ThermoSolutions CHAPTER08

8-1

Chapter 8 EXERGY – A MEASURE OF WORK POTENTIAL

Exergy, Irreversibility, Reversible Work, and Second-Law Efficiency

8-1C Reversible work differs from the useful work by irreversibilities. For reversible processes both are identical. Wu = Wrev -I.

8-2C Reversible work and irreversibility are identical for processes that involve no actual useful work.

8-3C The dead state.

8-4C Yes; exergy is a function of the state of the surroundings as well as the state of the system.

8-5C Useful work differs from the actual work by the surroundings work. They are identical for systemsthat involve no surroundings work such as steady-flow systems.

8-6C Yes.

8-7C No, not necessarily. The well with the higher temperature will have a higher exergy.

8-8C The system that is at the temperature of the surroundings has zero exergy. But the system that is at alower temperature than the surroundings has some exergy since we can run a heat engine between thesetwo temperature levels.

8-9C They would be identical.

8-10C The second-law efficiency is a measure of the performance of a device relative to its performanceunder reversible conditions. It differs from the first law efficiency in that it is not a conversion efficiency.

8-11C No. The power plant that has a lower thermal efficiency may have a higher second-law efficiency.

8-12C No. The refrigerator that has a lower COP may have a higher second-law efficiency.

8-13C A processes with Wrev = 0 is reversible if it involves no actual useful work. Otherwise it isirreversible.

8-14C Yes.

8-2

8-15 Windmills are to be installed at a location with steady winds to generate power. The minimumnumber of windmills that need to be installed is to be determined.Assumptions Air is at standard conditions of 1 atm and 25 CProperties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). Analysis The exergy or work potential of the blowing air is the kinetic energy it possesses,

Exergy = kJ/kg032.0s/m1000

kJ/kg12

m/s)8(2 22

22ke V

At standard atmospheric conditions (25 C, 101 kPa), the density and the mass flow rate of air are

PRT

101 kPa(0.287 kPa m kg K)(298 K)

m kg33

/. /118

and

Thus,kW23.74=kJ/kg)2kg/s)(0.03742(kePowerAvailable

kg/s742=m/s)8(m)10)(4/)(kg/m18.1(4

231

2

1

m

DAVm V

The minimum number of windmills that needs to be installed is

windmills263.25kW23.74

kW600total

WW

N

8-16 Water is to be pumped to a high elevation lake at times of low electric demand for use in a hydroelectric turbine at times of high demand. For a specified energy storage capacity, the minimumamount of water that needs to be stored in the lake is to be determined.Assumptions The evaporation of water from the lake is negligible.

75 mAnalysis The exergy or work potential of the water is the potentialenergy it possesses,

Thus,

kg102.45 10

s/kgkW1s/m1000

h1s3600

m)75)(m/s8.9(kWh105

mgh=PE=Exergy

22

2

6

ghPEm

8-3

8-17 A heat reservoir at a specified temperature can supply heat at a specified rate. The exergy of this heatsupplied is to be determined.Analysis The exergy of the supplied heat, in the rate form, is the amount of power that would be produced by a reversible heat engine,

kW33.4=kJ/s)3600/000,150)(8013.0(

Exergy

8013.0K1500K29811

inrevth,outrev,outmax,

0revth,maxth,

QWW

TT

H

WrevHE

1500 K

298 K

8-18 [Also solved by EES on enclosed CD] A heat engine receives heat from a source at a specified temperature at a specified rate, and rejects the waste heat to a sink. For a given power output, the reversiblepower, the rate of irreversibility, and the 2nd law efficiency are to be determined.Analysis (a) The reversible power is the power produced by a reversible heat engine operating between thespecified temperature limits,

kW550.7=kJ/s)700)(787.0(

787.0K1500K32011

inrevth,outrev,

revth,maxth,

QW

TT

H

L

700 kJ/s

320 kWHE

1500 K

320 K

(b) The irreversibility rate is the difference between the reversible powerand the actual power output:

kW230.73207.550outu,outrev, WWI

(c) The second law efficiency is determined from its definition,

58.1%kW7.550

kW320

outrev,

outu,II W

W

8-4

8-19 EES Problem 8-18 is reconsidered. The effect of reducing the temperature at which the waste heat is rejected on the reversible power, the rate of irreversibility, and the second law efficiency is to be studiedand the results are to be plotted.Analysis The problem is solved using EES, and the solution is given below.

"Input Data"T_H= 1500 [K] Q_dot_H= 700 [kJ/s] {T_L=320 [K]}W_dot_out = 320 [kW] T_Lsurr =25 [C]

"The reversible work is the maximum work done by the Carnot Enginebetween T_H and T_L:"Eta_Carnot=1 - T_L/T_H W_dot_rev=Q_dot_H*Eta_Carnot"The irreversibility is given as:"I_dot = W_dot_rev-W_dot_out

"The thermal efficiency is, in percent:"Eta_th = Eta_Carnot*Convert(, %)

"The second law efficiency is, in percent:"Eta_II = W_dot_out/W_dot_rev*Convert(, %)

II [%] I [kJ/s] Wrev [kJ/s] TL [K] 68.57 146.7 466.7 50067.07 157.1 477.1 477.665.63 167.6 487.6 455.164.25 178.1 498.1 432.762.92 188.6 508.6 410.261.65 199 519 387.860.43 209.5 529.5 365.359.26 220 540 342.958.13 230.5 550.5 320.457.05 240.9 560.9 298

275 320 365 410 455 500460

482

504

526

548

570

TL [K]

Wrev

[kJ/s]

8-5

275 320 365 410 455 50056

58

60

62

64

66

68

70

TL [K]

II[%]

275 320 365 410 455 500140

162

184

206

228

250

TL [K]

I[kJ/s]

8-6

8-20E The thermal efficiency and the second-law efficiency of a heat engine are given. The source temperature is to be determined.Analysis From the definition of the second law efficiency,

th = 36% II = 60%

HE

TH

530 R

Thus,

R1325=R)/0.40530()1/(1

60.060.036.0

revth,revth,

II

threvth,

revth,

thII

LHH

L TTTT

8-21 A body contains a specified amount of thermal energy at a specified temperature. The amount that can be converted to work is to be determined.Analysis The amount of heat that can be converted to work is simply the amount that a reversible heatengine can convert to work,

kJ62.75=kJ)100)(6275.0(

6275.0K800K29811

inrevth,outrev,outmax,

0revth,

QWW

TT

H100 kJ

HE

800 K

298 K

8-22 The thermal efficiency of a heat engine operating between specified temperature limits is given. The second-law efficiency of a engine is to be determined.Analysis The thermal efficiency of a reversible heat engine operating between the same temperaturereservoirs is

Thus,

49.9%801.040.0

801.0K2731200

K29311

revth,

thII

0revth,

HTT

th = 0.40HE

1200 C

20 C

8-7

8-23 A house is maintained at a specified temperature by electric resistance heaters. The reversible work for this heating process and irreversibility are to be determined.Analysis The reversible work is the minimum work required to accomplish this process, and theirreversibility is the difference between the reversible work and the actual electrical work consumed. Theactual power input is

kW22.22=kJ/h000,80outin HQQW

W·

80,000 kJ/h

15 C

House22 C

The COP of a reversible heat pump operating between thespecified temperature limits is

14.42295/2881

1/1

1COP revHP,HL TT

Thus,

and

kW21.69

kW0.53

53.022.22

14.42kW22.22

COP

inrev,inu,

revHP,inrev,

WWI

QW H

8-24E A freezer is maintained at a specified temperature by removing heat from it at a specified rate. The power consumption of the freezer is given. The reversible power, irreversibility, and the second-law efficiency are to be determined.Analysis (a) The reversible work is the minimum work required to accomplish this task, which is the work that a reversible refrigerator operating between the specified temperature limits would consume,

73.81480/535

11/

1COP revR,LH TT

75 Btu/min

0.70 hp R

75 F

Freezer20 F

hp0.20Btu/min42.41hp1

73.8Btu/min75

revR,inrev, COP

QW L

(b) The irreversibility is the difference between the reversiblework and the actual electrical work consumed,

hp0.5020.070.0inrev,inu, WWI

(c) The second law efficiency is determined from its definition,

28.9%hp7.0hp20.0rev

IIuW

W

8-8

8-25 It is to be shown that the power produced by a wind turbine is proportional to the cube of the windvelocity and the square of the blade span diameter.Analysis The power produced by a wind turbine is proportional to the kinetic energy of the wind, which isequal to the product of the kinetic energy of air per unit mass and the mass flow rate of air through theblade span area. Therefore,

2323

wind

22

wind

2

wind

)Constant(8

42)(

2=

air)ofrateflowssenergy)(May)(Kinetic(Efficienc=powerWind

DVDV

VDVAVV

which completes the proof that wind power is proportional to the cube of the wind velocity and to thesquare of the blade span diameter.

8-26 A geothermal power produces 14 MW power while the exergy destruction in the plant is 18.5 MW.The exergy of the geothermal water entering to the plant, the second-law efficiency of the plant, and theexergy of the heat rejected from the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Water properties are used for geothermal water.Analysis (a) The properties of geothermal water at the inlet of the plant and at the dead state are (Tables A-4 through A-6)

kJ/kg.K36723.0kJ/kg83.104

0C25

kJ/kg.K9426.1kJ/kg47.675

0C160

0

0

0

0

1

1

1

1

sh

xT

sh

xT

The exergy of geothermal water entering the plant is

MW44.53kW525,44kJ/kg.K)36723.09426.1)(K27325(0kJ/kg)83.104(675.47kg/s)440(

( 01001in ssThhmX

(b) The second-law efficiency of the plant is the ratio of power produced to the exergy input to the plant

0.314kW525,44kW000,14

in

out

XW

II

(c) The exergy of the heat rejected from the plant may be determined from an exergy balance on the plant

MW12.03kW025,12500,18000,14525,44destoutinoutheat, XWXX

8-9

Second-Law Analysis of Closed Systems

8-27C Yes.

8-28C Yes, it can. For example, the 1st law efficiency of a reversible heat engine operating between thetemperature limits of 300 K and 1000 K is 70%. However, the second law efficiency of this engine, likeall reversible devices, is 100%.

8-29 A cylinder initially contains air at atmospheric conditions. Air is compressed to a specified state and the useful work input is measured. The exergy of the air at the initial and final states, and the minimumwork input to accomplish this compression process, and the second-law efficiency are to be determinedAssumptions 1 Air is an ideal gas with constant specific heats. 2 The kinetic and potential energies are negligible.Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). The specific heats of air at the average temperature of (298+423)/2=360 K are cp = 1.009 kJ/kg·K and cv = 0.722 kJ/kg·K (Table A-2). Analysis (a) We realize that X1 = 1 = 0 since air initially is at the dead state. The mass of air is

kg00234.0K)K)(298kg/mkPa287.0(

)mkPa)(0.002100(3

3

1

11

RTP

mV

AIRV1 = 2 L P1 = 100 kPaT1 = 25 C

Also, L0.473=L)2(K)kPa)(298600(K)kPa)(423100(

112

212

1

11

2

22 VVVV

TPTP

TP

TP

and

KkJ/kg1608.0kPa100kPa600lnK)kJ/kg287.0(

K298K423lnK)kJ/kg009.1(

lnln0

2

0

2avg,02 P

PR

TT

css p

Thus, the exergy of air at the final state is

kJ0.171kPa]kJ/m[m0.002)-473kPa)(0.000100(

K)kJ/kgK)(-0.1608(298-298)K-K)(423kJ/kg(0.722kg)00234.0(

)()()(

33

02002002avg,22 VVv PssTTTcmX

(b) The minimum work input is the reversible work input, which can be determined from the exergybalance by setting the exergy destruction equal to zero,

kJ0.1710171.0=12inrev,

exergyinChange

system

ndestructioExergy

e)(reversibl0destroyed

massand work,heat,bynsferexergy traNetoutin

XXW

XXXX

(c) The second-law efficiency of this process is

14.3%kJ2.1

kJ171.0

inu,

inrev,II W

W

8-10

8-30 A cylinder is initially filled with R-134a at a specified state. The refrigerant is cooled and condensed at constant pressure. The exergy of the refrigerant at the initial and final states, and the exergy destroyedduring this process are to be determined.Assumptions The kinetic and potential energies are negligible. Properties From the refrigerant tables (Tables A-11 through A-13),

KkJ/kg0256.1kJ/kg01.274

kg/m034875.0

C60MPa7.0

1

1

31

1

1

su

TP

v

R-134a0.7 MPa

P = const. KkJ/kg31958.0=kJ/kg84.44=

kg/m0008261.0=

C24MPa7.0

C24@2

C24@2

3C24@2

2

2

f

f

f

ssuu

TP

vv

Q

KkJ/kg1033.1kJ/kg84.251

kg/m23718.0

C24MPa1.0

0

0

30

0

0

su

TP

v

Analysis (a) From the closed system exergy relation,

kJ125.1

}mkPa1

kJ1/kgm0.23718)875kPa)(0.034(100+

KkJ/kg1.1033)K)(1.0256(297kJ/kg251.84){(274.01kg)5(

)()()(

33

0100100111 vvPssTuumX

and,

kJ208.6

}mkPa1

kJ1/kgm0.23718)8261kPa)(0.000(100+

KkJ/kg1.1033)K)(0.31958(297-kJ/kg251.84){(84.44kg)(5

)()()(

33

0200200222 vvPssTuumX

(b) The reversible work input, which represents the minimum work input Wrev,in in this case can be determined from the exergy balance by setting the exergy destruction equal to zero,

kJ5.831.1256.20812inrev,

exergyinChange

system

ndestructioExergy



XXW

XXXX

Noting that the process involves only boundary work, the useful work input during this process is simplythe boundary work in excess of the work done by the surrounding air,

kJ1.102mkPa1

kJ1kg)/m0008261.0034875.0kPa)(100-kg)(7005(

))(()()()(

33

210

21021210ininsurr,ininu,

vv

vvVVVV

PPmmPPPWWWW

Knowing both the actual useful and reversible work inputs, the exergy destruction or irreversibility that is the difference between the two is determined from its definition to be

kJ18.65.831.102inrev,inu,destroyed WWIX

8-11

8-31 The radiator of a steam heating system is initially filled with superheated steam. The valves areclosed, and steam is allowed to cool until the pressure drops to a specified value by transferring heat to theroom. The amount of heat transfer to the room and the maximum amount of heat that can be supplied to theroom are to be determined.Assumptions Kinetic and potential energies are negligible.Properties From the steam tables (Tables A-4 through A-6), Qout

STEAM20 L

P1 = 200 kPa T1 = 200 C

KkJ/kg5081.7kJ/kg6.2654

kg/m0805.1

C200kPa200

1

1

31

1

1

su

TP

v

KkJ/kg1479.35355.63171.00756.1kJ/kg6.10156.21463171.097.334

3171.0001029.04053.3001029.00805.1

)(C80

22

22

22

12

2

fgf

fgf

fg

f

sxssuxuu

xT v

vv

vv

Analysis (a) The mass of the steam is

kg01851.0kg/m0805.1

m020.03

3

1vVm

The amount of heat transfer to the room is determined from an energy balance on the radiator expressed as

)(0)=PE=KE(since)(

21out

12out

energiesetc.potential,kinetic,internal,inChange

system

massand work,heat,bynsferenergy traNet

outin

uumQWuumUQ

EEE

or, kJ30.3=kJ/kg1015.6)kg)(2654.601851.0(outQ

(b) The reversible work output, which represents the maximum work output Wrev,out in this case can be determined from the exergy balance by setting the exergy destruction equal to zero,

2121outrev,12outrev,

exergyinChange

system

ndestructioExergy



XXWXXW

XXXX

Substituting the closed system exergy relation, the reversible work during this process is determined to be

kJ305.8Kkg3.1479)kJ/-K)(7.5081(273-kg1015.6)kJ/-(2654.6kg)01851.0()()(

)()()(

21021

20

1021021outrev,

ssTuumPssTuumW vv

When this work is supplied to a reversible heat pump, it will supply the room heat in the amount of

kJ116.3273/294-1

kJ305.8/1

COP revrevrevHP,

HLH TT

WWQ

Discussion Note that the amount of heat supplied to the room can be increased by about 3 times byeliminating the irreversibility associated with the irreversible heat transfer process.

8-12

8-32 EES Problem 8-31 is reconsidered. The effect of the final steam temperature in the radiator on the amount of actual heat transfer and the maximum amount of heat that can be transferred is to beinvestigated.Analysis The problem is solved using EES, and the solution is given below.

T_1=200 [C] P_1=200 [kPa] V=20 [L] T_2=80 [C] T_o=0 [C] P_o=100 [kPa]

"Conservation of energy for closed system is:"E_in - E_out = DELTAE DELTAE = m*(u_2 - u_1) E_in=0E_out= Q_outu_1 =intenergy(steam_iapws,P=P_1,T=T_1)v_1 =volume(steam_iapws,P=P_1,T=T_1)s_1 =entropy(steam_iapws,P=P_1,T=T_1)v_2 = v_1 u_2 = intenergy(steam_iapws, v=v_2,T=T_2)s_2 = entropy(steam_iapws, v=v_2,T=T_2)m=V*convert(L,m^3)/v_1W_rev=-m*(u_2 - u_1 -(T_o+273.15)*(s_2-s_1)+P_o*(v_1-v_2))

"When this work is supplied to a reversible heat pump, the heat pump will supply the room heat in the amount of :"

Q_H = COP_HP*W_rev COP_HP = T_H/(T_H-T_L) T_H = 294 [K] T_L = 273 [K]

QH

[kJ]Qout

[kJ]T2

[C]Wrev

[kJ]155.4 46.66 21 11.1153.9 45.42 30 11151.2 43.72 40 10.8146.9 41.55 50 10.49140.3 38.74 60 10.02130.4 35.09 70 9.318116.1 30.34 80 8.293

20 30 40 50 60 70 8020

40

60

80

100

120

140

160

T2 [C]HeatTransfertoRoom[kJ]

Maximum

Actual

8-13

8-33E An insulated rigid tank contains saturated liquid-vapor mixture of water at a specified pressure. An electric heater inside is turned on and kept on until all the liquid is vaporized. The exergy destruction andthe second-law efficiency are to be determined.Assumptions Kinetic and potential energies are negligible.Properties From the steam tables (Tables A-4 through A-6)

Rlbm/Btu70751.030632.125.038093.0lbm/Btu47.44319.86225.092.227

lbm/ft9880.2)01708.0901.11(25.001708.0

0.25psia35

11

11

311

1

1

fgf

fgf

fgf

sxssuxuu

x

xP

vvv

RBtu/lbm1.5692Btu/lbm1110.9=

vaporsat. /lbmft2.9880=@2

/lbmft2.9880=@212

3

3

g

g

g

g

ssuu

v

vvv

We

H2O35 psia

Analysis (a) The irreversibility can be determined from its definitionXdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on the tank, which is an insulated closed system,

)( 12systemgen

entropyinChange

system

generationEntropy

gen

massandheatbyansferentropy trNet

outin

ssmSS

SSSS

Substituting,

Btu2766=Ru/lbm0.70751)Bt-R)(1.5692lbm)(5356(

)( 120gen0destroyed ssmTSTX

(b) Noting that V = constant during this process, the W and Wu are identical and are determined from the energy balance on the closed system energy equation,

)( 12ine,


system


outin

uumUW

EEE

or,Btu4005=/lbm443.47)Btu-9lbm)(1110.6(ine,W

Then the reversible work during this process and the second-law efficiency becomeBtu123927664005destroyedinu,inrev, XWW

Thus,

30.9%Btu4005Btu1239rev

IIuW

W

8-14

8-34 A rigid tank is divided into two equal parts by a partition. One part is filled with compressed liquidwhile the other side is evacuated. The partition is removed and water expands into the entire tank. Theexergy destroyed during this process is to be determined.Assumptions Kinetic and potential energies are negligible.Analysis The properties of the water are (Tables A-4 through A-6)

Vacuum

1.5 kg 300 kPa

60 CWATERKkJ/kg8313.0=

kJ/kg251.16=kg/m001017.0=

C60kPa300

C60@1

C60@1

3C60@1

1

1

f

f

f

ssuu

TP

vv

Noting that ,kg/m002034.0001017.022 312 vv

KkJ/kg7556.02522.70001017.07549.0kJ/kg15.2261.22220001017.093.225

0001017.0001014.002.10

001014.0002034.0

002034.0vkPa15

22

22

22

2

2

fgf

fgf

fg

f

sxssuxuu

xP v

vv

Taking the direction of heat transfer to be to the tank, the energy balance on this closed system becomes

)( 12in


system


outin

uumUQ

EEE

or,kJ51.37kJ-37.51=kg251.16)kJ/-kg)(226.155.1( outin QQ

The irreversibility can be determined from its definition Xdestroyed = T0Sgen where the entropy generation isdetermined from an entropy balance on an extended system that includes the tank and its immediatesurroundings so that the boundary temperature of the extended system is the temperature of thesurroundings at all times,

surr

out12gen

12systemgenoutb,

out

entropyinChange

system

generationEntropy

gen


outin

)(

)(

TQ

ssmS

ssmSSTQ

SSSS

Substituting,

kJ3.67=K298kJ37.51+Kkg0.8313)kJ/-kg)(0.7556(1.5K)298(

)(surr

out120gen0destroyed T

QssmTSTX

8-15

8-35 EES Problem 8-34 is reconsidered. The effect of final pressure in the tank on the exergy destroyedduring the process is to be investigated.Analysis The problem is solved using EES, and the solution is given below.

T_1=60 [C] P_1=300 [kPa] m=1.5 [kg] P_2=15 [kPa] T_o=25 [C] P_o=100 [kPa] T_surr = T_o

"Conservation of energy for closed system is:"E_in - E_out = DELTAE DELTAE = m*(u_2 - u_1) E_in=0E_out= Q_outu_1 =intenergy(steam_iapws,P=P_1,T=T_1)v_1 =volume(steam_iapws,P=P_1,T=T_1)s_1 =entropy(steam_iapws,P=P_1,T=T_1)v_2 = 2*v_1 u_2 = intenergy(steam_iapws, v=v_2,P=P_2)kJ

15 17.2 19.4 21.6 23.8 26-4

-3

-2

-1

0

1

2

3

4

P2 [kPa]

Xdestroyed

[]

s_2 = entropy(steam_iapws, v=v_2,P=P_2)S_in -S_out+S_gen=DELTAS_sys S_in=0 [kJ/K]S_out=Q_out/(T_surr+273)DELTAS_sys=m*(s_2 - s_1)X_destroyed = (T_o+273)*S_gen

P2

[kPa]Xdestroyed

[kJ]Qout

[kJ]15 3.666 37.44

16.11 2.813 28.0717.22 1.974 19.2518.33 1.148 10.8919.44 0.336 2.9520.56 -0.4629 -4.61221.67 -1.249 -11.8422.78 -2.022 -18.7523.89 -2.782 -25.39

25 -3.531 -31.77

15 17.2 19.4 21.6 23.8 26-40

-30

-20

-10

0

10

20

30

40

P2 [kPa]

Qout

8-16

8-36 An insulated cylinder is initially filled with saturated liquid water at a specified pressure. The water isheated electrically at constant pressure. The minimum work by which this process can be accomplished and the exergy destroyed are to be determined.Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulatedand thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 Thecompression or expansion process is quasi-equilibrium.Analysis (a) From the steam tables (Tables A-4 through A-6),

We

SaturatedLiquidH2O

P = 150 kPa

KkJ/kg1.4337=kJ/kg467.13=

/kgm0.001053=kg/kJ466.97=

liquidsat.kPa150

kPa150@1

kPa150@1

3kPa150@1

kPa150@1

1

f

f

f

f

sshh

uuP vv

The mass of the steam is

kg899.1kg/m001053.0

m002.03

3

1vVm

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

)( 12ine,outb,ine,


system


outin hhmWUWWEEE

since U + Wb = H during a constant pressure quasi-equilibrium process. Solving for h2,

kJ/kg1625.1kg1.899kJ220013.467,

12 mW

hh ine

Thus,

/kgm6037.0)001053.01594.1(5202.0001053.0kJ/kg6.15343.20525202.097.466

KkJ/kg4454.47894.55202.04337.1

5202.00.2226

13.4671.1625

kJ/kg1.1625kPa150

322

22

22

22

2

2

fgf

fgf

fgf

fg

f

xuxuusxss

hhh

x

hP

vvv

The reversible work input, which represents the minimum work input Wrev,in in this case can be determinedfrom the exergy balance by setting the exergy destruction equal to zero,

12inrev,

exergyinChange

system

ndestructioExergy


massand work,heat,bynsferexergy traNetoutin XXWXXXX

Substituting the closed system exergy relation, the reversible work input during this process becomes

kJ437.7]}mkPakJ/11[kg/m0.6037)053kPa)(0.001(100+

KkJ/kg4.4454)K)(1.4337(298kJ/kg1534.6)7kg){(466.9899.1()()()(

33

21021021inrev, vvPssTuumW

(b) The exergy destruction (or irreversibility) associated with this process can be determined from its definition Xdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on thecylinder, which is an insulated closed system,

)( 12systemgen

entropyinChange

system

generationEntropy

gen


outin

ssmSS

SSSS

Substituting,kJ1705=Kkg1.4337)kJ/kg)(4.4454K)(1.899298()( 120gen0destroyed ssmTSTX

8-17

8-37 EES Problem 8-36 is reconsidered. The effect of the amount of electrical work on the minimum work and the exergy destroyed is to be investigated.Analysis The problem is solved using EES, and the solution is given below.

x_1=0P_1=150 [kPa] V=2 [L] P_2=P_1{W_Ele = 2200 [kJ]}T_o=25 [C] P_o=100 [kPa]

"Conservation of energy for closed system is:"E_in - E_out = DELTAE DELTAE = m*(u_2 - u_1) E_in=W_EleE_out= W_bW_b = m*P_1*(v_2-v_1) u_1 =intenergy(steam_iapws,P=P_1,x=x_1)v_1 =volume(steam_iapws,P=P_1,x=x_1)s_1 =entropy(steam_iapws,P=P_1,x=x_1)u_2 = intenergy(steam_iapws, v=v_2,P=P_2)s_2 = entropy(steam_iapws, v=v_2,P=P_2)m=V*convert(L,m^3)/v_1W_rev_in=m*(u_2 - u_1 -(T_o+273.15) *(s_2-s_1)+P_o*(v_2-v_1))

"Entropy Balance:"S_in - S_out+S_gen = DELTAS_sys DELTAS_sys = m*(s_2 - s_1) S_in=0 [kJ/K]S_out= 0 [kJ/K]

"The exergy destruction or irreversibility is:"

0 450 900 1350 1800 22500

50

100

150

200

250

300

350

400

450

WEle [kJ]

Wrev,in

[kJ]

X_destroyed = (T_o+273.15)*S_gen

WEle

[kJ]Wrev,in

[kJ]Xdestroyed

[kJ]0 0 0

244.4 48.54 189.5488.9 97.07 379.1733.3 145.6 568.6977.8 194.1 758.21222 242.7 947.71467 291.2 11371711 339.8 13271956 388.3 15162200 436.8 1706

0 450 900 1350 1800 22500

200

400

600

800

1000

1200

1400

1600

1800

WEle [kJ]

Xdestroyed

[kJ/K]

8-18

8-38 An insulated cylinder is initially filled with saturated R-134a vapor at a specified pressure. The refrigerant expands in a reversible manner until the pressure drops to a specified value. The change in theexergy of the refrigerant during this process and the reversible work are to be determined.Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulatedand thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 Theprocess is stated to be reversible.Analysis This is a reversible adiabatic (i.e., isentropic) process, and thus s2 = s1. From the refrigerant tables(Tables A-11 through A-13),

KkJ/kg0.9183=kJ/kg246.79=

kg/m0.02562=

vaporsat.MPa8.0

MPa8.0@1

MPa8.0@1

3MPa8.0@1

1

g

g

g

ssuu

Pvv

The mass of the refrigerant is

kg952.1kg/m02562.0

m05.03

3

1vVm

kJ/kg88.21921.1869753.028.38/kgm09741.0)0007533.0099867.0(099867.00007533.0

9753.078316.0

15457.09183.0

MPa2.0

22

322

22

12

2

fgf

fgf

fg

f

uxuux

sss

x

ssP

vvv

R-134a0.8 MPa

Reversible

The reversible work output, which represents the maximum work output Wrev,out can be determined fromthe exergy balance by setting the exergy destruction equal to zero,

21

21outrev,

12outrev,

exergyinChange

system

ndestructioExergy



-XXWXXW

XXXX

Therefore, the change in exergy and the reversible work are identical in this case. Using the definition ofthe closed system exergy and substituting, the reversible work is determined to be

kJ38.5]mkJ/kPa[kg/m0.09741)62kPa)(0.025(100+kJ/kg)88.2199kg)[(246.7952.1(

)()()vv()()(33

210212102102121outrev,0

vvPuumPssTuumW

8-19

8-39E Oxygen gas is compressed from a specified initial state to a final specified state. The reversiblework and the increase in the exergy of the oxygen during this process are to be determined.Assumptions At specified conditions, oxygen can be treated as an ideal gas with constant specific heats. Properties The gas constant of oxygen is R = 0.06206 Btu/lbm.R (Table A-1E). The constant-volumespecific heat of oxygen at the average temperature is

RBtu/lbm164.0F3002/)52575(2/)( avg,21avg vcTTT

Analysis The entropy change of oxygen is

RBtu/lbm0.02894/lbmft12/lbmft1.5lnR)Btu/lbm(0.06206

R535R985lnR)Btu/lbm(0.164

lnln

3

31

2

1

2avg12 v

vv, R

TT

css O212 ft3/lbm

75 F

The reversible work input, which represents the minimum work input Wrev,in in this case can be determinedfrom the exergy balance by setting the exergy destruction equal to zero,

12inrev,

exergyinChange

system

ndestructioExergy



Therefore, the change in exergy and the reversible work are identical in this case. Substituting the closedsystem exergy relation, the reversible work input during this process is determined to be

Btu/lbm60.7]}ftpsia9[Btu/5.403/lbmft1.5)psia)(12(14.7+

R)Btu/lbmR)(0.02894(535R985)-R)(535Btu/lbm(0.164{)]()()[(

33

2102102112inrev, vvPssTuuw

Also, the increase in the exergy of oxygen is Btu/lbm60.7inrev,12 w

8-20

8-40 An insulated tank contains CO2 gas at a specified pressure and volume. A paddle-wheel in the tankstirs the gas, and the pressure and temperature of CO2 rises. The actual paddle-wheel work and the minimum paddle-wheel work by which this process can be accomplished are to be determined.Assumptions 1 At specified conditions, CO2 can be treated as an ideal gas with constant specific heats at the averagetemperature. 2 The surroundings temperature is 298 K.

