Thermodynamics 8th Edition Solutions

Thermodynamics an engineering approach 8th edition solution manual pdfMain reasons why You Should Consider Getting a Test Bank for Thermodynamics An Engineering Approach 8th by Cengel getbooksolutions.

1-

junto de is simply an educational electronic manual that is a supplement to a certain textbook with a range of questions that correspond to the right answers the textbook authors have developed.Test Bank for thermodynamics an engineering approach 8th by Cengel is read in engineering universities.Thermodynamics an engineering approach 8th edition pdf by Yunus A.

Solutions Manual for

Thermodynamics: An Engineering Approach

8th Edition

Yunus A. Cengel, Michael A. Boles

McGraw-Hill, 2015

Chapter 1

INTRODUCTION AND BASIC CONCEPTS

PROPRIETARY AND CONFIDENTIAL

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Cengel and Michael A. Boles teaches you basic thermo dynamics, properties of pure substance, entropy and energy.Thermodynamics: An Engineering Approach solution manual (Yunus, cengel).

1-

1-1 Chapter 1 INTRODUCTION AND BASIC CONCEPTS Thermodynamics 1-1C. solution manual thermodynamics cengel 7th thermodynamics solution manual cengel 8th edition thermodynamics cengel solution manual pdf heat transfer.

Select your edition Below. Test Bank for Thermodynamics An Engineering Approach 8th. Test Bank for Thermodynamics An Engineering Approach 8thTest Bank for thermodynamics an engineering approach 8th by Cengel is read in engineering universities.Thermodynamics an engineering approach 8th edition pdf by Yunus A.

Cengel and Michael A. Boles teaches you basic thermo dynamics, properties of pure substance, entropy and energy.Main reasons why You Should Consider Getting a Test Bank for Thermodynamics An Engineering Approach 8th by Cengel getbooksolutions.

junto de is simply an educational electronic manual that is a supplement to a certain textbook with a range of questions that correspond to the right answers the textbook authors have developed.Thermodynamics: An Engineering Approach solution manual (Yunus, cengel).

1-1 Chapter 1 INTRODUCTION AND BASIC CONCEPTS Thermodynamics 1-1C. solution manual thermodynamics cengel 7th thermodynamics solution manual cengel 8th edition thermodynamics cengel solution manual pdf heat transfer.

1991 N

Select your edition Below.

N 1

Test Bank for Thermodynamics An Engineering Approach 8thTest Bank for thermodynamics an engineering approach 8th by Cengel is read in engineering universities.Thermodynamics an engineering approach 8th edition pdf by Yunus A.

Cengel and Michael A.

Boles teaches you basic thermo dynamics, properties of pure substance, entropy and energy.Thermodynamics: An Engineering Approach solution manual (Yunus, cengel). 1-1 Chapter 1 INTRODUCTION AND BASIC CONCEPTS Thermodynamics 1-1C.

solution manual thermodynamics cengel 7th thermodynamics solution manual cengel 8th edition thermodynamics cengel solution manual pdf heat transfer.

Select your edition Below.Main reasons why You Should Consider Getting a Test Bank for Thermodynamics An Engineering Approach 8th by Cengel getbooksolutions.

junto de is simply an educational electronic manual that is a supplement to a certain textbook with a range of questions that correspond to the right answers the textbook authors have developed.

1 kJ/kgK

 



 
 
  
 

 
 


 

4 kJ/kg C

1 Btu/lbm F(1 kJ/kg C)

4 kJ

1 kcal(1 kJ/kg C)

1000 g

1 kg1 kJ

1000 J

(1 kJ/kg C)

1 kJ/kg C

1 kJ/kg K(1 kJ/kg C)

p

p

p

p

c

c

c

c

m tank = 3 kgV =0 m 3

H 2 O

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1-

PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. Limited 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 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

29 N.

1 kgm/s

N 1

(3 kg)(9 m/s 2 ) 2  

W  mg 

Then the net force that acts on the rock is

F net F up F down 200  29. 37  170 N 6.

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

 56 m/s 2 

 
 
N 1

1 kg m/s3 kg

170 2

m

F

a

Stone

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PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. Limited 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- 99

4- 13 2 Problem 4- 13 1 is reconsidered. The effect of the environment temperature on the final pressure and the heattransfer as the environment temperature varies from 0°C to 50°C is to be investigated. The final results are to be plottedagainst the environment temperature.Analysis The problem is solved using EES, and the solution is given below.

