Chapter5 A

ηWorkQH

= Whence QHWorkη

:=

QH 1.583 105× kW= Ans.

QC QH Work−:= QC 6.333 104× kW= Ans.

(b) η 0.35:= QHWorkη

:= QH 2.714 105× kW= Ans.

QC QH Work−:= QC 1.764 105× kW= Ans.

5.4 (a) TC 303.15 K⋅:= TH 623.15 K⋅:=

ηCarnot 1TC

TH−:= η 0.55 ηCarnot⋅:= η 0.282= Ans.

Chapter 5 - Section A - Mathcad Solutions

5.2 Let the symbols Q and Work represent rates in kJ/s. Then by Eq. (5.8)

ηWorkQH

= 1TC

TH−=

TC 323.15 K⋅:= TH 798.15 K⋅:= QH 250kJs

⋅:=

Work QH 1TC

TH−

⎛⎜⎝

⎞

⎠⋅:= Work 148.78

kJs

=

or Work 148.78kW= which is the power. Ans.

By Eq. (5.1), QC QH Work−:= QC 101.22kJs

= Ans.

5.3 (a) Let symbols Q and Work represent rates in kJ/s

TH 750 K⋅:= TC 300 K⋅:= Work 95000− kW⋅:=

By Eq. (5.8): η 1TC

TH−:= η 0.6=

But

123

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QC 3.202 106× kW=

Work QCTH

TC1−

⎛⎜⎝

⎞

⎠⋅:= Work 5.336 106× kW= Ans.

QH QC Work+:= QH 8.538 106× kW= Ans.

5.8 Take the heat capacity of water to be constant at the valueCP 4.184kJ

kg K⋅⋅:=

(a) T1 273.15 K⋅:= T2 373.15 K⋅:= Q CP T2 T1−( )⋅:= Q 418.4kJkg

=

∆SH2O CP lnT2

T1

⎛⎜⎝

⎞

⎠⋅:= ∆SH2O 1.305

kJkg K⋅

=

∆SresQ−

T2:= ∆Sres 1.121−

kJkg K⋅

= Ans.

(b) η 0.35:= ηCarnotη

0.55:= ηCarnot 0.636=

By Eq. (5.8), THTC

1 ηCarnot−:= TH 833.66K= Ans.

5.7 Let the symbols represent rates where appropriate. Calculate mass rate ofLNG evaporation:

V 9000m3

s⋅:= P 1.0133 bar⋅:= T 298.15 K⋅:=

molwt 17gmmol

:= mLNGP V⋅R T⋅

molwt⋅:= mLNG 6254kgs

=

Maximum power is generated by a Carnot engine, for which

WorkQC

QH QC−

QC=

QH

QC1−=

TH

TC1−=

TH 303.15 K⋅:= TC 113.7 K⋅:=

QC 512kJkg⋅ mLNG⋅:=

124



Q 15000 J⋅:=

(a) Const.-V heating; ∆U Q W+= Q= n CV⋅ T2 T1−( )⋅=

T2 T1Q

n CV⋅+:= T2 1 103× K=

By Eq. (5.18), ∆S n CP lnT2

T1

⎛⎜⎝

⎞

⎠⋅ R ln

P2

P1

⎛⎜⎝

⎞

⎠⋅−

⎛⎜⎝

⎞

⎠⋅=

ButP2

P1

T2

T1= Whence ∆S n CV⋅ ln

T2

T1

⎛⎜⎝

⎞

⎠⋅:= ∆S 20.794

JK

= Ans.

(b) The entropy change of the gas is the same as in (a). The entropychange of the surroundings is zero. Whence

∆Stotal 10.794JK⋅= Ans.

The stirring process is irreversible.

∆Stotal ∆SH2O ∆Sres+:= ∆Stotal 0.184kJ

kg K⋅= Ans.

(b) The entropy change of the water is the same as in (a), and the totalheat transfer is the same, but divided into two halves.

∆SresQ−2

1323.15 K⋅

1373.15 K⋅

+⎛⎜⎝

⎞⎠

⋅:= ∆Sres 1.208−kJ

kg K⋅=

∆Stotal ∆Sres ∆SH2O+:= ∆Stotal 0.097kJ

kg K⋅= Ans.