Wpw

1.2 m3

2.13 kg CO2

100 kPa

Analysis (a) The initial and final temperature of CO2 are

K9.357)Kkg/mkPakg)(0.188913.2(

)mkPa)(1.2120(

K2.298)Kkg/mkPakg)(0.188913.2(

)mkPa)(1.2100(

3

322

2

3

311

1

mRP

T

mRP

T

V

V

KkJ/kg684.0K3282/)9.3572.298(2/)( avg,21avg vcTTT

The actual paddle-wheel work done is determined from the energy balance on the CO gas in the tank,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

)( 12inpw,


system


outin

TTmcUW

EEE

v

or,kJ87.0=K)2.298K)(357.9kJ/kgkg)(0.68413.2(inpw,W

(b) The minimum paddle-wheel work with which this process can be accomplished is the reversible work, which can be determined from the exergy balance by setting the exergy destruction equal to zero,

12inrev,

exergyinChange

system

ndestructioExergy



Substituting the closed system exergy relation, the reversible work input for this process is determined to be

kJ7.74K)kJ/kg(0.1253)2.298(K)2.298K)(357.9kJ/kg(0.684kg)(2.13

)()()()()(

12012avg,

12012012inrev,0

ssTTTcmPssTuumW

v

vv

since

KkJ/kg1253.0K298.2K357.9lnK)kJ/kg684.0(lnln 0

1

2

1

2avg,12 v

vv R

TT

css

8-21

8-41 An insulated cylinder initially contains air at a specified state. A resistance heater inside the cylinderis turned on, and air is heated for 15 min at constant pressure. The exergy destruction during this process isto be determined.Assumptions 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 The mass of the air and the electrical work done during this process are

AIR120 kPa P = const

kg0.0418K)K)(300/kgmkPa(0.287

)mkPa)(0.03(1203

3

1

11

RTP

mV

We

W W te e ( .0 05 15 kJ / s)(5 60 s) kJ

Also,

T h s1 1300 30019 1 70202 K kJ / kg and kJ / kg K1o. .

The energy balance for this stationary closed system can be expressed as

)( 12ine,

outb,ine,


system


outin

hhmWUWW

EEE

since U + Wb = H during a constant pressure quasi-equilibrium process. Thus,

KkJ/kg49364.2K650

kJ/kg04.659kg0.0418

kJ1519.300 o2

2ine,12 s

Tm

Whh

Also, KkJ/kg79162.070202.149364.2ln o1

o2

0

1

2o1

o212 ss

PP

Rssss

The exergy destruction (or irreversibility) associated with this process can be determined from its definition Xdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on thecylinder, which is an insulated closed system,

)( 12systemgen

entropyinChange

system

generationEntropy

gen


outin

ssmSS

SSSS

Substituting,kJ9.9=K)kJ/kg2kg)(0.7916K)(0.0418300()( 120gen0destroyed ssmTSTX

8-22

8-42 A fixed mass of helium undergoes a process from a specified state to another specified state. Theincrease in the useful energy potential of helium is to be determined.Assumptions 1 At specified conditions, helium can be treated as an idealgas. 2 Helium has constant specific heats at room temperature.Properties The gas constant of helium is R = 2.0769 kJ/kg.K (Table A-1). The constant volume specific heat of helium is cv = 3.1156 kJ/kg.K (Table A-2).

He8 kg

288 K

Analysis From the ideal-gas entropy change relation,

KkJ/kg3.087=/kgm3/kgm0.5lnK)kJ/kg(2.0769

K288K353lnK)kJ/kg(3.1156

lnln

3

31

2

1

2avg,12 v

vv R

TT

css

The increase in the useful potential of helium during this process is simply the increase in exergy,

kJ6980]}mkJ/kPa[kg/m)5.0kPa)(3(100+

K)kJ/kgK)(3.087(298K353)K)(288kJ/kg6kg){(3.1158()()()(

33

2102102112 vvPssTuum

8-43 One side of a partitioned insulated rigid tank contains argon gas at a specified temperature and pressure while the other side is evacuated. The partition is removed, and the gas fills the entire tank. Theexergy destroyed during this process is to be determined.Assumptions Argon is an ideal gas with constant specific heats, and thus ideal gas relations apply.Properties The gas constant of argon is R = 0.208 kJ/kg.K (Table A-1). Analysis Taking the entire rigid tank as the system, the energy balance can be expressed as

E E E

U m u uu u T T

in out Net energy transfer

by heat, work, and mass

system

Change in internal, kinetic, potential, etc. energies

0 2 1

2 1 2 1

( ) VacuumArgon

300 kPa 70 C

since u = u(T) for an ideal gas.The exergy destruction (or irreversibility) associated with this process can be determined from its

definition Xdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on theentire tank, which is an insulated closed system,

)( 12systemgen

entropyinChange

system

generationEntropy

gen


outin

ssmSS

SSSS

where

kJ/K0.433=(2)lnK)kJ/kgkg)(0.2083(

lnlnln)(1

2

1

2

1

02

avg,12system V

V

V

Vv mRR

TT

cmssmS

Substituting,kJ129=kJ/K)K)(0.433298()( 120gen0destroyed ssmTSTX

8-23

8-44E A hot copper block is dropped into water in an insulated tank. The final equilibrium temperature of the tank and the work potential wasted during this process are to be determined.Assumptions 1 Both the water and the copper block are incompressible substances with constant specificheats at room temperature. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The tank is well-insulated and thus there is no heat transfer.Properties The density and specific heat of water at the anticipated average temperature of 90 F are = 62.1 lbm/ft3 and cp = 1.00 Btu/lbm. F. The specific heat of copper at the anticipated average temperature of 100 F is cp = 0.0925 Btu/lbm. F (Table A-3E). Analysis We take the entire contents of the tank, water + copper block, as the system, which is a closed system. The energy balance for this system can be expressed as

U

EEE

0energiesetc.potential,

kinetic,internal,inChange

system


outin

Copper250 F

Water75 F

or,0waterCu UU

0)]([)]([ water12Cu12 TTmcTTmc

where

lbm15.93)ft5.1)(lbm/ft1.62( 33Vwm

Substituting,

R4.546F)75F)(Btu/lbmlbm)(1.015.93()F250F)(TBtu/lbm5lbm)(0.09270(0

2

22

F86.4TT

The wasted work potential is equivalent to the exergy destruction (or irreversibility), and it can bedetermined from its definition Xdestroyed = T0Sgen where the entropy generation is determined from an entropybalance on the system, which is an insulated closed system,

copperwatersystemgen

entropyinChange

system

generationEntropy

gen


outin

SSSS

SSSS

where

Btu/R960.1R535R546.4lnR)Btu/lbmlbm)(1.015.93(ln

Btu/R696.1R710R546.4lnR)Btu/lbmlbm)(0.09270(ln

1

2avgwater

1

2avgcopper

TT

mcS

TT

mcS

Substituting,Btu140.9=Btu/R)960.1696.1R)(535(destroyedX

8-24

8-45 A hot iron block is dropped into water in an insulated tank that is stirred by a paddle-wheel. The massof the iron block and the exergy destroyed during this process are to be determined.Assumptions 1 Both the water and the iron block are incompressible substances with constant specificheats at room temperature. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The tank is well-insulated and thus there is no heat transfer.Properties The density and specific heat of water at 25 C are = 997 kg/m3 and cp = 4.18 kJ/kg. F. The specific heat of iron at room temperature (the only value available in the tables) is cp = 0.45 kJ/kg. C(Table A-3). Analysis We take the entire contents of the tank, water + iron block, as the system, which is a closedsystem. The energy balance for this system can be expressed as

waterironinpw,


system


outin

UUUW

EEE

Wpw

Water

100 L 20 C

Iron85 C

W water12iron12inpw, )]([)]([ TTmcTTmc

where

kJ240)s6020)(kJ/s2.0(

kg7.99)m1.0)(kg/m997(

inpw,pw

33water

tWW

m V

Substituting,

kg52.0iron

iron C)20C)(24kJ/kgkg)(4.187.99(C)85C)(24kJ/kg45.0(=kJ240m

m

(b) The exergy destruction (or irreversibility) can be determined from its definition Xdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on the system, which is an insulated closedsystem,

waterironsystemgen

entropyinChange

system

generationEntropy

gen


outin

SSSS

SSSS

where

kJ/K651.5K293K297lnK)kJ/kgkg)(4.187.99(ln

kJ/K371.4K358K297lnK)kJ/kgkg)(0.450.52(ln

1

2avgwater

1

2avgiron

TT

mcS

TT

mcS

Substituting,kJ375.0=kJ/K)651.5371.4K)(293(gen0destroyed STX

8-25

8-46 An iron block and a copper block are dropped into a large lake where they cool to lake temperature.The amount of work that could have been produced is to be determined.Assumptions 1 The iron and copper blocks and water are incompressible substances with constant specificheats at room temperature. 2 Kinetic and potential energies are negligible.Properties The specific heats of iron and copper at room temperature are cp, iron = 0.45 kJ/kg. C and cp,copper

= 0.386 kJ/kg. C (Table A-3). Analysis The thermal-energy capacity of the lake is very large, and thus the temperatures of both the ironand the copper blocks will drop to the lake temperature (15 C) when the thermal equilibrium is established.

We take both the iron and the copper blocks as the system, which is a closed system. The energybalance for this system can be expressed as

copperironout


system


outin

UUUQ

EEE

CopperIron

Lake15 C

Iron85 C

or,Q [ copperout TTmcTTmc )]([)]( 21iron21

Substituting,

kJ1964K288353KkJ/kg0.386kg20K288353KkJ/kg0.45kg50outQ

The work that could have been produced is equal to the wasted work potential. It is equivalent to theexergy destruction (or irreversibility), and it can be determined from its definition Xdestroyed = T0Sgen . The entropy generation is determined from an entropy balance on an extended system that includes the blocksand the water in their immediate surroundings so that the boundary temperature of the extended system isthe temperature of the lake water at all times,

lake

outcopperirongen

copperironsystemgenoutb,

out

entropyinChange

system

generationEntropy

gen


outin

TQ

SSS

SSSSTQ

SSSS

where

kJ/K1.571K353K288

lnKkJ/kg0.386kg20ln

kJ/K4.579K353K288

lnKkJ/kg0.45kg50ln

1

2avgcopper

1

2avgiron

TT

mcS

TT

mcS

Substituting,

kJ196=kJ/KK288kJ1964

571.1579.4K)293(gen0destroyed STX

8-26

8-47E A rigid tank is initially filled with saturated mixture of R-134a. Heat is transferred to the tank from asource until the pressure inside rises to a specified value. The amount of heat transfer to the tank from the source and the exergy destroyed are to be determined.Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 There is no heat transfer with the environment.Properties From the refrigerant tables (Tables A-11E through A-13E),

lbm/ft65234.016368.155.001232.0vRlbm/Btu1436.017580.055.004688.0

lbm/Btu76.63307.7755.0246.21

0.55psia40

311

11

11

1

1

fgf

fgf

fgf

xsxssuxuu

xP

vv

Btu/lbm03.88360.738191.0939.27RBtu/lbm1922.016098.08191.006029.0

8191.001270.079361.001270.065234.0

)(psia60

22

22

22

12

2

fgf

fgf

fg

f

uxuusxss

xP v

vv

vv

Analysis (a) The mass of the refrigerant is

lbm40.18lbm/ft65234.0

ft123

3

1vVm

Q

Source120 CR-134a

40 psia We take the tank as the system, which is a closed system. The energy balance for this stationary closed system can be expressed as

)( 12in


system


outin

uumUQ

EEE

Substituting,Btu446.3=Btu/lbm63.76)-.0388lbm)(40.18()( 12in uumQ

(b) The exergy destruction (or irreversibility) can be determined from its definition Xdestroyed = T0Sgen . The entropy generation is determined from an entropy balance on an extended system that includes the tank and the region in its immediate surroundings so that the boundary temperature of the extended system where heat transfer occurs is the source temperature,

source

in12gen

12systemgeninb,

in

entropyinChange

system

generationEntropy

gen


outin

)(

)(

TQ

ssmS

ssmSSTQ

SSSS

,

Substituting,

Btu66.5=R580Btu3.446RBtu/lbm)1436.01922.0lbm)((18.40R)535(gen0destroyed STX

8-27

8-48 Chickens are to be cooled by chilled water in an immersion chiller that is also gaining heat from thesurroundings. The rate of heat removal from the chicken and the rate of exergy destruction during thisprocess are to be determined.Assumptions 1 Steady operating conditions exist. 2 Thermal properties of chickens and water are constant.3 The temperature of the surrounding medium is 25 C.Properties The specific heat of chicken is given to be 3.54 kJ/kg.°C. The specific heat of water at roomtemperature is 4.18 kJ/kg. C (Table A-3). Analysis (a) Chickens are dropped into the chiller at a rate of 500 per hour. Therefore, chickens can be considered to flow steadily through the chiller at a mass flow rate of

mchicken (500 chicken / h)(2.2 kg / chicken) 1100 kg / h = 0.3056kg / sTaking the chicken flow stream in the chiller as the system, the energy balance for steadily flowingchickens can be expressed in the rate form as

)(

0)peke(since

0

21chickenchickenout

2out1

outin

energiesetc.potential,kinetic,internal,inchangeofRate

(steady)0system

massand work,heat,bynsferenergy tranetofRate

outin

TTcmQQ

hmQhm

EEEEE

p

Then the rate of heat removal from the chickens as they are cooled from 15 C to 3ºC becomeskW13.0C3)ºC)(15kJ/kg.ºkg/s)(3.54(0.3056)( chickenchicken TcmQ p

The chiller gains heat from the surroundings as a rate of 200 kJ/h = 0.0556 kJ/s. Then the total rate of heat gain by the water is

. . .Q Q Qwater chicken heat gain kW13 0 0 056 13 056Noting that the temperature rise of water is not to exceed 2ºC as it flows through the chiller, the mass flow rate of water must be at least

kg/s1.56C)C)(2ºkJ/kg.º(4.18

kW13.056)( water

waterwater Tc

Qm

p

(b) The exergy destruction can be determined from its definition Xdestroyed = T0Sgen. The rate of entropygeneration during this chilling process is determined by applying the rate form of the entropy balance on anextended system that includes the chiller and the immediate surroundings so that the boundary temperatureis the surroundings temperature:

surr

in34water12chickengen

gensurr

in4water2chicken3water1chicken

gensurr

in43223311

entropyofchangeofRate

(steady)0system

generationentropyofRate

gen

massandheatbyansferentropy trnetofRate

outin

)()(

0

0

TQ

ssmssmS

STQ

smsmsmsm

STQ

smsmsmsm

SSSS

Noting that both streams are incompressible substances, the rate of entropy generation is determined to be

kW/K0.00128K298kW0556.0

273.5275.5lnkJ/kg.K)kg/s)(4.1856.1(

288276lnkJ/kg.K)kg/s)(3.543056.0(

lnlnsurr3

4water

1

2chickengen T

QTT

cmTT

cmS inpp

Finally, kW0.381/ =K)kWK)(0.00128298(gen0destroyed STX

8-28

8-49 An egg is dropped into boiling water. The amount of heat transfer to the egg by the time it is cooked and the amount of exergy destruction associated with this heat transfer process are to be determined.Assumptions 1 The egg is spherical in shape with a radius of r0 = 2.75 cm. 2 The thermal properties of theegg are constant. 3 Energy absorption or release associated with any chemical and/or phase changes withinthe egg is negligible. 4 There are no changes in kinetic and potential energies. 5 The temperature of the surrounding medium is 25 C.Properties The density and specific heat of the egg are given to be = 1020 kg/m3 and cp = 3.32 kJ/kg. C.Analysis We take the egg as the system. This is a closed system since no mass enters or leaves the egg. The energy balance for this closed system can be expressed as

)()( 1212eggin


system


outin

TTmcuumUQ

EEE

Then the mass of the egg and the amount of heat transfer become

kJ18.3C)870)(CkJ/kg.32.3)(kg0889.0()(

kg0889.06

m)055.0()kg/m1020(6

12in

33

3

TTmcQ

Dm

p

V

The exergy destruction can be determined from its definition Xdestroyed = T0Sgen. The entropy generatedduring this process can be determined by applying an entropy balance on an extended system that includesthe egg and its immediate surroundings so that the boundary temperature of the extended system is at 97 Cat all times:

systemin

gensystemgenin

entropyinChange

system

generationEntropy

gen


outin

STQ

SSSTQ

SSSS

bb

where

kJ/K0588.0273+8273+70lnkJ/kg.K)32.3)(kg0889.0(ln)(

1

2avg12system T

TmcssmS

Substituting,

egg)(perkJ/K0.00934kJ/K0588.0K370kJ18.3

systemin

gen STQ

Sb

Finally,kJ2.78/ =K)kJK)(0.00934298(gen0destroyed STX

8-29

8-50 Stainless steel ball bearings leaving the oven at a uniform temperature of 900 C at a rate of 1400 /min are exposed to air and are cooled to 850 C before they are dropped into the water for quenching. The rate of heat transfer from the ball to the air and the rate of exergy destruction due to this heat transfer are to be determined.Assumptions 1 The thermal properties of the bearing balls are constant. 2 The kinetic and potential energychanges of the balls are negligible. 3 The balls are at a uniform temperature at the end of the process. Properties The density and specific heat of the ball bearings are given to be = 8085 kg/m3 and cp = 0.480 kJ/kg. C.Analysis (a) We take a single bearing ball as the system. The energy balance for this closed system can be expressed as

)()(

21out

12ballout


system


outin

TTmcQuumUQ

EEE

The total amount of heat transfer from a ball is

kJ/ball1756.0C)850900(C)kJ/kg.480.0)(kg007315.0()(

kg007315.06

m)012.0()kg/m8085(

621out

33

3

TTmcQ

Dm V

Then the rate of heat transfer from the balls to the air becomes

kW4.10kJ/min245.8)kJ/ball1756.0(balls/min)1400(ball)(peroutballtotal QnQ

Therefore, heat is lost to the air at a rate of 4.10 kW.(b) The exergy destruction can be determined from its definition Xdestroyed = T0Sgen. The entropy generatedduring this process can be determined by applying an entropy balance on an extended system that includesthe ball and its immediate surroundings so that the boundary temperature of the extended system is at 30 Cat all times:

systemout

gensystemgenout

entropyinChange

system

generationEntropy

gen


outin

ST

QSSS

TQ

SSSS

bb

where

kJ/K0001530.0273+900273+850lnkJ/kg.K)480.0)(kg007315.0(ln)(

1

2avg12system T

TmcssmS

Substituting,

ball)(perkJ/K0004265.0kJ/K0001530.0K303

kJ0.1756system

outgen S

TQ

Sb

Then the rate of entropy generation becomes

kW/K0.00995=kJ/min.K0.597=balls/min)ball)(1400kJ/K0004265.0(ballgengen nSS

Finally,

kW/K3.01/ =K)kWK)(0.00995303(gen0destroyed STX

8-30

8-51 Carbon steel balls are to be annealed at a rate of 2500/h by heating them first and then allowing themto cool slowly in ambient air at a specified rate. The total rate of heat transfer from the balls to the ambientair and the rate of exergy destruction due to this heat transfer are to be determined.Assumptions 1 The thermal properties of the balls are constant. 2 There are no changes in kinetic and potential energies. 3 The balls are at a uniform temperature at the end of the process. Properties The density and specific heat of the balls are given to be = 7833 kg/m3 and cp = 0.465 kJ/kg. C.Analysis (a) We take a single ball as the system. The energy balance for this closed system can be expressed as

)()(

21out

12ballout


system


outin

TTmcQuumUQ

EEE

The amount of heat transfer from a single ball is

ball)(perkJ0.781=J781C)100900)(CkJ/kg.465.0)(kg0021.0()(

kg00210.06

m)008.0()kg/m7833(6

21out

33

3

TTmcQ

Dm

p

V

Then the total rate of heat transfer from the balls to the ambient air becomes

W260kJ/h936)kJ/ball781.0(balls/h)1200(outballout QnQ

(b) The exergy destruction (or irreversibility) can be determined from its definition Xdestroyed = T0Sgen. The entropy generated during this process can be determined by applying an entropy balance on an extendedsystem that includes the ball and its immediate surroundings so that the boundary temperature of theextended system is at 35 C at all times:

systemout

gensystemgenout

entropyinChange

system

generationEntropy

gen


outin

ST

QSSS

TQ

SSSS

bb

where

kJ/K00112.0273+900273+100kJ/kg.K)ln465.0)(kg00210.0(ln)(

1

2avg12system T

TmcssmS

Substituting,

ball)(perkJ/K00142.0kJ/K00112.0K308kJ0.781

systemout

gen ST

QS

b


kW/K0.000473=kJ/h.K1.704=balls/h)ball)(1200kJ/K00142.0(ballgengen nSS

Finally,

W146/ =kW146.0K)kW3K)(0.00047308(gen0destroyed STX

8-31

8-52 A tank containing hot water is placed in a larger tank. The amount of heat lost to the surroundings and the exergy destruction during the process are to be determined.Assumptions 1 Kinetic and potential energy changes are negligible. 2 Air is an ideal gas with constantspecific heats. 3 The larger tank is well-sealed.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 (Table A-2). The properties of water at room temperature are = 1000 kg/m3, cw = 4.18 kJ/kg.K.Analysis (a) The final volume of the air in the tank is

312 m025.0015.004.0waa VVV

The mass of the air in the room is

kg04724.0K)273K)(22/kgmkPa(0.287

)mkPa)(0.04(1003

3

1

11

a

aa RT

Pm

V

The pressure of air at the final state is

kPa9.171m0.025

K)273K)(44/kgmkPakg)(0.287(0.047243

3

2

22

a

aaa

RTmP

V

QWater85 C15 L

Air, 22 C

The mass of water is

kg53.14)m)(0.015kg/m(1000 33wwwm V

An energy balance on the system consisting of water and air is used to determine heat lost to thesurroundings

kJ2489)224kJ/kg.K)(4kg)(0.71804724.0()854kJ/kg.K)(4kg)(4.1853.14(

)()( 1212out aawww TTcmTTcmQ v

(b) An exergy balance written on the (system + immediate surroundings) can be used to determine exergydestruction. But we first determine entropy and internal energy changes

kJ/K3873.7K273)(44K273)(85kJ/kg.K)lnkg)(4.18(14.53ln

2

1

TT

cmS wwww

kJ/K003931.0kPa171.9

kPa100kJ/kg.K)ln(0.287K273)(44K273)(22

kJ/kg.K)ln(1.005kg)(0.04724

lnln2

1

2

1

PP

RTT

cmS aapaa

kJ7462.044)K-2kJ/kg.K)(2kg)(0.718(0.04724)(kJ249044)K-5kJ/kg.K)(8kg)(4.18(14.53)(

21

21

TTcmUTTcmU

aaa

wwww

v

kJ308.8kJ/K)9313K)(0.00(295kJ)7462.0(-kJ/K)K)(7.3873(295kJ2490

00

dest

aaww

aw

STUSTUXXX

8-32

8-53 Heat is transferred to a piston-cylinder device with a set of stops. The work done, the heat transfer,the exergy destroyed, and the second-law efficiency are to be determined.Assumptions 1 The device is stationary and kinetic and potential energy changes are zero. 2 There is no friction between the piston and the cylinder.Analysis (a) The properties of the refrigerant at theinitial and final states are (Tables A-11 through A-13)

Q

R-134a1.4 kg

140 kPa 20 C

kJ/kg.K3118.1kJ/kg96.331

/kgm17563.0

C120kPa180

kJ/kg.K0624.1kJ/kg22.248

/kgm16544.0

C20kPa120

2

2

32

2

2

1

1

31

1

1

suT

P

suT

P

v

v

The boundary work is determined to be

kJ2.57/kg0.16544)m63kPa)(0.175kg)(1804.1()( 3122outb, vvmPW

(b) The heat transfer can be determined from an energy balance on the systemkJ119.8kJ2.57kg248.22)kJ/kg)(331.964.1()( outb,12in WuumQ

(c) The exergy difference between the inlet and exit states is

kJ61.14/kg0.16544)m63kPa)(0.175(100K1.0624)kg.K)(1.3118(298kJ/kg)22.24896.331(kg)(1.4

)()(3

12012012 vvPssTuumX

The useful work output for the process is

kJ14.1/kgm)16544.063kPa)(0.175kg)(1004.1(kJ57.2)( 3120outb,outu, vvmPWW

The exergy destroyed is the difference between the exergy difference and the useful work outputkJ13.4714.161.14outu,dest WXX

(d) The second-law efficiency for this process is

0.078kJ61.14kJ14.1outu,

II XW

8-33

Second-Law Analysis of Control Volumes

8-54 Steam is throttled from a specified state to a specified pressure. The wasted work potential during thisthrottling process is to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The temperature of the surroundings is given to be 25 C. 4 Heat transfer is negligible.

Properties The properties of steam before and after the throttling process are (Tables A-4 through A-6)

KkJ/kg5579.6kJ/kg3.3273

C450MPa8

1

1

1

1

sh

TP

Steam

2

1KkJ/kg6806.6

MPa62

12

2 shh

P

Analysis The wasted work potential is equivalent to the exergy destruction (or irreversibility). It can be determined from an exergy balance or directly from itsdefinition Xdestroyed = T0Sgen where the entropy generation is determined from anentropy balance on the device, which is an adiabatic steady-flow system,

or)(0

0

12gen12gengen21


0system


gen


outin

sssssmSSsmsm

SSSS

Substituting,

kJ/kg36.6=KkJ/kg)5579.6K)(6.6806298()( 120gen0destroyed ssTsTx

Discussion Note that 36.6 kJ/kg of work potential is wasted during this throttling process.

8-34

8-55 [Also solved by EES on enclosed CD] Air is compressed steadily by an 8-kW compressor from a specified state to another specified state. The increase in the exergy of air and the rate of exergydestruction are to be determined.

Assumptions 1 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 = 0.287 kJ/kg.K (Table A-1). From the air table (Table A-17)

8 kWAIR

600 kPa 167 C

T hs

T hs

1 1

2 2

290 29016166802

440 441612 0887

K kJ / kg K

K kJ / kg K

1o

2o

..

..

kJ / kg

kJ / kg

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

100 kPa 17 C

)()()]()[(

exergyinIncrease

12012

12012

1200

ssThhssTpekehh

where

KkJ/kg09356.0kPa100kPa600lnK)kJ/kg(0.287-KkJ/kg)66802.10887.2(

ln)(1

2o1

o212 P

PRssss

Substituting,

kJ/kg178.6K)kJ/kg09356.0K)((290-kJ/kg)290.16(441.61

exergyinIncrease 12

Then the reversible power input becomes

kW6.25kJ/kg)6kg/s)(178.60/1.2()( 12inrev, mW

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

kW1.7525.68inrev,indestroyed WWX

Discussion Note that 1.75 kW of power input is wasted during this compression process.

8-35

8-56 EES Problem 8-55 is reconsidered. The problem is to be solved and the actual heat transfer, its direction, the minimum power input, and the compressor second-law efficiency are to be determined.

