.005 kJ/kg ⋅ K )(1400 − 669.2 )K = 44,067 kJ/s

⎛ 1 kJ/kg = (60 kg/s )(913.2 − 320 )m/s(320 m/s )⎜ ⎜ 1000 m 2 /s 2 ⎝ = 11,390 kW

m& fuel =

⎞ ⎟ ⎟ ⎠

Q& in 44,067 kJ/s = = 1.03 kg/s HV 42,700 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-111

9-145 A turbojet aircraft that has a pressure rate of 9 is stationary on the ground. The force that must be applied on the brakes to hold the plane stationary is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the nozzle exit. Properties The properties of air are given in Table A-17. Analysis (a) Using variable specific heats for air,

T

Compressor:

qin

T1 = 290 K ⎯ ⎯→ h1 = 290.16 kJ/kg

4

Pr1 = 1.2311 Pr2 =

2 5

P2 Pr = (9)(1.2311) = 11.08 ⎯ ⎯→ h2 = 544.07 kJ/kg P1 1

1

Q& in = m& fuel × HV = (0.5 kg/s )(42,700 kJ/kg ) = 21,350 kJ/s q in =

3

s

Q& in 21,350 kJ/s = = 1067.5 kJ/kg m& 20 kg/s

q in = h3 − h2 ⎯ ⎯→ h3 = h2 + q in = 544.07 + 1067.5 = 1611.6 kJ/kg ⎯ ⎯→ Pr3 = 568.5

Turbine: wcomp,in = wturb,out ⎯ ⎯→ h2 − h1 = h3 − h4

or h4 = h3 − h2 + h1 = 1611.6 − 544.07 + 290.16 = 1357.7 kJ/kg

Nozzle: ⎛P ⎞ ⎛1⎞ Pr5 = Pr3 ⎜⎜ 5 ⎟⎟ = (568.5)⎜ ⎟ = 63.17 ⎯ ⎯→ h5 = 888.56 kJ/kg ⎝9⎠ ⎝ P3 ⎠ E& in − E& out = ∆E& system ©0 (steady) E& in = E& out h4 + V42 / 2 = h5 + V52 / 2 0 = h5 − h4 +

V52 − V42 2

or V5 = 2(h4 − h5 ) =

/s 2 1 kJ/kg

(2)(1357.7 − 888.56)kJ/kg⎜⎜ 1000 m ⎝

2

⎞ ⎟ = 968.6 m/s ⎟ ⎠

⎛ 1N Brake force = Thrust = m& (Vexit − Vinlet ) = (20 kg/s )(968.6 − 0)m/s⎜ ⎜ 1 kg ⋅ m/s 2 ⎝

⎞ ⎟ = 19,370 N ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-112

9-146 Problem 9-145 is reconsidered. The effect of compressor inlet temperature on the force that must be applied to the brakes to hold the plane stationary is to be investigated. Analysis Using EES, the problem is solved as follows: P_ratio =9 T_1 = 7 [C] T[1] = T_1+273 "[K]" P[1]= 95 [kPa] P[5]=P[1] Vel[1]=0 [m/s] V_dot[1] = 18.1 [m^3/s] HV_fuel = 42700 [kJ/kg] m_dot_fuel = 0.5 [kg/s] Eta_c = 1.0 Eta_t = 1.0 Eta_N = 1.0 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) v[1]=volume(Air,T=T[1],P=P[1]) m_dot = V_dot[1]/v[1] "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) Q_dot_in = m_dot_fuel*HV_fuel m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" {P_ratio= P[3] /P[4]} T_s[4]=TEMPERATURE(Air,h=h_s[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" {h_s[4]=ENTHALPY(Air,T=T_s[4])} "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" T[4]=TEMPERATURE(Air,h=h[4]) P[4]=pressure(Air,s=s_s[4],h=h_s[4]) "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" W_dot_net = 0 [kW] "Exit nozzle analysis:" s[4]=entropy('air',T=T[4],P=P[4]) s_s[5]=s[4] "For the ideal case the entropies are constant across the nozzle" T_s[5]=TEMPERATURE(Air,s=s_s[5], P=P[5]) "T_s[5] is the isentropic value of T[5] at nozzle exit" h_s[5]=ENTHALPY(Air,T=T_s[5]) Eta_N=(h[4]-h[5])/(h[4]-h_s[5]) m_dot*h[4] = m_dot*(h_s[5] + Vel_s[5]^2/2*convert(m^2/s^2,kJ/kg)) PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-113

m_dot*h[4] = m_dot*(h[5] + Vel[5]^2/2*convert(m^2/s^2,kJ/kg)) T[5]=TEMPERATURE(Air,h=h[5]) s[5]=entropy('air',T=T[5],P=P[5]) "Brake Force to hold the aircraft:" Thrust = m_dot*(Vel[5] - Vel[1]) "[N]" BrakeForce = Thrust "[N]" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) s[2]=entropy('air',T=T[2],P=P[2]) m [kg/s]

T3 [K]

T1 [C]

21250

23.68 23.22 22.78 22.35 21.94 21.55 21.17 20.8 20.45 20.1 19.77

1284 1307 1330 1352 1375 1398 1420 1443 1466 1488 1510

-20 -15 -10 -5 0 5 10 15 20 25 30

20800

BrakeForce [N]

Brake Force [N] 21232 21007 20788 20576 20369 20168 19972 19782 19596 19415 19238

20350

19900

19450

19000 -20

-10

0

10

20

30

T1 [C] Air

855 kPa 103

1500 3

95 kPa 4s 103

T [K]

1000 5 2s

5x 102

500

1 0 4.5

5.0

5.5

6.0

6.5

7.0

7.5

8.0

0x 100 8.5

s [kJ/kg-K]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-114

9-147 Air enters a turbojet engine. The thrust produced by this turbojet engine is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. Properties The properties of air are given in Table A-17. Analysis We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 300 m/s. Taking the entire engine as our control volume and writing the steady-flow energy balance yield T1 = 280 K

⎯ ⎯→

h1 = 28013 . kJ / kg

T2 = 700 K

⎯ ⎯→

h2 = 713.27 kJ / kg

15,000 kJ/s

7°C 300 m/s 16 kg/s

E& in = E& out Q& in + m& (h1 + V12 / 2) = m& (h2 + V 22 / 2) ⎛ V 2 − V12 Q& in = m& ⎜ h2 − h1 + 2 ⎜ 2 ⎝

427°C 1

2

⎞ ⎟ ⎟ ⎠

⎡ V 2 − (300 m/s )2 15,000 kJ/s = (16 kg/s )⎢713.27 − 280.13 + 2 2 ⎢⎣

⎛ 1 kJ/kg ⎜ ⎜ 1000 m 2 /s 2 ⎝

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

It gives V 2 = 1048 m/s Thus,

Fp = m& (V2 − V1 ) = (16 kg/s )(1048 − 300)m/s = 11,968 N

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-115

Second-Law Analysis of Gas Power Cycles

9-148 The process with the highest exergy destruction for an ideal Otto cycle described in Prob. 9-36 is to be determined. Analysis From Prob. 9-36, qin = 582.5 kJ/kg, qout = 253.6 kJ/kg, T1 = 288 K, T2 = 661.7 K, T3 = 1473 K, and T4 = 641.2 K. The exergy destruction during a process of the cycle is ⎛ q q x dest = T0 s gen = T0 ⎜⎜ ∆s − in + out T T source sink ⎝

⎞ ⎟ ⎟ ⎠

P

3

Application of this equation for each process of the cycle gives

4

x dest,1- 2 = 0 (isentropic process)

2

1

T v s 3 − s 2 = s 4 − s1 = cv ln 3 + Rln 3 T2 v2

v

1473 K = (0.718 kJ/kg ⋅ K ) ln + 0 = 0.5746 kJ/kg ⋅ K 661.7 K ⎞ ⎛ q x dest,2-3 = T0 ⎜⎜ s 3 − s 2 − in ⎟⎟ Tsource ⎠ ⎝ 582.5 kJ/kg ⎞ ⎛ = (288 K)⎜ 0.5746 kJ/kg ⋅ K − ⎟ 1473 K ⎠ ⎝ = 51.59 kJ/kg

T

3

qin 2

4 1

x dest,3-4 = 0 (isentropic process)

qout s

⎛ q ⎞ x dest,4-1 = T0 ⎜⎜ s1 − s 4 + out ⎟⎟ T sink ⎠ ⎝ 253.6 kJ/kg ⎞ ⎛ = (288 K)⎜ − 0.5746 kJ/kg ⋅ K + ⎟ 288 K ⎠ ⎝ = 88.12 kJ/kg

The largest exergy destruction in the cycle occurs during the heat-rejection process.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-116

9-149E The exergy destruction associated with the heat rejection process of the Diesel cycle described in Prob. 9-55E and the exergy at the end of the expansion stroke are to be determined. Analysis From Prob. 9-55E, qout = 158.9 Btu/lbm, T1 = 540 R, P1 = 14.7 psia, T4 = 1420.6 R, P4 = 38.62 psia and v 4 = v 1.

The entropy change during process 4-1 is s1 − s 4 = s1o @540R − s 4o @1420.6 R − R ln( P1 / P4 ) = 0.60078 − 0.83984 − (0.06855) ln(14.7 / 38.62) = −0.1728 Btu/lbm ⋅ R Thus,

q R ,41 ⎞ ⎛ ⎛ 158.9 Btu/lbm ⎞ ⎟ = (540R )⎜ − 0.1728 Btu/lbm ⋅ R + ⎟⎟ = 65.6 Btu/lbm x destroyed, 41 = T0 ⎜⎜ s1 − s 4 + ⎜ ⎟ 540 R TR ⎠ ⎝ ⎠ ⎝ Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke (state 4) is determined from

φ 4 = (u 4 − u 0 ) − T0 (s 4 − s 0 ) + P0 (v 4 − v 0 ) where u 4 − u 0 = u 4 − u1 = q out = 158.9 Btu/lbm ⋅ R

v 4 −v 0 = v 4 −v1 = 0 s 4 − s 0 = s 4 − s1 = 0.1741 Btu/lbm ⋅ R

Thus,

φ 4 = (158.9 Btu/lbm ) − (540R )(0.1728 Btu/lbm ⋅ R ) + 0 = 65.6 Btu/lbm Discussion Note that the exergy at state 4 is identical to the exergy destruction for the process 4-1 since state 1 is identical to the dead state, and the entire exergy at state 4 is wasted during process 4-1.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-117

9-150 The exergy loss of each process for an ideal dual cycle described in Prob. 9-63 is to be determined. Analysis From Prob. 9-63, qin,x-3 = 114.6 kJ/kg, T1 = 291 K, T2 = 1037 K, Tx = 1141 K, T3 = 1255 K, and T4 = 494.8 K. Also, q in , 2 − x = cv (T x − T2 ) = (0.718 kJ/kg ⋅ K )(1141 − 1037 ) K = 74.67 kJ/kg

q out = cv (T4 − T1 ) = (0.718 kJ/kg ⋅ K )(494.8 − 291)K = 146.3 kJ/kg

P

x

The exergy destruction during a process of the cycle is x dest = T0 s gen

⎛ q q = T0 ⎜⎜ ∆s − in + out Tsource Tsink ⎝

⎞ ⎟ ⎟ ⎠

2

Application of this equation for each process of the cycle gives

3 qin

4 qout 1

v

T P s 2 − s1 = c p ln 2 − Rln 2 T1 P1 = (1.005 kJ/kg ⋅ K ) ln

1037 K 5148 kPa − (0.287 kJ/kg ⋅ K ) ln 291 K 90 kPa

= 0.1158 kJ/kg ⋅ K x dest, 1-2 = T0 ( s 2 − s1 ) = (291 K)(0.1158 kJ/kg ⋅ K) = 33.7 kJ/kg

s x − s 2 = cv ln

Tx v + Rln x T2 v2

= (0.718 kJ/kg ⋅ K ) ln

1141 K + 0 = 0.06862 kJ/kg ⋅ K 1037 K

q in, 2- x ⎛ x dest, 2- x = T0 ⎜⎜ s x − s 2 − Tsource ⎝ s 3 − s x = c p ln

T3 P 1255 K − Rln 3 = (1.005 kJ/kg ⋅ K ) ln − 0 = 0.09571 kJ/kg ⋅ K Tx 1141 K Px

q in, x -3 ⎛ x dest, x -3 = T0 ⎜⎜ s 3 − s x − Tsource ⎝ s 4 − s 3 = cv ln

⎞ 74.67 kJ/kg ⎞ ⎟ = (291 K)⎛⎜ 0.06862 kJ/kg ⋅ K − ⎟ = 2.65 kJ/kg ⎟ 1255 K ⎠ ⎝ ⎠

⎞ 114.6 kJ/kg ⎞ ⎟ = (291 K)⎛⎜ 0.09571 kJ/kg ⋅ K − ⎟ = 1.28 kJ/kg ⎟ 1255 K ⎠ ⎝ ⎠

T4 T v r + Rln 4 = cv ln 4 + Rln T3 v3 T3 rc

= (0.718 kJ/kg ⋅ K ) ln

494.8 K 18 + (0.287 kJ/kg ⋅ K ) ln = 0.1339 kJ/kg ⋅ K 1255 K 1 .1

x dest, 3-4 = T0 ( s 4 − s 3 ) = ( 291 K)(0.1339 kJ/kg ⋅ K) = 39.0 kJ/kg

s1 − s 4 = cv ln

T1 v 291 K + Rln 1 = (0.718 kJ/kg ⋅ K ) ln + 0 = −0.3811 kJ/kg ⋅ K T4 494.8 K v4

⎛ q x dest,4-1 = T0 ⎜⎜ s1 − s 4 + out Tsink ⎝

⎞ 146.3 kJ/kg ⎞ ⎛ ⎟ = (291 K)⎜ − 0.3811 kJ/kg ⋅ K + ⎟ = 35.4 kJ/kg ⎟ 291 K ⎠ ⎝ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-118

9-151 The exergy loss of each process for an air-standard Stirling cycle described in Prob. 9-81 is to be determined. Analysis From Prob. 9-81, qin = 1275 kJ/kg, qout = 212.5 kJ/kg, T1 = T2 = 1788 K, T3 = T4 = 298 K. The exergy destruction during a process of the cycle is ⎛ q q x dest = T0 s gen = T0 ⎜⎜ ∆s − in + out T T source sink ⎝

⎞ ⎟ ⎟ ⎠

T 1

qin

2

Application of this equation for each process of the cycle gives s 2 − s1 = cv ln

T2 v + Rln 2 T1 v1

= 0 + (0.287 kJ/kg ⋅ K ) ln(12) = 0.7132 kJ/kg ⋅ K ⎛ q x dest, 1-2 = T0 ⎜⎜ s 2 − s1 − in T source ⎝

⎞ ⎟ ⎟ ⎠

4

3 qout s

1275 kJ/kg ⎞ ⎛ = (298 K)⎜ 0.7132 kJ/kg ⋅ K − ⎟ = 0.034 kJ/kg ≈ 0 1788 K ⎠ ⎝ s 4 − s 3 = cv ln

v T4 + Rln 4 T3 v3

⎛ 1⎞ = 0 + (0.287 kJ/kg ⋅ K ) ln⎜ ⎟ = −0.7132 kJ/kg ⋅ K ⎝ 12 ⎠

⎛ q x dest, 3- 4 = T0 ⎜⎜ s 4 − s 3 + out T sink ⎝

⎞ ⎟ ⎟ ⎠

212.5 kJ/kg ⎞ ⎛ = (298 K)⎜ − 0.7132 kJ/kg ⋅ K + ⎟ = −0.034 kJ/kg ≈ 0 298 K ⎠ ⎝

These results are not surprising since Stirling cycle is totally reversible. Exergy destructions are not calculated for processes 2-3 and 4-1 because there is no interaction with the surroundings during these processes to alter the exergy destruction.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-119

9-152 The exergy destruction associated with each of the processes of the Brayton cycle described in Prob. 9-89 is to be determined. Analysis From Prob. 9-89, qin = 698.3 kJ/kg, qout = 487.9 kJ/kg, and

⎯ ⎯→ s1o = 1.68515 kJ/kg ⋅ K

T1 = 295 K

h2 = 626.60 kJ/kg ⎯ ⎯→ s 2o = 2.44117 kJ/kg ⋅ K T3 = 1240 K

⎯ ⎯→ s3o = 3.21751 kJ/kg ⋅ K

h4 = 783.04 kJ/kg ⎯ ⎯→ s 4o = 2.66807 kJ/kg ⋅ K Thus, ⎛ P ⎞ x destroyed,12 = T0 s gen,12 = T0 (s 2 − s1 ) = T0 ⎜⎜ s 2o − s1o − Rln 2 ⎟⎟ = P1 ⎠ ⎝ = (310 K )(2.44117 − 1.68515 − (0.287 kJ/kg ⋅ K )ln (10 )) = 29.51 kJ/kg ⎛ ⎞ P ©0 − q in ⎟ = T0 ⎜ s 3o − s 2o − Rln 3 + ⎟ ⎜ P2 TH ⎠ ⎝ ⎛ 698.3 kJ/kg ⎞ ⎟ = 105.4 kJ/kg = (310 K )⎜⎜ 3.21751 − 2.44117 − 1600 K ⎟⎠ ⎝

q R ,23 ⎛ x destroyed, 23 = T0 s gen,23 = T0 ⎜⎜ s 3 − s 2 + TR ⎝

⎞ ⎟ ⎟ ⎠

⎛ P ⎞ x destroyed, 34 = T0 s gen,34 = T0 (s 4 − s 3 ) = T0 ⎜⎜ s 4o − s 3o − Rln 4 ⎟⎟ = P3 ⎠ ⎝ = (310 K )(2.66807 − 3.21751 − (0.287 kJ/kg ⋅ K )ln (1/10 )) = 34.53 kJ/kg ⎛ q R ,41 ⎞ ⎛ P ©0 q out ⎟ = T0 ⎜ s1o − s 4o − Rln 1 + x destroyed, 41 = T0 s gen,41 = T0 ⎜⎜ s1 − s 4 + ⎜ TR ⎟⎠ P4 TL ⎝ ⎝

⎞ ⎟ ⎟ ⎠

⎛ 487.9 kJ/kg ⎞ ⎟ = 183.2 kJ/kg = (310 K )⎜⎜1.68515 − 2.66807 + 310 K ⎟⎠ ⎝

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-120

9-153 Exergy analysis is to be used to answer the question in Prob. 9-94. Analysis From Prob. 9-94, T1 = 288 K, T2s = 585.8 K, T2 = 618.9 K, T3 = 873 K, T4s = 429.2 K, T4 = 473.6 K, rp = 12. The exergy change of a flow stream between an inlet and exit state is given by ∆ψ = he − hi − T0 ( s e − s i )

This is also the expression for reversible work. Application of this equation for isentropic and actual compression processes gives s 2 s − s1 = c p ln

T 873 K

qin

3

2s 2

T2 s P − Rln 2 T1 P1

585.8 K − (0.287 kJ/kg ⋅ K ) ln(12) 288 K = 0.0003998 kJ/kg ⋅ K

= (1.005 kJ/kg ⋅ K ) ln

288 K

1

4

qout 4s s

w rev, 1− 2 s = c p (T2 s − T1 ) − T0 ( s 2s − s1 ) = (1.005 kJ/kg ⋅ K )(585.8 − 288)K − (288 K )(0.0003998 kJ/kg ⋅ K ) = 299.2 kJ/kg

s 2 − s1 = c p ln

T2 P − Rln 2 T1 P1

= (1.005 kJ/kg ⋅ K ) ln

618.9 K − (0.287 kJ/kg ⋅ K ) ln(12) = 0.05564 kJ/kg ⋅ K 288 K

w rev, 1− 2 = c p (T2 − T1 ) − T0 ( s 2 − s1 ) = (1.005 kJ/kg ⋅ K )(618.9 − 288)K − (288 K )(0.05564 kJ/kg ⋅ K ) = 316.5 kJ/kg

The irreversibilities therefore increase the minimum work that must be supplied to the compressor by ∆w rev,C = w rev, 1− 2 − w rev, 1− 2 s = 316.5 − 299.2 = 17.3 kJ/kg

Repeating the calculations for the turbine, s 3 − s 4 s = c p ln

P T3 − Rln 3 P4 T4 s

= (1.005 kJ/kg ⋅ K ) ln

873 K − (0.287 kJ/kg ⋅ K ) ln(12) = 0.0003944 kJ/kg ⋅ K 429.2K

w rev, 3− 4 s = c p (T3 − T4 s ) − T0 ( s 3 − s 4 s ) = (1.005 kJ/kg ⋅ K )(873 − 429.2)K − (288 K )(0.0003944 kJ/kg ⋅ K ) = 445.9 kJ/kg

s 3 − s 4 = c p ln

T3 P − Rln 3 T4 P4

= (1.005 kJ/kg ⋅ K ) ln

873 K − (0.287 kJ/kg ⋅ K ) ln(12) = −0.09854 kJ/kg ⋅ K 473.6K

w rev, 3− 4 = c p (T3 − T4 s ) − T0 ( s 3 − s 4 s ) = (1.005 kJ/kg ⋅ K )(873 − 473.6)K − (288 K )(−0.09854 kJ/kg ⋅ K ) = 429.8 kJ/kg ∆w rev,T = w rev, 3− 4 s − w rev, 3− 4 = 445 .9 − 429.8 = 16.1 kJ/kg

Hence, it is clear that the compressor is a little more sensitive to the irreversibilities than the turbine.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-121

9-154 The total exergy destruction associated with the Brayton cycle described in Prob. 9-116 and the exergy at the exhaust gases at the turbine exit are to be determined. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis From Prob. 9-116, qin = 480.82, qout = 372.73 kJ/kg, and

T1 = 310 K

T

⎯ ⎯→ s1o = 1.73498 kJ/kg ⋅ K

T3 = 1150 K

qin

1150 K

h2 = 618.26 kJ/kg ⎯ ⎯→ s 2o = 2.42763 kJ/kg ⋅ K

5

⎯ ⎯→ s 3o = 3.12900 kJ/kg ⋅ K

h4 = 803.14 kJ/kg ⎯ ⎯→ s 4o = 2.69407 kJ/kg ⋅ K h5 = 738.43 kJ/kg ⎯ ⎯→

s 5o

3

2s

= 2.60815 kJ/kg ⋅ K

310 K

4 4s

2

1

and, from an energy balance on the heat exchanger,

6 s

h5 − h2 = h4 − h6 ⎯ ⎯→ h6 = 803.14 − (738.43 − 618.26) = 682.97 kJ/kg ⎯ ⎯→ s 6o = 2.52861 kJ/kg ⋅ K Thus, ⎛ P ⎞ x destroyed,12 = T0 s gen,12 = T0 (s 2 − s1 ) = T0 ⎜⎜ s 2o − s1o − Rln 2 ⎟⎟ P1 ⎠ ⎝ = (290 K )(2.42763 − 1.73498 − (0.287 kJ/kg ⋅ K )ln (7 )) = 38.91 kJ/kg ⎛ P ⎞ x destroyed, 34 = T0 s gen,34 = T0 (s 4 − s 3 ) = T0 ⎜⎜ s 4o − s 3o − Rln 4 ⎟⎟ P3 ⎠ ⎝ = (290 K )(2.69407 − 3.12900 − (0.287kJ/kg ⋅ K )ln (1/7 )) = 35.83 kJ/kg

[(

) (

x destroyed, regen = T0 s gen,regen = T0 [(s 5 − s 2 ) + (s 6 − s 4 )] = T0 s 5o − s 2o + s 6o − s 4o

)]

= (290 K )(2.60815 − 2.42763 + 2.52861 − 2.69407 ) = 4.37 kJ/kg

⎛ ⎞ P ©0 q in ⎟ = T0 ⎜ s 3o − s 5o − Rln 3 − ⎟ ⎜ P5 TH ⎠ ⎝ ⎛ 480.82 kJ/kg ⎞ ⎟ = 58.09 kJ/kg = (290 K )⎜⎜ 3.12900 − 2.60815 − 1500 K ⎟⎠ ⎝

q R ,53 ⎛ x destroyed, 53 = T0 s gen,53 = T0 ⎜⎜ s 3 − s 5 − TR ⎝

⎞ ⎟ ⎟ ⎠

⎛ q R ,61 ⎞ ⎛ P ©0 q out ⎟ = T0 ⎜ s1o − s 6o − Rln 1 + x destroyed, 61 = T0 s gen,61 = T0 ⎜⎜ s1 − s 6 + ⎜ TR ⎟⎠ P6 TL ⎝ ⎝

⎞ ⎟ ⎟ ⎠

⎛ 372.73 kJ/kg ⎞ ⎟⎟ = 142.6 kJ/kg = (290 K )⎜⎜1.73498 − 2.52861 + 290 K ⎝ ⎠

Noting that h0 = [email protected] 290 K = 290.16 kJ/kg and T0 = 290 K the regenerator (state 6) is determined from V2 φ6 = (h6 − h0 ) − T0 (s6 − s0 ) + 6 2

⎯ ⎯→ s1o = 1.66802 kJ/kg ⋅ K , the stream exergy at the exit of

where s 6 − s 0 = s 6 − s1 = s 6o − s1o − R ln

P6 P1

= 2.52861 − 1.66802 = 0.86059 kJ/kg ⋅ K

Thus,

φ 6 = 682.97 − 290.16 − (290 K )(0.86059 kJ/kg ⋅ K ) = 143.2 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-122

9-155 Prob. 9-154 is reconsidered. The effect of the cycle pressure on the total irreversibility for the cycle and the exergy of the exhaust gas leaving the regenerator is to be investigated. Analysis Using EES, the problem is solved as follows: "Given" T[1]=310 [K] P[1]=100 [kPa] Ratio_P=7 P[2]=Ratio_P*P[1] T[3]=1150 [K] eta_C=0.75 eta_T=0.82 epsilon=0.65 T_H=1500 [K] T0=290 [K] P0=100 [kPa] "Analysis for Problem 9-154" q_in=h[3]-h[5] q_out=h[6]-h[1] h[5]-h[2]=h[4]-h[6] s[2]=entropy(Fluid\$, P=P[2], h=h[2]) s[4]=entropy(Fluid\$, h=h[4], P=P[4]) s[5]=entropy(Fluid\$, h=h[5], P=P[5]) P[5]=P[2] s[6]=entropy(Fluid\$, h=h[6], P=P[6]) P[6]=P[1] h[0]=enthalpy(Fluid\$, T=T0) s[0]=entropy(Fluid\$, T=T0, P=P0) x_destroyed_12=T0*(s[2]-s[1]) x_destroyed_34=T0*(s[4]-s[3]) x_destroyed_regen=T0*(s[5]-s[2]+s[6]-s[4]) x_destroyed_53=T0*(s[3]-s[5]-q_in/T_H) x_destroyed_61=T0*(s[1]-s[6]+q_out/T0) x_total=x_destroyed_12+x_destroyed_34+x_destroyed_regen+x_destroyed_53+x_destroyed_61 x6=h[6]-h[0]-T0*(s[6]-s[0]) "since state 0 and state 1 are identical" "Analysis for Problem 9-116" Fluid\$='air' "(a)" h[1]=enthalpy(Fluid\$, T=T[1]) s[1]=entropy(Fluid\$, T=T[1], P=P[1]) s_s[2]=s[1] "isentropic compression" h_s[2]=enthalpy(Fluid\$, P=P[2], s=s_s[2]) eta_C=(h_s[2]-h[1])/(h[2]-h[1]) h[3]=enthalpy(Fluid\$, T=T[3]) s[3]=entropy(Fluid\$, T=T[3], P=P[3]) P[3]=P[2] s_s[4]=s[3] "isentropic expansion" h_s[4]=enthalpy(Fluid\$, P=P[4], s=s_s[4]) P[4]=P[1] eta_T=(h[3]-h[4])/(h[3]-h_s[4]) q_regen=epsilon*(h[4]-h[2]) "(b)" w_C_in=(h[2]-h[1]) w_T_out=h[3]-h[4] PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-123

w_net_out=w_T_out-w_C_in q_in=(h[3]-h[2])-q_regen eta_th=w_net_out/q_in

Ratio_P

xtotal [kJ/kg] 270.1 280 289.9 299.5 308.8 317.8 326.6 335.1 343.3

6 7 8 9 10 11 12 13 14

x6 [kJ/kg] 137.2 143.5 149.6 155.5 161.1 166.6 171.9 177.1 182.1

350 330

xtotal,dest

310 290

x [kJ/kg]

270 250 230 210 190

x6

170 150 130 6

7

8

9

10

11

12

13

14

RatioP

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-124

9-156 The exergy loss of each process for a regenerative Brayton cycle with three stages of reheating and intercooling described in Prob. 9-135 is to be determined. T

Analysis From Prob. 9-135,

qout,14-1 = 206.9 kJ/kg, qout,2-3 = qout,4-5 = 141.6 kJ/kg,

9

T1 = T3 = T5 = 290 K , T2 = T4 = T6 = 430.9 K T7 = 520.7 K,

T8 = 819.2 K,

T10 = 849.8 K,

T11 = 571.9 K,

T13 = 585.7 K,

T14 = 495.9 K

12

10

8

rp = 4, qin,7-8 = qin,9-10 = qin,11-12 = 300 kJ/kg,

6

T9 = 551.3 K T12 = 870.4 K,

5

4

2

3

1

13

11

7

14

s

The exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q x dest = T0 s gen = T0 ⎜⎜ s e − s i − in + out Tsource Tsink ⎝

⎞ ⎟ ⎟ ⎠

Application of this equation for each process of the cycle gives ⎛ T P ⎞ x dest, 1-2 = x dest, 3-4 = x dest, 5-6 = T0 ⎜⎜ c p ln 2 − Rln 2 ⎟⎟ T1 P1 ⎠ ⎝ 430.9 ⎡ ⎤ = (290) ⎢(1.005)ln − (0.287) ln(4)⎥ = 0.03 kJ/kg ≈ 0 290 ⎣ ⎦ q in,7-8 ⎛ T P x dest, 7-8 = T0 ⎜⎜ c p ln 8 − Rln 8 − T7 P7 Tsource ⎝

⎞ 300 ⎤ 819.2 ⎟ = (290) ⎡⎢(1.005)ln = 32.1 kJ/kg −0− ⎟ 870 .4 ⎥⎦ 520.7 ⎣ ⎠

q in,9-10 ⎛ T P x dest, 9-10 = T0 ⎜⎜ c p ln 10 − Rln 8 − T9 P7 Tsource ⎝

⎞ 300 ⎤ 849.8 ⎟ = (290) ⎡⎢(1.005)ln = 26.2 kJ/kg −0− ⎟ 870 .4 ⎥⎦ 551.3 ⎣ ⎠

q in,11-12 ⎛ T P x dest, 11-12 = T0 ⎜⎜ c p ln 12 − Rln 12 − T11 P11 Tsource ⎝

⎞ 300 ⎤ 870.4 ⎟ = (290) ⎡⎢(1.005)ln = 22.5 kJ/kg −0− ⎟ 870.4 ⎥⎦ 571.9 ⎣ ⎠

⎛ T P ⎞ ⎡ 551.3 ⎛ 1 ⎞⎤ x dest, 8-9 = T0 ⎜⎜ c p ln 9 − Rln 9 ⎟⎟ = (290) ⎢(1.005)ln − (0.287) ln⎜ ⎟⎥ = −0.05 kJ/kg ≈ 0 T8 P8 ⎠ 819.2 ⎝ 4 ⎠⎦ ⎣ ⎝ ⎛ ⎡ T P ⎞ 571.9 ⎛ 1 ⎞⎤ x dest, 10-11 = T0 ⎜⎜ c p ln 11 − Rln 11 ⎟⎟ = (290) ⎢(1.005)ln − (0.287) ln⎜ ⎟⎥ = −0.04 kJ/kg ≈ 0 T P 849.8 ⎝ 4 ⎠⎦ ⎣ 10 10 ⎠ ⎝ ⎛ T P ⎞ ⎡ 585.7 ⎛ 1 ⎞⎤ x dest, 12-13 = T0 ⎜⎜ c p ln 13 − Rln 13 ⎟⎟ = (290) ⎢(1.005)ln − (0.287) ln⎜ ⎟⎥ = −0.08 kJ/kg ≈ 0 T12 P12 ⎠ 870.4 ⎝ 4 ⎠⎦ ⎣ ⎝ q out,14-1 ⎞ ⎛ T P 206.9 ⎤ 290 ⎟ = (290) ⎡⎢(1.005)ln x dest, 14-1 = T0 ⎜⎜ c p ln 1 − Rln 1 + = 50.6 kJ/kg −0+ ⎟ 290 ⎥⎦ T14 P14 Tsink ⎠ 495.9 ⎣ ⎝ q out,2-3 ⎛ T P x dest, 2-3 = x dest, 4-5 = T0 ⎜⎜ c p ln 3 − Rln 3 + T2 P2 Tsink ⎝

⎞ 141.6 ⎤ 290 ⎟ = (290) ⎡⎢(1.005)ln = 26.2 kJ/kg −0+ ⎟ 290 ⎥⎦ 430.9 ⎣ ⎠

⎛ T T ⎞ x dest,regen = T0 (∆s 6−7 + ∆s13−14 ) = T0 ⎜⎜ c p ln 7 + c p ln 14 ⎟⎟ T T13 ⎠ 6 ⎝ 495.9 ⎤ 520.7 ⎡ = (290) ⎢(1.005)ln + (1.005)ln ⎥ = 6.66 kJ/kg 585.7 430.9 ⎦ ⎣

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-125

9-157 A gas-turbine plant uses diesel fuel and operates on simple Brayton cycle. The isentropic efficiency of the compressor, the net power output, the back work ratio, the thermal efficiency, and the second-law efficiency are to be determined. Diesel fuel Assumptions 1 The air-standard assumptions are Combustion applicable. 2 Kinetic and potential energy changes are chamber negligible. 3 Air is an ideal gas with constant specific 3 heats. 700 kPa 2 Properties The properties of air at 500ºC = 773 K are 260°C cp = 1.093 kJ/kg·K, cv = 0.806 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.357 (Table A-2b). Turbine Compress. Analysis (a) The isentropic efficiency of the compressor may be determined if we first calculate 100 kPa 4 1 the exit temperature for the isentropic case 30°C ( k −1) / k (1.357-1)/1.357 ⎛P ⎞ ⎛ 700 kPa ⎞ = (303 K )⎜ T2 s = T1 ⎜⎜ 2 ⎟⎟ = 505.6 K ⎟ ⎝ 100 kPa ⎠ ⎝ P1 ⎠

ηC =

T2 s − T1 (505.6 − 303)K = 0.881 = T2 − T1 (533 − 303)K

(b) The total mass flowing through the turbine and the rate of heat input are m& 12.6 kg/s m& t = m& a + m& f = m& a + a = 12.6 kg/s + = 12.6 kg/s + 0.21 kg/s = 12.81 kg/s 60 AF Q& = m& q η = (0.21 kg/s)(42,000 kJ/kg)(0.97) = 8555 kW in

f

HV c

The temperature at the exit of combustion chamber is Q& in = m& c p (T3 − T2 ) ⎯ ⎯→ 8555 kJ/s = (12.81 kg/s)(1.093 kJ/kg.K)(T3 − 533)K ⎯ ⎯→ T3 = 1144 K The temperature at the turbine exit is determined using isentropic efficiency relation T4 s

⎛P = T3 ⎜⎜ 4 ⎝ P3

ηT =

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛ 100 kPa ⎞ = (1144 K )⎜ ⎟ ⎝ 700 kPa ⎠

(1.357-1)/1.357

= 685.7 K

T3 − T4 (1144 − T4 )K ⎯ ⎯→ 0.85 = ⎯ ⎯→ T4 = 754.4 K T3 − T4 s (1144 − 685.7)K

The net power and the back work ratio are W& = m& c (T − T ) = (12.6 kg/s)(1.09 3 kJ/kg.K)(5 33 − 303)K = 3168 kW C, in

a p

2

1

W&T, out = m& c p (T3 − T4 ) = (12.81 kg/s)(1.09 3 kJ/kg.K)(1 144 − 754.4)K = 5455 kW

W& net = W&T, out − W&C,in = 5455 − 3168 = 2287 kW W& 3168 kW rbw = & C,in = = 0.581 WT, out 5455 kW

(c) The thermal efficiency is W& 2287 kW η th = net = = 0.267 & 8555 kW Qin The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is, T 303 K η max = 1 − 1 = 1 − = 0.735 T3 1144 K and

η II =

η th 0.267 = = 0.364 η max 0.735

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-126

9-158 A modern compression ignition engine operates on the ideal dual cycle. The maximum temperature in the cycle, the net work output, the thermal efficiency, the mean effective pressure, the net power output, the second-law efficiency of the cycle, and the rate of exergy of the exhaust gases are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 1000 K are cp = 1.142 kJ/kg·K, cv = 0.855 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.336 (Table A-2b). Analysis (a) The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are

r=

Vc +Vd V + 0.0018 m 3 ⎯ ⎯→16 = c ⎯ ⎯→V c = 0.00012 m 3 = V 2 = V x Vc Vc

V1 = V c + V d = 0.00012 + 0.0018 = 0.00192 m 3 = V 4

P

Process 1-2: Isentropic compression ⎛v T2 = T1 ⎜⎜ 1 ⎝v2 ⎛v P2 = P1 ⎜⎜ 1 ⎝v2

⎞ ⎟⎟ ⎠

k −1

= (343 K )(16)

1.336-1

3

x = 870.7 K 2

k

Qin

⎞ ⎟⎟ = (95 kPa )(16)1.336 = 3859 kPa ⎠

4

Process 2-x and x-3: Constant-volume and constant pressure heat addition processes: T x = T2

Qout 1

Px 7500 kPa = (870.7 K) = 1692 K P2 3859 kPa

V

q 2- x = cv (T x − T2 ) = (0.855 kJ/kg.K)(1692 − 870.7)K = 702.6 kJ/kg

q 2− x = q x-3 = c p (T3 − Tx ) ⎯ ⎯→ 702.6 kJ/kg = (0.855 kJ/kg.K)(T3 − 1692)K ⎯ ⎯→ T3 = 2308 K (b)

q in = q 2− x + q x -3 = 702.6 + 702.6 = 1405 kJ/kg

V3 =V x

T3 2308 K = (0.00012 m 3 ) = 0.0001636 m 3 Tx 1692 K

Process 3-4: isentropic expansion. k −1

1.336-1

⎛V T4 = T3 ⎜⎜ 3 ⎝V 4

⎞ ⎟⎟ ⎠

⎛ 0.0001636 m 3 = (2308 K )⎜ ⎜ 0.00192 m 3 ⎝

⎞ ⎟ ⎟ ⎠

⎛V P4 = P3 ⎜⎜ 3 ⎝V 4

⎛ 0.0001636 m 3 ⎞ ⎟⎟ = (7500 kPa )⎜ ⎜ 0.00192 m 3 ⎠ ⎝

⎞ ⎟ ⎟ ⎠

k

= 1009 K 1.336

= 279.4 kPa

Process 4-1: constant voume heat rejection. q out = cv (T4 − T1 ) = (0.855 kJ/kg ⋅ K )(1009 − 343)K = 569.3 kJ/kg

The net work output and the thermal efficiency are wnet,out = q in − q out = 1405 − 569.3 = 835.8 kJ/kg

η th =

wnet,out q in

=

835.8 kJ/kg = 0.5948 = 59.5% 1405 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-127

(c) The mean effective pressure is determined to be m=

P1V1 (95 kPa)(0.00192 m 3 ) = = 0.001853 kg RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (343 K )

(

MEP =

mwnet,out

V1 − V 2

)

=

(0.001853 kg)(835.8kJ/kg) ⎛ kPa ⋅ m 3 ⎜ kJ (0.00192 − 0.00012)m 3 ⎜⎝

⎞ ⎟ = 860.4 kPa ⎟ ⎠

(d) The power for engine speed of 3500 rpm is 2200 (rev/min) ⎛ 1 min ⎞ n& W& net = mwnet = (0.001853 kg)(835.8 kJ/kg) ⎜ ⎟ = 28.39 kW (2 rev/cycle) ⎝ 60 s ⎠ 2

Note that there are two revolutions in one cycle in four-stroke engines. (e) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). We take the dead state temperature and pressure to be 25ºC and 100 kPa.

η max = 1 −

T0 (25 + 273) K =1− = 0.8709 T3 2308 K

and

η II =

η th 0.5948 = = 0.683 = 68.3% η max 0.8709

The rate of exergy of the exhaust gases is determined as follows ⎡ T P ⎤ x 4 = u 4 − u 0 − T0 ( s 4 − s 0 ) = cv (T4 − T0 ) − T0 ⎢c p ln 4 − R ln 4 ⎥ T0 P0 ⎦ ⎣ 279.4 ⎤ 1009 ⎡ = 285.0 kJ/kg = (0.855)(1009 − 298)− (298) ⎢(1.142 kJ/kg.K)ln − (0.287 kJ/kg.K ) ln 100 ⎥⎦ 298 ⎣

n& 2200 (rev/min) ⎛ 1 min ⎞ X& 4 = mx4 = (0.001853 kg)(285.0 kJ/kg) ⎜ ⎟ = 9.683 kW 2 (2 rev/cycle) ⎝ 60 s ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-128

Review Problems

9-159 An Otto cycle with a compression ratio of 7 is considered. The thermal efficiency is to be determined using constant and variable specific heats. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg·K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis (a) Constant specific heats:

η th = 1 −

1 r

k −1

=1−

1 7

1.4 −1

= 0.5408 = 54.1%

P

3 4

(b) Variable specific heats: (using air properties from Table A-17) Process 1-2: isentropic compression. T1 = 288 K ⎯ ⎯→

v r2 =

u1 = 205.48 kJ/kg

v r1 = 688.1

2

1

v

v2 1 1 v r 2 = v r 2 = (688.1) = 98.3 ⎯ ⎯→ u 2 = 447.62 kJ/kg v1 r 7

Process 2-3: v = constant heat addition. T3 = 1273 K ⎯ ⎯→

u 3 = 998.51 kJ/kg

v r 3 = 12.045

q in = u 3 − u 2 = 998.51 − 447.62 = 550.89 kJ/kg

Process 3-4: isentropic expansion.

v r4 =

v4 v r 3 = rv r 3 = (7)(12.045) = 84.32 ⎯ ⎯→ u 4 = 475.54 kJ/kg v3

Process 4-1: v = constant heat rejection. q out = u 4 − u1 = 475.54 − 205.48 = 270.06 kJ/kg

η th = 1 −

q out 270.06 kJ/kg =1− = 0.5098 = 51.0% q in 550.89 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-129

9-160E An ideal diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency is to be determined using constant and variable specific heats. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, R = 0.06855 Btu/lbm·R, and k = 1.4 (Table A-2Ea). Analysis (a) Constant specific heats:

P

Process 1-2: isentropic compression. ⎛V T2 = T1 ⎜⎜ 1 ⎝V 2

⎞ ⎟⎟ ⎠

2

qin

3

k −1

= (505 R)(20) 0.4 = 1673.8 R 4 qout

Process 2-3: P = constant heat addition.

1

V P3V 3 P2V 2 T 2260 R = ⎯ ⎯→ 3 = 3 = = 1.350 V 2 T2 1673.8 R T3 T2

v

Process 3-4: isentropic expansion. ⎛V T4 = T3 ⎜⎜ 3 ⎝V 4

⎞ ⎟⎟ ⎠

k −1

⎛ 1.350V 2 = T3 ⎜⎜ ⎝ V4

⎞ ⎟⎟ ⎠

k −1

⎛ 1.350 ⎞ = T3 ⎜ ⎟ ⎝ r ⎠

k −1

⎛ 1.350 ⎞ = (2260 R)⎜ ⎟ ⎝ 20 ⎠

0.4

= 768.8 R

q in = h3 − h2 = c p (T3 − T2 ) = (0.240 Btu/lbm ⋅ R )(2260 − 1673.8)R = 140.7 Btu/lbm q out = u 4 − u1 = cv (T4 − T1 ) = (0.171 Btu/lbm ⋅ R )(768.8 − 505)R = 45.11 Btu/lbm wnet,out = q in − q out = 140.7 − 45.11 = 95.59 Btu/lbm

η th =

wnet,out q in

=

95.59 Btu/lbm = 0.6794 = 67.9% 140.7 Btu/lbm

(b) Variable specific heats: (using air properties from Table A-17) Process 1-2: isentropic compression. T1 = 505 R ⎯ ⎯→

v r2 =

u1 = 86.06 Btu/lbm

v r1 = 170.82

T = 1582.3 R v2 1 1 ⎯→ 2 v r1 = v r1 = (170.82) = 8.541 ⎯ h2 = 391.01 Btu/lbm 20 v1 r

Process 2-3: P = constant heat addition.

v P3v 3 P2v 2 T 2260 R = ⎯ ⎯→ 3 = 3 = = 1.428 T3 T2 v 2 T2 1582.3 R ⎯→ T3 = 2260 R ⎯

h3 = 577.52 Btu/lbm

v r 3 = 2.922

q in = h3 − h2 = 577.52 − 391.01 = 186.51 Btu/lbm

Process 3-4: isentropic expansion.

v r4 =

v4 v4 20 r v r3 = v r3 = v r3 = (2.922) = 40.92 ⎯ ⎯→ u 4 = 152.65 Btu/lbm v3 1.428v 2 1.428 1.428

Process 4-1: v = constant heat rejection. q out = u 4 − u1 = 152.65 − 86.06 = 66.59 Btu/lbm

Then

η th = 1 −

q out 66.59 Btu/lbm =1− = 0.6430 = 64.3% 186.51 Btu/lbm q in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-130

9-161E A simple ideal Brayton cycle with air as the working fluid operates between the specified temperature limits. The net work is to be determined using constant and variable specific heats. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea). Analysis (a) Constant specific heats: T2 = T1 r p( k −1) / k = ( 480 R)(12) 0.4/1.4 = 976.3 R

⎛ 1 T4 = T3 ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

⎛1⎞ = (1460 R)⎜ ⎟ ⎝ 12 ⎠

0.4/1.4

= 717.8 R

wnet = wturb − wcomp = c p (T3 − T4 ) − c p (T2 − T1 ) = c p (T3 − T4 + T1 − T2 )

T 3

1460 R

qin 2

480 R

4 1

qout s

= (0.240 Btu/lbm ⋅ R )(1460 − 717.8 + 480 − 976.3)R = 59.0 Btu/lbm

(b) Variable specific heats: (using air properties from Table A-17E) T1 = 480 R ⎯ ⎯→

Pr 2 =

h1 = 114.69 Btu/lbm Pr1 = 0.9182

P2 Pr1 = (12)(0.9182) = 11.02 ⎯ ⎯→ h2 = 233.63 Btu/lbm P1

T3 = 1460 R ⎯ ⎯→ Pr 4 =

h3 = 358.63 Btu/lbm Pr 3 = 50.40

P4 ⎛1⎞ Pr 3 = ⎜ ⎟(50.40) = 4.12 ⎯ ⎯→ h4 = 176.32 Btu/lbm P3 ⎝ 12 ⎠

wnet = w turb − wcomp = (h3 − h4 ) − (h2 − h1 ) = (358.63 − 176.32) − (233.63 − 114.69) = 63.4 Btu/lbm

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-131

9-162 A turbocharged four-stroke V-16 diesel engine produces 3500 hp at 1200 rpm. The amount of power produced per cylinder per mechanical and per thermodynamic cycle is to be determined. Analysis Noting that there are 16 cylinders and each thermodynamic cycle corresponds to 2 mechanical cycles (revolutions), we have

(a) Total power produced (No. of cylinders)(No. of mechanical cycles)

wmechanical =

⎛ 42.41 Btu/min ⎞ 3500 hp ⎟⎟ ⎜ (16 cylinders)(1200 rev/min) ⎜⎝ 1 hp ⎠

=

= 7.73 Btu/cyl ⋅ mech cycle (= 8.16 kJ/cyl ⋅ mech cycle)

(b) wthermodynamic =

Total power produced (No. of cylinders)(No. of thermodynamic cycles)

=

⎛ 42.41 Btu/min ⎞ 3500 hp ⎜ ⎟⎟ (16 cylinders)(1200/2 rev/min) ⎜⎝ 1 hp ⎠

= 15.46 Btu/cyl ⋅ therm cycle (= 16.31 kJ/cyl ⋅ therm cycle)

9-163 A simple ideal Brayton cycle operating between the specified temperature limits is considered. The pressure ratio for which the compressor and the turbine exit temperature of air are equal is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The specific heat ratio of air is k =1.4 (Table A-2). Analysis We treat air as an ideal gas with constant specific heats. Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as ⎛P ⎞ T2 = T1⎜⎜ 2 ⎟⎟ ⎝ P1 ⎠

(k −1) / k

⎛P ⎞ T4 = T3 ⎜⎜ 4 ⎟⎟ ⎝ P3 ⎠

( )

T

= T1 rp (k −1) / k

(k −1) / k

⎛1⎞ = T3 ⎜ ⎟ ⎜ rp ⎟ ⎝ ⎠

3

T3

(k −1) / k

qin 2

Setting T2 = T4 and solving for rp gives ⎛T r p = ⎜⎜ 3 ⎝ T1

⎞ ⎟⎟ ⎠

k / 2(k −1)

⎛ 1500 K ⎞ ⎟⎟ = ⎜⎜ ⎝ 300 K ⎠

4 T1

1.4/0.8

= 16.7

1

qout

s

Therefore, the compressor and turbine exit temperatures will be equal when the compression ratio is 16.7.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-132

9-164 A four-cylinder spark-ignition engine with a compression ratio of 8 is considered. The amount of heat supplied per cylinder, the thermal efficiency, and the rpm for a net power output of 60 kW are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The properties of air are given in Table A-17. Analysis (a) Process 1-2: isentropic compression.

⎯→ u1 = 221.25 kJ/kg T1 = 310 K ⎯

P

v r1 = 572.3 v r2 =

v2 1 1 (572.3) = 54.50 vr = vr = v 1 1 r 1 10.5

⎯ ⎯→ u 2 = 564.29 kJ/kg

1800 K 3 Qin 4

Process 2-3: v = constant heat addition.

1

T3 = 2100 K ⎯ ⎯→ u 3 = 1775.3 kJ/kg

v r3 = 2.356 m=

(

Qout

2

v

)

P1V1 (98 kPa ) 0.0004 m 3 = = 4.406 × 10 − 4 kg RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (310 K )

(

(

)

)

Qin = m(u 3 − u 2 ) = 4.406 × 10 − 4 kg (1775.3 − 564.29 )kJ/kg = 0.5336 kJ

(b) Process 3-4: isentropic expansion.

v r4 =

v4 v r = rv r3 = (10.5)(2.356 ) = 24.74 ⎯ ⎯→ u 4 = 764.05 kJ/kg v3 3

Process 4-1: v = constant heat rejection.