(c) The reversible heating of the water requires an infinite number of heatreservoirs covering the range of temperatures from 273.15 to 373.15 K,each one exchanging an infinitesimal quantity of heat with the water andraising its temperature by a differential increment.

5.9 P1 1 bar⋅:= T1 500 K⋅:= V 0.06 m3⋅:=

nP1 V⋅

R T1⋅:= n 1.443mol= CV

52

R⋅:=

125



∆SA 8.726J

mol K⋅= ∆SB 8.512−

Jmol K⋅

= Ans.

∆Stotal ∆SA ∆SB+:= ∆Stotal 0.214J

mol K⋅= Ans.

5.16 By Eq. (5.8), dWdQ

1TσT

−= dW dQ TσdQT

⋅−=

dW dQ Tσ dS⋅−=Since dQ/T = dS,

Integration gives the required result.

T1 600 K⋅:= T2 400 K⋅:= Tσ 300 K⋅:=

Q CP T2 T1−( )⋅:= Q 5.82− 103×J

mol=

5.10 (a) The temperature drop of the second stream (B) in eithercase is the same as the temperature rise of the first stream(A), i.e., 120 degC. The exit temperature of the secondstream is therefore 200 degC. In both cases we thereforehave:

CP72

R:=

∆SA CP ln463.15343.15⎛⎜⎝

⎞⎠

⋅:= ∆SB CP ln473.15593.15⎛⎜⎝

⎞⎠

⋅:=

∆SA 8.726J

mol K⋅= ∆SB 6.577−

Jmol K⋅

= Ans.

(b) For both cases:

∆Stotal ∆SA ∆SB+:= ∆Stotal 2.149J

mol K⋅= Ans.

(c) In this case the final temperature of steam B is 80 degC, i.e., there isa 10-degC driving force for heat transfer throughout the exchanger.Now

∆SA CP ln463.15343.15⎛⎜⎝

⎞⎠

⋅:= ∆SB CP ln353.15473.15⎛⎜⎝

⎞⎠

⋅:=

126



WQC2

TH2 TC2−

TC2=

Equate the two work quantities and solve for the required ratio of the heatquantities:

rTC2

TH1

TH1 TC1−

TH2 TC2−⎛⎜⎝

⎞

⎠⋅:= r 2.5= Ans.

5.18 (a) T1 300K:= P1 1.2bar:= T2 450K:= P2 6bar:= Cp72

R:=

∆H Cp T2 T1−( )⋅:= ∆H 4.365 103×J

mol= Ans.

∆S Cp lnT2

T1

⎛⎜⎝

⎞

⎠⋅ R ln

P2

P1

⎛⎜⎝

⎞

⎠⋅−:= ∆S 1.582−

Jmol K⋅

= Ans.

(b) ∆H 5.82 103⋅J

mol= ∆S 1.484

Jmol K⋅

=

∆S CP lnT2

T1

⎛⎜⎝

⎞

⎠⋅:= ∆S 11.799−

Jmol K⋅

=

Work Q Tσ ∆S⋅−:= Work 2280−J

mol= Ans.

Qσ Q Work−:= Qσ 3540J

mol= Ans.

∆SreservoirQσ

Tσ:= ∆Sreservoir 11.8

Jmol K⋅

= Ans.

∆S ∆Sreservoir+ 0J

mol K⋅= Process is reversible.

5.17 TH1 600 K⋅:= TC1 300 K⋅:= TH2 300 K⋅:= TC2 250 K⋅:=

For the Carnot engine, use Eq. (5.8): WQH1

TH1 TC1−

TH1=

The Carnot refrigerator is a reverse Carnot engine. Combine Eqs. (5.8) & (5.7) to get:

127



For isobaric step 2 to 3:P2

T2

P3

T3=

Solving these 4 equations for T4 yields: T4 T1T2

T3

⎛⎜⎝

⎞

⎠

γ−

⋅=

Cp72

R:= Cv52

R:= γCp

Cv:= γ 1.4=

T1 200 273.15+( )K:= T2 1000 273.15+( )K:= T3 1700 273.15+( )K:=

T4 T1T2

T3

⎛⎜⎝

⎞

⎠

γ−

⋅:= T4 873.759K=

Eq. (A) p. 306 η 11γ

T4 T1−

T3 T2−⎛⎜⎝

⎞

⎠⋅−:= η 0.591= Ans.