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

Function Direction$(Q)If Q<0 then Direction$='out' else Direction$='in'endFunction Violation$(eta)If eta>1 then Violation$='You have violated the 2nd Law!!!!!' else Violation$=''end

{"Input Data from the Diagram Window"T_1=17 [C] P_1=100 [kPa] W_dot_c = 8 [kW] P_2=600 [kPa] S_dot_gen=0Q_dot_net=0}{"Special cases" T_2=167 [C] m_dot=2.1 [kg/min]}T_o=T_1P_o=P_1m_dot_in=m_dot*Convert(kg/min, kg/s) "Steady-flow conservation of mass"m_dot_in = m_dot_out "Conservation of energy for steady-flow is:"E_dot_in - E_dot_out = DELTAE_dot DELTAE_dot = 0 E_dot_in=Q_dot_net + m_dot_in*h_1 +W_dot_c"If Q_dot_net < 0, heat is transferred from the compressor"E_dot_out= m_dot_out*h_2h_1 =enthalpy(air,T=T_1)h_2 = enthalpy(air, T=T_2)W_dot_net=-W_dot_cW_dot_rev=-m_dot_in*(h_2 - h_1 -(T_1+273.15)*(s_2-s_1))"Irreversibility, entropy generated, second law efficiency, and exergy destroyed:"s_1=entropy(air, T=T_1,P=P_1) s_2=entropy(air,T=T_2,P=P_2)s_2s=entropy(air,T=T_2s,P=P_2)s_2s=s_1"This yields the isentropic T_2s for an isentropic process bewteen T_1, P_1 and P_2"I_dot=(T_o+273.15)*S_dot_gen"Irreversiblility for the Process, KW"S_dot_gen=(-Q_dot_net/(T_o+273.15) +m_dot_in*(s_2-s_1)) "Entropy generated, kW"Eta_II=W_dot_rev/W_dot_net"Definition of compressor second law efficiency, Eq. 7_6"h_o=enthalpy(air,T=T_o)s_o=entropy(air,T=T_o,P=P_o)Psi_in=h_1-h_o-(T_o+273.15)*(s_1-s_o) "availability function at state 1"Psi_out=h_2-h_o-(T_o+273.15)*(s_2-s_o) "availability function at state 2"X_dot_in=Psi_in*m_dot_inX_dot_out=Psi_out*m_dot_outDELTAX_dot=X_dot_in-X_dot_out"General Exergy balance for a steady-flow system, Eq. 7-47"(1-(T_o+273.15)/(T_o+273.15))*Q_dot_net-W_dot_net+m_dot_in*Psi_in - m_dot_out*Psi_out=X_dot_dest"For the Diagram Window"

8-36

Text$=Direction$(Q_dot_net)Text2$=Violation$(Eta_II)

II I [kW] Xdest [kW] T2s [C] T2 [C] Qnet [kW]0.7815 1.748 1.748 209.308 167 -2.70.8361 1.311 1.311 209.308 200.6 -1.5010.8908 0.874 0.874 209.308 230.5 -0.42520.9454 0.437 0.437 209.308 258.1 0.5698

1 1.425E-13 5.407E-15 209.308 283.9 1.506

5.0 5.5 6.0 6.50

50

100

150

200

250

s [kJ/kg-K]

T[C] 100 kPa

600 kPa

1

2

2s

ideal

actual

How can entropy decrease?

0.75 0.80 0.85 0.90 0.95 1.00160

180

200

220

240

260

280

300

0.0

0.5

1.0

1.5

2.0

II

T2

Xdest

0.75 0.80 0.85 0.90 0.95 1.00-3

-2

-1

0

1

2

0.0

0.5

1.0

1.5

2.0

II

Qnet

Xdest

8-37

8-57 Refrigerant-124a is throttled from a specified state to a specified pressure. The reversible work and the exergy destroyed during this process are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Heat transfer is negligible.

Properties The properties of R-134a before and after the throttling process are (Tables A-11 through A-13)

KkJ/kg1031.1kJ/kg06.335

C100MPa1

1

1

1

1

sh

TP

R-134a

2

1KkJ/kg1198.1MPa8.0

212

2 shh

P

Analysis The exergy destruction (or irreversibility) can be determined from an exergybalance or directly from its definition Xdestroyed = T0Sgen where the entropy generationis determined from an entropy balance on the system, which is an adiabatic steady-flow device,

or)(0

0

12gen12gengen21


0system


gen


outin

sssssmSSsmsm

SSSS

Substituting,

kJ/kg5.04=KkJ/kg)1031.1K)(1.1198303()( 120gen0destroyed ssTsTx

This process involves no actual work, and thus the reversible work and irreversibility are identical,

kJ/kg5.04destroyedoutrev,0

outact,outrev,destroyed xwwwx


8-38

8-58 EES Problem 8-57 is reconsidered. The effect of exit pressure on the reversible work and exergydestruction is to be investigated.


T_1=100"[C]"P_1=1000"[kPa]"{P_2=800"[kPa]"}T_o=298"[K]""Steady-flow conservation of mass""m_dot_in = m_dot_out""Conservation of energy for steady-flow per unit mass is:"e_in - e_out = DELTAe DELTAe = 0"[kJ/kg]"E_in=h_1"[kJ/kg]"E_out= h_2 "[kJ/kg]"h_1 =enthalpy(R134a,T=T_1,P=P_1) "[kJ/kg]"T_2 = temperature(R134a, P=P_2,h=h_2) "[C]""Irreversibility, entropy generated, and exergy destroyed:"s_1=entropy(R134a, T=T_1,P=P_1)"[kJ/kg-K]"s_2=entropy(R134a,P=P_2,h=h_2)"[kJ/kg-K]"I=T_o*s_gen"[kJ/kg]" "Irreversiblility for the Process, KJ/kg"s_gen=s_2-s_1"]kJ/kg-K]" "Entropy generated, kW"x_destroyed = I"[kJ/kg]"w_rev_out=x_destroyed"[kJ/kg]"

P2 [kPa] wrev,out [kJ/kg] xdestroyed [kJ/kg]100 53.82 53.82200 37.22 37.22300 27.61 27.61400 20.86 20.86500 15.68 15.68600 11.48 11.48700 7.972 7.972800 4.961 4.961900 2.33 2.33

1000 -4.325E-10 -4.325E-10

100 200 300 400 500 600 700 800 900 10000

10

20

30

40

50

60

P2 [kPa]

wrev,out

[kJ/kg]

exergydestroyed[kJ/kg]

or

8-39

8-59 Air is accelerated in a nozzle while losing some heat to the surroundings. The exit temperature of air and the exergy destroyed during the process are to be determined.Assumptions 1 Air is an ideal gas with variable specific heats. 2 The nozzle operates steadily. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The properties of air at the nozzleinlet are (Table A-17) 4 kJ/kg

AIR 300 m/s

T hs

1 1360 360 58188543

K kJ / kg K1

o.

.kJ / kg

50 m/s Analysis (a) We take the nozzle as the system, which is acontrol volume. The energy balance for this steady-flowsystem can be expressed in the rate form as

out222

211outin


(steady)0system


outin +/2)V+()2/(0 QhmVhmEEEEE

or2

+02

12

212out

VVhhq

Therefore,

kJ/kg83.312s/m1000

kJ/kg12

m/s)50(m/s)300(458.360

2 22

2221

22

out12VV

qhh

At this h2 value we read, from Table A-17, T KkJ/kg74302.1and 022 sC39.5=K312.5

S(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition where the entropy generation SX T gendestroyed 0 gen is determined from an entropy balance on

an extended system that includes the device and its immediate surroundings so that the boundary temperature of the extended system is Tsurr at all times. It gives

surr

out12gengen

surrb,

out21


0system


gen


outin 00TQ

ssmSSTQ

smsmSSSS

where

KkJ/kg1876.0kPa300

kPa95lnK)kJ/kg(0.287KkJ/kg)88543.174302.1(ln1

2o1

o2air P

PRsss

Substituting, the entropy generation and exergy destruction per unit mass of air are determined to be

kJ/kg58.4K290

kJ/kg4+KkJ/kg0.1876K)290(1200destroyedsurr

surrgensurrgen T

qssTsTsTx

Alternative solution The exergy destroyed during a process can be determined from an exergy balanceapplied on the extended system that includes the device and its immediate surroundings so that theboundary temperature of the extended system is environment temperature T0 (or Tsurr) at all times. Notingthat exergy transfer with heat is zero when the temperature at the point of transfer is the environment temperature, the exergy balance for this steady-flow system can be expressed as

8-40

gen00

out120

12outout120

121200

21021

2121destroyed

exergyofchangeofRate

(steady)0system

ndestructioexergyofRate

destroyed

massand work,heat,bynsferexergy tranetofRate

outin

)(

balance,energyfromsince,])([)]()([])()[(

)(0

STT

QssmT

kehhqqssTmkehhssTmpekessThhm

mmmXXXXXXX outin

Therefore, the two approaches for the determination of exergy destruction are identical.

8-41

8-60 EES Problem 8-59 is reconsidered. The effect of varying the nozzle exit velocity on the exittemperature and exergy destroyed is to be investigated.


"Knowns:"WorkFluid$ = 'Air'P[1] = 300 [kPa] T[1] =87 [C] P[2] = 95 [kPa] Vel[1] = 50 [m/s]{Vel[2] = 300 [m/s]}T_o = 17 [C] T_surr = T_oq_loss = 4 [kJ/kg]"Conservation of Energy - SSSF energy balance for nozzle -- neglecting the change in potentialenergy:"h[1]=enthalpy(WorkFluid$,T=T[1])s[1]=entropy(WorkFluid$,P=P[1],T=T[1])ke[1] = Vel[1]^2/2 ke[2]=Vel[2]^2/2h[1]+ke[1]*convert(m^2/s^2,kJ/kg) = h[2] + ke[2]*convert(m^2/s^2,kJ/kg)+q_lossT[2]=temperature(WorkFluid$,h=h[2])s[2]=entropy(WorkFluid$,P=P[2],h=h[2])"The entropy generated is detemined from the entropy balance:"s[1] - s[2] - q_loss/(T_surr+273) + s_gen = 0 x_destroyed = (T_o+273)*s_gen

T2 [C] Vel2 [m/s] xdestroyed

[kJ/kg]79.31 100 93.4174.55 140 89.4368.2 180 84.04

60.25 220 77.1750.72 260 68.739.6 300 58.49

100 140 180 220 260 30035

40

45

50

55

60

65

70

75

80

Vel[2]

T[2][C]

100 140 180 220 260 30055

60

65

70

75

80

85

90

95

Vel[2]

xdestroedy

[kJ/kg]

8-42

8-61 Steam is decelerated in a diffuser. The mass flow rate of steam and the wasted work potential during the process are to be determined.

Assumptions 1 The diffuser operates steadily. 2 The changes in potential energies are negligible.

Properties The properties of steam at the inlet and the exit of the diffuser are (Tables A-4 through A-6)

KkJ/kg1741.8kJ/kg0.2592

C50kPa10

1

1

1

1

sh

TP

H2O300 m/s

/kgm026.12KkJ/kg0748.8

kJ/kg3.2591

sat.vaporC50

32

2

22

v

sh

T70 m/s

Analysis (a) The mass flow rate of the steam is

kg/s17.46=m/s)70)(m3(kg/m026.12

11 2322

2VAm

v

(b) We take the diffuser to be the system, which is a control volume. Assuming the direction of heattransfer to be from the stem, the energy balance for this steady-flow system can be expressed in the rateform as

2

+/2)V+()2/(

0

21

22

12out

out222

211

outin


(steady)0system


outin

VVhhmQ

QhmVhm

EE

EEE

Substituting,

kJ/s8.754s/m1000

kJ/kg12

m/s)300(m/s)70(0.25922591.3kg/s)46.17(22

22

outQ

The wasted work potential is equivalent to exergy destruction. The exergy destroyed during a process can be determined from an exergy balance or directly from its definition where the entropy

generation S

X T gendestroyed 0S

gen is determined from an entropy balance on an extended system that includes the device and its immediate surroundings so that the boundary temperature of the extended system is Tsurr at all times. It gives

surr

out12gengen

surrb,

out21


0system


gen


outin

0

0

TQ

ssmSSTQ

smsm

SSSS

Substituting, the exergy destruction is determined to be

8-43

kW238.3K298kW754.8+KkJ/kg8.1741)-48kg/s)(8.07(17.46K)298(

)(0


QssmTSTX

8-44

8-62E Air is compressed steadily by a compressor from a specified state to another specified state. The minimum power input required for the compressor is to be determined.


Properties The gas constant of air is R = 0.06855 Btu/lbm.R (Table A-1E). From the air table (Table A-17E)

RBtu/lbm73509.0Btu/lbm11.226R940

RBtu/lbm59173.0Btu/lbm27.124R520

o2

22

o1

11

shT

shT

14.7 psia 60 F

AIR15 lbm/min

100 psia 480 F

Analysis The reversible (or minimum) power input is determinedfrom the rate form of the exergy balance applied on the compressor and setting the exergy destruction term equal to zero,

])()[()(

0

001201212inrev,

2inrev,1

outin


(steady)0system




outin

pekessThhmmW

mWm

XX

XXXX

where

RBtu/lbm01193.0psia14.7psia100

lnR)Btu/lbm(0.06855RBtu/lbm)59173.073509.0(

ln1

2o1

o2air P

PRsss

Substituting,

hp49.6=Btu/s35.1R)Btu/lbmR)(0.01193(520Btu/lbm124.27)(226.11lbm/s)(22/60inrev,W

Discussion Note that this is the minimum power input needed for this compressor.

8-45

8-63 Steam expands in a turbine from a specified state to another specified state. The actual power outputof the turbine is given. The reversible power output and the second-law efficiency are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The temperature of the surroundings is given to be 25 C.

Properties From the steam tables (Tables A-4 through A-6)

KkJ/kg1693.7kJ/kg8.3658

C600MPa6

1

1

1

1

sh

TP

KkJ/kg6953.7kJ/kg4.2682

C100kPa50

2

2

2

2

sh

TP

Analysis (b) There is only one inlet and one exit, and thus m m m1 2 . We take the turbine as the system,which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed in the rate form as

outin


(steady)0system


outin 0

EE

EEE 80 m/s 6 MPa 600 C

50 kPa 100 C

140 m/s

STEAM

2

)2/()2/(2

22

121out

222out

211

VVhhmW

VhmWVhm

5 MW

Substituting,

kg/s5.156s/m1000

kJ/kg12

m/s)140(m/s)80(4.26828.3658kJ/s5000

22

22

m

m

The reversible (or maximum) power output is determined from the rate form of the exergy balance appliedon the turbine and setting the exergy destruction term equal to zero,

]peke)()[()(

0

002102121outrev,

2outrev,1

outin


(steady)0system




outin

ssThhmmW

mWm

XX

XXXX

Substituting,

kW5842KkJ/kg)7.6953K)(7.1693(2984.26823658.8kg/s)156.5()]()[( 21021outrev, ssThhmW

(b) The second-law efficiency of a turbine is the ratio of the actual work output to the reversible work,

85.6%MW842.5

MW5

outrev,

outII W

W

8-46

Discussion Note that 14.4% percent of the work potential of the steam is wasted as it flows through theturbine during this process.

8-47

8-64 Steam is throttled from a specified state to a specified pressure. The decrease in the exergy of thesteam during this throttling process is to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The temperature of the surroundings is given to be 25 C. 4 Heat transfer is negligible.

Properties The properties of steam before and after throttling are (Tables A-4 through A-6)

KkJ/kg6603.6kJ/kg4.3387

C500MPa9

1

1

1

1

sh

TP

Steam

2

1KkJ/kg7687.6MPa7

212

2 shh

P

Analysis The decrease in exergy is of the steam is the difference between the

inlet and exit flow exergies,

kJ/kg32.3KkJ/kg)6603.6K)(6.7687(298

)()]([exergyinDecrease 12021021000

ssTssTpekeh


8-48

8-65 Combustion gases expand in a turbine from a specified state to another specified state. The exergy of the gases at the inlet and the reversible work output of the turbine are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The temperature of the surroundings is given to be 25 C. 4 The combustion gases are ideal gases with constant specific heats.

Properties The constant pressure specific heat and the specific heat ratio are given to be cp = 1.15 kJ/kg.Kand k = 1.3. The gas constant R is determined from

KkJ/kg265.01/1.3)K)(1kJ/kg15.1()/11(/ kckccccR pppp v

Analysis (a) The exergy of the gases at the turbine inlet is simply the flow exergy,

400 kPa 650 C

800 kPa 900 C0

1

21

010011 2)( gz

VssThh

GASTURBINE

where

KkJ/kg1.025kPa100kPa800K)lnkJ/kg(0.265

K298K1173K)lnkJ/kg(1.15

lnln0

1

0

101 P

PR

TT

css p

Thus,

kJ/kg705.8

22

2

1s/m1000

kJ/kg12m/s)(100+K)kJ/kgK)(1.025298(C25)00kJ/kg.K)(915.1(

(b) The reversible (or maximum) work output is determined from an exergy balance applied on the turbineand setting the exergy destruction term equal to zero,

]peke)()[()(

0

02102121outrev,

2outrev,1

outin


(steady)0system




outin

ssThhmmW

mWm

XX

XXXX

where

kJ/kg2.19s/m1000

kJ/kg12

m/s)100(m/s)220(2

ke22

2221

22 VV

and


K1173K923K)lnkJ/kg(1.15

lnln1

2

1

212 P

PR

TT

css p

Then the reversible work output on a unit mass basis becomes

8-49

kJ/kg240.9kJ/kg2.19K)kJ/kg09196.0K)((298+C)650K)(900kJ/kg15.1(

ke)()(ke)( 1202112021outrev, ssTTTcssThhw p

8-50

8-66E Refrigerant-134a enters an adiabatic compressor with an isentropic efficiency of 0.80 at a specifiedstate with a specified volume flow rate, and leaves at a specified pressure. The actual power input and thesecond-law efficiency to the compressor are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.Properties From the refrigerant tables (Tables A-11E through A-13E)

/lbmft1.5492=RBtu/lbm0.2238=

lbm/Btu105.32=

sat.vaporpsia30

3psia30@1

psia30@1

psia30@11

g

g

g

sshh

P

vv

30 psia sat. vapor

R-134a20 ft3/min

70 psia s2 = s1

Btu/lbm80.112psia70

212

2s

sh

ssP

Analysis From the isentropic efficiency relation,

Btu/lbm67.11480.0/)32.10580.112(32.105

/)( 121212

12csa

a

sc hhhh

hhhh

Then,

Btu/lbm2274.067.114

psia702

2

2 shP

a

Also, lbm/s2152.0lbm/ft5492.1

s/ft60/203

3

1

1

v

Vm

There is only one inlet and one exit, and thus m m m1 2 . We take the actual compressor as the system, which is a control volume. The energy balance for this steady-flow system can be expressed as

0

outin


(steady)0system


outin

EE

EEE

( )

W mh mh Q ke pe

W m h ha

a

,in

,in

(since 0)1 2

2 1

Substituting, the actual power input to the compressor becomes

hp2.85Btu/s0.7068

hp1Btu/lbm)32.105.67lbm/s)(1142152.0(ina,W

(b) The reversible (or minimum) power input is determined from the exergy balance applied on thecompressor and setting the exergy destruction term equal to zero,

]peke)()[()(

0

001201212inrev,

21inrev,

outin


(steady)0system




outin

ssThhmmW

mmW

XX

XXXX

Substituting,Btu/s)0.7068=hp1(since27=Btu/s.6061

RBtu/lbm0.2238)2274.0R)((535Btu/lbm105.32)67.114(lbm/s)(0.2152inrev,

hp2.W

8-51

Thus, 79.8%hp85.2hp27.2

inact,

inrev,II W

W

8-52

8-67 Refrigerant-134a is compressed by an adiabatic compressor from a specified state to another specifiedstate. The isentropic efficiency and the second-law efficiency of the compressor are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.

Properties From the refrigerant tables (Tables A-11E through A-13E)

/kgm14605.0KkJ/kg97236.0

kJ/kg36.246

C10kPa140

31

1

1

1

1

v

sh

TP

0.5 kW

700 kPa 60 C

R-134aKkJ/kg0256.1

kJ/kg42.298C60kPa700

2

2

2

2

sh

TP

kJ/kg16.281kPa700

212

2s

s

s hss

P

140 kPa -10 C

Analysis (a) The isentropic efficiency is

66.8%668.036.24642.29836.24616.281

12

12

hhhh

a

sc

(b) There is only one inlet and one exit, and thus m m m1 2 . We take the actual compressor as thesystem, which is a control volume. The energy balance for this steady-flow system can be expressed as

outin


(steady)0system


outin 0

EE

EEE

( )

W mh mh Q ke pe

W m h ha

a

,in

,in

(since 0)1 2

2 1

Then the mass flow rate of the refrigerant becomes

kg/s009603.0kJ/kg)36.24642.298(

kJ/s5.0

12

ina,

hhW

ma

The reversible (or minimum) power input is determined from the exergy balance applied on thecompressor and setting the exergy destruction term equal to zero,

]peke)()[()(

0

001201212inrev,

21inrev,

outin


(steady)0system




outin

ssThhmmW

mmW

XX

XXXX

Substituting,

W kW347.0KkJ/kg)0.97236K)(1.0256300(kJ/kg)246.36(298.42kg/s)(0.009603inrev,

and

8-53

69.3%kW5.0

kW347.0

ina,

inrev,II W

W

8-54

8-68 Air is compressed steadily by a compressor from a specified state to another specified state. The increase in the exergy of air and the rate of exergy destruction are to be determined.


Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). From the air table (Table A-17)

600 kPa 277 C

95 kPa 27 C

AIR0.06 kg/s

KkJ/kg318.2kJ/kg74.555K550

KkJ/kg702.1kJ/kg19.300K300

o2

22

o1

11

shT

shT

Analysis The reversible (or minimum) power input is determinedfrom the rate form of the exergy balance applied on the compressor and setting the exergy destruction term equal to zero,

]peke)()[()(

0

001201212inrev,

2inrev,1

outin


(steady)0system




outin

ssThhmmW

mWm

XX

XXXX

where

s s s s R PP

o o2 1 2 1

2

1ln

(2.318 1.702) kJ / kg K (0.287 kJ / kg K)ln 600 kPa95 kPa

0.0870 kJ / kg K

Substituting,

kW13.7K)kJ/kg0870.0K)((298-kJ/kg)300.19(555.74kg/s)06.0(inrev,W

Discussion Note that a minimum of 13.7 kW of power input is needed for this compression process.

8-55

8-69 EES Problem 8-68 is reconsidered. The effect of compressor exit pressure on reversible power is tobe investigated.


T_1=27 [C] P_1=95 [kPa] m_dot = 0.06 [kg/s] {P_2=600 [kPa]}T_2=277 [C] T_o=25 [C] P_o=100 [kPa] m_dot_in=m_dot

"Steady-flow conservation of mass"m_dot_in = m_dot_out h_1 =enthalpy(air,T=T_1)h_2 = enthalpy(air, T=T_2)W_dot_rev=m_dot_in*(h_2 - h_1 -(T_1+273.15)*(s_2-s_1))s_1=entropy(air, T=T_1,P=P_1) s_2=entropy(air,T=T_2,P=P_2)

P2 [kPa] Wrev [kW]200 8.025250 9.179300 10.12350 10.92400 11.61450 12.22500 12.76550 13.25600 13.7

200 250 300 350 400 450 500 550 6008

9

10

11

12

13

14

P2 [kPa]

Wrev

[kW]

8-56

8-70 Argon enters an adiabatic compressor at a specified state, and leaves at another specified state. The reversible power input and irreversibility are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Argon is an ideal gas withconstant specific heats.

Properties For argon, the gas constant is R = 0.2081 kJ/kg.K; the specific heat ratio is k = 1.667; theconstant pressure specific heat is cp = 0.5203 kJ/kg.K (Table A-2).

Analysis The mass flow rate, the entropy change, and the kinetic energy change of argon during thisprocess are

kg/s0.495=m/s)20)(m0130.0(kg/m5255.0

11

/m5255.0kPa)120(

K)K)(303kg/mkPa2081.0(

2311

1

33

1

11

Vm

kgP

RT

Av

v 1.2 MPa530 C80 m/s

120 kPa 30 C

20 m/s

ARGON


K303K803K)lnkJ/kg(0.5203

lnln1

2

1

212 P

PR

TT

css p

and

kJ/kg0.3s/m1000

kJ/kg12

m/s)20(m/s)80(2

ke22

2221

22 VV

The reversible (or minimum) power input is determined from the rate form of the exergy balance appliedon the compressor, and setting the exergy destruction term equal to zero,

]peke)()[()(

0

01201212inrev,

2inrev,1

outin


(steady)0system




outin

ssThhmmW

mWm

XX

XXXX

Substituting,

kW1263.0+K)kJ/kgK)(0.02793298(K)30K)(530kJ/kg(0.5203kg/s)(0.495

ke)()( 12012inrev, ssTTTcmW p

The exergy destruction (or irreversibility) can be determined from an exergy balance or directly from its definition Xdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on thesystem, which is an adiabatic steady-flow device,

)(0

0

12gengen21


0system


gen


outin

ssmSSsmsm

SSSS

Substituting,

8-57

kW4.12K)kJ/kg793kg/s)(0.02K)(0.495298(=)( 120destroyed ssmTX

8-58

8-71 Steam expands in a turbine steadily at a specified rate from a specified state to another specified state. The power potential of the steam at the inlet conditions and the reversible power output are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The temperature of the surroundings is given to be 25 C.

Properties From the steam tables (Tables A-4 through 6)

KkJ/kg5579.6kJ/kg3.3273

C450MPa8

1

1

1

1

sh

TP 8 MPa

450 C

50 kPa sat. vapor

STEAM15,000 kg/h

KkJ/kg5931.7kJ/kg2.2645

vaporsat.kPa50

2

22

shP

KkJ/kg36723.0kJ/kg83.104

C25kPa100

C25@0

C25@0

0

0

f

f

sshh

TP

Analysis (a) The power potential of the steam at the inlet conditions is equivalent to its exergy at the inlet state,

kW5515KkJ/kg0.36723)-K)(6.5579298(kJ/kg)83.104(3273.3kg/s)3600/000,15(

)(2

)( 010011

021

0100110

ssThhmgzV

ssThhmm

(b) The power output of the turbine if there were no irreversibilities is the reversible power, is determinedfrom the rate form of the exergy balance applied on the turbine and setting the exergy destruction termequal to zero,

]peke)()[()(

0

002102121outrev,

2outrev,1

outin


(steady)0system




outin

ssThhmmW

mWm

XX

XXXX

Substituting,

kW3902KkJ/kg)5931.7K)(6.5579(298kJ/kg)2.2645(3273.3kg/s)00(15,000/36

)]()[( 21021outrev, ssThhmW

8-59

8-72E Air is compressed steadily by a 400-hp compressor from a specified state to another specified statewhile being cooled by the ambient air. The mass flow rate of air and the part of input power that is used tojust overcome the irreversibilities are to be determined.Assumptions 1 Air is an ideal gas with variable specific heats. 2 Potential energy changes are negligible. 3 Thetemperature of the surroundings is given to be 60 F.

15 psia 60 F

AIR

350 ft/s150 psia620 FProperties The gas constant of air is R = 0.06855 Btu/lbm.R

(Table A-1E). From the air table (Table A-17E) 1,500 Btu/min

RBtu/lbm59173.0Btu/lbm27.124

psia15R520

o1

1

1

1

sh

PT

400 hp


psia150R1080

o1

2

2

2

sh

PT

Analysis (a) There is only one inlet and one exit, and thus m m m1 2 .We take the actual compressor as the system, which is a control volume.The energy balance for this steady-flow system can be expressed as

outin


(steady)0system


outin 0

EE

EEE

2)2/()2/(

21

22

12outin,out2

222

11in,VV

hhmQWQVhmVhmW aa

Substituting, the mass flow rate of the refrigerant becomes

22

2

s/ft25,037Btu/lbm1

20ft/s)350(27.12497.260Btu/s)60/1500(

hp1Btu/s0.7068hp)400( m

It yields m 1.852 lbm / s(b) The portion of the power output that is used just to overcome the irreversibilities is equivalent to exergydestruction, which can be determined from an exergy balance or directly from its definition

where the entropy generation Sgen0destroyed STX gen is determined from an entropy balance on an

extended system that includes the device and its immediate surroundings. It gives

0

out12gengen

surrb,

out21


0system


gen


outin

0

0

TQ

ssmSSTQ

smsm

SSSS

where

Btu/lbm.R02007.0psia15psia150lnBtu/lbm.R)(0.06855Btu/lbm)59173.076964.0(ln

1

201

0212 P

PRssss


8-60

hp62.72Btu/s0.7068

hp1R520Btu/s60/1500+R)Btu/lbm2007lbm/s)(0.0(1.852R)520(

)(0


QssmTSTX

8-61

8-73 Hot combustion gases are accelerated in an adiabatic nozzle. The exit velocity and the decrease in the exergy of the gases are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 The combustion gases are ideal gases with constant specific heats.