(

)

Qout = m(u 4 − u1 ) = 4.406 × 10 -4 kg (764.05 − 221.25)kJ/kg = 0.2392 kJ Wnet = Qin − Qout = 0.5336 − 0.2392 = 0.2944 kJ

η th =

(c)

n& = 2

Wnet 0.2944 kJ = = 0.5517 = 55.2% Qin 0.5336 kJ

⎛ 60 s ⎞ W& net 45 kJ/s ⎜ ⎟ = 4586 rpm = (2 rev/cycle) ncyl Wnet,cyl 4 × (0.2944 kJ/cycle) ⎜⎝ 1 min ⎟⎠

Note that for four-stroke cycles, there are two revolutions per cycle.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-133

9-165 Problem 9-164 is reconsidered. The effect of the compression ratio net work done and the efficiency of the cycle is to be investigated. Also, the T-s and P-v diagrams for the cycle are to be plotted. Analysis Using EES, the problem is solved as follows: "Input Data" T[1]=(37+273) [K] P[1]=98 [kPa] T[3]= 2100 [K] V_cyl=0.4 [L]*Convert(L, m^3) r_v=10.5 "Compression ratio" W_dot_net = 45 [kW] N_cyl=4 "number of cyclinders" v[1]/v[2]=r_v "The first part of the solution is done per unit mass." "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] s[2]=entropy(air, T=T[2], v=v[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] "Conservation of energy for process 1 to 2: no heat transfer (s=const.) with work input" w_in = DELTAu_12 DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" s[3]=entropy(air, T=T[3], P=P[3]) {P[3]*v[3]/T[3]=P[2]*v[2]/T[2]} P[3]*v[3]=R*T[3] v[3]=v[2] "Conservation of energy for process 2 to 3: the work is zero for v=const, heat is added" q_in = DELTAu_23 DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=entropy(air,T=T[4],P=P[4]) s[4]=s[3] P[4]*v[4]/T[4]=P[3]*v[3]/T[3] {P[4]*v[4]=R*T[4]} "Conservation of energy for process 3 to 4: no heat transfer (s=const) with work output" - w_out = DELTAu_34 DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant volume heat rejection" v[4]=v[1] "Conservation of energy for process 2 to 3: the work is zero for v=const; heat is rejected" - q_out = DELTAu_41 DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) w_net = w_out - w_in Eta_th=w_net/q_in*Convert(, %) "Thermal efficiency, in percent" "The mass contained in each cylinder is found from the volume of the cylinder:" V_cyl=m*v[1] "The net work done per cycle is:" W_dot_net=m*w_net"kJ/cyl"*N_cyl*N_dot"mechanical cycles/min"*1"min"/60"s"*1"thermal cycle"/2"mechanical cycles"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-134

ηth [%] 42 45.55 48.39 50.74 52.73 54.44 55.94

5 6 7 8 9 10 11

wnet [kJ/kg] 568.3 601.9 625.7 642.9 655.5 664.6 671.2

Air Otto Cycle P-v Diagram

104 3 2 3

s = const

10

P [kPa]

rv

4 102

1 2100 K 310 K

101 10-2

10-1

100 v [m3/kg]

101

102

Air Otto Cycle T-s Diagram 3000 6971 kPa

2500

3

T [K]

2000

98 kPa

1500 4

1000 2 500

v = const 1

0 4.0

4.5

5.0

5.5

6.0

6.5

7.0

7.5

8.0

8.5

s [kJ/kg-K]

56 680

54 660

wnet [kJ/kg]

η th [%]

52 50 48 46 44

640 620 600 580

42 5

6

7

8

rv

9

10

11

560 5

6

7

8

rv

9

10

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11

9-135

9-166 An ideal gas Carnot cycle with helium as the working fluid is considered. The pressure ratio, compression ratio, and minimum temperature of the energy source are to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 Helium is an ideal gas with constant specific heats. Properties The specific heat ratio of helium is k = 1.667 (Table A-2a). Analysis From the definition of the thermal efficiency of a Carnot heat engine,

η th,Carnot

T TL (15 + 273) K = 1− L ⎯ ⎯→ T H = = = 576 K 1 − η th, Carnot 1 − 0.50 TH

An isentropic process for an ideal gas is one in which Pvk remains constant. Then, the pressure ratio is P2 ⎛ T2 =⎜ P1 ⎜⎝ T1

⎞ ⎟⎟ ⎠

k /( k −1)

⎛ 576 K ⎞ =⎜ ⎟ ⎝ 288 K ⎠

T TH

288 K

qin 2

1

4

qout

1.667 /(1.667 −1)

3 s

= 5.65

Based on the process equation, the compression ratio is

v 1 ⎛ P2 ⎞ =⎜ ⎟ v 2 ⎜⎝ P1 ⎟⎠

1/ k

= (5.65)1 / 1.667 = 2.83

9-167E An ideal gas Carnot cycle with helium as the working fluid is considered. The pressure ratio, compression ratio, and minimum temperature of the energy-source reservoir are to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 Helium is an ideal gas with constant specific heats. Properties The specific heat ratio of helium is k = 1.667 (Table A-2Ea). Analysis From the definition of the thermal efficiency of a Carnot heat engine,

η th,Carnot

T TL (60 + 460) R = 1− L ⎯ ⎯→ T H = = = 1300 R 1 − η th,Carnot 1 − 0.60 TH

An isentropic process for an ideal gas is one in which Pvk remains constant. Then, the pressure ratio is P2 ⎛ T2 ⎞ =⎜ ⎟ P1 ⎜⎝ T1 ⎟⎠

k /( k −1)

⎛ 1300 R ⎞ =⎜ ⎟ ⎝ 520 R ⎠

1.667 /(1.667 −1)

= 9.88

T

qin

TH

1

520 R

4

2

qout

3 s

Based on the process equation, the compression ratio is

v 1 ⎛ P2 ⎞ =⎜ ⎟ v 2 ⎜⎝ P1 ⎟⎠

1/ k

= (9.88)1 / 1.667 = 3.95

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-136

9-168 The compression ratio required for an ideal Otto cycle to produce certain amount of work when consuming a given amount of fuel is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. 4 The combustion efficiency is 100 percent. Properties The properties of air at room temperature are k = 1.4 (Table A-2). Analysis The heat input to the cycle for 0.043 grams of fuel consumption is

Qin = m fuel q HV = (0.035 ×10

−3

P

kg)(43,000 kJ/kg) = 1.505 kJ

qin

The thermal efficiency is then

η th =

3

2

4 qout 1

W net 1 kJ = = 0.6645 Qin 1.505 kJ

v

From the definition of thermal efficiency, we obtain the required compression ratio to be

η th = 1 −

1 r

k −1

⎯ ⎯→ r =

1 (1 − η th )

1 /( k −1)

=

1 (1 − 0.6645)1 /(1.4 −1)

= 15.3

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-137

9-169 An ideal Otto cycle with air as the working fluid with a compression ratio of 9.2 is considered. The amount of heat transferred to the air, the net work output, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The properties of air are given in Table A-17. Analysis (a) Process 1-2: isentropic compression.

P

T1 = 300 K ⎯ ⎯→ u1 = 214.07 kJ/kg

v r1 = 621.2 v r2 =

3

v2 1 1 v r1 = v r1 = (621.2) = 67.52 ⎯⎯→ T2 = 708.3 K r 9.2 v1

qin 4

u 2 = 518.9 kJ/kg

qout

2 1

⎛ 708.3 K ⎞ P2v 2 P1v 1 v T ⎟⎟(98 kPa ) = 2129 kPa = ⎯ ⎯→ P2 = 1 2 P1 = (9.2 )⎜⎜ T2 T1 v 2 T1 ⎝ 300 K ⎠

v

Process 2-3: v = constant heat addition. P3v 3 P2v 2 P = ⎯ ⎯→ T3 = 3 T2 = 2T2 = (2 )(708.3) = 1416.6 K ⎯ ⎯→ u 3 = 1128.7 kJ/kg T3 T2 P2 v r3 = 8.593 q in = u 3 − u 2 = 1128.7 − 518.9 = 609.8 kJ/kg

(b) Process 3-4: isentropic expansion.

v r4 =

v4 v r = rv r3 = (9.2)(8.593) = 79.06 ⎯⎯→ u 4 = 487.75 kJ/kg v3 3

Process 4-1: v = constant heat rejection. q out = u 4 − u1 = 487.75 − 214.07 = 273.7 kJ/kg w net = q in − q out = 609.8 − 273.7 = 336.1 kJ/kg

wnet 336.1 kJ/kg = = 55.1% 609.8 kJ/kg q in

(c)

η th =

(d)

v max = v 1 = v min = v 2 = MEP =

(

)

RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (300 K ) = = 0.879 m 3 /kg P1 98 kPa

v max r

⎛ 1 kPa ⋅ m 3 w net w net 336.1 kJ/kg ⎜ = = v 1 − v 2 v 1 (1 − 1 / r ) 0.879 m 3 /kg (1 − 1/9.2 ) ⎜⎝ 1 kJ

(

)

⎞ ⎟ = 429 kPa ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-138

9-170 An ideal Otto cycle with air as the working fluid with a compression ratio of 9.2 is considered. The amount of heat transferred to the air, the net work output, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.

P

Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).

qin

Analysis (a) Process 1-2 is isentropic compression: ⎛v T2 = T1 ⎜⎜ 1 ⎝v2

⎞ ⎟⎟ ⎠

k −1

3 4 qout

2

= (300 K )(9.2 )0.4 = 728.8 K

1

v

⎛ 728.8 K ⎞ P2v 2 P1v 1 v T ⎟⎟(98 kPa ) = 2190 kPa = ⎯ ⎯→ P2 = 1 2 P1 = (9.2 )⎜⎜ T2 T1 v 2 T1 ⎝ 300 K ⎠

Process 2-3: v = constant heat addition.

P3v 3 P2v 2 P = ⎯ ⎯→ T3 = 3 T2 = 2T2 = (2 )(728.8) = 1457.6 K T3 T2 P2 q in = u 3 − u 2 = cv (T3 − T2 ) = (0.718 kJ/kg ⋅ K )(1457.6 − 728.8)K = 523.3 kJ/kg (b) Process 3-4: isentropic expansion.

⎛v T4 = T3 ⎜⎜ 3 ⎝v 4

⎞ ⎟⎟ ⎠

k −1

⎛ 1 ⎞ = (1457.6 K )⎜ ⎟ ⎝ 9.2 ⎠

0.4

= 600.0 K

Process 4-1: v = constant heat rejection. q out = u 4 − u1 = cv (T4 − T1 ) = (0.718 kJ/kg ⋅ K )(600 − 300 )K = 215.4 kJ/kg w net = q in − q out = 523.3 − 215.4 = 307.9 kJ/kg

wnet 307.9 kJ/kg = = 58.8% 523.3 kJ/kg q in

(c)

η th =

(d)

v max = v 1 = v min = v 2 = MEP =

(

)

RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (300 K ) = = 0.879 m 3 /kg P1 98 kPa

v max r

⎛ 1 kPa ⋅ m 3 w net w net 307.9 kJ/kg ⎜ = = v 1 − v 2 v 1 (1 − 1 / r ) 0.879 m 3 /kg (1 − 1/9.2 ) ⎜⎝ 1 kJ

(

)

⎞ ⎟ = 393 kPa ⎟ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-139

9-171E An ideal dual cycle with air as the working fluid with a compression ratio of 12 is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm.R, cv = 0.171 Btu/lbm.R, and k = 1.4 (Table A-2E). Analysis The mass of air is m=

(

)

P1V1 (14.7 psia ) 98/1728 ft 3 = = 0.003881 lbm RT1 0.3704 psia ⋅ ft 3 /lbm ⋅ R (580 R )

(

)

P x

Process 1-2: isentropic compression.

⎛V T2 = T1 ⎜⎜ 1 ⎝V 2

⎞ ⎟⎟ ⎠

k −1

1.1 Btu

2

3

0.6 Btu 4 Qout

= (580 R )(14)0.4 = 1667 R

1

v

Process 2-x: v = constant heat addition, Q2− x ,in = m(u x − u 2 ) = mcv (T x − T2 )

0.6 Btu = (0.003881 lbm )(0.171 Btu/lbm ⋅ R )(Tx − 1667 )R ⎯ ⎯→ T x = 2571 R

Process x-3: P = constant heat addition. Q x −3,in = m(h3 − h x ) = mc p (T3 − T x ) 1.1 Btu = (0.003881 lbm )(0.240 Btu/lbm ⋅ R )(T3 − 2571)R ⎯ ⎯→ T3 = 3752 R P3V 3 PxV x V T 3752 R = ⎯ ⎯→ rc = 3 = 3 = = 1.459 T3 Tx V x T x 2571 R

Process 3-4: isentropic expansion.

⎛V T4 = T3 ⎜⎜ 3 ⎝V 4

⎞ ⎟⎟ ⎠

k −1

⎛ 1.459V1 ⎞ ⎟⎟ = T3 ⎜⎜ ⎝ V4 ⎠

k −1

⎛ 1.459 ⎞ = T3 ⎜ ⎟ ⎝ r ⎠

k −1

⎛ 1.459 ⎞ = (3752 R )⎜ ⎟ ⎝ 14 ⎠

0.4

= 1519 R

Process 4-1: v = constant heat rejection. Qout = m(u 4 − u1 ) = mcv (T4 − T1 ) = (0.003881 lbm )(0.171 Btu/lbm ⋅ R )(1519 − 580 )R = 0.6229 Btu

η th = 1 −

Qout 0.6229 Btu =1− = 0.6336 = 63.4% 1.7 Btu Qin

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-140

9-172 An ideal Stirling cycle with air as the working fluid is considered. The maximum pressure in the cycle and the net work output are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).

T

qin = 900 kJ/kg

Analysis (a) The entropy change during process 1-2 is q 900 kJ/kg s 2 − s1 = 12 = = 0.5 kJ/kg ⋅ K TH 1800 K

350 K

2

1

1800 K

4

3 qout s

and s 2 − s1 = cv ln

T2 T1

+ R ln

v2 v v ⎯ ⎯→ 0.5 kJ/kg ⋅ K = (0.287 kJ/kg ⋅ K ) ln 2 ⎯ ⎯→ 2 = 5.710 v1 v1 v1

⎛ 1800 K ⎞ P3v 3 P1v 1 v T v T ⎟⎟ = 5873 kPa = ⎯ ⎯→ P1 = P3 3 1 = P3 2 1 = (200 kPa )(5.710 )⎜⎜ T3 T1 v 1 T3 v 1 T3 ⎝ 350 K ⎠

(b) The net work output is ⎛ T w net = η th q in = ⎜⎜1 − L ⎝ TH

⎞ ⎛ 350 K ⎞ ⎟⎟q in = ⎜⎜1 − ⎟⎟(900 kJ/kg ) = 725 kJ/kg ⎝ 1800 K ⎠ ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-141

9-173 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17.

T

Analysis The properties at various states are T1 = 300 K

⎯ ⎯→

3′

h1 = 300.19 kJ / kg Pr 1 = 1.386

T3 = 1300 K ⎯ ⎯→

2′

3

qin

h3 = 1395.97 kJ / kg Pr 3 = 330.9

2 4

For rp = 6,

1 P ⎯→ h2 = 501.40 kJ/kg Pr 2 = 2 Pr1 = (6 )(1.386 ) = 8.316 ⎯ P1 Pr 4 =

qout s

P4 ⎛1⎞ ⎯→ h4 = 855.3 kJ/kg Pr3 = ⎜ ⎟(330.9) = 55.15 ⎯ P3 ⎝6⎠

q in = h3 − h2 = 1395.97 − 501.40 = 894.57 kJ/kg q out = h4 − h1 = 855.3 − 300.19 = 555.11 kJ/kg wnet = q in − q out = 894.57 − 555.11 = 339.46 kJ/kg

η th =

wnet 339.46 kJ/kg = = 37.9% 894.57 kJ/kg q in

For rp = 12, Pr2 =

P2 ⎯→ h2 = 610.6 kJ/kg Pr = (12 )(1.386 ) = 16.63 ⎯ P1 1

Pr4 =

P4 ⎛1⎞ Pr = ⎜ ⎟(330.9 ) = 27.58 ⎯ ⎯→ h4 = 704.6 kJ/kg P3 3 ⎝ 12 ⎠

q in = h3 − h2 = 1395.97 − 610.60 = 785.37 kJ/kg q out = h4 − h1 = 704.6 − 300.19 = 404.41 kJ/kg w net = q in − q out = 785.37 − 404.41 = 380.96 kJ/kg

η th =

w net 380.96 kJ/kg = = 48.5% q in 785.37 kJ/kg

Thus, (a) (b)

∆w net = 380.96 − 339.46 = 41.5 kJ/kg (increase) ∆η th = 48.5% − 37.9% = 10.6%

(increase)

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-142

9-174 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis Processes 1-2 and 3-4 are isentropic. Therefore, For rp = 6, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

⎛P T4 = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

(k −1) / k

= (300 K )(6 )0.4/1.4 = 500.6 K

T 3′

(k −1) / k

⎛1⎞ = (1300 K )⎜ ⎟ ⎝6⎠

2

0.4/1.4

= 779.1 K

q in = h3 − h2 = c p (T3 − T2 )

2

q out = h4 − h1 = c p (T4 − T1 )

1

= (1.005 kJ/kg ⋅ K )(1300 − 500.6)K = 803.4 kJ/kg

= (1.005 kJ/kg ⋅ K )(779.1 − 300)K = 481.5 kJ/kg

3

qin

4 qout s

wnet = q in − q out = 803.4 − 481.5 = 321.9 kJ/kg

η th =

wnet 321.9 kJ/kg = = 40.1% q in 803.4 kJ/kg

For rp = 12, ⎛P T2 = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

⎛P T4 = T3 ⎜⎜ 4 ⎝ P3

⎞ ⎟ ⎟ ⎠

(k −1) / k

(k −1) / k

= (300 K )(12 )0.4/1.4 = 610.2 K ⎛1⎞ = (1300 K )⎜ ⎟ ⎝ 12 ⎠

0.4/1.4

= 639.2 K

q in = h3 − h2 = c p (T3 − T2 )

= (1.005 kJ/kg ⋅ K )(1300 − 610.2)K = 693.2 kJ/kg

q out = h4 − h1 = c p (T4 − T1 )

= (1.005 kJ/kg ⋅ K )(639.2 − 300)K = 340.9 kJ/kg

wnet = q in − q out = 693.2 − 340.9 = 352.3 kJ/kg

η th =

wnet 352.3 kJ/kg = = 50.8% q in 693.2 kJ/kg

Thus, (a)

∆wnet = 352.3 − 321.9 = 30.4 kJ/kg (increase)

(b)

∆η th = 50.8% − 40.1% = 10.7%

(increase)

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-143

9-175 A regenerative gas-turbine engine operating with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.

T 6

Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. (k −1) / k

⎛P = T1 ⎜⎜ 2 ⎝ P1

⎞ ⎟⎟ ⎠

⎛P T9 s = T7 s = T6 ⎜⎜ 7 ⎝ P6

⎞ ⎟ ⎟ ⎠

T4 s = T 2 s

ηC =

ηT =

ε=

= (300 K )(4 )0.4/1.4 = 445.8 K

5 4

4

3

h2 s − h1 c p (T2 s − T1 ) ⎯ ⎯→ T4 = T2 = T1 + (T2 s − T1 ) / η C = h2 − h1 c p (T2 − T1 ) = 300 + (445.8 − 300 ) / (0.78) = 486.9 K (k −1) / k

⎛1⎞ = (1400 K )⎜ ⎟ ⎝4⎠

2s

8

7 9 7s 9

2 1 s

0.4/1.4

= 942.1 K

c p (T6 − T7 ) h6 − h7 ⎯ ⎯→ T9 = T7 = T6 − η T (T6 − T7 s ) = h6 − h7 s c p (T6 − T7 s ) = 1400 − (0.86 )(1400 − 942.1) = 1006 K h5 − h4 c p (T5 − T4 ) ⎯ ⎯→ T5 = T4 + ε (T9 − T4 ) = h9 − h4 c p (T9 − T4 ) = 486.9 + (0.75)(1006 − 486.9 ) = 876.4 K

wC,in = 2(h2 − h1 ) = 2c p (T2 − T1 ) = 2(1.005 kJ/kg ⋅ K )(486.9 − 300)K = 375.7 kJ/kg wT,out = 2(h6 − h7 ) = 2c p (T6 − T7 ) = 2(1.005 kJ/kg ⋅ K )(1400 − 1006 )K = 791.5 kJ/kg

Thus,

rbw =

wC,in wT,out

=

375.7 kJ/kg = 0.475 791.5 kJ/kg

q in = (h6 − h5 ) + (h8 − h7 ) = c p [(T6 − T5 ) + (T8 − T7 )]

= (1.005 kJ/kg ⋅ K )[(1400 − 876.4) + (1400 − 1006)]K = 922.0 kJ/kg

wnet = wT,out − wC,in = 791.5 − 375.7 = 415.8 kJ/kg

η th =

wnet 415.8 kJ/kg = = 0.451 = 45.1% 922.0 kJ/kg q in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-144

9-176 Problem 9-175 is reconsidered. The effect of the isentropic efficiencies for the compressor and turbine and regenerator effectiveness on net work done and the heat supplied to the cycle is to be investigated. Also, the T-s diagram for the cycle is to be plotted. Analysis Using EES, the problem is solved as follows: "Input data" T[6] = 1400 [K] T[8] = T[6] Pratio = 4 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = T[1] Eta_reg = 0.75 "Regenerator effectiveness" Eta_c =0.78 "Compressor isentorpic efficiency" Eta_t =0.86 "Turbine isentropic efficiency" "LP Compressor:" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = w_compisen_LP/w_comp_LP "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the LP compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" h[1] + w_compisen_LP = h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" h[1] + w_comp_LP = h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "HP Compressor:" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the HP compressor" P[4] = Pratio*P[3] P[3] = P[2] s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4]) "T_s[4] is the isentropic value of T[4] at compressor exit" Eta_c = w_compisen_HP/w_comp_HP "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" h[3] + w_compisen_HP = h_s[4] h[3]=ENTHALPY(Air,T=T[3]) h_s[4]=ENTHALPY(Air,T=T_s[4]) "Actual compressor analysis:" h[3] + w_comp_HP = h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-145

"Intercooling heat loss:" h[2] = q_out_intercool + h[3] "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" h[4] + q_in_noreg = h[6] h[6]=ENTHALPY(Air,T=T[6]) P[6]=P[4]"process 4-6 is SSSF constant pressure" "HP Turbine analysis" s[6]=ENTROPY(Air,T=T[6],P=P[6]) s_s[7]=s[6] "For the ideal case the entropies are constant across the turbine" P[7] = P[6] /Pratio s_s[7]=ENTROPY(Air,T=T_s[7],P=P[7])"T_s[7] is the isentropic value of T[7] at HP turbine exit" Eta_t = w_turb_HP /w_turbisen_HP "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" h[6] = w_turbisen_HP + h_s[7] h_s[7]=ENTHALPY(Air,T=T_s[7]) "Actual Turbine analysis:" h[6] = w_turb_HP + h[7] h[7]=ENTHALPY(Air,T=T[7]) s[7]=ENTROPY(Air,T=T[7], P=P[7]) "Reheat Q_in:" h[7] + q_in_reheat = h[8] h[8]=ENTHALPY(Air,T=T[8]) "HL Turbine analysis" P[8]=P[7] s[8]=ENTROPY(Air,T=T[8],P=P[8]) s_s[9]=s[8] "For the ideal case the entropies are constant across the turbine" P[9] = P[8] /Pratio s_s[9]=ENTROPY(Air,T=T_s[9],P=P[9])"T_s[9] is the isentropic value of T[9] at LP turbine exit" Eta_t = w_turb_LP /w_turbisen_LP "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" h[8] = w_turbisen_LP + h_s[9] h_s[9]=ENTHALPY(Air,T=T_s[9]) "Actual Turbine analysis:" h[8] = w_turb_LP + h[9] h[9]=ENTHALPY(Air,T=T[9]) s[9]=ENTROPY(Air,T=T[9], P=P[9]) "Cycle analysis" w_net=w_turb_HP+w_turb_LP - w_comp_HP - w_comp_LP q_in_total_noreg=q_in_noreg+q_in_reheat Eta_th_noreg=w_net/(q_in_total_noreg)*Convert(, %) "[%]" "Cycle thermal efficiency" Bwr=(w_comp_HP + w_comp_LP)/(w_turb_HP+w_turb_LP)"Back work ratio" "With the regenerator, the heat added in the external heat exchanger is" h[5] + q_in_withreg = h[6] h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[4] PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-146

"The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[4])/(h[9]-h[4]) "Energy balance on regenerator gives h[10] and thus T[10] as:" h[4] + h[9]=h[5] + h[10] h[10]=ENTHALPY(Air, T=T[10]) s[10]=ENTROPY(Air,T=T[10], P=P[10]) P[10]=P[9] "Cycle thermal efficiency with regenerator" q_in_total_withreg=q_in_withreg+q_in_reheat Eta_th_withreg=w_net/(q_in_total_withreg)*Convert(, %) "[%]" "The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6] s_s[8]=s[8] T_s[8]=T[8] s_s[10]=s[10] T_s[10]=T[10]

ηreg

ηC

ηt

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

0.78 0.78 0.78 0.78 0.78 0.78 0.78 0.78 0.78

0.86 0.86 0.86 0.86 0.86 0.86 0.86 0.86 0.86

ηth,noreg [%] 30.57 30.57 30.57 30.57 30.57 30.57 30.57 30.57 30.57

qin,total,noreg [kJ/kg] 1434 1434 1434 1434 1434 1434 1434 1434 1434

qin,total,withreg [kJ/kg] 1062 1031 1000 969.1 938.1 907.1 876.1 845.1 814.1

wnet [kJ/kg] 438.5 438.5 438.5 438.5 438.5 438.5 438.5 438.5 438.5

Air 6

a

8

10 0

kP

1600

1200 7

9

a

5

40 0

kP

800

400

4

3 0 4.5

10

2

16 00

kP a

T [K]

ηth,withreg [%] 41.29 42.53 43.85 45.25 46.75 48.34 50.06 51.89 53.87

5.0

5.5

1

6.0

6.5

7.0

7.5

8.0

8.5

s [kJ/kg-K]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-147

52

η th [%]

48 44

With regeneration

40 36 No regeneration

32 28 0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

0.9

0.95

1

0.9

0.95

1

η re g 55 50 45

With regeneration

η th [%]

40 35 30 25 No regeneration 20 15 10 0.6

0.65

0.7

0.75

0.8

ηt

0.85

600 550

wnet [kJ/kg]

500 450 400 350 300 250 200 150 0.6

0.65

0.7

0.75

0.8

0.85

ηt

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-148

1500 1400 1300

qin,total [kJ/kg]

No regeneration

1200 1100 With regeneration

1000 900 800 0.6

0.65

0.7

0.75

0.8

ηt

0.85

0.9

0.95

1

0.9

0.95

1

55 50

η th [%]

45 With regeneration 40 35

No regeneration

30 25 0.6

0.65

0.7

0.75

0.8

0.85

ηc

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-149

9-177 A regenerative gas-turbine engine operating with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Helium is an ideal gas with constant specific heats.

T qin

Properties The properties of helium at room temperature are cp = 5.1926 kJ/kg.K and k = 1.667 (Table A-2).

5

Analysis The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. ⎛P T4 s = T2 s = T1 ⎜⎜ 2 ⎝ P1

ηC =

ε=

4 = (300 K )(4 )0.667/1.667 = 522.4 K

4

3

c p (T2 s − T1 )

2s

8

7 9 7s 9

2 1 s

h2 s − h1 ⎯ ⎯→ T4 = T2 = T1 + (T2 s − T1 ) / η C = h2 − h1 c p (T2 − T1 ) = 300 + (522.4 − 300 ) / (0.78) = 585.2 K

⎛P T9 s = T7 s = T6 ⎜⎜ 7 ⎝ P6

ηT =

⎞ ⎟⎟ ⎠

(k −1) / k

6

⎞ ⎟ ⎟ ⎠

(k −1) / k

⎛1⎞ = (1400 K )⎜ ⎟ ⎝4⎠

0.667/1.667

= 804.0 K

c p (T6 − T7 ) h6 − h7 ⎯ ⎯→ T9 = T7 = T6 − η T (T6 − T7 s ) = h6 − h7 s c p (T6 − T7 s ) = 1400 − (0.86 )(1400 − 804.0 ) = 887.4 K h5 − h4 c p (T5 − T4 ) ⎯ ⎯→ T5 = T4 + ε (T9 − T4 ) = h9 − h4 c p (T9 − T4 ) = 585.2 + (0.75)(887.4 − 585.2) = 811.8 K

wC,in = 2(h2 − h1 ) = 2c p (T2 − T1 ) = 2(5.1926 kJ/kg ⋅ K )(585.2 − 300 )K = 2961 kJ/kg wT,out = 2(h6 − h7 ) = 2c p (T6 − T7 ) = 2(5.1926 kJ/kg ⋅ K )(1400 − 887.4)K = 5323 kJ/kg

Thus,

rbw =

wC,in wT,out

=

2961 kJ/kg = 0.556 5323 kJ/kg

q in = (h6 − h5 ) + (h8 − h7 ) = c p [(T6 − T5 ) + (T8 − T7 )]

= (5.1926 kJ/kg ⋅ K )[(1400 − 811.8) + (1400 − 887.4)]K = 5716 kJ/kg

wnet = wT,out − wC,in = 5323 − 2961 = 2362 kJ/kg

η th =

wnet 2362 kJ/kg = = 0.4133 = 41.3% 5716 kJ/kg q in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-150

9-178 An ideal gas-turbine cycle with one stage of compression and two stages of expansion and regeneration is considered. The thermal efficiency of the cycle as a function of the compressor pressure ratio and the high-pressure turbine to compressor inlet temperature ratio is to be determined, and to be compared with the efficiency of the standard regenerative cycle. Analysis The T-s diagram of the cycle is as shown in the figure. If the overall pressure ratio of the cycle is rp, which is the pressure ratio across the compressor, then the pressure ratio across each turbine stage in the ideal case becomes √ rp. Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as ⎛P ⎞ T5 = T2 = T1⎜⎜ 2 ⎟⎟ ⎝ P1 ⎠

(k −1) / k

⎛P ⎞ T7 = T4 = T3 ⎜⎜ 4 ⎟⎟ ⎝ P3 ⎠ ⎛P ⎞ T6 = T5 ⎜⎜ 6 ⎟⎟ ⎝ P5 ⎠

(k −1) / k

3

7

( )

⎛ 1 ⎞ ⎟ = T3 ⎜ ⎜ r ⎟ p ⎝ ⎠

⎛ 1 ⎞ ⎟ = T5 ⎜ ⎜ r ⎟ ⎝ p⎠

Then,

qin

4

(k −1) / k

= T1 rp (k −1) / k

T

(k −1) / k

= T3rp (1− k ) / 2 k

2

5

1

6 qout

s

(k −1) / k

= T2rp (1− k ) / 2 k = T1rp (k −1) / k rp (1− k ) / 2 k = T1rp (k −1) / 2 k

( − T ) = c T (r (

) − 1)

q in = h3 − h7 = c p (T3 − T7 ) = c p T3 1 − r p (1− k ) / 2 k q out = h6 − h1 = c p (T6

and thus

η th

1

( (

p 1

p

k −1) / 2 k

) )

c p T1 r p (k −1) / 2 k − 1 q out = 1− = 1− q in c p T3 1 − r p (1− k ) / 2 k

which simplifies to

η th = 1 −

T1 (k −1) / 2 k rp T3

The thermal efficiency of the single stage ideal regenerative cycle is given as

η th = 1 −

T1 (k −1) / k rp T3

Therefore, the regenerative cycle with two stages of expansion has a higher thermal efficiency than the standard regenerative cycle with a single stage of expansion for any given value of the pressure ratio rp.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-151

9-179 A gas-turbine plant operates on the regenerative Brayton cycle with reheating and intercooling. The back work ratio, the net work output, the thermal efficiency, the second-law efficiency, and the exergies at the exits of the combustion chamber and the regenerator are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K. Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.

Optimum intercooling and reheating pressure is P2 = P1 P4 = (100)(1200) = 346.4 kPa

T

Process 1-2, 3-4: Compression ⎯→ h1 = 300.43 kJ/kg T1 = 300 K ⎯

4 2 4s 2s

P2 = 346.4 kPa

3

ηC =

7

6 8 6s 8s

9

T1 = 300 K ⎫ ⎬s1 = 5.7054 kJ/kg ⋅ K P1 = 100 kPa ⎭

⎫ ⎬h2 s = 428.79 kJ/kg s 2 = s1 = 5.7054 kJ/kg.K ⎭

5

qin

10

1 s

h2 s − h1 428.79 − 300.43 ⎯ ⎯→ 0.80 = ⎯ ⎯→ h2 = 460.88 kJ/kg h2 − h1 h2 − 300.43

T3 = 350 K ⎯ ⎯→ h3 = 350.78 kJ/kg T3 = 350 K

⎫ ⎬s 3 = 5.5040 kJ/kg ⋅ K P3 = 346.4 kPa ⎭ P4 = 1200 kPa

⎫ ⎬h4 s = 500.42 kJ/kg s 4 = s 3 = 5.5040 kJ/kg.K ⎭

ηC =

h4 s − h3 500.42 − 350.78 ⎯ ⎯→ 0.80 = ⎯ ⎯→ h4 = 537.83 kJ/kg h4 − h3 h4 − 350.78

Process 6-7, 8-9: Expansion T6 = 1400 K ⎯ ⎯→ h6 = 1514.9 kJ/kg T6 = 1400 K ⎫ ⎬s 6 = 6.6514 kJ/kg ⋅ K P6 = 1200 kPa ⎭ P7 = 346.4 kPa

⎫ ⎬h7 s = 1083.9 kJ/kg s 7 = s 6 = 6.6514 kJ/kg.K ⎭

ηT =

1514.9 − h7 h6 − h7 ⎯ ⎯→ h7 = 1170.1 kJ/kg ⎯ ⎯→ 0.80 = 1514.9 − 1083.9 h6 − h7 s

T8 = 1300 K ⎯ ⎯→ h8 = 1395.6 kJ/kg T8 = 1300 K

⎫ ⎬s 8 = 6.9196 kJ/kg ⋅ K P8 = 346.4 kPa ⎭

P9 = 100 kPa

⎫ ⎬h9 s = 996.00 kJ/kg s 9 = s 8 = 6.9196 kJ/kg.K ⎭

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-152

ηT =

h8 − h9 1395.6 − h9 ⎯ ⎯→ 0.80 = ⎯ ⎯→ h9 = 1075.9 kJ/kg h8 − h9 s 1395.6 − 996.00

Cycle analysis: wC,in = h2 − h1 + h4 − h3 = 460.88 − 300.43 + 537 .83 − 350.78 = 347.50 kJ/kg

wT, out = h6 − h7 + h8 − h9 = 1514 .9 − 1170 .1 + 1395.6 − 1075.9 = 664.50 kJ/kg

rbw =

wC,in wT,out

=

347.50 = 0.523 664.50

w net = wT, out − wC,in = 664.50 − 347.50 = 317.0 kJ/kg

Regenerator analysis:

ε regen =

h9 − h10 1075.9 − h10 ⎯ ⎯→ 0.75 = ⎯ ⎯→ h10 = 672.36 kJ/kg h9 − h4 1075.9 − 537.83

h10 = 672.36 K ⎫ ⎬s10 = 6.5157 kJ/kg ⋅ K P10 = 100 kPa ⎭ q regen = h9 − h10 = h5 − h4 ⎯ ⎯→ 1075.9 − 672.36 = h5 − 537.83 ⎯ ⎯→ h5 = 941.40 kJ/kg

(b)

q in = h6 − h5 = 1514.9 − 941.40 = 573.54 kJ/kg

η th =

wnet 317.0 = = 0.553 573.54 q in

(c) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,

η max = 1 −

T1 300 K =1− = 0.786 T6 1400 K

and

η II =

η th 0.553 = = 0.704 η max 0.786

(d) The exergies at the combustion chamber exit and the regenerator exit are x 6 = h 6 − h 0 − T0 ( s 6 − s 0 ) = (1514.9 − 300.43)kJ/kg − (300 K )(6.6514 − 5.7054)kJ/kg.K = 930.7 kJ/kg x10 = h10 − h0 − T0 ( s10 − s 0 ) = (672.36 − 300.43)kJ/kg − (300 K )(6.5157 − 5.7054)kJ/kg.K = 128.8 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-153

9-180 The thermal efficiency of a two-stage gas turbine with regeneration, reheating and intercooling to that of a threestage gas turbine is to be compared. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis

Two Stages:

T

The pressure ratio across each stage is 873 K

r p = 16 = 4

6

8

7

9

5

The temperatures at the end of compression and expansion are Tc = Tmin r p( k −1) / k = ( 283 K)(4) 0.4/1.4 = 420.5 K

⎛ 1 Te = Tmax ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

283 K ⎛1⎞ = (873 K)⎜ ⎟ ⎝4⎠

0.4/1.4

4

2

3

1

10

= 587.5 K

s

The heat input and heat output are

q in = 2c p (Tmax − Te ) = 2(1.005 kJ/kg ⋅ K)(873 − 587.5) K = 573.9 kJ/kg q out = 2c p (Tc − Tmin ) = 2(1.005 kJ/kg ⋅ K)(420.5 − 283) K = 276.4 kJ/kg The thermal efficiency of the cycle is then

η th = 1 −

q out 276.4 = 1− = 0.518 573 .9 q in

Three Stages: The pressure ratio across each stage is r p = 16

1/ 3

T

= 2.520

8 7

The temperatures at the end of compression and expansion are

9 11

Tc = Tmin r p( k −1) / k = ( 283 K)(2.520) 0.4/1.4 = 368.5 K

⎛ 1 Te = Tmax ⎜ ⎜ rp ⎝

⎞ ⎟ ⎟ ⎠

( k −1) / k

6 ⎛ 1 ⎞ = (873 K)⎜ ⎟ ⎝ 2.520 ⎠

4

2

3

1

0.4/1.4

= 670.4 K

10

5

12 13

14 s

The heat input and heat output are

q in = 3c p (Tmax − Te ) = 3(1.005 kJ/kg ⋅ K)(873 − 670.4) K = 610.8 kJ/kg

q out = 3c p (Tc − Tmin ) = 3(1.005 kJ/kg ⋅ K)(368.5 − 283) K = 257.8 kJ/kg The thermal efficiency of the cycle is then

η th = 1 −

q out 257.8 = 1− = 0.578 610.8 q in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-154

9-181E A pure jet engine operating on an ideal cycle is considered. The thrust force produced per unit mass flow rate is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1E), cp = 0.24 Btu/lbm⋅R and k = 1.4 (Table A-2Ea). Analysis (a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 1200 ft/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0).

T 4 qin

Diffuser:

5 E& in − E& out = ∆E& system

⎯ ⎯→ E& in = E& out

⎯→ 0 = h2 − h1 + h1 + V12 / 2 = h2 + V22 / 2 ⎯

V22

2

0 = c p (T2 − T1 ) − V12 / 2 T2 = T1 +

3 − V12 2

1

⎞ ⎟⎟ ⎠

k / (k −1)

⎛ 609.8 R ⎞ = (10 psia )⎜ ⎟ ⎝ 490 R ⎠

qout s

⎛ 1 Btu/lbm V12 (1200 ft/s)2 ⎜ = 490 R + (2 )(0.24 BtuJlbm ⋅ R ) ⎜⎝ 25,037 ft 2 /s 2 2c p

⎛T P2 = P1 ⎜⎜ 2 ⎝ T1

6

⎞ ⎟⎟ = 609.8 R ⎠

1.4/0.4

= 21.5 psia

Compressor:

( )

P3 = P4 = rp (P2 ) = (9 )(21.5 psia ) = 193.5 psia ⎛P T3 = T2 ⎜⎜ 3 ⎝ P2

⎞ ⎟⎟ ⎠

(k −1) / k

= (609.8 R )(9 )0.4/1.4 = 1142.4 R

Turbine:

wcomp,in = wturb,out ⎯ ⎯→ h3 − h2 = h4 − h5 ⎯ ⎯→ c p (T3 − T2 ) = c p (T4 − T5 ) or

T5 = T4 − T3 + T2 = 1160 − 1142.4 + 609.8 = 627.4 R

Nozzle: ⎛P T6 = T4 ⎜⎜ 6 ⎝ P4

⎞ ⎟⎟ ⎠

(k −1) / k

⎛ 10 psia ⎞ ⎟⎟ = (1160 R )⎜⎜ ⎝ 193.5 psia ⎠

0.4/1.4

= 497.5 R

E& in − E& out = ∆E& system©0 (steady) ⎯ ⎯→ E& in = E& out h5 + V52 / 2 = h6 + V62 / 2 0 = h6 − h5 +

or,

V6 = Vexit =

V62 − V52 2

⎯ ⎯→ 0 = c p (T6 − T5 ) + V62 / 2

/s 2 ⎝ 1 Btu/lbm

(2)(0.24 Btu/lbm ⋅ R )(627.4 − 497.5)R ⎜⎜ 25,037 ft

2

⎞ ⎟ = 1249 ft/s ⎟ ⎠

The specific impulse is then F = V exit − Vinlet = 1249 − 1200 = 49 m/s m&

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-155

9-182 The electricity and the process heat requirements of a manufacturing facility are to be met by a cogeneration plant consisting of a gas-turbine and a heat exchanger for steam production. The mass flow rate of the air in the cycle, the back work ratio, the thermal efficiency, the rate at which steam is produced in the heat exchanger, and the utilization efficiency of the cogeneration plant are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.

325°C 5 Combustion chamber

Analysis (a) For this problem, we use the properties of air from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.

1 MPa

Compress.

⎯→ h1 = 293.5 kJ/kg T1 = 20°C ⎯ ⎫ ⎬s1 = 5.682 kJ/kg ⋅ K P1 = 100 kPa ⎭

Heat exchanger 3

2

Process 1-2: Compression T1 = 20°C

15°C

1

4 Turbine

450°C Sat. vap. 200°C

100 kPa 20°C

P2 = 1000 kPa

⎫ ⎬h2 s = 567.2 kJ/kg s 2 = s1 = 5.682 kJ/kg.K ⎭

ηC =

h2 s − h1 567.2 − 293.5 ⎯ ⎯→ 0.86 = ⎯ ⎯→ h2 = 611.8 kJ/kg h2 − h1 h2 − 293.5

Process 3-4: Expansion T4 = 450°C ⎯ ⎯→ h4 = 738.5 kJ/kg

ηT =

h3 − h4 h − 738.5 ⎯ ⎯→ 0.88 = 3 h3 − h4 s h3 − h4 s

We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with the isentropic efficiency relation, we find h3 = 1262 kJ/kg, T3 = 913.2ºC, s3 = 6.507 kJ/kg.K. The solution by hand would require a trialerror approach. h_3=enthalpy(Air, T=T_3) s_3=entropy(Air, T=T_3, P=P_2) h_4s=enthalpy(Air, P=P_1, s=s_3)

Also, T5 = 325°C ⎯ ⎯→ h5 = 605.4 kJ/kg

The inlet water is compressed liquid at 15ºC and at the saturation pressure of steam at 200ºC (1555 kPa). This is not available in the tables but we can obtain it in EES. The alternative is to use saturated liquid enthalpy at the given temperature. Tw1 = 15°C ⎫ ⎬hw1 = 64.47 kJ/kg P1 = 1555 kPa ⎭ Tw2 = 200°C⎫ ⎬hw2 = 2792 kJ/kg x2 = 1 ⎭ The net work output is wC,in = h2 − h1 = 611.8 − 293.5 = 318.2 kJ/kg wT,out = h3 − h4 = 1262 − 738.5 = 523.4 kJ/kg PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-156

w net = wT, out − wC,in = 523.4 − 318.2 = 205.2 kJ/kg

The mass flow rate of air is W& net 1500 kJ/s = = 7.311 kg/s wnet 205.2 kJ/kg

m& a =

(b) The back work ratio is

rbw =

wC,in wT,out

=

318.2 = 0.608 523.4

The rate of heat input and the thermal efficiency are

Q& in = m& a (h3 − h2 ) = (7.311 kg/s)(1262 − 611.8)kJ/kg = 4753 kW

η th =

W& net 1500 kW = = 0.3156 = 31.6% 4753 kW Q& in

(c) An energy balance on the heat exchanger gives m& a (h4 − h5 ) = m& w (hw2 − hw1 ) (7.311 kg/s)(738.5 − 605.4)kJ/kg = m& w (2792 − 64.47)kJ/kg ⎯ ⎯→ m& w = 0.3569 kg/s

(d) The heat supplied to the water in the heat exchanger (process heat) and the utilization efficiency are Q& p = m& w (hw2 − hw1 ) = (0.3569 kg/s)(2792 − 64.47)kJ/kg = 973.5 kW

εu =

W& net + Q& p 1500 + 973.5 = = 0.5204 = 52.0% 4753 kW Q& in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-157

9-183 A turbojet aircraft flying is considered. The pressure of the gases at the turbine exit, the mass flow rate of the air through the compressor, the velocity of the gases at the nozzle exit, the propulsive power, and the propulsive efficiency of the cycle are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.

T

Diffuser, Process 1-2:

qin

4 5

3

T1 = −35°C ⎯ ⎯→ h1 = 238.23 kJ/kg

2

6

qout 1 V12 V2 s = h2 + 2 2 2 (900/3.6 m/s) 2 ⎛ 1 kJ/kg ⎞ (15 m/s) 2 ⎛ 1 kJ/kg ⎞ h = + ⎯→ h2 = 269.37 kJ/kg (238.23 kJ/kg) + ⎜ ⎟ ⎜ ⎟⎯ 2 2 2 2 2 2 2 ⎝ 1000 m /s ⎠ ⎝ 1000 m /s ⎠ h1 +

h2 = 269.37 kJ/kg ⎫ ⎬s 2 = 5.7951 kJ/kg ⋅ K P2 = 50 kPa ⎭

Compressor, Process 2-3: P3 = 450 kPa

⎫ ⎬h3s = 505.19 kJ/kg s 3 = s 2 = 5.7951 kJ/kg.K ⎭

ηC =

h3s − h2 505.19 − 269.37 ⎯ ⎯→ 0.83 = ⎯ ⎯→ h3 = 553.50 kJ/kg h3 − h2 h3 − 269.37

Turbine, Process 3-4: T4 = 950°C ⎯⎯→ h4 = 1304.8 kJ/kg h3 − h2 = h4 − h5 ⎯ ⎯→ 553.50 − 269.37 = 1304.8 − h5 ⎯ ⎯→ h5 = 1020.6 kJ/kg

where the mass flow rates through the compressor and the turbine are assumed equal.

ηT =

h4 − h5 1304.8 − 1020.6 ⎯ ⎯→ 0.83 = ⎯ ⎯→ h5 s = 962.45 kJ/kg h4 − h5 s 1304.8 − h5 s

T4 = 950°C ⎫ ⎬s 4 = 6.7725 kJ/kg ⋅ K P4 = 450 kPa ⎭ h5 s = 962.45 kJ/kg

⎫ ⎬ P5 = 147.4 kPa s 5 = s 4 = 6.7725 kJ/kg ⋅ K ⎭

(b) The mass flow rate of the air through the compressor is

m& =

W& C 500 kJ/s = = 1.760 kg/s h3 − h2 (553.50 − 269.37) kJ/kg

(c) Nozzle, Process 5-6: h5 = 1020.6 kJ/kg ⎫ ⎬s 5 = 6.8336 kJ/kg ⋅ K P5 = 147.4 kPa ⎭

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-158

P6 = 40 kPa

⎫ ⎬h6 s = 709.66 kJ/kg s 6 = s 5 = 6.8336 kJ/kg.K ⎭

ηN =

h5 − h6 1020.6 − h6 ⎯ ⎯→ 0.83 = ⎯ ⎯→ h6 = 762.52 kJ/kg h5 − h6 s 1020.6 − 709.66 h5 +

V52 V2 = h6 + 6 2 2

(1020.6 kJ/kg) + 0 = 762.52 kJ/kg +

V62 ⎛ 1 kJ/kg ⎞ ⎯→V6 = 718.5 m/s ⎜ ⎟⎯ 2 ⎝ 1000 m 2 /s 2 ⎠

where the velocity at nozzle inlet is assumed zero. (d) The propulsive power and the propulsive efficiency are ⎛ 1 kJ/kg ⎞ W& p = m& (V6 − V1 )V1 = (1.76 kg/s) (718.5 m/s − 250 m/s)(250 m/s)⎜ ⎟ = 206.1 kW 2 2 ⎝ 1000 m /s ⎠

Q& in = m& (h4 − h3 ) = (1.76 kg/s)(1304.8 − 553.50)kJ/kg = 1322 kW

ηp =

W& p 206.1 kW = = 0.156 Q& in 1322 kW

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-159

9-184 The three processes of an air standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the expressions for back work ratio and the thermal efficiency are to be obtained. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Analysis (a) The P-v and T-s diagrams for this cycle are as shown.

(b) The work of compression is found by the first law for process 1-2:

q1− 2 − w1− 2 = ∆u1− 2 q1− 2 = 0(isentropic process )

P

w1− 2 = −∆u1− 2 = −Cv (T2 − T1 )

3

2

wcomp = − w1− 2 = Cv (T2 − T1 ) The expansion work is found by

1

3

wexp = w2 −3 = ∫ Pdv = P ( v3 − v2 ) = R (T3 − T2 )

v

2

The back work ratio is

wcomp wexp

=

Cv (T3 − T1 ) R (T3 − T2 )

C T (T / T − 1) = v 1 3 1 R T2 (T3 / T2 − 1)

T 3 2

Process 1-2 is isentropic; therefore,

T1 ⎛ V2 ⎞ =⎜ ⎟ T2 ⎝ V1 ⎠

k −1

=

1 r k −1

k

P ⎛V ⎞ k and 2 = ⎜ 1 ⎟ = r P1 ⎝ V2 ⎠

1 s

Process 2-3 is constant pressure; therefore,

PV T V V PV 3 3 = 2 2 ⇒ 3 = 3 = 1 =r T3 T2 T2 V2 V2 Process 3-1 is constant volume; therefore,

PV T P P PV 3 3 = 1 1 ⇒ 3 = 3 = 2 = rk T3 T1 T1 P1 P1 The back work ratio becomes (Cv=R/(k-1))

wcomp wexp

=

1 1 r k −1 − 1 k − 1 r k −1 r − 1

(c) Apply first law to the closed system for processes 2-3 and 3-1 to show:

qin = C p (T3 − T2 ) qout = Cv (T3 − T1 ) The cycle thermal efficiency is given by

ηth = 1 −

C (T − T ) qout 1 T1 (T3 / T1 − 1) = 1− v 3 1 = 1− qin C p (T3 − T2 ) k T2 (T3 / T2 − 1)

The efficiency becomes

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-160

ηth = 1 −

1 1 r −1 k r k −1 r − 1 k

(d) Determine the value of the back work ratio and efficiency as r goes to unity.

wcomp wexp lim r →1

lim r →1

wcomp wexp wcomp wexp

=

1 1 r k −1 − 1 k − 1 r k −1 r − 1

( k − 1) r k −2 ⎪⎫ 1 ⎧⎪ 1 r k −1 − 1 ⎫⎪ 1 ⎧ 1 ⎪⎧ r k −1 − 1 ⎫ lim lim lim = = ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ k −1 ⎪ r k −1 r − 1 ⎪ k − 1 ⎩ r →1 r k − r k −1 ⎭ k − 1 ⎩⎪ r →1 kr k −1 − ( k − 1) r k − 2 ⎭⎪ ⎩ r →1 ⎭ 1 ⎧ k −1 ⎫ 1 ⎧ k −1⎫ = ⎨ ⎬= ⎨ ⎬ =1 k − 1 ⎩ k − k + 1⎭ k − 1 ⎩ 1 ⎭ =

1 1 r k −1 ηth = 1 − k −1 kr r −1 ⎫⎪ 1⎧ 1 r k − 1⎫ 1⎧ r k −1 ⎫ 1 ⎧⎪ kr k −1 = 1 − ⎨lim k = − limηth = 1 − ⎨lim k −1 1 lim ⎬ ⎬ ⎨ ⎬ r →1 k ⎩ r →1 r r −1 ⎭ k ⎩ r →1 r − r k −1 ⎭ k ⎩⎪ r →1 kr k −1 − ( k − 1) r k − 2 ⎭⎪ 1⎧ k 1 ⎧k ⎫ ⎫ = 1− ⎨ ⎬ = 0 limηth = 1 − ⎨ ⎬ r →1 k ⎩ k − k + 1⎭ k ⎩1 ⎭ These results show that if there is no compression (i.e. r = 1), there can be no expansion and no net work will be done even though heat may be added to the system.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-161

9-185 The three processes of an air standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the expressions for back work ratio and the thermal efficiency are to be obtained. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Analysis (a) The P-v and T-s diagrams for this cycle are as shown.