(c) ∆H 3.118− 103⋅J

mol= ∆S 4.953

Jmol K⋅

=

(d) ∆H 3.741− 103⋅J

mol= ∆S 2.618

Jmol K⋅

=

(e) ∆H 6.651− 103⋅J

mol= ∆S 3.607−

Jmol K⋅

=

5.19 This cycle is the same as is shown in Fig. 8.10 on p. 305. The equivalent statesare A=3, B=4, C=1, and D=2. The efficiency is given by Eq. (A) on p. 305.

Temperature T4 is not given and must be calaculated. The following equationsare used to derive and expression for T4.

For adiabatic steps 1 to 2 and 3 to 4:

T1 V1γ 1−⋅ T2 V2

γ 1−⋅= T3 V3γ 1−⋅ T4 V4

γ 1−⋅=

For constant-volume step 4 to 1: V1 V4=

128



∆S 2.914J

mol K⋅= Ans.

5.25 P 4:= T 800:=

Step 1-2: Volume decreases at constant P. Heat flows out of the system. Work is done on the system.

W12 P V2 V1−( )⋅⎡⎣ ⎤⎦−= R T2 T1−( )⋅⎡⎣ ⎤⎦−=

Step 2-3: Isothermal compression. Work is done on the system. Heat flowsout of the system.

W23 R T2⋅ lnP3

P2

⎛⎜⎝

⎞

⎠⋅= R T2⋅ ln

P3

P1

⎛⎜⎝

⎞

⎠⋅=

Step 3-1: Expansion process that produces work. Heat flows into thesystem. Since the PT product is constant,

P dT⋅ T dP⋅+ 0= TdPP

⋅ dT−= (A)

P V⋅ R T⋅= P dV⋅ V dP⋅+ R dT⋅=

P dV⋅ R dT⋅ V dP⋅−= R dT⋅ R T⋅dPP

⋅−=

5.21 CV CP R−:= P1 2 bar⋅:= P2 7 bar⋅:= T1 298.15 K⋅:=

γCP

CV:= γ 1.4=

With the reversible work given by Eq. (3.34), we get for the actual W:

Work 1.35R T1⋅

γ 1−⋅

P2

P1

⎛⎜⎝

⎞

⎠

γ 1−γ

1−

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦

⋅:= Work 3.6 103×J

mol=

But Q = 0, and W ∆U= CV T2 T1−( )⋅= Whence T2 T1WorkCV

+:=

T2 471.374K=

∆S CP lnT2

T1

⎛⎜⎝

⎞

⎠⋅ R ln

P2

P1

⎛⎜⎝

⎞

⎠⋅−:=

129



Ans.η 0.068=ηW12 W23+ W31+

Q31:=

Q31 1.309 104×J

mol=Q31 CP R+( ) T1 T2−( )⋅:=

W31 5.82− 103×J

mol=W31 2− R⋅ T1 T2−( )⋅:=

W23 2.017 103×J

mol=W23 R T2⋅ ln

P3

P1

⎛⎜⎝

⎞

⎠⋅:=

W12 2.91 103×J

mol=W12 R T2 T1−( )⋅⎡⎣ ⎤⎦−:=

P3 P1T1

T2⋅:=P1 1.5 bar⋅:=

T2 350 K⋅:=T1 700 K⋅:=CP72

R⋅:=

ηWnet

Qin=

W12 W23+ W31+

Q31=

Q31 CV 2 R⋅+( ) T1 T3−( )⋅= CP R+( ) T1 T2−( )⋅=

Q31 ∆U31 W31−= CV T1 T3−( )⋅ 2 R⋅ T1 T3−( )⋅+=

W31V3

V1

VP⌠⎮⌡

d−= 2− R⋅ T1 T3−( )⋅= 2− R⋅ T1 T2−( )⋅=

P3 P1T1

T3⋅= P1

T1

T2⋅=Moreover,

P dV⋅ R dT⋅ R dT⋅+= 2 R⋅ dT⋅=

In combination with (A) this becomes

130



Ans.

∆Stotal ∆S ∆Sres+:= ∆Stotal 6.02J

mol K⋅= Ans.

