Solution Manual Thermodynamics 8th Edition

  0 Btu 25 , 037 ft /s. ( 50 )ft/s( 10 lbm)2. (60m).1( 25 kg/m)( 10 m/s)4. 1 kW( 120 kg/s)(1 kJ/kg).

Yunus A. Cengel, Michael A. Boles

Wmax Emech emmech. .0 0245 kJ/kg1000 m /s. .0 050 kJ/kg1000 m /s. .1( 25 kg/m)(7 m/s)(1m )(0 kJ/kg) 0 kW. .1( 25 kg/m)(10 m/s)(1m )(0 kJ/kg) 0 kW. .0( 2144 kW)(3000 h/yr) (per m flow area).

returned unopened to McGraw-Hill Education: This Manual is being provided only to authorized

reproduced, displayed or distributed in any form or by any means, electronic or otherwise,

8th Edition

E 1 max, W t11 max,   643 kWh/yr. .0( 625 kW)(1500 h/yr) (per m flow area). E 2 max, W 2 max, t 2   938 kWh/yr.

PROPRIETARY AND CONFIDENTIAL

1 Btu( 144 , 000 lbf ft). lbf 13( 2000 lbm)( 32. 1 Btu.0 001909 lbf ft .0( 001909 lbf ft). 2 ( ) .0(2 005 lbf/ft /3() 12 ft) /5( 12 ft). 1 kJ(12,500 kg)(9 m/s)(200 m)22. 1 m/s( 10 km/h) .

.2 778 m/s 0- . 1 kJ/kg(12,500 kg) .2( 778 m/s) 02. 1 kJ/kg/ (12,500 kg)(9 m/s)(1 m)22. in sh,in out 2 1.

Change in internal, kinetic,.

by heat, work,and mass. Netenergy transfer. 30 kJ kJ5 kJ5 10 kJ. Change in internal, kinetic,. by heat, work,and mass. Netenergy transfer. 65 Btu 5 Btu 8 Btu. potential, etc. Rateof changein internal, kinetic,. 0 (steady)system.

by heat, work, mass and. Rateofnet energy transfer. in out      .

professors and instructors for use in preparing for the classes using the affiliated textbook. No

1 Btu8( ft /s)( 70 15 )psia.

W 2 max, Emech, 2  2 em mech, 2 V 2 Ake 2  

(Power consumed per lamp) (No.

in out  Ein Eout

of lamps)=(400 6 110 W)=264,000 264 kW.

(Power consumed per lamp) (No. of lamps)=(200 12 110 W)=264,000 264 kW. lighting, total lighting, classroom lighting, offices. lighting, offices. lighting, classroom.   \$55,757/yr.

Cost savings (Energy savings)(Unit cost of energy) (506,880 kWh/yr)(\$0/kWh).

Energy savings (Elighting, total)(Unoccupied hours) (528 kW)(960 h/yr) 506,880 kWh.

Annual cost savings.

Implementation cost.

specified rate of heat loss, the required rated power of resistance heaters is to be determined.

potential, etc.

Rateof change in internal, kinetic,.

by heat, work,and mass.

Rateof netenergy transfer.

in lights TV refrig iron.

potential, etc.

W 5( W)

Rateof change in internal, kinetic,.

by heat, work,and mass.

Rateof netenergy transfer.

outelect,in air out air.

.0( 075 lbm/ft)(3 3 ft )(22 ft/s) 14.

  .0 1435 Btu/s 25,037 ft /s.

( 22 ft/s)1(4 lbm/s)2 22.

potential, etc.

Rateof changein internal, kinetic,.

by heat, work, mass and.

Rateofnet energy transfer.

in out   .

1 kJ/kg(3750 kg)(9 m/s )(0 m/s)sin22. 1 kJ/kg(3750 kg)(9 m/s)(1 m/s)sin22. potential, etc. Rateof changein internal, kinetic,.