(b) The work of expansion is found by the first law for process 2-3:

q2−3 − w2−3 = ∆u2−3

P

q2−3 = 0(isentropic process)

2

w2−3 = −∆u2−3 = −Cv (T3 − T2 ) wexp = w2−3 = Cv (T2 − T3 )

3

1

v

The compression work is found by 1

wcomp = − w3−1 = − ∫ Pdv = − P ( v1 − v3 ) = R (T3 − T1 ) 3

T 2

The back work ratio is

wcomp wexp

=

Cv (T3 − T1 ) R (T2 − T3 )

=

Cv T3 (1 − T1 / T3 ) Cv (1 − T1 / T3 ) = R T3 (T2 / T3 − 1) R (T2 / T3 − 1)

3 1

Process 3-1 is constant pressure; therefore,

s

PV 1 PV T V V 3 3 = 1 1⇒ 1 = 1 = 2 = T3 T1 T3 V3 V3 r Process 2-3 is isentropic; therefore,

T2 ⎛ V2 ⎞ =⎜ ⎟ T3 ⎝ V2 ⎠

k −1

k

=r

k −1

P ⎛V ⎞ k and 2 = ⎜ 3 ⎟ = r P3 ⎝ V2 ⎠

The back work ratio becomes (Cv=R/(k-1))

wcomp wexp

1 k −1 r −1 = ( k − 1) k −1 r = r −1 r r k −1 − 1 1−

(c) Apply first law to the closed system for processes 1-2 and 3-1 to show:

qin = Cv (T2 − T1 ) qout = C p (T3 − T1 ) The cycle thermal efficiency is given by

ηth = 1 −

C p (T3 − T1 ) T (T / T − 1) qout = 1− = 1− k 1 3 1 qin Cv (T2 − T1 ) T1 (T2 / T1 − 1)

Process 1-2 is constant volume; therefore,

PV PV T P P 2 2 = 1 1 ⇒ 2 = 2 = 2 = rk T2 T1 T1 P1 P3 The efficiency becomes PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-162

ηth = 1 − k

r −1 r k −1

(d) Determine the value of the back work ratio and efficiency as r goes to unity.

wcomp wexp lim r →1

lim r →1

wcomp wexp wcomp wexp

1 k −1 r −1 = ( k − 1) k −1 r = r −1 r r k −1 − 1 1−

⎧ ⎧ r −1 ⎫ 1 ⎫ = ( k − 1) ⎨lim k ⎬ = ( k − 1) ⎨lim ⎬ r →1 r − r r →1 kr k −1 − 1 ⎩ ⎭ ⎩ ⎭ ⎧ 1 ⎫ = ( k − 1) ⎨ ⎬ =1 ⎩ k − 1⎭

r −1 r k −1 ⎧ ⎧ r −1 ⎫ 1 ⎫ limηth = 1 − k ⎨lim k ⎬ = 1 − k ⎨lim k −1 ⎬ r →1 ⎩ r →1 r − 1 ⎭ ⎩ r →1 kr ⎭

ηth = 1 − k

⎧1 ⎫ limηth = 1 − k ⎨ ⎬ = 0 r →1 ⎩k ⎭ These results show that if there is no compression (i.e. r = 1), there can be no expansion and no net work will be done even though heat may be added to the system.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-163

9-186 The four processes of an air-standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams; an expression for the cycle thermal efficiency is to be obtained; and the limit of the efficiency as the volume ratio during heat rejection approaches unity is to be evaluated. Analysis (a) The P-v and T-s diagrams of the cycle are shown in the figures.

(b) Apply first law to the closed system for processes 2-3 and 4-1 to show:

qin = Cv (T3 − T2 ) qout = C p (T4 − T1 )

P 3

The cycle thermal efficiency is given by

ηth = 1 −

C p (T4 − T1 ) T (T / T − 1) qout = 1− = 1− k 1 4 1 qin Cv (T3 − T2 ) T2 (T3 / T2 − 1)

T1 ⎛ V2 ⎞ Process 1-2 is isentropic; therefore, =⎜ ⎟ T2 ⎝ V1 ⎠

k −1

T3 ⎛ V4 ⎞ Process 3-4 is isentropic; therefore, =⎜ ⎟ T4 ⎝ V3 ⎠

k −1

2 4 1

=

r k −1 3 k −1 e

=r

T 4

Process 4-1 is constant pressure; therefore, 2

PV PV T V 4 4 = 1 1 ⇒ 4 = 4 = rp T4 T1 T1 V1

T3 T3 T4 T1 1 ⎛r ⎞ = = rek −1rp k −1 = ⎜ e ⎟ T2 T4 T1 T2 r ⎝r⎠

v

1

k −1

1

rp

s

Since process 2-3 is constant volume and V3 = V2,

re =

V4 V4 V4 V1 = = = rp r V3 V2 V1 V2

T3 ⎛ rp r ⎞ =⎜ ⎟ T2 ⎝ r ⎠

k −1

rp = rpk

The efficiency becomes

ηth = 1 − k

1 rp − 1 r k −1 rpk − 1

(c) In the limit as rp approaches unity, the cycle thermal efficiency becomes

limηth = 1 − k

rp − 1 ⎫⎪ 1 ⎧⎪ 1 ⎧⎪ 1 ⎫⎪ lim ⎬ = 1 − k k −1 ⎨lim ⎬ k −1 ⎨ r →1 k r ⎩⎪ p rp − 1 ⎭⎪ r ⎩⎪ rp →1 krpk −1 ⎪⎭

limηth = 1 − k

1 ⎧1 ⎫ 1 = 1 − k −1 = ηth Otto k −1 ⎨ ⎬ r ⎩k ⎭ r

rp →1

rp →1

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-164

9-187 The four processes of an air-standard cycle are described. The back work ratio and its limit as rp goes to unity are to be determined, and the result is to be compared to the expression for the Otto cycle. Analysis The work of compression for process 1-2 is found by the first law:

q1− 2 − w1−2 = ∆u1− 2 q1− 2 = 0(isentropic process )

w1− 2 = −∆u1− 2 = −Cv (T2 − T1 ) wcomp ,1− 2 = − w1− 2 = Cv (T2 − T1 )

The work of compression for process 4-1 is found by 1

wcomp , 4−1 = − w4−1 = − ∫ Pdv = − P ( v1 − v4 ) = R (T4 − T1 ) 4

The work of expansion for process 3-4 is found by the first law:

q3− 4 − w3− 4 = ∆u3− 4 q3− 4 = 0(isentropic process)

w3− 4 = −∆u3− 4 = −Cv (T4 − T3 ) wexp,3− 4 = − w3− 4 = Cv (T3 − T4 )

The back work ratio is

wcomp wexp

R (T / T − 1) + (T2 / T1 − 1) R (T4 − T1 ) + Cv (T2 − T1 ) T1 Cv 4 1 = = Cv (T3 − T4 ) T3 (1 − T4 / T3 )

Using data from the previous problem and Cv = R/(k-1)

wcomp wexp

lim

rp →1

lim

rp →1

(

)

k −1 T1 (k − 1) ( rp − 1) + r − 1 = T3 ⎛ 1 ⎞ ⎜⎜1 − k −1 k −1 ⎟⎟ ⎝ rp r ⎠

wcomp wexp

wcomp wexp

⎧ ⎪ ⎪ (k − 1) ( rp − 1) + r k −1 − 1 T1 ⎪ = ⎨lim T3 ⎪ rp →1 ⎛ 1 ⎞ ⎜⎜1 − k −1 k −1 ⎟⎟ ⎪ ⎝ rp r ⎠ ⎪⎩

(

T = 1 T3

)

⎫ ⎧ ⎪ ⎪ k −1 ⎪ ⎪ T1 ⎪ ( k − 1)( 0 ) + r − 1 ⎬= ⎨ 1 ⎪ T3 ⎪ 1 − k −1 r ⎪ ⎪ ⎩ ⎪⎭

(

)

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

⎧ ⎫ ⎪ r k −1 − 1 ⎪ ⎨ ⎬ ⎪1 − 1k −1 ⎪ ⎩ r ⎭

This result is the same expression for the back work ratio for the Otto cycle.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-165

9-188 The effects of compression ratio on the net work output and the thermal efficiency of the Otto cycle for given operating conditions is to be investigated. Analysis Using EES, the problem is solved as follows: "Input Data" T[1]=300 [K] P[1]=100 [kPa] T[3] = 2000 [K] r_comp = 12 "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for process 1 to 2" q_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" v[3]=v[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process 2 to 3" q_23 - w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for process 3 to 4" q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant volume heat rejection" V[4] = V[1] "Conservation of energy for process 4 to 1" q_41 - w_41 = DELTAu_41 w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-166

ηth [%] 45.83 48.67 51.03 53.02 54.74 56.24 57.57 58.75 59.83 60.8

rcomp

wnet [kJ/kg] 567.4 589.3 604.9 616.2 624.3 630 633.8 636.3 637.5 637.9

6 7 8 9 10 11 12 13 14 15

640 630

w net [kJ/kg]

620 610 600 590 580 570 560 6

7

8

9

10

r

11

12

13

14

15

com p

62.5

59

η th [%]

55.5

52

48.5

45 6

7

8

9

10

11

12

13

14

15

r comp

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-167

9-189 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cycle is to be investigated. The pressure ratios at which the net work output and the thermal efficiency are maximum are to be determined. Analysis Using EES, the problem is solved as follows: P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 1800 [K] m_dot = 1 [kg/s] Eta_c = 100/100 Eta_t = 100/100 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature(air,h=h[2]) T[4]=temperature(air,h=h[4]) s[2]=entropy(air,T=T[2],P=P[2]) s[4]=entropy(air,T=T[4],P=P[4])

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-168

Bwr

η

Pratio

0.254 0.2665 0.2776 0.2876 0.2968 0.3052 0.313 0.3203 0.3272 0.3337 0.3398 0.3457 0.3513 0.3567 0.3618 0.3668 0.3716 0.3762 0.3806 0.385 0.3892 0.3932 0.3972 0.401 0.4048 0.4084 0.412 0.4155 0.4189 0.4222

0.3383 0.3689 0.3938 0.4146 0.4324 0.4478 0.4615 0.4736 0.4846 0.4945 0.5036 0.512 0.5197 0.5269 0.5336 0.5399 0.5458 0.5513 0.5566 0.5615 0.5663 0.5707 0.575 0.5791 0.583 0.5867 0.5903 0.5937 0.597 0.6002

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

Wnet [kW] 516.3 553.7 582.2 604.5 622.4 637 649 659.1 667.5 674.7 680.8 685.9 690.3 694.1 697.3 700 702.3 704.3 705.9 707.2 708.3 709.2 709.8 710.3 710.6 710.7 710.8 710.7 710.4 710.1

Wc [kW] 175.8 201.2 223.7 244.1 262.6 279.7 295.7 310.6 324.6 337.8 350.4 362.4 373.9 384.8 395.4 405.5 415.3 424.7 433.8 442.7 451.2 459.6 467.7 475.5 483.2 490.7 498 505.1 512.1 518.9

725

Wt [kW] 692.1 754.9 805.9 848.5 885 916.7 944.7 969.6 992.1 1013 1031 1048 1064 1079 1093 1106 1118 1129 1140 1150 1160 1169 1177 1186 1194 1201 1209 1216 1223 1229

Qin [kW] 1526 1501 1478 1458 1439 1422 1406 1392 1378 1364 1352 1340 1328 1317 1307 1297 1287 1277 1268 1259 1251 1243 1234 1227 1219 1211 1204 1197 1190 1183

0.65

700

0.6

675

Wnet [kW]

625

0.5

600

0.45

575

η th

0.55

650

0.4

550 0.35

525 500 5

10

15

20

25

30

0.3 35

Pratio

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-169

9-190 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cycle is to be investigated assuming adiabatic efficiencies of 85 percent for both the turbine and the compressor. The pressure ratios at which the net work output and the thermal efficiency are maximum are to be determined. Analysis Using EES, the problem is solved as follows: P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 1800 [K] m_dot = 1 [kg/s] Eta_c = 80/100 Eta_t = 80/100 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature(air,h=h[2]) T[4]=temperature(air,h=h[4]) s[2]=entropy(air,T=T[2],P=P[2]) s[4]=entropy(air,T=T[4],P=P[4])

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-170

Bwr

η

Pratio

0.3515 0.3689 0.3843 0.3981 0.4107 0.4224 0.4332 0.4433 0.4528 0.4618 0.4704 0.4785 0.4862 0.4937 0.5008 0.5077 0.5143 0.5207 0.5268 0.5328

0.2551 0.2764 0.2931 0.3068 0.3182 0.3278 0.3361 0.3432 0.3495 0.355 0.3599 0.3643 0.3682 0.3717 0.3748 0.3777 0.3802 0.3825 0.3846 0.3865

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Wnet [kW] 381.5 405 421.8 434.1 443.3 450.1 455.1 458.8 461.4 463.2 464.2 464.7 464.7 464.4 463.6 462.6 461.4 460 458.4 456.6

Wc [kW] 206.8 236.7 263.2 287.1 309 329.1 347.8 365.4 381.9 397.5 412.3 426.4 439.8 452.7 465.1 477.1 488.6 499.7 510.4 520.8

470

0.38

W net

η th

440

0.36 0.34

430 0.32 0.3

410 400 390 380 5

9

13

17

P

P ratio for

0.28

W

0.26

net,m ax 21

th

420

η

W net [kW ]

Qin [kW] 1495 1465 1439 1415 1393 1373 1354 1337 1320 1305 1290 1276 1262 1249 1237 1225 1214 1202 1192 1181

0.4

460 450

Wt [kW] 588.3 641.7 685 721.3 752.2 779.2 803 824.2 843.3 860.6 876.5 891.1 904.6 917.1 928.8 939.7 950 959.6 968.8 977.4

0.24 25

ratio

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-171

9-191 The effects of pressure ratio, maximum cycle temperature, and compressor and turbine inefficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid is to be investigated. Constant specific heats at room temperature are to be used. Analysis Using EES, the problem is solved as follows: Procedure ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) "For Air:" C_V = 0.718 [kJ/kg-K] k = 1.4 T2 = T[1]*r_comp^(k-1) P2 = P[1]*r_comp^k q_in_23 = C_V*(T[3]-T2) T4 = T[3]*(1/r_comp)^(k-1) q_out_41 = C_V*(T4-T[1]) Eta_th_ConstProp = (1-q_out_41/q_in_23)*Convert(, %) "[%]" "The Easy Way to calculate the constant property Otto cycle efficiency is:" Eta_th_easy = (1 - 1/r_comp^(k-1))*Convert(, %) "[%]" END "Input Data" T[1]=300 [K] P[1]=100 [kPa] {T[3] = 1000 [K]} r_comp = 12 "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for process 1 to 2" q_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" v[3]=v[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process 2 to 3" q_23 - w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for process 3 to 4" q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant volume heat rejection" V[4] = V[1] "Conservation of energy for process 4 to 1" q_41 - w_41 = DELTAu_41 PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-172

w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent" Call ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) PerCentError = ABS(Eta_th - Eta_th_ConstProp)/Eta_th*Convert(, %) "[%]"

PerCentError [%] 3.604 6.681 9.421 11.64

ηth [%] 60.8 59.04 57.57 56.42

rcomp 12 12 12 12

P e rc e n t E rro r = | η

th

ηth,ConstProp [%] 62.99 62.99 62.99 62.99

ηth,easy [%] 62.99 62.99 62.99 62.99

T3 [K] 1000 1500 2000 2500

- η |/η th ,C o n s tP ro p th

4 .3

PerCentError [%]

4 .2

T

m ax

= 1000 K

4 .1 4 3 .9 3 .8 3 .7 3 .6 6

9

8

7

r

10

12

11

com p

15

PerCentError [%]

r

comp

=6

12.8

=12 10.6

8.4

6.2

4 1000

1200

1400

1600

1800

2000

2200

2400

2600

T[3] [K]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-173

9-192 The effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid is to be investigated. Variable specific heats are to be used. Analysis Using EES, the problem is solved as follows: "Input data - from diagram window" {P_ratio = 8} {T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 800 [K] m_dot = 1 [kg/s] Eta_c = 75/100 Eta_t = 82/100} "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy(air,T=T[2],P=P[2]) s[4]=entropy(air,T=T[4],P=P[4])

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-174

Bwr

η

Pratio

0.5229 0.6305 0.7038 0.7611 0.8088 0.85 0.8864 0.9192 0.9491 0.9767

0.1 0.1644 0.1814 0.1806 0.1702 0.1533 0.131 0.1041 0.07272 0.03675

2 4 6 8 10 12 14 16 18 20

Wc [kW] 1818 4033 5543 6723 7705 8553 9304 9980 10596 11165

Wnet [kW] 1659 2364 2333 2110 1822 1510 1192 877.2 567.9 266.1

Wt [kW] 3477 6396 7876 8833 9527 10063 10496 10857 11164 11431

Qin [kW] 16587 14373 12862 11682 10700 9852 9102 8426 7809 7241

1500 Air Standard Brayton Cycle Pressure ratio = 8 and T max = 1160K 3

T [K]

1000

4

2 2s

4s

500 800 kPa 100 kPa

0 5.0

1

5.5

6.0

6.5

7.0

7.5

s [kJ/kg-K]

0.25

2500

η

Cycle efficiency,

η

W

0.15

T

0.00 2

t c

0.10

0.05

1500

η η

Note P

4

ratio

6

2000 net

m ax

= 0.82 = 0.75

1000

=1160 K 500

for m axim um w ork and η

8

10

12 P

14

W net [kW ]

0.20

16

18

0 20

ratio

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-175

9-193 The effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with helium as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: Function hFunc(WorkFluid\$,T,P) "The EES functions teat helium as a real gas; thus, T and P are needed for helium's enthalpy." IF WorkFluid\$ = 'Air' then hFunc:=enthalpy(Air,T=T) ELSE hFunc: = enthalpy(Helium,T=T,P=P) endif END Procedure EtaCheck(Eta_th:EtaError\$) If Eta_th < 0 then EtaError\$ = 'Why are the net work done and efficiency < 0?' Else EtaError\$ = '' END "Input data - from diagram window" {P_ratio = 8} {T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 800 [K] m_dot = 1 [kg/s] Eta_c = 0.8 Eta_t = 0.8 WorkFluid\$ = 'Helium'} "Inlet conditions" h[1]=hFunc(WorkFluid\$,T[1],P[1]) s[1]=ENTROPY(WorkFluid\$,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(WorkFluid\$,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=hFunc(WorkFluid\$,T_s[2],P[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=hFunc(WorkFluid\$,T[3],P[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(WorkFluid\$,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(WorkFluid\$,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=hFunc(WorkFluid\$,T_s[4],P[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta_th=W_dot_net/Q_dot_in"Cycle thermal efficiency" Call EtaCheck(Eta_th:EtaError\$) Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature(air,h=h[2]) T[4]=temperature(air,h=h[4]) s[2]=entropy(air,T=T[2],P=P[2]) s[4]=entropy(air,T=T[4],P=P[4]) PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-176

Bwr

η

Pratio

0.5229 0.6305 0.7038 0.7611 0.8088 0.85 0.8864 0.9192 0.9491 0.9767

0.1 0.1644 0.1814 0.1806 0.1702 0.1533 0.131 0.1041 0.07272 0.03675

2 4 6 8 10 12 14 16 18 20

Wc [kW] 1818 4033 5543 6723 7705 8553 9304 9980 10596 11165

Wnet [kW] 1659 2364 2333 2110 1822 1510 1192 877.2 567.9 266.1

Wt [kW] 3477 6396 7876 8833 9527 10063 10496 10857 11164 11431

Qin [kW] 16587 14373 12862 11682 10700 9852 9102 8426 7809 7241

1500 B r a yto n C yc le P r e s s u r e r a tio = 8 a n d T

m ax

= 1160K 3

T [K]

1000

4

2 2

4

s

s

500 800 kP a 100 kPa

0 5 .0

1

6 .0

5 .5

7 .5

7 .0

6 .5

s [k J /k g -K ]

0.25

Brayton Cycle using Air m air = 20 kg/s

0.20

2000

η

η

Wnet 0.15

0.10

0.05

0.00 2

1500

η = 0.82 t η = 0.75 c

1000

Tmax=1160 K

500

Note Pratio for maximum work and η

4

6

8

10 12 Pratio

14

Wnet [kW]

Cycle efficiency,

2500

16

18

0 20

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-177

9-194 The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and air as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: "Input data for air" C_P = 1.005 [kJ/kg-K] k = 1.4 "Nstages is the number of compression and expansion stages" Nstages = 1 T_6 = 1200 [K] Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa] Eta_reg = 1.0 "regenerator effectiveness" Eta_c =1.0 "Compressor isentorpic efficiency" Eta_t =1.0 "Turbine isentropic efficiency" R_p = Pratio^(1/Nstages) "Isentropic Compressor anaysis" T_2s = T_1*R_p^((k-1)/k) P_2 = R_p*P_1 "T_2s is the isentropic value of T_2 at compressor exit" Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_2s-T_1) "Actual compressor analysis:" w_comp = C_P*(T_2 - T_1) "Since intercooling is assumed to occur such that T_3 = T_1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is:" w_comp_total = Nstages*w_comp "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" "The heat added in the external heat exchanger + the reheat between turbines is" q_in_total = C_P*(T_6 - T_5) +(Nstages - 1)*C_P*(T_8 - T_7) "Reheat is assumed to occur until:" T_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p "T_7s is the isentropic value of T_7 at turbine exit" T_7s = T_6*(1/R_p)^((k-1)/k) "Turbine adiabatic efficiency, w_turbisen > w_turb" Eta_t = w_turb /w_turbisen "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" w_turbisen = C_P*(T_6 - T_7s) "Actual Turbine analysis:" w_turb = C_P*(T_6 - T_7) w_turb_total = Nstages*w_turb "Cycle analysis" w_net=w_turb_total-w_comp_total "[kJ/kg]" Bwr=w_comp/w_turb "Back work ratio" PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-178

P_4=P_2 P_5=P_4 P_6=P_5 T_4 = T_2 "The regenerator effectiveness gives T_5 as:" Eta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "Energy balance on regenerator gives T_10 as:" T_4 + T_9=T_5 + T_10 "Cycle thermal efficiency with regenerator" Eta_th_regenerative=w_net/q_in_total*Convert(, %) "[%]" "The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures, T_6 and T_1 for this problem." Eta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]"

ηth,Ericksson [%] 75 75 75 75 75 75 75 75

ηth,Regenerative [%] 49.15 64.35 68.32 70.14 72.33 73.79 74.05 74.18

Nstages 1 2 3 4 7 15 19 22

80

70

Ericsson η th [%]

60

Ideal Regenerative Brayton

50

40 0

2

4

6

8

10

12

14

16

18

20

22

24

Nstages

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-179

9-195 The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and helium as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: "Input data for Helium" C_P = 5.1926 [kJ/kg-K] k = 1.667 "Nstages is the number of compression and expansion stages" {Nstages = 1} T_6 = 1200 [K] Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa] Eta_reg = 1.0 "regenerator effectiveness" Eta_c =1.0 "Compressor isentorpic efficiency" Eta_t =1.0 "Turbine isentropic efficiency" R_p = Pratio^(1/Nstages) "Isentropic Compressor anaysis" T_2s = T_1*R_p^((k-1)/k) P_2 = R_p*P_1 "T_2s is the isentropic value of T_2 at compressor exit" Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_2s-T_1) "Actual compressor analysis:" w_comp = C_P*(T_2 - T_1) "Since intercooling is assumed to occur such that T_3 = T_1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is:" w_comp_total = Nstages*w_comp "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" "The heat added in the external heat exchanger + the reheat between turbines is" q_in_total = C_P*(T_6 - T_5) +(Nstages - 1)*C_P*(T_8 - T_7) "Reheat is assumed to occur until:" T_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p "T_7s is the isentropic value of T_7 at turbine exit" T_7s = T_6*(1/R_p)^((k-1)/k) "Turbine adiabatic efficiency, w_turbisen > w_turb" Eta_t = w_turb /w_turbisen "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" w_turbisen = C_P*(T_6 - T_7s) "Actual Turbine analysis:" w_turb = C_P*(T_6 - T_7) w_turb_total = Nstages*w_turb "Cycle analysis" w_net=w_turb_total-w_comp_total PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-180

Bwr=w_comp/w_turb "Back work ratio" P_4=P_2 P_5=P_4 P_6=P_5 T_4 = T_2 "The regenerator effectiveness gives T_5 as:" Eta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "Energy balance on regenerator gives T_10 as:" T_4 + T_9=T_5 + T_10 "Cycle thermal efficiency with regenerator" Eta_th_regenerative=w_net/q_in_total*Convert(, %) "[%]" "The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures, T_6 and T_1 for this problem." Eta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]"

ηth,Ericksson [%] 75 75 75 75 75 75 75 75

ηth,Regenerative [%] 32.43 58.9 65.18 67.95 71.18 73.29 73.66 73.84

Nstages 1 2 3 4 7 15 19 22

80

70

Ericsson

η th [%]

60

Ideal Regenerative Brayton

50

40

30 0

2

4

6

8

10

12

14

16

18

20

22

24

Nstages

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-181

Fundamentals of Engineering (FE) Exam Problems

9-196 An Otto cycle with air as the working fluid has a compression ratio of 10.4. Under cold air standard conditions, the thermal efficiency of this cycle is (a) 10%

(b) 39%

(c) 61%

(d) 79%

(e) 82%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). r=10.4 k=1.4 Eta_Otto=1-1/r^(k-1) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/r "Taking efficiency to be 1/r" W2_Eta = 1/r^(k-1) "Using incorrect relation" W3_Eta = 1-1/r^(k1-1); k1=1.667 "Using wrong k value"

9-197 For specified limits for the maximum and minimum temperatures, the ideal cycle with the lowest thermal efficiency is (a) Carnot

(b) Stirling

(c) Ericsson

(d) Otto

(e) All are the same

9-198 A Carnot cycle operates between the temperatures limits of 300 K and 2000 K, and produces 600 kW of net power. The rate of entropy change of the working fluid during the heat addition process is (a) 0

(b) 0.300 kW/K

(c) 0.353 kW/K

(d) 0.261 kW/K

(e) 2.0 kW/K

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=300 "K" TH=2000 "K" Wnet=600 "kJ/s" Wnet= (TH-TL)*DS "Some Wrong Solutions with Common Mistakes:" W1_DS = Wnet/TH "Using TH instead of TH-TL" W2_DS = Wnet/TL "Using TL instead of TH-TL" W3_DS = Wnet/(TH+TL) "Using TH+TL instead of TH-TL"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-182

9-199 Air in an ideal Diesel cycle is compressed from 2 L to 0.13 L, and then it expands during the constant pressure heat addition process to 0.30 L. Under cold air standard conditions, the thermal efficiency of this cycle is (a) 41%

(b) 59%

(c) 66%

(d) 70%

(e) 78%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V1=2 "L" V2= 0.13 "L" V3= 0.30 "L" r=V1/V2 rc=V3/V2 k=1.4 Eta_Diesel=1-(1/r^(k-1))*(rc^k-1)/k/(rc-1) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1-(1/r1^(k-1))*(rc^k-1)/k/(rc-1); r1=V1/V3 "Wrong r value" W2_Eta = 1-Eta_Diesel "Using incorrect relation" W3_Eta = 1-(1/r^(k1-1))*(rc^k1-1)/k1/(rc-1); k1=1.667 "Using wrong k value" W4_Eta = 1-1/r^(k-1) "Using Otto cycle efficiency"

9-200 Helium gas in an ideal Otto cycle is compressed from 20°C and 2.5 L to 0.25 L, and its temperature increases by an additional 700°C during the heat addition process. The temperature of helium before the expansion process is (a) 1790°C

(b) 2060°C

(c) 1240°C

(d) 620°C

(e) 820°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 V1=2.5 V2=0.25 r=V1/V2 T1=20+273 "K" T2=T1*r^(k-1) T3=T2+700-273 "C" "Some Wrong Solutions with Common Mistakes:" W1_T3 =T22+700-273; T22=T1*r^(k1-1); k1=1.4 "Using wrong k value" W2_T3 = T3+273 "Using K instead of C" W3_T3 = T1+700-273 "Disregarding temp rise during compression" W4_T3 = T222+700-273; T222=(T1-273)*r^(k-1) "Using C for T1 instead of K"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-183 3

9-201 In an ideal Otto cycle, air is compressed from 1.20 kg/m and 2.2 L to 0.26 L, and the net work output of the cycle is 440 kJ/kg. The mean effective pressure (MEP) for this cycle is (a) 612 kPa

(b) 599 kPa

(c) 528 kPa

(d) 416 kPa

(e) 367 kPa

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho1=1.20 "kg/m^3" k=1.4 V1=2.2 V2=0.26 m=rho1*V1/1000 "kg" w_net=440 "kJ/kg" Wtotal=m*w_net MEP=Wtotal/((V1-V2)/1000) "Some Wrong Solutions with Common Mistakes:" W1_MEP = w_net/((V1-V2)/1000) "Disregarding mass" W2_MEP = Wtotal/(V1/1000) "Using V1 instead of V1-V2" W3_MEP = (rho1*V2/1000)*w_net/((V1-V2)/1000); "Finding mass using V2 instead of V1" W4_MEP = Wtotal/((V1+V2)/1000) "Adding V1 and V2 instead of subtracting"

9-202 In an ideal Brayton cycle, air is compressed from 95 kPa and 25°C to 1100 kPa. Under cold air standard conditions, the thermal efficiency of this cycle is (a) 45%

(b) 50%

(c) 62%

(d) 73%

(e) 86%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=95 "kPa" P2=1100 "kPa" T1=25+273 "K" rp=P2/P1 k=1.4 Eta_Brayton=1-1/rp^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/rp "Taking efficiency to be 1/rp" W2_Eta = 1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-1/rp^((k1-1)/k1); k1=1.667 "Using wrong k value"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-184

9-203 Consider an ideal Brayton cycle executed between the pressure limits of 1200 kPa and 100 kPa and temperature limits of 20°C and 1000°C with argon as the working fluid. The net work output of the cycle is (a) 68 kJ/kg

(b) 93 kJ/kg

(c) 158 kJ/kg

(d) 186 kJ/kg

(e) 310 kJ/kg

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=100 "kPa" P2=1200 "kPa" T1=20+273 "K" T3=1000+273 "K" rp=P2/P1 k=1.667 Cp=0.5203 "kJ/kg.K" Cv=0.3122 "kJ/kg.K" T2=T1*rp^((k-1)/k) q_in=Cp*(T3-T2) Eta_Brayton=1-1/rp^((k-1)/k) w_net=Eta_Brayton*q_in "Some Wrong Solutions with Common Mistakes:" W1_wnet = (1-1/rp^((k-1)/k))*qin1; qin1=Cv*(T3-T2) "Using Cv instead of Cp" W2_wnet = (1-1/rp^((k-1)/k))*qin2; qin2=1.005*(T3-T2) "Using Cp of air instead of argon" W3_wnet = (1-1/rp^((k1-1)/k1))*Cp*(T3-T22); T22=T1*rp^((k1-1)/k1); k1=1.4 "Using k of air instead of argon" W4_wnet = (1-1/rp^((k-1)/k))*Cp*(T3-T222); T222=(T1-273)*rp^((k-1)/k) "Using C for T1 instead of K"

9-204 An ideal Brayton cycle has a net work output of 150 kJ/kg and a backwork ratio of 0.4. If both the turbine and the compressor had an isentropic efficiency of 85%, the net work output of the cycle would be (a) 74 kJ/kg

(b) 95 kJ/kg

(c) 109 kJ/kg

(d) 128 kJ/kg

(e) 177 kJ/kg

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). wcomp/wturb=0.4 wturb-wcomp=150 "kJ/kg" Eff=0.85 w_net=Eff*wturb-wcomp/Eff "Some Wrong Solutions with Common Mistakes:" W1_wnet = Eff*wturb-wcomp*Eff "Making a mistake in Wnet relation" W2_wnet = (wturb-wcomp)/Eff "Using a wrong relation" W3_wnet = wturb/eff-wcomp*Eff "Using a wrong relation"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-185

9-205 In an ideal Brayton cycle, air is compressed from 100 kPa and 25°C to 1 MPa, and then heated to 927°C before entering the turbine. Under cold air standard conditions, the air temperature at the turbine exit is (a) 349°C

(b) 426°C

(c) 622°C

(d) 733°C

(e) 825°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=100 "kPa" P2=1000 "kPa" T1=25+273 "K" T3=900+273 "K" rp=P2/P1 k=1.4 T4=T3*(1/rp)^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp "Using wrong relation" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T1+800-273 "Disregarding temp rise during compression"

9-206 In an ideal Brayton cycle with regeneration, argon gas is compressed from 100 kPa and 25°C to 400 kPa, and then heated to 1200°C before entering the turbine. The highest temperature that argon can be heated in the regenerator is (a) 246°C

(b) 846°C

(c) 689°C

(d) 368°C

(e) 573°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 Cp=0.5203 "kJ/kg.K" P1=100 "kPa" P2=400 "kPa" T1=25+273 "K" T3=1200+273 "K" "The highest temperature that argon can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2=T1*rp^((k-1)/k) T4=T3/rp^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T2-273 "Taking compressor exit temp as the answer"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-186

9-207 In an ideal Brayton cycle with regeneration, air is compressed from 80 kPa and 10°C to 400 kPa and 175°C, is heated to 450°C in the regenerator, and then further heated to 1000°C before entering the turbine. Under cold air standard conditions, the effectiveness of the regenerator is (a) 33%

(b) 44%

(c) 62%

(d) 77%

(e) 89%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 Cp=1.005 "kJ/kg.K" P1=80 "kPa" P2=400 "kPa" T1=10+273 "K" T2=175+273 "K" T3=1000+273 "K" T5=450+273 "K" "The highest temperature that the gas can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2check=T1*rp^((k-1)/k) "Checking the given value of T2. It checks." T4=T3/rp^((k-1)/k) Effective=(T5-T2)/(T4-T2) "Some Wrong Solutions with Common Mistakes:" W1_eff = (T5-T2)/(T3-T2) "Using wrong relation" W2_eff = (T5-T2)/(T44-T2); T44=(T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_eff = (T5-T2)/(T444-T2); T444=T3/rp "Using wrong relation for T4"

9-208 Consider a gas turbine that has a pressure ratio of 6 and operates on the Brayton cycle with regeneration between the temperature limits of 20°C and 900°C. If the specific heat ratio of the working fluid is 1.3, the highest thermal efficiency this gas turbine can have is (a) 38%

(b) 46%

(c) 62%

(d) 58%

(e) 97%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.3 rp=6 T1=20+273 "K" T3=900+273 "K" Eta_regen=1-(T1/T3)*rp^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1-((T1-273)/(T3-273))*rp^((k-1)/k) "Using C for temperatures instead of K" W2_Eta = (T1/T3)*rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-(T1/T3)*rp^((k1-1)/k1); k1=1.4 "Using wrong k value (the one for air)"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-187

9-209 An ideal gas turbine cycle with many stages of compression and expansion and a regenerator of 100 percent effectiveness has an overall pressure ratio of 10. Air enters every stage of compressor at 290 K, and every stage of turbine at 1200 K. The thermal efficiency of this gas-turbine cycle is (a) 36%

(b) 40%

(c) 52%

(d) 64%

(e) 76%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 rp=10 T1=290 "K" T3=1200 "K" Eff=1-T1/T3 "Some Wrong Solutions with Common Mistakes:" W1_Eta = 100 W2_Eta = 1-1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-(T1/T3)*rp^((k-1)/k) "Using wrong relation" W4_Eta = T1/T3 "Using wrong relation"

9-210 Air enters a turbojet engine at 320 m/s at a rate of 30 kg/s, and exits at 650 m/s relative to the aircraft. The thrust developed by the engine is (a) 5 kN

(b) 10 kN

(c) 15 kN

(d) 20 kN

(e) 26 kN

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Vel1=320 "m/s" Vel2=650 "m/s" Thrust=m*(Vel2-Vel1)/1000 "kN" m= 30 "kg/s" "Some Wrong Solutions with Common Mistakes:" W1_thrust = (Vel2-Vel1)/1000 "Disregarding mass flow rate" W2_thrust = m*Vel2/1000 "Using incorrect relation"

9-211 ··· 9-219 Design and Essay Problems.

KJ PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-1

Solutions Manual for

Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011

Chapter 10 VAPOR AND COMBINED POWER CYCLES

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-2

Carnot Vapor Cycle 10-1C The Carnot cycle is not a realistic model for steam power plants because (1) limiting the heat transfer processes to two-phase systems to maintain isothermal conditions severely limits the maximum temperature that can be used in the cycle, (2) the turbine will have to handle steam with a high moisture content which causes erosion, and (3) it is not practical to design a compressor that will handle two phases.

10-2E A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the quality at the end of the heat rejection process, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We note that

TH = Tsat @ 250 psia = 401°F = 861 R

T

TL = Tsat @ 40psia = 267.2°F = 727.2 R and

η th,C = 1 −

TL 727.2 R =1− = 0.1553 = 15.5% 861 R TH

1 250 psia 2 qin

(b) Noting that s4 = s1 = sf @ 250 psia = 0.56784 Btu/lbm·R,

x4 =

s4 − s f s fg

=

0.56784 − 0.3921 = 0.137 1.2845

40 psia 4

3 s

(c) The enthalpies before and after the heat addition process are h1 = h f @ 250 psia = 376.09 Btu/lbm

h2 = h f + x 2 h fg = 376.09 + (0.95)(825.47 ) = 1160.3 Btu/lbm

Thus, q in = h2 − h1 = 1160.3 − 376.09 = 784.2 Btu/lbm

and

wnet = η th q in = (0.1553)(784.2 Btu/lbm ) = 122 Btu/lbm

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-3

10-3 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the amount of heat rejected, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH = 250°C = 523 K and TL = Tsat @ 20 kPa = 60.06°C = 333.1 K, the thermal efficiency becomes

η th,C = 1 −

TL 333.1 K = 1− = 0.3632 = 36.3% 523 K TH

T

(b) The heat supplied during this cycle is simply the enthalpy of vaporization, q in = h fg @ 250oC = 1715.3 kJ/kg

250°C

2

1 qin

Thus, q out = q L =

⎛ 333.1 K ⎞ TL ⎟⎟(1715.3 kJ/kg ) = 1092.3 kJ/kg q in = ⎜⎜ TH ⎝ 523 K ⎠

20 kPa 4

qout

3 s

(c) The net work output of this cycle is wnet = η th q in = (0.3632 )(1715.3 kJ/kg ) = 623.0 kJ/kg

10-4 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the amount of heat rejected, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH = 250°C = 523 K and TL = Tsat @ 10 kPa = 45.81°C = 318.8 K, the thermal efficiency becomes

η th, C = 1 −

TL 318.8 K =1− = 39.04% 523 K TH

T

(b) The heat supplied during this cycle is simply the enthalpy of vaporization, q in = h fg @ 250°C = 1715.3 kJ/kg

250°C

qin

Thus, q out = q L =

⎛ 318.8 K ⎞ TL ⎟⎟(1715.3 kJ/kg ) = 1045.6 kJ/kg q in = ⎜⎜ TH ⎝ 523 K ⎠

(c) The net work output of this cycle is wnet = η th q in = (0.3904)(1715.3 kJ/kg ) = 669.7 kJ/kg

2

1

10 kPa 4

qout

3 s

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-4

10-5 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the pressure at the turbine inlet, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The thermal efficiency is determined from

η th, C = 1 −

TL 60 + 273 K = 1− = 46.5% TH 350 + 273 K

T

(b) Note that s2 = s3 = sf + x3sfg

350°C

1

2

4

3

= 0.8313 + 0.891 × 7.0769 = 7.1368 kJ/kg·K Thus,

60°C T2 = 350°C

⎫ ⎬ P2 ≅ 1.40 MPa (Table A-6) s 2 = 7.1368 kJ/kg ⋅ K ⎭

s

(c) The net work can be determined by calculating the enclosed area on the T-s diagram,

s 4 = s f + x 4 s fg = 0.8313 + (0.1)(7.0769) = 1.5390 kJ/kg ⋅ K Thus,

wnet = Area = (TH − TL )(s3 − s 4 ) = (350 − 60)(7.1368 − 1.5390) = 1623 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-5

The Simple Rankine Cycle 10-6C The four processes that make up the simple ideal cycle are (1) Isentropic compression in a pump, (2) P = constant heat addition in a boiler, (3) Isentropic expansion in a turbine, and (4) P = constant heat rejection in a condenser.

10-7C Heat rejected decreases; everything else increases.

10-8C Heat rejected decreases; everything else increases.

10-9C The pump work remains the same, the moisture content decreases, everything else increases.

10-10C The actual vapor power cycles differ from the idealized ones in that the actual cycles involve friction and pressure drops in various components and the piping, and heat loss to the surrounding medium from these components and piping.

10-11C The boiler exit pressure will be (a) lower than the boiler inlet pressure in actual cycles, and (b) the same as the boiler inlet pressure in ideal cycles.

10-12C We would reject this proposal because wturb = h1 - h2 - qout, and any heat loss from the steam will adversely affect the turbine work output.

10-13C Yes, because the saturation temperature of steam at 10 kPa is 45.81°C, which is much higher than the temperature of the cooling water.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-6

10-14 A simple ideal Rankine cycle with R-134a as the working fluid operates between the specified pressure limits. The mass flow rate of R-134a for a given power production and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), h1 = h f @ 0.4 MPa = 63.94 kJ/kg

v 1 = v f @ 0.4 MPa = 0.0007907 m 3 /kg

T

w p,in = v 1 ( P2 − P1 )

⎛ 1 kJ ⎞ = (0.0007907 m 3 /kg )(1600 − 400)kPa ⎜ ⎟ 1 kPa ⋅ m 3 ⎠ ⎝ = 0.95 kJ/kg

h2 = h1 + w p,in = 63.94 + 0.95 = 64.89 kJ/kg

P3 = 1.6 MPa ⎫ h3 = 305.07 kJ/kg ⎬ T3 = 80°C ⎭ s 3 = 0.9875 kJ/kg ⋅ K P4 = 0.4 MPa ⎫ ⎬ h4 = 273.21 kJ/kg s 4 = s3 ⎭

1.6 MPa

3

qin

2

0.4 MPa

1

qout

4 s

Thus, q in = h3 − h2 = 305.07 − 64.89 = 240.18 kJ/kg q out = h4 − h1 = 273.21 − 63.94 = 209.27 kJ/kg w net = q in − q out = 240.18 − 209.27 = 30.91 kJ/kg

The mass flow rate of the refrigerant and the thermal efficiency of the cycle are then m& =

W& net 750 kJ/s = = 24.26 kg/s wnet 30.91 kJ/kg

η th = 1 −

q out 209.27 = 1− = 0.129 240.18 q in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-7

10-15 A simple ideal Rankine cycle with R-134a as the working fluid is considered. The turbine inlet temperature, the cycle thermal efficiency, and the back-work ratio of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = Psat @ 10°C = 414.89 kPa

T

h1 = h f @ 10°C = 65.43 kJ/kg

v 1 = v f @ 10°C = 0.0007930 m 3 /kg

1.4 MPa

qin

2

w p,in = v 1 ( P2 − P1 )

⎛ 1 kJ ⎞ = (0.0007930 m 3 /kg )(1400 − 414.89)kPa ⎜ ⎟ 1 kPa ⋅ m 3 ⎠ ⎝ = 0.78 kJ/kg h2 = h1 + w p,in = 65.43 + 0.78 = 66.21 kJ/kg

3

10°C

1

qout

4 s

T4 = 10°C ⎫ h4 = h f + x 4 h fg = 65.43 + (0.98)(190.73) = 252.35 kJ/kg ⎬ x 4 = 0.98 ⎭ s 4 = s f + x 4 s fg = 0.25286 + (0.98)(0.67356) = 0.91295 kJ/kg ⋅ K P3 = 1400 kPa

⎫ h3 = 276.91 kJ/kg ⎬ s 3 = s 4 = 0.91295 kJ/kg ⋅ K ⎭ T3 = 53.0°C

Thus, q in = h3 − h2 = 276.91 − 66.21 = 210.70 kJ/kg q out = h4 − h1 = 252.35 − 65.43 = 186.92 kJ/kg

The thermal efficiency of the cycle is

η th = 1 −

q out 186.92 = 1− = 0.113 210.70 q in

The back-work ratio is determined from rbw =

wP,in WT,out

=

wP,in h3 − h 4

=

0.78 kJ/kg = 0.0318 (276.91 − 252.35) kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-8

10-16 A simple ideal Rankine cycle with water as the working fluid is considered. The work output from the turbine, the heat addition in the boiler, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6),

P1 = Psat @ 40°C = 7.385 kPa

T

P2 = Psat @ 300°C = 8588 kPa h1 = h f @ 40°C = 167.53 kJ/kg 3

v 1 = v f @ 40°C = 0.001008 m /kg

300°C

qin

2

3

40°C

wp,in = v 1 ( P2 − P1 )

⎛ 1 kJ ⎞ = (0.001008 m /kg)(8588 − 7.385)kPa ⎜ ⎟ 1 kPa ⋅ m 3 ⎠ ⎝ = 8.65 kJ/kg h2 = h1 + wp,in = 167.53 + 8.65 = 176.18 kJ/kg 3

1

qout 4 s

T3 = 300°C ⎫ h3 = 2749.6 kJ/kg ⎬ x3 = 1 ⎭ s 3 = 5.7059 kJ/kg ⋅ K s4 − s f 5.7059 − 0.5724 T4 = 40°C ⎫ x 4 = = = 0.6681 s 7.6832 ⎬ fg s 4 = s3 ⎭ h = h + x h = 167.53 + (0.6681)(2406.0) = 1775.1 kJ/kg 4 f 4 fg

Thus, wT, out = h3 − h4 = 2749.6 − 1775.1 = 974.5 kJ/kg q in = h3 − h2 = 2749.6 − 176.18 = 2573.4 kJ/kg q out = h4 − h1 = 1775.1 − 167.53 = 1607.6 kJ/kg

The thermal efficiency of the cycle is

η th = 1 −

q out 1607.6 = 1− = 0.375 2573.4 q in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-9

10-17E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The rates of heat addition and rejection, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4E, A-5E, and A-6E),

h1 = h f @ 3 psia = 109.40 Btu/lbm

T

v 1 = v f @ 3 psia = 0.01630 ft 3 /lbm wp,in = v 1 ( P2 − P1 )

⎛ 1 Btu = (0.01630 ft 3 /lbm)(800 − 3)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 2.40 Btu/lbm

⎞ ⎟ ⎟ ⎠

800 psia 2

h2 = h1 + wp,in = 109.40 + 2.40 = 111.81 Btu/lbm 1 P3 = 800 psia ⎫ h3 = 1456.0 Btu/lbm ⎬ T3 = 900°F ⎭ s 3 = 1.6413 Btu/lbm ⋅ R s 4 − s f 1.6413 − 0.2009 P4 = 3 psia ⎫ x 4 = = = 0.8549 s fg 1.6849 ⎬ s 4 = s3 ⎭ h = h + x h = 109.40 + (0.8549)(1012.8) = 975.24 Btu/lbm 4 f 4 fg

3

qin 3 psia qout

4 s

Knowing the power output from the turbine the mass flow rate of steam in the cycle is determined from W& T,out 1750 kJ/s ⎛ 0.94782 Btu ⎞ W& T,out = m& (h3 − h4 ) ⎯ ⎯→ m& = = ⎜ ⎟ = 3.450 lbm/s h3 − h4 (1456.0 − 975.24)Btu/lbm ⎝ 1 kJ ⎠

The rates of heat addition and rejection are Q& in = m& (h3 − h2 ) = (3.450 lbm/s)(1456.0 − 111.81)Btu/lbm = 4637 Btu/s Q& out = m& (h4 − h1 ) = (3.450 lbm/s)(975.24 − 109.40)Btu/lbm = 2987 Btu/s

and the thermal efficiency of the cycle is

η th = 1 −

Q& out 2987 =1− = 0.3559 = 35.6% & 4637 Qin

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-10

10-18E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The turbine inlet temperature and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4E, A-5E, and A-6E),

T

h1 = h f @ 5 psia = 130.18 Btu/lbm

2500 psia

v 1 = v f @ 5 psia = 0.01641 ft 3 /lbm wp,in = v 1 ( P2 − P1 )

⎛ 1 Btu = (0.01641 ft 3 /lbm)(2500 − 5)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 7.58 Btu/lbm

qin

2 ⎞ ⎟ ⎟ ⎠

3

5 psia 1

h2 = h1 + wp,in = 130.18 + 7.58 = 137.76 Btu/lbm

qout

4 s

P4 = 5 psia ⎫ h4 = h f + x 4 h fg = 130.18 + (0.80)(1000.5) = 930.58 Btu/lbm ⎬ x 4 = 0.80 ⎭ s 4 = s f + x 4 s fg = 0.23488 + (0.80)(1.60894) = 1.52203 Btu/lbm ⋅ R P3 = 2500 psia

⎫ h3 = 1450.8 Btu/lbm ⎬ s 3 = s 4 = 1.52203 Btu/lbm ⋅ R ⎭ T3 = 989.2 °F

Thus, q in = h3 − h2 = 1450.8 − 137.76 = 1313.0 Btu/lbm q out = h4 − h1 = 930.58 − 130.18 = 800.4 Btu/lbm

The thermal efficiency of the cycle is

η th = 1 −

q out 800.4 = 1− = 0.390 1313.0 q in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-11

10-19E A simple steam Rankine cycle operates between the specified pressure limits. The mass flow rate, the power produced by the turbine, the rate of heat addition, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4E, A-5E, and A-6E),

h1 = h f @ 1 psia = 69.72 Btu/lbm

T

v 1 = v f @ 6 psia = 0.01614 ft 3 /lbm wp,in = v 1 ( P2 − P1 )

⎛ 1 Btu = (0.01614 ft 3 /lbm)(2500 − 1)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 7.46 Btu/lbm

⎞ ⎟ ⎟ ⎠

h2 = h1 + wp,in = 69.72 + 7.46 = 77.18 Btu/lbm

qin

2

1 psia 1

P3 = 2500 psia ⎫ h3 = 1302.0 Btu/lbm ⎬ T3 = 800°F ⎭ s 3 = 1.4116 Btu/lbm ⋅ R s 4 − s f 1.4116 − 0.13262 P4 = 1 psia ⎫ x 4 s = = = 0.6932 s fg 1.84495 ⎬ s 4 = s3 ⎭ h = h + x h = 69.72 + (0.6932)(1035.7) = 787.70 Btu/lbm 4s f 4 s fg