5.27 (a) By Eq. (5.14) with P = const. and Eq. (5.15), we get for the entropy changeof 10 moles

n 10 mol⋅:=

∆S n R⋅ ICPS 473.15K 1373.15K, 5.699, 0.640 10 3−⋅, 0.0, 1.015− 105⋅,( )⋅:=

∆S 536.1JK

= Ans.

(b) By Eq. (5.14) with P = const. and Eq. (5.15), we get for the entropychange of 12 moles

n 12 mol⋅:=

∆S n R⋅ ICPS 523.15K 1473.15K, 1.213, 28.785 10 3−⋅, 8.824− 10 6−⋅, 0.0,( )⋅:=

∆S 2018.7JK

= Ans.

5.26 T 403.15 K⋅:= P1 2.5 bar⋅:= P2 6.5 bar⋅:= Tres 298.15 K⋅:=

By Eq. (5.18), ∆S R− lnP2

P1

⎛⎜⎝

⎞

⎠⋅:= ∆S 7.944−

Jmol K⋅

= Ans.

With the reversible work given by Eq. (3.27), we get for the actual W:

Work 1.3 R⋅ T⋅ lnP2

P1

⎛⎜⎝

⎞

⎠⋅:= (Isothermal compresion) Work 4.163 103×

Jmol

=

Q Work−:= Q here is with respect to the system.

So for the heat reservoir, we have

∆SresQ−

Tres:= ∆Sres 13.96

Jmol K⋅

=

131



(guess)x 0.3:=

x CP⋅ T1 T0−( )⋅ 1 x−( ) CP⋅ T2 T0−( )⋅+ 0=

Temperature of warm airT2 348.15 K⋅:=

Temperature of chilled airT1 248.15 K⋅:=

Temperature of entering airT0 298.15 K⋅:=

The relative amounts of the two streams are determined by an energybalance. Since Q = W = 0, the enthalpy changes of the two streams mustcancel. Take a basis of 1 mole of air entering, and let x = moles of chilled air.Then 1 - x = the moles of warm air.

5.29

Ans.∆S 1.2436 106×JK

=

∆S n R⋅ ICPS 533.15K 1202.9K, 1.424, 14.394 10 3−⋅, 4.392− 10 6−⋅, 0.0,( )⋅:=

n 18140 mol⋅:=

The final temperature for this process was found in Pb. 4.2c to be 1202.9 K.The entropy change for 18.14 kg moles is then found as follows

(c)

Ans.∆S 2657.5JK

=

∆S n R⋅ ICPS 533.15K 1413.8K, 1.967, 31.630 10 3−⋅, 9.873− 10 6−⋅, 0.0,( )⋅:=

n 15 mol⋅:=

The final temperature for this process was found in Pb. 4.2b to be 1413.8 K.The entropy change for 15 moles is then found as follows:

(b)

Ans.∆S 900.86JK

=

∆S n R⋅ ICPS 473.15K 1374.5K, 1.424, 14.394 10 3−⋅, 4.392− 10 6−⋅, 0.0,( )⋅:=

n 10 mol⋅:=

The final temperature for this process was found in Pb. 4.2a to be 1374.5 K.The entropy change for 10 moles is then found as follows

(a)5.28

132



PROCESS IS POSSIBLE.∆Stotal 3.42J

mol K⋅=∆Stotal ∆S ∆Sres+:=

∆S 2.301−J

mol K⋅=∆S CP ln

T2

T1

⎛⎜⎝

⎞

⎠⋅ R ln

P2

P1

⎛⎜⎝

⎞

⎠⋅−:=

Q 1.733− 103×J

mol=∆Sres 5.718

Jmol K⋅

=∆SresQ−

Tres:=

Q CV T2 T1−( )⋅ Work−:=Q ∆U Work−=CV CP R−:=

CP72

R⋅:=Work 1800−J

mol⋅:=Tres 303.15 K⋅:=

P2 1 bar⋅:=

Givenx

1 x−

T2 T0−

T1 T0−⎛⎜⎝

⎞

⎠−= x Find x( ):= x 0.5=

Thus x = 0.5, and the process produces equal amounts of chilled and warmedair. The only remaining question is whether the process violates the secondlaw. On the basis of 1 mole of entering air, the total entropy change is asfollows.