"Knowns"Vol_A=0 [m^3]P_A[1]=400 [kPa]x_A[1]=0.T_B[1]=250 [C]P_B[1]=200 [kPa]Vol_B=0 [m^3]T_final=25 [C] "T_final = T_surroundings. To do the parametric studyor to solve the problem when Q_out = 0, place this statement in {}."{Q_out=0 [kJ]} "To determine the surroundings temperature thatmakes Q_out = 0, remove the {} and resolve the problem."

"Solution""Conservation of Energy for the combined tanks:"E_in-E_out=DELTAEE_in=E_out=Q_outDELTAE=m_A(u_A[2]-u_A[1])+m_B(u_B[2]-u_B[1])m_A=Vol_A/v_A[1]m_B=Vol_B/v_B[1]Fluid$='Steam_IAPWS'u_A[1]=INTENERGY(Fluid$,P=P_A[1], x=x_A[1])v_A[1]=volume(Fluid$,P=P_A[1], x=x_A[1])T_A[1]=temperature(Fluid$,P=P_A[1], x=x_A[1])u_B[1]=INTENERGY(Fluid$,P=P_B[1],T=T_B[1])v_B[1]=volume(Fluid$,P=P_B[1],T=T_B[1])"At the final state the steam has uniform properties through out the entire system."u_B[2]=u_finalu_A[2]=u_finalm_final=m_A+m_BVol_final=Vol_A+Vol_Bv_final=Vol_final/m_finalu_final=INTENERGY(Fluid$,T=T_final, v=v_final)P_final=pressure(Fluid$,T=T_final, v=v_final)

Pfinal [kPa] Qout [kJ] Tfinal [C] 0 2300 0 0 2274 5. 1 2247 11. 1 2218 16. 2 2187 22. 3 2153 27. 5 2116 33. 6 2075 38. 9 2030 44. 12 1978 50

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PROPRIETARY MATERIAL. © 20 15 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. Ifyou are a student using this Manual, you are using it without permission.

6-6- 27E An OTEC power plant operates between the temperature limits of 86F and 41F. The cooling water experiences atemperature rise of 6F in the condenser. The amount of power that can be generated by this OTEC plans is to bedetermined.Assumptions1 Steady operating conditions exist. 2 Water is anincompressible substance with constant properties.Properties The density and specific heat of water are taken  = 64.lbm/ft 3 and c = 1 Btu/lbm.F, respectively.Analysis The mass flow rate of the cooling water is

=113,790 lbm/min= 1897 lbm/s7

1 ft( 64 0. lbm/ft)(13,300 gal/min)

33water water 

m   V  

The rate of heat rejection to the cooling water isQ out m water ( TC out T in)( 1897 lbm/s)(1 Btu/lbm.F)(6F)=11,380 Btu/sNoting that the thermal efficiency of this plant is 2%, the power generation is determined to be

292 Btu/s= 308 kW ( 11 , 380 Btu/s)

0.

in out

 
 
  W
W
W
W Q
W
Q
W 
 

since 1 kW = 0 Btu/s.

Refrigerators and Heat Pumps

6-28C The difference between the two devices is one of purpose. The purpose of a refrigerator is to remove heat from acold medium whereas the purpose of a heat pump is to supply heat to a warm medium.

6- 29C The difference between the two devices is one of purpose. The purpose of a refrigerator is to remove heat from arefrigerated space whereas the purpose of an air-conditioner is remove heat from a living space.

6-30C No. Because the refrigerator consumes work to accomplish this task.

6-31C No. Because the heat pump consumes work to accomplish this task.

6-32C The coefficient of performance of a refrigerator represents the amount of heat removed from the refrigerated spacefor each unit of work supplied. It can be greater than unity.

6-33C The coefficient of performance of a heat pump represents the amount of heat supplied to the heated space for eachunit of work supplied. It can be greater than unity.

6-34C No. The heat pump captures energy from a cold medium and carries it to a warm medium. It does not create it.

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PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission.