Properties The constant pressure specific heat and the specific heat ratio are given to be cp = 1.15 kJ/kg.Kand k = 1.3. The gas constant R is determined from

KkJ/kg2654.01/1.3)K)(1kJ/kg15.1()/11(/ kckccccR pppp v

Analysis (a) There is only one inlet and one exit, and thus m m m1 2 . We take the nozzle as the system, which is a control volume. The energy balance for this steady-flow system can be expressed as

outin


(steady)0system


outin 0

EE

EEE

Comb.gases

260 kPa747 C80 /

70 kPa500 C

2

0)pe(since/2)V+()2/(2

12

212

222

211

VVhh

QWhmVhm

Then the exit velocity becomes

m/s758

222

21212

m/s)80(kJ/kg1

/sm1000K)500K)(747kJ/kg15.1(2

)(2 VTTcV p

(b) The decrease in exergy of combustion gases is simply the difference between the initial and finalvalues of flow exergy, and is determined to be

ke)()()(peke 1202112021rev210

ssTTTcssThhw p

where

kJ/kg1.284s/m1000

kJ/kg12

m/s)80(m/s)758(2

ke22

2221

22 VV

and

KkJ/kg02938.0kPa260

kPa70lnK)kJ/kg(0.2654K1020K773lnK)kJ/kg15.1(

lnln1

2

1

212 P

PR

TT

css p

Substituting,

kJ/kg8.56kJ/kg1.284K)kJ/kgK)(0.02938(293+C)500K)(747kJ/kg(1.15

exergyinDecrease 21

8-62

8-74 Steam is accelerated in an adiabatic nozzle. The exit velocity of the steam, the isentropic efficiency, and the exergy destroyed within the nozzle are to be determined.

Assumptions 1 The nozzle operates steadily. 2 The changes in potential energies are negligible.

Properties The properties of steam at the inlet and the exit of the nozzle are (Tables A-4 through A-6)

KkJ/kg8000.6kJ/kg4.3411

C500MPa7

1

1

1

1

sh

TP

STEAM7 MPa 500 C70 /

KkJ/kg8210.6kJ/kg2.3317

C450MPa5

2

2

2

2

sh

TP 5 MPa

450 C

kJ/kg0.3302MPa5

212

2s

s

s hss

P

Analysis (a) We take the nozzle to be the system, which is a control volume. The energy balance for thissteady-flow system can be expressed in the rate form as

outin


(steady)0system


outin 0

EE

EEE

20

0)pe(since/2)V+()2/(2

12

212

222

211

VVhh

QWhmVhm

Then the exit velocity becomes

m/s439.6V 222

21212 m/s)70(

kJ/kg1/sm1000kJ/kg)2.33174.3411(2)(2 hhV

(b) The exit velocity for the isentropic case is determined from

m/s472.9m/s)70(kJ/kg1

/sm1000kJ/kg)0.33024.3411(2)(2 222

21212 Vss hhV

Thus,

86.4%2/m/s)9.472(2/m/s)6.439(

2/2/

2

2

22

22

sN

VV

(c) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition where the entropy generation SX T gendestroyed 0S gen is determined from an entropy balance on

the actual nozzle. It gives

12gen12gengen21


0system


gen


outin

or0

0

sssssmSSsmsm

SSSS

Substituting, the exergy destruction in the nozzle on a unit mass basis is determined to be

kJ/kg6.28KkJ/kg)8000.6K)(6.8210298()( 120gen0destroyed ssTsTx

8-63

8-75 CO2 gas is compressed steadily by a compressor from a specified state to another specified state. The power input to the compressor if the process involved no irreversibilities is to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 CO2 is anideal gas with constant specific heats.

Properties At the average temperature of (300 + 450)/2 = 375 K, the constant pressure specific heat and thespecific heat ratio of CO2 are k = 1.261 and cp = 0.917 kJ/kg.K (Table A-2).

Analysis The reversible (or minimum) power input is determined from the exergy balance applied on thecompressor, and setting the exergy destruction term equal to zero,

]peke)()[(

)(

0

0012012

12inrev,

2inrev,1

outin


(steady)0system




outin

ssThhm

mW

mWm

XX

XXXX600 kPa450 K

CO2

0.2 kg/s

where 100 kPa 300 K

KkJ/kg03335.0kPa100kPa600lnK)kJ/kg(0.1889

K300K450lnK)kJ/kg9175.0(

lnln1

2

1

212 P

PR

TT

css p

Substituting,

kW25.5K)kJ/kgK)(0.03335(298K)300K)(450kJ/kg(0.917kg/s)(0.2inrev,W

Discussion Note that a minimum of 25.5 kW of power input is needed for this compressor.

8-64

8-76E A hot water stream is mixed with a cold water stream. For a specified mixture temperature, the mass flow rate of cold water stream and the rate of exergy destruction are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The mixing chamber is well-insulated so that heat lossto the surroundings is negligible. 3 Changes in the kinetic and potential energies of fluid streams are negligible.

Properties Noting that that T < Tsat@50 psia = 280.99 F, the water in all three streams exists as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus from Table A-4E,


F160psia50

F160@1

F160@1

1

1

f

f

sshh

TP

110 FMixing

ChamberH2O

50 psia 70 F

160 F4 lbm/s


F70psia50

F70@2

F70@2

2

2

f

f

sshh

TP


F110psia50

F110@3

F110@3

3

3

f

f

sshh

TP

Analysis (a) We take the mixing chamber as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as

Mass balance: 321(steady)0

systemoutin 0 mmmmmm

Energy balance:

0)peke(since

0

33221

outin


(steady)0system


outin

WQhmhmhm

EE

EEE

Combining the two relations gives 3212211 hmmhmhm

Solving for and substituting, the mass flow rate of cold water stream is determined to bem2

lbm/s5.0=lbm/s)0.4(Btu/lbm)08.3802.78(Btu/lbm)02.7800.128(

123

312 m

hhhh

m

Also,

m m m3 1 2 4 5 9 lbm / s

(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition where the entropy generation Sgen0destroyed STX gen is determined from an entropy balance on

the mixing chamber. It gives

221133gengen332211


0system


gen


outin

=0

0

smsmsmSSsmsmsm

SSSS


8-65

Btu/s14.7RBtu/s)23136.00.407459.00.50.14728R)(9.0535(

)( 1122330gen0destroyed smsmsmTSTX

8-66

8-77 Liquid water is heated in a chamber by mixing it with superheated steam. For a specified mixing temperature, the mass flow rate of the steam and the rate of exergy destruction are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions.

Properties Noting that T < Tsat @ 200 kPa = 120.23 C, the cold water and the exit mixture streams exist as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. From TablesA-4 through A-6,

KkJ/kg0.83130kJ/kg251.18

C60kPa200

KkJ/kg7.8941kJ/kg3072.1

C300kPa200

KkJ/kg0.29649kJ/kg83.91

C20kPa200

C60@3

C60@3

3

3

2

2

2

2

C20@1

C20@1

1

1

f

f

f

f

sshh

TP

sh

TP

sshh

TP

2

MIXINGCHAMBER

200 kPa

20 C2.5 kg/s

300 C

1

600 kJ/min

60 C 3

Analysis (a) We take the mixing chamber as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as

Mass balance: 321(steady)0

systemoutin 0 mmmmmm

Energy balance:

33out221

outin


(steady)0system


outin 0

hmQhmhm

EE

EEE

Combining the two relations gives 3223113212211out hhmhhmhmmhmhmQ

Solving for and substituting, the mass flow rate of the superheated steam is determined to bem2

kg/s0.148kJ/kg18.2513072.1

kJ/kg18.25183.91kg/s2.5kJ/s)(600/60

32

311out2 hh

hhmQm

Also, kg/s2.6480.1482.5213 mmm

(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition where the entropy generation Sgen0destroyed STX gen is determined from an entropy balance on

an extended system that includes the mixing chamber and its immediate surroundings. It gives

0

out221133gengen

surrb,

out332211


0system


gen


outin

0

0

TQ

smsmsmSSTQ

smsmsm

SSSS


8-67

kW96.4kW/K)298/1029649.05.28941.7148.00.83130K)(2.648298(,

1122330gen0destroyedsurrb

out

TQ

smsmsmTSTX

8-68

8-78 Refrigerant-134a is vaporized by air in the evaporator of an air-conditioner. For specified flow rates, the exit temperature of air and the rate of exergy destruction are to be determined for the cases of insulatedand uninsulated evaporator.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions. 4 Air is an ideal gas with constant specificheats at room temperature.Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The constant pressure specific heat of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). The properties of R-134a at the inlet and the exitstates are (Tables A-11 through A-13)

KkJ/kg34926.0)85503.0(3.009275.0kJ/kg83.8648.2143.049.22

3.0kPa120

11

11

1

1

fgf

fgf

sxsshxhh

xP

KkJ/kg94779.0kJ/kg97.236

vaporsat.kPa120

kPa120@2

kPa120@22

g

g

sshhT

sat. vapor

2 kg/min

2

1R-134a

AIR

4

36 m3/minAnalysis Air at specified conditions can be treated as an

ideal gas with specific heats at room temperature. The properties of the refrigerant are

kg/min6.97K300K/kgmkPa0.287

/minm6kPa1003

3

3

33air RT

Pm

V

(a) We take the entire heat exchanger as the system, which is a control volume. The mass and energybalances for this steady-flow system can be expressed in the rate form as: Mass balance ( for each fluid stream):

Rmmmmmmmmmmm 43air21outin(steady)0

systemoutin and0

Energy balance (for the entire heat exchanger):

0)peke(since

0

44223311

outin


(steady)0system


outin

WQhmhmhmhm

EE

EEE

Combining the two, 43air43air12 TTcmhhmhhm pR

Solving for T4,p

R

cmhhm

TTair

1234

Substituting, K257.1=K)kJ/kg005kg/min)(1.97.6(

kJ/kg)83.866.97kg/min)(23(2C274 C15.9T

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition where the entropy generation Sgen0destroyed STX gen is determined from an entropy balance on

the evaporator. Noting that the condenser is well-insulated and thus heat transfer is negligible, the entropybalance for this steady-flow system can be expressed as

8-69

0

)0(since0

gen4air23air1

gen44223311


(steady)0system


gen


outin

Ssmsmsmsm

QSsmsmsmsm

SSSS

RR

or, 34air12gen ssmssmS R

where KkJ/kg1551.0K300K257.1lnK)kJ/kg005.1(lnlnln

3

4

3

4

3

434

0

TT

cP

PR

TT

css pp


kW0.59kJ/min35.4KkJ/kg0.1551kg/min6.97KkJ/kg34926.00.94779kg/min2K)305(

)]()([ 34air120gen0destroyed ssmssmTSTX R

(b) When there is a heat gain from the surroundings, the steady-flow energy equation reduces to

34air12in TTcmhhmQ pR

Solving for T4,p

R

cmhhmQ

TTair

12in34

Substituting, K261.4K)kJ/kg005kg/min)(1.97.6(

kJ/kg)83.866.97kg/min)(23(2kJ/min)(30+C274 C11.6T

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition where the entropy generation SX T gendestroyed 0S gen is determined from an entropy balance on an

extended system that includes the evaporator and its immediate surroundings. It gives

0

0

gen4air23air10

in

gen44223311inb,

in


(steady)0system



in

SsmsmsmsmTQ

SsmsmsmsmTQ

SSSS

RR

genout

or0

in34air12gen T

QssmssmS R

where KkJ/kg1384.0K300K261.4lnK)kJ/kg005.1(lnln

3

4

3

434

0

PP

RTT

css p


8-70

kW0.68kJ/min40.9K305

kJ/min30KkJ/kg0.1384kg/min6.97KkJ/kg34926.00.94779kg/min2K)305(

)()(0

in34120gen0destroyed T

QssmssmTSTX airR

8-60

8-79 A rigid tank initially contains saturated R-134a vapor. The tank is connected to a supply line, and R-134a is allowed to enter the tank. The mass of the R-134a that entered the tank and the exergy destroyedduring this process are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Thedirection of heat transfer is to the tank (will be verified).Properties The properties of refrigerant are (Tables A-11 through A-13)

KkJ/kg0.91303=kJ/kg253.81=

kg/m0.01672=

vaporsat.MPa2.1

MPa2.1@1

MPa2.1@1

3MPa2.1@1

1

g

g

g

ssuu

Pvv

Q

R-134a

R-134a0.1 m3

1.2 MPa Sat. vapor

1.6 MPa 30 C

KkJ/kg0.45315=kJ/kg125.94=

kg/m0.0009166=

liquidsat.MPa4.1

MPa4.1@2

MPa4.1@2

3MPa4.1@2

2

f

f

f

ssuu

Tvv

KkJ/kg34554.0kJ/kg56.93

C30MPa6.1

i

i

i

i

sh

TP

Analysis We take the tank as the system, which is a control volume. Noting that the microscopic energiesof flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 12systemoutin mmmmmm i

Energy balance:

)0peke(since1122in


system


outin

WumumhmQ

EEE

ii

(a) The initial and the final masses in the tank are

kg109.10/kgm0.0009166

m0.1kg.9835

/kgm0.01672m0.1

3

3

2

223

3

1

11 v

VvV

mm

Then from the mass balance kg103.11983.510.10912 mmmi

The heat transfer during this process is determined from the energy balance to be

kJ2573kJ/kg253.81kg5.983kJ/kg125.94)10.109(kJ/kg)56.93(kg103.111122in umumhmQ ii

(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition . The entropy generation Sgen0destroyed STX gen in this case is determined from an entropybalance on an extended system that includes the tank and its immediate surroundings so that the boundarytemperature of the extended system is the surroundings temperature Tsurr at all times. It gives

0

in1122gen

tank1122tankgeninb,

in

entropyinChange

system

generationEntropy

gen


outin )(=

TQ

smsmsmS

smsmSSsmTQ

SSSS

ii

ii


kJ80.3

K)kJ)/(3182573(0.3455411.10391303.0983.545315.0109.10K)318(0


QsmsmsmTSTX ii

8-61

8-80 A rigid tank initially contains saturated liquid water. A valve at the bottom of the tank is opened, andhalf of mass in liquid form is withdrawn from the tank. The temperature in the tank is maintained constant.The amount of heat transfer, the reversible work, and the exergy destruction during this process are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remainsconstant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Thedirection of heat transfer is to the tank (will be verified).

H2O0.6 m3

170 CT = const.

QProperties The properties of water are (Tables A-4 through A-6)

KkJ/kg2.0417kJ/kg.08719

liquidsat.C017

KkJ/kg2.0417kJ/kg.20718

/kgm0.001114

liquidsat.C017

C017@

C017@

C017@1

C017@1

3C017@1

1

fe

fee

f

f

f

sshhT

ssuu

Tvv

me

Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e

Energy balance:

)0peke(since1122in


system


outin

WumumhmQ

EEE

ee

The initial and the final masses in the tank are

emmm

m

=kg.24269kg538.4721

21

kg.47538/kgm0.001114

m0.6

12

3

3

11 v

V

Now we determine the final internal energy and entropy,

KkJ/kg2.06304.62330.0046142.0417kJ/kg.777261857.50.004614.20718

004614.0C017

0.0046140.0011140.242600.0011140.002229

/kgm0.002229kg269.24

m0.6

22

22

2

2

22

33

22

fgf

fgf

fg

f

sxssuxuu

xT

x

m

v

vv

Vv

The heat transfer during this process is determined by substituting these values into the energy balanceequation,

kJ2545kJ/kg718.20kg538.47kJ/kg726.77kg269.24kJ/kg719.08kg269.24

1122in umumhmQ ee

8-62

(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition . The entropy generation Sgen0destroyed STX gen in this case is determined from an entropybalance on an extended system that includes the tank and the region between the tank and the source so thatthe boundary temperature of the extended system at the location of heat transfer is the source temperatureTsource at all times. It gives

source

in1122gen

tank1122tankgeninb,

in

entropyinChange

system

generationEntropy

gen


outin

)(=

SS

TQ

smsmsmS

smsmSSsmTQ

SS

ee

ee


kJ141.2

K)kJ)/(5232545(0417.2269.24+0417.247.5380630.224.269K)298(source


QsmsmsmTSTX ee

For processes that involve no actual work, the reversible work output and exergy destruction are identical.Therefore,

kJ141.2destroyedoutrev,outact,outrev,destroyed XWWWX

8-63

8-81E An insulated rigid tank equipped with an electric heater initially contains pressurized air. A valve is opened, and air is allowed to escape at constant temperature until the pressure inside drops to 30 psia. Theamount of electrical work done and the exergy destroys are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the exit temperature (and enthalpy) of airremains constant. 2 Kinetic and potential energies are negligible. 3 The tank is insulated and thus heattransfer is negligible. 4 Air is an ideal gas with variable specific heats. 5 The environment temperature is given to be 70 F.Properties The gas constant of air is R = 0.3704 psia.ft3/lbm.R (Table A-1E). The properties of air are (Table A-17E)

Btu/lbm34.102R600

Btu/lbm34.102R600Btu/lbm,47.143R600

22

11

uT

uThT ee

Analysis We take the tank as the system, which is a control volume. Noting that the microscopic energiesof flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e

We

AIR150 ft3

75 psia 140 F

Energy balance:

)0peke(since1122ine,


system


outin

QumumhmW

EEE

ee

The initial and the final masses of air in the tank are

lbm62.50)RR)(600/lbmftpsia3704.0(

)ftpsia)(15075(3

3

1

11 RT

Pm

V

lbm25.20)RR)(600/lbmftpsia3704.0(

)ftpsia)(15030(3

3

2

22 RT

Pm

V

Then from the mass and energy balances, m m me 1 2 50 62 20 25 30 37. . . lbm

Btu1249Btu/lbm)4lbm)(102.362.50(Btu/lbm)4lbm)(102.3(20.25Btu/lbm)7lbm)(143.437.30(

1122ine, umumhmW ee

(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition where the entropy generation Sgen0destroyed STX gen is determined from an entropy balance on the insulated tank. It gives

)()()(

)(=

11222111221122gen

tank1122tankgen

entropyinChange

system

generationEntropy

gen


outin

eeeee

ee

ssmssmsmmsmsmsmsmsmS

smsmSSsmSSSS

Assuming a constant average pressure of (75 + 30) / 2 = 52.5 psia for the exit stream, the entropy changes are determined to be

RBtu/lbm0.02445=psia52.5

psia75lnR)Btu/lbm06855.0(lnln

RBtu/lbm0.03836=psia52.5

psia30lnR)Btu/lbm06855.0(lnln

111

222

0

0

eepe

eepe

PP

RT

Tcss

PP

RT

Tcss


Btu1068R)Btu/lbm02445.0lbm)(62.50(R)Btu/lbm36lbm)(0.038(20.25R)530()]()([ 11220gen0destroyed ee ssmssmTSTX

8-64

8-82 A rigid tank initially contains saturated liquid-vapor mixture of refrigerant-134a. A valve at the bottom of the tank is opened, and liquid is withdrawn from the tank at constant pressure until no liquidremains inside. The final mass in the tank and the reversible work associated with this process are to bedetermined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess. It can be analyzed as a uniform-flow process since the state of fluid leaving the device remainsconstant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Thedirection of heat transfer is to the tank (will be verified).Properties The properties of R-134a are (Tables A-11 through A-13)

Source60 C

Q

R-134a800 kPa

P = const.

kJ/kg.K35404.0kJ/kg95.47

liquidsat.kPa800

KkJ/kg0.91835=kJ/kg246.79=

kg/m0.02562=

vaporsat.kPa800

kJ/kg.K91835.0=kJ/kg.K,43540.0

kJ/kg79.246=kJ/kg,79.94

/kgm0.025621=/kg,m0.0008458kPa800

kPa800@

kPa800@

kPa800@2

kPa800@2

3kPa800@2

2

331

fe

fee

g

g

g

gf

gf

gf

sshhP

ssuu

P

ss

uu

P

vv

vv

me

Analysis (b) We take the tank as the system, which is a control volume. Noting that the microscopicenergies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u,respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e

Energy balance:

)0peke(since1122in


system


outin

WumumhmQ

EEE

ee

The initial mass, initial internal energy, and final mass in the tank are

kg202.38732.2470.35kg/m025621.0

m7.01.0kg/m0008458.0

m3.01.03

3

3

3

1g

g

f

fgf mmm

v

V

v

V

kJ/K067.1591835.0732.235404.0470.35kJ4.403679.246732.279.94470.35

111

111

ggff

ggff

smsmsmSumumumU

kg3.903/kgm02562.0

m1.03

3

22 v

Vm

Then from the mass and energy balances, kg299.34903.3202.3821 mmme

kJ2201kJ4.4036kJ/kg)kg)(246.79(3.903+kJ/kg)kg)(95.47299.34(in .Q

(b) This process involves no actual work, thus the reversible work and exergy generation are identical since.destroyedoutrev,outact,outrev,destroyed XWWWX

The exergy destroyed during a process can be determined from an exergy balance or directly fromits definition . The entropy generation Sgen0destroyed STX gen in this case is determined from an entropybalance on an extended system that includes the tank and the region between the tank and the heat source so that the boundary temperature of the extended system at the location of heat transfer is the sourcetemperature Tsource at all times. It gives

8-65

source

in1122gen

tank1122tankgeninb,

in

entropyinChange

system

generationEntropy

gen


outin

)(=

TQ

smsmsmS

smsmSSsmTQ

SSSS

ee

ee

Substituting,

kJ16.87

)333/2.20135404.0299.34067.1591835.03.903K)298(source

in11220gen0destroyedoutrev, T

QsmsmsmTSTXW ee

That is, 16.87 kJ of work could have been produced during this process.

8-66

8-83 A cylinder initially contains helium gas at a specified pressure and temperature. A valve is opened,and helium is allowed to escape until its volume decreases by half. The work potential of the helium at theinitial state and the exergy destroyed during the process are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process by using constant average properties for thehelium leaving the tank. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved other than boundary work. 4 The tank is insulated and thus heat transfer is negligible. 5 Helium is an ideal gas with constant specific heats.Properties The gas constant of helium is R = 2.0769 kPa.m3/kg.K = 2.0769 kJ/kg.K. The specific heats of helium are cp = 5.1926 kJ/kg.K and cv = 3.1156 kJ/kg.K (Table A-2). Analysis (a) From the ideal gas relation, the initial and the final masses in the cylinder are determined to be

kg0493.0)KK)(293/kgmkPa0769.2(

)mkPa)(0.1300(3

3

1

11 RT

Pm

V

kg0247.02/0493.02/12 mmme

The work potential of helium at the initial state is simply the initial exergy of helium, and is determinedfrom the closed-system exergy relation,

)()()( 01001001111 vvPssTuumm

where

QHELIUM300 kPa 0.1 m3

20 C

/kgm405.6kPa95

)KK)(293/kgmkPa0769.2(

/kgm0284.2kPa300

)KK)(293/kgmkPa0769.2(

33

0

00

33

1

11

PRT

PRT

v

v

and

KkJ/kg28.2kPa100kPa300lnK)kJ/kg0769.2(

K293K293lnK)kJ/kg1926.5(=

lnln0

1

0

101 P

PR

TT

css p

Thus,

kJ12.44]}m[kJ/kPa/kgm)405.64kPa)(2.02895(+

K)kJ/kg28.2K)(293(C20)K)(20kJ/kg6kg){(3.1150493.0(33

1

(b) We take the cylinder as the system, which is a control volume. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the massand energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e

Energy balance:

1122inb,in


system


outin

umumWhmQ

EEE

ee

Combining the two relations gives

8-67

0)(

)()(

11221

112221

inb,112221in

hmmmmhmhmhmm

WumumhmmQ

e

e

since the boundary work and U combine into H for constant pressure expansion and compressionprocesses.

The exergy destroyed during a process can be determined from an exergy balance or directly fromits definition where the entropy generation Sgen0destroyed STX gen can be determined from an entropy balance on the cylinder. Noting that the pressure and temperature of helium in the cylinder are maintainedconstant during this process and heat transfer is zero, it gives

0

)()(

)(=

12112

211122

1122gen

cylinder1122cylindergen

entropyinChange

system

generationEntropy

gen


outin

smmmmsmmsmsm

smsmsmS

smsmSSsm

SSSS

e

ee

ee

since the initial, final, and the exit states are identical and thus se = s2 = s1. Therefore, this discharge process is reversible, and

0gen0destroyed STX

8-68

8-84 A rigid tank initially contains saturated R-134a vapor at a specified pressure. The tank is connected to a supply line, and R-134a is allowed to enter the tank. The amount of heat transfer with the surroundingsand the exergy destruction are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Thedirection of heat transfer is from the tank (will be verified).

R-134a

R-134a0.2 m3

1 MPaSat. vapor

1.4 MPa 60 C

Properties The properties of refrigerant are (Tables A-11 through A-13)

kg/m0.020313=KkJ/kg0.91558=

kJ/kg250.68=

sat.vaporMPa1

3MPa1@1

MPa1@1

MPa1@11

g

g

g

ssuu

P

vv

KkJ/kg93889.0kJ/kg47.285

C60MPa4.1

i

i

i

i

sh

TP Q

Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 12systemoutin mmmmmm i

Energy balance:

)0peke(since1122out


system


outin

WumumQhm

EEE

iiThe initial and the final masses in the tank are

kg846.9kg/m020313.0

m2.03

3

11 v

Vm

kg91.117983.593.111kg/m016715.0

m1.0kg/m0008934.0

m1.03

3

3

3

2g

g

f

fgf mmm

v

V

v

V

kJ/K967.5291303.0983.542441.093.111kJ581,1481.253983.570.11693.111

222

222

ggff

ggff

smsmsmSumumumU

Then from the mass and energy balances, kg06108.846.991.11712 mmmi

The heat transfer during this process is determined from the energy balance to be kJ18,73768.250.846914,58147.28506.1081122out umumhmQ ii

(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition . The entropy generation Sgen0destroyed STX gen in this case is determined from an entropybalance on an extended system that includes the cylinder and its immediate surroundings so that theboundary temperature of the extended system is the surroundings temperature Tsurr at all times. It gives

0

out1122gen

tank1122tankgenoutb,

out

entropyinChange

system

generationEntropy

gen


outin )(=

TQ

smsmsmS

smsmSSsmTQ

SSSS

ii

ii


kJ1599]298/737,180.9388906.10891558.0846.9K)[52.967298(0


QsmsmsmTSTX ii

8-69

8-85 An insulated cylinder initially contains saturated liquid-vapor mixture of water. The cylinder is connected to a supply line, and the steam is allowed to enter the cylinder until all the liquid is vaporized. The amount of steam that entered the cylinder and the exergy destroyed are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 The expansion process is quasi-equilibrium. 3 Kinetic and potential energies are negligible. 4 The device is insulated and thus heat transfer is negligible.Properties The properties of steam are (Tables A-4 through A-6)

KkJ/kg8883.45968.56.05302.1kJ/kg6.18256.22016.071.504

6.015/9kPa200

11

11

1

1

fgf

fgf

sxsshxhh

xP

KkJ/kg1270.7kJ/kg3.2706

sat.vaporkPa200

kPa200@2

kPa200@22

g

g

sshhP

1 MPa 400 C

H2O200 kPa

P = const.KkJ/kg4670.7

kJ/kg5.3264C400

MPa1

i

i

i

i

sh

TP

Analysis (a) We take the cylinder as the system, which is a controlvolume. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internalenergy u, respectively, the mass and energy balances for this unsteady-flow system can be expressed as Mass balance: 12systemoutin mmmmmm i

Energy balance:

)0peke(since1122outb,


system


outin

QumumWhm

EEE

ii

Combining the two relations gives 112212outb,0 umumhmmW i

or, 1122120 hmhmhmm i

since the boundary work and U combine into H for constant pressure expansion and compressionprocesses. Solving for m2 and substituting,

kg38.66=kg)15(kJ/kg)3.27065.3264(kJ/kg)6.18255.3264(

12

12 m

hhhh

mi

i

Thus,kg23.661566.3812 mmmi

(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition where the entropy generation Sgen0destroyed STX gen is determined from an entropy balance on the insulated cylinder,

iiii smsmsmSsmsmSSsm

SSSS

1122gen1122systemgen

entropyinChange

system

generationEntropy

gen


outin

=Substituting, the exergy destruction is determined to be

kJ7610)4670.766.238883.4151270.766.K)(38298(][ 11220gen0destroyed ii smsmsmTSTX

8-70

8-86 Each member of a family of four take a shower every day. The amount of exergy destroyed by thisfamily per year is to be determined.Assumptions 1 Steady operating conditions exist. 2 The kinetic and potential energies are negligible. 3Heat losses from the pipes, mixing section are negligible and thus Q 4 Showers operate at maximumflow conditions during the entire shower. 5 Each member of the household takes a shower every day. 6Water is an incompressible substance with constant properties at room temperature. 7 The efficiency of theelectric water heater is 100%.