ηT =

3

2500 psia

qout

4s 4 s

h3 − h 4 ⎯ ⎯→ h4 = h3 − η T (h3 − h4s ) = 1302.0 − (0.90)(1302.0 − 787.70) = 839.13 kJ/kg h3 − h 4 s

Thus, q in = h3 − h2 = 1302.0 − 77.18 = 1224.8 Btu/lbm q out = h4 − h1 = 839.13 − 69.72 = 769.41 Btu/lbm w net = q in − q out = 1224.8 − 769.41 = 455.39 Btu/lbm

The mass flow rate of steam in the cycle is determined from W& 1000 kJ/s ⎛ 0.94782 Btu ⎞ ⎯→ m& = net = W& net = m& wnet ⎯ ⎜ ⎟ = 2.081 lbm/s 1 kJ wnet 455.39 Btu/lbm ⎝ ⎠ The power output from the turbine and the rate of heat addition are 1 kJ ⎛ ⎞ W& T,out = m& (h3 − h4 ) = (2.081 lbm/s)(1302.0 − 839.13)Btu/lbm⎜ ⎟ = 1016 kW 0.94782 Btu ⎝ ⎠ Q& in = m& q in = (2.081 lbm/s)(1224.8 Btu/lbm) = 2549 Btu/s and the thermal efficiency of the cycle is

η th =

W& net 1000 kJ/s ⎛ 0.94782 Btu ⎞ = ⎜ ⎟ = 0.3718 2549 Btu/s ⎝ 1 kJ Q& in ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-12

10-20E A simple steam Rankine cycle operates between the specified pressure limits. The mass flow rate, the power produced by the turbine, the rate of heat addition, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4E, A-5E, and A-6E),

h1 = h f @ 1 psia = 69.72 Btu/lbm T

v 1 = v f @ 6 psia = 0.01614 ft 3 /lbm wp,in = v 1 ( P2 − P1 )

⎛ 1 Btu = (0.01614 ft 3 /lbm)(2500 − 1)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 7.46 Btu/lbm

⎞ ⎟ ⎟ ⎠

2500 psia 2

h2 = h1 + wp,in = 69.72 + 7.46 = 77.18 Btu/lbm 1 P3 = 2500 psia ⎫ h3 = 1302.0 Btu/lbm ⎬ T3 = 800°F ⎭ s 3 = 1.4116 Btu/lbm ⋅ R s 4 − s f 1.4116 − 0.13262 P4 = 1 psia ⎫ x 4 s = = = 0.6932 s fg 1.84495 ⎬ s 4 = s3 ⎭ h = h + x h = 69.72 + (0.6932)(1035.7) = 787.70 Btu/lbm 4s f 4 s fg

ηT =

3

qin 1 psia qout

4s 4 s

h3 − h 4 ⎯ ⎯→ h4 = h3 − η T (h3 − h4s ) = 1302.0 − (0.90)(1302.0 − 787.70) = 839.13 kJ/kg h3 − h 4 s

The mass flow rate of steam in the cycle is determined from

W& net 1000 kJ/s ⎛ 0.94782 Btu ⎞ ⎯→ m& = = W& net = m& (h3 − h4 ) ⎯ ⎜ ⎟ = 2.048 lbm/s h3 − h4 (1302.0 − 839.13) Btu/lbm ⎝ 1 kJ ⎠ The rate of heat addition is 1 kJ ⎛ ⎞ Q& in = m& ( h3 − h2 ) = ( 2.048 lbm/s)(1302.0 − 77.18) Btu/lbm⎜ ⎟ = 2508 Btu/s ⎝ 0.94782 Btu ⎠

and the thermal efficiency of the cycle is

η th =

W& net 1000 kJ/s ⎛ 0.94782 Btu ⎞ = ⎜ ⎟ = 0.3779 & 2508 Btu/s ⎝ 1 kJ Qin ⎠

The thermal efficiency in the previous problem was determined to be 0.3718. The error in the thermal efficiency caused by neglecting the pump work is then Error =

0.3779 − 0.3718 × 100 = 1.64% 0.3718

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-13

10-21 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

T

v 1 = v f @ 10 kPa = 0.00101 m 3 /kg

w p ,in = v 1 (P2 − P1 )

(

)

⎛ 1 kJ = 0.00101 m /kg (7,000 − 10 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 7.06 kJ/kg 3

⎞ ⎟ ⎟ ⎠

h2 = h1 + w p ,in = 191.81 + 7.06 = 198.87 kJ/kg P3 = 7 MPa ⎫ h3 = 3411.4 kJ/kg ⎬ T3 = 500°C ⎭ s 3 = 6.8000 kJ/kg ⋅ K

3 2

7 MPa qin 10 kPa

1

qout

4 s

s 4 − s f 6.8000 − 0.6492 P4 = 10 kPa ⎫ = = 0.8201 ⎬ x4 = s 4 = s3 s fg 7.4996 ⎭

h4 = h f + x 4 h fg = 191.81 + (0.8201)(2392.1) = 2153.6 kJ/kg

Thus, q in = h3 − h2 = 3411.4 − 198.87 = 3212.5 kJ/kg q out = h4 − h1 = 2153.6 − 191.81 = 1961.8 kJ/kg wnet = q in − q out = 3212.5 − 1961.8 = 1250.7 kJ/kg

and

η th = (b)

m& =

wnet 1250.7 kJ/kg = = 38.9% 3212.5 kJ/kg q in

45,000 kJ/s W&net = = 36.0 kg/s wnet 1250.7 kJ/kg

(c) The rate of heat rejection to the cooling water and its temperature rise are Q& out = m& q out = (35.98 kg/s )(1961.8 kJ/kg ) = 70,586 kJ/s Q& out 70,586 kJ/s ∆Tcooling water = = = 8.4°C (m& c) cooling water (2000 kg/s )(4.18 kJ/kg ⋅ °C )

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-14

10-22 A steam power plant operates on a simple nonideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

T

v 1 = v f @ 10 kPa = 0.00101 m 3 /kg w p ,in = v 1 (P2 − P1 ) / η p

(

)

⎛ 1 kJ = 0.00101 m 3 /kg (7,000 − 10 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 8.11 kJ/kg

⎞ ⎟ / (0.87 ) ⎟ ⎠

2

2

h2 = h1 + w p ,in = 191.81 + 8.11 = 199.92 kJ/kg P3 = 7 MPa ⎫ h3 = 3411.4 kJ/kg ⎬ T3 = 500°C ⎭ s 3 = 6.8000 kJ/kg ⋅ K

7 MPa qin

3

10 kPa 1

qout

4 4 s

s 4 − s f 6.8000 − 0.6492 P4 = 10 kPa ⎫ = = 0.8201 ⎬ x4 = s 4 = s3 7.4996 s fg ⎭

h4 s = h f + x 4 h fg = 191.81 + (0.820 )(2392.1) = 2153.6 kJ/kg

ηT =

h3 − h4 ⎯ ⎯→ h4 = h3 − ηT (h3 − h4 s ) h3 − h4 s = 3411.4 − (0.87 )(3411.4 − 2153.6) = 2317.1 kJ/kg

Thus, qin = h3 − h2 = 3411.4 − 199.92 = 3211.5 kJ/kg qout = h4 − h1 = 2317.1 − 191.81 = 2125.3 kJ/kg wnet = qin − qout = 3211.5 − 2125.3 = 1086.2 kJ/kg

and

η th = (b)

m& =

wnet 1086.2 kJ/kg = = 33.8% 3211.5 kJ/kg q in

W&net 45,000 kJ/s = = 41.43 kg/s wnet 1086.2 kJ/kg

(c) The rate of heat rejection to the cooling water and its temperature rise are

Q& out = m& q out = (41.43 kg/s )(2125.3 kJ/kg ) = 88,051 kJ/s ∆Tcooling water =

Q& out (m& c) cooling water

=

88,051 kJ/s = 10.5°C (2000 kg/s)(4.18 kJ/kg ⋅ °C)

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-15

10-23 A simple Rankine cycle with water as the working fluid operates between the specified pressure limits. The rate of heat addition in the boiler, the power input to the pumps, the net power, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), P1 = 50 kPa T1 = Tsat @ 50 kPa

⎫ h1 ≅ h f @ 75°C = 314.03 kJ/kg ⎬ − 6.3 = 81.3 − 6.3 = 75°C ⎭ v 1 = v f @ 75°C = 0.001026 m 3 /kg

w p,in = v 1 ( P2 − P1 )

⎛ 1 kJ ⎞ = (0.001026 m 3 /kg )(6000 − 50)kPa ⎜ ⎟ 1 kPa ⋅ m 3 ⎠ ⎝ = 6.10 kJ/kg

h2 = h1 + wp,in = 314.03 + 6.10 = 320.13 kJ/kg

T 6 MPa

2

P3 = 6000 kPa ⎫ h3 = 3302.9 kJ/kg ⎬ 1 T3 = 450°C ⎭ s 3 = 6.7219 kJ/kg ⋅ K s4 − s f 6.7219 − 1.0912 P4 = 50 kPa ⎫ x 4 s = = = 0.8660 6.5019 s fg ⎬ s 4 = s3 ⎭ h = h + x h = 340.54 + (0.8660)(2304.7) = 2336.4 kJ/kg 4s f 4 s fg

ηT =

3

qin 50 kPa qout 4s 4 s

h3 − h 4 ⎯ ⎯→ h4 = h3 − η T (h3 − h4s ) = 3302.9 − (0.94)(3302.9 − 2336.4) = 2394.4 kJ/kg h3 − h 4 s

Thus, Q& in = m& (h3 − h2 ) = (20 kg/s)(3302.9 − 320.13)kJ/kg = 59,660 kW W& T,out = m& (h3 − h4 ) = (20 kg/s)(3302.9 − 2394.4)kJ/kg = 18,170 kW W& P,in = m& wP,in = (20 kg/s)(6.10 kJ/kg) = 122 kW W& = W& − W& = 18,170 − 122 = 18,050 kW net

T, out

P,in

and

η th =

W& net 18,050 = = 0.3025 59,660 Q& in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-16

10-24 The change in the thermal efficiency of the cycle in Prob. 10-23 due to a pressure drop in the boiler is to be determined. Analysis We use the following EES routine to obtain the solution. "Given" P[2]=6000 [kPa] DELTAP=50 [kPa] P[3]=6000-DELTAP [kPa] T[3]=450 [C] P[4]=50 [kPa] Eta_T=0.94 DELTAT_subcool=6.3 [C] T[1]=temperature(Fluid\$, P=P[1], x=x[1])-DELTAT_subcool m_dot=20 [kg/s] "Analysis" Fluid\$='steam_iapws' P[1]=P[4] x[1]=0 h[1]=enthalpy(Fluid\$, P=P[1], T=T[1]) v[1]=volume(Fluid\$, P=P[1], T=T[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid\$, P=P[3], T=T[3]) s[3]=entropy(Fluid\$, P=P[3], T=T[3]) s[4]=s[3] h_s[4]=enthalpy(Fluid\$, P=P[4], s=s[4]) Eta_T=(h[3]-h[4])/(h[3]-h_s[4]) q_in=h[3]-h[2] q_out=h[4]-h[1] w_net=q_in-q_out Eta_th=1-q_out/q_in Solution DELTAP=50 [kPa] Eta_th=0.3022 h[2]=320.21 [kJ/kg] h_s[4]=2338.1 [kJ/kg] P[2]=6000 q_in=2983.4 [kJ/kg] s[4]=6.7265 [kJ/kg-K] v[1]=0.001026 [m^3/kg] x[1]=0

DELTAT_subcool=6.3 [C] Fluid\$='steam_iapws' h[3]=3303.64 [kJ/kg] m_dot=20 [kg/s] P[3]=5950 q_out=2081.9 [kJ/kg] T[1]=75.02 [C] w_net=901.5 [kJ/kg]

Eta_T=0.94 h[1]=314.11 [kJ/kg] h[4]=2396.01 [kJ/kg] P[1]=50 P[4]=50 s[3]=6.7265 [kJ/kg-K] T[3]=450 [C] w_p_in=6.104 [kJ/kg]

Discussion The thermal efficiency without a pressure drop was obtained to be 0.3025.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-17

10-25 The net work outputs and the thermal efficiencies for a Carnot cycle and a simple ideal Rankine cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Rankine cycle analysis: From the steam tables (Tables A-4, A-5, and A-6),

h1 = h f @ 50 kPa = 340.54 kJ/kg

v 1 = v f @ 50 kPa = 0.001030 m 3 /kg w p ,in = v 1 (P2 − P1 )

(

)

⎛ 1 kJ = 0.001030 m 3 /kg (5000 − 50 ) kPa ⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 5.10 kJ/kg

T

Rankine cycle

⎞ ⎟ ⎟ ⎠

h2 = h1 + w p ,in = 340.54 + 5.10 = 345.64 kJ/kg

3 2

P3 = 5 MPa ⎫ h3 = 2794.2 kJ/kg ⎬ x3 = 1 ⎭ s 3 = 5.9737 kJ/kg ⋅ K

4

1

s

s 4 − s f 5.9737 − 1.09120 P4 = 50 kPa ⎫ = = 0.7509 ⎬ x4 = s 4 = s3 6.5019 s fg ⎭ h4 = h f + x 4 h fg = 340.54 + (0.7509 )(2304.7 ) = 2071.2 kJ/kg q in = h3 − h2 = 2794.2 − 345.64 = 2448.6 kJ/kg q out = h4 − h1 = 2071.2 − 340.54 = 1730.7 kJ/kg wnet = q in − q out = 2448.6 − 1730.7 = 717.9 kJ/kg

η th = 1 −

q out 1730.7 =1− = 0.2932 = 29.3% q in 2448.6

(b) Carnot Cycle analysis: P3 = 5 MPa ⎫ h3 = 2794.2 kJ/kg ⎬ x3 = 1 ⎭ T3 = 263.9°C T2 = T3 = 263.9°C ⎫ h2 = 1154.5 kJ/kg ⎬ x2 = 0 ⎭ s 2 = 2.9207 kJ/kg ⋅ K x1 =

s1 − s f

=

2.9207 − 1.0912 = 0.2814 6.5019

P1 = 50 kPa ⎫ s fg ⎬h = h + x h s1 = s 2 f 1 fg ⎭ 1 = 340.54 + (0.2814)(2304.7) = 989.05 kJ/kg

T

Carnot cycle 2

1

3

4 s

q in = h3 − h2 = 2794.2 − 1154.5 = 1639.7 kJ/kg q out = h4 − h1 = 2071.2 − 340.54 = 1082.2 kJ/kg wnet = q in − q out = 1639.7 − 1082.2 = 557.5 kJ/kg

η th = 1 −

q out 1082.2 =1− = 0.3400 = 34.0% 1639.7 q in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-18

10-26 A 120-MW coal-fired steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The overall plant efficiency and the required rate of the coal supply are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

h1 = h f @ 15 kPa = 225.94 kJ/kg

T

v 1 = v f @ 15 kPa = 0.0010140 m 3 /kg w p ,in = v 1 (P2 − P1 )

(

)

⎛ 1 kJ = 0.001014 m 3 /kg (9000 − 15 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 9.11 kJ/kg

⎞ ⎟ ⎟ ⎠

h2 = h1 + w p ,in = 225.94 + 9.11 = 235.05 kJ/kg P3 = 9 MPa ⎫ h3 = 3512.0 kJ/kg ⎬ T3 = 550°C ⎭ s 3 = 6.8164 kJ/kg ⋅ K

3 9 MPa · Qin

2 1

15 kPa · Qout

4 s

s 4 − s f 6.8164 − 0.7549 P4 = 15 kPa ⎫ = = 0.8358 ⎬ x4 = s 4 = s3 s fg 7.2522 ⎭

h4 = h f + x 4 h fg = 225.94 + (0.8358)(2372.4) = 2208.8 kJ/kg

The thermal efficiency is determined from

qin = h3 − h2 = 3512.0 − 235.05 = 3276.9 kJ/kg q out = h4 − h1 = 2208.8 − 225.94 = 1982.9 kJ/kg and

η th = 1 − Thus,

q out 1982.9 =1− = 0.3949 q in 3276.9

η overall = η th × η comb × η gen = (0.3949)(0.75)(0.96) = 0.2843 = 28.4%

(b) Then the required rate of coal supply becomes

and

W& net 120,000 kJ/s = = 422,050 kJ/s Q& in = η overall 0.2843 m& coal =

Q& in 422,050 kJ/s = = 14.404 kg/s = 51.9 tons/h C coal 29,300 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-19

10-27 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The mass flow rate of steam through the turbine, the isentropic efficiency of the turbine, the power output from the turbine, and the thermal efficiency of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6)

T1 = 230°C⎫ ⎬ h1 = 990.14 kJ/kg x1 = 0 ⎭ h2 − h f P2 = 500 kPa ⎫ 990.14 − 640.09 = = 0.1661 ⎬x2 = h2 = h1 = 990.14 kJ/kg ⎭ h fg 2108

3 steam turbine separator

The mass flow rate of steam through the turbine is m& 3 = x 2 m& 1 = (0.1661)(230 kg/s) = 38.20 kg/s

2

4

condenser

(b) Turbine:

6

P3 = 500 kPa ⎫ h3 = 2748.1 kJ/kg ⎬ x3 = 1 ⎭ s 3 = 6.8207 kJ/kg ⋅ K P4 = 10 kPa ⎫ ⎬h4 s = 2160.3 kJ/kg s 4 = s3 ⎭

Flash chamber production well

1

5

reinjection well

P4 = 10 kPa ⎫ ⎬h4 = h f + x 4 h fg = 191.81 + (0.90)(2392.1) = 2344.7 kJ/kg x 4 = 0.90 ⎭

ηT =

h3 − h 4 2748.1 − 2344.7 = = 0.686 h3 − h4 s 2748.1 − 2160.3

(c) The power output from the turbine is W& T,out = m& 3 (h3 − h4 ) = (38.20 kJ/kg)(2748.1 − 2344.7)kJ/kg = 15,410 kW

(d) We use saturated liquid state at the standard temperature for dead state enthalpy T0 = 25°C⎫ ⎬ h0 = 104.83 kJ/kg x0 = 0 ⎭

E& in = m& 1 (h1 − h0 ) = (230 kJ/kg)(990.14 − 104.83)kJ/kg = 203,622 kW

η th =

W& T, out 15,410 = = 0.0757 = 7.6% & 203,622 E in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-20

10-28 A double-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The temperature of the steam at the exit of the second flash chamber, the power produced from the second turbine, and the thermal efficiency of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6)

T1 = 230°C⎫ ⎬ h1 = 990.14 kJ/kg x1 = 0 ⎭ P2 = 500 kPa ⎫ ⎬ x 2 = 0.1661 h2 = h1 = 990.14 kJ/kg ⎭ 3

m& 3 = x2 m& 1 = (0.1661)(230 kg/s) = 38.20 kg/s m& 6 = m& 1 − m& 3 = 230 − 0.1661 = 191.80 kg/s

P3 = 500 kPa ⎫ ⎬ h3 = 2748.1 kJ/kg x3 = 1 ⎭ P4 = 10 kPa ⎫ ⎬h4 = 2344.7 kJ/kg x 4 = 0.90 ⎭

8

steam turbine 4

separator 2

P6 = 500 kPa ⎫ ⎬ h6 = 640.09 kJ/kg x6 = 0 ⎭ P7 = 150 kPa ⎫ T7 = 111.35 °C ⎬ ⎭ x 7 = 0.0777

h7 = h6

P8 = 150 kPa ⎫ ⎬h8 = 2693.1 kJ/kg x8 = 1 ⎭

7

6 Flash chamber

separator Flash chamber

1 production well

9

condenser 5

reinjection well

(b) The mass flow rate at the lower stage of the turbine is m& 8 = x7 m& 6 = (0.0777)(191.80 kg/s) = 14.90 kg/s

The power outputs from the high and low pressure stages of the turbine are W&T1, out = m& 3 (h3 − h4 ) = (38.20 kJ/kg)(2748.1 − 2344.7)kJ/kg = 15,410 kW W&T2, out = m& 8 (h8 − h4 ) = (14.90 kJ/kg)(2693.1 − 2344.7)kJ/kg = 5191 kW

(c) We use saturated liquid state at the standard temperature for the dead state enthalpy T0 = 25°C⎫ ⎬ h0 = 104.83 kJ/kg x0 = 0 ⎭

E& in = m& 1 (h1 − h0 ) = (230 kg/s)(990.14 − 104.83)kJ/kg = 203,621 kW

η th =

W& T, out 15,410 + 5193 = = 0.101 = 10.1% 203,621 E& in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-21

10-29 A combined flash-binary geothermal power plant uses hot geothermal water at 230ºC as the heat source. The mass flow rate of isobutane in the binary cycle, the net power outputs from the steam turbine and the binary cycle, and the thermal efficiencies for the binary cycle and the combined plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6)

T1 = 230°C⎫ ⎬ h1 = 990.14 kJ/kg x1 = 0 ⎭ P2 = 500 kPa ⎫ ⎬ x 2 = 0.1661 h2 = h1 = 990.14 kJ/kg ⎭ m& 3 = x2 m& 1 = (0.1661)(230 kg/s) = 38.20 kg/s m& 6 = m& 1 − m& 3 = 230 − 38.20 = 191.80 kg/s P3 = 500 kPa ⎫ ⎬ h3 = 2748.1 kJ/kg x3 = 1 ⎭

3 separator

steam turbine

P4 = 10 kPa ⎫ ⎬h4 = 2344.7 kJ/kg x 4 = 0.90 ⎭

P6 = 500 kPa ⎫ ⎬ h6 = 640.09 kJ/kg x6 = 0 ⎭ T7 = 90°C ⎫ ⎬ h7 = 377.04 kJ/kg x7 = 0 ⎭ The isobutane properties are obtained from EES: P8 = 3250 kPa ⎫ ⎬ h8 = 755.05 kJ/kg T8 = 145°C ⎭ P9 = 400 kPa ⎫ ⎬ h9 = 691.01 kJ/kg T9 = 80°C ⎭

condenser

4

9

1

isobutane turbine

2

BINARY CYCLE

8

pump

heat exchanger flash chamber

1

5

air-cooled condenser

6

1

7

production well

reinjection well

P10 = 400 kPa ⎫ h10 = 270.83 kJ/kg ⎬ 3 x10 = 0 ⎭ v 10 = 0.001839 m /kg

w p ,in = v10 (P11 − P10 ) / η p

(

)

⎛ 1 kJ ⎞ ⎟ / 0.90 = 0.001819 m3/kg (3250 − 400 ) kPa ⎜⎜ 1 kPa ⋅ m3 ⎟⎠ ⎝ = 5.82 kJ/kg.

h11 = h10 + w p ,in = 270.83 + 5.82 = 276.65 kJ/kg An energy balance on the heat exchanger gives m& 6 (h6 − h7 ) = m& iso (h8 − h11 ) (191.81 kg/s)(640.09 - 377.04)kJ/kg = m& iso (755.05 - 276.65)kJ/kg ⎯ ⎯→ m& iso = 105.46 kg/s

(b) The power outputs from the steam turbine and the binary cycle are W&T,steam = m& 3 (h3 − h4 ) = (38.19 kJ/kg)(2748.1 − 2344.7)kJ/kg = 15,410 kW W& T,iso = m& iso (h8 − h9 ) = (105.46 kJ/kg)(755 .05 − 691.01)kJ/kg = 6753 kW W& net, binary = W& T,iso − m& iso w p ,in = 6753 − (105.46 kg/s )(5.82 kJ/kg ) = 6139 kW

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-22

(c) The thermal efficiencies of the binary cycle and the combined plant are Q& in, binary = m& iso ( h8 − h11 ) = (105.46 kJ/kg)(755 .05 − 276.65) kJ/kg = 50,454 kW

η th,binary =

W& net,binary 6139 = = 0.122 = 12.2% & 50 ,454 Qin, binary

T0 = 25°C⎫ ⎬ h0 = 104.83 kJ/kg x0 = 0 ⎭

E& in = m& 1 (h1 − h0 ) = (230 kJ/kg)(990.14 − 104.83)kJ/kg = 203,622 kW

η th, plant =

W& T,steam + W& net, binary 15,410 + 6139 = = 0.106 = 10.6% 203,622 E& in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-23

The Reheat Rankine Cycle

10-30C The pump work remains the same, the moisture content decreases, everything else increases.

10-31C The T-s diagram shows two reheat cases for the reheat Rankine cycle similar to the one shown in Figure 10-11. In the first case there is expansion through the high-pressure turbine from 6000 kPa to 4000 kPa between states 1 and 2 with reheat at 4000 kPa to state 3 and finally expansion in the low-pressure turbine to state 4. In the second case there is expansion through the high-pressure turbine from 6000 kPa to 500 kPa between states 1 and 5 with reheat at 500 kPa to state 6 and finally expansion in the low-pressure turbine to state 7. Increasing the pressure for reheating increases the average temperature for heat addition makes the energy of the steam more available for doing work, see the reheat process 2 to 3 versus the reheat process 5 to 6. Increasing the reheat pressure will increase the cycle efficiency. However, as the reheating pressure increases, the amount of condensation increases during the expansion process in the low-pressure turbine, state 4 versus state 7. An optimal pressure for reheating generally allows for the moisture content of the steam at the lowpressure turbine exit to be in the range of 10 to 15% and this corresponds to quality in the range of 85 to 90%.

SteamIAPWS

900

800

700

T [K]

1 600

3

6

2 6000 kPa 4000 kPa

500

5

500 kPa

400 0.2

300

0.4

20 kPa

0.6

0.8

4

7

200

0

20

40

60

80

100

120

140

160

180

s [kJ/kmol-K]

10-32C The thermal efficiency of the simple ideal Rankine cycle will probably be higher since the average temperature at which heat is added will be higher in this case.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-24

10-33 An ideal reheat steam Rankine cycle produces 5000 kW power. The rates of heat addition and rejection, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v 1 = v f @ 10 kPa = 0.001010 m 3 /kg

T 3

w p,in = v 1 ( P2 − P1 )

⎛ 1 kJ ⎞ = (0.001010 m 3 /kg )(8000 − 10)kPa ⎜ ⎟ 1 kPa ⋅ m 3 ⎠ ⎝ = 8.07 kJ/kg h2 = h1 + w p,in = 191.81 + 8.07 = 199.88 kJ/kg

5

8 MPa

2

500 kPa

P3 = 8000 kPa ⎫ h3 = 3273.3 kJ/kg 1 ⎬ T3 = 450°C ⎭ s 3 = 6.5579 kJ/kg ⋅ K s4 − s f 6.5579 − 1.8604 P4 = 500 kPa ⎫ x 4 = = = 0.9470 4.9603 s fg ⎬ s 4 = s3 ⎭ h = h + x h = 640.09 + (0.9470)(2108.0) = 2636.4 kJ/kg 4 f 4 fg

4 10 kPa 6

s

P5 = 500 kPa ⎫ h5 = 3484.5 kJ/kg ⎬ T5 = 500°C ⎭ s 5 = 8.0893 kJ/kg ⋅ K s6 − s f 8.0893 − 0.6492 P6 = 10 kPa ⎫ x 6 = = = 0.9921 7.4996 s fg ⎬ s6 = s5 ⎭ h = h + x h = 191.81 + (0.9921)(2392.1) = 2564.9 kJ/kg 6 f 6 fg

Thus, q in = (h3 − h2 ) + (h5 − h4 ) = 3273.3 − 199.88 + 3484.5 − 2636.4 = 3921.5 kJ/kg q out = h6 − h1 = 2564.9 − 191.81 = 2373.1 kJ/kg w net = q in − q out = 3921.5 − 2373.1 = 1548.5 kJ/kg

The mass flow rate of steam in the cycle is determined from

W& 5000 kJ/s W& net = m& (h3 − h4 ) ⎯ ⎯→ m& = net = = 3.229 kg/s wnet 1548.5 kJ/kg and the thermal efficiency of the cycle is

η th = 1 −

q out 2373.1 = 1− = 0.395 q in 3921.5

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-25

10-34 An ideal reheat steam Rankine cycle produces 2000 kW power. The mass flow rate of the steam, the rate of heat transfer in the reheater, the power used by the pumps, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6 or EES), h1 = h f @ 100 kPa = 417.51 kJ/kg

v 1 = v f @ 100 kPa = 0.001043 m 3 /kg

T 3

w p,in = v 1 ( P2 − P1 )

⎛ 1 kJ ⎞ = (0.001043 m 3 /kg )(15000 − 100)kPa ⎜ ⎟ 1 kPa ⋅ m 3 ⎠ ⎝ = 15.54 kJ/kg

h2 = h1 + w p,in = 417.51 + 15.54 = 433.05 kJ/kg

5

15 MPa

2

P3 = 15,000 kPa ⎫ h3 = 3157.9 kJ/kg 1 ⎬ T3 = 450°C ⎭ s 3 = 6.1434 kJ/kg ⋅ K s4 − s f 6.1434 − 2.4467 P4 = 2000 kPa ⎫ x 4 = = = 0.9497 3.8923 s fg ⎬ s 4 = s3 ⎭ h = h + x h = 908.47 + (0.9497)(1889.8) = 2703.3 kJ/kg 4 f 4 fg

2 MPa 4 100 kPa 6

s

P5 = 2000 kPa ⎫ h5 = 3358.2 kJ/kg ⎬ T5 = 450°C ⎭ s 5 = 7.2866 kJ/kg ⋅ K s6 − s f 7.2866 − 1.3028 P6 = 100 kPa ⎫ x 6 = = = 0.9880 6.0562 s fg ⎬ s6 = s5 ⎭ h = h + x h = 417.51 + (0.9880)(22257.5) = 2648.0 kJ/kg 6 f 6 fg

Thus, q in = (h3 − h2 ) + (h5 − h4 ) = 3157.9 − 433.05 + 3358.2 − 2703.3 = 3379.8 kJ/kg q out = h6 − h1 = 2648.0 − 417.51 = 2230.5 kJ/kg w net = q in − q out = 379.8 − 2230.5 = 1149.2 kJ/kg

The power produced by the cycle is

W& net = m& wnet = (1.74 kg/s)(1149.2 kJ/kg) = 2000 kW The rate of heat transfer in the rehetaer is Q& reheater = m& (h5 − h4 ) = (1.740 kg/s)(3358.2 − 2703.3) kJ/kg = 1140 kW W& P,in = m& wP,in = (1.740 kg/s)(15.54 kJ/kg) = 27 kW and the thermal efficiency of the cycle is

η th = 1 −

q out 2230.5 = 1− = 0.340 q in 3379.8

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-26

10-35 A steam power plant that operates on the ideal reheat Rankine cycle is considered. The turbine work output and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg

T

v 1 = v f @ 20 kPa = 0.001017 m 3 /kg w p ,in = v 1 (P2 − P1 )

(

)

⎛ 1 kJ = 0.001017 m 3 /kg (6000 − 20 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 6.08 kJ/kg

3 ⎞ ⎟ ⎟ ⎠

h2 = h1 + w p ,in = 251.42 + 6.08 = 257.50 kJ/kg P3 = 6 MPa ⎫ h3 = 3178.3 kJ/kg ⎬ T3 = 400°C ⎭ s 3 = 6.5432 kJ/kg ⋅ K

5

6 MPa 4 2 20 kPa 1

6

s

P4 = 2 MPa ⎫ ⎬ h4 = 2901.0 kJ/kg s 4 = s3 ⎭ P5 = 2 MPa ⎫ h5 = 3248.4 kJ/kg ⎬ T5 = 400°C ⎭ s 5 = 7.1292 kJ/kg ⋅ K s 6 − s f 7.1292 − 0.8320 = = 0.8900 P6 = 20 kPa ⎫ x 6 = s fg 7.0752 ⎬ s 6 = s5 ⎭ h6 = h f + x 6 h fg = 251.42 + (0.8900 )(2357.5) = 2349.7 kJ/kg

The turbine work output and the thermal efficiency are determined from wT,out = (h3 − h4 ) + (h5 − h6 ) = 3178.3 − 2901.0 + 3248.4 − 2349.7 = 1176 kJ/kg

and

q in = (h3 − h2 ) + (h5 − h4 ) = 3178.3 − 257.50 + 3248.4 − 2901.0 = 3268 kJ/kg

wnet = wT ,out − w p ,in = 1176 − 6.08 = 1170 kJ/kg Thus,

η th =

wnet 1170 kJ/kg = = 0.358 = 35.8% q in 3268 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-27

10-36 Problem 10-35 is reconsidered. The problem is to be solved by the diagram window data entry feature of EES by including the effects of the turbine and pump efficiencies and reheat on the steam quality at the low-pressure turbine exit Also, the T-s diagram is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "Input Data - from diagram window" {P[6] = 20 [kPa] P[3] = 6000 [kPa] T[3] = 400 [C] P[4] = 2000 [kPa] T[5] = 400 [C] Eta_t = 100/100 "Turbine isentropic efficiency" Eta_p = 100/100 "Pump isentropic efficiency"} "Pump analysis" function x6\$(x6) "this function returns a string to indicate the state of steam at point 6" x6\$='' if (x6>1) then x6\$='(superheated)' if (x6 w_comp_isen" h[8] + w_gas_comp_isen =hs9"SSSF conservation of energy for the isentropic compressor, assuming: adiabatic, ke=pe=0 per unit gas mass flow rate in kg/s" h[8]=ENTHALPY(Air,T=T[8]) hs9=ENTHALPY(Air,T=Ts9) h[8] + w_gas_comp = h[9]"SSSF conservation of energy for the actual compressor, assuming: adiabatic, ke=pe=0" T[9]=temperature(Air,h=h[9]) s[9]=ENTROPY(Air,T=T[9],P=P[9]) "Gas Cycle External heat exchanger analysis" h[9] + q_in = h[10]"SSSF conservation of energy for the external heat exchanger, assuming W=0, ke=pe=0" h[10]=ENTHALPY(Air,T=T[10]) P[10]=P[9] "Assume process 9-10 is SSSF constant pressure" Q_dot_in"MW"*1000"kW/MW"=m_dot_gas*q_in "Gas Turbine analysis" s[10]=ENTROPY(Air,T=T[10],P=P[10]) ss11=s[10] "For the ideal case the entropies are constant across the turbine" P[11] = P[10] /Pratio Ts11=temperature(Air,s=ss11,P=P[11])"Ts11 is the isentropic value of T[11] at gas turbine exit" Eta_gas_turb = w_gas_turb /w_gas_turb_isen "gas turbine adiabatic efficiency, w_gas_turb_isen > w_gas_turb" h[10] = w_gas_turb_isen + hs11"SSSF conservation of energy for the isentropic gas turbine, assuming: adiabatic, ke=pe=0" hs11=ENTHALPY(Air,T=Ts11) h[10] = w_gas_turb + h[11]"SSSF conservation of energy for the actual gas turbine, assuming: adiabatic, ke=pe=0" T[11]=temperature(Air,h=h[11]) s[11]=ENTROPY(Air,T=T[11],P=P[11]) PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-83

"Gas-to-Steam Heat Exchanger" "SSSF conservation of energy for the gas-to-steam heat exchanger, assuming: adiabatic, W=0, ke=pe=0" m_dot_gas*h[11] + m_dot_steam*h[4] = m_dot_gas*h[12] + m_dot_steam*h[5] h[12]=ENTHALPY(Air, T=T[12]) s[12]=ENTROPY(Air,T=T[12],P=P[12]) "STEAM CYCLE ANALYSIS" "Steam Condenser exit pump or Pump 1 analysis" Fluid\$='Steam_IAPWS' P[1] = P[7] P[2]=P[6] h[1]=enthalpy(Fluid\$,P=P[1],x=0) {Saturated liquid} v1=volume(Fluid\$,P=P[1],x=0) s[1]=entropy(Fluid\$,P=P[1],x=0) T[1]=temperature(Fluid\$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[1]+w_pump1= h[2] "Steady-flow conservation of energy" s[2]=entropy(Fluid\$,P=P[2],h=h[2]) T[2]=temperature(Fluid\$,P=P[2],h=h[2]) "Open Feedwater Heater analysis" y*h[6] + (1-y)*h[2] = 1*h[3] "Steady-flow conservation of energy" P[3]=P[6] h[3]=enthalpy(Fluid\$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]" T[3]=temperature(Fluid\$,P=P[3],x=0) s[3]=entropy(Fluid\$,P=P[3],x=0) "Boiler condensate pump or Pump 2 analysis" P[4] = P[5] v3=volume(Fluid\$,P=P[3],x=0) w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume" w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency" h[3]+w_pump2= h[4] "Steady-flow conservation of energy" s[4]=entropy(Fluid\$,P=P[4],h=h[4]) T[4]=temperature(Fluid\$,P=P[4],h=h[4]) w_steam_pumps = (1-y)*w_pump1+ w_pump2 "Total steam pump work input/ mass steam" "Steam Turbine analysis" h[5]=enthalpy(Fluid\$,T=T[5],P=P[5]) s[5]=entropy(Fluid\$,P=P[5],T=T[5]) ss6=s[5] hs6=enthalpy(Fluid\$,s=ss6,P=P[6]) Ts6=temperature(Fluid\$,s=ss6,P=P[6]) h[6]=h[5]-Eta_steam_turb*(h[5]-hs6)"Definition of steam turbine efficiency" T[6]=temperature(Fluid\$,P=P[6],h=h[6]) s[6]=entropy(Fluid\$,P=P[6],h=h[6]) ss7=s[5] hs7=enthalpy(Fluid\$,s=ss7,P=P[7]) Ts7=temperature(Fluid\$,s=ss7,P=P[7]) h[7]=h[5]-Eta_steam_turb*(h[5]-hs7)"Definition of steam turbine efficiency" T[7]=temperature(Fluid\$,P=P[7],h=h[7]) s[7]=entropy(Fluid\$,P=P[7],h=h[7]) "SSSF conservation of energy for the steam turbine: adiabatic, neglect ke and pe" h[5] = w_steam_turb + y*h[6] +(1-y)*h[7] "Steam Condenser analysis" (1-y)*h[7]=q_out+(1-y)*h[1]"SSSF conservation of energy for the Condenser per unit mass" Q_dot_out*Convert(MW, kW)=m_dot_steam*q_out "Cycle Statistics" MassRatio_gastosteam =m_dot_gas/m_dot_steam W_dot_net*Convert(MW, kW)=m_dot_gas*(w_gas_turb-w_gas_comp)+ m_dot_steam*(w_steam_turb w_steam_pumps)"definition of the net cycle work" Eta_th=W_dot_net/Q_dot_in*Convert(, %) "Cycle thermal efficiency, in percent" PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-84

Bwr=(m_dot_gas*w_gas_comp + m_dot_steam*w_steam_pumps)/(m_dot_gas*w_gas_turb + m_dot_steam*w_steam_turb) "Back work ratio" W_dot_net_steam = m_dot_steam*(w_steam_turb - w_steam_pumps) W_dot_net_gas = m_dot_gas*(w_gas_turb - w_gas_comp) NetWorkRatio_gastosteam = W_dot_net_gas/W_dot_net_steam

Pratio

MassRatio gastosteam

10 11 12 13 14 15 16 17 18 19 20

7.108 7.574 8.043 8.519 9.001 9.492 9.993 10.51 11.03 11.57 12.12

Wnetgas [kW] 342944 349014 354353 359110 363394 367285 370849 374135 377182 380024 382687

ηth [%] 59.92 60.65 61.29 61.86 62.37 62.83 63.24 63.62 63.97 64.28 64.57

Wnetsteam [kW] 107056 100986 95647 90890 86606 82715 79151 75865 72818 69976 67313

NetWorkRatio gastosteam

3.203 3.456 3.705 3.951 4.196 4.44 4.685 4.932 5.18 5.431 5.685

C o m b in ed G as an d S team P o w er C ycle 1600 1500

10

1400 1300

G as C ycle

1200 1100

T [K]

1000

S te am C yc le

900

11

800

9

700

5

600

8000 kP a

500 400

12

3,4

600 kP a

6

1,2 20 kP a

300 200 0.0

8 1.1

2.2

3.3

4.4

5.5

7 6.6

7.7

8.8

9.9

11.0

s [kJ/kg -K ]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-85 66

6 3 .8

η th [%]

6 1 .6

5 9 .4

5 7 .2

55 5

9

13

17

21

25

P ra tio

W 6.5

dot,gas

/W

dot,steam

vs Gas Pressure Ratio

NetW orkRatio gastosteam

6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 5

9

14

18

23

Pratio

Ratio of Gas Flow Rate to Steam Flow Rate vs Gas Pressure Ratio 14.0 13.0

M assRatio gastosteam

12.0 11.0 10.0 9.0 8.0 7.0 6.0 5.0 5

9

14

18

23

Pratio

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-86

10-85 A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The mass flow rate of air for a specified power output is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable fo Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis Working around the topping cycle gives the following results: T6 s

⎛P = T5 ⎜⎜ 6 ⎝ P5

⎞ ⎟ ⎟ ⎠

( k −1) / k

= (293 K)(8) 0.4/1.4 = 530.8 K

T 7

1373 K · Qin

c p (T6 s − T5 ) h −h η C = 6s 5 = h6 − h5 c p (T6 − T5 ) ⎯ ⎯→ T6 = T5 +

T6 s − T5

= 293 + ⎛P T8 s = T7 ⎜⎜ 8 ⎝ P7

⎞ ⎟ ⎟ ⎠

6 6s

ηC

⎛1⎞ = (1373 K)⎜ ⎟ ⎝8⎠

8s 6 MPa

530.8 − 293 = 572.8 K 0.85

( k −1) / k

GAS CYCLE

8 3 320°C

9

0.4/1.4

= 758.0 K

293 K

5

2 1

c p (T7 − T8 ) h −h ηT = 7 8 = ⎯ ⎯→ T8 = T7 − η T (T7 − T8 s ) h7 − h8 s c p (T7 − T8 s ) = 1373 − (0.90)(1373 − 758.0)

STEAM CYCLE 20 kPa · 4s Qout

4

s

= 819.5 K

T9 = Tsat @ 6000 kPa = 275.6°C = 548.6 K Fixing the states around the bottom steam cycle yields (Tables A-4, A-5, A-6): h1 = h f @ 20 kPa = 251.42 kJ/kg

v 1 = v f @ 20 kPa = 0.001017 m 3 /kg w p,in = v 1 ( P2 − P1 )

⎛ 1 kJ ⎞ = (0.001017 m 3 /kg )(6000 − 20)kPa ⎜ ⎟ 1 kPa ⋅ m 3 ⎠ ⎝ = 6.08 kJ/kg h2 = h1 + wp,in = 251.42 + 6.08 = 257.5 kJ/kg

P3 = 6000 kPa ⎫ h3 = 2953.6 kJ/kg ⎬ T3 = 320°C ⎭ s 3 = 6.1871 kJ/kg ⋅ K P4 = 20 kPa ⎫ ⎬ h4 s = 2035.8 kJ/kg s 4 = s3 ⎭

ηT =

h3 − h 4 ⎯ ⎯→h4 = h3 − η T (h3 − h4 s ) h3 − h 4 s = 2953.6 − (0.90)(2953.6 − 2035.8) = 2127.6 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-87

The net work outputs from each cycle are w net, gas cycle = wT, out − wC,in = c p (T7 − T8 ) − c p (T6 − T5 ) = (1.005 kJ/kg ⋅ K )(1373 − 819.5 − 572.7 + 293)K = 275.2 kJ/kg w net, steam cycle = wT,out − wP,in = (h3 − h4 ) − wP,in = (2953.6 − 2127.6) − 6.08 = 819.9 kJ/kg

An energy balance on the heat exchanger gives

m& a c p (T8 − T9 ) = m& w (h3 -h2 ) ⎯ ⎯→ m& w =

c p (T8 − T9 ) h3 -h2

m& a =

(1.005)(819.5 − 548.6) = 0.1010m& a 2953.6 − 257.5

That is, 1 kg of exhaust gases can heat only 0.1010 kg of water. Then, the mass flow rate of air is

m& a =

W& net 100,000 kJ/s = = 279.3 kg/s wnet (1× 275.2 + 0.1010 × 819.9) kJ/kg air

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-88

10-86 A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The mass flow rate of air for a specified power output is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable fo Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis With an ideal regenerator, the temperature of the air at the compressor exit will be heated to the to the temperature at the turbine exit. Representing this state by “6a”

T 7

1373 K · Qin

T6 a = T8 = 819.5 K

6a

The rate of heat addition in the cycle is

6

8s

6s

Q& in = m& a c p (T7 − T6 a )

6 MPa

= (279.3 kg/s)(1.005 kJ/kg ⋅ °C)(1373 − 819.5) K

η th =

W& net 100,000 kW = = 0.6436 155,370 kW Q& in

8 3 320°C

9

= 155,370 kW

The thermal efficiency of the cycle is then

GAS CYCLE

293 K

5

2 1

STEAM CYCLE 20 kPa · 4s Qout

4

s

Without the regenerator, the rate of heat addition and the thermal efficiency are Q& in = m& a c p (T7 − T6 ) = (279.3 kg/s) (1.005 kJ/kg ⋅ °C)(1373 − 572.7) K = 224,640 kW

η th =

W& net 100,000 kW = = 0.4452 224,640 kW Q& in

The change in the thermal efficiency due to using the ideal regenerator is ∆η th = 0.6436 − 0.4452 = 0.1984

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-89

10-87 The component of the combined cycle with the largest exergy destruction of the component of the combined cycle in Prob. 10-86 is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From Problem 10-86,

T 7

1373 K

Tsource, gas cycle = 1373 K

· Qin

Tsource, steam cycle = T8 = 819.5 K Tsink = 293 K s1 = s 2 = s f @ 20 kPa = 0.8320 kJ/kg ⋅ K

6

s 3 = 6.1871 kJ/kg ⋅ K

6s

8s 6 MPa

s 4 = 6.4627 kJ/kg ⋅ K q in,67 = c p (T7 − T6 ) = 804.3 kJ/kg

8 3 320°C

9

q in,23 = h3 − h2 = 2696.1 kJ/kg q out = h4 − h1 = 1876.2 kJ/kg m& w = 0.1010m& a = 0.1010(279.3) = 28.21 kg/s

GAS CYCLE

293 K

5

2 1

STEAM CYCLE 20 kPa · 4s Qout

4

X& destroyed, 12 = 0 (isentropic process)

s

X& destroyed, 34 = m& wT0 (s 4 − s 3 ) = ( 28.21 kg/s) ( 293 K )(6.4627 − 6.1871) = 2278 kW

⎛ q X& destroyed, 41 = m& w T0 ⎜⎜ s1 − s 4 + out T sink ⎝

⎞ ⎟ ⎟ ⎠

1876.2 kJ/kg ⎞ ⎛ = (28.21 kg/s)(293 K )⎜ 0.8320 − 6.1871 + ⎟ = 8665 kW 293 K ⎝ ⎠ ⎛ T ⎞ X& destroyed,heat exchanger = m& a T0 ∆s 89 + m& w T0 ∆s 23 = m& a T0 ⎜⎜ c p ln 9 ⎟⎟ + m& a T0 ( s 3 − s 2 ) T8 ⎠ ⎝ 548.6 ⎤ ⎡ = (279.3)(293) ⎢(1.005) ln + (28.21)(293)(6.1871 − 0.8320) 819.5 ⎥⎦ ⎣ = 11260 kW ⎛ T P ⎞ 572.7 ⎡ ⎤ X& destroyed, 56 = m& a T0 ⎜⎜ c p ln 6 − Rln 6 ⎟⎟ = (279.3)(293) ⎢(1.005)ln − (0.287) ln(8)⎥ = 6280 kW T5 P5 ⎠ 293 ⎣ ⎦ ⎝ ⎛ T q X& destroyed, 67 = m& a T0 ⎜⎜ c p ln 7 − in T6 Tsource ⎝

⎞ 1373 804.3 ⎤ ⎟ = (279.3)(293) ⎡⎢(1.005)ln − = 23,970 kW ⎟ 572.7 1373 ⎥⎦ ⎣ ⎠

⎛ T P ⎞ ⎡ 819.5 ⎛ 1 ⎞⎤ X& destroyed, 78 = m& a T0 ⎜⎜ c p ln 8 − Rln 8 ⎟⎟ = (279.3)(293) ⎢(1.005)ln − (0.287) ln⎜ ⎟⎥ = 6396 kW T P 1373 ⎝ 8 ⎠⎦ ⎣ 7 7 ⎠ ⎝

The largest exergy destruction occurs during the heat addition process in the combustor of the gas cycle.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-90

10-88 A 280-MW combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a nonideal Rankine cycle with an open feedwater heater. The mass flow rate of air to steam, the required rate of heat input in the combustion chamber, and the thermal efficiency of the combined cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Analysis (a) Using the properties of air from Table A-17, the analysis of gas cycle yields

T

T8 = 300 K ⎯ ⎯→ h8 = 300.19 kJ/kg Pr8 = 1.386 P9 Pr9 = Pr = (11)(1.386 ) = 15.25 ⎯ ⎯→ h9 s = 595.84 kJ/kg P8 8

ηC =

10 · Qin

h9 s − h8 ⎯ ⎯→ h9 = h8 + (h9 s − h8 ) / η C h9 − h8 = 300.19 + (595.84 − 300.19) / (0.82) = 660.74 kJ/kg

9s

T10 = 1100 K ⎯ ⎯→ h10 = 1161.07 kJ/kg Pr10 = 167.1 Pr11

ηT =

h10 − h11 ⎯ ⎯→ h11 = h10 − η T (h10 − h11s ) h10 − h11s = 1161.07 − (0.86 )(1161.07 − 595.18)

11 5

11s

9 4 12

P ⎛1⎞ = 11 Pr10 = ⎜ ⎟(167.1) = 15.19 ⎯ ⎯→ h11s = 595.18 kJ/kg P10 ⎝ 11 ⎠

GAS CYCLE

2 3 8 1

STEAM CYCLE 6s · Qout

6

7s 7

s

= 674.40 kJ/kg

T12 = 420 K ⎯ ⎯→ h12 = 421.26 kJ/kg

From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg v 1 = v f @ 10 kPa = 0.00101 m 3 /kg wpI,in = v 1 (P2 − P1 )

(

)

⎛ 1 kJ = 0.00101 m 3 /kg (800 − 10 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 0.80 kJ/kg

⎞ ⎟ ⎟ ⎠

h2 = h1 + wpI,in = 191.81 + 0.80 = 192.60 kJ/kg h3 = h f @ 0.8 MPa = 720.87 kJ/kg v 3 = v f @ 0.8 MPa = 0.001115 m 3 /kg wpII,in = v 3 (P4 − P3 )

(

)

⎛ 1 kJ = 0.001115 m 3 /kg (5000 − 800 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 4.68 kJ/kg

⎞ ⎟ ⎟ ⎠

h4 = h3 + wpI,in = 720.87 + 4.68 = 725.55 kJ/kg

P5 = 5 MPa ⎫ h5 = 3069.3 kJ/kg T5 = 350°C ⎬⎭ s 5 = 6.4516 kJ/kg ⋅ K

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-91

s 6 s − s f 6.4516 − 2.0457 P6 = 0.8 MPa ⎫ x 6 s = = = 0.9545 s fg 4.6160 ⎬ s 6s = s5 ⎭ h = h + x h = 720.87 + (0.9545)(2085.8) = 2675.1 kJ/kg 6s 6 s fg f