7- 1337- 176 A piston-cylinder device contains air that undergoes a reversible thermodynamic cycle composed of three processes.The work and heat transfer for each process are to be determined.Assumptions1 All processes are reversible. 2 Kinetic and potentialenergy changes are negligible. 3 Air is an ideal gas with variable specificheats.Properties The gas constant of air is R = 0 kPa 3 /kg (Table A-1).Analysis Using variable specific heats, the properties can be determinedusing the air table as follows

 

396 K 6.

283 kJ/kg1 .3 696150 kPa

400 kPa

.1 3860

.1 70203 kJ/kg

214 kJ/kg300 K

3

322

33

1 2

02

01

1 21 2

   
 
 
 
  
T

uPP

P
P
P P

s s

u uT T

r r

r r

The mass of the air and the volumes at the various states are

3

3

3

33

3

3

2

22

3

3

1

11

.0 3967 m400 kPa

(1 kg)(0 kPam /kgK)(396 K)

m 8.150 kPa

(1 kg)(0 kPa m/kgK)(300 K)

.1 394 kg(0 kPam/kgK)(300 K)

(400 kPa)(0 )

 
 
 
 
 
 
P

mRT

P

mRT

RT
P

m

V
V
V

Process 1-2: Isothermal expansion ( T 2 = T 1 )

.0 3924 kJ/kg400 kPa

150 kPaln (1 kg)(0 kJ/kg)ln1

2  21   P

P

S mR

Q in,1 2  T 1  S  21 ( 300 K)(0 kJ/K) 117 kJ

W out,1 2  Q in,1 2  117 kJ

Process 2-3: Isentropic (reversible-adiabatic) compression ( s 2 = s 1 )W in,2 3  ( um 3  u 2 )(1 kg)(283.71-214) kJ/kg 97 kJ

Q 2-3 = 0 kJProcess 3-1: Constant pressure compression process ( P 1 = P 3 )

W in,3 1  P (V 33 V 1 )( 400 kg)(0.3924-0) kJ/kg 37 kJ

Q out,3 1  W in,3 1  ( um 1  u 3 ) 37 0. kJ-(1 kg)(214.07-283) kJ/kg 135 kJ

1

T = const. 2

3

s = const.

P = const.

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PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission.

8- 38

8-51 Air is compressed steadily by an 8-kW compressor from a specified state to another specified state. Theincrease in the exergy of air and the rate of exergy destruction are to be determined.Assumptions1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible.Properties The gas constant of air is R = 0 kJ/kg (Table A-1). From the air table (Table A-17)

T hsT hs

1 1

2 2

290 290 16
1 66802
440 441 61
2 0887
   
 
   
 

K kJ /kg kJ /kgKK kJ /kgkJ /kgK

1

o

2

o

.
.
.
.

Analysis The increase in exergy is the difference between the exit andinlet flow exergies,

( ) ( )
[( ) ( )]

Increasein exergy

2 1 0 2 1

2 1 0 2 1

2 1 0 0

h h T s s

h h ke pe T s s   

     
 

 

 

where

.0 09356 kJ/kg K

100 kPa

600 kPa.2( 0887 .1 66802 )kJ/kg -K (0 kJ/kg K)ln

( ) ln 1

o 21

o2 1 2

 
   
   
P
P

s s s s R

Substituting,

 

 178 kJ/kg

   
 

(441 290)kJ/kg-(290K)( .0 09356 kJ/kgK)

Increasein exergy  2  1

Then the reversible power input is

W rev,in m ( 2  1 ) /1( 60 kg/s)(178. 6 kJ/kg)6 kW

( b ) The rate of exergy destruction (or irreversibility) is determined from its definition,

X destroyed W in W rev,in 8  .6 25  1 kW

Discussion Note that 1 kW of power input is wasted during this compression process.

100 kPa 17 C

AIR

600 kPa 167 C

8 kW

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PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. Limited 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- 919-1 12 An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator is to be developed.Assumptions1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at roomtemperature. 3 Kinetic and potential energy changes are negligible.Analysis The expressions for the isentropic compression and expansion processes are

T 2  1 rT pk  /)1( k

k k

rp

T T

/)1(4 3

1

For an ideal regenerator,

6 2

5 4T T

T T

The thermal efficiency of the cycle is

k kp

k kp

k kp

rT

T

r

rT

T
T T
T T
T
T
T T
T T
T
T
T T
T T

q

q

/)1(3

1

/)1(

/)1(

3

1

4 3

2 13

1

5 3

6 13

13 5

6 1in

outth

1
1
1
1
(1 / )
( / 1)
1
(1 / )
( / 1)
1 1 1



 
 
 
 
    

s

T
1
2
4

q in 3

5

6

q out

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11 -

0 0 0 0 0 0 1

9

12

15

18

21

24

27

30

 comp

W

in,net

[kW]

turb=1.

turb=0.

turb=0.

0 0 0 0 0 0 1

8

12

16

20

 comp

Q

Refrig

[kW]

turb=0.

turb=1.

turb=0.

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16 -

16 -49 Problem 16- 48 is reconsidered. The effect of the final temperature on the rate of heat supplied for the twocases is to be studied.Analysis The problem is solved using EES, and the solution is given below.