.0

Properties The density and specific heat of water are at room temperature are = 1 kg/L = 1000 kg/3 and c = 4.18 kJ/kg. C (Table A-3).Analysis The mass flow rate of water at the shower head is

kg/min10=L/min)kg/L)(10(1== Vm

The mass balance for the mixing chamber can be expressed in the rate form as

0 321outin(steady)0

systemoutin mmmmmmmm

where the subscript 1 denotes the cold water stream, 2 the hot water stream, and 3 the mixture.The rate of entropy generation during this process can be determined by applying the rate form of

the entropy balance on a system that includes the electric water heater and the mixing chamber (the T-elbow). Noting that there is no entropy transfer associated with work transfer (electricity) and there is noheat transfer, the entropy balance for this steady-flow system can be expressed as

221133gen

gen332211


(steady)0system


gen


outin

free)entropyis work and0(since0

smsmsmS

QSsmsmsm

SSSS

Noting from mass balance that m m m1 2 3 and s2 = s1 since hot water enters the system at the sametemperature as the cold water, the rate of entropy generation is determined to be

kJ/min.K746.3273+15273+42lnkJ/kg.K)18kg/min)(4.10(

ln)()(1

3313312133gen T

TcmssmsmmsmS p

Noting that 4 people take a 6-min shower every day, the amount of entropy generated per year is

year)(perkJ/K32,815=days/year)65persons)(3day)(4min/person6kJ/min.K)(746.3(

days)of.people)(NoofNo.()( gengen tSS

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition ,gen0destroyed STX

kJ9,779,000)kJ/KK)(32,815298(gen0destroyed STX

Discussion The value above represents the exergy destroyed within the water heater and the T-elbow in theabsence of any heat losses. It does not include the exergy destroyed as the shower water at 42 C is discarded or cooled to the outdoor temperature. Also, an entropy balance on the mixing chamber alone(hot water entering at 55 C instead of 15 C) will exclude the exergy destroyed within the water heater.

8-71

8-87 Air is compressed in a steady-flow device isentropically. The work done, the exit exergy of compressed air, and the exergy of compressed air after it is cooled to ambient temperature are to be determined.Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 The process is givento be reversible and adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply. 3The environment temperature and pressure are given to be 300 K and 100 kPa. 4 The kinetic and potentialenergies are negligible. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The constant pressure specific heatand specific heat ratio of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2). Analysis (a) From the constant specific heats ideal gas isentropic relations,

K2.579kPa100kPa1000

K3004.1/4.0/)1(

1

212

kk

PP

TT 1 MPa s2 = s1

AIR

For a steady-flow isentropic compression process, the work input isdetermined from

kJ/kg280.5

1/100)1000(11.4

K300KkJ/kg0.2871.4

11

0.4/1.4

/112

1incomp,

kkPPkkRT

w

100 kPa 300 K

(b) The exergy of air at the compressor exit is simply the flow exergy at the exit state,

kJ/kg280.6=300)K-79.2kJ/kg.K)(5005.1(

)(

)isentropicis2-0proccess the(since2

)(

02

2

22

0200220

0

0

TTc

gzV

ssThh

p

which is the same as the compressor work input. This is not surprising since the compression process isreversible.(c) The exergy of compressed air at 1 MPa after it is cooled to 300 K is again the flow exergy at that state,

)(

K)300(since)()(

2)(

030

0303003

3

23

030033

0

0

0

ssT

TTssTTTc

gzV

ssThh

p

where

kJ/kg.K661.0kPa100kPa1000K)lnkJ/kg(0.287lnlnln

0

3

0

3

0

303

0

PP

RPP

RTT

css p

Substituting,

3 300 0 661( ( . K) kJ / kg.K) 198 kJ / kg

Note that the exergy of compressed air decreases from 280.6 to 198 as it is cooled to ambient temperature.

8-72

8-88 Cold water is heated by hot water in a heat exchanger. The rate of heat transfer and the rate of exergy destruction within the heat exchanger are to be determined.Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to thecold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluidproperties are constant. 5 The temperature of the environment is 25 C.Properties The specific heats of cold and hot water are given to be 4.18 and 4.19 kJ/kg. C, respectively. Analysis We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flowsystem can be expressed in the rate form as

45 C

Cold water15 C

0.25 kg/sHot water

100 C3 kg/s

)(

0)peke(since

0

12in

21in

outin


(steady)0system


outin

TTcmQ

hmhmQ

EE

EEE

p

Then the rate of heat transfer to the cold water in this heat exchanger becomes

kW31.35=C)15CC)(45kJ/kg.kg/s)(4.1825.0()]([ watercoldinoutin TTcmQ p

Noting that heat gain by the cold water is equal to the heat loss by the hot water, the outlet temperature ofthe hot water is determined to be

C97.5C)kJ/kg.kg/s)(4.193(

kW35.31C100

)]([ inouthot wateroutinp

p cmQTTTTcmQ

(b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of theentropy balance on the entire heat exchanger:

)()(

0

)0(since0

34hot12coldgen

gen4hot2cold3hot1cold

gen43223311


(steady)0system


gen


outin

ssmssmS

Ssmsmsmsm

QSsmsmsmsm

SSSS

Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation isdetermined to be

kW/K0.0190273+100273+97.5kJ/kg.K)lnkg/s)(4.193(

273+15273+45kJ/kg.K)lnkg/s)(4.1825.0(

lnln3

4hot

1

2coldgen T

Tcm

TT

cmS pp


kW5.66)kW/KK)(0.019298(gen0destroyed STX

8-73

8-89 Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer and the rate of exergy destruction in the heat exchanger are to be determined.Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to thecold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluidproperties are constant.Properties The specific heats of air and combustion gases are given to be 1.005 and 1.10 kJ/kg. C,respectively. The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). Analysis We take the exhaust pipes as the system, which is a control volume. The energy balance for this steady-flowsystem can be expressed in the rate form as

Air95 kPa 20 C

0.8 m3/s

)(

0)peke(since

0

21out

2out1

outin


(steady)0system


outin

TTCmQ

hmQhm

EE

EEE

p Exhaust gases 1.1 kg/s, 95 C

Then the rate of heat transfer from the exhaust gases becomes

kW102.85=C)95CC)(180kJ/kg.kg/s)(1.11.1()]([ gas.outin TTcmQ p

The mass flow rate of air is

kg/s904.0K293/kg.K)kPa.m287.0(

/s)mkPa)(0.8(953

3

RTPm V

Noting that heat loss by exhaust gases is equal to the heat gain by the air, the air exit temperature becomes

C133.2C)kJ/kg.5kg/s)(1.00904.0(

kW85.102C20)( inoutairinoutp

p cmQTTTTCmQ

The rate of entropy generation within the heat exchanger is determined by applying the rate form of theentropy balance on the entire heat exchanger:

)()(

0

)0(since0

34air12exhaustgen

gen4air2exhaust3air1exhaust

gen43223311


(steady)0system


gen


outin

ssmssmS

Ssmsmsmsm

QSsmsmsmsm

SSSS

Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is

kW/K0.0453273+20

273+133.2kJ/kg.K)ln5kg/s)(1.00904.0(273+180273+95kJ/kg.K)lnkg/s)(1.11.1(

lnln3

4air

1

2exhaustgen T

Tcm

TT

cmS pp


kW13.3)kW/KK)(0.0453293(gen0destroyed STX

8-74

8-90 Water is heated by hot oil in a heat exchanger. The outlet temperature of the oil and the rate of exergydestruction within the heat exchanger are to be determined.Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to thecold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluidproperties are constant.Properties The specific heats of water and oil are given to be 4.18 and 2.3 kJ/kg. C, respectively. Analysis We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

)(

0)peke(since

0

12in

21in

outin


(steady)0system


outin

TTcmQ

hmhmQ

EE

EEE

p

Then the rate of heat transfer to the cold water in this heatexchanger becomes

kW5940=C)20CC)(70kJ/kg.kg/s)(4.185.4()]([ waterinout .TTcmQ p

(12 tube passes)

Water20 C4.5

Oil170 C10 kg/s

70 C

Noting that heat gain by the water is equal to the heat loss by the oil, the outlet temperature of the hot water is determined from

C129.1C)kJ/kg.kg/s)(2.310(

kW5.940C170)]([ inoutoiloutinp

p cmQTTTTcmQ


)()(

0

)0(since0

34oil12watergen

gen4oil2water3oil1water

gen43223311


(steady)0system


gen


outin

ssmssmS

Ssmsmsmsm

QSsmsmsmsm

SSSS


kW/K0.736273+170273+129.1lnkJ/kg.K)kg/s)(2.310(

273+20273+70lnkJ/kg.K)kg/s)(4.185.4(

lnln3

4oil

1

2watergen T

Tcm

TT

cmS pp


kW219)kW/KK)(0.736298(gen0destroyed STX

8-75

8-91E Steam is condensed by cooling water in a condenser. The rate of heat transfer and the rate of exergydestruction within the heat exchanger are to be determined.Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to thecold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid propertiesare constant. 5 The temperature of the environment is 77 F.Properties The specific heat of water is 1.0 Btu/lbm. F (Table A-3E). The enthalpy and entropy of vaporization of water at 120 F are 1025.2 Btu/lbm and sfg = 1.7686 Btu/lbm.R (Table A-4E). Analysis We take the tube-side of the heat exchanger where cold water is flowing as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

)(

0)peke(since

0

12in

21in

outin


(steady)0system


outin

TTcmQ

hmhmQ

EE

EEE

p

Then the rate of heat transfer to the cold water in this heatexchanger becomes

Btu/s1499=F)60FF)(73Btu/lbm.lbm/s)(1.03.115(

)]([ waterinout TTcmQ p

Noting that heat gain by the water is equal to the heat loss by the condensing steam,the rate of condensation of the steam in the heat exchanger is determined from

120

73 F

60 F

Water

Steam120 F

lbm/s1.462Btu/lbm2.1025Btu/s1499)( steamsteam

fgfg h

QmhmQ


)()(

0

)0(since0

34steam12watergen

gen4steam2water3steam1water

gen44223311


(steady)0system


gen


outin

ssmssmS

Ssmsmsmsm

QSsmsmsmsm

SSSS

Noting that water is an incompressible substance and steam changes from saturated vapor to saturatedliquid, the rate of entropy generation is determined to be

Btu/s.R0.2613Btu/lbm.R)686lbm/s)(1.7462.1(460+60460+73lnBtu/lbm.R)lbm/s)(1.03.115(

ln)(ln steam1

2watersteam

1

2watergen fgpgfp sm

TT

cmssmTT

cmS

gen0destroyed STXThe exergy destroyed during a process can be determined from an exergy balance or directly from its definition ,

Btu/s140.3)Btu/s.RR)(0.2613537(gen0destroyed STX

8-76

8-92 Steam expands in a turbine, which is not insulated. The reversible power, the exergy destroyed, thesecond-law efficiency, and the possible increase in the turbine power if the turbine is well insulated are tobe determined.Assumptions 1 Steady operating conditions exist. 2 Potential energy change is negligible.Analysis (a) The properties of the steam at the inlet and exit of the turbine are (Tables A-4 through A-6)

kJ/kg.K5535.7kJ/kg1.2491

95.0kPa20

kJ/kg.K6554.6kJ/kg7.3481

C550MPa12

2

2

2

2

1

1

1

1

sh

xP

sh

TP

The enthalpy at the dead state is

kJ/kg83.1040

C250

0 hxT

The mass flow rate of steam may be determined from an energy balance on the turbine

kg/s693.2kW2500kW150

/sm1000kJ/kg1

2m/s)(130

kJ/kg1.2491/sm1000

kJ/kg12m/s)(60

kJ/kg7.3481

22

22

2

22

2

out

22

2

21

1

m

mm

WQV

hmV

hm a

20 kPa 130 m/s x=0.95

Q

Steam12 MPa 550 C, 60 m/s

Turbine

The reversible power may be determined from

kW3371

22

22

22

21

21021rev

/sm1000kJ/kg1

2m/s)(130m/s)(607.5535)-)(6.6554298()1.24917.3481()693.2(

2-

)(VV

ssThhmW

(b) The exergy destroyed in the turbine is

kW87125003371arevdest WWX

(c) The second-law efficiency is

0.742kW3371kW2500

rev

a

WW

II

(d) The energy of the steam at the turbine inlet in the given dead state is

kW9095kg104.83)kJ/-.7kg/s)(3481693.2()( 01 hhmQ

The fraction of energy at the turbine inlet that is converted to power is

2749.0kW9095kW2500a

QW

f

Assuming that the same fraction of heat loss from the turbine could have been converted to work, thepossible increase in the power if the turbine is to be well-insulated becomes

kW41.2kW)150)(2749.0(outincrease QfW

8-77

8-93 Air is compressed in a compressor that is intentionally cooled. The actual and reversible power inputs,the second law efficiency, and the mass flow rate of cooling water are to be determined.Assumptions 1 Steady operating conditions exist. 2 Potential energychange is negligible. 3 Air is an ideal gas with constant specific heats.Properties The gas constant of air is R = 0.287 kJ/kg.K and thespecific heat of air at room is cp = 1.005 kJ/kg.K. the specific heat ofwater at room temperature is cw = 4.18 kJ/kg.K (Tables A-2, A-3). Analysis (a) The mass flow rate of air is

kg/s351.5/s)m5.4(K)2730kJ/kg.K)(2287.0(

kPa)100( 31

1

11 VV

RTP

m

The power input for a reversible-isothermal process is given by

kW988.8kPa100kPa900K)ln2730kJ/kg.K)(27kg/s)(0.28351.5(ln

1

21rev P

PRTmW

Air100 kPa

20 C

Q

900 kPa 60 C

80 m/s

Compressor

Given the isothermal efficiency, the actual power may be determined from

kW14130.70

kW8.988revactual

T

WW

(b) The given isothermal efficiency is actually the second-law efficiency of the compressor0.70TII

(c) An energy balance on the compressor gives

kW1181

kW1413/sm1000

kJ/kg12m/s)80(0C0)6C)(20kJ/kg.(1.005kg/s)351.5(

2)(

22

2

inactual,

22

21

21out WVV

TTCmQ p

The mass flow rate of the cooling water is

kg/s28.25C)C)(10kJ/kg.(4.18

kW1181out

TcQ

mw

w

8-78

8-94 Water is heated in a chamber by mixing it with saturated steam. The temperature of the steam entering the chamber, the exergy destruction, and the second-law efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Heat loss from the chamber is negligible.

Mixture45 C

Sat. vap.0.23 kg/s

Mixingchamber

Water15 C4.6 kg/s

Analysis (a) The properties of water are (Tables A-4 through A-6)

kJ/kg.K63862.0kJ/kg44.188

0C45

kJ/kg.K22447.0kJ/kg98.62

0C15

3

3

1

3

01

01

1

1

sh

xT

sshh

xT

An energy balance on the chamber gives

kJ/kg5.2697kJ/kg)44.188)(kg/s23.06.4(kg/s)23.0(kJ/kg)98.62(kg/s)6.4(

)(

2

2

321332211

hh

hmmhmhmhm

The remaining properties of the saturated steam are

kJ/kg.K1907.71kJ/kg5.2697

2

2

2

2

sT

xh C114.3

(b) The specific exergy of each stream is 01

kJ/kg28.628)kJ/kg.K22447.0K)(7.1907273(15kJ/kg)98.625.2697()( 020022 ssThh

kJ/kg18.6)kJ/kg.K22447.0K)(0.63862273(15kJ/kg)98.6244.188()( 030033 ssThh

The exergy destruction is determined from an exergy balance on the chamber to be

kW114.7kJ/kg)18.6)(kg/s23.06.4(kJ/kg)28kg/s)(628.23.0(0

)( 3212211dest mmmmX

(c) The second-law efficiency for this mixing process may be determined from

0.207kJ/kg)28kg/s)(628.23.0(0kJ/kg)18.6)(kg/s23.06.4()(

2211

321II mm

mm

8-79

Review Problems

8-95 Refrigerant-134a is expanded adiabatically in an expansion valve. The work potential of R-134a atthe inlet, the exergy destruction, and the second-law efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The properties of the refrigerant at the inlet and exit of the valve and at dead state are (Tables A-11 through A-13)

180 kPaR-134a1.2 MPa

40 C

kJ/kg.K0918.1kJ/kg17.272

C20kPa100

kJ/kg.K42271.0kJ/kg23.108

kPa180

kJ/kg.K39424.0kJ/kg23.108

C40MPa2.1

0

0

0

0

212

2

1

1

1

1

sh

TP

shh

P

sh

TP

The specific exergy of the refrigerant at the inlet and exit of the valve are kJ/kg40.55kg.K1.0918)kJ/-K)(0.39424273.15(20-kJ/kg)17.27223.108()( 010011 ssThh

kJ/kg32.20kJ/kg.K1.0918K)(0.42271273.15(20kJ/kg)17.27223.108()( 020022 ssThh

(b) The exergy destruction is determined to be kJ/kg8.34/kg.K0.39424)kJ-K)(0.42271273.15(20)( 120dest ssTx

(c) The second-law efficiency for this process may be determined from

0.794kJ/kg55.40kJ/kg20.32

1

2II

8-80

8-96 Steam is accelerated in an adiabatic nozzle. The exit velocity, the rate of exergy destruction, and the second-law efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Potential energy changes are negligible.Analysis (a) The properties of the steam at the inlet and exit of the turbine and at the dead state are (Tables A-4 through A-6)

kJ/kg.K2678.0kJ/kg54.75

0C18

kJ/kg.K6753.6kJ/kg9.2919

C250kPa6.1

kJ/kg.K4484.6kJ/kg4.2978

C300MPa5.3

0

00

2

2

2

2

1

1

1

1

sh

xT

sh

TP

sh

TP

Steam3.5 MPa300 C

1.6 MPa 250 CVel2

The exit velocity is determined from an energy balance on the nozzle

m/s342.02

22

22

22

2

22

2

21

1

/sm1000kJ/kg1

2V

kJ/kg9.2919/sm1000

kJ/kg12

m/s)(0kJ/kg4.2978

22

V

Vh

Vh

(b) The rate of exergy destruction is the exergy decrease of the steam in the nozzle

kW26.41kJ/kg.K)4484.66753.6)(K291(

/sm1000kJ/kg1

20m/s)(342kg2978.4)kJ/(2919.9

kg/s)4.0(

(2

22

2

120

21

22

12dest ssTVV

hhmX

(c) The exergy of the refrigerant at the inlet is

kW72.441kJ/kg.K)2678.04484.6)(K291(0kJ/kg)54.75(2978.4kg/s)4.0(

(2 010

21

011 ssTVhhmX

The second-law efficiency for this device may be defined as the exergy output divided by the exergy input:

0.940kW72.441

kW41.26111

dest

1

2II X

XXX

8-81

8-97 An electrical radiator is placed in a room and it is turned on for a period of time. The time period for which the heater was on, the exergy destruction, and the second-law efficiency are to be determined.Assumptions 1 Kinetic and potential energy changes are negligible. 2 Air is an ideal gas with constantspecific heats. 3 The room is well-sealed. 4 Standard atmospheric pressure of 101.3 kPa is assumed.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 (Table A-2). The properties of oil are given to be = 950 kg/m3, coil = 2.2 kJ/kg.K.Analysis (a) The masses of air and oil are

kg36.62K)273K)(10/kgmkPa(0.287

)mkPa)(50(101.33

3

1

1

RTP

maV

kg50.28)m)(0.030kg/m(950 33oiloiloil Vm

An energy balance on the system can be used to determine time period for which the heater was kept on

min34s2038tt

TTmcTTmctQW a

C)10C)(50kJ/kg.kg)(2.250.28(C)10C)(20kJ/kg.kg)(0.71836.62(kW)35.08.1()()()( oil1212outin v

Room

Radiator

Q10 C

(b) The pressure of the air at the final state is

kPa9.104m50

K)273K)(20/kgmkPakg)(0.287(62.363

32

2 Vaa

aRTm

P

The amount of heat transfer to the surroundings iskJ5.713s)kJ/s)(2038(0.35outout tQQ

The entropy generation is the sum of the entropy changes of air, oil, and the surroundings

kJ/K5548.1kPa101.3kPa104.9kJ/kg.K)ln(0.287

K273)(10K273)(20

kJ/kg.K)ln(1.005kg)(62.36

lnln1

2

1

2

PP

RTT

cmS pa

kJ/K2893.8K273)(10K273)(50kJ/kg.K)lnkg)(2.2(28.50ln

1

2oil T

TmcS

kJ/K521.2K273)(10

kJ713.5

surr

outsurr T

QS

kJ/K365.12521.22893.85548.1surroilagen SSSS

The exergy destruction is determined fromkJ3500kJ/K)K)(12.365273(10gen0dest STX

(c) The second-law efficiency may be defined in this case as the ratio of the exergy recovered to the exergyinput. That is,

kJ729.7kJ/K)K)(1.554827310(C10)-C)(20kJ/kg.(0.718kg)(62.36)( 0122, aa STTTcmX v

kJ13.162kJ/K)K)(8.289327310(C10)-C)(50kJ/kg.(2.2kg)(28.50)( 0122,oil aSTTTCmX

4.6%0.046s)kJ/s)(2038(1.8kJ162.13)(7.729

in

oil,2,2

supplied

recovered

tW

XXXX a

II

8-82

8-98 Hot exhaust gases leaving an internal combustion engine is to be used to obtain saturated steam in an adiabatic heat exchanger. The rate at which the steam is obtained, the rate of exergy destruction, and thesecond-law efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air properties are used for exhaust gases. 4 Pressure drops in the heat exchanger are negligible.Properties The gas constant of air is R = 0.287 kJkg.K. The specific heat of air at the average temperatureof exhaust gases (650 K) is cp = 1.063 kJ/kg.K (Table A-2). Analysis (a) We denote the inlet and exit states of exhaust gases by (1) and (2) and that of the water by (3) and (4). The properties of water are (Table A-4)

350 CHeatExchanger

Sat. vap.200 C

Water20 C

Exh. gas400 C150 kPa

kJ/kg.K4302.6kJ/kg0.2792

1C200

kJ/kg.K29649.0kJ/kg91.83

0C20

4

4

4

4

3

3

3

3

sh

xT

sh

xT

An energy balance on the heat exchanger gives

kg/s0.01570w

w

wpa

wawa

mm

hhmTTcmhmhmhmhm

kJ/kg)91.830.2792(C)350400(C)kJ/kg3kg/s)(1.068.0(

)()( 3421

4231

(b) The specific exergy changes of each stream as it flows in the heat exchanger is

kJ/kg.K08206.0K273)(400K273)(350kJ/kg.K)ln3kg/s)(1.06(0.8ln

1

2

TT

cs pa

kJ/kg106.29kJ/kg.K)6K)(-0.0820273(20C400)-C)(350kJ/kg.063.1(

)( 012 apa sTTTc

kJ/kg913.910)kJ/kg.K29649.0K)(6.4302273(20kJ/kg)91.830.2792(

)( 34034 ssThhw

The exergy destruction is determined from an exergy balance on the heat exchanger to be

or

kW8.98dest

dest kW98.8kJ/kg)913.910)(kg/s01570.0(kJ/kg)106kg/s)(-29.8.0(

X

mmX wwaa

(c) The second-law efficiency for a heat exchanger may be defined as the exergy increase of the cold fluiddivided by the exergy decrease of the hot fluid. That is,

0.614kJ/kg)106kg/s)(-29.8.0(

kJ/kg)913.910)(kg/s01570.0(II

aa

ww

mm

8-83

8-99 The inner and outer surfaces of a brick wall are maintained at specified temperatures. The rate of exergy destruction is to be determined.Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain

constant at the specified values. 2 The environment temperature is given to be T0 = 0 C.Analysis We take the wall to be the system, which is a closedsystem. Under steady conditions, the rate form of the entropybalance for the wall simplifies to

30

QBrick Wall

W/K0.1660K278

W900K293

W900

0

0

wallgen,wallgen,

wallgen,outb,

out

inb,

in


0system


gen


outin

SS

STQ

TQ

SSSS

5 C20 C


W45.3) W/K166.0(K)273(gen0destroyed STX

8-100 A 1000-W iron is left on the iron board with its base exposed to air. The rate of exergy destruction insteady operation is to be determined.Assumptions Steady operating conditions exist.Analysis The rate of total entropy generation during this process is determined by applying the entropybalance on an extended system that includes the iron and its immediate surroundings so that the boundarytemperature of the extended system is 20 C at all times. It gives

0

0

genoutb,

out


0system


gen


outin

STQ

SSSS

Therefore,

W/K3.413K293 W1000

0outb,

outgen T

QTQ

S


W1000) W/K413.3(K)293(gen0destroyed STX

Discussion The rate of entropy generation within the iron can be determined by performing an entropybalance on the iron alone (it gives 2.21 W/K). Therefore, about one-third of the entropy generation and thus exergy destruction occurs within the iron. The rest occurs in the air surrounding the iron as thetemperature drops from 150 C to 20 C without serving any useful purpose.

8-84

8-101 The heating of a passive solar house at night is to be assisted by solar heated water. The amount ofheating this water will provide to the house at night and the exergy destruction during this heat transfer process are to be determined.Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored in the glass containers themselves is negligible relative to the energy stored in water. 3 The house ismaintained at 22°C at all times. 4 The outdoor temperature is given to be 5 C.Properties The density and specific heat of water at room temperature are = 997 kg/m3 and c = 4.18 kJ/kg·°C (Table A-3).Analysis The total mass of water is

kg95.348m0.350kg/m997 33Vwm

water45 C

22 C

The amount of heat this water storage system can provide isdetermined from an energy balance on the 350-L water storage system

water12systemout


system


outin

)( TTmcUQ

EEE

Substituting,kJ33,548=C22)-C)(45kJ/kgkg)(4.18(348.95outQ

The entropy generated during this process is determined by applying the entropybalance on an extended system that includes the water and its immediatesurroundings so that the boundary temperature of the extended system is theenvironment temperature at all times. It gives

watergenoutb,

out

entropyinChange

system

generationEntropy

gen


outin

SSTQ

SSSS

Substituting,

KkJ215.4K295

kJ33,548K318K295

lnKkJ/kg4.18kg348.95

lnroom

out

water1

2

outb,

outwatergen

/

TQ

TT

mcTQ

SS


kJ1172)kJ/K215.4(K)278(gen0destroyed STX

8-85

8-102 The inner and outer surfaces of a window glass are maintained at specified temperatures. The amount of heat loss and the amount of exergy destruction in 5 h are to be determinedAssumptions Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values.Analysis We take the glass to be the system, which is a closed system. The amount of heat loss isdetermined from

Jk57,600s)3600kJ/s)(5(3.2tQQUnder steady conditions, the rate form of the entropy balance for the glass simplifies to

W/K0.28680K276W3200

K283W3200

0

0

glassgen,wallgen,

glassgen,outb,

out

inb,

in


0system


gen


outin

SS

STQ

TQ

SSSSGlass

3 C10 CThen the amount of entropy generation over a period of 5 h becomes

J/K5162s)3600K)(5/W2868.(0glassgen,glassgen, tSS



Discussion The total entropy generated during this process can be determined by applying the entropybalance on an extended system that includes the glass and its immediate surroundings on both sides so thatthe boundary temperature of the extended system is the room temperature on one side and the environmenttemperature on the other side at all times. Using this value of entropy generation will give the total exergydestroyed during the process, including the temperature gradient zones on both sides of the window.

8-103 Heat is transferred steadily to boiling water in the pan through its bottom. The inner and outersurface temperatures of the bottom of the pan are given. The rate of exergy destruction within the bottomplate is to be determined.Assumptions Steady operating conditions exist since the surface temperatures of the pan remain constant at the specified values.Analysis We take the bottom of the pan to be the system, which is a closed system. Under steadyconditions, the rate form of the entropy balance for this system can be expressed as

W/K0.005610K377W800

K378W800

0

0

systemgen,systemgen,

systemgen,outb,

out

inb,

in


0system


gen


outin

SS

STQ

TQ

SSSS

800 W

104 C


105 C


8-86

8-104 A elevation, base area, and the depth of a crater lake are given. The maximum amount of electricitythat can be generated by a hydroelectric power plant is to be determined.Assumptions The evaporation of water from the lake is negligible.Analysis The exergy or work potential of the water is the potential energy it possesses relative to theground level,

mgh=PE=Exergy

d

z

12 m

140 m

Therefore,

kWh109.55 4

2222

2243

21

22

s/m1000kJ/kg1

s3600h1)m140(m)152(

)m/s81.9)(m102)(kg/m1000(5.0

2/)(

)(Exergy

2

1

zzAgzdzAg

AdzgzdmgzdPEPE

z

z

8-105E The 2nd-law efficiency of a refrigerator and the refrigeration rate are given. The power input to therefrigerator is to be determined.Analysis From the definition of the second law efficiency, the COP of the refrigerator is determined to be

57.5375.1245.0COPCOPCOP

COP

375.121495/535

11/

1COP

revR,IIRrevR,

R

revR,

II

LH TT

II = 0.45

200 Btu/min

R

75 F

35 F

Thus the power input is

hp0.85Btu/min42.41hp1

57.5Btu/min020

COPRin

LQW

8-87

8-106 Writing energy and entropy balances, a relation for the reversible work is to be obtained for a closedsystem that exchanges heat with surroundings at T0 in the amount of Q0 as well as a heat reservoir at temperature TR in the amount QR.Assumptions Kinetic and potential changes are negligible.Analysis We take the direction of heat transfers to be to the system (heat input) and the direction of work transfer to be from the system (work output). The result obtained is still general since quantities wit opposite directions can be handled the same way by using negative signs. The energy and entropy balancesfor this stationary closed system can be expressed as

Energy balance: (1) RR QQUUWUUWQQEEE 021120systemoutin

Entropy balance:0

012gensystemgenoutin )(

TQ

TQ

SSSSSSSR

R (2)

Solving for Q0 from (2) and substituting in (1) yields

QR

SourceTR

System

gen00

21021 1)()( STTT

QSSTUUWR

R

The useful work relation for a closed system is obtained from

)(1)()( 120gen00

21021

surr

VVPSTTT

QSSTUU

WWW

RR

u

Then the reversible work relation is obtained by substituting Sgen = 0,

RR T

TQPSSTUUW 0

21021021rev 1)()()( VV

A positive result for Wrev indicates work output, and a negative result work input. Also, the QR is a positivequantity for heat transfer to the system, and a negative quantity for heat transfer from the system.