ηT =

h5 − h6 ⎯ ⎯→ h6 = h5 − η T (h5 − h6 s ) = 3069.3 − (0.86 )(3069.3 − 2675.1) = 2730.3 kJ/kg h5 − h6 s

s 7 − s f 6.4516 − 0.6492 P7 = 10 kPa ⎫ x 7 s = = = 0.7737 s fg 7.4996 ⎬ s 7 = s5 ⎭ h = h + x h = 191.81 + (0.7737 )(2392.1) = 2042.5 kJ/kg 7s 7 fg f

ηT =

h5 − h7 ⎯ ⎯→ h7 = h5 − η T (h5 − h7 s ) = 3069.3 − (0.86)(3069.3 − 2042.5) = 2186.3 kJ/kg h5 − h7 s

Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 for the heat exchanger, the steady-flow energy balance equation yields E& in − E& out = ∆E& system ©0 (steady) = 0 E& in = E& out

∑ m& h = ∑ m& h i i

e e

⎯ ⎯→ m& s (h5 − h4 ) = m& air (h11 − h12 )

m& air h − h4 3069.3 − 725.55 = 5 = = 9.259 kg air / kg steam m& s h11 − h12 674.40 − 421.26

(b) Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 for the open FWH, the steady-flow energy balance equation yields E& in − E& out = ∆E& system ©0 (steady) = 0 → E& in = E& out

∑ m& h = ∑ m& h i i

e e

⎯ ⎯→ m& 2 h2 + m& 6 h6 = m& 3 h3 ⎯ ⎯→ yh6 + (1 − y )h2 = (1)h3

Thus, y=

h3 − h2 720.87 − 192.60 = = 0.2082 h6 − h2 2730.3 − 192.60

(the

fraction of steam extracted )

wT = η T [h5 − h6 + (1 − y )(h6 − h7 )] = (0.86 )[3069.3 − 2730.3 + (1 − 0.2082 )(2730.3 − 2186.3)] = 769.8 kJ/kg

wnet,steam = wT − wp,in = wT − (1 − y )wp,I − wp,II = 769.8 − (1 − 0.2082)(0.80 ) − 4.68 = 764.5 kJ/kg wnet,gas = wT − wC ,in = (h10 − h11 ) − (h9 − h8 ) = 1161.07 − 674.40 − (660.74 − 300.19 ) = 126.12 kJ/kg

The net work output per unit mass of gas is

wnet = wnet,gas + m& air =

and (c)

1 1 (764.5) = 208.69 kJ/kg wnet,steam = 126.12 + 9.259 6.425

W& net 280,000 kJ/s = = 1341.7 kg/s wnet 208.69 kJ/kg

Q& in = m& air (h10 − h9 ) = (1341.7 kg/s)(1161.07 − 660.74) kJ/kg = 671,300 kW

η th =

W& net 280,000 kW = = 0.4171 = 41.7% 671,300 kW Q& in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-92

10-89 Problem 10-88 is reconsidered. The effect of the gas cycle pressure ratio on the ratio of gas flow rate to steam flow rate and cycle thermal efficiency is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input data" T[8] = 300 [K] P[8] = 100 [kPa] "Pratio = 11" T[10] = 1100 [K] T[12] = 420 [K] P[12] = P[8] W_dot_net=280 [MW] Eta_comp = 0.82 Eta_gas_turb = 0.86 Eta_pump = 1.0 Eta_steam_turb = 0.86 P[5] = 5000 [kPa] T[5] =(350+273.15) "K" P[6] = 800 [kPa] P[7] = 10 [kPa]

"Gas compressor inlet" "Assumed air inlet pressure" "Pressure ratio for gas compressor" "Gas turbine inlet" "Gas exit temperature from Gas-to-steam heat exchanger " "Assumed air exit pressure"

"Steam turbine inlet" "Steam turbine inlet" "Extraction pressure for steam open feedwater heater" "Steam condenser pressure"

"GAS POWER CYCLE ANALYSIS" "Gas Compressor anaysis" s[8]=ENTROPY(Air,T=T[8],P=P[8]) ss9=s[8] "For the ideal case the entropies are constant across the compressor" P[9] = Pratio*P[8] Ts9=temperature(Air,s=ss9,P=P[9])"Ts9 is the isentropic value of T[9] at compressor exit" Eta_comp = w_gas_comp_isen/w_gas_comp "compressor adiabatic efficiency, w_comp > w_comp_isen" h[8] + w_gas_comp_isen =hs9"SSSF conservation of energy for the isentropic compressor, assuming: adiabatic, ke=pe=0 per unit gas mass flow rate in kg/s" h[8]=ENTHALPY(Air,T=T[8]) hs9=ENTHALPY(Air,T=Ts9) h[8] + w_gas_comp = h[9]"SSSF conservation of energy for the actual compressor, assuming: adiabatic, ke=pe=0" T[9]=temperature(Air,h=h[9]) s[9]=ENTROPY(Air,T=T[9],P=P[9]) "Gas Cycle External heat exchanger analysis" h[9] + q_in = h[10]"SSSF conservation of energy for the external heat exchanger, assuming W=0, ke=pe=0" h[10]=ENTHALPY(Air,T=T[10]) P[10]=P[9] "Assume process 9-10 is SSSF constant pressure" Q_dot_in"MW"*1000"kW/MW"=m_dot_gas*q_in "Gas Turbine analysis" s[10]=ENTROPY(Air,T=T[10],P=P[10]) ss11=s[10] "For the ideal case the entropies are constant across the turbine" P[11] = P[10] /Pratio Ts11=temperature(Air,s=ss11,P=P[11])"Ts11 is the isentropic value of T[11] at gas turbine exit" Eta_gas_turb = w_gas_turb /w_gas_turb_isen "gas turbine adiabatic efficiency, w_gas_turb_isen > w_gas_turb" h[10] = w_gas_turb_isen + hs11"SSSF conservation of energy for the isentropic gas turbine, assuming: adiabatic, ke=pe=0" hs11=ENTHALPY(Air,T=Ts11) h[10] = w_gas_turb + h[11]"SSSF conservation of energy for the actual gas turbine, assuming: adiabatic, ke=pe=0" T[11]=temperature(Air,h=h[11]) s[11]=ENTROPY(Air,T=T[11],P=P[11]) PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-93

"Gas-to-Steam Heat Exchanger" "SSSF conservation of energy for the gas-to-steam heat exchanger, assuming: adiabatic, W=0, ke=pe=0" m_dot_gas*h[11] + m_dot_steam*h[4] = m_dot_gas*h[12] + m_dot_steam*h[5] h[12]=ENTHALPY(Air, T=T[12]) s[12]=ENTROPY(Air,T=T[12],P=P[12]) "STEAM CYCLE ANALYSIS" "Steam Condenser exit pump or Pump 1 analysis" Fluid\$='Steam_IAPWS' P[1] = P[7] P[2]=P[6] h[1]=enthalpy(Fluid\$,P=P[1],x=0) {Saturated liquid} v1=volume(Fluid\$,P=P[1],x=0) s[1]=entropy(Fluid\$,P=P[1],x=0) T[1]=temperature(Fluid\$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[1]+w_pump1= h[2] "Steady-flow conservation of energy" s[2]=entropy(Fluid\$,P=P[2],h=h[2]) T[2]=temperature(Fluid\$,P=P[2],h=h[2]) "Open Feedwater Heater analysis" y*h[6] + (1-y)*h[2] = 1*h[3] "Steady-flow conservation of energy" P[3]=P[6] h[3]=enthalpy(Fluid\$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]" T[3]=temperature(Fluid\$,P=P[3],x=0) s[3]=entropy(Fluid\$,P=P[3],x=0) "Boiler condensate pump or Pump 2 analysis" P[4] = P[5] v3=volume(Fluid\$,P=P[3],x=0) w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume" w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency" h[3]+w_pump2= h[4] "Steady-flow conservation of energy" s[4]=entropy(Fluid\$,P=P[4],h=h[4]) T[4]=temperature(Fluid\$,P=P[4],h=h[4]) w_steam_pumps = (1-y)*w_pump1+ w_pump2 "Total steam pump work input/ mass steam" "Steam Turbine analysis" h[5]=enthalpy(Fluid\$,T=T[5],P=P[5]) s[5]=entropy(Fluid\$,P=P[5],T=T[5]) ss6=s[5] hs6=enthalpy(Fluid\$,s=ss6,P=P[6]) Ts6=temperature(Fluid\$,s=ss6,P=P[6]) h[6]=h[5]-Eta_steam_turb*(h[5]-hs6)"Definition of steam turbine efficiency" T[6]=temperature(Fluid\$,P=P[6],h=h[6]) s[6]=entropy(Fluid\$,P=P[6],h=h[6]) ss7=s[5] hs7=enthalpy(Fluid\$,s=ss7,P=P[7]) Ts7=temperature(Fluid\$,s=ss7,P=P[7]) h[7]=h[5]-Eta_steam_turb*(h[5]-hs7)"Definition of steam turbine efficiency" T[7]=temperature(Fluid\$,P=P[7],h=h[7]) s[7]=entropy(Fluid\$,P=P[7],h=h[7]) "SSSF conservation of energy for the steam turbine: adiabatic, neglect ke and pe" h[5] = w_steam_turb + y*h[6] +(1-y)*h[7] "Steam Condenser analysis" (1-y)*h[7]=q_out+(1-y)*h[1]"SSSF conservation of energy for the Condenser per unit mass" Q_dot_out*Convert(MW, kW)=m_dot_steam*q_out "Cycle Statistics" MassRatio_gastosteam =m_dot_gas/m_dot_steam W_dot_net*Convert(MW, kW)=m_dot_gas*(w_gas_turb-w_gas_comp)+ m_dot_steam*(w_steam_turb w_steam_pumps)"definition of the net cycle work" Eta_th=W_dot_net/Q_dot_in*Convert(, %) "Cycle thermal efficiency, in percent" PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-94

Bwr=(m_dot_gas*w_gas_comp + m_dot_steam*w_steam_pumps)/(m_dot_gas*w_gas_turb + m_dot_steam*w_steam_turb) "Back work ratio" W_dot_net_steam = m_dot_steam*(w_steam_turb - w_steam_pumps) W_dot_net_gas = m_dot_gas*(w_gas_turb - w_gas_comp) NetWorkRatio_gastosteam = W_dot_net_gas/W_dot_net_steam

MassRatiogastosteam

10 11 12 13 14 15 16 17 18 19 20

8.775 9.262 9.743 10.22 10.7 11.17 11.64 12.12 12.59 13.07 13.55

ηth [%] 42.03 41.67 41.22 40.68 40.08 39.4 38.66 37.86 36.99 36.07 35.08

Combined Gas and Steam Power Cycle 1600 1500

10

1400 1300

Gas Cycle

1200 1100

Steam Cycle

1000

T [K]

Pratio

900

11

800

9

700

5

600

5000 kPa

500 400

12

3,4

800 kPa 10 kPa

300 200 0.0

6

1,2 8 1.1

2.2

3.3

4.4

5.5

7 6.6

7.7

8.8

9.9

11.0

s [kJ/kg-K] 14

MassRatiogastosteam

13 12 11 10 9 8 10

12

14

16

18

20

16

18

20

Pratio

43 42

η th [%]

41 40 39 38 37 36 35 10

12

14

Pratio

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-95

10-90 A combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a nonideal reheat Rankine cycle. The moisture percentage at the exit of the low-pressure turbine, the steam temperature at the inlet of the high-pressure turbine, and the thermal efficiency of the combined cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Analysis (a) We obtain the air properties from EES. The analysis of gas cycle is as follows ⎯→ h7 = 288.50 kJ/kg T7 = 15°C ⎯ T7 = 15°C

⎫ ⎬s 7 = 5.6648 kJ/kg P7 = 100 kPa ⎭ P8 = 700 kPa ⎫ ⎬h8 s = 503.47 kJ/kg s8 = s 7 ⎭

ηC =

Combustion chamber

8

9 Compressor

h8 s − h7 ⎯ ⎯→ h8 = h7 + (h8 s − h7 ) / η C h8 − h7 = 290.16 + (503.47 − 290.16 ) / (0.80 ) = 557.21 kJ/kg

Gas turbine

7 11

10

Heat exchanger 3

⎯→ h9 = 1304.8 kJ/kg T9 = 950°C ⎯ T9 = 950°C ⎫ ⎬s 9 = 6.6456 kJ/kg P9 = 700 kPa ⎭

Steam turbine 4 6

P10 = 100 kPa ⎫ ⎬h10 s = 763.79 kJ/kg ⎭ h9 − h10 ⎯ ⎯→ h10 = h9 − η T (h9 − h10 s ) ηT = h9 − h10 s = 1304.8 − (0.80 )(1304.8 − 763.79 )

5

s10 = s 9

Condenser

pump

2

1

= 871.98 kJ/kg

T11 = 200 °C ⎯ ⎯→ h11 = 475.62 kJ/kg

From the steam tables (Tables A-4, A-5, and A-6 or from EES), h1 = h f v1 = v f

T 9

950°C · Qin

= 191.81 kJ/kg 3 @ 10 kPa = 0.00101 m /kg

@ 10 kPa

wpI,in = v1 (P2 − P1 ) / η p

(

)

⎛ 1 kJ ⎞ ⎟ / 0.80 = 0.00101 m3 /kg (6000 − 10 kPa )⎜⎜ 1 kPa ⋅ m3 ⎟⎠ ⎝ = 7.56 kJ/kg

8s

P6 = 10 kPa ⎫ x 6 s ⎬ s 6s = s5 ⎭h 6s

15°C

7

2

1 7.4670 − 0.6492 = = = 0.9091 s fg 7.4996 = h f + x 6 s h fg = 191.81 + (0.9091)(2392.1) = 2366.4 kJ/kg s 6s − s f

10 10s 3

8

h2 = h1 + wpI,in = 191.81 + 7.65 = 199.37 kJ/kg P5 = 1 MPa ⎫ h5 = 3264.5 kJ/kg T5 = 400°C ⎬⎭ s 5 = 7.4670 kJ/kg ⋅ K

GAS CYCLE

6 MPa

1 MPa 5

11 STEAM 4 CYCLE 4s 10 kPa · 6s 6 Qout

s

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-96

ηT =

h5 − h 6 ⎯ ⎯→ h6 = h5 − η T (h5 − h6 s ) h5 − h 6 s = 3264.5 − (0.80 )(3264.5 − 2366.4 ) = 2546.0 kJ/kg

P6 = 10 kPa ⎫ x = 0.9842 h6 = 2546.5 kJ/kg ⎬⎭ 6 Moisture Percentage = 1 − x 6 = 1 − 0.9842 = 0.0158 = 1.6%

(b) Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 for the heat exchanger, the steady-flow energy balance equation yields E& in = E& out

∑ m& h = ∑ m& h i i

e e

m& s (h3 − h2 ) + m& s (h5 − h4 ) = m& air (h10 − h11 )

(1.15)[(3346.5 − 199.37) + (3264.5 − h4 )] = (10)(871.98 − 475.62) ⎯ ⎯→ h4 = 2965.0 kJ/kg

Also, P3 = 6 MPa ⎫ h3 = ⎬ T3 = ? ⎭ s3 =

ηT =

P4 = 1 MPa ⎫ ⎬ h4 s = s 4s = s3 ⎭

h3 − h 4 ⎯ ⎯→ h4 = h3 − η T (h3 − h4 s ) h3 − h 4 s

The temperature at the inlet of the high-pressure turbine may be obtained by a trial-error approach or using EES from the above relations. The answer is T3 = 468.0ºC. Then, the enthalpy at state 3 becomes: h3 = 3346.5 kJ/kg (c)

W& T, gas = m& air (h9 − h10 ) = (10 kg/s )(1304 .8 − 871.98) kJ/kg = 4328 kW W& C,gas = m& air (h8 − h7 ) = (10 kg/s )(557.21 − 288.50 ) kJ/kg = 2687 kW W& net,gas = W& T,gas − W& C,gas = 4328 − 2687 = 1641 kW

W& T,steam = m& s (h3 − h4 + h5 − h6 ) = (1.15 kg/s )(3346.5 − 2965.0 + 3264.5 − 2546.0 ) kJ/kg = 1265 kW W& P,steam = m& s w pump = (1.15 kg/s )(7.564 ) kJ/kg = 8.7 kW

W& net,steam = W& T,steam − W& P,steam = 1265 − 8.7 = 1256 kW W& net, plant = W& net,gas + W& net,steam = 1641 + 1256 = 2897 kW

(d)

Q& in = m& air (h9 − h8 ) = (10 kg/s)(1304.8 − 557.21) kJ/kg = 7476 kW

η th =

W& net, plant 2897 kW = = 0.388 = 38.8% 7476 kW Q& in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-97

Special Topic: Binary Vapor Cycles 10-91C In binary vapor power cycles, both cycles are vapor cycles. In the combined gas-steam power cycle, one of the cycles is a gas cycle.

10-92C Binary power cycle is a cycle which is actually a combination of two cycles; one in the high temperature region, and the other in the low temperature region. Its purpose is to increase thermal efficiency.

10-93C Steam is not an ideal fluid for vapor power cycles because its critical temperature is low, its saturation dome resembles an inverted V, and its condenser pressure is too low.

10-94C Because mercury has a high critical temperature, relatively low critical pressure, but a very low condenser pressure. It is also toxic, expensive, and has a low enthalpy of vaporization.

10-95 Consider the heat exchanger of a binary power cycle. The working fluid of the topping cycle (cycle A) enters the heat exchanger at state 1 and leaves at state 2. The working fluid of the bottoming cycle (cycle B) enters at state 3 and leaves at state 4. Neglecting any changes in kinetic and potential energies, and assuming the heat exchanger is wellinsulated, the steady-flow energy balance relation yields E& in − E& out = ∆E& system©0 (steady) = 0 E& in = E& out

∑ m& h = ∑ m& h e e

i i

m& A h2 + m& B h4 = m& A h1 + m& B h3 or m& A (h2 − h1 ) = m& B (h3 − h4 ) Thus, m& A h3 − h4 = m& B h2 − h1

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-98

Review Problems 10-96 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The thermal efficiency of the cycle is to be compared when it is operated so that the liquid enters the pump as a saturated liquid against that when the liquid enters as a subcooled liquid. determined power produced by the turbine and consumed by the pump are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 50 kPa = 340.54 kJ/kg

T

3

v 1 = v f @ 20 kPa = 0.001030 m /kg 6 MPa

wp,in = v 1 ( P2 − P1 )

⎛ 1 kJ ⎞ = (0.001030 m 3 /kg )(6000 − 50)kPa ⎜ ⎟ 1 kPa ⋅ m 3 ⎠ ⎝ = 6.13 kJ/kg h2 = h1 + wp,in = 340.54 + 6.13 = 346.67 kJ/kg

3

qin

2

50 kPa

1 P3 = 6000 kPa ⎫ h3 = 3658.8 kJ/kg ⎬ T3 = 600°C ⎭ s 3 = 7.1693 kJ/kg ⋅ K s4 − s f 7.1693 − 1.0912 P4 = 50 kPa ⎫ x 4 = = = 0.9348 s fg 6.5019 ⎬ s 4 = s3 ⎭ h = h + x h = 340.54 + (0.9348)(2304.7) = 2495.0 kJ/kg 4 f 4 fg

qout

4 s

Thus, q in = h3 − h2 = 3658.8 − 346.67 = 3312.1 kJ/kg q out = h4 − h1 = 2495.0 − 340.54 = 2154.5 kJ/kg

and the thermal efficiency of the cycle is

η th = 1 −

q out 2154.5 = 1− = 0.3495 q in 3312.1

When the liquid enters the pump 11.3°C cooler than a saturated liquid at the condenser pressure, the enthalpies become P1 = 50 kPa T1 = Tsat @ 50 kPa

⎫ h1 ≅ h f @ 70°C = 293.07 kJ/kg ⎬ − 11.3 = 81.3 − 11.3 = 70°C ⎭ v 1 ≅ v f @ 70°C = 0.001023 m 3 /kg

wp,in = v 1 ( P2 − P1 )

⎛ 1 kJ ⎞ = (0.001023 m 3 /kg )(6000 − 50)kPa ⎜ ⎟ 1 kPa ⋅ m 3 ⎠ ⎝ = 6.09 kJ/kg

h2 = h1 + wp,in = 293.07 + 6.09 = 299.16 kJ/kg Then, q in = h3 − h2 = 3658.8 − 299.16 = 3359.6 kJ/kg q out = h4 − h1 = 2495.0 − 293.09 = 2201.9 kJ/kg

η th = 1 −

q out 2201.9 = 1− = 0.3446 q in 3359.6

The thermal efficiency slightly decreases as a result of subcooling at the pump inlet.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-99

10-97E A geothermal power plant operating on the simple Rankine cycle using an organic fluid as the working fluid is considered. The exit temperature of the geothermal water from the vaporizer, the rate of heat rejection from the working fluid in the condenser, the mass flow rate of geothermal water at the preheater, and the thermal efficiency of the Level I cycle of this plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The exit temperature of geothermal water from the vaporizer is determined from the steady-flow energy balance on the geothermal water (brine), Q& brine = m& brine c p (T2 − T1 )

− 22,790,000 Btu/h = (384,286 lbm/h )(1.03 Btu/lbm ⋅ °F)(T2 − 325°F) T2 = 267.4°F

(b) The rate of heat rejection from the working fluid to the air in the condenser is determined from the steady-flow energy balance on air, Q& air = m& air c p (T9 − T8 )

= (4,195,100 lbm/h )(0.24 Btu/lbm ⋅ °F )(84.5 − 55°F)

= 29.7 MBtu/h

(c) The mass flow rate of geothermal water at the preheater is determined from the steady-flow energy balance on the geothermal water, Q& geo = m& geo c p (Tout − Tin )

− 11,140,000 Btu/h = m& geo (1.03 Btu/lbm ⋅ °F)(154.0 − 211.8°F ) m& geo = 187,120 lbm/h

(d) The rate of heat input is , ,000 Q& in = Q& vaporizer + Q& reheater = 22,790,000 + 11140 and

= 33,930,000 Btu / h

W& net = 1271 − 200 = 1071 kW Then,

η th =

⎛ 3412.14 Btu ⎞ W& net 1071 kW ⎟ = 10.8% ⎜ = 33,930,000 Btu/h ⎜⎝ 1 kWh ⎟⎠ Q& in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-100

10-98 A steam power plant operating on an ideal Rankine cycle with two stages of reheat is considered. The thermal efficiency of the cycle and the mass flow rate of the steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

h1 = h f @ 30 kPa = 289.18 kJ/kg v 1 = v f @ 30 kPa = 0.001022 m 3 /kg wp,in = v 1 (P2 − P1 )

(

T

4 MPa

)

⎛ 1 kJ ⎞ ⎟ = 0.001022 m 3 /kg (10,000 − 30 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎟ ⎝ ⎠ = 10.19 kJ/kg

h2 = h1 + wp,in = 289.18 + 10.19 = 299.37 kJ/kg P3 = 10 MPa ⎫ h3 = 3500.9 kJ/kg T3 = 550°C ⎬⎭ s 3 = 6.7561 kJ/kg ⋅ K P4 = 4 MPa ⎫ ⎬ h4 = 3204.9 kJ/kg s 4 = s3 ⎭

10 MPa

2 MPa

3

5

4

6

7

2

1

30 kPa

8

s

P5 = 4 MPa ⎫ h5 = 3559.7 kJ/kg T5 = 550°C ⎬⎭ s 5 = 7.2335 kJ/kg ⋅ K P6 = 2 MPa ⎫ ⎬ h6 = 3321.1 kJ/kg s 6 = s5 ⎭ P7 = 2 MPa ⎫ h7 = 3578.4 kJ/kg T7 = 550°C ⎬⎭ s 7 = 7.5706 kJ/kg ⋅ K s8 − s f 7.5706 − 0.9441 = = 0.9711 P8 = 30 kPa ⎫ x8 = s fg 6.8234 ⎬ s8 = s 7 ⎭ h8 = h f + x8 h fg = 289.27 + (0.9711)(2335.3) = 2557.1 kJ/kg

Then, q in = (h3 − h2 ) + (h5 − h4 ) + (h7 − h6 ) q out

= 3500.9 − 299.37 + 3559.7 − 3204.9 + 3578.4 − 3321.1 = 3813.7 kJ/kg = h8 − h1 = 2557.1 − 289.18 = 2267.9 kJ/kg

wnet = q in − q out = 3813.7 − 2267.9 = 1545.8 kJ/kg

Thus,

η th =

wnet 1545.8 kJ/kg = = 0.4053 = 40.5% q in 3813.7 kJ/kg

(b) The mass flow rate of the steam is then m& =

W& net 75,000 kJ/s = = 48.5 kg/s wnet 1545.8 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-101

10-99 A steam power plant operating on the ideal Rankine cycle with reheating is considered. The reheat pressures of the cycle are to be determined for the cases of single and double reheat. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Single Reheat: From the steam tables (Tables A-4, A-5, and A-6),

P6 = 10 kPa x6 = 0.92

⎫ h6 = h f + x6h fg = 191.81 + (0.92)(2392.1) = 2392.5 kJ/kg ⎬ s = s + x s = 0.6492 + (0.92)(7.4996) = 7.5488 kJ/kg ⋅ K f 6 fg ⎭ 6

T5 = 600°C ⎫ ⎬ P5 = 2780 kPa s5 = s6 ⎭

T 600°C

(b) Double Reheat: P3 = 25 MPa ⎫ s = 6.3637 kJ/kg ⋅ K T3 = 600°C ⎬⎭ 3 P4 = Px P = Px and 5 T5 = 600°C s 4 = s3

SINGLE 3 25 MPa

5

T

DOUBLE 3

600°C

25 MPa

4

2

7

6

2 10 kPa

1

4

5

10 kPa 6

1

8

s

s Any pressure Px selected between the limits of 25 MPa and 2.78 MPa will satisfy the requirements, and can be used for the double reheat pressure.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-102

10-100 An 150-MW steam power plant operating on a regenerative Rankine cycle with an open feedwater heater is considered. The mass flow rate of steam through the boiler, the thermal efficiency of the cycle, and the irreversibility associated with the regeneration process are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis

T 5 Turbine

Boiler

qin 1-y

6

7

y Open fwh

4

Condenser

P II

10 MPa 6

y

3 0.5 MPa

6s

2 10 kPa

2

3

4

5

1 PI

1

qout

7s

1-y 7 s

(a) From the steam tables (Tables A-4, A-5, and A-6),

h1 = hf @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.00101 m3/kg wpI,in = v1(P2 − P1) / ηp

(

)

⎛ 1 kJ ⎞ ⎟ /(0.95) = 0.00101m3/kg (500−10 kPa)⎜⎜ 1 kPa⋅ m3 ⎟⎠ ⎝ = 0.52 kJ/kg

h2 = h1 + wpI,in = 191.81+ 0.52 = 192.33 kJ/kg P3 = 0.5 MPa ⎫ h3 = hf @ 0.5 MPa = 640.09 kJ/kg ⎬v = v 3 sat.liquid f @ 0.5 MPa = 0.001093 m /kg ⎭ 3 wpII,in = v 3 (P4 − P3 ) / η p

(

)

⎛ 1 kJ ⎞ ⎟ / (0.95) = 0.001093 m3/kg (10,000 − 500 kPa )⎜⎜ 1 kPa ⋅ m 3 ⎟⎠ ⎝ = 10.93 kJ/kg

h4 = h3 + wpII,in = 640.09 + 10.93 = 651.02 kJ/kg

P5 = 10 MPa ⎫ h5 = 3375.1 kJ/kg T5 = 500°C ⎬⎭ s 5 = 6.5995 kJ/kg ⋅ K s6s − s f

6.5995 − 1.8604 = = 0.9554 4.9603 s fg P6 s = 0.5 MPa ⎫ ⎬ h6 s = h f + x 6 s h fg = 640.09 + (0.9554 )(2108.0) s 6s = s5 ⎭ = 2654.1 kJ/kg x6s =

ηT =

h5 − h6 ⎯ ⎯→ h6 = h5 − ηT (h5 − h6 s ) h5 − h6 s = 3375.1 − (0.80)(3375.1 − 2654.1) = 2798.3 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-103

s7s − s f

6.5995 − 0.6492 = 0.7934 7.4996 s fg P7 s = 10 kPa ⎫ ⎬ h7 s = h f + x 7 s h fg = 191.81 + (0.7934 )(2392.1) s7s = s5 ⎭ = 2089.7 kJ/kg x7s =

ηT =

=

h5 − h7 ⎯ ⎯→ h7 = h5 − ηT (h5 − h7 s ) h5 − h7 s = 3375.1 − (0.80)(3375.1 − 2089.7 ) = 2346.8 kJ/kg

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 , E& in − E& out = ∆E& system©0 (steady) = 0 E& in = E& out

∑ m& h = ∑ m& h i i

e e

⎯ ⎯→ m& 6 h6 + m& 2 h2 = m& 3 h3 ⎯ ⎯→ yh6 + (1 − y )h2 = 1(h3 )

&6 / m & 3 ). Solving for y, where y is the fraction of steam extracted from the turbine ( = m

y=

Then,

h3 − h2 640.09 − 192.33 = = 0.1718 h6 − h2 2798.3 − 192.33

qin = h5 − h4 = 3375.1 − 651.02 = 2724.1 kJ/kg

qout = (1 − y )(h7 − h1 ) = (1 − 0.1718 )(2346.8 − 191.81) = 1784.7 kJ/kg wnet = qin − qout = 2724.1 − 1784.7 = 939.4 kJ/kg

and m& =

W& net 150,000 kJ/s = = 159.7 kg/s wnet 939.4 kJ/kg

(b) The thermal efficiency is determined from

η th = 1 −

q out 1784.7 kJ/kg = 1− = 34.5% 2724.1 kJ/kg q in

Also, P6 = 0.5 MPa

⎫ ⎬ s6 = 6.9453 kJ/kg ⋅ K h6 = 2798.3 kJ/kg ⎭

s3 = s f @ 0.5 MPa = 1.8604 kJ/kg ⋅ K s2 = s1 = s f @ 10 kPa = 0.6492 kJ/kg ⋅ K

Then the irreversibility (or exergy destruction) associated with this regeneration process is ⎛ q ©0 ⎞⎟ i regen = T0 s gen = T0 ⎜ me s e − m i s i + surr = T0 [s 3 − ys 6 − (1 − y )s 2 ] ⎜ ⎟ TL ⎝ ⎠ = (303 K )[1.8604 − (0.1718)(6.9453) − (1 − 0.1718)(0.6492 )] = 39.25 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-104

10-101 An 150-MW steam power plant operating on an ideal regenerative Rankine cycle with an open feedwater heater is considered. The mass flow rate of steam through the boiler, the thermal efficiency of the cycle, and the irreversibility associated with the regeneration process are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis

T 5 Turbine

Boiler

qin 1-y

6 Open fwh

4

Condenser 2

3

P II

10 MPa

4

7

y

y

3 0.5 MPa

6 1-y

2 10 kPa

1

1

PI

5

qout

7 s

(a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f

@ 10 kPa

= 191.81 kJ/kg

v1 = v f

@ 10 kPa

= 0.00101 m 3 /kg

w pI,in = v 1 (P2 − P1 )

(

)

⎛ 1 kJ = 0.00101 m 3 /kg (500 − 10 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ h2 = h1 + wpI,in = 191.81 + 0.50 = 192.30 kJ/kg

P3 = 0.5 MPa sat.liquid

⎞ ⎟ = 0.50 kJ/kg ⎟ ⎠

⎫ h3 = h f @ 0.5 MPa = 640.09 kJ/kg ⎬v = v 3 f @ 0.5 MPa = 0.001093 m /kg ⎭ 3

wpII,in = v 3 (P4 − P3 )

(

)

⎛ 1 kJ ⎞ ⎟ = 10.38 kJ/kg = 0.001093 m3/kg (10,000 − 500 kPa )⎜⎜ 3⎟ ⎝ 1 kPa ⋅ m ⎠ h4 = h3 + wpII,in = 640.09 + 10.38 = 650.47 kJ/kg

P5 = 10 MPa ⎫ h5 = 3375.1 kJ/kg T5 = 500°C ⎬⎭ s5 = 6.5995 kJ/kg ⋅ K s6 − s f 6.5995 − 1.8604 = = 0.9554 P6 = 0.5 MPa ⎫ x6 = s fg 4.9603 ⎬ s6 = s5 ⎭ h6 = h f + x6 h fg = 640.09 + (0.9554 )(2108.0 ) = 2654.1 kJ/kg s7 − s f 6.5995 − 0.6492 = = 0.7934 P7 = 10 kPa ⎫ x7 = s 7.4996 ⎬ fg s7 = s5 ⎭ h7 = h f + x7 h fg = 191.81 + (0.7934 )(2392.1) = 2089.7 kJ/kg

The fraction of steam extracted is determined from the steady-flow energy equation applied to the feedwater heaters. Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 , E& in − E& out = ∆E& system ©0 (steady) = 0 → E& in = E& out

∑ m& h = ∑ m& h i i

e e

⎯ ⎯→ m& 6 h6 + m& 2 h2 = m& 3 h3 ⎯ ⎯→ yh6 + (1 − y )h2 = 1(h3 )

&6 / m & 3 ). Solving for y, where y is the fraction of steam extracted from the turbine ( = m PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-105

y=

Then,

h3 − h2 640.09 − 192.31 = = 0.1819 h6 − h2 2654.1 − 192.31

qin = h5 − h4 = 3375.1 − 650.47 = 2724.6 kJ/kg

qout = (1 − y )(h7 − h1 ) = (1 − 0.1819 )(2089.7 − 191.81) = 1552.7 kJ/kg wnet = qin − qout = 2724.6 − 1552.7 = 1172.0 kJ/kg

and

m& =

W&net 150,000 kJ/s = = 128.0 kg/s wnet 1171.9 kJ/kg

(b) The thermal efficiency is determined from

η th = 1 −

q out 1552.7 kJ/kg = 1− = 43.0% 2724.7 kJ/kg q in

Also, s 6 = s 5 = 6.5995 kJ/kg ⋅ K s3 = s f

@ 0.5 MPa

s 2 = s1 = s f

= 1.8604 kJ/kg ⋅ K

@ 10 kPa

= 0.6492 kJ/kg ⋅ K

Then the irreversibility (or exergy destruction) associated with this regeneration process is ⎛ q ©0 ⎞⎟ i regen = T0 s gen = T0 ⎜ me s e − m i s i + surr = T0 [s 3 − ys 6 − (1 − y )s 2 ] ⎜ ⎟ TL ⎝ ⎠ = (303 K )[1.8604 − (0.1819 )(6.5995) − (1 − 0.1819 )(0.6492 )] = 39.0 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-106

10-102 An ideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. The fraction of steam extracted for regeneration and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

h1 = h f

@ 15 kPa

= 225.94 kJ/kg

v1 = v f

@ 15 kPa

= 0.001014 m 3 /kg

wpI,in = v 1 (P2 − P1 )

(

)

⎛ 1 kJ = 0.001014 m 3 /kg (600 − 15 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 0.59 kJ/kg h2 = h1 + wpI,in = 225.94 + 0.59 = 226.53 kJ/kg

5

⎞ ⎟ ⎟ ⎠

7

(

Open fwh

4

P II

)

⎛ 1 kJ ⎞ ⎟ = 0.001101 m3/kg (10,000 − 600 kPa )⎜⎜ 1 kPa ⋅ m3 ⎟⎠ ⎝ = 10.35 kJ/kg

h4 = h3 + wpII,in = 670.38 + 10.35 = 680.73 kJ/kg P5 = 10 MPa ⎫ h5 = 3375.1 kJ/kg T5 = 500°C ⎬⎭ s5 = 6.5995 kJ/kg ⋅ K

P8 = 0.6 MPa ⎫ ⎬ h8 = 3310.2 kJ/kg s8 = s7 ⎭

9 Condens.

2

1

3

PI

1 MPa 5 7

T 10 MPa

P6 = 1.0 MPa ⎫ ⎬ h6 = 2783.8 kJ/kg s6 = s5 ⎭ P7 = 1.0 MPa ⎫ h7 = 3479.1 kJ/kg T7 = 500°C ⎬⎭ s7 = 7.7642 kJ/kg ⋅ K

1-y

8 y

P3 = 0.6 MPa ⎫ h3 = h f @ 0.6 MPa = 670.38 kJ/kg ⎬v = v 3 sat. liquid f @ 0.6 MPa = 0.001101 m /kg ⎭ 3 wpII,in = v 3 (P4 − P3 )

Turbine

6

Boiler

4 3

6

8

0.6 MPa

2 15 kPa 1

9

s

s9 − s f

7.7642 − 0.7549 = = 0.9665 P9 = 15 kPa ⎫ x9 = s 7.2522 ⎬ fg s9 = s7 ⎭ h9 = h f + x9 h fg = 225.94 + (0.9665)(2372.3) = 2518.8 kJ/kg

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 , E& in − E& out = ∆E& system ©0 (steady) = 0 → E& in = E& out

∑ m& h = ∑ m& h i i

e e

⎯ ⎯→ m& 8 h8 + m& 2 h2 = m& 3 h3 ⎯ ⎯→ yh8 + (1 − y )h2 = 1(h3 )

where y is the fraction of steam extracted from the turbine ( = m& 8 / m& 3 ). Solving for y, y=

h3 − h2 670.38 − 226.53 = 0.144 = h8 − h2 3310.2 − 226.53

(b) The thermal efficiency is determined from q in = (h5 − h4 ) + (h7 − h6 ) = (3375.1 − 680.73) + (3479.1 − 2783.8) = 3389.7 kJ/kg

q out = (1 − y )(h9 − h1 ) = (1 − 0.1440 )(2518.8 − 225.94 ) = 1962.7 kJ/kg

and

η th = 1 −

q out 1962.7 kJ/kg = 1− = 42.1% 3389.7 kJ/kg q in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-107

10-103 A nonideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. The fraction of steam extracted for regeneration and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis

T

5 Boiler

5

Turbine

6 7

8

1-y

4

9

3

y Open fwh

4

P II

2

Condenser

2

3

1

1

PI

7

6 6s 8s 8 y 1-y

9s 9

s

(a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 15 kPa = 225.94 kJ/kg

v 1 = v f @ 15 kPa = 0.001014 m 3 /kg w pI ,in = v 1 (P2 − P1 )

(

)

⎛ 1 kJ = 0.001014 m 3 /kg (600 − 15 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 0.59 kJ/kg

⎞ ⎟ ⎟ ⎠

h2 = h1 + w pI ,in = 225.94 + 0.59 = 226.54 kJ/kg P3 = 0.6 MPa ⎫ h3 = h f @ 0.6 MPa = 670.38 kJ/kg ⎬v = v 3 sat. liquid f @ 0.6 MPa = 0.001101 m /kg ⎭ 3 wpII,in = v 3 (P4 − P3 )

(

)

⎛ 1 kJ ⎞ ⎟ = 0.001101 m3/kg (10,000 − 600 kPa )⎜⎜ 1 kPa ⋅ m3 ⎟⎠ ⎝ = 10.35 kJ/kg

h4 = h3 + wpII,in = 670.38 + 10.35 = 680.73 kJ/kg P5 = 10 MPa ⎫ h5 = 3375.1 kJ/kg T5 = 500°C ⎬⎭ s5 = 6.5995 kJ/kg ⋅ K P6 s = 1.0 MPa ⎫ ⎬ h6 s = 2783.8 kJ/kg s6 s = s5 ⎭

ηT =

h5 − h6 ⎯ ⎯→ h6 = h5 − ηT (h5 − h6 s ) h5 − h6 s = 3375.1 − (0.84)(3375.1 − 2783.8) = 2878.4 kJ/kg

P7 = 1.0 MPa ⎫ h7 = 3479.1 kJ/kg T7 = 500°C ⎬⎭ s7 = 7.7642 kJ/kg ⋅ K P8 s = 0.6 MPa ⎫ ⎬ h8 s = 3310.2 kJ/kg s8 s = s7 ⎭ ηT =

h7 − h8 ⎯ ⎯→ h8 = h7 − ηT (h7 − h8 s ) = 3479.1 − (0.84)(3479.1 − 3310.2 ) h7 − h8 s = 3337.2 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-108

s 9 s − s f 7.7642 − 0.7549 = = 0.9665 P9 s = 15 kPa ⎫ x 9 s = s fg 7.2522 ⎬ s9s = s7 ⎭ h9 s = h f + x 9 s h fg = 225.94 + (0.9665)(2372.3) = 2518.8 kJ/kg

ηT =

h7 − h9 ⎯ ⎯→ h9 = h7 − ηT (h7 − h9 s ) = 3479.1 − (0.84)(3479.1 − 2518.8) h7 − h9 s = 2672.5 kJ/kg

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 , E& in − E& out = ∆E& system©0 (steady) = 0 E& in = E& out

∑ m& h = ∑ m& h i i

e e

⎯ ⎯→ m& 8 h8 + m& 2 h2 = m& 3 h3 ⎯ ⎯→ yh8 + (1 − y )h2 = 1(h3 )

where y is the fraction of steam extracted from the turbine ( = m& 8 / m& 3 ). Solving for y, y=

h3 − h2 670.38 − 226.53 = 0.1427 = h8 − h2 3335.3 − 226.53

(b) The thermal efficiency is determined from qin = (h5 − h4 ) + (h7 − h6 ) = (3375.1 − 680.73) + (3479.1 − 2878.4 ) = 3295.1 kJ/kg qout = (1 − y )(h9 − h1 ) = (1 − 0.1427 )(2672.5 − 225.94 ) = 2097.2 kJ/kg

and

η th = 1 −

q out 2097.2 kJ/kg = 1− = 36.4% 3295.1 kJ/kg q in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-109

10-104 A steam power plant operating on the ideal reheat-regenerative Rankine cycle with three feedwater heaters is considered. Various items for this system per unit of mass flow rate through the boiler are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

High-P Turbine

Low-P Turbine

13 Boiler

16 z 14

15

12 Closed FWH I 10

11

9

P IV

8 6

Closed FWH II 7

5

n

17

y

x

19

P II 4

Open FWH 3

m

18

Condenser

PI 2

1

P III

Analysis The compression processes in the pumps and the expansion processes in the turbines are isentropic. Also, the state of water at the inlet of pumps is saturated liquid. Then, from the steam tables (Tables A-4, A-5, and A-6), h1 h2 h3 h4 h5 h6 h9 h10

= 168.75 kJ/kg = 168.84 kJ/kg = 417.51 kJ/kg = 419.28 kJ/kg = 884.46 kJ/kg = 885.86 kJ/kg = 1008.3 kJ/kg = 1011.8 kJ/kg

h13 = 3423.1 kJ/kg h14 = 3204.5 kJ/kg h15 = 3063.6 kJ/kg h16 = 2871.0 kJ/kg h17 = 3481.3 kJ/kg h18 = 2891.5 kJ/kg h19 = 2454.7 kJ/kg

For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure. Then, P7 = 1800 kPa

⎫ ⎬ h7 = 884.91 kJ/kg T7 = T5 = 207.1°C ⎭ P11 = 3000 kPa ⎫ ⎬ h11 = 1008.8 kJ/kg T11 = T9 = 233.9°C ⎭ Enthalpies at other states and the fractions of steam extracted from the turbines can be determined from mass and energy balances on cycle components as follows: Mass Balances: x + y + z =1 m+n = z

Open feedwater heater: mh18 + nh2 = zh3

Closed feedwater heater-II: zh4 + yh15 = zh7 + yh5 PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-110

Closed feedwater heater-I: ( y + z )h8 + xh14 = ( y + z )h11 + xh9

Mixing chamber after closed feedwater heater II: zh7 + yh6 = ( y + z )h8

Mixing chamber after closed feedwater heater I: xh10 + ( y + z )h11 = 1h12

Substituting the values and solving the above equations simultaneously using EES, we obtain h8 = 885.08 kJ/kg h12 = 1009.0 kJ/kg x = 0.05334 y = 0.1667 z = 0.78000 m = 0.07124 n = 0.70882 Note that these values may also be obtained by a hand solution by using the equations above with some rearrangements and substitutions. Other results of the cycle are wT,out,HP = x(h13 − h14 ) + y (h13 − h15 ) + z (h13 − h16 ) = 502.3 kJ/kg wT,out,LP = m(h17 − h18 ) + n(h17 − h19 ) = 769.6 kJ/kg q in = h13 − h12 + z (h17 − h16 ) = 2890 kJ/kg q out = n(h19 − h1 ) = 1620 kJ/kg

η th = 1 −

q out 1620 =1− = 0.4394 = 43.9% q in 2890

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-111

10-105 The optimum bleed pressure for the open feedwater heater that maximizes the thermal efficiency of the cycle is to be determined using EES. Analysis The EES program used to solve this problem as well as the solutions are given below. "Given" P_boiler=6000 [kPa] P_cfwh1=3000 [kPa] P_cfwh2=1800 [kPa] P_reheat=800 [kPa] "P_ofwh=100 [kPa]" P_condenser=7.5 [kPa] T_turbine=500 [C] "Analysis" Fluid\$='steam_iapws' "turbines" h[13]=enthalpy(Fluid\$, P=P_boiler, T=T_turbine) s[13]=entropy(Fluid\$, P=P_boiler, T=T_turbine) h[14]=enthalpy(Fluid\$, P=P_cfwh1, s=s[13]) h[15]=enthalpy(Fluid\$, P=P_cfwh2, s=s[13]) h[16]=enthalpy(Fluid\$, P=P_reheat, s=s[13]) h[17]=enthalpy(Fluid\$, P=P_reheat, T=T_turbine) s[17]=entropy(Fluid\$, P=P_reheat, T=T_turbine) h[18]=enthalpy(Fluid\$, P=P_ofwh, s=s[17]) h[19]=enthalpy(Fluid\$, P=P_condenser, s=s[17]) "pump I" h[1]=enthalpy(Fluid\$, P=P_condenser, x=0) v[1]=volume(Fluid\$, P=P_condenser, x=0) w_pI_in=v[1]*(P_ofwh-P_condenser) h[2]=h[1]+w_pI_in "pump II" h[3]=enthalpy(Fluid\$, P=P_ofwh, x=0) v[3]=volume(Fluid\$, P=P_ofwh, x=0) w_pII_in=v[3]*(P_cfwh2-P_ofwh) h[4]=h[3]+w_pII_in "pump III" h[5]=enthalpy(Fluid\$, P=P_cfwh2, x=0) T[5]=temperature(Fluid\$, P=P_cfwh2, x=0) v[5]=volume(Fluid\$, P=P_cfwh2, x=0) w_pIII_in=v[5]*(P_cfwh1-P_cfwh2) h[6]=h[5]+w_pIII_in "pump IV" h[9]=enthalpy(Fluid\$, P=P_cfwh1, x=0) T[9]=temperature(Fluid\$, P=P_cfwh1, x=0) v[9]=volume(Fluid\$, P=P_cfwh1, x=0) w_p4_in=v[5]*(P_boiler-P_cfwh1) h[10]=h[9]+w_p4_in "Mass balances" x+y+z=1 m+n=z

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-112

"Open feedwater heater" m*h[18]+n*h[2]=z*h[3] "closed feedwater heater 2" T[7]=T[5] h[7]=enthalpy(Fluid\$, P=P_cfwh1, T=T[7]) z*h[4]+y*h[15]=z*h[7]+y*h[5] "closed feedwater heater 1" T[11]=T[9] h[11]=enthalpy(Fluid\$, P=P_boiler, T=T[11]) (y+z)*h[8]+x*h[14]=(y+z)*h[11]+x*h[9] "Mixing chamber after closed feedwater heater 2" z*h[7]+y*h[6]=(y+z)*h[8] "Mixing chamber after closed feedwater heater 1" x*h[10]+(y+z)*h[11]=1*h[12] "cycle" w_T_out_high=x*(h[13]-h[14])+y*(h[13]-h[15])+z*(h[13]-h[16]) w_T_out_low=m*(h[17]-h[18])+n*(h[17]-h[19]) q_in=h[13]-h[12]+z*(h[17]-h[16]) q_out=n*(h[19]-h[1]) Eta_th=1-q_out/q_in

ηth 0.429828 0.433780 0.435764 0.436978 0.437790 0.438359 0.438768 0.439065 0.439280 0.439432 0.439536 0.439602 0.439638 0.439647 0.439636 0.439608 0.439565 0.439509 0.439442 0.439367 0.439283 0.439192 0.439095 0.438993 0.438887 0.438776 0.438662 0.438544 0.438424 0.438301

0.44 0.438 0.436

η th

P open fwh [kPa] 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300

0.434 0.432 0.43 0.428 0

50

100

150

200

250

300

Pofwh [kPa]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-113

10-106 A cogeneration plant is to produce power and process heat. There are two turbines in the cycle: a high-pressure turbine and a low-pressure turbine. The temperature, pressure, and mass flow rate of steam at the inlet of high-pressure turbine are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), P4 = 1.4 MPa ⎫ h4 = h g @ 1.4 MPa = 2788.9 kJ/kg ⎬s = s sat. vapor g @ 1.4 MPa = 6.4675 kJ/kg ⋅ K ⎭ 4 x5s =

s 4s − s f

=

3

6.4675 − 0.6492 = 0.7758 7.4996

s fg P5 = 10 kPa ⎫ h h = ⎬ 5s f + x 5 s h fg s5s = s 4 ⎭ = 191.81 + (0.7758)(2392.1) = 2047.6 kJ/kg

ηT =

T

h4 − h5 ⎯ ⎯→ h5 = h4 − ηT (h4 − h5 s ) h4 − h5 s = 2788.9 − (0.60 )(2788.9 − 2047.6 )

2 1

= 2344.1 kJ/kg

4

4

5 5

s

and wturb,low = h4 − h5 = 2788.9 − 2344.1 = 444.8 kJ/kg m& low turb =

W& turb,II wturb,low

=

800 kJ/s = 1.799 kg/s = 107.9 kg/min 444.8 kJ/kg

Therefore , m& total = 1000 + 108 = 1108 kg/min = 18.47 kg/s wturb,high =

W& turb, I m& high,turb

=

1000 kJ/s = 54.15 kJ/kg = h3 − h4 18.47 kg/s

h3 = wturb,high + h4 = 54.15 + 2788.9 = 2843.0 kJ/kg

ηT =

h3 − h4 ⎯ ⎯→ h4 s = h3 − (h3 − h4 ) / ηT h3 − h4 s = 2843.0 − (2843.0 − 2788.9) / (0.75) = 2770.8 kJ/kg

h4 s − h f 2770.8 − 829.96 = = 0.9908 P4 s = 1.4 MPa ⎫ x 4 s = 1958.9 h fg ⎬ s 4s = s3 ⎭ s 4 s = s f + x 4 s s fg = 2.2835 + (0.9908)(4.1840 ) = 6.4289 kJ/kg ⋅ K

Then from the tables or the software, the turbine inlet temperature and pressure becomes h3 = 2843.0 kJ/kg ⎫ P3 = 2 MPa ⎬ s 3 = 6.4289 kJ/kg ⋅ K ⎭ T3 = 227.5°C

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-114

10-107 A cogeneration plant is to generate power and process heat. Part of the steam extracted from the turbine at a relatively high pressure is used for process heating. The rate of process heat, the net power produced, and the utilization factor of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

T 6 Boiler

6

Turbine

7 8 Process heater

5

3 PI 4

2

Conden.