"Given"T1=298 [K]T2=3000 [K]"P=1 [atm]"m_dot=0 [kg/min]T0=298 [K]

"The equilibrium constant for these two reactions at 3000 K are determined from Table A-28"K_p1=exp(-3)K_p2=exp(-2)

"Properties"MM_H2O=molarmass(H2O)

"Analysis""(a)""Actual reaction: H2O = N_H2O H2O + N_H2 H2 + N_O2 O2 + N_OH OH"2=2N_H2O+2N_H2+N_OH "H balance"1=N_H2O+2N_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=nu_H2_1=nu_O2_1=1/"Stoichiometric reaction 2: H2O = 1/2 H2 + OH""Stoichiometric coefficients for reaction 2"nu_H2O_2=nu_H2_2=1/nu_OH_2="K_p relations are"K_p1=(N_H2^nu_H2_1*N_O2^nu_O2_1)/N_H2O^nu_H2O_1(P/N_total)^(nu_H2_1+nu_O2_1-nu_H2O_1)K_p2=(N_H2^nu_H2_2N_OH^nu_OH_2)/N_H2O^nu_H2O_2(P/N_total)^(nu_H2_2+nu_OH_2-nu_H2O_2)

"Enthalpy of formation data from Table A-26"h_f_OH="Enthalpies of products"h_H2O_R=enthalpy(H2O, T=T1)h_H2O_P=enthalpy(H2O, T=T2)h_H2=enthalpy(H2, T=T2)h_O2=enthalpy(O2, T=T2)h_OH=98763 "at T2 from the ideal gas tables in the text""Standard state enthalpies"h_o_OH=9188 "at T0 from the ideal gas tables in the text"

"Heat transfer"H_P=N_H2Oh_H2O_P+N_H2h_H2+N_O2h_O2+N_OH(h_f_OH+h_OH-h_o_OH)H_R=N_H2O_Rh_H2O_RN_H2O_R=Q_in_a=H_P-H_RQ_dot_in_a=(m_dot/MM_H2O)*Q_in_a"(b)"Q_in_b=N_H2O_R(h_H2O_P-h_H2O_R)Q_dot_in_b=(m_dot/MM_H2O)*Q_in_b

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18- 29

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18 -64E A wind turbine is to generate power with two specified wind speeds. The amount of electricity that can be producedby the turbine and the blade tip speed are to be determined.Assumptions Wind flows steadily at the specified speeds.Properties The density of air is given to be  = 0 lbm/ft 3.Analysis The blade span area is

A  D 2 4/ ( 185 ft) 2 4/  26 , 880 ft 2

The wind power potential at the wind speed of 16 ft/s is

174 0. kW 23,730 lbmft /s

1 kW(0 lbm/ft )(26,880 ft )(16 ft/s)2

1
2
1

32

3 3 2 3available, 11 

W    AV 

The wind power potential at the wind speed of 24 ft/s is

587 2. kW 23,730 lbm ft /s

1 kW(0 lbm/ft )(26,880 ft )(24 ft/s)2

1
2
1

32

3 3 2 3available, 22 

W    AV 

The overall wind turbine efficiency at a wind speed of 16 ft/s is

wt,overall,1wt,1gen .0( 30 )( .0 93 ) .0 2790

The overall wind turbine efficiency at a wind speed of 24 ft/s is

wt,overall,2wt,2gen .0( 35 )( .0 93 ) .0 3255

The electric power generated at a wind speed of 16 ft/s is

W electric,1wt,overall,1 W available, 1  .0( 2790 )(174 kW)48 kW

The electric power generated at a wind speed of 24 ft/s is

W electric,2wt,overall,2 W available, 2  .0( 3255 )(587 kW) 191 1. kW

The amount of electricity produced at a wind speed of 16 ft/s is

W electric,1 W electric,1Operating hours 1 ( 48. 54 kW)(3000 h)145,630 kWh

The amount of electricity produced at a wind speed of 24 ft/s is

W electric,2 W electric,2Operating hours 2 ( 191 1. kW)(4000 h) 764 ,560 kWh

The total amount of electricity produced isW electric,total W electric,1 W electric,2 145 , 560  764 , 560  910,000 kWh

Noting that the tip of blade travels a distance of  D per revolution, the tip velocity of the turbine blade for a rotational speedof n  becomes

 99mph

 

1/s

1 mi/h60 s

1 min

V tip  nD  ( 185 (ft) 15 /min)

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