8-88

8-107 Writing energy and entropy balances, a relation for the reversible work is to be obtained for a steady-flow system that exchanges heat with surroundings at T0 at a rate of Q as well as a heat reservoir

at temperature T0

R in the amount Q .R

Analysis We take the direction of heat transfers to be to the system (heat input) and the direction of work transfer to be from the system (work output). The result obtained is still general since quantities wit opposite directions can be handled the same way by using negative signs. The energy and entropy balancesfor this stationary closed system can be expressed as

Energy balance: outinsystemoutin EEEEE

)2

()2

(22

0 ii

iiee

eeR gzV

hmgzV

hmWQQ

Systemor Re

eeei

iii QQgz

Vhmgz

VhmW 0

22)

2()

2( (1)

Entropy balance: inoutgensystemgenoutin SSSSSSS

0

0gen T

QTQsmsmSR

Riiee (2)

Solving for from (2) and substituting in (1) yieldsQ0

RRee

eeeii

iii T

TQSTsTgz

VhmsTgz

VhmW 0

gen00

2

0

21)

2()

2(


RRee

eeeii

iii T

TQsTgz

VhmsTgz

VhmW 0

0

2

0

2

rev 1)2

()2

(


8-89

8-108 Writing energy and entropy balances, a relation for the reversible work is to be obtained for a uniform-flow system that exchanges heat with surroundings at T0 in the amount of Q0 as well as a heat reservoir at temperature TR in the amount QR.Assumptions Kinetic and potential changes are negligible.Analysis We take the direction of heat transfers to be to the system (heat input) and the direction of work transfer to be from the system (work output). The result obtained is still general since quantities wit opposite directions can be handled the same way by using negative signs. The energy and entropy balancesfor this stationary closed system can be expressed as Energy balance: systemoutin EEE

cvii

iiee

eeR UUgzV

hmgzV

hmWQQ )()2

()2

( 12

22

0

or, Rcvee

eeii

ii QQUUgzV

hmgzV

hmW 012

22)()

2()

2( (1)

Entropy balance: systemgenoutin SSSS

SourceTR

QSystem0

012gen )(

TQ

TQsmsmSSSR

Riieecv (2)

Solving for Q0 from (2) and substituting in (1) yields

RRgencv

eee

eeiii

ii

TT

QSTSSTUU

sTgzV

hmsTgzV

hmW

0021021

0

2

0

2

1)()(

)2

()2

(

me

The useful work relation for a closed system is obtained from

)(1)()(

)2

()2

(

1200

gen021021

0

2

0

2

surr

VVPTT

QSTSSTUU

sTgzV

hmsTgzV

hmWWW

RRcv

eee

eeiii

iiu


RRcv

eee

eeiii

ii

TT

QPSSTUU

sTgzV

hmsTgzV

hmW

021021021

0

2

0

2

rev

1)()()(

)2

()2

(

VV


8-90

8-109 An electric resistance heater is immersed in water. The time it will take for the electric heater toraise the water temperature to a specified temperature, the minimum work input, and the exergy destroyedduring this process are to be determined.Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored in the container itself and the heater is negligible. 3 Heat loss from the container is negligible. 4 Theenvironment temperature is given to be T0 = 20 C.Properties The specific heat of water at room temperature is c = 4.18 kJ/kg·°C (Table A-3).Analysis Taking the water in the container as the system, which is a closed system, the energy balance can be expressed as

water12ine,

waterine,


system


outin

)(

)(

TTmctW

UW

EEE

Heater

Water40 kg

Substituting, (800 J/s) t = (40 kg)(4180 J/kg· C)(80 - 20) CSolving for t gives

t = 12,544 s = 209.1 min = 3.484 hAgain we take the water in the tank to be the system. Noting that no heat or mass crosses the

boundaries of this system and the energy and entropy contents of the heater are negligible, the entropybalance for it can be expressed as

watergen

entropyinChange

system

generationEntropy

gen


outin

0 SS

SSSS

Therefore, the entropy generated during this process is

kJ/K31.18K293K353

lnKkJ/kg4.184kg40ln1

2watergen T

TmcSS



The actual work input for this process is

kJ042,10=s)52kJ/s)(12,58.0(inact,inact, tWW

Then the reversible (or minimum required )work input becomeskJ9069136042,10destroyedinact,inrev, XWW

8-91

8-110 A hot water pipe at a specified temperature is losing heat to the surrounding air at a specified rate.The rate at which the work potential is wasted during this process is to be determined.Assumptions Steady operating conditions exist.Analysis We take the air in the vicinity of the pipe (excluding the pipe) as our system, which is a closedsystem.. The system extends from the outer surface of the pipe to a distance at which the temperature drops to the surroundings temperature. In steady operation, the rate form of the entropy balance for this systemcan be expressed as

W/K0.03440K278

W45K353

W45

0

0

systemgen,systemgen,

systemgen,outb,

out

inb,

in


0system


gen


outin

SS

STQ

TQ

SSSS 80 C

L = 10 m

D = 5 cm

Q Air, 5 C



8-92

8-111 Two rigid tanks that contain water at different states are connected by a valve. The valve is openedand steam flows from tank A to tank B until the pressure in tank A drops to a specified value. Tank B losesheat to the surroundings. The final temperature in each tank and the work potential wasted during this process are to be determined.Assumptions 1 Tank A is insulated and thus heat transfer is negligible. 2 The water that remains in tank A undergoes a reversible adiabatic process. 3 The thermal energy stored in the tanks themselves is negligible.4 The system is stationary and thus kinetic and potential energy changes are negligible. 5 There are no work interactions.Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s2 = s1. From the steam tables (Tables A-4 through A-6),

KkJ/kg7.7100kJ/kg2731.4

/kgm1.1989

C250kPa200

kJ/kg2125.9kJ/kg1982.17895.011.561/kgm0.47850001073.060582.07895.0001073.0

7895.03200.5

6717.18717.5

mixturesat.

kPa300

KkJ/kg5.87171191.58.07765.1kJ/kg2163.39.19488.022.604

/kgm0.37015001084.046242.08.0001084.0

8.0kPa400

,1

,1

3,1

1

1

,2,2

3,2,2

,2,2

300@,2

12

2

1,1

1,1

31,1

1

1

B

B

B

fgAfA

fgAfA

fg

fAA

kPasatA

fgfA

fgfA

fgfA

su

TP

uxuux

sss

x

TT

ssP

sxssuxuu

x

xP

v

vvv

vvv

:BTank

: ATank

C133.52

900 kJ

Bm = 3 kg

steamT = 250 C

P = 200 kPa

AV = 0.2 m3

steamP = 400 kPa

x = 0.8

The initial and the final masses in tank A are

and

kg0.4180/kgm0.479

m0.2

kg0.5403/kgm0.37015

m0.2

3

3

,2,2

3

3

,1,1

A

AA

A

AA

m

m

v

V

v

V

Thus, 0.540 - 0.418 = 0.122 kg of mass flows into tank B. Then, m mB B2 1 0122 3 0122, , . . 3.122 kg

The final specific volume of steam in tank B is determined from

/kgm1.152m3.122

/kgm1.1989kg3 33

3

,2

11

,2,2

B

B

B

BB m

vmm

vV

We take the entire contents of both tanks as the system, which is a closed system. The energy balance forthis stationary closed system can be expressed as

BA

BA

umumumumQWUUUQ

EEE

11221122out

out


system


outin

0)=PE=KE(since)()(

Substituting,

8-93

kJ/kg2425.94.27313122.33.21635403.09.2125418.0900

,2

,2

B

B

uu

Thus,

KkJ/kg.97726kJ/kg2425.9

/kgm1.152

,2

,2

,2

3,2

B

B

B

B

sT

u

C110.1v

(b) The total entropy generation during this process is determined by applying the entropy balance on an extended system that includes both tanks and their immediate surroundings so that the boundarytemperature of the extended system is the temperature of the surroundings at all times. It gives

BAgensurrb,

out

entropyinChange

system

generationEntropy

gen


outin

SSSTQ

SSSS

Rearranging and substituting, the total entropy generated during this process is determined to be

kJ/K234.1K273kJ900

7100.739772.6122.38717.55403.08717.5418.0

surrb,

out11221122

surrb,

outgen T

Qsmsmsmsm

TQ

SSS BABA

The work potential wasted is equivalent to the exergy destroyed during a process, which can be determinedfrom an exergy balance or directly from its definition ,gen0destroyed STX


8-94

8-112E A cylinder initially filled with helium gas at a specified state is compressed polytropically to aspecified temperature and pressure. The actual work consumed and the minimum useful work input needed are to be determined.Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The cylinder is stationary and thus thekinetic and potential energy changes are negligible. 3 The thermal energy stored in the cylinder itself isnegligible. 4 The compression or expansion process is quasi-equilibrium. 5 The environment temperature is 70 F.Properties The gas constant of helium is R = 2.6805 psia.ft3/lbm.R = 0.4961 Btu/lbm.R (Table A-1E). Thespecific heats of helium are cv = 0.753 and cv = 1.25 Btu/lbm.R (Table A-2E). Analysis (a) Helium at specified conditions can be treated as an ideal gas. The mass of helium is

lbm0.264)R530)(R/lbmftpsia2.6805(

)ft15)(psia25(3

3

1

11

RTP

mV

Q

HELIUM15 ft3

PVn = const

The exponent n and the boundary work for this polytropic process aredetermined to be

539.1682.715

2570

ft7.682)ft15()psia70)(R530()psia25)(R760(

2

1

1

21122

331

2

1

1

22

2

22

1

11

nPP

PP

PP

TT

TP

TP

nnnn

V

VVV

VVVV

Then the boundary work for this polytropic process can be determined from

Btu55.91.5391

R530760RBtu/lbm0.4961lbm0.26411

1211222

1inb, n

TTmRnPP

dPWVV

V

Also,

Thus,Btu36.09.199.55

Btu9.19ftpsia5.4039

Btu1ft5)12psia)(7.687.14()(

insurr,inb,inu,

33

120insurr,

WWW

PW VV

(b) We take the helium in the cylinder as the system, which is a closed system. Taking the direction of heattransfer to be from the cylinder, the energy balance for this stationary closed system can be expressed as

)()(

)(

12inb,out

inb,12out

12inb,out


system


outin

TTmcWQWuumQuumUWQ

EEE

v

Substituting,Btu10.2R530760RBtu/lbm0.753lbm0.264Btu55.9outQ

The total entropy generation during this process is determined by applying the entropy balance on an extended system that includes the cylinder and its immediate surroundings so that the boundary temperatureof the extended system is the temperature of the surroundings at all times. It gives

8-95

sysgensurrb,

out

entropyinChange

system

generationEntropy

gen


outin

SSTQ

SSSS

where the entropy change of helium is

RBtu01590psia25psia70

ln)RBtu/lbm0.4961(R530R760

ln)RBtu/lbm1.25()lbm0.264(

lnln1

2

1

2avg,heliumsys

/.

PP

RTT

cmSS p

Rearranging and substituting, the total entropy generated during this process is determined to be

Btu/R0033450R530

Btu10.2Btu/R)0159.0(

0

outheliumgen .

TQ

SS

The work potential wasted is equivalent to the exergy destroyed during a process, which can be determinedfrom an exergy balance or directly from its definition ,gen0destroyed STX

Btu1.77)Btu/R003345.0(R)530(gen0destroyed STX

The minimum work with which this process could be accomplished is the reversible work input, Wrev, in.which can be determined directly from

Btu34.2377.10.36destroyedinact,inrev, XWW

Discussion The reversible work input, which represents the minimum work input Wrev,in in this case can be determined from the exergy balance by setting the exergy destruction term equal to zero,

12inrev,

exergyinChange

system

ndestructioExergy



Substituting the closed system exergy relation, the reversible work input during this process is determinedto be

Btu34.24]ftpsiaBtu/5.4039[ft)15682.7psia)((14.7+

Btu/R)R)(-0.0159530(F70)R)(300Btu/lbmlbm)(0.753264.0()()()(

33

12012012rev VVPSSTUUW

8-96

8-113 A well-insulated room is heated by a steam radiator, and the warm air is distributed by a fan. The average temperature in the room after 30 min, the entropy changes of steam and air, and the exergydestruction during this process are to be determined.Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 The kinetic and potential energy changes are negligible. 3 The air pressure in the room remains constant and thus the airexpands as it is heated, and some warm air escapes. 4 The environment temperature is given to be T0 = 10 C.Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Also, cp = 1.005 kJ/kg.K for airat room temperature (Table A-2).Analysis We first take the radiator as the system. This is a closed system since no mass enters or leaves. The energy balance for this closed system can be expressed as

)(0)=PE=KE(since)(

21out

12out


system


outin

uumQWuumUQ

EEE

10 C4 m 4 m 5 m

Steamradiator

Using data from the steam tables (Tables A-4 through A-6), some properties are determined to be

kJ/kg.K0562.6kJ/kg.K,3028.1

kJ/kg2088.2,40.417/kgm1.6941,001043.0kPa100

kJ/kg.K5081.7kJ/kg2654.6

/kgm1.0805

C200kPa200

3

12

2

1

1

31

1

1

fgf

fgf

gf

ss

uuP

su

TP

vv

vv

v

kg0.01388/kgm1.0805

m0.015

kJ/kg.K1639.50562.60.63763028.1

kJ/kg1748.72088.20.6376417.40

6376.0001043.06941.1001043.00805.1

3

3

1

1

22

22

22

v

V

v

vv

m

sxss

uxuu

x

fgf

fgf

fg

f

Substituting,Qout = (0.01388 kg)( 2654.6 - 1748.7)kJ/kg = 12.58 kJ

The volume and the mass of the air in the room are V = 4 4 5 = 80 m3

and kg98.5K283K/kgmkPa0.2870

m80kPa1003

3

1

11air RT

Pm

V

The amount of fan work done in 24 min is

kJ216s)60kJ/s)(24150.0(infan,infan, tWW

We now take the air in the room as the system. The energy balance for this closed system is expressed as

)( 12infan,in

outb,infan,in

systemoutin

TTmcHWQUWWQ

EEE

p

8-97

since the boundary work and U combine into H for a constant pressure expansion or compressionprocess. It can also be expressed as

)()( 12avg,infan,in TTmctWQ p

Substituting, (12.58 kJ) + (216 kJ) = (98.5 kg)(1.005 kJ/kg C)(T2 - 10) Cwhich yields T2 = 12.3 CTherefore, the air temperature in the room rises from 10 C to 12.3 C in 24 minutes.(b) The entropy change of the steam is

kJ/K0.0325KkJ/kg7.50815.1639kg0.0138812steam ssmS

(c) Noting that air expands at constant pressure, the entropy change of the air in the room is

kJ/K0.8012K283K285.3

lnKkJ/kg1.005kg98.5lnln0

1

2

1

2air P

PmR

TT

mcS p

(d) We take the contents of the room (including the steam radiator) as our system, which is a closedsystem. Noting that no heat or mass crosses the boundaries of this system, the entropy balance for it can beexpressed as

airsteamgen

entropyinChange

system

generationEntropy

gen


outin

0 SSS

SSSS

Substituting, the entropy generated during this process is determined to beKkJ0.76878012.00325.0airsteamgen /SSS



Alternative Solution In the solution above, we assumed the air pressure in the room to remain constant.This is an extreme case, and it is commonly used in practice since it gives higher results for heat loads, and thus allows the designer to be conservative results. The other extreme is to assume the house to be airtight,and thus the volume of the air in the house to remain constant as the air is heated. There is no expansion inthis case and thus boundary work, and cv is used in energy change relation instead of cp. It gives thefollowing results:

C13.22T

kJ/K0.0325KkJ/kg7.50815.1639kg0.0138812steam ssmS

kJ/K0.7952=K283K286.2lnK)kJ/kgkg)(0.7185.98(lnln

1

2

1

2air

0

V

Vv Rm

TT

mcS

kJ/K7627.07952.00325.0airsteamgen SSS

andkJ216=kJ/K)K)(0.7627283(gen0destroyed STX

The actual value in practice will be between these two limits.

8-98

8-114 The heating of a passive solar house at night is to be assisted by solar heated water. The length oftime that the electric heating system would run that night, the exergy destruction, and the minimum work input required that night are to be determined.Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored in the glass containers themselves is negligible relative to the energy stored in water. 3 The house ismaintained at 22°C at all times. 4 The environment temperature is given to be T0 = 5 C.Properties The density and specific heat of water at room temperature are = 1 kg/L and c = 4.18 kJ/kg·°C (Table A-3).Analysis (a) The total mass of water is

kg1000L2050kg/L1Vwm

80 Cwater

50,000 kJ/h

22 C

Taking the contents of the house, including the water as oursystem, the energy balance relation can be written as

water12water

0airwateroutine,


system


outin

)()()()(

TTmcUUUUQW

EEE

or, W water12, )]([ TTmcQt outine

Substituting,(15 kJ/s) t - (50,000 kJ/h)(10 h) = (1000 kg)(4.18 kJ/kg· C)(22 - 80) C

It givest = 17,170 s = 4.77 h

(b) We take the house as the system, which is a closed system. The entropy generated during this process isdetermined by applying the entropy balance on an extended system that includes the house and itsimmediate surroundings so that the boundary temperature of the extended system is the temperature of thesurroundings at all times. The entropy balance for the extended system can be expressed as

water0

airwatergenoutb,

out

entropyinChange

system

generationEntropy

gen


outin

SSSSTQ

SSSS

since the state of air in the house remains unchanged. Then the entropy generated during the 10-h periodthat night is

kJ/K1048K278

kJ500,000K353K295

lnKkJ/kg4.18kg1000

ln0

out

water1

2

outb,

outwatergen T

QTT

mcTQ

SS


kJ291,400=kJ/K)K)(1048278(gen0destroyed STX(c) The actual work input during this process is

kJ550,257=s)70kJ/s)(17,115(inact,inact, tWWThe minimum work with which this process could be accomplished is the reversible work input, Wrev, in.which can be determined directly from

kWh9.40kJ33,850=

kJ-33,850=400,291550,257

outrev,

destroyedinact,inrev,

W

XWW

That is, 9.40 kWh of electricity could be generated while heating the house by the solar heated water(instead of consuming electricity) if the process was done reversibly.

8-99

8-115 Steam expands in a two-stage adiabatic turbine from a specified state to specified pressure. Somesteam is extracted at the end of the first stage. The wasted power potential is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The turbine is adiabatic and thus heat transfer is negligible. 4 Theenvironment temperature is given to be T0 = 25 C.Analysis The wasted power potential is equivalent to the rate of exergy destruction during a process, whichcan be determined from an exergy balance or directly from its definition .gen0destroyed STX

The total rate of entropy generation during this process is determined by taking the entire turbine,which is a control volume, as the system and applying the entropy balance. Noting that this is a steady-flowprocess and there is no heat transfer,

]1.09.0[09.01.0

0

0

1231gengen312111

gen332211


0system


gen


outin

sssmSSsmsmsm

Ssmsmsm

SSSS

And ]1.09.0[ 12310gen0destroyed sssmTSTX9 MPa 500 CFrom the steam tables (Tables A-4 through 6)

Kkg/kJ6603.6kg/kJ4.3387

C500MPa9

1

1

1

1

sh

TP

STEAM13.5 kg/s

STEAM15 kg/s

kg/kJ4.2882MPa4.1

212

2s

sh

ssP

I II1.4 MPa

and, 50 kPa

kJ/kg0.2943)4.28824.3387(88.04.3387

)( 211221

21sT

sT hhhh

hhhh 90%

10%

Kkg/kJ7776.6kJ/kg0.2943

MPa4.12

2

2 shP

kJ/kg6.23147.23048565.054.340

8565.05019.6

0912.16603.6xkPa50

33

33

13

3

fgsfs

fg

fss

s hxhhs

ss

ssP

and

kJ/kg3.2443)6.23144.3387(88.04.3387

)( 311331

31sT

sT hhhh

hhhh

KkJ/kg0235.75019.69124.00912.1

9124.07.2304

54.3403.2443xkJ/kg3.2443

kPa50

33

33

3

3

fgf

fg

f

sxssh

hh

hP

Substituting, the wasted work potential is determined to be

kW1514kJ/kg)6603.66.77760.1+7.0235kg/s)(0.9K)(15298(gen0destroyed STX

8-100

8-116 Steam expands in a two-stage adiabatic turbine from a specified state to another specified state. Steam is reheated between the stages. For a given power output, the reversible power output and the rate ofexergy destruction are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energychanges are negligible. 3 The turbine is adiabatic andthus heat transfer is negligible. 4 The environmenttemperature is given to be T0 = 25 C.

2 MPa 500 C

Heat

30 kPa x = 97%

2 MPa350 C

Stage II

8 MPa 500 C

Stage I

Properties From the steam tables (Tables A-4 through 6)

Kkg/kJ7266.6kg/kJ5.3399

C500MPa8

1

1

1

1

sh

TP

5 MW

Kkg/kJ9583.6kg/kJ7.3137

C350MPa2

2

2

2

2

sh

TP

Kkg/kJ4337.7kg/kJ3.3468

C500MPa2

3

3

3

3

sh

TP

KkJ/kg5628.78234.697.09441.0kJ/kg5.25543.233597.027.289

97.0kPa30

44

44

4

4

fgf

fgf

sxsshxhh

xP

Analysis We take the entire turbine, excluding the reheat section, as the system, which is a control volume.The energy balance for this steady-flow system can be expressed in the rate form as

outin


(steady)0system


outin 0

EE

EEE

)]()[( 4321outout4231 hhhhmWWhmhmhmhmSubstituting, the mass flow rate of the steam is determined from the steady-flow energy equation applied tothe actual process,

kg/s4.253kJ/kg)5.25543.34687.31375.3399(

kJ/s5000

4321

out

hhhhW

m

The reversible (or maximum) power output is determined from the rate form of the exergy balance appliedon the turbine and setting the exergy destruction term equal to zero,

]peke)()[(

]peke)()[(

)()(

0

0034043

0012021

4321outrev,

outrev,4231

outin


(steady)0system




outin

ssThhm

ssThhm

mmW

Wmmmm

XX

XXXX

Then the reversible power becomes

kW5457K]kJ/kg)4337.75628.77266.6K)(6.9583298(

kJ/kg)5.25543.34687.31379.5kg/s)[(339253.4()( 341204321outrev, ssssThhhhmW

Then the rate of exergy destruction is determined from its definition,kW45750005457outoutrev,destroyed WWX

8-101

8-117 One ton of liquid water at 80°C is brought into a room. The final equilibrium temperature in the room and the entropy generated are to be determined.Assumptions 1 The room is well insulated and well sealed. 2 The thermal properties of water and air are constant at room temperature. 3 The system is stationary and thus the kinetic and potential energy changes are zero. 4 There are no work interactions involved.Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). The constant volume specificheat of water at room temperature is cv = 0.718 kJ/kg C (Table A-2). The specific heat of water at roomtemperature is c = 4.18 kJ/kg C (Table A-3).

HeatWater80 C

4 m 5 m 6 m

ROOM22 C

100 kPa

Analysis The volume and the mass of the air in the room are V = 4x5x6 = 120 m³

kg141.74)K295)(K/kgmkPa0.2870(

)m120)(kPa100(3

3

1

11air RT

Pm

V

Taking the contents of the room, including the water, asour system, the energy balance can be written as

airwater


system


outin 0 UUUEEE

or 0air12water12 TTmcTTmc v

Substituting, 0C22CkJ/kg0.718kg141.74C80CkJ/kg4.18kg1000 ff TT

It gives the final equilibrium temperature in the room to beTf = 78.6 C

(b) We again take the room and the water in it as the system, which is a closed system. Considering thatthe system is well-insulated and no mass is entering and leaving, the entropy balance for this system can beexpressed as

waterairgen

entropyinChange

system

generationEntropy

gen


outin

0 SSS

SSSS

where

KkJ16.36K353K351.6

lnKkJ/kg4.18kg1000ln

KkJ17.87K295K351.6

lnKkJ/kg0.718kg141.74lnln

1

2water

0

1

2

1

2air

/

/

TT

mcS

mRTT

mcSV

Vv

Substituting, the entropy generation is determined to beSgen = 17.87 - 16.36 = 1.51 kJ/K


kJ427kJ/K)K)(1.51283(gen0destroyed STX

(c) The work potential (the maximum amount of work that can be produced) during a process is simply thereversible work output. Noting that the actual work for this process is zero, it becomes

kJ427destroyedoutrev,outact,outrev,destroyed XWWWX

8-102

8-118 An insulated cylinder is divided into two parts. One side of the cylinder contains N2 gas and theother side contains He gas at different states. The final equilibrium temperature in the cylinder and thewasted work potential are to be determined for the cases of piston being fixed and moving freely.Assumptions 1 Both N2 and He are ideal gases with constant specific heats. 2 The energy stored in the container itself is negligible. 3 The cylinder is well-insulated and thus heat transfer is negligible.Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m3/kg.K is cv = 0.743 kJ/kg·°C for N2, and R = 2.0769 kPa.m3/kg.K is cv = 3.1156 kJ/kg·°C for He (Tables A-1 and A-2)Analysis The mass of each gas in the cylinder is

kg0.808K298K/kgmkPa2.0769

m1kPa500


m1kPa500

3

3

He1

11He

3

3

N1

11N

2

2

RTP

m

RTP

m

V

V

He1 m3

500 kPa25 C

N21 m3

500 kPa 80 C

Taking the entire contents of the cylinder as our system, the 1stlaw relation can be written as

He12N12

HeN


system


outin

)]([)]([0

0

2

2

TTmcTTmc

UUU

EEE

vv

Substituting,0C25CkJ/kg3.1156kg0.808C80CkJ/kg0.743kg4.77 ff TT

It gives Tf = 57.2 Cwhere Tf is the final equilibrium temperature in the cylinder.