P II

8 MPa · Qin 2 MPa 4 3 · Qprocess 5

1

1

2

· Qout

7

20 kPa 8s 8 s

Analysis From the steam tables (Tables A-4, A-5, and A-6),

h1 = h f @ 20 kPa = 251.42 kJ/kg

v1 = v f

@ 20 kPa

= 0.001017 m 3 /kg

wpI,in = v 1 (P2 − P1 ) / η p

(

)

⎛ 1 kJ = 0.001017 m 3 /kg (2000 − 20 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 2.29 kJ/kg

⎞ ⎟ / 0.88 ⎟ ⎠

h2 = h1 + wpI,in = 251.42 + 2.29 = 253.71 kJ/kg h3 = h f

@ 2 MPa

= 908.47 kJ/kg

Mixing chamber: m& 3 h3 + m& 2 h2 = m& 4 h4 (4 kg/s)(908.47 kJ/kg) + (11 − 4 kg/s)(253.71 kJ/kg)) = (11 kg/s) h4 ⎯ ⎯→ h4 = 491.81 kJ/kg

v4 ≅ v f

w pII,in = v 4 (P5 − P4 ) / η p

(

= 0.001058 m 3/kg

@ h f = 491.81 kJ/kg

)

⎛ 1 kJ ⎞ ⎟ / 0.88 = 0.001058 m3 /kg (8000 − 2000 kPa )⎜⎜ 3⎟ ⋅ 1 kPa m ⎝ ⎠ = 7.21 kJ/kg

h5 = h4 + w pII,in = 491.81 + 7.21 = 499.02 kJ/kg P6 = 8 MPa ⎫ h6 = 3399.5 kJ/kg ⎬ T6 = 500°C ⎭ s 6 = 6.7266 kJ/kg ⋅ K P7 = 2 MPa ⎫ ⎬ h7 s = 3000.4 kJ/kg s7 = s6 ⎭

ηT =

h6 − h7 ⎯ ⎯→ h7 = h6 − η T (h6 − h7 s ) = 3399.5 − (0.88)(3399.5 − 3000.4 ) = 3048.3 kJ/kg h6 − h7 s

P8 = 20 kPa ⎫ ⎬ h8 s = 2215.5 kJ/kg s8 = s 6 ⎭ PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-115

ηT =

h6 − h8 ⎯ ⎯→ h8 = h6 − η T (h6 − h8 s ) = 3399.5 − (0.88)(3399.5 − 2215.5) = 2357.6 kJ/kg h6 − h8 s

Then, Q& process = m& 7 (h7 − h3 ) = (4 kg/s )(3048.3 − 908.47 ) kJ/kg = 8559 kW

(b) Cycle analysis: W& T,out = m& 7 (h6 − h7 ) + m& 8 (h6 − h8 ) = (4 kg/s )(3399.5 − 3048.3)kJ/kg + (7 kg/s )(3399.5 − 2357.6 )kJ/kg = 8698 kW W& p,in = m& 1 w pI,in + m& 4 wpII,in = (7 kg/s )(2.29 kJ/kg ) + (11 kg/s )(7.21 kJ/kg ) = 95 kW W& net = W& T,out − W& p,in = 8698 − 95 = 8603 kW

(c) Then,

Q& in = m& 5 (h6 − h5 ) = (11 kg/s )(3399.5 − 499.02) = 31,905 kW and

εu =

W& net + Q& process 8603 + 8559 = = 0.538 = 53.8% 31,905 Q& in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-116

10-108E A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The thermal efficiency of the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable for Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea). Analysis Working around the topping cycle gives the following results: T6 s

⎛P = T5 ⎜⎜ 6 ⎝ P5

ηC =

⎞ ⎟ ⎟ ⎠

( k −1) / k

7

2560 R

h6 s − h5 c p (T6 s − T5 ) = h 6 − h5 c p (T6 − T5 )

⎯ ⎯→ T6 = T5 +

⎛P T8 s = T7 ⎜⎜ 8 ⎝ P7

⎞ ⎟ ⎟ ⎠

· Qin

T6 s − T5

= 540 +

ηT =

T

= (540 R)(10) 0.4/1.4 = 1043 R

ηC

6

1043 − 540 = 1099 R 0.90

( k −1) / k

⎛1⎞ = (2560 R)⎜ ⎟ ⎝ 10 ⎠

GAS CYCLE

6s

8s 800 psia

0.4/1.4

8 3 600°F

9

= 1326 R

c p (T7 − T8 ) h7 − h8 = ⎯ ⎯→ T8 = T7 − η T (T7 − T8 s ) h7 − h8 s c p (T7 − T8 s ) = 2560 − (0.90)(2560 − 1326)

540 R

5

2 1

STEAM CYCLE 5 psia · 4s Qout

4

s

= 1449 R

T9 = Tsat @ 800 psia + 50 = 978.3 R + 50 = 1028 R Fixing the states around the bottom steam cycle yields (Tables A-4E, A-5E, A-6E): h1 = h f @ 5 psia = 130.18 Btu/lbm

v 1 = v f @ 5 psia = 0.01641 ft 3 /lbm wp,in = v 1 ( P2 − P1 )

⎛ 1 Btu = (0.01641 ft 3 /lbm)(800 − 5)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 2.41 Btu/lbm

⎞ ⎟ ⎟ ⎠

h2 = h1 + wp,in = 130.18 + 2.41 = 132.59 Btu/lbm P3 = 800 psia ⎫ h3 = 1270.9 Btu/lbm ⎬ T3 = 600°F ⎭ s 3 = 1.4866 Btu/lbm ⋅ R P4 = 5 psia ⎫ ⎬ h4 s = 908.6 Btu/lbm s 4 = s3 ⎭

ηT =

h3 − h 4 ⎯ ⎯→h4 = h3 − η T (h3 − h4 s ) h3 − h 4 s = 1270.9 − (0.95)(1270.9 − 908.6) = 926.7 Btu/lbm

The net work outputs from each cycle are w net, gas cycle = wT, out − wC,in = c p (T7 − T8 ) − c p (T6 − T5 ) = (0.240 Btu/lbm ⋅ R )(2560 − 1449 − 1099 + 540)R = 132.5 Btu/lbm PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-117

w net, steam cycle = wT,out − wP,in = (h3 − h4 ) − w P,in = (1270.9 − 926.7) − 2.41 = 341.8 Btu/lbm

An energy balance on the heat exchanger gives m& a c p (T8 − T9 ) = m& w (h3 -h2 ) ⎯ ⎯→ m& w =

c p (T8 − T9 ) h3 -h2

m& a =

(0.240)(1449 − 1028) = 0.08876m& a 1270.9 − 132.59

That is, 1 lbm of exhaust gases can heat only 0.08876 lbm of water. Then the heat input, the heat output and the thermal efficiency are q in = q out =

m& a c p (T7 − T6 ) = (0.240 Btu/lbm ⋅ R )(2560 − 1099)R = 350.6 Btu/lbm m& a m& m& a c p (T9 − T5 ) + w (h4 − h1 ) & m& a ma

= 1× (0.240 Btu/lbm ⋅ R )(1028 − 540)R + 0.08876 × (926.7 − 130.18) Btu/lbm = 187.8 Btu/lbm

η th = 1 −

q out 187.8 = 1− = 0.4643 q in 350.6

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-118

10-109E A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The thermal efficiency of the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable fo Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea).

T

Analysis Working around the topping cycle gives the following results: ⎛P T6 s = T5 ⎜⎜ 6 ⎝ P5

ηC =

⎞ ⎟ ⎟ ⎠

( k −1) / k

= (540 R)(10) 0.4/1.4 = 1043 R

⎯ ⎯→ T6 = T5 +

⎞ ⎟ ⎟ ⎠

6s

8 3 600°F

9

1043 − 540 = 1099 R 0.90 ⎛1⎞ = (2560 R)⎜ ⎟ ⎝ 10 ⎠

8s 800 psia

ηC

( k −1) / k

GAS CYCLE

6

T6 s − T5

= 540 +

ηT =

· Qin

h6 s − h5 c p (T6 s − T5 ) = h 6 − h5 c p (T6 − T5 )

⎛P T8 s = T7 ⎜⎜ 8 ⎝ P7

7

2560 R

540 R

0.4/1.4

= 1326 R

5

2 1

STEAM CYCLE 10 psia · 4s Qout

4

s

c p (T7 − T8 ) h7 − h8 = ⎯ ⎯→ T8 = T7 − η T (T7 − T8 s ) h7 − h8 s c p (T7 − T8 s ) = 2560 − (0.90)(2560 − 1326) = 1449 R

T9 = Tsat @ 800 psia + 50 = 978.3 R + 50 = 1028 R Fixing the states around the bottom steam cycle yields (Tables A-4E, A-5E, A-6E): h1 = h f @ 10 psia = 161.25 Btu/lbm

v 1 = v f @ 10 psia = 0.01659 ft 3 /lbm wp,in = v 1 ( P2 − P1 )

⎛ 1 Btu = (0.01659 ft 3 /lbm)(800 − 10)psia ⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ = 2.43 Btu/lbm

⎞ ⎟ ⎟ ⎠

h2 = h1 + wp,in = 161.25 + 2.43 = 163.7 Btu/lbm P3 = 800 psia ⎫ h3 = 1270.9 Btu/lbm ⎬ T3 = 600°F ⎭ s 3 = 1.4866 Btu/lbm ⋅ R P4 = 10 psia ⎫ ⎬ h4 s = 946.6 Btu/lbm s 4 = s3 ⎭

ηT =

h3 − h 4 ⎯ ⎯→h4 = h3 − η T (h3 − h4 s ) h3 − h 4 s = 1270.9 − (0.95)(1270.9 − 946.6) = 962.8 Btu/lbm

The net work outputs from each cycle are

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-119

w net, gas cycle = wT, out − wC,in = c p (T7 − T8 ) − c p (T6 − T5 ) = (0.240 Btu/lbm ⋅ R )(2560 − 1449 − 1099 + 540)R = 132.5 Btu/lbm

w net, steam cycle = wT,out − wP,in = (h3 − h4 ) − wP,in = (1270.9 − 962.8) − 2.43 = 305.7 Btu/lbm

An energy balance on the heat exchanger gives

m& a c p (T8 − T9 ) = m& w (h3 − h2 ) ⎯ ⎯→ m& w =

c p (T8 − T9 ) h3 − h2

m& a =

(0.240)(1449 − 1028) = 0.09126m& a 1270.9 − 163.7

That is, 1 lbm of exhaust gases can heat only 0.09126 lbm of water. Then the heat input, the heat output and the thermal efficiency are q in = q out =

m& a c p (T7 − T6 ) = (0.240 Btu/lbm ⋅ R )(2560 − 1099)R = 350.6 Btu/lbm m& a m& m& a c p (T9 − T5 ) + w (h4 − h1 ) m& a m& a

= 1× (0.240 Btu/lbm ⋅ R )(1028 − 540)R + 0.09126 × (962.8 − 161.25) Btu/lbm = 190.3 Btu/lbm

η th = 1 −

q out 190.3 = 1− = 0.4573 q in 350.6

When the condenser pressure is increased from 5 psia to 10 psia, the thermal efficiency is decreased from 0.4643 to 0.4573.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-120

10-110E A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The cycle supplies a specified rate of heat to the buildings during winter. The mass flow rate of air and the net power output from the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 The airstandard assumptions are applicable to Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.

T 7

2560 R

Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea).

· Qin

Analysis The mass flow rate of water is m& w =

Q& buildings h4 − h1

2 × 10 6 Btu/h = = 2495 lbm/h (962.8 - 161.25) Btu/lbm

6 6s

8 3 600°F

9

m& w 2495 = = 27,340 lbm/h 0.09126 0.09126

The power outputs from each cycle are

8s 800 psia

The mass flow rate of air is then m& a =

GAS CYCLE

540 R

W& net, gas cycle = m& a ( wT, out − wC,in )

5

STEAM CYCLE

2 1

10 psia

· Qout

= m& a c p (T7 − T8 ) − m& a c p (T6 − T5 )

4s

4

s

1 kW ⎛ ⎞ = (27,340 lbm/h)(0.240 Btu/lbm ⋅ R )(2560 − 1449 − 1099 + 540)R ⎜ ⎟ ⎝ 3412.14 Btu/h ⎠ = 1062 kW W& net, steam cycle = m& a ( wT,out − wP,in ) = m& a (h3 − h4 − wP,in ) 1 kW ⎛ ⎞ = (2495 lbm/h)(1270.9 − 962.8 − 2.43)⎜ ⎟ 3412.14 Btu/h ⎝ ⎠ = 224 kW The net electricity production by this cycle is then

W& net = 1062 + 224 = 1286 kW

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-121

10-111 A combined gas-steam power plant is considered. The topping cycle is an ideal gas-turbine cycle and the bottoming cycle is an ideal reheat Rankine cycle. The mass flow rate of air in the gas-turbine cycle, the rate of total heat input, and the thermal efficiency of the combined cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Analysis (a) The analysis of gas cycle yields T ⎯→ h7 = 310.24 kJ/kg T7 = 310 K ⎯ 9 1400 K Pr7 = 1.5546 P ⎯→ h8 = 630.18 kJ/kg Pr8 = 8 Pr7 = (12)(1.5546 ) = 18.66 ⎯ · Qin P7 GAS CYCLE ⎯→ h9 = 1515.42 kJ/kg T9 = 1400 K ⎯ Pr9 = 450.5 10 3 P10 ⎛1⎞ 8 2.5 MPa ⎯→ h10 = 768.38 kJ/kg Pr = ⎜ ⎟(450.5) = 37.54 ⎯ Pr10 = 12.5 MPa P9 9 ⎝ 12 ⎠ 5 ⎯→ h11 = 523.63 kJ/kg T11 = 520 K ⎯ From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f

@ 10 kPa

310 K

7

1

= 0.00101 m 3 /kg

(

2

)

⎛ 1 kJ wpI,in = v 1 (P2 − P1 ) = 0.00101 m 3 /kg (12,500 − 10 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ h2 = h1 + wpI,in = 191.81 + 12.62 = 204.42 kJ/kg

11 STEAM CYCLE 4 10 kPa · Qout

6 s

⎞ ⎟ = 12.62 kJ/kg ⎟ ⎠

P3 = 12.5 MPa ⎫ h3 = 3343.6 kJ/kg ⎬ s = 6.4651 kJ/kg ⋅ K T3 = 500°C ⎭ 3 P4 = 2.5 MPa ⎫ ⎬h4 = 2909.6 kJ/kg s 4 = s3 ⎭ P5 = 2.5 MPa ⎫ h5 = 3574.4 kJ/kg T5 = 550°C ⎬⎭ s 5 = 7.4653 kJ/kg ⋅ K s 6 − s f 7.4653 − 0.6492 = = 0.9089 P6 = 10 kPa ⎫ x 6 = s fg 7.4996 ⎬ s 6 = s5 ⎭ h = h + x h = 191.81 + (0.9089 )(2392.1) = 2365.8 kJ/kg 6 f 6 fg

Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 for the heat exchanger, the steady-flow energy balance equation yields E& in = E& out ⎯ ⎯→

∑ m& h = ∑ m& h i i

e e

⎯ ⎯→ m& s (h3 − h2 ) = m& air (h10 − h11 )

h3 − h2 3343.6 − 204.42 (12 kg/s ) = 153.9 kg/s m& s = h10 − h11 768.38 − 523.63 (b) The rate of total heat input is Q& in = Q& air + Q& reheat = m& air (h9 − h8 ) + m& reheat (h5 − h4 ) = (153.9 kg/s )(1515.42 − 630.18) kJ/kg + (12 kg/s )(3574.4 − 2909.6 )kJ/kg = 144,200 kW ≅ 1.44 × 10 5 kW (c) The rate of heat rejection and the thermal efficiency are then Q& out = Q& out,air + Q& out,steam = m& air (h11 − h7 ) + m& s (h6 − h1 ) = (153.9 kg/s )(523.63 − 310.24) kJ/kg + (12 kg/s )(2365.8 − 191.81) kJ/kg = 58,930 kW Q& 58,930 kW η th = 1 − out = 1 − = 0.5913 = 59.1% & 144,200 kW Qin m& air =

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-122

10-112 A combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a nonideal reheat Rankine cycle. The mass flow rate of air in the gas-turbine cycle, the rate of total heat input, and the thermal efficiency of the combined cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Analysis (a) The analysis of gas cycle yields (Table A-17) T7 = 290 K ⎯ ⎯→ h7 = 290.16 kJ/kg Pr 7 = 1.2311

T 9

1400 K · Qin

P = 8 s Pr 7 = (8)(1.2311) = 9.849 ⎯ ⎯→ h8 s = 526.12 kJ/kg P7

Pr8 s

h − h7 ηC = 8 s ⎯ ⎯→ h8 = h7 + (h8 s − h7 ) / ηC h8 − h7 = 290.16 + (526.12 − 290.16 ) / (0.80 ) = 585.1 kJ/kg T9 = 1400 K ⎯ ⎯→ h9 = 1515.42 kJ/kg Pr 9 = 450.5 Pr10 s

P ⎛1⎞ ⎯→ h10 s = 860.35 kJ/kg = 10 s Pr 9 = ⎜ ⎟(450.5) = 56.3 ⎯ P9 ⎝8⎠

8

290 K

7

GAS CYCLE 1 10s 3

8 15 MPa

2 1

3 MPa 5

11 STEAM 4 CYCLE 4 10 kPa · 6s 6 Qout

h −h ηT = 9 10 ⎯ ⎯→ h10 = h9 − ηT (h9 − h10 s ) h9 − h10 s = 1515.42 − (0.85)(1515.42 − 860.35) = 958.4 kJ/kg T11 = 520 K ⎯ ⎯→ h11 = 523.63 kJ/kg

From the steam tables (Tables A-4, A-5, and A-6), h1 = h f

@ 10 kPa

v1 = v f

@ 10 kPa

wpI,in = v1 (P2 − P1 )

(

= 191.81 kJ/kg = 0.00101 m 3 /kg

)

⎛ 1 kJ ⎞ ⎟ = 0.00101 m 3 /kg (15,000 − 10 kPa )⎜⎜ 1 kPa ⋅ m 3 ⎟⎠ ⎝ = 15.14 kJ/kg

h2 = h1 + wpI,in = 191.81 + 15.14 = 206.95 kJ/kg P3 = 15 MPa ⎫ h3 = 3157.9 kJ/kg T3 = 450°C ⎬⎭ s3 = 6.1428 kJ/kg ⋅ K s4 s − s f 6.1434 − 2.6454 = = 0.9880 P4 = 3 MPa ⎫ x4 s = s fg 3.5402 ⎬ s4 s = s3 ⎭ h4 s = h f + x4 s h fg = 1008.3 + (0.9879)(1794.9 ) = 2781.7 kJ/kg ηT =

h3 − h4 ⎯ ⎯→ h4 = h3 − ηT (h3 − h4 s ) h3 − h4 s = 3157.9 − (0.85)(3157.9 − 2781.7 ) = 2838.1 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

s

10-123

P5 = 3 MPa ⎫ h5 = 3457.2 kJ/kg T5 = 500°C ⎬⎭ s5 = 7.2359 kJ/kg ⋅ K s6 s − s f 7.2359 − 0.6492 = = 0.8783 P6 = 10 kPa ⎫ x6 s = 7.4996 s fg ⎬ s6 s = s5 ⎭ h = h + x h = 191.81 + (0.8782 )(2392.1) = 2292.8 kJ/kg 6s f 6 s fg ηT =

h5 − h6 ⎯ ⎯→ h6 = h5 − ηT (h5 − h6 s ) h5 − h6 s = 3457.2 − (0.85)(3457.2 − 2292.8) = 2467.5 kJ/kg

Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 for the heat exchanger, the steady-flow energy balance equation yields E& in − E& out = ∆E& system©0 (steady) = 0 E& in = E& out

∑ m& h = ∑ m& h i i

m& air =

e e

⎯ ⎯→ m& s (h3 − h2 ) = m& air (h10 − h11 )

h3 − h2 3157.9 − 206.95 m& s = (30 kg/s ) = 203.6 kg/s h10 − h11 958.4 − 523.63

(b)

Q& in = Q& air + Q& reheat = m& air (h9 − h8 ) + m& reheat (h5 − h4 ) = (203.6 kg/s )(1515.42 − 585.1) kJ/kg + (30 kg/s )(3457.2 − 2838.1) kJ/kg = 207,986 kW

(c)

Q& out = Q& out,air + Q& out,steam = m& air (h11 − h7 ) + m& s (h6 − h1 ) = (203.6 kg/s )(523.63 − 290.16 ) kJ/kg + (30 kg/s )(2467.5 − 191.81) kJ/kg = 115,805 kW Q& 115,805 kW η th = 1 − out = 1 − = 44.3% 207,986 kW Q& in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-124

10-113 A Rankine steam cycle modified with two closed feedwater heaters and one open feed water heater is considered. The T-s diagram for the ideal cycle is to be sketched. The fraction of mass extracted for the open feedwater heater y and the cooling water flow temperature rise are to be determined. Also, the rate of heat rejected in the condenser and the thermal efficiency of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (b) Using the data from the problem statement, the enthalpies at various states are h1 = h f

@ 20 kPa

h15 = h3 = h14 = h f h4 = h f

@ 620 kPa

h6 = h12 = h f

5 MPa

= 251 kJ/kg @ 140 kPa

7

= 458 kJ/kg

620 kPa

= 676 kJ/kg

@ 1910 kPa

1910 kPa

T

= 898 kJ/kg

where z is the fraction of steam extracted from the low-pressure turbine. Solving for z, h − h3 676 − 458 y= 4 = = 0.08086 h9 − h3 3154 − 458

w

9

y z

14

3

140 kPa

12

6 5 4

An energy balance on the open feedwater heater gives yh9 + (1 − y )h3 = 1h4

8

w

10

20 kPa

13 w+z 1-y-z-w

2 1

15

11

s

(c) An energy balance on the condenser gives

m& 7 [(1 − w − y − z )h11 + ( w + z )h15 − (1 − y)h1 ] = m& w (hw2 − hw2 ) = m& w c pw ∆Tw Solving for the temperature rise of cooling water, and substituting with correct units, ∆Tw =

m& 7 [(1 − w − y − z )h11 + ( w + z )h15 − (1 − y )h1 ] m& w c pw

(100)[(1 − 0.0830 − 0.08086 − 0.0655)(2478) + (0.0830 + 0.0655)(458) − (1 − 0.08086)(251)] (4200)(4.18) = 9.95°C =

(d) The rate of heat rejected in the condenser is Q& out = m& w c pw ∆Tw = ( 4200 kg/s)(4.18 kJ/kg ⋅ °C)(9.95°C) = 174,700 kW

The rate of heat input in the boiler is

Q& in = m& (h7 − h6 ) = (100 kg/s)(3900 − 898) kJ/kg = 300,200 kW The thermal efficiency is then

η th = 1 −

Q& out 174,700 kW =1− = 0.418 = 41.8% & 300,200 kW Q in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-125

10-114 A Rankine steam cycle modified for reheat, two closed feedwater heaters and a process heater is considered. The T-s diagram for the ideal cycle is to be sketched. The fraction of mass, w, that is extracted for the closed feedwater heater is to be determined. Also, the mass flow rate through the boiler, the rate of process heat supplied, and the utilization efficiency of this cogeneration plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (b) Using the data from the problem statement, the enthalpies at various states are h1 = h f

@ 20 kPa

= 251.4 kJ/kg

v1 = v f

@ 20 kPa

= 0.00102 m 3 /kg

wpI,in = v 1 (P2 − P1 )

(

)

⎛ 1 kJ = 0.00102 m 3 /kg (5000 − 20 kPa )⎜ ⎜ 1 kPa ⋅ m 3 ⎝ = 5.1 kJ/kg h2 = h1 + wpI,in = 251.4 + 5.1 = 256.5 kJ/kg

h13 = h12 = h f

@ 245 kPa

h4 = h15 = h14 = h f h3 = h16 = h f

⎞ ⎟ ⎟ ⎠

T

= 830 kJ/kg

= 467 kJ/kg

An energy balance on the closed feedwater heater gives

1.4 MPa

5

4

= 533 kJ/kg

@ 1400 kPa

@ 150 kPa

5 MPa

y

14

16 13

2 1

8

245 kPa

7 9

z

12 3

1.2 MPa

6

15

w

10

1-y-z-w

17

11

150 kPa 20 kPa

s

1h2 + wh10 + zh13 + yh15 = 1h3 + ( y + z + w)h16

where w is the fraction of steam extracted from the low-pressure turbine. Solving for z, ( h − h2 ) + ( y + z )h16 − zh13 − yh15 w= 3 h10 − h16 (467 − 256.5) + (0.1160 + 0.15)(467) − (0.15)(533) − (0.1160)(830) 3023 − 467 = 0.0620 (c) The work output from the turbines is =

wT,out = h5 − yh6 − (1 − y )h7 + (1 − y )h8 − zh9 − wh10 − (1 − y − z − w)h11 = 3894 − (0.1160)(3400) − (1 − 0.1160)(3349) + (1 − 0.1160)(3692) − (0.15)(3154) − (0.0620)(3023) − (1 − 0.1160 − 0.15 − 0.0620)(2620) = 1381.6 kJ/kg

The net work output from the cycle is w net = wT, out − w P,in = 1381 .6 − 5.1 = 1376 .5 kJ/kg

The mass flow rate through the boiler is W& 300,000 kW m& = net = = 217.9 kg/s wnet 1376.5 kJ/kg The rate of heat input in the boiler is Q& in = m& (h5 − h4 ) + (1 − y )m& (h8 − h7 ) = (217.9 kg/s )(3894 − 830) kJ/kg + (1 − 0.1160)(217.9 kg/s )(3692 − 3349) kJ/kg = 733,700 kW

The rate of process heat and the utilization efficiency of this cogeneration plant are Q& = zm& ( h − h ) = (0.15)(217.9 kg/s )(3154 − 533) kJ/kg = 85,670 kW process

εu =

9

12

W& net + Q& process (300,000 + 85,670) kW = = 0.526 = 52.6% 733,700 kW Q& in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-126

10-115

The effect of the condenser pressure on the performance a simple ideal Rankine cycle is to be investigated.

Analysis The problem is solved using EES, and the solution is given below. function x4\$(x4) "this function returns a string to indicate the state of steam at point 4" x4\$='' if (x4>1) then x4\$='(superheated)' if (x41) then x4\$='(superheated)' if (x41) then x6\$='(superheated)' if (x6imax) END "NoRHStages = 2" P[6] = 10"kPa" P[3] = 15000"kPa" P_extract = P[6] "Select a lower limit on the reheat pressure" T[3] = 500"C" T[5] = 500"C" Eta_t = 1.0 "Turbine isentropic efficiency" Eta_p = 1.0 "Pump isentropic efficiency" Pratio = P[3]/P_extract P[4] = P[3]*(1/Pratio)^(1/(NoRHStages+1))"kPa" Fluid\$='Steam_IAPWS' PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-131

"Pump analysis" P[1] = P[6] P[2]=P[3] x[1]=0 "Sat'd liquid" h[1]=enthalpy(Fluid\$,P=P[1],x=x[1]) v[1]=volume(Fluid\$,P=P[1],x=x[1]) s[1]=entropy(Fluid\$,P=P[1],x=x[1]) T[1]=temperature(Fluid\$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" W_p=W_p_s/Eta_p h[2]=h[1]+W_p "SSSF First Law for the pump" v[2]=volume(Fluid\$,P=P[2],h=h[2]) s[2]=entropy(Fluid\$,P=P[2],h=h[2]) T[2]=temperature(Fluid\$,P=P[2],h=h[2]) "High Pressure Turbine analysis" h[3]=enthalpy(Fluid\$,T=T[3],P=P[3]) s[3]=entropy(Fluid\$,T=T[3],P=P[3]) v[3]=volume(Fluid\$,T=T[3],P=P[3]) s_s[4]=s[3] hs[4]=enthalpy(Fluid\$,s=s_s[4],P=P[4]) Ts[4]=temperature(Fluid\$,s=s_s[4],P=P[4]) Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency" T[4]=temperature(Fluid\$,P=P[4],h=h[4]) s[4]=entropy(Fluid\$,h=h[4],P=P[4]) v[4]=volume(Fluid\$,s=s[4],P=P[4]) h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine" "Low Pressure Turbine analysis" Call Reheat(P[3],T[3],T[5],h[4],NoRHStages,Pratio,Eta_t:Q_in_reheat,W_t_lp,h6) h[6]=h6 {P[5]=P[4] s[5]=entropy(Fluid\$,T=T[5],P=P[5]) h[5]=enthalpy(Fluid\$,T=T[5],P=P[5]) s_s[6]=s[5] hs[6]=enthalpy(Fluid\$,s=s_s[6],P=P[6]) Ts[6]=temperature(Fluid\$,s=s_s[6],P=P[6]) vs[6]=volume(Fluid\$,s=s_s[6],P=P[6]) Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency" h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine" x[6]=QUALITY(Fluid\$,h=h[6],P=P[6]) W_t_lp_total = NoRHStages*W_t_lp Q_in_reheat = NoRHStages*(h[5] - h[4])} "Boiler analysis" Q_in_boiler + h[2]=h[3]"SSSF First Law for the Boiler" Q_in = Q_in_boiler+Q_in_reheat "Condenser analysis" h[6]=Q_out+h[1]"SSSF First Law for the Condenser" T[6]=temperature(Fluid\$,h=h[6],P=P[6]) s[6]=entropy(Fluid\$,h=h[6],P=P[6]) x[6]=QUALITY(Fluid\$,h=h[6],P=P[6]) x6s\$=x6\$(x[6]) "Cycle Statistics" W_net=W_t_hp+W_t_lp - W_p Eta_th=W_net/Q_in

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-132

0.4097 0.4122 0.4085 0.4018 0.3941 0.386 0.3779 0.3699 0.3621 0.3546

NoRH Stages 1 2 3 4 5 6 7 8 9 10

Qin [kJ/kg] 4085 4628 5020 5333 5600 5838 6058 6264 6461 6651

2400

Wnet [kJ/kg] 1674 1908 2051 2143 2207 2253 2289 2317 2340 2358

2300

2200

W net [kJ/kg]

ηth

2100

2000

1900

1800

1700

1600 1

2

3

4

5

6

7

8

9

10

NoRHStages 0.42 0.41

ηth

0.4 0.39 0.38 0.37 0.36 0.35 1

2

3

4

5

6

7

8

9

10

8

9

10

NoRHStages 7000 6500

Q in [kJ/kg]

6000 5500 5000 4500 4000 1

2

3

4

5

6

7

NoRHStages

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-133

10-118 The effect of number of regeneration stages on the performance an ideal regenerative Rankine cycle with one open feedwater heater is to be investigated. Analysis The problem is solved using EES, and the solution is given below. Procedure Reheat(NoFwh,T[5],P[5],P_cond,Eta_turb,Eta_pump:q_in,w_net) Fluid\$='Steam_IAPWS' Tcond = temperature(Fluid\$,P=P_cond,x=0) Tboiler = temperature(Fluid\$,P=P[5],x=0) P[7] = P_cond s[5]=entropy(Fluid\$, T=T[5], P=P[5]) h[5]=enthalpy(Fluid\$, T=T[5], P=P[5]) h[1]=enthalpy(Fluid\$, P=P[7],x=0) P4[1] = P[5] "NOTICE THIS IS P4[i] WITH i = 1" DELTAT_cond_boiler = Tboiler - Tcond If NoFWH = 0 Then "the following are h7, h2, w_net, and q_in for zero feedwater heaters, NoFWH = 0" h7=enthalpy(Fluid\$, s=s[5],P=P[7]) h2=h[1]+volume(Fluid\$, P=P[7],x=0)*(P[5] - P[7])/Eta_pump w_net = Eta_turb*(h[5]-h7)-(h2-h[1]) q_in = h[5] - h2 else i=0 REPEAT i=i+1 "The following maintains the same temperature difference between any two regeneration stages." T_FWH[i] = (NoFWH +1 - i)*DELTAT_cond_boiler/(NoFWH + 1)+Tcond"[C]" P_extract[i] = pressure(Fluid\$,T=T_FWH[i],x=0)"[kPa]" P3[i]=P_extract[i] P6[i]=P_extract[i] If i > 1 then P4[i] = P6[i - 1] UNTIL i=NoFWH P4[NoFWH+1]=P6[NoFWH] h4[NoFWH+1]=h[1]+volume(Fluid\$, P=P[7],x=0)*(P4[NoFWH+1] - P[7])/Eta_pump i=0 REPEAT i=i+1 "Boiler condensate pump or the Pumps 2 between feedwater heaters analysis" h3[i]=enthalpy(Fluid\$,P=P3[i],x=0) v3[i]=volume(Fluid\$,P=P3[i],x=0) w_pump2_s=v3[i]*(P4[i]-P3[i])"SSSF isentropic pump work assuming constant specific volume" w_pump2[i]=w_pump2_s/Eta_pump "Definition of pump efficiency" h4[i]= w_pump2[i] +h3[i] "Steady-flow conservation of energy" s4[i]=entropy(Fluid\$,P=P4[i],h=h4[i]) T4[i]=temperature(Fluid\$,P=P4[i],h=h4[i]) Until i = NoFWH i=0 REPEAT PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-134

i=i+1 "Open Feedwater Heater analysis:" {h2[i] = h6[i]} s5[i] = s[5] ss6[i]=s5[i] hs6[i]=enthalpy(Fluid\$,s=ss6[i],P=P6[i]) Ts6[i]=temperature(Fluid\$,s=ss6[i],P=P6[i]) h6[i]=h[5]-Eta_turb*(h[5]-hs6[i])"Definition of turbine efficiency for high pressure stages" If i=1 then y[1]=(h3[1] - h4[2])/(h6[1] - h4[2]) "Steady-flow conservation of energy for the FWH" If i > 1 then js = i -1 j=0 sumyj = 0 REPEAT j = j+1 sumyj = sumyj + y[ j ] UNTIL j = js y[i] =(1- sumyj)*(h3[i] - h4[i+1])/(h6[i] - h4[i+1]) ENDIF T3[i]=temperature(Fluid\$,P=P3[i],x=0) "Condensate leaves heater as sat. liquid at P[3]" s3[i]=entropy(Fluid\$,P=P3[i],x=0) "Turbine analysis" T6[i]=temperature(Fluid\$,P=P6[i],h=h6[i]) s6[i]=entropy(Fluid\$,P=P6[i],h=h6[i]) yh6[i] = y[i]*h6[i] UNTIL i=NoFWH ss[7]=s6[i] hs[7]=enthalpy(Fluid\$,s=ss[7],P=P[7]) Ts[7]=temperature(Fluid\$,s=ss[7],P=P[7]) h[7]=h6[i]-Eta_turb*(h6[i]-hs[7])"Definition of turbine efficiency for low pressure stages" T[7]=temperature(Fluid\$,P=P[7],h=h[7]) s[7]=entropy(Fluid\$,P=P[7],h=h[7]) sumyi = 0 sumyh6i = 0 wp2i = W_pump2[1] i=0 REPEAT i=i+1 sumyi = sumyi + y[i] sumyh6i = sumyh6i + yh6[i] If NoFWH > 1 then wp2i = wp2i + (1- sumyi)*W_pump2[i] UNTIL i = NoFWH "Condenser Pump---Pump_1 Analysis:" P[2] = P6 [ NoFWH] P[1] = P_cond h[1]=enthalpy(Fluid\$,P=P[1],x=0) {Sat'd liquid} v1=volume(Fluid\$,P=P[1],x=0) s[1]=entropy(Fluid\$,P=P[1],x=0) T[1]=temperature(Fluid\$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[2]=w_pump1+ h[1] "Steady-flow conservation of energy" s[2]=entropy(Fluid\$,P=P[2],h=h[2]) T[2]=temperature(Fluid\$,P=P[2],h=h[2]) "Boiler analysis" q_in = h[5] - h4[1]"SSSF conservation of energy for the Boiler" w_turb = h[5] - sumyh6i - (1- sumyi)*h[7] "SSSF conservation of energy for turbine" PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-135

"Condenser analysis" q_out=(1- sumyi)*(h[7] - h[1])"SSSF First Law for the Condenser" "Cycle Statistics" w_net=w_turb - ((1- sumyi)*w_pump1+ wp2i) endif END "Input Data" NoFWH = 2 P[5] = 10000 [kPa] T[5] = 500 [C] P_cond=10 [kPa] Eta_turb= 1.0 "Turbine isentropic efficiency" Eta_pump = 1.0 "Pump isentropic efficiency" P[1] = P_cond P[4] = P[5] "Condenser exit pump or Pump 1 analysis" Call Reheat(NoFwh,T[5],P[5],P_cond,Eta_turb,Eta_pump:q_in,w_net) Eta_th=w_net/q_in

ηth 0.4019 0.4311 0.4401 0.4469 0.4513 0.4544 0.4567 0.4585 0.4599 0.4611 0.462

wnet [kJ/kg] 1275 1125 1061 1031 1013 1000 990.5 983.3 977.7 973.1 969.4

qin [kJ/kg] 3173 2609 2411 2307 2243 2200 2169 2145 2126 2111 2098

0.47 0.46 0.45

η th

No FWH 0 1 2 3 4 5 6 7 8 9 10

0.44 0.43 0.42 0.41 0.4 0

1

2

3

4

5

6

7

8

9

10

NoFwh 1300

3200

1250

2800

1150

qin [kJ/kg]

wnet [kJ/kg]

3000

1200

1100 1050 1000

2600 2400 2200

950 0

1

2

3

4

5

NoFwh

6

7

8

9

10

2000 0

1

2

3

4

5

6

7

8

9

NoFwh

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10

10-136

10-119 It is to be demonstrated that the thermal efficiency of a combined gas-steam power plant ηcc can be expressed as ηcc = ηg + ηs − ηgηs where ηg = Wg / Qin and ηs = Ws / Qg,out are the thermal efficiencies of the gas and steam cycles, respectively, and the efficiency of a combined cycle is to be obtained. Analysis The thermal efficiencies of gas, steam, and combined cycles can be expressed as

η cc = ηg = ηs =

Wtotal Q = 1 − out Qin Qin Wg Qin

= 1−

Qg,out Qin

Ws Q = 1 − out Qg,out Qg,out

where Qin is the heat supplied to the gas cycle, where Qout is the heat rejected by the steam cycle, and where Qg,out is the heat rejected from the gas cycle and supplied to the steam cycle. Using the relations above, the expression η g + η s − η gη s can be expressed as ⎛

Q

⎞ ⎛

Q

⎞ ⎛ Qg,out ⎞⎛ Q ⎟ − ⎜1 − ⎟⎜1 − out ⎟ ⎟ ⎜ ⎜ Qin ⎠⎝ Qg,out ⎠ ⎝ Qg,out Q Q −1+ + out − out Qin Qg,out Qin

g,out ⎟ + ⎜1 − out η g + η s − η gη s = ⎜⎜1 − Qin ⎟⎠ ⎜⎝ Qg,out ⎝

= 1− = 1−

Qg,out Qin

+1−

Qout Qg,out

⎞ ⎟ ⎟ ⎠

Qout Qin

= η cc

Therefore, the proof is complete. Using the relation above, the thermal efficiency of the given combined cycle is determined to be

η cc = η g + η s − η gη s = 0.4 + 0.30 − 0.40 × 0.30 = 0.58

10-120 The thermal efficiency of a combined gas-steam power plant ηcc can be expressed in terms of the thermal efficiencies of the gas and the steam turbine cycles as ηcc = ηg + ηs − ηgηs . It is to be shown that the value of ηcc is greater than either of ηg or ηs . Analysis By factoring out terms, the relation ηcc = ηg + ηs − ηgηs can be expressed as

η cc = η g + η s − η gη s = η g + η s (1 − η g ) > η g 14243 Positive since η g η s 14243 Positive since η s = 1 then SolMeth\$ = '>= 100%, the solution assumes complete combustion.' {MolCO = 0 MolCO2 = x} PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-124

w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth\$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END {"Input data from the diagram window" T_air = 298 [K] Theo_air = 120 [%] Fuel\$='CH4(g)'} T_fuel = 298 [K] Call Fuel(Fuel\$,T_fuel:x,y,z,h_fuel,Name\$) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth\$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" HP=(x-w)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x+y/4z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2 SOLUTION for the sample calculation A_th=5 fuel\$='C3H8(l)' HP=-119035 [kJ/kg] HR=-119035 [kJ/kg] h_fuel=-118910 Moles_CO=0.000 Moles_CO2=3.000 Moles_H2O=4 Moles_N2=22.560 Moles_O2=1.000 MolO2=1 Name\$='propane(liq)' SolMeth\$='>= 100%, the solution assumes complete combustion.' Theo_air=120 [%] Th_air=1.200 T_air=298 [K] T_fuel=298 [K] T_prod=2112 [K] w=0 x=3 y=8 z=0

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-125

15-125 The minimum percent of excess air that needs to be used for the fuels CH4(g), C2H2(g), CH3OH(g), C3H8(g), and C8H18(l) if the adiabatic flame temperature is not to exceed 1500 K is to be determined. Analysis The problem is solved using EES, and the solution is given below. Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) xCO2 + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + (y/4 + x-z/2) (Theo_air/100 - 1) O2" {"For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " 1+ 2*A_th=1*2+2*1"theoretical O balance"} "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel\$,T_fuel:x,y,z,h_fuel,Name\$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel\$='C2H2(g)' then x=2;y=2; z=0 Name\$='acetylene' h_fuel = 226730 else If fuel\$='C3H8(g)' then x=3; y=8; z=0 Name\$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) else If fuel\$='C8H18(l)' then x=8; y=18; z=0 Name\$='octane' h_fuel = -249950 else if fuel\$='CH4(g)' then x=1; y=4; z=0 Name\$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel\$='CH3OH(g)' then x=1; y=4; z=1 Name\$='methyl alcohol' h_fuel = -200670 endif; endif; endif; endif; endif end {"Input data from the diagram window" T_air = 298 [K] Fuel\$='CH4(g)'} T_fuel = 298 [K] Excess_air=Theo_air - 100 "[%]" Call Fuel(Fuel\$,T_fuel:x,y,z,h_fuel,Name\$) A_th = y/4 + x-z/2 Th_air = Theo_air/100 HR=h_fuel+ (y/4 + x-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(y/4 + x-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-126

HP=x*enthalpy(CO2,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(y/4 + x-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+(y/4 + x-z/2) *(Theo_air/100 - 1)*enthalpy(O2,T=T_prod) Moles_O2=(y/4 + x-z/2) *(Theo_air/100 - 1) Moles_N2=3.76*(y/4 + x-z/2)* (Theo_air/100) Moles_CO2=x Moles_H2O=y/2 T[1]=T_prod; xa[1]=Theo_air SOLUTION for a sample calculation A_th=2.5 fuel\$='C2H2(g)' HR=226596 [kJ/kg] Moles_CO2=2 Moles_N2=24.09 Name\$='acetylene' Th_air=2.563 T_air=298 [K] T_prod=1500 [K] xa[1]=256.3 z=0

Excess_air=156.251 [%] HP=226596 [kJ/kg] h_fuel=226730 Moles_H2O=1 Moles_O2=3.906 Theo_air=256.3 [%] T[1]=1500 [K] T_fuel=298 [K] x=2 y=2

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-127

15-126 The minimum percentages of excess air that need to be used for the fuels CH4(g), C2H2(g), CH3OH(g), C3H8(g), and C8H18(l) AFOR adiabatic flame temperatures of 1200 K, 1750 K, and 2000 K are to be determined. Analysis The problem is solved using EES, and the solution is given below. Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) xCO2 + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + (y/4 + x-z/2) (Theo_air/100 - 1) O2" {"For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " 1+ 2*A_th=1*2+2*1"theoretical O balance"} "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel\$,T_fuel:x,y,z,h_fuel,Name\$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel\$='C2H2(g)' then x=2;y=2; z=0 Name\$='acetylene' h_fuel = 226730 else If fuel\$='C3H8(g)' then x=3; y=8; z=0 Name\$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) else If fuel\$='C8H18(l)' then x=8; y=18; z=0 Name\$='octane' h_fuel = -249950 else if fuel\$='CH4(g)' then x=1; y=4; z=0 Name\$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel\$='CH3OH(g)' then x=1; y=4; z=1 Name\$='methyl alcohol' h_fuel = -200670 endif; endif; endif; endif; endif end {"Input data from the diagram window" T_air = 298 [K] Fuel\$='CH4(g)'} T_fuel = 298 [K] Excess_air=Theo_air - 100 "[%]" Call Fuel(Fuel\$,T_fuel:x,y,z,h_fuel,Name\$) A_th = y/4 + x-z/2 Th_air = Theo_air/100 HR=h_fuel+ (y/4 + x-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(y/4 + x-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-128

HP=x*enthalpy(CO2,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(y/4 + x-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+(y/4 + x-z/2) *(Theo_air/100 - 1)*enthalpy(O2,T=T_prod) Moles_O2=(y/4 + x-z/2) *(Theo_air/100 - 1) Moles_N2=3.76*(y/4 + x-z/2)* (Theo_air/100) Moles_CO2=x Moles_H2O=y/2 T[1]=T_prod; xa[1]=Theo_air SOLUTION for a sample calculation A_th=5 fuel\$='C3H8(g)' HR=-103995 [kJ/kg] Moles_CO2=3 Moles_N2=24.7 Name\$='propane' Th_air=1.314 T_air=298 [K] T_prod=2000 [K] xa[1]=131.4 z=0

Excess_air=31.395 [%] HP=-103995 [kJ/kg] h_fuel=-103858 Moles_H2O=4 Moles_O2=1.570 Theo_air=131.4 [%] T[1]=2000 [K] T_fuel=298 [K] x=3 y=8

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-129

15-127 The adiabatic flame temperature of CH4(g) is to be determined when both the fuel and the air enter the combustion chamber at 25°C for the cases of 0, 20, 40, 60, 80, 100, 200, 500, and 1000 percent excess air. Analysis The problem is solved using EES, and the solution is given below. Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) xCO2 + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + (y/4 + x-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Adiabatic, Incomplete Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) (x-w)CO2 +wCO + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + ((y/4 + x-z/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel\$,T_fuel:x,y,z,h_fuel,Name\$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel\$='C2H2(g)' then x=2;y=2; z=0 Name\$='acetylene' h_fuel = 226730 else If fuel\$='C3H8(g)' then x=3; y=8; z=0 Name\$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) else If fuel\$='C8H18(l)' then x=8; y=18; z=0 Name\$='octane' h_fuel = -249950 else if fuel\$='CH4(g)' then x=1; y=4; z=0 Name\$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel\$='CH3OH(g)' then x=1; y=4; z=1 Name\$='methyl alcohol' h_fuel = -200670 endif; endif; endif; endif; endif end Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth\$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100 IF Th_air >= 1 then SolMeth\$ = '>= 100%, the solution assumes complete combustion.' {MolCO = 0 MolCO2 = x} w=0 MolO2 = A_th*(Th_air - 1) PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-130

GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth\$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END {"Input data from the diagram window" T_air = 298 [K] Theo_air = 200 [%] Fuel\$='CH4(g)'} T_fuel = 298 [K] Call Fuel(Fuel\$,T_fuel:x,y,z,h_fuel,Name\$) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth\$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" HP=(x-w)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x+y/4z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2 Product temperature vs % excess air for CH4 3000 Tprod [K] 2329 2071 1872 1715 1587 1480 1137 749.5 553

2500 Tprod [K]

Theoair [%] 100 120 140 160 180 200 300 600 1100

2000 1500 1000 500 0 100 200 300 400 500 600 700 800 900 1000 1100 Theoair [%]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-131

15-128 The fuel among CH4(g), C2H2(g), C2H6(g), C3H8(g), and C8H18(l) that gives the highest temperature when burned completely in an adiabatic constant-volume chamber with the theoretical amount of air is to be determined. Analysis The problem is solved using EES, and the solution is given below. Adiabatic Combustion of fuel CnHm with Stoichiometric Air at T_fuel =T_air=T_reac in a constant volume, closed system: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> xCO2 + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + (x+y/4-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Adiabatic, Incomplete Combustion of fuel CnHm with Stoichiometric Air at T_fuel =T_air=T_reac in a constant volume, closed system: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> (x-w)CO2 +wCO + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + ((x+y/4-z/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel\$,T_fuel:x,y,z,h_fuel,Name\$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel\$='C2H2(g)' then x=2;y=2; z=0 Name\$='acetylene' h_fuel = 226730"Table A.26" else If fuel\$='C3H8(g)' then x=3; y=8; z=0 Name\$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) else If fuel\$='C8H18(l)' then x=8; y=18; z=0 Name\$='octane' h_fuel = -249950"Table A.26" else if fuel\$='CH4(g)' then x=1; y=4; z=0 Name\$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel\$='CH3OH(g)' then x=1; y=4; z=1 Name\$='methyl alcohol' h_fuel = -200670"Table A.26" endif; endif; endif; endif; endif end Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth\$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100 IF Th_air >= 1 then SolMeth\$ = '>= 100%, the solution assumes complete combustion.' w=0 MolO2 = A_th*(Th_air - 1) PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-132

GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth\$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END {"Input data from the diagram window" Theo_air = 200 [%] Fuel\$='CH4(g)'} T_reac = 298 [K] T_air = T_reac T_fuel = T_reac R_u = 8.314 [kJ/kmol-K] Call Fuel(Fuel\$,T_fuel:x,y,z,h_fuel,Name\$) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth\$) UR=(h_fuel-R_u*T_fuel)+ (x+y/4-z/2) *(Theo_air/100) *(enthalpy(O2,T=T_air)-R_u*T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *(enthalpy(N2,T=T_air)-R_u*T_air) UP=(x-w)*(enthalpy(CO2,T=T_prod)-R_u*T_prod)+w*(enthalpy(CO,T=T_prod)R_u*T_prod)+(y/2)*(enthalpy(H2O,T=T_prod)-R_u*T_prod)+3.76*(x+y/4-z/2)* (Theo_air/100)*(enthalpy(N2,T=T_prod)-R_u*T_prod)+MolO2*(enthalpy(O2,T=T_prod)-R_u*T_prod) UR =UP "Adiabatic, constant volume conservation of energy" Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2

SOLUTION for CH4 A_th=2 fuel\$='CH4(g)' Moles_CO=0.000 Moles_CO2=1.000 Moles_N2=7.520 Moles_O2=0.000 Name\$='methane' R_u=8.314 [kJ/kmol-K] SolMeth\$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_prod=2824 [K] T_fuel=298 [K] UP=-100981 UR=-100981 x=1 y=4 SOLUTION for C2H2 A_th=2.5 fuel\$='C2H2(g)' Moles_CO=0.000 Moles_CO2=2.000 Moles_N2=9.400 Moles_O2=0.000 Name\$='acetylene' R_u=8.314 [kJ/kmol-K] SolMeth\$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_prod=3535 [K] T_fuel=298 [K] UP=194717 UR=194717 x=2 y=2

h_fuel=-74875 Moles_H2O=2 MolO2=0 T_air=298 [K] T_reac=298 [K] w=0 z=0 h_fuel=226730 Moles_H2O=1 MolO2=0 T_air=298 [K] T_reac=298 [K] w=0 z=0