CP72

R⋅:= P0 5 bar⋅:= P 1 bar⋅:=

∆Stotal x CP⋅ lnT1

T0

⎛⎜⎝

⎞

⎠⋅ 1 x−( ) CP⋅ ln

T2

T0

⎛⎜⎝

⎞

⎠⋅+ R ln

PP0

⎛⎜⎝

⎞⎠

⋅−:=

∆Stotal 12.97J

mol K⋅= Ans.

Since this is positive, there is no violation of the second law.

5.30 T1 523.15 K⋅:= T2 353.15 K⋅:= P1 3 bar⋅:=

133



By Eq. (5.28): WdotWdotideal

η t:= Wdot 951.6kW= Ans.

5.34 E 110 volt⋅:= i 9.7 amp⋅:= Tσ 300 K⋅:=

Wdotmech 1.25− hp⋅:= Wdotelect i E⋅:= Wdotelect 1.067 103× W=

At steady state: Qdot Wdotelect+ Wdotmech+tUtd

d= 0=

QdotTσ

SdotG+tStd

d= 0=

Qdot Wdotelect− Wdotmech−:= Qdot 134.875− W=

SdotGQdot−Tσ

:= SdotG 0.45WK

= Ans.

5.33 For the process of cooling the brine:

CP 3.5kJ

kg K⋅⋅:= ∆T 40− K⋅:= mdot 20

kgsec⋅:= η t 0.27:=

T1 273.15 25+( ) K⋅:= T1 298.15K=

T2 273.15 15−( ) K⋅:= T2 258.15K=

Tσ 273.15 30+( ) K⋅:= Tσ 303.15K=

∆H CP ∆T⋅:= ∆H 140−kJkg

=

∆S CP lnT2

T1

⎛⎜⎝

⎞

⎠⋅:= ∆S 0.504−

kJkg K⋅

=

Eq. (5.26): Wdotideal mdot ∆H Tσ ∆S⋅−( )⋅:= Wdotideal 256.938kW=

134



∆SR

T1

T2

TCp

R1T

⌠⎮⎮⌡

d lnP2

P1

⎛⎜⎝

⎞

⎠−= Eq. (5.14)

∆S72

R lnT2

T1

⎛⎜⎝

⎞

⎠⋅ R ln

P2

P1

⎛⎜⎝

⎞

⎠⋅−:= ∆S 17.628

Jmol K⋅

= Ans.

(c) SdotG mdot ∆S⋅:= SdotG 48.966WK

= Ans.

(d) Tσ 20 273.15+( )K:= Wlost Tσ ∆S⋅:= Wlost 5.168 103×J

mol= Ans.

5.39 (a) T1 500K:= P1 6bar:= T2 371K:= P2 1.2bar:= Cp72

R:=

Tσ 300K:= Basis: 1 mol n 1mol:=

∆H n Cp⋅ T2 T1−( )⋅:= Ws ∆H:= Ws 3753.8− J= Ans.

5.35 Ω 25 ohm⋅:= i 10 amp⋅:= Tσ 300 K⋅:=

Wdotelect i2 Ω⋅:= Wdotelect 2.5 103× W=

At steady state: Qdot Wdotelect+tUtd

d= 0= Qdot Wdotelect−:=

QdotTσ

SdotG+tStd

d= 0= SdotG

Qdot−Tσ

:=

Qdot 2.5− 103× watt= SdotG 8.333wattK

= Ans.

5.38 mdot 10kmol

hr:= T1 25 273.15+( )K:= P1 10bar:= P2 1.2bar:=

Cp72

R:= Cv Cp R−:= γCp

Cv:= γ

75

=

(a) Assuming an isenthalpic process: T2 T1:= T2 298.15K= Ans.