The answer would be the same if the piston were not free to move since it would effect onlypressure, and not the specific heats. (b) We take the entire cylinder as our system, which is a closed system. Noting that the cylinder is well-insulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as

HeNgen

entropyinChange

system

generationEntropy

gen


outin

20 SSS

SSSS

But first we determine the final pressure in the cylinder:

kPa510.6m2

K330.2K/kmolmkPa8.314kmol0.372

kmol0.372kg/kmol4

kg0.808kg/kmol28

kg4.77

3

3

total

total2

HeNHeNtotal

2

2

VTRN

P

Mm

MmNNN

u

Then,

kJ/K0.361kPa500kPa510.6

lnKkJ/kg0.2968K353K330.2

lnKkJ/kg1.039kg4.77

lnln2

2N1

2

1

2N P

PR

TT

cmS p

8-103

KkJ0340395.0361.0

kJ/K0.395kPa500kPa510.6

lnKkJ/kg2.0769K298K330.2


lnln

HeNgen

He1

2

1

2He

2/.SSS

PP

RTT

cmS p

The wasted work potential is equivalent to the exergy destroyed during a process, and it can be determinedfrom an exergy balance or directly from its definition ,X T gendestroyed 0S

kJ10.1kJ/K)K)(0.034298(0destroyed genSTX

If the piston were not free to move, we would still have T2 = 330.2 K but the volume of each gas wouldremain constant in this case:

kJ/K0.021258.0237.0

kJ/K0.258K298K330.2


kJ/K0.237K353K330.2


HeNgen

He

0

1

2

1

2He

N

0

1

2

1

2N

2

2

2

SSS

RTT

cmS

RTT

cmS

V

V

V

V

v

v

and kJ6.26kJ/K)K)(0.021298(gen0destroyed STX

8-104

8-119 An insulated cylinder is divided into two parts. One side of the cylinder contains N2 gas and theother side contains He gas at different states. The final equilibrium temperature in the cylinder and thewasted work potential are to be determined for the cases of piston being fixed and moving freely.Assumptions 1 Both N2 and He are ideal gases with constant specific heats. 2 The energy stored in the container itself, except the piston, is negligible. 3 The cylinder is well-insulated and thus heat transfer is negligible. 4 Initially, the piston is at the average temperature of the two gases.Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m3/kg.K is cv = 0.743 kJ/kg·°C for N2, and R = 2.0769 kPa.m3/kg.K is cv = 3.1156 kJ/kg·°C for He (Tables A-1 and A-2). The specific heat of copper piston is c = 0.386 kJ/kg·°C (Table A-3). Analysis The mass of each gas in the cylinder is


m1kPa500


m1kPa500

3

3

He1

11He

3

3

N1

11N

2

2

RTP

m

RTP

m

V

V

Copper

He1 m3

500 kPa25 C

N21 m3

500 kPa 80 C

Taking the entire contents of the cylinder as our system, the 1st law relation can be written as

)]([)]([)]([0

0

Cu12He12N12

CuHeN


system


outin

2

2

TTmcTTmcTTmc

UUUU

EEE

vv

whereT1, Cu = (80 + 25) / 2 = 52.5 C

Substituting,

0C52.5CkJ/kg0.386kg5.0

C25CkJ/kg3.1156kg0.808C80CkJ/kg0.743kg4.77

f

ff

T

TT

It givesTf = 56.0 C

where Tf is the final equilibrium temperature in the cylinder.The answer would be the same if the piston were not free to move since it would effect only

pressure, and not the specific heats. (b) We take the entire cylinder as our system, which is a closed system. Noting that the cylinder is well-insulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as

pistonHeNgen

entropyinChange

system

generationEntropy

gen


outin

20 SSSS

SSSS

But first we determine the final pressure in the cylinder:

kPa508.8m2

K329K/kmolmkPa8.314kmol0.372

kmol0.372kg/kmol4

kg0.808kg/kmol28

kg4.77

3

3

total

total2

HeNHeNtotal

2

2

V

TRNP

Mm

MmNNN

u

Then,

8-105

kJ/K0.0334021.0386.0374.0

kJ/K0.021K325.5

K329lnKkJ/kg0.386kg5ln

kJ/K0.386kPa500kPa508.8

lnKkJ/kg2.0769K353K329


lnln

kJ/K0.374kPa500kPa508.8

lnKkJ/kg0.2968K353K329

lnKkJ/kg1.039kg4.77

lnln

pistonHeNgen

piston1

2piston

He1

2

1

2He

N1

2

1

2N

2

2

2

SSSSTT

mcS

PP

RTT

cmS

PP

RTT

cmS

p

p

The wasted work potential is equivalent to the exergy destroyed during a process, and it can be determinedfrom an exergy balance or directly from its definition ,gen0destroyed STX

kJ9.83kJ/K)K)(0.033298(gen0destroyed STX

If the piston were not free to move, we would still have T2 = 330.2 K but the volume of each gas wouldremain constant in this case:

kJ/K0.020021.0249.0250.0

kJ/K0.249K353K329


kJ/K0.250K353K329


pistonHeNgen

He

0

1

2

1

2He

N

0

1

2

1

2N

2

2

2

SSSS

RTT

cmS

RTT

cmS

V

V

V

V

v

v

and kJ6.0kJ/K)K)(0.020298(gen0destroyed STX

8-106

8-120E Argon enters an adiabatic turbine at a specified state with a specified mass flow rate, and leaves ata specified pressure. The isentropic efficiency of turbine is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Argon is an ideal gas with constant specific heats.Properties The specific heat ratio of argon is k = 1.667. The constant pressure specific heat of argon is cp = 0.1253 Btu/lbm.R (Table A-2E). Analysis There is only one inlet and one exit, and thus m m m1 2 . We take the isentropic turbine as thesystem, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

outin


(steady)0system


outin 0

EE

EEE1

370 kWAr

T)(

0)peke(since

21out,

2out,1

ss

ss

hhmW

QhmWhm

From the isentropic relations,

R5.917psia200

psia30R)1960(667.1/667.0/)1(

1

212

kks

s PPTT 2

Then the power output of the isentropic turbine becomes

hp123.2=Btu/min42.41hp1R)5.917R)(1960Btu/lbm.1253lbm/min)(040(21out, sps TTcmW

Then the isentropic efficiency of the turbine is determined from

77.1%771.0hp2.123

hp95

out,

out,

s

aT W

W

(b) Using the steady-flow energy balance relation 21out, TTcm paW above, the actual turbine exittemperature is determined to be

R1.1156F1.696hp1Btu/min41.42

R)Btu/lbm.1253lbm/min)(0(40hp95

1500out,12

p

a

cmW

TT

The entropy generation during this process can be determined from an entropy balance on the turbine,)(00 12gengen21


0system


gen


outin ssmSSsmsmSSSS

where

Btu/lbm.R02816.0psia200

psia30lnR)Btu/lbm(0.04971R1960R1156.1lnR)Btu/lbm1253.0(lnln

1

2

1

212 P

PR

TT

css p


hp3.14Btu/min42.41hp1R)Btu/lbmR)(0.0281637lbm/min)(540()( 120gen0destroyed ssTmSTX

Then the reversible power and second-law efficiency becomehp3.1093.1495destroyedout,outrev, XWW a

and 86.9%hp3.109

hp95

revII W

W

8-107

8-121 [Also solved by EES on enclosed CD] The feedwater of a steam power plant is preheated using steamextracted from the turbine. The ratio of the mass flow rates of the extracted steam and the feedwater are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Heat loss from the device to the surroundings is negligible and thus heattransfer from the hot fluid is equal to the heat transfer to the cold fluid. Properties The properties of steam and feedwater are (Tables A-4 through A-6)

KkJ/kg2.0417kJ/kg719.08

C170C10MPa2.5

KkJ/kg0.7038kJ/kg209.34

C50MPa2.5

C179.88KkJ/kg2.1381

kJ/kg762.51

liquidsat.MPa1

KkJ/kg6.6956kJ/kg2828.3

C200MPa1

C170@4

C170@4

24

4

C50@3

C50@3

3

3

2

MPa1@2

MPa1@22

1

1

1

1

f

f

f

f

f

f

sshh

TTP

sshh

TP

Tss

hhP

sh

TP

Analysis (a) We take the heat exchanger as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as follows:

2.5MPa

Feedwater

1 MPa200 C

Steamfromturbine

4

3

2

1

sat. liquidMass balance (for each fluid stream):

fws mmmmmmmmmmm 4321outin(steady)0

systemoutin and0

Energy balance (for the heat exchanger):

0)peke(since

0

44223311

outin


(steady)0system


outin

WQhmhmhmhm

EEEEE

Combining the two, 4312 hhmhhm fws

Dividing by and substituting,mfw

0.247kJ/kg2828.3762.51kJ/kg719.08209.34

12

43

hhhh

mm

fw

s

(b) The entropy generation during this process per unit mass of feedwater can be determined from anentropy balance on the feedwater heater expressed in the rate form as

0)()(

0

0

gen4321

gen44332211


0system


gen


outin

Sssmssm

Ssmsmsmsm

SSSS

fws

fwkgkJ/K0.2137038.00417.26956.61381.2247.03412gen ssss

mm

mS

fw

s

fw

Noting that this process involves no actual work, the reversible work and exergy destruction becomeequivalent since The exergy destroyed during a

process can be determined from an exergy balance or directly from its definition ,

.destroyedoutrev,outact,outrev,destroyed XWWWX

gen0destroyed STX

feedwaterkJ/kg63.5/. fw)kgKkJ213K)(0298(gen0destroyed STX

8-108

8-122 EES Problem 8-121 is reconsidered. The effect of the state of the steam at the inlet of the feedwaterheater on the ratio of mass flow rates and the reversible power is to be investigated.Analysis Using EES, the problem is solved as follows:

"Input Data""Steam (let st=steam data):"Fluid$='Steam_IAPWS'T_st[1]=200 [C]{P_st[1]=1000 [kPa]}P_st[2] = P_st[1]x_st[2]=0 "saturated liquid, quality = 0%"T_st[2]=temperature(steam, P=P_st[2], x=x_st[2])

"Feedwater (let fw=feedwater data):"T_fw[1]=50 [C] P_fw[1]=2500 [kPa] P_fw[2]=P_fw[1] "assume no pressure drop for the feedwater"T_fw[2]=T_st[2]-10

"Surroundings:"T_o = 25 [C] P_o = 100 [kPa] "Assumed value for the surrroundings pressure""Conservation of mass:""There is one entrance, one exit for both the steam and feedwater.""Steam: m_dot_st[1] = m_dot_st[2]""Feedwater: m_dot_fw[1] = m_dot_fw[2]""Let m_ratio = m_dot_st/m_dot_fw""Conservation of Energy:""We write the conservation of energy for steady-flow control volumehaving two entrances and two exits with the above assumptions. Sinceneither of the flow rates is know or can be found, write the conservationof energy per unit mass of the feedwater."E_in - E_out =DELTAE_cvDELTAE_cv=0 "Steady-flow requirement" E_in = m_ratio*h_st[1] + h_fw[1] h_st[1]=enthalpy(Fluid$, T=T_st[1], P=P_st[1])h_fw[1]=enthalpy(Fluid$,T=T_fw[1], P=P_fw[1])E_out = m_ratio*h_st[2] + h_fw[2] h_fw[2]=enthalpy(Fluid$, T=T_fw[2], P=P_fw[2])h_st[2]=enthalpy(Fluid$, x=x_st[2], P=P_st[2])

"The reversible work is given by Eq. 7-47, where the heat transfer is zero(the feedwater heater is adiabatic) and the Exergy destroyed is set equalto zero"W_rev = m_ratio*(Psi_st[1]-Psi_st[2]) +(Psi_fw[1]-Psi_fw[2])Psi_st[1]=h_st[1]-h_st_o -(T_o + 273)*(s_st[1]-s_st_o) s_st[1]=entropy(Fluid$,T=T_st[1], P=P_st[1])h_st_o=enthalpy(Fluid$, T=T_o, P=P_o) s_st_o=entropy(Fluid$, T=T_o, P=P_o) Psi_st[2]=h_st[2]-h_st_o -(T_o + 273)*(s_st[2]-s_st_o) s_st[2]=entropy(Fluid$,x=x_st[2], P=P_st[2])Psi_fw[1]=h_fw[1]-h_fw_o -(T_o + 273)*(s_fw[1]-s_fw_o) h_fw_o=enthalpy(Fluid$, T=T_o, P=P_o) s_fw[1]=entropy(Fluid$,T=T_fw[1], P=P_fw[1])s_fw_o=entropy(Fluid$, T=T_o, P=P_o)

8-109

Psi_fw[2]=h_fw[2]-h_fw_o -(T_o + 273)*(s_fw[2]-s_fw_o) s_fw[2]=entropy(Fluid$,T=T_fw[2], P=P_fw[2])

mratio [kg/kg] Wrev [kJ/kg] Pst,1 [kPa]0.06745 12.9 1000.1067 23.38 2000.1341 31.24 3000.1559 37.7 4000.1746 43.26 5000.1912 48.19 6000.2064 52.64 7000.2204 56.72 8000.2335 60.5 9000.246 64.03 1000

100 200 300 400 500 600 700 800 900 10000.04

0.08

0.12

0.16

0.2

0.24

0.28

Pst[1] [kPa]

mratio

[kgst

/kg,fw]

100 200 300 400 500 600 700 800 900 100010

20

30

40

50

60

70

Pst[1] [kPa]

Wrev

[kJ/kgfw

]

8-110

8-123 A 1-ton (1000 kg) of water is to be cooled in a tank by pouring ice into it. The final equilibriumtemperature in the tank and the exergy destruction are to be determined.Assumptions 1 Thermal properties of the ice and water are constant. 2 Heat transfer to the water tank is negligible. 3 There is no stirring by hand or a mechanical device (it will add energy).Properties The specific heat of water at room temperature is c = 4.18 kJ/kg·°C, and the specific heat of iceat about 0 C is c = 2.11 kJ/kg·°C (Table A-3). The melting temperature and the heat of fusion of ice at 1 atm are 0 C and 333.7 kJ/kg..Analysis (a) We take the ice and the water as the system, and disregard any heat transfer between the system and the surroundings. Then theenergy balance for this process can be written as

ice-5 C80 kg

WATER1 ton

waterice


system


outin

00

UUU

EEE

0)]([])C0()C0([ water12iceliquid2solid1 TTmcTmcmhTmc if

Substituting,

(

( )( )( )

80 2

2

kg){(2.11 kJ / kg C)[0 (-5)] C + 333.7 kJ / kg + (4.18 kJ / kg C)( 0) C}

1000 kg 4.18 kJ / kg C 20 C 0

T

T

It gives T2 = 12.42 Cwhich is the final equilibrium temperature in the tank.(b) We take the ice and the water as our system, which is a closed system .Considering that the tank iswell-insulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as

waterSSS

SSSS

icegen

entropyinChange

system

generationEntropy

gen


outin

0

where

kJ/K115.783K273

K285.42lnKkJ/kg4.18

K273kJ/kg333.7

K268K273

lnKkJ/kg2.11kg80

lnln

kJ/K109.590K293

K285.42lnKkJ/kg4.18kg1000ln

iceliquid1

2

meltingsolid1

melting

iceliquidmeltingsolidice

water1

2water

TT

mcTmh

TT

mc

SSSS

TT

mcS

ig

Then, kJ/K.1936783.115590.109icewatergen SSS


kJ1815=kJ/K)K)(6.193293(gen0destroyed STX

8-111

8-124 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fillthe bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanicalequilibrium is established and the amount of exergy destroyed are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Air is an ideal gas. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The direction of heat transfer is to the air in the bottle (will be verified).Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1).Analysis We take the bottle as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances can be expressed as Mass balance: )0(since initialout2systemoutin mmmmmmm i

Energy balance:

)0peke(since initialout22in


system


outin

EEWumhmQ

EEE

ii

Combining the two balances:

ihumQ 22in100 kPa

17 C

12 L Evacuated

where

kJ/kg206.91kJ/kg290.16

K290


m0.012kPa100

2

17-ATable2

3

3

2

22

uh

TT

RTPm

ii

V

Substituting,Qin = (0.0144 kg)(206.91 - 290.16) kJ/kg = - 1.2 kJ Qout = 1.2 kJ

Note that the negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reversed the direction.

The entropy generated during this process is determined by applying the entropy balance on anextended system that includes the bottle and its immediate surroundings so that the boundary temperatureof the extended system is the temperature of the surroundings at all times. The entropy balance for it can beexpressed as

220

1122tankgeninb,

out

entropyinChange

system

generationEntropy

gen


outin

smsmsmSSTQ

sm

SSSS

ii

Therefore, the total entropy generated during this process is

kJ/K0.00415K290

kJ1.2

surr

out

outb,

out022

outb,

out22gen T

QTQ

ssmTQ

smsmS iii


kJ1.2=kJ/K)K)(0.00415290(gen0destroyed STX

8-112

8-125 A heat engine operates between two tanks filled with air at different temperatures. The maximumwork that can be produced and the final temperatures of the tanks are to be determined.Assumptions Air is an ideal gas with constant specific heats at room temperature.Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The constant volume specific heat of air at room temperature is cv = 0.718 kJ/kg.K (Table A-2). Analysis For maximum power production, the entropy generation must be zero. We take the two tanks(the heat source and heat sink) and the heat engine as the system. Noting that the system involves no heatand mass transfer and that the entropy change for cyclic devices is zero, the entropy balance can be expressed as

0engineheatsinktank,sourcetank,

0gen

entropyinChange

system

generationEntropy

0gen


outin

0 SSSS

SSSS

QH

AIR30 kg 900 K

QL

AIR30 kg 300 K

HE

BABA

TTTTT

TT

mRTT

mcmRTT

mc

SS

112

21

2

1

2

sink

0

1

2

1

2

source

0

1

2

1

2

sinktank,sourcetank,

0ln

0lnlnlnln

0

V

V

V

Vvv W

where T1A and T1B are the initial temperatures of the source and the sink,respectively, and T2 is the common final temperature. Therefore, the final temperature of the tanks for maximum power production is

K519.6K)K)(300900(112 BATTT

The energy balance for the source and sink can be expressed as follows:E E Ein out system

Source: )()( 21outsource,12outsource, TTmcQTTmcUQ AA vv

kJ8193=519.6)KK)(900kJ/kgkg)(0.71830(=)( 21outsource, TTmcQ Av

Sink: kJ4731=300)KK)(519.6kJ/kgkg)(0.718(30=)( 12insink, BTTmcQ v

Then the work produced in this case becomeskJ346347318193insink,outsource,outmax, QQQQW LH

Therefore, a maximum of 3463 kJ of work can be produced during this process.

8-113

8-126 A heat engine operates between two constant-pressure cylinders filled with air at differenttemperatures. The maximum work that can be produced and the final temperatures of the cylinders are to be determined.Assumptions Air is an ideal gas with constant specific heats at room temperature.Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The constant pressure specific heat of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). Analysis For maximum power production, the entropy generation must be zero. We take the two cylinders(the heat source and heat sink) and the heat engine as the system. Noting that the system involves no heatand mass transfer and that the entropy change for cyclic devices is zero, the entropy balance can be expressed as

0engineheatsinkcylinder,sourcecylinder,

0gen

entropyinChange

system

generationEntropy

0gen


outin

0 SSSS

SSSS

W

QH

AIR30 kg 900 K

QL

AIR30 kg 300 K

HE

BABA

pp

TTTTT

TT

PP

mRTT

mcPP

mRTT

mc

SS

112

21

2

1

2

sink

0

1

2

1

2

source

0

1

2

1

2

sinkcylinder,sourcecylinder,

0ln

0lnln0lnln

0

where T1A and T1B are the initial temperatures of the source and thesink, respectively, and T2 is the common final temperature.Therefore, the final temperature of the tanks for maximum power production is

K519.6K)K)(300900(112 BATTT

The energy balance for the source and sink can be expressed as follows:E E Ein out system

Source: )( 21outsource,,outsource, TTmcHQUWQ Apinb

kJ11,469=519.6)KK)(900kJ/kgkg)(1.00530(=)( 21outsource, TTmcQ Ap

Sink: )( 12insink,,insink, Apoutb TTmcHQUWQ

kJ6621=300)KK)(519.6kJ/kgkg)(1.005(30=)( 12insink, Bp TTmcQ

Then the work produced becomeskJ48476621469,11insink,outsource,outmax, QQQQW LH

Therefore, a maximum of 4847 kJ of work can be produced during this process

8-114

8-127 A pressure cooker is initially half-filled with liquid water. It is kept on the heater for 30 min. The amount water that remained in the cooker and the exergy destroyed are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of water vapor leaving the cooker remains constant. 2 Kinetic and potential energies are negligible. 3 Heat loss from the cooker isnegligible.Properties The properties of water are (Tables A-4 through A-6)

kJ/kg.K1716.7kJ/kg.K,4850.1

kJ/kg5.2524kJ/kg,82.486

/kgm1.0037/kg,m0.001057kPa517 331

gf

gf

gf

ss

uu

P vv

750 W

4 L 175 kPa

KkJ/kg1716.7kJ/kg2.2700

vaporsat.kPa175

kPa175@

kPa175@

ge

gee

sshhP

Analysis (a) We take the cooker as the system, which is a controlvolume since mass crosses the boundary. Noting that the microscopicenergies of flowing and nonflowing fluids are represented by enthalpyh and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e

Energy balance:

)0peke(since1122ine,


system


outin

QumumhmW

EEE

ee

The initial mass, initial internal energy, initial entropy, and final mass in the tank are

kJ/K8239.21716.7002.04850.1892.1

kJ6.9265.2524002.082.486893.1

kg8945.1002.0893.1/kgm0037.1

m002.0/kgm001057.0

m002.0

m002.0=L2

111

111

3

3

3

3

1

3

ggff

ggff

g

g

f

fg

gf

smsmsmS

umumumU

mmmv

V

v

V

VV

f

2

3

22

m004.0vv

Vm

The amount of electrical energy supplied during this process iskJ900=s)60kJ/s)(20750.0(,, tWW ineine

Then from the mass and energy balances,

kJ6.926))(004.0(+kJ/kg)2.2700)(004.0(1.894=kJ900

004.0894.1

222

221

u

mmme

vv

v

Substituting u , and solving for xfgffgf xuxu vvv 2222 and 2 yields

x2 = 0.001918 Thus,

8-115

Kkg/kJ5642.16865.5001918.04850.1

kg/m002654.0)001057.00037.1(001918.0001057.0

22

322

fgf

fgf

sxss

x vvv

and kg1.507kg/m002654.0

m004.03

3

22 v

Vm

(b) The entropy generated during this process is determined by applying the entropy balance on the cooker. Noting that there is no heat transfer and some mass leaves, the entropy balance can be expressed as

1122gen

1122sysgen

entropyinChange

system

generationEntropy

gen


outin

smsmsmS

smsmSSsm

SSSS

ee

ee

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition . Using the Sgen0destroyed STX gen relation obtained above and substituting,

kJ689

8239.25642.1507.11716.7)507.1(1.894K)298(

)( 11220gen0destroyed smsmsmTSTX ee

8-116

8-128 A pressure cooker is initially half-filled with liquid water. Heat is transferred to the cooker for 30 min. The amount water that remained in the cooker and the exergy destroyed are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of water vapor leaving the cooker remains constant. 2 Kinetic and potential energies are negligible. 3 Heat loss from the cooker isnegligible.Properties The properties of water are (Tables A-4 through A-6)

kJ/kg.K1716.7kJ/kg.K,4850.1

kJ/kg5.2524kJ/kg,82.486

/kgm1.0037/kg,m0.001057175 331

gf

gf

gf

ss

uu

kPaP vv

750 W

4 L 175 kPa KkJ/kg1716.7

kJ/kg2.2700vaporsat.

kPa175

kPa175@

kPa175@

ge

gee

sshhP

Analysis (a) We take the cooker as the system, which is a controlvolume since mass crosses the boundary. Noting that the microscopicenergies of flowing and nonflowing fluids are represented by enthalpyh and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: 21systemoutin mmmmmm e

Energy balance:

)0peke(since1122in


system


outin

WumumhmQ

EEE

ee

The initial mass, initial internal energy, initial entropy, and final mass in the tank are

kJ/K8239.21716.7002.04850.1892.1

kJ6.9265.2524002.082.486893.1

kg8945.1002.0893.1/kgm0037.1

m002.0/kgm001057.0

m002.0

m002.0=L2

111

111

3

3

3

3

1

3

ggff

ggff

g

g

f

fgf

gf

smsmsmS

umumumU

mmmv

V

v

V

VV

2

3

22

m004.0vv

Vm

The amount of heat transfer during this process iskJ900=s)60kJ/s)(20750.0(tQQ

Then from the mass and energy balances,

kJ6.926))(004.0(+kJ/kg)2.2700)(004.0(1.894=kJ900

004.0894.1

222

221

u

mmme

vv

v

Substituting u , and solving for xfgffgf xuxu vvv 2222 and 2 yields

x2 = 0.001918 Thus,

8-117

Kkg/kJ5642.16865.5001918.04850.1

kg/m002654.0)001057.00037.1(001918.0001057.0

22

322

fgf

fgf

sxss

x vvv

and kg1.507kg/m002654.0

m004.03

3

22 v

Vm

(b) The entropy generated during this process is determined by applying the entropy balance on anextended system that includes the cooker and its immediate surroundings so that the boundary temperatureof the extended system at the location of heat transfer is the heat source temperature, Tsource = 180 C at all times. The entropy balance for it can be expressed as

source

in1122gen

1122sysgeninb,

in

entropyinChange

system

generationEntropy

gen


outin

TQ

smsmsmS

smsmSSsmTQ

SSSS

ee

ee

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition . Using the Sgen0destroyed STX gen relation obtained above and substituting,

kJ96.8

453/9008239.25642.1507.11716.7)507.1(1.894K)298(

source


QsmsmsmTSTX ee

Note that the exergy destroyed is much less when heat is supplied from a heat source rather than an electric resistance heater.

8-118

8-129 A heat engine operates between a nitrogen tank and an argon cylinder at different temperatures. The maximum work that can be produced and the final temperatures are to be determined.Assumptions Nitrogen and argon are ideal gases with constant specific heats at room temperature.Properties The constant volume specific heat of nitrogen at room temperature is cv = 0.743 kJ/kg.K. The constant pressure specific heat of argon at room temperature is cp = 0.5203 kJ/kg.K (Table A-2). Analysis For maximum power production, the entropy generation must be zero. We take the tank, thecylinder (the heat source and the heat sink) and the heat engine as the system. Noting that the systeminvolves no heat and mass transfer and that the entropy change for cyclic devices is zero, the entropybalance can be expressed as

0engineheatsinkcylinder,sourcetank,

0gen

entropyinChange

system

generationEntropy

0gen


outin

0 SSSS

SSSS

0lnln0lnln

0)()(

sink

0

1

2

1

2

source

0

1

2

1

2

sinksource

PP

mRTT

mcmRTT

mc

SS

pV

Vv

Substituting,

( ln ( ln201000

10300

02 2 kg)(0.743 kJ / kg K) K

kg)(0.5203 kJ / kg K)K

T T

W

QH

N220 kg

1000 K

QL

Ar10 kg 300 K

HE

Solving for T2 yieldsT2 = 731.8 K

where T2 is the common final temperature of the tanks for maximum power production.The energy balance for the source and sink can be expressed as follows:systemoutin EEE


kJ3985=)K8.731K)(1000kJ/kgkg)(0.74320(=)( 21outsource, TTmcQ Av

Sink: )( 12insink,outb,insink, Ap TTmcHQUWQ

kJ2247=)K300K)(731.8kJ/kgkg)(0.520310(=)( 12insink, ATTmcQ v

Then the work produced becomeskJ173922473985insink,outsource,outmax, QQQQW LH

Therefore, a maximum of 1739 kJ of work can be produced during this process

8-119

8-130 A heat engine operates between a tank and a cylinder filled with air at different temperatures. Themaximum work that can be produced and the final temperatures are to be determined.Assumptions Air is an ideal gas with constant specific heats at room temperature.Properties The specific heats of air are cv = 0.718 kJ/kg.K and cp = 1.005 kJ/kg.K (Table A-2). Analysis For maximum power production, the entropy generation must be zero. We take the tank, thecylinder (the heat source and the heat sink) and the heat engine as the system. Noting that the systeminvolves no heat and mass transfer and that the entropy change for cyclic devices is zero, the entropybalance can be expressed as

0engineheatsinkcylinder,sourcetank,

0gen

entropyinChange

system

generationEntropy

0gen


outin

0 SSSS

SSSS

W

QH

Air20 kg 800 K

QL

Air20 kg 290 K

HE

)1/(1112

1

2

1

2

1

2

1

2

sink

0

1

2

1

2

source

0

1

2

1

2

sinksource

10lnln

0lnln0lnln

0)()(

kkBA

k

BAB

p

A

p

TTTTT

TT

TT

cc

TT

PP

mRTT

mcmRTT

m

SS

v

v V

Vc

where T1A and T1B are the initial temperatures of the source and the sink,respectively, and T2 is the common final temperature. Therefore, the finaltemperature of the tanks for maximum power production is

K442.64.21

1.42 K)K)(290800(T


kJ5132=)K6.442K)(800kJ/kgkg)(0.71820(=)( 21outsource, TTmcQ Av

Sink: )( 12insink,outb,insink, Ap TTmcHQUWQ

kJ3068=)K290K)(442.6kJ/kgkg)(1.00520(=)( 12insink, ATTmcQ v

Then the work produced becomeskJ206430685132insink,outsource,outmax, QQQQW LH

Therefore, a maximum of 2064 kJ of work can be produced during this process.

8-120

8-131 Using an incompressible substance as an example, it is to be demonstrated if closed system and flowexergies can be negative. Analysis The availability of a closed system cannot be negative. However, the flow availability can be negative at low pressures. A closed system has zero availability at dead state, and positive availability at any other state since we can always produce work when there is a pressure or temperature differential.

To see that the flow availability can be negative, consider an incompressible substance. The flowavailability can be written as

h h T s su u P P T s s

P P

0 0 0

0 0 0

0

( )( ) ( ) (

( )v

v0 )

The closed system availability is always positive or zero, and the flow availability can be negative whenP << P0.

8-132 A relation for the second-law efficiency of a heat engine operating between a heat source and a heatsink at specified temperatures is to be obtained.Analysis The second-law efficiency is defined as the ratio of the availability recovered to availability supplied during a process. The work W produced is theavailability recovered. The decrease in the availability of the heat supplied QH is the availability supplied or invested. QH

SourceTH

QL

TLSink

HE

Therefore,

)(11 00II

WQTT

QTT

W

HL

HH

W

Note that the first term in the denominator is the availability of heat supplied to the heat engine whereas the second term is the availability of the heat rejected by the heat engine. The difference between the two is the availability consumedduring the process.