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-133

SOLUTION for CH3OH A_th=1.5 fuel\$='CH3OH(g)' Moles_CO=0.000 Moles_CO2=1.000 Moles_N2=5.640 Moles_O2=0.000 Name\$='methyl alcohol' R_u=8.314 [kJ/kmol-K] SolMeth\$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_prod=2817 [K] T_fuel=298 [K] UP=-220869 UR=-220869 x=1 y=4 SOLUTION for C3H8 A_th=5 fuel\$='C3H8(g)' Moles_CO=0.000 Moles_CO2=3.000 Moles_N2=18.800 Moles_O2=0.000 Name\$='propane' R_u=8.314 [kJ/kmol-K] SolMeth\$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_prod=2909 [K] T_fuel=298 [K] UP=-165406 UR=-165406 x=3 y=8 SOLUTION for C8H18 A_th=12.5 fuel\$='C8H18(l)' Moles_CO=0.000 Moles_CO2=8.000 Moles_N2=47.000 Moles_O2=0.000 Name\$='octane' R_u=8.314 [kJ/kmol-K] SolMeth\$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_prod=2911 [K] T_fuel=298 [K] UP=-400104 UR=-400104 x=8 y=18

h_fuel=-200670 Moles_H2O=2 MolO2=0 T_air=298 [K] T_reac=298 [K] w=0 z=1 h_fuel=-103858 Moles_H2O=4 MolO2=0 T_air=298 [K] T_reac=298 [K] w=0 z=0 h_fuel=-249950 Moles_H2O=9 MolO2=0 T_air=298 [K] T_reac=298 [K] w=0 z=0

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-134

Fundamentals of Engineering (FE) Exam Problems

15-129 A fuel is burned with 70 percent theoretical air. This is equivalent to (a) 30% excess air (b) 70% excess air (c) 30% deficiency of air (d) 70% deficiency of air (e) stoichiometric amount of air Answer (c) 30% deficiency ofair Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). air_th=0.7 "air_th=air_access+1" air_th=1-air_deficiency

15-130 Propane C3H8 is burned with 150 percent theoretical air. The air-fuel mass ratio for this combustion process is

(a) 5.3

(b) 10.5

(c) 15.7

(d) 23.4

(e) 39.3

Answer (d) 23.4 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). n_C=3 n_H=8 m_fuel=n_H*1+n_C*12 a_th=n_C+n_H/4 coeff=1.5 "coeff=1 for theoretical combustion, 1.5 for 50% excess air" n_O2=coeff*a_th n_N2=3.76*n_O2 m_air=n_O2*32+n_N2*28 AF=m_air/m_fuel

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-135

15-131 One kmol of methane (CH4) is burned with an unknown amount of air during a combustion process. If the combustion is complete and there are 1 kmol of free O2 in the products, the air-fuel mass ratio is

(a) 34.6

(b) 25.7

(c) 17.2

(d) 14.3

(e) 11.9

Answer (b) 25.7 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). n_C=1 n_H=4 m_fuel=n_H*1+n_C*12 a_th=n_C+n_H/4 (coeff-1)*a_th=1 "O2 balance: Coeff=1 for theoretical combustion, 1.5 for 50% excess air" n_O2=coeff*a_th n_N2=3.76*n_O2 m_air=n_O2*32+n_N2*28 AF=m_air/m_fuel "Some Wrong Solutions with Common Mistakes:" W1_AF=1/AF "Taking the inverse of AF" W2_AF=n_O2+n_N2 "Finding air-fuel mole ratio" W3_AF=AF/coeff "Ignoring excess air"

15-132 A fuel is burned steadily in a combustion chamber. The combustion temperature will be the highest except when (a) the fuel is preheated. (b) the fuel is burned with a deficiency of air. (c) the air is dry. (d) the combustion chamber is well insulated. (e) the combustion is complete. Answer (b) the fuel is burned with a deficiency of air.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-136

15-133 An equimolar mixture of carbon dioxide and water vapor at 1 atm and 60°C enter a dehumidifying section where the entire water vapor is condensed and removed from the mixture, and the carbon dioxide leaves at 1 atm and 60°C. The entropy change of carbon dioxide in the dehumidifying section is

(a) –2.8 kJ/kg⋅K

(b) –0.13 kJ/kg⋅K

(c) 0

(d) 0.13 kJ/kg⋅K

(e) 2.8 kJ/kg⋅K

Answer (b) –0.13 kJ/kg⋅K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Cp_CO2=0.846 R_CO2=0.1889 T1=60+273 "K" T2=T1 P1= 1 "atm" P2=1 "atm" y1_CO2=0.5; P1_CO2=y1_CO2*P1 y2_CO2=1; P2_CO2=y2_CO2*P2 Ds_CO2=Cp_CO2*ln(T2/T1)-R_CO2*ln(P2_CO2/P1_CO2) "Some Wrong Solutions with Common Mistakes:" W1_Ds=0 "Assuming no entropy change" W2_Ds=Cp_CO2*ln(T2/T1)-R_CO2*ln(P1_CO2/P2_CO2) "Using pressure fractions backwards"

15-134 Methane (CH4) is burned completely with 80% excess air during a steady-flow combustion process. If both the reactants and the products are maintained at 25°C and 1 atm and the water in the products exists in the liquid form, the heat transfer from the combustion chamber per unit mass of methane is

(a) 890 MJ/kg

(b) 802 MJ/kg

(c) 75 MJ/kg

(d) 56 MJ/kg

(e) 50 MJ/kg

Answer (d) 56 MJ/kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T= 25 "C" P=1 "atm" EXCESS=0.8 "Heat transfer in this case is the HHV at room temperature," HHV_CH4 =55.53 "MJ/kg" LHV_CH4 =50.05 "MJ/kg" "Some Wrong Solutions with Common Mistakes:" W1_Q=LHV_CH4 "Assuming lower heating value" W2_Q=EXCESS*hHV_CH4 "Assuming Q to be proportional to excess air"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-137

15-135 The higher heating value of a hydrocarbon fuel CnHm with m = 8 is given to be 1560 MJ/kmol of fuel. Then its lower heating value is

(a) 1384 MJ/kmol

(b) 1208 MJ/kmol

(c) 1402 MJ/kmol

(d) 1540 MJ/kmol

(e) 1550 MJ/kmol

Answer (a) 1384 MJ/kmol Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). HHV=1560 "MJ/kmol fuel" h_fg=2.4423 "MJ/kg, Enthalpy of vaporization of water at 25C" n_H=8 n_water=n_H/2 m_water=n_water*18 LHV=HHV-h_fg*m_water "Some Wrong Solutions with Common Mistakes:" W1_LHV=HHV - h_fg*n_water "Using mole numbers instead of mass" W2_LHV= HHV - h_fg*m_water*2 "Taking mole numbers of H2O to be m instead of m/2" W3_LHV= HHV - h_fg*n_water*2 "Taking mole numbers of H2O to be m instead of m/2, and using mole numbers"

15-136 Acetylene gas (C2H2) is burned completely during a steady-flow combustion process. The fuel and the air enter the combustion chamber at 25°C, and the products leave at 1500 K. If the enthalpy of the products relative to the standard reference state is –404 MJ/kmol of fuel, the heat transfer from the combustion chamber is

(a) 177 MJ/kmol

(b) 227 MJ/kmol

(c) 404 MJ/kmol

(d) 631 MJ/kmol

(e) 751 MJ/kmol

Answer (d) 631 MJ/kmol Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). hf_fuel=226730/1000 "MJ/kmol fuel" H_prod=-404 "MJ/kmol fuel" H_react=hf_fuel Q_out=H_react-H_prod "Some Wrong Solutions with Common Mistakes:" W1_Qout= -H_prod "Taking Qout to be H_prod" W2_Qout= H_react+H_prod "Adding enthalpies instead of subtracting them"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-138

15-137 Benzene gas (C6H6) is burned with 95 percent theoretical air during a steady-flow combustion process. The mole fraction of the CO in the products is

(a) 8.3%

(b) 4.7%

(c) 2.1%

(d) 1.9%

(e) 14.3%

Answer (c) 2.1% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). n_C=6 n_H=6 a_th=n_C+n_H/4 coeff=0.95 "coeff=1 for theoretical combustion, 1.5 for 50% excess air" "Assuming all the H burns to H2O, the combustion equation is C6H6+coeff*a_th(O2+3.76N2)----- (n_CO2) CO2+(n_CO)CO+(n_H2O) H2O+(n_N2) N2" n_O2=coeff*a_th n_N2=3.76*n_O2 n_H2O=n_H/2 n_CO2+n_CO=n_C 2*n_CO2+n_CO+n_H2O=2*n_O2 "Oxygen balance" n_prod=n_CO2+n_CO+n_H2O+n_N2 "Total mole numbers of product gases" y_CO=n_CO/n_prod "mole fraction of CO in product gases" "Some Wrong Solutions with Common Mistakes:" W1_yCO=n_CO/n1_prod; n1_prod=n_CO2+n_CO+n_H2O "Not including N2 in n_prod" W2_yCO=(n_CO2+n_CO)/n_prod "Using both CO and CO2 in calculations"

15-138 A fuel is burned during a steady-flow combustion process. Heat is lost to the surroundings at 300 K at a rate of 1120 kW. The entropy of the reactants entering per unit time is 17 kW/K and that of the products is 15 kW/K. The total rate of exergy destruction during this combustion process is

(a) 520 kW

(b) 600 kW

(c) 1120 kW

(d) 340 kW

(e) 739 kW

Answer (a) 520 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). To=300 "K" Q_out=1120 "kW" S_react=17 "kW'K" S_prod= 15 "kW/K" S_react-S_prod-Q_out/To+S_gen=0 "Entropy balance for steady state operation, Sin-Sout+Sgen=0" X_dest=To*S_gen "Some Wrong Solutions with Common Mistakes:" W1_Xdest=S_gen "Taking Sgen as exergy destruction" W2_Xdest=To*S_gen1; S_react-S_prod-S_gen1=0 "Ignoring Q_out/To"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-139

15-139 ··· 15-144 Design and Essay Problems

15-139 A certain industrial process generates a liquid solution of ethanol and water as the waste product. The solution is to be burned using methane. A combustion process is to be developed to accomplish this incineration process with minimum amount of methane. Analysis The mass flow rate of the liquid ethanol-water solution is given to be 10 kg/s. Considering that the mass fraction of ethanol in the solution is 0.2, m& ethanol = (0.2 )(10 kg/s ) = 2 kg/s m& water = (0.8)(10 kg/s ) = 8 kg/s

Noting that the molar masses Methanol = 46 and Mwater = 18 kg/kmol and that mole numbers N = m/M, the mole flow rates become m& 2 kg/s = 0.04348 kmol/s N& ethanol = ethanol = M ethanol 46 kg/kmol m& 8 kg/s = 0.44444 kmol/s N& water = water = M water 18 kg/kmol

Note that N& water 0.44444 = = 10.222 kmol H 2 O/kmol C 2 H 5 OH & N ethanol 0.04348

That is, 10.222 moles of liquid water is present in the solution for each mole of ethanol. Assuming complete combustion, the combustion equation of C2H5OH (l) with stoichiometric amount of air is C 2 H 5 OH(l ) + a th (O 2 + 3.76N 2 ) ⎯ ⎯→ 2CO 2 + 3H 2 O + 3.76a th N 2

where ath is the stoichiometric coefficient and is determined from the O2 balance, 1 + 2a th = 4 + 3 ⎯ ⎯→ a th = 3

Thus, C 2 H 5 OH(l ) + 3(O 2 + 3.76N 2 ) ⎯ ⎯→ 2CO 2 + 3H 2 O + 11.28N 2

Noting that 10.222 kmol of liquid water accompanies each kmol of ethanol, the actual combustion equation can be written as C 2 H 5 OH(l ) + 3(O 2 + 3.76N 2 ) + 10.222H 2 O(l ) ⎯ ⎯→ 2CO 2 + 3H 2 O(g ) + 11.28N 2 + 10.222H 2 O(l )

The heat transfer for this combustion process is determined from the steady-flow energy balance equation with W = 0, Q=

∑ N (h P

o f

+h −ho

) − ∑ N (h P

R

o f

+h −ho

)

R

Assuming the air and the combustion products to be ideal gases, we have h = h(T). We assume all the reactants to enter the combustion chamber at the standard reference temperature of 25°C. Furthermore, we assume the products to leave the combustion chamber at 1400 K which is a little over the required temperature of 1100°C. From the tables,

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

15-140

Substance

C2H5OH (l) CH4 O2 N2 H2O (g) H2O (l) CO2

hfo

h 298 K

h1400 K

kJ/kmol -277,690 -74,850 0 0 -241,820 -285,830 -393,520

kJ/kmol ----8682 8669 9904 --9364

kJ/kmol ----45,648 43,605 53,351 --65,271

Thus, Q = (2 )(−393,520 + 65,271 − 9364) + (3)(−241,820 + 53,351 − 9904) + (11.28)(0 + 43,605 − 8669) − (1)(− 277,690) − 0 − 0 + (10.222)(− 241,820 + 53,351 − 9904) − (10.222)(− 285,830) = 295,409 kJ/kmol of C2H5OH The positive sign indicates that 295,409 kJ of heat must be supplied to the combustion chamber from another source (such as burning methane) to ensure that the combustion products will leave at the desired temperature of 1400 K. Then the rate of heat transfer required for a mole flow rate of 0.04348 kmol C2H5OH/s CO becomes

Q& = N& Q = (0.04348 kmol/s)(295,409 kJ/kmol) = 12,844 kJ/s Assuming complete combustion, the combustion equation of CH4(g) with stoichiometric amount of air is CH 4 + a th (O 2 + 3.76N 2 ) ⎯ ⎯→ CO 2 + 2H 2 O + 3.76a th N 2

where ath is the stoichiometric coefficient and is determined from the O2 balance, Thus, a th = 1 + 1 ⎯ ⎯→ a th = 2 CH 4 + 2(O 2 + 3.76N 2 ) ⎯ ⎯→ CO 2 + 2H 2 O + 7.52N 2 The heat transfer for this combustion process is determined from the steady-flow energy balance Ein − E out = ∆E system equation as shown above under the same assumptions and using the same mini table: Q = (1)(−393,520 + 65,271 − 9364 ) + (2 )(−241,820 + 53,351 − 9904 ) + (7.52 )(0 + 43,605 − 8669 ) − (1)(− 74,850 ) − 0 − 0 = −396,790 kJ/kmol of CH 4

That is, 396,790 kJ of heat is supplied to the combustion chamber for each kmol of methane burned. To supply heat at the required rate of 12,844 kJ/s, we must burn methane at a rate of

or,

12,844 kJ/s Q& N& CH 4 = = = 0.03237 kmolCH 4 /s Q 396,790 kJ/kmol m& CH 4 = M CH 4 N& CH 4 = (16 kg/kmol )(0.03237 kmolCH 4 /s ) = 0.5179 kg/s

Therefore, we must supply methane to the combustion chamber at a minimum rate 0.5179 kg/s in order to maintain the temperature of the combustion chamber above 1400 K.

KJ PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-1

Solutions Manual for

Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011

Chapter 16 CHEMICAL AND PHASE EQUILIBRIUM

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-2 Kp and Equilibrium Composition of Ideal Gases 16-1C No, the wooden table is NOT in chemical equilibrium with the air. With proper catalyst, it will reach with the oxygen in the air and burn.

16-2C They are PC C PDν D v

Kp =

PAν A PBvB

, K p = e − ∆G*(T ) / RuT

ν

and

Kp =

N CC N νDD ⎛ P ⎜ ⎜ N νAA N νBB ⎝ N total

⎞ ⎟ ⎟ ⎠

∆ν

where ∆ν = ν C + ν D −ν A −ν B . The first relation is useful in partial pressure calculations, the second in determining the Kp from gibbs functions, and the last one in equilibrium composition calculations.

16-3C (a) No, because Kp depends on temperature only. (b) In general, the total mixture pressure affects the mixture composition. The equilibrium constant for the reaction N 2 + O 2 ⇔ 2NO can be expressed as ν

Kp =

NO N NO

ν

ν

N NN 2 N OO 2 2

2

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(ν NO −ν N 2 −ν O 2 )

The value of the exponent in this case is 2-1-1 = 0. Therefore, changing the total mixture pressure will have no effect on the number of moles of N2, O2 and NO.

16-4C (a) The equilibrium constant for the reaction CO + 12 O 2 ⇔ CO 2 can be expressed as ν

Kp =

CO 2 N CO 2

ν

ν

CO N CO N OO 2 2

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(ν CO 2 −ν CO −ν O 2 )

Judging from the values in Table A-28, the Kp value for this reaction decreases as temperature increases. That is, the indicated reaction will be less complete at higher temperatures. Therefore, the number of moles of CO2 will decrease and the number moles of CO and O2 will increase as the temperature increases. (b) The value of the exponent in this case is 1-1-0.5=-0.5, which is negative. Thus as the pressure increases, the term in the brackets will decrease. The value of Kp depends on temperature only, and therefore it will not change with pressure. Then to keep the equation balanced, the number of moles of the products (CO2) must increase, and the number of moles of the reactants (CO, O2) must decrease.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-3 16-5C (a) The equilibrium constant for the reaction N 2 ⇔ 2N can be expressed as ν

N N ⎛ P K p = νN ⎜⎜ N NN 2 ⎝ N total 2

⎞ ⎟ ⎟ ⎠

(ν N −ν N 2 )

Judging from the values in Table A-28, the Kp value for this reaction increases as the temperature increases. That is, the indicated reaction will be more complete at higher temperatures. Therefore, the number of moles of N will increase and the number moles of N2 will decrease as the temperature increases. (b) The value of the exponent in this case is 2-1 = 1, which is positive. Thus as the pressure increases, the term in the brackets also increases. The value of Kp depends on temperature only, and therefore it will not change with pressure. Then to keep the equation balanced, the number of moles of the products (N) must decrease, and the number of moles of the reactants (N2) must increase.

16-6C The equilibrium constant for the reaction CO + 12 O 2 ⇔ CO 2 can be expressed as ν

Kp =

CO 2 N CO 2

ν

ν

CO N CO N OO 2 2

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(ν CO 2 −ν CO −ν O 2 )

Adding more N2 (an inert gas) at constant temperature and pressure will increase Ntotal but will have no direct effect on other terms. Then to keep the equation balanced, the number of moles of the products (CO2) must increase, and the number of moles of the reactants (CO, O2) must decrease.

16-7C The values of the equilibrium constants for each dissociation reaction at 3000 K are, from Table A-28, N 2 ⇔ 2N ⇔ ln K p = −22.359 H 2 ⇔ 2H ⇔ ln K p = −3.685

(greater than - 22.359)

Thus H2 is more likely to dissociate than N2.

16-8C (a) This reaction is the reverse of the known CO reaction. The equilibrium constant is then 1/ KP (b) This reaction is the reverse of the known CO reaction at a different pressure. Since pressure has no effect on the equilibrium constant, 1/ KP (c) This reaction is the same as the known CO reaction multiplied by 2. The quilibirium constant is then

K P2 (d) This is the same as reaction (c) occurring at a different pressure. Since pressure has no effect on the equilibrium constant,

K P2

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-4 16-9C (a) This reaction is the reverse of the known H2O reaction. The equilibrium constant is then 1/ KP (b) This reaction is the reverse of the known H2O reaction at a different pressure. Since pressure has no effect on the equilibrium constant, 1/ KP (c) This reaction is the same as the known H2O reaction multiplied by 3. The quilibirium constant is then

K P3 (d) This is the same as reaction (c) occurring at a different pressure. Since pressure has no effect on the equilibrium constant,

K P3

16-10 The partial pressures of the constituents of an ideal gas mixture is given. The Gibbs function of the nitrogen in this mixture at the given mixture pressure and temperature is to be determined. Analysis The partial pressure of nitrogen is PN2 = 110 kPa = (110 / 101.325) = 1.086 atm

The Gibbs function of nitrogen at 293 K and 1.086 atm is g (293 K, 1.086 atm) = g * (293 K, 1 atm) + Ru T ln PN2 = 0 + (8.314 kJ/kmol.K)(293 K)ln(1.086 atm) = 200 kJ/kmol

N2 ,CO2, NO PN2 = 110 kPa 293 K

16-11 The mole fractions of the constituents of an ideal gas mixture is given. The Gibbs function of the N2 in this mixture at the given mixture pressure and temperature is to be determined. Analysis From Tables A-18 and A-26, at 1 atm pressure,

[

g * (600 K, 1 atm) = g of + ∆ h (T ) − Ts o (T )

]

= 0 + (17,563 − 600 × 212.066) − (8669 − 298 × 191.502) = −61,278 kJ/kmol

The partial pressure of N2 is

30% N2 30% O2 40% H2O 5 atm 600 K

PCO = y N2 P = (0.30)(5 atm) = 1.5 atm

The Gibbs function of N2 at 600 K and 1.5 atm is

g (600 K, 1.5 atm) = g * (600 K, 1 atm) + Ru T ln PCO = −61,278 kJ/kmol + (8.314 kJ/kmol)(600 K)ln(1.5 atm) = −59,260 kJ/kmol

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-5 16-12 The temperature at which 0.2 percent of diatomic oxygen dissociates into monatomic oxygen at two pressures is to be determined. Assumptions 1 The equilibrium composition consists of N2 and N. 2 The constituents of the mixture are ideal gases. Analysis (a) The stoichiometric and actual reactions can be written as

Stoichiometric:

N 2 ⇔ 2 N (thus ν N2 = 1 and ν N = 2)

Actual:

N 2 ⇔ 0.998 N 2 + 0.004 N 424 3 1424 3 1 prod.

react.

N2 ↔ 2N 0.2 % 1 kPa

The equilibrium constant Kp can be determined from Kp =

N νNN ⎛ P ⎜⎜ N νN2N2 ⎝ N total

ν N −ν N2

⎞ ⎟⎟ ⎠

=

0.004 2 ⎛ 1 / 101.325 ⎞ ⎜ ⎟ 0.998 ⎝ 0.998 + 0.004 ⎠

2−1

= 1.579 × 10 −7

and

ln K p = −15.66 From Table A-28, the temperature corresponding to this lnKp value is T = 3628 K (b) At 10 kPa, Kp =

N νNN ⎛ P ⎜ ⎜ ν N N2N2 ⎝ N total

ν N −ν N2

⎞ ⎟ ⎟ ⎠

=

0.004 2 0.998

⎛ 10 / 101.325 ⎞ ⎜ ⎟ ⎝ 0.998 + 0.004 ⎠

2 −1

= 1.579 × 10 −6

ln K p = −13.36 From Table A-28, the temperature corresponding to this lnKp value is T = 3909 K

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-6 16-13 The equilibrium constant of the reaction H 2 O ⇔ H 2 + 12 O 2 is to be determined using Gibbs function. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e − ∆G *( T )/ Ru T or ln K p = − ∆G * ( T ) / Ru T

where

H2O ↔ H2 + ½O2

∆G * (T )

∗ ∗ ∗ = ν H2 g H2 (T ) +ν O2 g O2 (T ) −ν H2O g H2O (T )

500 K

At 500 K, ∗ ∗ ∗ ∆G * (T ) = ν H2 g H2 (T ) + ν O2 g O2 (T ) − ν H2O g H2O (T )

= ν H2 (h − Ts ) H2 + ν O2 (h − Ts ) O2 − ν H2O (h − Ts ) H2O = ν H2 [(h f + h500 − h298 ) − Ts ] H2 + ν O2 [(h f + h500 − h298 ) − Ts ] O2 − ν H2O [(h f + h500 − h298 ) − Ts ] H2O = 1 × (0 + 14,350 − 8468 − 500 × 145.628) + 0.5 × (0 + 14,770 − 8682 − 500 × 220.589) − 1 × (−241,820 + 16,828 − 9904 − 500 × 206.413) = 219,067 kJ/kmol

Substituting,

ln K p = −(219,067 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(500 K)] = −52.70 or K p = 1.30 × 10 −23 (Table A - 28 : ln K p = −52.70)

At 2000 K, ∗ ∗ ∗ ∆G * (T ) = ν H2 g H2 (T ) + ν O2 g O2 (T ) − ν H2O g H2O (T )

= ν H2 (h − Ts ) H2 + ν O2 (h − Ts ) O2 − ν H2O (h − Ts ) H2O = ν H2 [(h f + h2000 − h298 ) − Ts ] H2 + ν O2 [(h f + h2000 − h298 ) − Ts ] O2 − ν H2O [(h f + h2000 − h298 ) − Ts ] H2O = 1 × (0 + 61,400 − 8468 − 2000 × 188.297) + 0.5 × (0 + 67,881 − 8682 − 2000 × 268.655) − 1 × (−241,820 + 82,593 − 9904 − 2000 × 264.571) = 135,556 kJ/kmol

Substituting,

ln K p = −(135,556 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(2000 K)] = −8.15 or K p = 2.88 × 10 −4 (Table A - 28 : ln K p = −8.15)

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-7 16-14 The reaction C + O2 ⇔ CO2 is considered. The mole fraction of the carbon dioxide produced when this reaction occurs at a1 atm and 3800 K are to be determined. Assumptions 1 The equilibrium composition consists of CO2, C and O2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are

Stoichiometric:

C + O 2 ⇔ CO 2 (thus ν C = 1, ν O2 = 1, and ν CO2 = 1)

Actual:

C + O2 ⎯ ⎯→ xC + yO 2 + zCO 2 14243 123 react.

products

C balance:

1= x + z ⎯ ⎯→ z = 1 − x

O balance:

2 = 2 y + 2z ⎯ ⎯→ y = 1 − z = 1 − (1 − x) = x

Total number of moles:

N total = x + y + z = 1 + x

C + O2 ⇔ CO2 3800 K 1 atm

The equilibrium constant relation can be expressed as ν

Kp =

CO2 N CO2

ν

ν

N CC N O2O2

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(ν CO2 −ν C −ν O2 )

From the problem statement at 3800 K, ln K p = −0.461 . Then,

K p = exp(−0.461) = 0.6307 Substituting,

0.6307 =

(1 − x) ⎛ 1 ⎞ ⎜ ⎟ ( x)( x) ⎝ 1 + x ⎠

1−1−1

Solving for x, x = 0.7831 Then, y = x = 0.7831 z = 1 − x = 0.2169 Therefore, the equilibrium composition of the mixture at 3800 K and 1 atm is 0.7831 C + 0.7831 O 2 + 0.2169 CO 2

The mole fraction of carbon dioxide is y CO2 =

N CO2 0.2169 = = 0.1216 N total 1 + 0.7831

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-8 16-15 The reaction C + O2 ⇔ CO2 is considered. The mole fraction of the carbon dioxide produced when this reaction occurs at a1 atm and 3800 K and 700 kPa and 3800 K are to be determined. Assumptions 1 The equilibrium composition consists of CO2, C and O2. 2 The constituents of the mixture are ideal gases. Analysis We first solve the problem for 1 atm pressure: The stoichiometric and actual reactions in this case are Stoichiometric: C + O 2 ⇔ CO 2 (thus ν C = 1, ν O2 = 1, and ν CO2 = 1) C + O2 ⎯ ⎯→ xC + yO 2 + zCO 2 14243 123

Actual:

react.

products

C balance:

1= x + z ⎯ ⎯→ z = 1 − x

O balance:

2 = 2 y + 2z ⎯ ⎯→ y = 1 − z = 1 − (1 − x) = x

Total number of moles:

N total = x + y + z = 1 + x

C + O2 ⇔ CO2 3800 K 1 atm

The equilibrium constant relation can be expressed as ν

Kp =

CO2 N CO2

ν

ν

N CC N O2O2

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(ν CO2 −ν C −ν O2 )

From the problem statement at 3800 K, ln K p = −0.461 . Then,

K p = exp(−0.461) = 0.6307 Substituting, 0.6307 =

(1 − x) ⎛ 1 ⎞ ⎜ ⎟ ( x)( x) ⎝ 1 + x ⎠

1−1−1

Solving for x, x = 0.7831 Then, y = x = 0.7831 z = 1 − x = 0.2169 Therefore, the equilibrium composition of the mixture at 3800 K and 1 atm is 0.7831 C + 0.7831 O 2 + 0.2169 CO 2 The mole fraction of carbon dioxide is N 0.2169 y CO2 = CO2 = = 0.1216 N total 1 + 0.7831 We repeat the calculations at 700 kPa pressure: The pressure in this case is 700 kPa/(101.325 kPa/atm) = 6.908 atm. Then, ν

Kp =

CO2 N CO2

ν

ν

N CC N O2O2

0.6307 =

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(1 − x) ⎛ 6.908 ⎞ ⎜ ⎟ ( x)( x) ⎝ 1 + x ⎠

(ν CO2 −ν C −ν O2 )

1−1−1

x = 0.4320 y = x = 0.4320 z = 1 − x = 0.5680 Therefore, the equilibrium composition of the mixture at 3800 K and 700 kPa is 0.4320 C + 0.4320 O 2 + 0.5680 CO 2

C + O2 ⇔ CO2 3800 K 700 kPa

The mole fraction of carbon dioxide is N 0.5680 y CO2 = CO2 = = 0.3966 N total 1 + 0.4320

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-9 16-16 The reaction C + O2 ⇔ CO2 is considered. The mole fraction of the carbon dioxide produced when this reaction occurs at a1 atm and 3800 K and 700 kPa and 3800 K are to be determined. Assumptions 1 The equilibrium composition consists of CO2, C and O2. 2 The constituents of the mixture are ideal gases. Analysis We first solve the problem for the reaction C + O2 ⇔ CO2:

The stoichiometric and actual reactions in this case are Stoichiometric:

C + O 2 ⇔ CO 2 (thus ν C = 1, ν O2 = 1, and ν CO2 = 1)

Actual:

C + O2 ⎯ ⎯→ xC + yO 2 + zCO 2 14243 123 react.

products

C balance:

1= x + z ⎯ ⎯→ z = 1 − x

O balance:

2 = 2 y + 2z ⎯ ⎯→ y = 1 − z = 1 − (1 − x) = x

Total number of moles:

N total = x + y + z = 1 + x

C + O2 ⇔ CO2 3800 K 1 atm

The equilibrium constant relation can be expressed as ν

Kp =

CO2 N CO2

ν

ν

N CC N O2O2

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(ν CO2 −ν C −ν O2 )

From the problem statement at 3800 K, ln K p = −0.461 . Then,

K p = exp(−0.461) = 0.6307 Substituting, 0.6307 =

(1 − x) ⎛ 1 ⎞ ⎜ ⎟ ( x)( x) ⎝ 1 + x ⎠

1−1−1

Solving for x, x = 0.7831 Then, y = x = 0.7831 z = 1 − x = 0.2169 Therefore, the equilibrium composition of the mixture at 3800 K and 1 atm is 0.7831 C + 0.7831 O 2 + 0.2169 CO 2

The mole fraction of carbon dioxide is y CO2 =

N CO2 0.2169 = = 0.1216 N total 1 + 0.7831

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-10 If the reaction is: C + (O2 + 3.76 N2) ⇔ CO2 + 3.76 N2

The stoichiometric and actual reactions in this case are Stoichiometric: C + (O 2 + 3.76 N 2 ) ⇔ CO 2 + 3.76 N 2 (thus ν C = 1, ν O2 = 1,ν N2 = 3.76, ν CO2 = 1, and ν N2 = 3.76) C + ( O 2 + 3. 76 N 2 ) ⎯ ⎯→ xC + yO 2 + zCO 2 + 3.76 N 2 14243 1442443

Actual:

react.

products

C balance:

1= x + z ⎯ ⎯→ z = 1 − x

O balance:

2 = 2 y + 2z ⎯ ⎯→ y = 1 − z = 1 − (1 − x) = x

Total number of moles:

N total = x + y + z + 3.76 = 4.76 + x

C+(O2+3.76N2) ⇔CO2+3.76N2

3800 K 1 atm

The equilibrium constant relation can be expressed as Kp =

or

Kp =

CO2 N νCO2 N νN2N2

N νCC N νO2O2 N νN2N2 CO2 N νCO2

ν

ν

N CC N O2O2

⎛ P ⎜⎜ ⎝ N total

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

⎞ ⎟⎟ ⎠

(ν CO2 +ν N2 −ν C −ν O2 −ν N2 )

(ν CO2 −ν C −ν O2

From the problem statement at 3800 K, ln K p = 12.49 . Then,

K p = exp(12.49) = 265,670 Substituting,

265,670 =

(1 − x) ⎛ 1 ⎞ ⎜ ⎟ ( x)( x) ⎝ 4.76 + x ⎠

−1

Solving for x, x = 0.004226 Then, y = x = 0.004226 z = 1 − x = 0.9958 Therefore, the equilibrium composition of the mixture at 3800 K and 1 atm is 0.004226 C + 0.004226 O 2 + 3.76 N 2 + 0.9958 CO 2 + 3.76 N 2

The mole fraction of carbon dioxide is y CO2 =

N CO2 0.9958 = = 0.2090 N total 4.76 + 0.004226

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-11 16-17 A gaseous mixture consisting of methane and carbon dioxide is heated. The equilibrium composition (by mole fraction) of the resulting mixture is to be determined. Assumptions 1 The equilibrium composition consists of CH4, C, H2, and CO2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are

Stoichiometric:

CH 4 ⇔ C + 2H 2 (thus ν CH4 = 1, ν C = 1, and ν H2 = 2)

Actual:

0.3CH 4 + 0.7CO 2 ⎯ ⎯→ xCH 4 + yC + zH 2 + 0.7CO 2 123 14243 1 424 3 react.

products

C balance:

0 .3 = x + y ⎯ ⎯→ y = 0.3 − x

H balance:

1.2 = 4 x + 2 z ⎯ ⎯→ z = 0.6 − 2 x

Total number of moles:

N total = x + y + z + 1 = 1.6 − 2 x

inert

CH4, CO2 1200 K 1 atm

The equilibrium constant relation can be expressed as ν

Kp =

CH4 N CH4

ν

ν

N CC N H2H2

⎛ P ⎜ ⎜N ⎝ total

ν CH4 −ν C −ν H2

⎞ ⎟ ⎟ ⎠

From the problem statement at 1200 K, ln K p = 4.147 . Then, K p = exp( 4.147 ) = 63.244

For the reverse reaction that we consider,

K p = 1 / 63.244 = 0.01581 Substituting, 0.01581 =

1 ⎞ ⎛ ⎟ ⎜ 2 1 .6 − 2 x (0.3 − x)(0.6 − 2 x) ⎝ ⎠ x

1−1− 2

Solving for x, x = 0.0006637 Then, y = 0.3 − x = 0.2993 z = 0.6 − 2x = 0.5987 Therefore, the equilibrium composition of the mixture at 1200 K and 1 atm is 0.0006637 CH 4 + 0.2993 C + 0.5987 H 2 + 0.7 CO 2

The mole fractions are y CH4 =

N CH4 0.0006637 0.0006637 = = = 0.000415 1.599 N total 1.6 − 2 × 0.0006637

yC =

NC 0.2993 = = 0.1872 1.599 N total

y H2 =

N H2 0.5987 = = 0.3745 1.599 N total

y CO2 =

N CO2 0.7 = = 0.4379 N total 1.599

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-12 16-18 The dissociation reaction CO2 ⇔ CO + O is considered. The composition of the products at given pressure and temperature is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, and O. 2 The constituents of the mixture are ideal gases. Analysis For the stoichiometric reaction CO 2 ⇔ CO + 12 O 2 , from Table A-28, at 2500 K

ln K p = −3.331 For the oxygen dissociation reaction 0.5O 2 ⇔ O , from Table A-28, at 2500 K,

ln K p = −8.509 / 2 = −4.255

CO2 2500 K 1 atm

For the desired stoichiometric reaction CO 2 ⇔ CO + O (thus ν CO2 = 1, ν CO = 1 and ν O = 1) ,

ln K p = −3.331 − 4.255 = −7.586 and

K p = exp(−7.586) = 0.0005075 CO 2 ⎯ ⎯→ xCO 2 + yCO + zO 123 14243

Actual:

products

react.

C balance:

1= x+ y ⎯ ⎯→ y = 1 − x

O balance:

2 = 2x + y + z ⎯ ⎯→ z = 1 − x

Total number of moles:

N total = x + y + z = 2 − x

The equilibrium constant relation can be expressed as ν

ν

N CO N O ⎛ P K p = COν O ⎜⎜ CO2 N CO2 ⎝ N total

ν CO +ν O −ν CO2

⎞ ⎟ ⎟ ⎠

Substituting, 0.0005075 =

(1 − x)(1 − x) ⎛ 1 ⎞ ⎜ ⎟ x ⎝ 2− x⎠

1+1−1

Solving for x, x = 0.9775 Then, y = 1 − x = 0.0225 z = 1 − x = 0.0225 Therefore, the equilibrium composition of the mixture at 2500 K and 1 atm is 0.9775 CO 2 + 0.0225 CO + 0.0225 O

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-13 16-19 The dissociation reaction CO2 ⇔ CO + O is considered. The composition of the products at given pressure and temperature is to be determined when nitrogen is added to carbon dioxide. Assumptions 1 The equilibrium composition consists of CO2, CO, O, and N2. 2 The constituents of the mixture are ideal gases. Analysis For the stoichiometric reaction CO 2 ⇔ CO + 12 O 2 , from Table A-28, at 2500 K

ln K p = −3.331

CO2, 3N2

For the oxygen dissociation reaction 0.5O 2 ⇔ O , from Table A-28, at 2500 K,

ln K p = −8.509 / 2 = −4.255

2500 K 1 atm

For the desired stoichiometric reaction CO 2 ⇔ CO + O (thus ν CO2 = 1, ν CO = 1 and ν O = 1) ,

ln K p = −3.331 − 4.255 = −7.586 and

K p = exp(−7.586) = 0.0005075 CO 2 + 3N 2 ⎯ ⎯→ xCO 2 + yCO + zO + 3N 123 14243 {2

Actual:

products

react.

C balance:

1= x+ y ⎯ ⎯→ y = 1 − x

O balance:

2 = 2x + y + z ⎯ ⎯→ z = 1 − x

Total number of moles:

N total = x + y + z + 3 = 5 − x

inert

The equilibrium constant relation can be expressed as ν

ν

N CO N O ⎛ P K p = COν O ⎜⎜ CO2 N CO2 ⎝ N total

ν CO +ν O −ν CO2

⎞ ⎟ ⎟ ⎠

Substituting, 0.0005075 =

(1 − x)(1 − x) ⎛ 1 ⎞ ⎜ ⎟ x ⎝5− x ⎠

1+1−1

Solving for x, x = 0.9557 Then, y = 1 − x = 0.0443 z = 1 − x = 0.0443 Therefore, the equilibrium composition of the mixture at 2500 K and 1 atm is 0.9557 CO 2 + 0.0443 CO + 0.0443 O + 3N 2

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-14 16-20 The reaction N2 + O2 ⇔ 2NO is considered. The equilibrium mole fraction of NO 1600 K and 1 atm is to be determined. Assumptions 1 The equilibrium composition consists of N2, O2, and NO. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are

Stoichiometric:

N 2 + O 2 ⇔ 2 NO (thus ν N2 = 1, ν O2 = 1, and ν NO = 2)

Actual:

N2 + O2 ⎯ ⎯→ xN 2 + yO 2 + { zNO 14243 products react.

N balance:

2 = 2x + z ⎯ ⎯→ z = 2 − 2 x

O balance:

2 = 2y + z ⎯ ⎯→ y = x

Total number of moles:

N total = x + y + z = 2

N2, O2 1600 K 1 atm

The equilibrium constant relation can be expressed as ν

Kp =

NO N NO

ν

ν

N N2N2 N O2O2

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(ν NO −ν N2 −ν O2 )

From Table A-28, at 1600 K, ln K p = −5.294 . Since the stoichiometric reaction being considered is double this reaction, K p = exp(−2 × 5.294) = 2.522 × 10 −5

Substituting, 2.522 × 10 −5 =

(2 − 2 x) 2 ⎛ 1 ⎞ ⎜ ⎟ x2 ⎝2⎠

2−1−1

Solving for x, x = 0.9975 Then, y = x = 0.9975 z = 2 − 2x = 0.005009 Therefore, the equilibrium composition of the mixture at 1000 K and 1 atm is 0.9975 N 2 + 0.9975 O 2 + 0.005009 NO

The mole fraction of NO is then y NO =

N NO 0.005009 = = 0.002505 2 N total

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-15 16-21E The equilibrium constant of the reaction H2 + 1/2O2 ↔ H2O is listed in Table A-28 at different temperatures. The data are to be verified at two temperatures using Gibbs function data. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −∆G *( T )/ Ru T or ln K p = − ∆G * ( T ) / Ru T

where

H2 + ½O2 ↔ H2O ∆G * (T ) = ν H 2 O g H∗ 2O (T ) −ν H 2 g H∗ 2 (T ) −ν O 2 g O∗ 2 (T )

537 R

At 537 R, ∆G * ( T ) = 1( −98,350) − 1( 0) − 0.5( 0) = −98,350 Btu / lbmol

Substituting,

ln K p = −( −98,350 Btu / lbmol) / [(1.986 Btu / lbmol ⋅ R)(537 R)] = 92.22 or K p = 1.12 × 10 40 (Table A - 28: ln K p = 92.21)

(b) At 4320 R, ∆G * (T ) = ν H 2O g H∗ 2O (T ) − ν H 2 g H∗ 2 (T ) − ν O 2 g O∗ 2 (T ) = ν H 2 O ( h − Ts ) H 2 O − ν H 2 ( h − Ts ) H 2 − ν O 2 ( h − Ts ) O 2 = ν H 2O [(h f + h4320 − h537 ) − Ts ] H 2O − ν H 2 [(h f + h4320 − h298 ) − Ts ] H 2 − ν O 2 [(h f + h4320 − h298 ) − Ts ] O 2 = 1 × (−104,040 + 44,533 − 4258 − 4320 × 65.504) − 1 × (0 + 32,647.2 − 3640.3 − 4320 × 46.554) − 0.5 × (0 + 35,746 − 3725.1 − 4320 × 65.831) = −48,451 Btu/lbmol

Substituting,

ln K p = −(−48,451 Btu/lbmol)/[(1.986 Btu/lbmol.R)(4320 R)] = 5.647 or K p = 283 (Table A - 28 : ln K p = 5.619

Discussion Solving this problem using EES with the built-in ideal gas properties give Kp = 1.04×1040 for part (a) and Kp = 278 for part (b).

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-16 16-22 The equilibrium constant of the reaction CO + 1/2O2 ↔ CO2 at 298 K and 2000 K are to be determined, and compared with the values listed in Table A-28. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −∆G *( T )/ Ru T or ln K p = − ∆G * ( T ) / Ru T

where ∗ ∗ ∗ ∆G * (T ) = ν CO2 g CO2 (T ) −ν CO g CO (T ) −ν O2 g O2 (T )

At 298 K,

CO + 1 O 2 ⇔ CO 2 2

298 K

∆G * (T ) = 1(−394,360) − 1(−137,150) − 0.5(0) = −257,210 kJ/kmol

where the Gibbs functions are obtained from Table A-26. Substituting, ln K p = −

(−257,210 kJ/kmol) = 103.81 (8.314 kJ/kmol ⋅ K)(298 K)

ln K p = 103.76

From Table A-28: (b) At 2000 K,

∗ ∗ ∗ ∆G * (T ) = ν CO2 g CO2 (T ) −ν CO g CO (T ) −ν O2 g O2 (T )

= ν CO2 (h − Ts ) CO2 −ν CO (h − Ts ) CO −ν O2 (h − Ts ) O2

= 1[(−302,128) − (2000)(309.00)] − 1[(−53,826) − (2000)(258.48)] − 0.5[(59,193) − (2000)(268.53)] = −110,409 kJ/kmol

The enthalpies at 2000 K and entropies at 2000 K and 101.3 kPa (1 atm) are obtained from EES. Substituting, ln K p = −

(−110,409 kJ/kmol) = 6.64 (8.314 kJ/kmol ⋅ K)(2000 K)

From Table A-28:

ln K p = 6.635

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-17

16-23

The effect of varying the percent excess air during the steady-flow combustion of hydrogen is to be studied.

Analysis The combustion equation of hydrogen with stoichiometric amount of air is H 2 + 0.5[O 2 + 3.76N 2 ] ⎯ ⎯→ H 2 O + 0.5(3.76) N 2

For the incomplete combustion with 100% excess air, the combustion equation is H 2 + (1 + Ex)(0.5)[O 2 + 3.76N 2 ] ⎯ ⎯→ 0.97 H 2 O + a H 2 + b O 2 + c N 2

The coefficients are to be determined from the mass balances Hydrogen balance:

2 = 0.97 × 2 + a × 2 ⎯ ⎯→ a = 0.03

Oxygen balance:

(1 + Ex) × 0.5 × 2 = 0.97 + b × 2

Nitrogen balance: (1 + Ex) × 0.5 × 3.76 × 2 = c × 2 Solving the above equations, we find the coefficients (Ex = 1, a = 0.03 b = 0.515, c = 3.76) and write the balanced reaction equation as H 2 + [O 2 + 3.76N 2 ] ⎯ ⎯→ 0.97 H 2 O + 0.03 H 2 + 0.515 O 2 + 3.76 N 2

Total moles of products at equilibrium are N tot = 0.97 + 0.03 + 0.515 + 3.76 = 5.275

The assumed equilibrium reaction is H 2 O ←⎯→ H 2 + 0.5O 2

The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −∆G *( T )/ Ru T or ln K p = − ∆G * ( T ) / Ru T

where ∗ ∗ ∗ ∆G * (T ) = ν H2 g H2 (Tprod ) + ν O2 g O2 (Tprod ) −ν H2O g H2O (Tprod )

and the Gibbs functions are defined as ∗ g H2 (Tprod ) = (h − Tprod s ) H2 ∗ g O2 (Tprod ) = (h − Tprod s ) O2 ∗ g H2O (Tprod ) = (h − Tprod s ) H2O

The equilibrium constant is also given by ⎛ P K p = ⎜⎜ ⎝ N tot

and

⎞ ⎟ ⎟ ⎠

1+ 0.5 −1

ab 0.5

⎛ 1 ⎞ =⎜ ⎟ 1 0.97 ⎝ 5.275 ⎠

0.5

(0.03)(0.515) 0.5 = 0.009664 0.97

ln K p = ln(0.009664) = −4.647

The corresponding temperature is obtained solving the above equations using EES to be

Tprod = 2600 K This is the temperature at which 97 percent of H2 will burn into H2O. The copy of EES solution is given next. "Input Data from parametric table:" {PercentEx = 10} Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3"[kPa]" PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-18 R_u=8.314 "[kJ/kmol-K]" "The combustion equation of H2 with stoichiometric amount of air is H2 + 0.5(O2 + 3.76N2)=H2O +0.5(3.76)N2" "For the incomplete combustion with 100% excess air, the combustion equation is H2 + (1+EX)(0.5)(O2 + 3.76N2)=0.97 H2O +aH2 + bO2+cN2" "Specie balance equations give the values of a, b, and c." "H, hydrogen" 2 = 0.97*2 + a*2 "O, oxygen" (1+Ex)*0.5*2=0.97 + b*2 "N, nitrogen" (1+Ex)*0.5*3.76 *2 = c*2 N_tot =0.97+a +b +c "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is H2O=H2+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each H2mponent in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_H2O=Enthalpy(H2O,T=T_prod )-T_prod *Entropy(H2O,T=T_prod ,P=101.3) g_H2=Enthalpy(H2,T=T_prod )-T_prod *Entropy(H2,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_H2+0.5*g_O2-1*g_H2O "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P = (P/N_tot)^(1+0.5-1)*(a^1*b^0.5)/(0.97^1)" sqrt(P/N_tot )*a *sqrt(b )=K_P *0.97 lnK_p = ln(k_P) 2625

-5.414 -5.165 -5.019 -4.918 -4.844 -4.786 -4.739 -4.7 -4.667 -4.639

PercentEx [%] 10 20 30 40 50 60 70 80 90 100

Tprod [K] 2440 2490 2520 2542 2557 2570 2580 2589 2596 2602

2585

2545 Tprod

ln Kp

2505

2465

2425 10

20

30

40

50 60 PercentEx

70

80

90

100

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-19 16-24 The equilibrium constant of the reaction CH4 + 2O2 ↔ CO2 + 2H2O at 25°C is to be determined. Analysis The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e − ∆G *( T )/ Ru T or ln K p = − ∆G * ( T ) / Ru T

CH4 + 2O2 ↔ CO2 + 2H2O

where ∗ ∗ ∆G * (T ) = ν CO 2 g CO (T ) + ν H 2 O g H∗ 2 O (T ) − ν CH 4 g CH (T ) − ν O 2 g O∗ 2 (T ) 2 4

25°C

At 25°C, ∆G * ( T ) = 1( −394,360) + 2( −228,590) − 1( −50,790) − 2( 0) = −800,750 kJ / kmol

Substituting, ln K p = −( −800,750 kJ/kmol)/[ (8.314 kJ/kmol ⋅ K)(298 K)] = 323.04

or

K p = 1.96 × 10 140

16-25 The equilibrium constant of the reaction CO2 ↔ CO + 1/2O2 is listed in Table A-28 at different temperatures. It is to be verified using Gibbs function data. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e − ∆G*(T ) / RuT or ln K p = −∆G * (T ) / Ru T where

∗ ∗ ∆G * (T ) = ν CO g CO (T ) + ν O 2 g O∗ 2 (T ) − ν CO 2 g CO (T ) 2

At 298 K,

CO2 ↔ CO + ½O2 298 K

∆G * (T ) = 1(−137,150) + 0.5(0) − 1(−394,360) = 257,210 kJ/kmol

Substituting, ln K p = −( 257,210 kJ/kmol)/[ (8.314 kJ/kmol ⋅ K)(298 K)] = -103.81

or

K p = 8.24 × 10 -46 (Table A - 28 : ln K p = −103.76)

(b) At 1800 K, ∗ ∗ ∆G * (T ) = ν CO g CO (T ) + ν O 2 g O∗ 2 (T ) −ν CO 2 g CO (T ) 2

= ν CO (h − Ts ) CO + ν O 2 (h − Ts ) O 2 −ν CO 2 (h − Ts ) CO 2 = ν CO [(h f + h1800 − h298 ) − Ts ] CO + ν O 2 [(h f + h1800 − h298 ) − Ts ] O 2 −ν CO 2 [(h f + h1800 − h298 ) − Ts ] CO 2 = 1× (−110,530 + 58,191 − 8669 − 1800 × 254.797) + 0.5 × (0 + 60,371 − 8682 − 1800 × 264.701) − 1× (−393,520 + 88,806 − 9364 − 1800 × 302.884) = 127,240.2 kJ/kmol

Substituting, or

ln K p = −(127,240.2 kJ/kmol)/[ (8.314 kJ/kmol ⋅ K)(1800 K)] = −8.502

K p = 2.03 × 10 -4 (Table A - 28 : ln K p = −8.497 )

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-20 16-26 Carbon monoxide is burned with 100 percent excess air. The temperature at which 93 percent of CO burn to CO2 is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis Assuming N2 to remain as an inert gas, the stoichiometric and actual reactions can be written as

Stoichiometric:

CO + 12 O 2 ⇔ CO 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )

Actual:

CO + 1(O 2 + 3.76 N 2 ) ⎯ ⎯→ 0.93CO 2 + 0.07CO + 0.535O 2 + 3.76 N 2 1424 3 14442444 3 1 424 3 product

reactants

inert

The equilibrium constant Kp can be determined from ν

Kp =

CO 2 N CO 2

ν

ν

CO N CO N OO 2 2

=

⎛ P ⎜ ⎜N ⎝ total

0.93

0.07 × 0.535 0.5 = 41.80

⎞ ⎟ ⎟ ⎠

(ν CO 2 −ν CO −ν O 2 )

1 ⎛ ⎞ ⎜ ⎟ + + + 0 . 93 0 . 07 0 . 535 3 . 76 ⎝ ⎠

1−1.5

CO + ½O2 ↔ CO2 93 % 1 atm

and

ln K p = 3.733 From Table A-28, the temperature corresponding to this Kp value is T = 2424 K

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-21 16-27 Problem 16-26 is reconsidered. The effect of varying the percent excess air during the steady-flow process from 0 to 200 percent on the temperature at which 93 percent of CO burn into CO2 is to be studied. Analysis The problem is solved using EES, and the solution is given below. "To solve this problem, we need to give EES a guess value for T_prop other than the default value of 1. Set the guess value of T_prod to 1000 K by selecting Variable Information in the Options menu. Then press F2 or click the Calculator icon." "Input Data from the diagram window:" {PercentEx = 100} Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3 [kPa] R_u=8.314 [kJ/kmol-K] f=0.93 "The combustion equation of CO with stoichiometric amount of air is CO + 0.5(O2 + 3.76N2)=CO2 +0.5(3.76)N2" "For the incomplete combustion with 100% excess air, the combustion equation is CO + (1+EX)(0.5)(O2 + 3.76N2)=0.97 CO2 +aCO + bO2+cN2" "Specie balance equations give the values of a, b, and c." "C, Carbon" 1=f+a "O, oxygen" 1 +(1+Ex)*0.5*2=f*2 + a *1 + b*2 "N, nitrogen" (1+Ex)*0.5*3.76 *2 = c*2 N_tot =f+a +b +c "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P = (P/N_tot)^(1+0.5-1)*(a^1*b^0.5)/(0.97^1)" sqrt(P/N_tot )*a *sqrt(b )=K_P *f lnK_p = ln(k_P) "Compare the value of lnK_p calculated by EES with the value of lnK_p from table A-28 in the text."