Mass = (50 persons)(75 kg/person) = 375 0 kg

by heat, work, mass and. Rateofnet energy transfer. in out   . (110/3 m/s) -(70/3 m/s)(1400 kg)2 22. (110/3 m/s) -(70/3 m/s)(700 kg)2 t. mech,out mech,inpump-motor pumpmotorW.

Mechanical energy extracted from the fluid | |. Mechanical energy output. elect,outgeneratorMechanical power input.

Electricity

Electrical power output.

Vm for

mech,in mech,out.

At 10 km/h, it will take

elect,outturbinegen- turbinegeneratorE.

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 

1 kg m/s3 kg

m

since 1 kW = 1 kJ/s. Then the maximum electric power generations per year become

a

Stone

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

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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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 

Analysis Kinetic energy is the only form of harvestable mechanical

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)

in out

the world. Hydrogen can be obtained from water by using another energy source, such as solar or nuclear energy, and then

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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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

 

h at  at  

283 kJ/kg1 .3 696150 kPa

400 kPa

.1 70203 kJ/kg

214 kJ/kg300 K

3

322

33

1 2

02

01

1 21 2

uPP

Discussion Note the power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the

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 )

mRT

mRT

m

Assumptions 1 The room is well sealed, and heat loss from the room is negligible. 2 All the appliances are kept on.

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

since no energy is leaving the room in any form, and thus Eout 0. Also,

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 )

2- 31E The engine of a car develops 450 hp at 3000 rpm. The torque transmitted through the shaft is to be determined.

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

T = const. 2

EinEout  dE /dt  dEroom/dtEin

s = const.

P = const.

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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

2- 23C This is neither a heat nor a work interaction since no energy is crossing the system boundary. This is simply the

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

1

o

2

o

2- 22C It is a heat interaction since it is due to the temperature difference between the sun and the room.

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

Analysis The total kinetic energy of the object is given by

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   

 

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

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

s s s s R

Substituting,

 

 178 kJ/kg

Properties The density of air is given to be  = 1 kg/m

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

2- 19C (a) From the perspective of the contents, heat must be removed in order to reduce and maintain the content's

Then the reversible power input is

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

( 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

Analysis A fan motor converts electrical energy to mechanical shaft energy, and the fan transmits the mechanical energy of

600 kPa 167 C

8 kW

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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

/)1(4 3

No work is produced since there is no motion of the forces acting at the interface between the tire and road.

For an ideal regenerator,

6 2

5 4T T

 0 Btu

The thermal efficiency of the cycle is

k kp

k kp

k kp

rT

r

rT

q

q

/)1(3

1

/)1(

/)1(

3

1

4 3

2 13

1

5 3

6 13

13 5

6 1in

outth



the hydrogen obtained can be used as a fuel to power cars or generators. Therefore, it is more proper to view hydrogen is an

s

q in 3

5

fan motor is to be determined.

q out

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11 -

0 0 0 0 0 0 1

9

12

15

18

21

24

27

30

 comp

in,net

[kW]

turb=1.

turb=0.

turb=0.

0 0 0 0 0 0 1

8

12

16

20

 comp

are kept on. The amounts of electricity and money the campus will save per year if the lights are turned off during

Refrig

[kW]

turb=0.

turb=1.

turb=0.

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Discussion Note that if all the lamps were burned out today and are replaced by high-efficiency lamps instead of the

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

2- 41E The heat loss from a house is to be made up by heat gain from people, lights, appliances, and resistance heaters. For a

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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

8 kJ/kg

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

rate. Considering a unit flow area (A = 1 m

32

3 3 2 3available, 11 

), the maximum wind

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

= [(\$2-\$1)/lamp](700 lamps)

32

3 3 2 3available, 22 

= (4 kW)(1)(2800 h/year)

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

energy balance on this system gives

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

Properties The density of air is given to be  = 0 lbm/ft

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

Assumptions Water jet flows steadily at the specified speed and flow rate.

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

acceleration will be

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