8-121

8-133E Large brass plates are heated in an oven at a rate of 300/min. The rate of heat transfer to the platesin the oven and the rate of exergy destruction associated with this heat transfer process are to be determined.Assumptions 1 The thermal properties of the plates are constant. 2 The changes in kinetic and potentialenergies are negligible. 3 The environment temperature is 75 F.Properties The density and specific heat of the brass are given to be = 532.5 lbm/ft3 and cp = 0.091 Btu/lbm. F.Analysis We take the plate to be the system. The energy balance for this closed system can be expressed as

)()( 1212platein


system


outin

TTmcuumUQ

EEE

The mass of each plate and the amount of heat transfer to each plate is

lbm213ft)]ft)(22)(ft12/2.1)[(lbm/ft5.532( 3LAVm

Btu/plate930,17F)751000(F)Btu/lbm.091.0)(lbm/plate213()( 12in TTmcQ

Then the total rate of heat transfer to the plates becomes

Q Btu/s89,650=Btu/min5,379,000)Btu/plate930,17()plates/min300(plateperin,platetotal Qn

We again take a single plate as the system. The entropy generated during this process can be determined byapplying an entropy balance on an extended system that includes the plate and its immediate surroundingsso that the boundary temperature of the extended system is at 1300 F at all times:

systemin

gensystemgenin

entropyinChange

system

generationEntropy

gen


outin

STQ

SSSTQ

SSSS

bb

where

Btu/R46.19R460)+(75

R460)+(1000lnBtu/lbm.R)091.0)(lbm213(ln)(1

2avg12system T

TmcssmS

Substituting,

plate)(perBtu/R272.9Btu/R46.19R460+1300

Btu17,930system

ingen S

TQ

Sb


Btu/s.R46.35=Btu/min.R2781=)plates/minplate)(300Btu/R272.9(ballgengen nSS


Btu/s24,797=Btu/s.R)R)(46.35535(gen0destroyed STX

8-122

8-134 Long cylindrical steel rods are heat-treated in an oven. The rate of heat transfer to the rods in theoven and the rate of exergy destruction associated with this heat transfer process are to be determined.Assumptions 1 The thermal properties of the rods are constant. 2 The changes in kinetic and potentialenergies are negligible. 3 The environment temperature is 30 C.Properties The density and specific heat of the steel rods are given to be = 7833 kg/m3 and cp = 0.465 kJ/kg. C.Analysis Noting that the rods enter the oven at a velocity of 3 m/min and exit at the same velocity, we cansay that a 3-m long section of the rod is heated in the oven in 1 min. Then the mass of the rod heated in 1 minute is

m V LA L D( / ) ( )( [ ( . ) / .2 24 7833 3 01 4 184 6 kg / m m) m ] kg3

We take the 3-m section of the rod in the oven as the system. The energy balancefor this closed system can be expressed as

)()( 1212rodin


system


outin

TTmcuumUQ

EEE

Substituting,kJ512,57C)30700(C)kJ/kg.465.0)(kg6.184()( 12in TTmcQ

Noting that this much heat is transferred in 1 min, the rate of heat transfer to the rod becomes

kW958.5=kJ/min57,512=min)kJ)/(1512,57(/inin tQQ

We again take the 3-m long section of the rod as the system The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the rod and itsimmediate surroundings so that the boundary temperature of the extended system is at 900 C at all times:

systemin

gensystemgenin

entropyinChange

system

generationEntropy

gen


outin

STQ

SSSTQ

SSSS

bb

where

kJ/K1.100273+30273+700lnkJ/kg.K)465.0)(kg6.184(ln)(

1

2avg12system T

TmcssmS

Substituting,

kJ/K1.51kJ/K1.100R273)+(900

kJ57,512system

ingen S

TQ

Sb

Noting that this much entropy is generated in 1 min, the rate of entropy generation becomes

kW/K0.852=kJ/min.K1.51min1kJ/K1.51gen

gen tS

S


kW254=kW/K)K)(0.852298(gen0destroyed STX

8-123

8-135 Steam is condensed by cooling water in the condenser of a power plant. The rate of condensation of steam and the rate of exergy destruction are to be determined.Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to thecold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid propertiesare constant. Properties The enthalpy and entropy of vaporization of water at 60 C are hfg =2357.7 kJ/kg and sfg=7.0769 kJ/kg.K (Table A-4). The specific heat of water at room temperature is cp = 4.18 kJ/kg. C (TableA-3).Analysis (a) We take the cold water tubes as the system, which is a control volume.The energy balance for this steady-flow system can be expressed in the rate form as

)(

0)peke(since

0

12in

21in

outin


(steady)0system


outin

TTCmQ

hmhmQ

EE

EEE

p

25 C

60 C

15 C

Water

Steam60 C

Then the heat transfer rate to the cooling water in thecondenser becomes

kJ/s5852=C)15CC)(25kJ/kg.kg/s)(4.18(140

)]([ watercoolinginout TTCmQ p

The rate of condensation of steam is determined to be

kg/s2.482kJ/kg7.2357kJ/s5852)( steamsteam

fgfg h

QmhmQ

(b) The rate of entropy generation within the condenser during this process can be determined by applyingthe rate form of the entropy balance on the entire condenser. Noting that the condenser is well-insulatedand thus heat transfer is negligible, the entropy balance for this steady-flow system can be expressed as

)()(

0

)0(since0

34steam12watergen

gen4steam2water3steam1water

gen44223311


(steady)0system


gen


outin

ssmssmS

Ssmsmsmsm

QSsmsmsmsm

SSSS

Noting that water is an incompressible substance and steam changes from saturated vapor to saturatedliquid, the rate of entropy generation is determined to be

kW/K.4092kJ/kg.K)69kg/s)(7.07482.2(273+15273+25kJ/kg.K)lnkg/s)(4.18140(

ln)(ln steam1

2watersteam

1

2watergen fgpgfp sm

TT

cmssmTT

cmS

Then the exergy destroyed can be determined directly from its definition to begen0destroyed STX

kW694=kW/K)K)(2.409288(gen0destroyed STX

8-124

8-136 Water is heated in a heat exchanger by geothermal water. The rate of heat transfer to the water andthe rate of exergy destruction within the heat exchanger are to be determined.Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to thecold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid propertiesare constant. 5 The environment temperature is 25 C.Properties The specific heats of water and geothermal fluid are given to be 4.18 and 4.31 kJ/kg. C,respectively.Analysis (a) We take the cold water tubes as the system,which is a control volume. The energy balance for thissteady-flow system can be expressed in the rate form as

60 C

Brine140 C

Water25 C

)(

0)peke(since

0

12in

21in

outin


(steady)0system


outin

TTcmQ

hmhmQ

EE

EEE

p

Then the rate of heat transfer to the cold water in the heat exchanger becomes

kW58.52=C)25CC)(60kJ/kg.kg/s)(4.184.0()]([ waterinoutwaterin, TTcmQ p

Noting that heat transfer to the cold water is equal to the heat loss from the geothermal water, the outlettemperature of the geothermal water is determined from

C7.94C)kJ/kg.kg/s)(4.313.0(

kW52.58C140)]([ outinoutgeooutinout

pp cm

QTTTTcmQ


)()(

0

)0(since0

34geo12watergen

gen4geo2water3geo1water

gen44223311


(steady)0system


gen


outin

ssmssmS

Ssmsmsmsm

QSsmsmsmsm

SSSS

Noting that both fresh and geothermal water are incompressible substances, the rate of entropy generationis determined to be

kW/K0.0356273+140273+94.7kJ/kg.K)lnkg/s)(4.313.0(

273+25273+60kJ/kg.K)lnkg/s)(4.184.0(

lnln3

4geo

1

2watergen T

Tcm

TT

cmS pp

gen0destroyed STXThe exergy destroyed during a process can be determined from an exergy balance or directly from its definition ,

kW10.61=kW/K)K)(0.0356298(gen0destroyed STX

8-125

8-137 Ethylene glycol is cooled by water in a heat exchanger. The rate of heat transfer and the rate of exergy destruction within the heat exchanger are to be determined.Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to thecold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid propertiesare constant. 5 The environment temperature is 20 C.Properties The specific heats of water and ethylene glycol are given to be 4.18 and 2.56 kJ/kg. C,respectively.Analysis (a) We take the ethylene glycol tubes as the system,which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

40 C

Cold water20 C

Hot Glycol

80 C2 kg/s

)(

0)peke(since

0

21out

2out1

outin


(steady)0system


outin

TTCmQ

hmQhm

EE

EEE

p

Then the rate of heat transfer becomes

kW204.8=C)40CC)(80kJ/kg.kg/s)(2.562()]([ glycoloutinout TTcmQ p

The rate of heat transfer from water must be equal to the rate of heat transfer to the glycol. Then,

kg/s1.4=C)20C)(55kJ/kg.(4.18

kJ/s8.204)(

)]([inout

inwaterwaterinoutin TTc

QmTTcmQ

pp


)()(

0

)0(since0

34water12glycolgen

gen4water2glycol3water1glycol

gen43223311


(steady)0system


gen


outin

ssmssmS

Ssmsmsmsm

QSsmsmsmsm

SSSS


kW/K0.0446273+20273+55lnkJ/kg.K)kg/s)(4.184.1(

273+80273+40lnkJ/kg.K)kg/s)(2.562(

lnln3

4water

1

2glycolgen T

Tcm

TT

cmS pp



8-126

8-138 Oil is to be cooled by water in a thin-walled heat exchanger. The rate of heat transfer and the rate ofexergy destruction within the heat exchanger are to be determined.Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to thecold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluidproperties are constant.Properties The specific heats of water and oil are given to be 4.18 and 2.20 kJ/kg. C, respectively. Analysis We take the oil tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Hot oil150 C2 kg/s

Cold water

22 C1.5 kg/s

)(

0)peke(since

0

21out

2out1

outin


(steady)0system


outin

TTcmQ

hmQhm

EE

EEE

p

Then the rate of heat transfer from the oil becomes

=C)40CC)(150kJ/kg.kg/s)(2.22()]([ oiloutinout kW484TTcmQ p

Noting that heat lost by the oil is gained by the water, the outlet temperature of water is determined from

C2.99C)kJ/kg.kg/s)(4.185.1(

kW484C22)]([ inoutwateroutinp

p cmQTTTTcmQ


)()(

0

)0(since0

34water12oilgen

gen4water2oil3water1oil

gen43223311


(steady)0system


gen


outin

ssmssmS

Ssmsmsmsm

QSsmsmsmsm

SSSS


kW/K0.132273+22273+99.2lnkJ/kg.K)kg/s)(4.185.1(

273+150273+40lnkJ/kg.K)kg/s)(2.22(

lnln3

4water

1

2oilgen T

Tcm

TT

cmS pp



8-127

8-139 A regenerator is considered to save heat during the cooling of milk in a dairy plant. The amounts of fuel and money such a generator will save per year and the rate of exergy destruction within theregenerator are to be determined.Assumptions 1 Steady operating conditions exist. 2 The properties of the milk are constant. 5 Theenvironment temperature is 18 C.Properties The average density and specific heat of milk can be taken to be milk water 1 kg/L and cp,milk= 3.79 kJ/kg. C (Table A-3). Analysis The mass flow rate of the milk is

kg/h43,200=kg/s12L/s)kg/L)(121(milkmilk VmTaking the pasteurizing section as the system, the energy balance for this steady-flow system can be expressed in the rate form as

)(

0)peke(since

0

12milkin

21in

outin


(steady)0system


outin

TTcmQ

hmhmQ

EEEEE

p

Therefore, to heat the milk from 4 to 72 C as being done currently, heat must be transferred to the milk at arate of

kJ/s3093C4)C)(72kJ/kg.kg/s)(3.79(12)]([ milkionrefrigerationpasturizatcurrent TTcmQ p

The proposed regenerator has an effectiveness of = 0.82, and thus it will save 82 percent of this energy.Therefore,

( . )(Q Qsaved current kJ / s) = 2536 kJ / s0 82 3093Noting that the boiler has an efficiency of boiler = 0.82, the energy savings above correspond to fuelsavings of

Fuel Saved (2536 kJ / s)(0.82)

(1therm)(105,500 kJ)

0.02931therm / ssaved

boiler

Q

Noting that 1 year = 365 24=8760 h and unit cost of natural gas is $1.04/therm, the annual fuel and moneysavings will be

Fuel Saved = (0.02931 therms/s)(8760 3600 s) = 924,450 therms/yr r$961,430/y=rm)($1.04/the therm/yr)(924,450=fuel)ofcosttsaved)(Uni(Fuel=savedMoney

The rate of entropy generation during this process is determined by applying the rate form of the entropybalance on an extended system that includes the regenerator and the immediate surroundings so that theboundary temperature is the surroundings temperature, which we take to be the cold water temperature of18 C.:

inoutgen


(steady)0system


gen


outin SSSSSSS

Disregarding entropy transfer associated with fuel flow, the only significant difference between the two cases is the reduction is the entropy transfer to water due to the reduction in heat transfer to water, and is determined to be

kW/K715.8273+18kJ/s2536

surr

saved

surr

reductionout,reductionout,reductiongen, T

QT

QSS

year)(perkJ/K102.75=s/year)360060kJ/s.K)(87715.8( 8reductiongen,reductiongen, tSS


kJ108.00 10=kJ/K)10K)(2.75291( 8reductiongen,0reductiondestroyed, STX (per year)

8-128

8-140 Exhaust gases are expanded in a turbine, which is not well-insulated. Tha actual and reversiblepower outputs, the exergy destroyed, and the second-law efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Potential energychange is negligible. 3 Air is an ideal gas with constant specific heats.

630 C500 kPa

Q

Exh. gas 750 C

1.2 MPa

Turbine

Properties The gas constant of air is R = 0.287 kJ/kg.K and thespecific heat of air at the average temperature of (750+630)/2 =690ºC is cp = 1.134 kJ/kg.ºC (Table A-2). Analysis (a) The enthalpy and entropy changes of air across the turbine are

kJ/kg08.136C0)63C)(750kJ/kg.(1.134)( 21 TTch p

kJ/kg.K005354.0kPa500kPa1200lnkJ/kg.K)(0.287

K273)(630K273)(750kJ/kg.K)ln(1.134

lnln2

1

2

1

PP

RTT

cs p

The actual and reversible power outputs from the turbine are

kW516.9kW432.7

kJ/kg.K)0.005354K)(27325(kJ/kg)08kg/s)(136.4.3()(

kW30kJ/kg)08kg/s)(136.4.3(

0rev

outa

sThmW

QhmW

(b) The exergy destroyed in the turbine is

kW84.27.4329.516arevdest WWX


0.837kW9.516kW7.432

rev

aII W

W

8-129

8-141 Refrigerant-134a is compressed in an adiabatic compressor, whose second-law efficiency is given.The actual work input, the isentropic efficiency, and the exergy destruction are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The properties of the refrigerant at the inlet of the compressor are (Tables A-11 through A-13)

kJ/kg.K95153.0kJ/kg60.243

C)360.15(kPa160

C60.15

1

1

1

1

kPasat@160

sh

TP

T1 MPa

R-134a160 kPa

CompressorThe enthalpy at the exit for if the process was isentropic is

kJ/kg41.282kJ/kg.K95153.0

MPa12

12

2sh

ssP

The expressions for actual and reversible works are kJ/kg)60.243( 212a hhhw

/kg.K0.95153)kJK)(273(25kJ/kg)60.243()( 2212012rev shssThhw

Substituting these into the expression for the second-law efficiency

60.2430.95153)(298)(60.243

80.02

22

a

revII h

shw

w

The exit pressure is given (1 MPa). We need one more property to fix the exit state. By a trial-errorapproach or using EES, we obtain the exit temperature to be 60ºC. The corresponding enthalpy and entropyvalues satisfying this equation are

kJ/kg.K98492.0kJ/kg36.293

2

2

sh

Then,49.76kJ/kg60.24336.29312a hhw

kJ/kg81.39Kkg0.9515)kJ/K)(0.98492273(25kJ/kg)60.24336.293()( 12012rev ssThhw

(b) The isentropic efficiency is determined from its definition

0.780kJ/kg)60.24336.293(kJ/kg)60.24341.282(

12

12s

hhhh

s

(b) The exergy destroyed in the compressor is kJ/kg9.9581.3976.49revadest wwx

8-130

8-142 The isentropic efficiency of a water pump is specified. The actual power output, the rate of frictionalheating, the exergy destruction, and the second-law efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) Using saturated liquid properties at thegiven temperature for the inlet state (Table A-4)

4 MPaPUMP

Water100 kPa

30 C1.35 kg/s/kgm001004.0

kJ/kg.K43676.0kJ/kg74.125

0C30

31

1

1

1

1

v

sh

xT

The power input if the process was isentropic is

kW288.5kPa)100/kg)(4000m1004kg/s)(0.0035.1()( 3121s PPmW v

Given the isentropic efficiency, the actual power may be determined to be

kW7.55470.0

kW288.5

s

sa

WW

(b) The difference between the actual and isentropic works is the frictional heating in the pump

kW2.266288.5554.7frictional sa WWQ

(c) The enthalpy at the exit of the pump for the actual process can be determined from

kJ/kg33.131kJ/kg)74.125kg/s)(35.1(kW555.7)( 2212a hhhhmW

The entropy at the exit is

kJ/kg.K4420.0kJ/kg33.131

MPa42

2

2 shP

The reversible power and the exergy destruction are

kW487.5/kg.K0.95153)kJK)(0.4420273(20kJ/kg)60.24333.131(kg/s)35.1()( 12012rev ssThhmW

kW2.068487.5555.7revadest WWX

(d) The second-law efficiency is

0.726kW555.7kW487.5

a

revII W

W

8-131

8-143 Argon gas is expanded adiabatically in an expansion valve. The exergy of argon at the inlet, theexergy destruction, and the second-law efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are zero. 3 Argon is an ideal gas with constant specific heats.Properties The properties of argon gas are R = 0.2081 kJ/kg.K, cp = 0.5203 kJ/kg.ºC (Table A-2). Analysis (a) The exergy of the argon at the inlet is

kJ/kg224.7kPa100kPa3500kJ/kg.K)ln(0.2081

K298K373kJ/kg.K)ln(0.5203K)298(C25)00kJ/kg.K)(1(0.5203

lnln)(

)(

0

1

0

1001

010011

PP

RTT

cTTTc

ssThhx

pp

500 kPaArgon

3.5 MPa100 C

(b) Noting that the temperature remains constant in a throttling process, the exergy destruction is determined from

kJ/kg120.7kPa3500kPa500kJ/kg.K)ln(0.2081K)298(ln

)(

0

10

120

gen0dest

PP

RT

ssT

sTx


0.463kJ/kg7.224

kJ/kg)7.1207.224(

1

dest1II x

xx

8-132

8-144 Heat is lost from the air flowing in a diffuser. The exit temperature, the rate of exergy destruction,and the second law efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Potential energy change is negligible. 3 Nitrogen is anideal gas with variable specific heats.Properties The gas constant of nitrogen is R = 0.2968 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. At the inlet of the diffuser and at the dead state, we have

KkJ/kg2006.7kJ/kg08.130

kPa100K423C15

1

1

1

1

sh

PT

q

110 kPa25 m/s

Nitrogen100 kPa150 C

180 m/sKkJ/kg8426.6

kJ/kg93.1kPa100K300

0

0

1

1

sh

PT

An energy balance on the diffuser gives

kJ/kg47.141

kJ/kg5.4/sm1000

kJ/kg12m/s)(25

/sm1000kJ/kg1

2m/s)(180

kJ/kg08.

22

2

22

2

222

2

out

22

2

21

1

h

h

qV

hV

h

130

The corresponding properties at the exit of the diffuser are

KkJ/kg1989.7K9.433

kPa110kJ/kg47.141

2

2

1

2

sT

Ph C160.9

(b) The mass flow rate of the nitrogen is determined to be

kg/s281.1m/s)25()m06.0(K)33.9kJ/kg.K)(42968.0(

kPa110 222

2

2222 VA

RTP

VAm

The exergy destruction in the nozzle is the exergy difference between the inlet and exit of the diffuser

kW5.11kJ/kg.K)1989.72006.7)(K300(

/sm1000kJ/kg1

2m/s)(25m/s)(180)kJ/kg47.141(130.08

kg/s)281.1(

)(2

22

22

210

22

21

21dest ssTVV

hhmX

(c) The second-law efficiency for this device may be defined as the exergy output divided by the exergyinput:

kW35.47

kJ/kg.K)8426.62006.7)(K300(/sm1000

kJ/kg12m/s)180(kJ/kg)93.1(130.08kg/s)281.1(

)(2

22

2

010

21

011 ssTVhhmX

0.892kW35.47

kW11.5111

dest

1

2II X

XXX

8-133

Fundamentals of Engineering (FE) Exam Problems

8-145 Heat is lost through a plane wall steadily at a rate of 800 W. If the inner and outer surface temperatures of the wall are 20 C and 5 C, respectively, and the environment temperature is 0 C, the rate of exergy destruction within the wall is(a) 40 W (b) 17,500 W (c) 765 W (d) 32,800 W (e) 0 W

Answer (a) 40 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 numericalvalues).

Q=800 "W"T1=20 "C"T2=5 "C"To=0 "C""Entropy balance S_in - S_out + S_gen= DS_system for the wall for steady operation gives"Q/(T1+273)-Q/(T2+273)+S_gen=0 "W/K"X_dest=(To+273)*S_gen "W"

"Some Wrong Solutions with Common Mistakes:"Q/T1-Q/T2+Sgen1=0; W1_Xdest=(To+273)*Sgen1 "Using C instead of K in Sgen"Sgen2=Q/((T1+T2)/2); W2_Xdest=(To+273)*Sgen2 "Using avegage temperature in C for Sgen"Sgen3=Q/((T1+T2)/2+273); W3_Xdest=(To+273)*Sgen3 "Using avegage temperature in K"W4_Xdest=To*S_gen "Using C for To"

8-146 Liquid water enters an adiabatic piping system at 15 C at a rate of 5 kg/s. It is observed that thewater temperature rises by 0.5 C in the pipe due to friction. If the environment temperature is also 15 C,the rate of exergy destruction in the pipe is(a) 8.36 kW (b) 10.4 kW (c) 197 kW (d) 265 kW (e) 2410 kW

Answer (b) 10.4 kW


Cp=4.18 "kJ/kg.K"m=5 "kg/s"T1=15 "C"T2=15.5 "C"To=15 "C"S_gen=m*Cp*ln((T2+273)/(T1+273)) "kW/K"X_dest=(To+273)*S_gen "kW"

"Some Wrong Solutions with Common Mistakes:"W1_Xdest=(To+273)*m*Cp*ln(T2/T1) "Using deg. C in Sgen"W2_Xdest=To*m*Cp*ln(T2/T1) "Using deg. C in Sgen and To"W3_Xdest=(To+273)*Cp*ln(T2/T1) "Not using mass flow rate with deg. C"W4_Xdest=(To+273)*Cp*ln((T2+273)/(T1+273)) "Not using mass flow rate with K"

8-134

8-147 A heat engine receives heat from a source at 1500 K at a rate of 600 kJ/s and rejects the waste heat to a sink at 300 K. If the power output of the engine is 400 kW, the second-law efficiency of this heatengine is(a) 42% (b) 53% (c) 83% (d) 67% (e) 80%

Answer (c) 83%


Qin=600 "kJ/s"W=400 "kW"TL=300 "K"TH=1500 "K"Eta_rev=1-TL/THEta_th=W/QinEta_II=Eta_th/Eta_rev

"Some Wrong Solutions with Common Mistakes:"W1_Eta_II=Eta_th1/Eta_rev; Eta_th1=1-W/Qin "Using wrong relation for thermal efficiency"W2_Eta_II=Eta_th "Taking second-law efficiency to be thermal efficiency"W3_Eta_II=Eta_rev "Taking second-law efficiency to be reversible efficiency"W4_Eta_II=Eta_th*Eta_rev "Multiplying thermal and reversible efficiencies instead of dividing"

8-148 A water reservoir contains 100 tons of water at an average elevation of 60 m. The maximum amountof electric power that can be generated from this water is(a) 8 kWh (b) 16 kWh (c) 1630 kWh (d) 16,300 kWh (e) 58,800 kWh

Answer (b) 16 kWh


m=100000 "kg"h=60 "m"g=9.81 "m/s^2""Maximum power is simply the potential energy change,"W_max=m*g*h/1000 "kJ"W_max_kWh=W_max/3600 "kWh"

"Some Wrong Solutions with Common Mistakes:"W1_Wmax =m*g*h/3600 "Not using the conversion factor 1000"W2_Wmax =m*g*h/1000 "Obtaining the result in kJ instead of kWh"W3_Wmax =m*g*h*3.6/1000 "Using worng conversion factor"W4_Wmax =m*h/3600"Not using g and the factor 1000 in calculations"

8-135

8-149 A house is maintained at 25 C in winter by electric resistance heaters. If the outdoor temperature is2 C, the second-law efficiency of the resistance heaters is(a) 0% (b) 7.7% (c) 8.7% (d) 13% (e) 100%

Answer (b) 7.7%


TL=2+273 "K"TH=25+273 "K"To=TLCOP_rev=TH/(TH-TL)COP=1Eta_II=COP/COP_rev

"Some Wrong Solutions with Common Mistakes:"W1_Eta_II=COP/COP_rev1; COP_rev1=TL/(TH-TL) "Using wrong relation for COP_rev"W2_Eta_II=1-(TL-273)/(TH-273) "Taking second-law efficiency to be reversible thermal efficiency with C for temp"W3_Eta_II=COP_rev "Taking second-law efficiency to be reversible COP"W4_Eta_II=COP_rev2/COP; COP_rev2=(TL-273)/(TH-TL) "Using C in COP_rev relation instead of K, and reversing"

8-150 A 10-kg solid whose specific heat is 2.8 kJ/kg. C is at a uniform temperature of -10 C. For anenvironment temperature of 25 C, the exergy content of this solid is(a) Less than zero (b) 0 kJ (c) 22.3 kJ (d) 62.5 kJ (e) 980 kJ

Answer (d) 62.5 kJ


m=10 "kg"Cp=2.8 "kJ/kg.K"T1=-10+273 "K"To=25+273 "K""Exergy content of a fixed mass is x1=u1-uo-To*(s1-so)+Po*(v1-vo)" ex=m*(Cp*(T1-To)-To*Cp*ln(T1/To))

"Some Wrong Solutions with Common Mistakes:"W1_ex=m*Cp*(To-T1) "Taking the energy content as the exergy content"W2_ex=m*(Cp*(T1-To)+To*Cp*ln(T1/To)) "Using + for the second term instead of -"W3_ex=Cp*(T1-To)-To*Cp*ln(T1/To) "Using exergy content per unit mass"W4_ex=0 "Taking the exergy content to be zero"

8-136

8- 151 Keeping the limitations imposed by the second-law of thermodynamics in mind, choose the wrong statement below:

(a) A heat engine cannot have a thermal efficiency of 100%. (b) For all reversible processes, the second-law efficiency is 100%. (c) The second-law efficiency of a heat engine cannot be greater than its thermal efficiency.(d) The second-law efficiency of a process is 100% if no entropy is generated during that process. (e) The coefficient of performance of a refrigerator can be greater than 1.

Answer (c) The second-law efficiency of a heat engine cannot be greater than its thermal efficiency.

8-152 A furnace can supply heat steadily at a 1600 K at a rate of 800 kJ/s. The maximum amount of power that can be produced by using the heat supplied by this furnace in an environment at 300 K is(a) 150 kW (b) 210 kW (c) 325 kW (d) 650 kW (e) 984 kW

Answer (d) 650 kW


Q_in=800 "kJ/s"TL=300 "K"TH=1600 "K"W_max=Q_in*(1-TL/TH) "kW"

"Some Wrong Solutions with Common Mistakes:"W1_Wmax=W_max/2 "Taking half of Wmax"W2_Wmax=Q_in/(1-TL/TH) "Dividing by efficiency instead of multiplying by it"W3_Wmax =Q_in*TL/TH "Using wrong relation"W4_Wmax=Q_in "Assuming entire heat input is converted to work"

8-153 Air is throttled from 50 C and 800 kPa to a pressure of 200 kPa at a rate of 0.5 kg/s in an environment at 25 C. The change in kinetic energy is negligible, and no heat transfer occurs during theprocess. The power potential wasted during this process is(a) 0 (b) 0.20 kW (c) 47 kW (d) 59 kW (e) 119 kW

Answer (d) 59 kW


R=0.287 "kJ/kg.K"Cp=1.005 "kJ/kg.K"m=0.5 "kg/s"T1=50+273 "K"P1=800 "kPa"To=25 "C"P2=200 "kPa""Temperature of an ideal gas remains constant during throttling since h=const and h=h(T)"

8-137

T2=T1ds=Cp*ln(T2/T1)-R*ln(P2/P1)X_dest=(To+273)*m*ds "kW"

"Some Wrong Solutions with Common Mistakes:"W1_dest=0 "Assuming no loss"W2_dest=(To+273)*ds "Not using mass flow rate"W3_dest=To*m*ds "Using C for To instead of K"W4_dest=m*(P1-P2) "Using wrong relations"

8-154 Steam enters a turbine steadily at 4 MPa and 400 C and exits at 0.2 MPa and 150 C in anenvironment at 25 C. The decrease in the exergy of the steam as it flows through the turbine is(a) 58 kJ/kg (b) 445 kJ/kg (c) 458 kJ/kg (d) 518 kJ/kg (e) 597 kJ/kg

Answer (e) 597 kJ/kg


P1=4000 "kPa"T1=400 "C"P2=200 "kPa"T2=150 "C"To=25 "C"h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1)s1=ENTROPY(Steam_IAPWS,T=T1,P=P1)h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2)s2=ENTROPY(Steam_IAPWS,T=T2,P=P2)"Exergy change of s fluid stream is Dx=h2-h1-To(s2-s1)"-Dx=h2-h1-(To+273)*(s2-s1)

"Some Wrong Solutions with Common Mistakes:"-W1_Dx=0 "Assuming no exergy destruction"-W2_Dx=h2-h1 "Using enthalpy change"-W3_Dx=h2-h1-To*(s2-s1) "Using C for To instead of K"-W4_Dx=(h2+(T2+273)*s2)-(h1+(T1+273)*s1) "Using wrong relations for exergy"

8- 155 … 8- 158 Design and Essay Problems

ThermoSolutions CHAPTER08

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