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-22 Tprod [K] 2247 2342 2377 2398 2411 2421 2429 2435 2440 2444 2447

2450

2400

Tprod [K]

PercentEx [%] 0 20 40 60 80 100 120 140 160 180 200

2350

2300

2250 0

40

80

120

160

200

PercentEx [% ]

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-23 16-28E Carbon monoxide is burned with 100 percent excess air. The temperature at which 93 percent of CO burn to CO2 is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis Assuming N2 to remain as an inert gas, the stoichiometric and actual reactions can be written as

Stoichiometric:

CO + 12 O 2 ⇔ CO 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )

Actual:

CO + 1(O 2 + 3.76 N 2 ) ⎯ ⎯→ 0.93CO 2 + 0.07CO + 0.535O 2 + 3.76 N 2 1424 3 14442444 3 1 424 3 product

reactants

inert

The equilibrium constant Kp can be determined from ν

Kp =

CO 2 N CO 2

ν

ν

CO N CO N OO 2 2

⎛ P ⎜ ⎜N ⎝ total

0.93

=

0.07 × 0.535 0.5 = 41.80

⎞ ⎟ ⎟ ⎠

(ν CO 2 −ν CO −ν O 2 )

1 ⎛ ⎞ ⎜ ⎟ ⎝ 0.93 + 0.07 + 0.535 + 3.76 ⎠

CO + ½O2 ↔ CO2 93 % 1 atm

1−1.5

and

ln K p = 3.733 From Table A-28, the temperature corresponding to this Kp value is T = 2424 K = 4363 R

16-29 Hydrogen is burned with 150 percent theoretical air. The temperature at which 98 percent of H2 will burn to H2O is to be determined. Assumptions 1 The equilibrium composition consists of H2O, H2, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis Assuming N2 to remain as an inert gas, the stoichiometric and actual reactions can be written as

Stoichiometric:

H 2 + 21 O 2 ⇔ H 2 O (thus ν H 2 O = 1, ν H 2 = 1, and ν O 2 = 21 )

Actual:

H 2 + 0.75(O 2 + 3.76 N 2 )

⎯ ⎯→

0.98 H 2 O + 0.02 H 2 + 0.26O 2 + 2.82 N 2 1424 3 144 42444 3 12 4 4 3 product

reactants

inert

The equilibrium constant Kp can be determined from ν

Kp = =

N HHO2O 2

ν H2

ν O2

2

2

NH NO

⎛ P ⎜ ⎜N ⎝ total

0.98

0.02 × 0.26 0.5 = 194.11

⎞ ⎟ ⎟ ⎠

(ν H 2 O −ν H 2 −ν O 2 )

1 ⎛ ⎞ ⎜ ⎟ + + + 0 . 98 0 . 02 0 . 26 2 . 82 ⎝ ⎠

H2 Combustion chamber

1−1.5

H2O, H2 O2, N2

Air

From Table A-28, the temperature corresponding to this Kp value is T = 2472 K.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-24 16-30 Air is heated to a high temperature. The equilibrium composition at that temperature is to be determined. Assumptions 1 The equilibrium composition consists of N2, O2, and NO. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are N 2 + 12 O 2 ⇔ NO (thus ν NO = 1, ν N 2 = 12 , and ν O 2 = 12 )

Stoichiometric:

1 2

Actual:

3.76 N 2 + O 2

⎯ ⎯→

x NO + y N + z O 123 1422432 prod.

reactants

N balance:

7.52 = x + 2y or y = 3.76 - 0.5x

O balance:

2 = x + 2z or z = 1 - 0.5x

Total number of moles:

Ntotal = x + y + z = x + 4.76- x = 4.76

AIR 2000 K 2 atm

The equilibrium constant relation can be expressed as

Kp =

NO N νNO

ν

ν

N NN2 2 N OO2 2

⎛ P ⎜⎜ ⎝ N total

⎞ ⎟⎟ ⎠

(ν NO −ν N 2 −ν O 2 )

From Table A-28, ln Kp = -3.931 at 2000 K. Thus Kp = 0.01962. Substituting, 0.01962 =

x (3.76 − 0.5 x) 0.5 (1 − 0.5 x) 0.5

⎛ 2 ⎞ ⎜ ⎟ ⎝ 4.76 ⎠

1−1

Solving for x, x = 0.0376 Then, y = 3.76-0.5x = 3.7412 z = 1-0.5x = 0.9812 Therefore, the equilibrium composition of the mixture at 2000 K and 2 atm is 0.0376NO + 3.7412N 2 + 0.9812O 2

The equilibrium constant for the reactions O 2 ⇔ 2O (ln Kp = -14.622) and N 2 ⇔ 2 N (ln Kp = -41.645) are much smaller than that of the specified reaction (ln Kp = -3.931). Therefore, it is realistic to assume that no monatomic oxygen or nitrogen will be present in the equilibrium mixture. Also the equilibrium composition is in this case is independent of pressure since ∆ν = 1 − 0.5 − 0.5 = 0 .

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-25 16-31 Hydrogen is heated to a high temperature at a constant pressure. The percentage of H2 that will dissociate into H is to be determined. Assumptions 1 The equilibrium composition consists of H2 and H. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions can be written as

Stoichiometric:

H 2 ⇔ 2H (thus ν H 2 = 1 and ν H = 2)

Actual:

H2 ⎯ ⎯→ { xH 2 + { yH react.

prod.

H balance:

2 = 2x + y or y = 2 − 2x

Total number of moles:

Ntotal = x + y = x + 2 − 2x = 2 − x

H2 4000 K 5 atm

The equilibrium constant relation can be expressed as Kp =

ν H −ν H 2

NνHH ⎛ P ⎞ ⎟ ⎜ ν ⎟ ⎜ N H 2 ⎝ N total ⎠ H2

From Table A-28, ln Kp = 0.934 at 4000 K. Thus Kp = 2.545. Substituting, 2.545 =

(2 − 2 x) 2 x

⎛ 5 ⎞ ⎜ ⎟ ⎝2− x⎠

2 −1

Solving for x, x = 0.664 Thus the percentage of H2 which dissociates to H at 3200 K and 8 atm is 1 − 0.664 = 0.336 or 33.6%

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-26 16-32E A mixture of CO, O2, and N2 is heated to a high temperature at a constant pressure. The equilibrium composition is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are

Stoichiometric:

CO + 12 O 2 ⇔ CO 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )

Actual:

2 CO + 2 O 2 + 6 N 2

⎯ ⎯→

x CO + y CO + z O 2 + 6 N 2 1232 14244 3 products

C balance:

2= x+ y

⎯ ⎯→

O balance:

6 = 2x + y + 2z

Total number of moles:

N total = x + y + z + 6 = 10 − 0.5x

reactants

:

inert

2 CO 2 O2 6 N2 4320 R 3 atm

y = 2− x ⎯ ⎯→

z = 2 − 0.5x

The equilibrium constant relation can be expressed as ν

Kp =

CO 2 N CO 2

ν

ν

CO N CO N OO 2 2

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

(ν CO 2 −ν CO −ν O 2 )

From Table A-28, ln K p = 3.860 at T = 4320 R = 2400 K. Thus K p = 47.465. Substituting, 47.465 =

x ( 2 − x)(2 − 0.5 x) 0.5

3 ⎛ ⎞ ⎜ ⎟ 10 0 . 5 − x ⎝ ⎠

1−1.5

Solving for x, x = 1.930 Then, y = 2 - x = 0.070 z = 2 - 0.5x = 1.035 Therefore, the equilibrium composition of the mixture at 2400 K and 3 atm is 1.930CO 2 + 0.070CO + 1.035O 2 + 6N 2

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-27 16-33 A mixture of N2, O2, and Ar is heated to a high temperature at a constant pressure. The equilibrium composition is to be determined. Assumptions 1 The equilibrium composition consists of N2, O2, Ar, and NO. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are N 2 + 12 O 2 ⇔ NO (thus ν NO = 1, ν N 2 = 12 , and ν O 2 = 12 )

Stoichiometric:

1 2

Actual:

3 N 2 + O 2 + 01 . Ar

⎯ ⎯→

x NO + y N + z O + 0.1 Ar 123 1422432 123 prod.

reactants

N balance:

6 = x + 2y

⎯ ⎯→

y = 3 − 0.5x

O balance:

2 = x + 2z

⎯ ⎯→

z = 1 − 0.5x

Total number of moles:

N total = x + y + z + 01 . = 41 .

inert

3 N2 1 O2 0.1 Ar 2400 K 10 atm

The equilibrium constant relation becomes, Kp =

ν NO N NO ν

ν

N NN 2 N OO2 2

2

⎛ P ⎜⎜ ⎝ N total

⎞ ⎟⎟ ⎠

(ν NO − ν N 2 − νO2 )

=

x y 0.5 z 0.5

⎛ P ⎜⎜ ⎝ N total

⎞ ⎟⎟ ⎠

1−0 .5−0 .5

From Table A-28, ln K p = −3.019 at 2400 K. Thus K p = 0.04885. Substituting, 0.04885 =

x ×1 (3 − 0.5 x) (1 − 0.5 x)0.5 0.5

Solving for x, x = 0.0823 Then, y = 3 - 0.5x = 2.9589 z = 1 - 0.5x = 0.9589 Therefore, the equilibrium composition of the mixture at 2400 K and 10 atm is 0.0823NO + 2.9589N 2 + 0.9589O 2 + 0.1Ar

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-28 +

-

16-34 The mole fraction of sodium that ionizes according to the reaction Na ⇔ Na + e at 2000 K and 1.5 atm is to be determined. Assumptions All components behave as ideal gases. Analysis The stoichiometric and actual reactions can be written as

Stoichiometric:

Na ⇔ Na + + e - (thus ν Na = 1, ν Na + = 1 and ν e - = 1)

Actual:

Na ⎯ ⎯→ { x Na + y Na + + y e − 142 4 43 4 react.

Na ⇔ Na+ + e2000 K 1.5 atm

products

1 = x + y or y = 1 − x

Na balance:

N total = x + 2 y = 2 − x

Total number of moles:

The equilibrium constant relation becomes, ν

Kp =

N NaNa N ν

N NaNa

ν

e-

e-

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

Na +

+ν - −ν Na ) e

=

y2 x

⎛ P ⎜ ⎜N ⎝ total

⎞ ⎟ ⎟ ⎠

1+1−1

Substituting, 0.668 =

(1 − x) 2 ⎛ 1.5 ⎞ ⎜ ⎟ x ⎝2− x⎠

Solving for x, x = 0.4449 Thus the fraction of Na which dissociates into Na+ and e- is 1 − 0.4449 = 0.555 or 55.5%

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-29 16-35 Oxygen is heated from a specified state to another state. The amount of heat required is to be determined without and with dissociation cases. Assumptions 1 The equilibrium composition consists of O2 and O. 2 The constituents of the mixture are ideal gases. Analysis (a) Obtaining oxygen properties from table A-19, an energy balance gives E −E 1in424out 3

=

Net energy transfer by heat, work, and mass

∆E system 1 424 3

Change in internal, kinetic, potential, etc. energies

q in = u 2 − u1 = 57,192 − 6203 = 50,989 kJ/kmol

(b) The stoichiometric and actual reactions in this case are Stoichiometric:

O 2 ⇔ 2O (thus ν O2 = 1 and ν O = 2)

Actual:

O2 ⎯ ⎯→ { xO 2 + { yO react.

O2 2200 K 1 atm

products

O balance:

2 = 2x + y ⎯ ⎯→ y = 2 − 2 x

Total number of moles:

N total = x + y = 2 − x

The equilibrium constant relation can be expressed as ν

Kp =

N OO ⎛ P ⎜ ⎜ ν N O2O2 ⎝ N total

ν O −ν O2

⎞ ⎟ ⎟ ⎠

From Table A-28, at 2200 K, ln K p = −11.827 . Then, K p = exp( −11.827 ) = 7.305 × 10 −6

Substituting, 7.305 × 10 −6 =

(2 − 2 x) 2 ⎛ 1 ⎞ ⎜ ⎟ x ⎝ 2− x⎠

2 −1

Solving for x, x = 0.99865 Then, y = 2 − 2x = 0.0027 Therefore, the equilibrium composition of the mixture at 2200 K and 1 atm is 0.99865 O 2 + 0.0027 O

Hence, the oxygen ions are negligible and the result is same as that in part (a), q in = 50,989 kJ/kmol

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-30 16-36 Air is heated from a specified state to another state. The amount of heat required is to be determined without and with dissociation cases. Assumptions 1 The equilibrium composition consists of O2 and O, and N2. 2 The constituents of the mixture are ideal gases. Analysis (a) Obtaining air properties from table A-17, an energy balance gives E −E 1in424out 3

=

Net energy transfer by heat, work, and mass

∆E system 1 424 3

Change in internal, kinetic, potential, etc. energies

q in = u 2 − u1

O2, 3.76N2 2200 K 1 atm

= 1872.4 − 212.64 = 1660 kJ/kg

(b) The stoichiometric and actual reactions in this case are Stoichiometric:

O 2 ⇔ 2O (thus ν O2 = 1 and ν O = 2)

Actual:

O 2 + 3.76 N 2 ⎯ ⎯→ { xO 2 + { yO + 3.76 N 2 1 424 3 react.

products

inert

O balance:

2 = 2x + y ⎯ ⎯→ y = 2 − 2 x

Total number of moles:

N total = x + y + 3.76 = 5.76 − x

The equilibrium constant relation can be expressed as ν

Kp =

N OO ⎛ P ⎜ ⎜ ν N O2O2 ⎝ N total

ν O −ν O2

⎞ ⎟ ⎟ ⎠

From Table A-28, at 2200 K, ln K p = −11.827 . Then, K p = exp( −11.827 ) = 7.305 × 10 −6

Substituting, 7.305 × 10 −6 =

(2 − 2 x) 2 ⎛ 1 ⎞ ⎜ ⎟ x ⎝ 5.76 − x ⎠

2 −1

Solving for x, x = 0.99706 Then, y = 2 − 2x = 0.00588 Therefore, the equilibrium composition of the mixture at 2200 K and 1 atm is 0.99706 O 2 + 0.00588 O + 3.76 N 2

Hence, the atomic oxygen is negligible and the result is same as that in part (a), q in = 1660 kJ/kg

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-31 16-37 Liquid propane enters a combustion chamber. The equilibrium composition of product gases and the rate of heat transfer from the combustion chamber are to be determined. Assumptions 1 The equilibrium composition consists of CO2, H2O, CO, N2, and O2. 2 The constituents of the mixture are ideal gases. Analysis (a) Considering 1 kmol of C3H8, the stoichiometric combustion equation can be written as

C3H8 25°C Air

C 3 H 8 (l) + a th (O 2 + 3.76N 2 ) ⎯ ⎯→ 3CO 2 + 4H 2 O + 3.76a th N 2

Combustion chamber 2 atm

12°C

CO 1200 K CO2 H2O O2 N2

where ath is the stoichiometric coefficient and is determined from the O2 balance, 2.5a th = 3 + 2 + 1.5a th

⎯ ⎯→

a th = 5

Then the actual combustion equation with 150% excess air and some CO in the products can be written as C3H 8 ( l ) + 12.5( O 2 + 3.76 N 2 )

⎯ ⎯→

xCO 2 + (3 − x )CO + (9 − 0.5x )O 2 + 4H 2 O + 47N 2

After combustion, there will be no C3 H8 present in the combustion chamber, and H2O will act like an inert gas. The equilibrium equation among CO2, CO, and O2 can be expressed as CO 2 ⇔ CO + 12 O 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )

and ν

ν

CO N CO N OO 2 ⎛ P 2 ⎜ Kp = ⎜N ν CO 2 ⎝ total N CO 2

⎞ ⎟ ⎟ ⎠

(ν CO +ν O 2 −ν CO 2 )

where N total = x + (3 − x ) + (9 − 0.5x ) + 4 + 47 = 63 − 0.5x

From Table A-28, ln K p = −17.871 at 1200 K. Thus K p = 1.73 × 10 −8 . Substituting, 1.73 × 10 −8 =

(3 − x)(9 − 0.5 x) 0.5 ⎛ 2 ⎞ ⎜ ⎟ x ⎝ 63 − 0.5 x ⎠

1.5−1

Solving for x, x = 2.9999999 ≅ 3.0

Therefore, the amount CO in the product gases is negligible, and it can be disregarded with no loss in accuracy. Then the combustion equation and the equilibrium composition can be expressed as C 3 H 8 ( l) + 12.5(O 2 + 3.76N 2 ) ⎯ ⎯→ 3CO 2 + 7.5O 2 + 4 H 2 O + 47N 2

and 3CO 2 + 7.5O 2 + 4H 2 O + 47N 2

(b) The heat transfer for this combustion process is determined from the steady-flow energy balance Ein − E out = ∆E system on the combustion chamber with W = 0, − Qout =

∑ N (h P

o f

+ h −ho

) − ∑ N (h P

R

o f

+ h −ho

)

R

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, (The h fo of liquid

propane is obtained by adding the hfg at 25°C to h fo of gaseous propane).

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-32

h fo

h 285 K

h 298 K

h1200 K

kJ/kmol

kJ/kmol

kJ/kmol

kJ/kmol

C3H8 (l)

-118,910

---

---

---

O2

0

8696.5

8682

38,447

N2

0

8286.5

8669

36,777

H2O (g)

-241,820

---

9904

44,380

CO2

-393,520

---

9364

53,848

Substance

Substituting, − Qout = 3( −393,520 + 53,848 − 9364) + 4( −241,820 + 44,380 − 9904) + 7.5( 0 + 38,447 − 8682) + 47( 0 + 36,777 − 8669) − 1( −118,910 + h298 − h298 ) − 12.5( 0 + 8296.5 − 8682) − 47( 0 + 8186.5 − 8669) = −185,764 kJ / kmol of C3H 8

or Qout = 185,764 kJ / kmol of C3H 8

The mass flow rate of C3H8 can be expressed in terms of the mole numbers as & 12 . kg / min m N& = = = 0.02727 kmol / min M 44 kg / kmol

Thus the rate of heat transfer is

Q& out = N& × Qout = (0.02727 kmol/min)(185,746 kJ/kmol) = 5066 kJ/min The equilibrium constant for the reaction

1 2

N 2 + 12 O 2 ⇔ NO is ln Kp = -7.569, which is very small. This indicates that the

amount of NO formed during this process will be very small, and can be disregarded.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-33 16-38 Problem 16-37 is reconsidered. It is to be investigated if it is realistic to disregard the presence of NO in the product gases. Analysis The problem is solved using EES, and the solution is given below. "To solve this problem, the Gibbs function of the product gases is minimized. Click on the Min/Max icon." For this problem at 1200 K the moles of CO are 0.000 and moles of NO are 0.000, thus we can disregard both the CO and NO. However, try some product temperatures above 1286 K and observe the sign change on the Q_out and the amout of CO and NO present as the product temperature increases." "The reaction of C3H8(liq) with excess air can be written: C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO The coefficients A_th and EX are the theoretical oxygen and the percent excess air on a decimal basis. Coefficients a, b, c, d, e, and f are found by minimiming the Gibbs Free Energy at a total pressure of the product gases P_Prod and the product temperature T_Prod. The equilibrium solution can be found by applying the Law of Mass Action or by minimizing the Gibbs function. In this problem, the Gibbs function is directly minimized using the optimization capabilities built into EES. To run this program, click on the Min/Max icon. There are six compounds present in the products subject to four specie balances, so there are two degrees of freedom. Minimize the Gibbs function of the product gases with respect to two molar quantities such as coefficients b and f. The equilibrium mole numbers a, b, c, d, e, and f will be determined and displayed in the Solution window." PercentEx = 150 [%] Ex = PercentEx/100 "EX = % Excess air/100" P_prod =2*P_atm T_Prod=1200 [K] m_dot_fuel = 0.5 [kg/s] Fuel\$='C3H8' T_air = 12+273 "[K]" T_fuel = 25+273 "[K]" P_atm = 101.325 [kPa] R_u=8.314 [kJ/kmol-K] "Theoretical combustion of C3H8 with oxygen: C3H8 + A_th O2 = 3 C02 + 4 H2O " 2*A_th = 3*2 + 4*1 "Balance the reaction for 1 kmol of C3H8" "C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO" b_max = 3 f_max = (1+Ex)*A_th*3.76*2 e_guess=Ex*A_th 1*3 = a*1+b*1 "Carbon balance" 1*8=c*2 "Hydrogen balance" (1+Ex)*A_th*2=a*2+b*1+c*1+e*2+f*1 "Oxygen balance" (1+Ex)*A_th*3.76*2=d*2+f*1 "Nitrogen balance"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-34 "Total moles and mole fractions" N_Total=a+b+c+d+e+f y_CO2=a/N_Total; y_CO=b/N_Total; y_H2O=c/N_Total; y_N2=d/N_Total; y_O2=e/N_Total; y_NO=f/N_Total "The following equations provide the specific Gibbs function for each component as a function of its molar amount" g_CO2=Enthalpy(CO2,T=T_Prod)-T_Prod*Entropy(CO2,T=T_Prod,P=P_Prod*y_CO2) g_CO=Enthalpy(CO,T=T_Prod)-T_Prod*Entropy(CO,T=T_Prod,P=P_Prod*y_CO) g_H2O=Enthalpy(H2O,T=T_Prod)-T_Prod*Entropy(H2O,T=T_Prod,P=P_Prod*y_H2O) g_N2=Enthalpy(N2,T=T_Prod)-T_Prod*Entropy(N2,T=T_Prod,P=P_Prod*y_N2) g_O2=Enthalpy(O2,T=T_Prod)-T_Prod*Entropy(O2,T=T_Prod,P=P_Prod*y_O2) g_NO=Enthalpy(NO,T=T_Prod)-T_Prod*Entropy(NO,T=T_Prod,P=P_Prod*y_NO) "The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance" Gibbs=a*g_CO2+b*g_CO+c*g_H2O+d*g_N2+e*g_O2+f*g_NO "For the energy balance, we adjust the value of the enthalpy of gaseous propane given by EES:" h_fg_fuel = 15060"[kJ/kmol]" "Table A.27" h_fuel = enthalpy(Fuel\$,T=T_fuel)-h_fg_fuel "Energy balance for the combustion process:" "C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO" HR =Q_out+HP HR=h_fuel+ (1+Ex)*A_th*(enthalpy(O2,T=T_air)+3.76*enthalpy(N2,T=T_air)) HP=a*enthalpy(CO2,T=T_prod)+b*enthalpy(CO,T=T_prod)+c*enthalpy(H2O,T=T_prod)+d*enthalpy(N2,T=T_pro d)+e*enthalpy(O2,T=T_prod)+f*enthalpy(NO,T=T_prod) "The heat transfer rate is:" Q_dot_out=Q_out/molarmass(Fuel\$)*m_dot_fuel "[kW]" SOLUTION a=3.000 [kmol] A_th=5 b=0.000 [kmol] b_max=3 c=4.000 [kmol] d=47.000 [kmol] e=7.500 [kmol] Ex=1.5 e_guess=7.5 f=0.000 [kmol] Fuel\$='C3H8' f_max=94 Gibbs=-17994897 [kJ] g_CO=-703496 [kJ/kmol]

g_CO2=-707231 [kJ/kmol] g_H2O=-515974 [kJ/kmol] g_N2=-248486 [kJ/kmol] g_NO=-342270 [kJ/kmol] g_O2=-284065 [kJ/kmol] HP=-330516.747 [kJ/kmol] HR=-141784.529 [kJ/kmol] h_fg_fuel=15060 [kJ/kmol] h_fuel=-118918 [kJ/kmol] m_dot_fuel=0.5 [kg/s] N_Total=61.5 [kmol/kmol_fuel] PercentEx=150 [%] P_atm=101.3 [kPa] P_prod=202.7 [kPa]

Q_dot_out=2140 [kW] Q_out=188732 [kJ/kmol_fuel] R_u=8.314 [kJ/kmol-K] T_air=285 [K] T_fuel=298 [K] T_Prod=1200.00 [K] y_CO=1.626E-15 y_CO2=0.04878 y_H2O=0.06504 y_N2=0.7642 y_NO=7.857E-08 y_O2=0.122

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-35 16-39 Oxygen is heated during a steady-flow process. The rate of heat supply needed during this process is to be determined for two cases. Assumptions 1 The equilibrium composition consists of O2 and O. 2 All components behave as ideal gases. Analysis (a) Assuming some O2 dissociates into O, the dissociation equation can be written as O2

⎯ ⎯→

x O 2 + 2(1 − x )O

Q&

The equilibrium equation among O2 and O can be expressed as

O 2 ⇔ 2O (thus ν O 2 = 1 and ν O = 2) Assuming ideal gas behavior for all components, the equilibrium constant relation can be expressed as ν

Kp =

N OO ⎛ P ⎜ ⎜ ν N OO 2 ⎝ N total 2

where

O2

O2, O

298 K

3000 K

ν O −ν O 2

⎞ ⎟ ⎟ ⎠

N total = x + 2(1 − x ) = 2 − x

From Table A-28, ln K p = −4.357 at 3000 K. Thus K p = 0.01282. Substituting, 0.01282 =

(2 − 2 x) 2 ⎛ 1 ⎞ ⎜ ⎟ x ⎝2− x⎠

2−1

Solving for x gives x = 0.943 Then the dissociation equation becomes O2

⎯ ⎯→

0.943 O 2 + 0114 . O

The heat transfer for this combustion process is determined from the steady-flow energy balance Ein − E out = ∆E system on the combustion chamber with W = 0, Qin =

∑ N (h P

o f

+ h −ho

) − ∑ N (h P

R

o f

+ h −ho

)

R

Assuming the O2 and O to be ideal gases, we have h = h(T). From the tables,

hfo

h 298 K

h 3000 K

kJ/kmol

kJ/kmol

kJ/kmol

O

249,190

6852

63,425

O2

0

8682

106,780

Substance

Substituting, Qin = 0.943( 0 + 106,780 − 8682) + 0114 . ( 249,190 + 63,425 − 6852) − 0 = 127,363 kJ / kmol O 2

The mass flow rate of O2 can be expressed in terms of the mole numbers as 0.5 kg/min m& N& = = = 0.01563 kmol/min M 32 kg/kmol

Thus the rate of heat transfer is

Q& in = N& × Qin = (0.01563 kmol/min)(127,363 kJ/kmol) = 1990 kJ/min (b) If no O2 dissociates into O, then the process involves no chemical reactions and the heat transfer can be determined from the steady-flow energy balance for nonreacting systems to be

Q& in = m& (h2 − h1 ) = N& (h2 − h1 ) = (0.01563 kmol/min)(106,780 - 8682) kJ/kmol = 1533 kJ/min

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-36 16-40 The equilibrium constant, Kp is to be estimated at 3000 K for the reaction CO + H2O = CO2 + H2. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e − ∆G*(T ) / RuT or ln K p = −∆G * (T ) / Ru T where ∗ ∗ ∗ ∗ ∆G * (T ) = ν CO2 g CO2 (T ) +ν H2 g H2 (T ) −ν CO g CO (T ) −ν H2O g H2O (T )

At 3000 K, ∗ ∗ ∗ ∗ ∆G * (T ) = ν CO2 g CO2 (T ) + ν H2 g H2 (T ) − ν CO g CO (T ) − ν H2O g H2O (T )

= ν CO2 (h − Ts ) CO2 + ν H2 (h − Ts ) H2 − ν CO (h − Ts ) CO − ν H2O (h − Ts ) H2O

= 1[(−393,520 + 162,226 − 9364) − (3000)(334.084)] + 1[(0 + 97,211 − 8468) − (3000)(202.778)] − 1[(−110,530 + 102,210 − 8669) − (3000)(273.508] − 1[(−241,820 + 136,264 − 9904) − (3000)(286.273] = 49,291 kJ/kmol

Substituting,

ln K p = −

49,291 kJ/kmol = −1.9762 ⎯ ⎯→ K p = 0.1386 (8.314 kJ/kmol ⋅ K)(3000 K)

The equilibrium constant may be estimated using the integrated van't Hoff equation: ⎛ K p ,est ⎞ ⎟= ln⎜ ⎜ K p1 ⎟ ⎠ ⎝ K ⎛ p ,est ⎞ ⎟= ln⎜⎜ ⎟ ⎝ 0.2209 ⎠

hR Ru

⎛ 1 1⎞ ⎜⎜ − ⎟⎟ ⎝ TR T ⎠

1 ⎞ − 26,176 kJ/kmol ⎛ 1 − ⎯→ K p ,est = 0.1307 ⎜ ⎟⎯ 8.314 kJ/kmol.K ⎝ 2000 K 3000 K ⎠

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-37 16-41 A constant volume tank contains a mixture of H2 and O2. The contents are ignited. The final temperature and pressure in the tank are to be determined. Analysis The reaction equation with products in equilibrium is H2 + O2 ⎯ ⎯→ a H 2 + b H 2 O + c O 2

The coefficients are determined from the mass balances Hydrogen balance:

2 = 2a + 2b

Oxygen balance:

2 = b + 2c

The assumed equilibrium reaction is H 2 O ←⎯→ H 2 + 0.5O 2

The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e − ∆G *( T )/ Ru T or ln K p = − ∆G * ( T ) / Ru T

where ∗ ∗ ∗ ∆G * (T ) = ν H2 g H2 (Tprod ) + ν O2 g O2 (Tprod ) − ν H2O g H2O (Tprod )

and the Gibbs functions are given by ∗ (Tprod ) = (h − Tprod s ) H2 g H2 ∗ (Tprod ) = (h − Tprod s ) O2 g O2 ∗ (Tprod ) = (h − Tprod s ) H2O g H2O

The equilibrium constant is also given by Kp =

a 1 c 0.5 ⎛ P ⎜ b 1 ⎜⎝ N tot

⎞ ⎟ ⎟ ⎠

1+ 0.5 −1

=

ac 0.5 b

⎛ P2 / 101.3 ⎞ ⎟⎟ ⎜⎜ ⎝ a+b+c ⎠

0.5

An energy balance on the tank under adiabatic conditions gives UR =UP

where U R = 1([email protected]°C − Ru Treac ) + 1([email protected]°C − Ru Treac ) = 0 − (8.314 kJ/kmol.K)(298.15 K) + 0 − (8.314 kJ/kmol.K)(298.15 K) = −4958 kJ/kmol

U P = a([email protected] Tprod − Ru Tprod ) + b([email protected] Tprod − Ru Tprod ) + c([email protected] Tprod − Ru Tprod ) The relation for the final pressure is

P2 =

N tot Tprod ⎛ a + b + c ⎞⎛⎜ Tprod ⎞⎟ P1 = ⎜ (101.3 kPa) ⎟⎜ N 1 Treac 2 ⎝ ⎠⎝ 298.15 K ⎟⎠

Solving all the equations simultaneously using EES, we obtain the final temperature and pressure in the tank to be Tprod = 3857 K P2 = 1043 kPa

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-38 16-42 It is to be shown that as long as the extent of the reaction, α, for the disassociation reaction X2 ⇔ 2X is smaller than

one, α is given by α =

KP 4+ KP

Assumptions The reaction occurs at the reference temperature. Analysis The stoichiometric and actual reactions can be written as

Stoichiometric:

X 2 ⇔ 2X (thus ν X2 = 1 and ν X = 2)

Actual:

X 2 ⇔ (1 − α )X 2 + 2{ αX 1424 3 prod. react.

The equilibrium constant Kp is given by ν

N X ⎛ P K p = νX ⎜⎜ N X2X2 ⎝ N total

ν X −ν X2

⎞ ⎟ ⎟ ⎠

=

( 2α ) 2 ⎛ 1 ⎞ ⎜ ⎟ (1 − α ) ⎝ α + 1 ⎠

2 −1

=

4α 2 (1 − α )(1 + α )

Solving this expression for α gives

α=

KP 4+ KP

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-39 Simultaneous Reactions 16-43C It can be expresses as “(dG)T,P = 0 for each reaction.” Or as “the Kp relation for each reaction must be satisfied.”

16-44C The number of Kp relations needed to determine the equilibrium composition of a reacting mixture is equal to the difference between the number of species present in the equilibrium mixture and the number of elements.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-40 16-45 Two chemical reactions are occurring in a mixture. The equilibrium composition at a specified temperature is to be determined. Assumptions 1 The equilibrium composition consists of H2O, OH, O2, and H2. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as H 2O

⎯ ⎯→

H 2O ⇒

x H 2O + y H 2 + z O 2 + w OH

2 = 2x + 2 y + w

(1)

O balance:

1 = x + 2z + w

(2)

O 2 ,H 2

3400 K 1 atm

Mass balances for hydrogen and oxygen yield H balance:

H 2O,OH

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the Kp relations) to determine the equilibrium composition of the mixture. They are

H 2 O ⇔ H 2 + 12 O 2

(reaction 1)

H 2 O ⇔ 12 H 2 + OH

(reaction 2)

The equilibrium constant for these two reactions at 3400 K are determined from Table A-28 to be ln K P1 = −1891 .

⎯ ⎯→

K P1 = 015092 .

ln K P 2 = −1576 .

⎯ ⎯→

K P 2 = 0.20680

The Kp relations for these two simultaneous reactions are ν

K P1 =

ν

N HH 2 N OO 2 ⎛ P 2 2 ⎜ ⎜N ν H 2O ⎝ total NH O 2

where

⎞ ⎟ ⎟ ⎠

ν

(ν H 2 +ν O 2 −ν H 2 O )

and

K P2 =

ν

OH N HH 2 N OH ⎛ P 2 ⎜ ⎜N ν H 2O ⎝ total NH O 2

⎞ ⎟ ⎟ ⎠

(ν H 2 +ν OH −ν H 2 O )

N total = N H2O + N H2 + N O2 + N OH = x + y + z + w

Substituting, 1/ 2

0.15092 =

( y )( z )1 / 2 x

⎛ ⎞ 1 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠

0.20680 =

( w)( y )1/ 2 x

⎛ ⎞ 1 ⎜⎜ ⎟⎟ x + y + z + w ⎝ ⎠

(3) 1/ 2

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 0.574

y = 0.308

z = 0.095

w = 0.236

Therefore, the equilibrium composition becomes 0.574H 2 O + 0.308H 2 + 0.095O 2 + 0.236OH

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-41 16-46 Two chemical reactions are occurring in a mixture. The equilibrium composition at a specified temperature is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and O. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as ⎯ ⎯→

2 CO 2 + O 2

x CO 2 + y CO + z O 2 + w O

CO2, CO, O2, O 2000 K 4 atm

Mass balances for carbon and oxygen yield C balance:

2= x+ y

(1)

O balance:

6 = 2x + y + 2z + w

(2)

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the KP relations) to determine the equilibrium composition of the mixture. They are

CO 2 ⇔ CO + 12 O 2

(reaction 1)

O 2 ⇔ 2O

(reaction 2)

The equilibrium constant for these two reactions at 2000 K are determined from Table A-28 to be ln K P1 = −6.635 ⎯ ⎯→ K P1 = 0.001314 ln K P 2 = −14.622 ⎯ ⎯→ K P 2 = 4.464 × 10 −7 The KP relations for these two simultaneous reactions are ν

ν

K P1

CO N CO N OO 2 ⎛ P 2 ⎜ = ⎜N ν CO 2 ⎝ total N CO 2

νO

K P2

N ⎛ P = νO ⎜⎜ O2 N O ⎝ N total 2

⎞ ⎟ ⎟ ⎠

(ν CO +ν O 2 −ν CO 2 )

ν O −ν O 2

⎞ ⎟ ⎟ ⎠

where

N total = N CO2 + N O2 + N CO + N O = x + y + z + w Substituting, ( y )( z )1 / 2 0.001314 = x 4.464 × 10

−7

w2 = z

⎛ ⎞ 4 ⎜⎜ ⎟⎟ x + y + z + w ⎝ ⎠

⎛ ⎞ 4 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠

1/ 2

(3)

2 −1

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously using an equation solver such as EES for the four unknowns x, y, z, and w yields x = 1.998

y = 0.002272

z = 1.001

w = 0.000579

Thus the equilibrium composition is 1.998CO 2 + 0.002272CO + 1.001O 2 + 0.000579O

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-42 16-47 Two chemical reactions are occurring at high-temperature air. The equilibrium composition at a specified temperature is to be determined. Assumptions 1 The equilibrium composition consists of O2, N2, O, and NO. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as ⎯ ⎯→

O 2 + 3.76 N 2

Heat

x N 2 + y NO + z O 2 + w O

Mass balances for nitrogen and oxygen yield

AIR

N balance:

7.52 = 2 x + y

(1)

O balance:

2 = y + 2z + w

(2)

Reaction chamber, 2 atm

O2, N2, O, NO 3000 K

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the Kp relations) to determine the equilibrium composition of the mixture. They are 1 2

N 2 + 21 O 2 ⇔ NO

(reaction 1)

O 2 ⇔ 2O

(reaction 2)

The equilibrium constant for these two reactions at 3000 K are determined from Table A-28 to be ln K P1 = −2.114

⎯ ⎯→

K P1 = 012075 .

ln K P 2 = −4.357

⎯ ⎯→

K P 2 = 0.01282

The KP relations for these two simultaneous reactions are ν

K P1 =

ν

ν

N NN 2 N OO 2 2

2

K P2 =

⎛ P ⎜ ⎜N ⎝ total

NO N NO

νO

NO ⎛ P ⎜ ⎜ ν N OO 2 ⎝ N total 2

where

⎞ ⎟ ⎟ ⎠

(ν NO −ν N 2 −ν O 2 )

ν O −ν O 2

⎞ ⎟ ⎟ ⎠

N total = N N 2 + N NO + N O 2 + N O = x + y + z + w

Substituting, 0.12075 =

0.01282 =

y x 0 .5 z 0 .5 w2 z

⎛ ⎞ 2 ⎜⎜ ⎟⎟ ⎝ x+ y+ z+w⎠

⎛ ⎞ 2 ⎜⎜ ⎟⎟ ⎝ x+ y+ z+w⎠

1− 0.5 − 0.5

(3)

2 −1

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously using EES for the four unknowns x, y, z, and w yields x = 3.656

y = 0.2086

z = 0.8162

w = 0.1591

Thus the equilibrium composition is 3.656N 2 + 0.2086NO + 0.8162O 2 + 0.1591O

The equilibrium constant of the reaction N 2 ⇔ 2N at 3000 K is lnKP = -22.359, which is much smaller than the KP values of the reactions considered. Therefore, it is reasonable to assume that no N will be present in the equilibrium mixture.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-43 16-48E determined.

Two chemical reactions are occurring in air. The equilibrium composition at a specified temperature is to be

Assumptions 1 The equilibrium composition consists of O2, N2, O, and NO. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as ⎯ ⎯→

O 2 + 3.76 N 2

Heat

x N 2 + y NO + z O 2 + w O

Mass balances for nitrogen and oxygen yield

AIR

N balance:

7.52 = 2 x + y

(1)

O balance:

2 = y + 2z + w

(2)

Reaction chamber, 1 atm

O2, N2, O, NO 5400 R

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the Kp relations) to determine the equilibrium composition of the mixture. They are 1 2

N 2 + 12 O 2 ⇔ NO

(reaction 1)

O 2 ⇔ 2O

(reaction 2)

The equilibrium constant for these two reactions at T = 5400 R = 3000 K are determined from Table A-28 to be ln K P1 = −2.114

⎯ ⎯→

K P1 = 012075 .

ln K P 2 = −4.357

⎯ ⎯→

K P 2 = 0.01282

The KP relations for these two simultaneous reactions are ν

K P1 =

⎛ P ⎜ ⎜N ⎝ total

NO N NO

ν

ν

N NN 2 N OO 2 2

2

νO

K P2

N ⎛ P = νO ⎜⎜ O2 N O ⎝ N total 2

where

⎞ ⎟ ⎟ ⎠

(ν NO −ν N 2 −ν O 2 )

ν O −ν O 2

⎞ ⎟ ⎟ ⎠

N total = N N 2 + N NO + N O2 + N O = x + y + z + w

Substituting, 0.12075 =

y x 0 .5 z 0 .5

w2 0.01282 = z

⎛ ⎞ 1 ⎜⎜ ⎟⎟ x + y + z + w ⎝ ⎠

⎛ ⎞ 1 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠

1− 0.5− 0.5

(3)

2 −1

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 3.658

y = 0.2048

z = 0.7868

w = 0.2216

Thus the equilibrium composition is 3.658N 2 + 0.2048NO + 0.7868O 2 + 0.2216O

The equilibrium constant of the reaction N 2 ⇔ 2N at 5400 R is lnKP = -22.359, which is much smaller than the KP values of the reactions considered. Therefore, it is reasonable to assume that no N will be present in the equilibrium mixture.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-44 14-49E Problem 16-48E is reconsidered. Using EES (or other) software, the equilibrium solution is to be obtained by minimizing the Gibbs function by using the optimization capabilities built into EES. This solution technique is to be compared with that used in the previous problem. Analysis The problem is solved using EES, and the solution is given below. "This example illustrates how EES can be used to solve multi-reaction chemical equilibria problems by directly minimizing the Gibbs function. 0.21 O2+0.79 N2 = a O2+b O + c N2 + d NO Two of the four coefficients, a, b, c, and d, are found by minimiming the Gibbs function at a total pressure of 1 atm and a temperature of 5400 R. The other two are found from mass balances. The equilibrium solution can be found by applying the Law of Mass Action to two simultaneous equilibrium reactions or by minimizing the Gibbs function. In this problem, the Gibbs function is directly minimized using the optimization capabilities built into EES. To run this program, select MinMax from the Calculate menu. There are four compounds present in the products subject to two elemental balances, so there are two degrees of freedom. Minimize Gibbs with respect to two molar quantities such as coefficients b and d. The equilibrium mole numbers of each specie will be determined and displayed in the Solution window. Minimizing the Gibbs function to find the equilibrium composition requires good initial guesses." "Data from Data Input Window" {T=5400 "R" P=1 "atm" } AO2=0.21; BN2=0.79 "Composition of air" AO2*2=a*2+b+d "Oxygen balance" BN2*2=c*2+d "Nitrogen balance" "The total moles at equilibrium are" N_tot=a+b+c+d y_O2=a/N_tot; y_O=b/N_tot; y_N2=c/N_tot; y_NO=d/N_tot "The following equations provide the specific Gibbs function for three of the components." g_O2=Enthalpy(O2,T=T)-T*Entropy(O2,T=T,P=P*y_O2) g_N2=Enthalpy(N2,T=T)-T*Entropy(N2,T=T,P=P*y_N2) g_NO=Enthalpy(NO,T=T)-T*Entropy(NO,T=T,P=P*y_NO) "EES does not have a built-in property function for monatomic oxygen so we will use the JANAF procedure, found under Options/Function Info/External Procedures. The units for the JANAF procedure are kgmole, K, and kJ so we must convert h and s to English units." T_K=T*Convert(R,K) "Convert R to K" Call JANAF('O',T_K:Cp`,h`,S`) "Units from JANAF are SI" S_O=S`*Convert(kJ/kgmole-K, Btu/lbmole-R) h_O=h`*Convert(kJ/kgmole, Btu/lbmole) "The entropy from JANAF is for one atmosphere so it must be corrected for partial pressure." g_O=h_O-T*(S_O-R_u*ln(Y_O)) R_u=1.9858 "The universal gas constant in Btu/mole-R " "The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance." Gibbs=a*g_O2+b*g_O+c*g_N2+d*g_NO

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-45 d [lbmol] 0.002698 0.004616 0.007239 0.01063 0.01481 0.01972 0.02527 0.03132 0.03751 0.04361

b [lbmol] 0.00001424 0.00006354 0.0002268 0.000677 0.001748 0.004009 0.008321 0.01596 0.02807 0.04641

Gibbs [Btu/lbmol] -162121 -178354 -194782 -211395 -228188 -245157 -262306 -279641 -297179 -314941

yO2

yO

yNO

yN2

0.2086 0.2077 0.2062 0.2043 0.2015 0.1977 0.1924 0.1849 0.1748 0.1613

0.0000 0.0001 0.0002 0.0007 0.0017 0.0040 0.0083 0.0158 0.0277 0.0454

0.0027 0.0046 0.0072 0.0106 0.0148 0.0197 0.0252 0.0311 0.0370 0.0426

0.7886 0.7877 0.7863 0.7844 0.7819 0.7786 0.7741 0.7682 0.7606 0.7508

T [R] 3000 3267 3533 3800 4067 4333 4600 4867 5133 5400

Mole fraction of NO and O

0.050

0.040

0.030

NO

0.020

O 0.010

0.000 3000

3500

4000

4500

5000

5500

T [R] Discussion The equilibrium composition in the above table are based on the reaction in which the reactants are 0.21 kmol O2 and 0.79 kmol N2. If you multiply the equilibrium composition mole numbers above with 4.76, you will obtain equilibrium composition for the reaction in which the reactants are 1 kmol O2 and 3.76 kmol N2.This is the case in problem 16-43E.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-46 16-50 Water vapor is heated during a steady-flow process. The rate of heat supply for a specified exit temperature is to be determined for two cases. Assumptions 1 The equilibrium composition consists of H2O, OH, O2, and H2. 2 The constituents of the mixture are ideal gases. Analysis (a) Assuming some H2O dissociates into H2, O2, and O, the dissociation equation can be written as ⎯ ⎯→

H 2O

x H 2O + y H 2 + z O 2 + w OH

H2O

Mass balances for hydrogen and oxygen yield H balance:

2 = 2x + 2 y + w

(1)

O balance:

1 = x + 2z + w

(2)

Q

298 K

H2O, H2, O2, OH 2400 K

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the KP relations) to determine the equilibrium composition of the mixture. They are

H 2 O ⇔ H 2 + 12 O 2

(reaction 1)

H 2 O ⇔ 12 H 2 + OH

(reaction 2)

The equilibrium constant for these two reactions at 2400 K are determined from Table A-28 to be

ln K P1 = −5.619 ⎯ ⎯→ K P1 = 0.003628 ln K P 2 = −5.832 ⎯ ⎯→ K P 2 = 0.002932 The KP relations for these three simultaneous reactions are ν

K P1

ν

N HH 2 N OO 2 ⎛ P 2 2 ⎜ = ⎜N ν H 2O ⎝ total NH O 2

ν H2

K P2

OH N H N νOH ⎛ P 2 ⎜ = ⎜N ν H 2O ⎝ total NH O 2

⎞ ⎟ ⎟ ⎠

(ν H 2 +ν O 2 −ν H 2 O )

⎞ ⎟ ⎟ ⎠

(ν H 2 +ν OH −ν H 2 O )

where

N total = N H 2O + N H 2 + N O 2 + N OH = x + y + z + w Substituting, 0.003628 =

( y )( z )1/ 2 x

0.002932 =

( w)( y )1 / 2 x

⎛ ⎞ 1 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠

1/ 2

⎛ ⎞ 1 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠

(3) 1/ 2

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 0.960

y = 0.03204

z = 0.01205

w = 0.01588

Thus the balanced equation for the dissociation reaction is H2O ⎯ ⎯→ 0.960H 2 O + 0.03204H 2 + 0.01205O 2 + 0.01588OH

The heat transfer for this dissociation process is determined from the steady-flow energy balance E in − E out = ∆E system with W = 0, Qin =

∑ N (h P

o f

+h −ho

) − ∑ N (h P

R

o f

+h −ho

)

R

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-47 Assuming the O2 and O to be ideal gases, we have h = h(T). From the tables,

hfo

h 298 K

h 3000 K

kJ/kmol

kJ/kmol

kJ/kmol

H2O

-241,820

9904

103,508

H2

0

8468

75,383

O2

0

8682

83,174

OH

39,460

9188

77,015

Substance

Substituting, Qin = 0.960(−241,820 + 103,508 − 9904) + 0.03204(0 + 75,383 − 8468) + 0.01205(0 + 83,174 − 8682) + 0.01588(39,460 + 77,015 − 9188) − (−241,820) = 103,380 kJ/kmol H 2 O

The mass flow rate of H2O can be expressed in terms of the mole numbers as 0.6 kg/min m& N& = = = 0.03333 kmol/min M 18 kg/kmol

Thus,

Q& in = N& × Qin = (0.03333 kmol/min)(103,380 kJ/kmol) = 3446 kJ/min (b) If no dissociates takes place, then the process involves no chemical reactions and the heat transfer can be determined from the steady-flow energy balance for nonreacting systems to be Q& in = m& (h2 − h1 ) = N& (h2 − h1 ) = (0.03333 kmol/min)(103,508 − 9904) kJ/kmol = 3120 kJ/min

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

16-48 16-51 Problem 16-50 is reconsidered. The effect of the final temperature on the rate of heat supplied for the two cases is to be studied. Analysis The problem is solved using EES, and the solution is given below. "Given" T1=298 [K] "T2=2400 [K]" P=1 [atm] m_dot=0.6 [kg/min] T0=298 [K] "The equilibrium constant for these two reactions at 2400 K are determined from Table A-28" K_p1=exp(-5.619) K_p2=exp(-5.832) "Properties" MM_H2O=molarmass(H2O) "Analysis" "(a)" "Actual reaction: H2O = N_H2O H2O + N_H2 H2 + N_O2 O2 + N_OH OH" 2=2*N_H2O+2*N_H2+N_OH "H balance" 1=N_H2O+2*N_O2+N_OH "O balance" N_total=N_H2O+N_H2+N_O2+N_OH "Stoichiometric reaction 1: H2O = H2 + 1/2 O2" "Stoichiometric coefficients for reaction 1" nu_H2O_1=1 nu_H2_