(b)

135



(d) 3853.5− J 4952.4− J 1098.8J 3.663JK

(e) 3055.4− J 4119.2− J 1063.8J 3.546JK

5.41 P1 2500kPa:= P2 150kPa:= Tσ 300K:= mdot 20molsec

:=

∆S R− lnP2

P1

⎛⎜⎝

⎞

⎠⋅:=

ηactual 0.45=ηactualWQH

:=

TC 298.15K=TC 25 273.15+( )K:=

TH 523.15K=TH 250 273.15+( )K:=W 0.45kJ:=QH 1kJ:=5.42

Ans.Wdotlost 140.344kW=Wdotlost Tσ SdotG⋅:=

Ans.SdotG 0.468kJ

sec K⋅=SdotG mdot ∆S⋅:=

∆S 0.023kJ

mol K⋅=

Wideal Ws

Ans.SG 4.698JK

=SGWlost

Tσ:=Eq. (5.39)

Ans.Wlost 1409.3 J=Wlost Wideal Ws−:=Eq. (5.30)

Ans.Wideal 5163− J=Wideal ∆H Tσ ∆S⋅−( ):=Eq. (5.27)

∆S 4.698JK

=∆S n Cp lnT2

T1

⎛⎜⎝

⎞

⎠⋅ R ln

P2

P1

⎛⎜⎝

⎞

⎠⋅−

⎛⎜⎝

⎞

⎠⋅:=

3.767JK

1130J4193.7− J3063.7− J(c)

1.643JK

493J2953.9− J2460.9− J(b)

4.698JK

1409.3J5163− J3753.8− J(a)

SGWlost

136



TC 293.15K=

(a) ηmax 1TC

TH−:= ηmax 0.502= Ans.

QdotHWdotηmax

:= QdotC QdotH Wdot−:=

QdotC 745.297MW= (minimum value)

(b) η 0.6 ηmax⋅:= QdotHWdotη

:= QdotH 2.492 109× W=

QdotC QdotH Wdot−:= QdotC 1.742 103× MW= (actual value)

River temperature rise: Vdot 165m3

s⋅:= ρ 1

gm

cm3⋅:=

Cp 1cal

gm K⋅⋅:= ∆T

QdotCVdot ρ⋅ Cp⋅

:= ∆T 2.522K= Ans.

ηmax 1TC

TH−:= ηmax 0.43=

Since ηactual>ηmax, the process is impossible.

5.43 QH 150− kJ⋅:= Q1 50 kJ⋅:= Q2 100 kJ⋅:=

TH 550 K⋅:= T1 350 K⋅:= T2 250 K⋅:= Tσ 300 K⋅:=

(a) SGQH

TH

Q1

T1+

Q2

T2+:= SG 0.27

kJK

= Ans.

(b) Wlost Tσ SG⋅:= Wlost 81.039kJ= Ans.

5.44 Wdot 750− MW⋅:= TH 315 273.15+( ) K⋅:= TC 20 273.15+( ) K⋅:=

TH 588.15K=

137



Wideal 1.776hp=

Wideal ndot R ICPH T1 T2, 3.355, 0.575 10 3−⋅, 0, 0.016− 105⋅,( )⋅

Tσ− R ICPS T1 T2, 3.355, 0.575 10 3−⋅, 0, 0.016− 105⋅,( )⋅( )⋅+

...⎡⎢⎢⎣

⎤⎥⎥⎦

⋅:=

Calculate ideal work using Eqn. (5.26)

ndot 258.555lbmol

hr=ndot

P Vdot⋅R T1⋅

:=

Assume air is an Ideal Gas

Tσ 70 459.67+( )rankine:=P 1atm:=

T2 20 459.67+( )rankine:=T1 70 459.67+( )rankine:=Vdot 100000ft3

hr:=a)

5.47

Since SG 0≥ , this process is possible. SG 0.013kJ

mol K⋅=

SG67

R⋅ ICPS T1 T2, 3.355, 0.575 10 3−⋅, 0, 0.016− 105⋅,( )⋅

17

R⋅ ICPS T1 T3, 3.355, 0.575 10 3−⋅, 0, 0.016− 105⋅,( )⋅ R lnP2

P1

⎛⎜⎝

⎞

⎠⋅−+

...:=

Calculate the rate of entropy generation using Eqn. (5.23)

∆H is essentially zero so the first law is satisfied.∆H 8.797 10 4−×kJ

mol=

∆H67

R⋅ ICPH T1 T2, 3.355, 0.575 10 3−⋅, 0, 0.016− 105⋅,( )⋅

17

ICPH T1 T3, 3.355, 0.575 10 3−⋅, 0, 0.016− 105⋅,( ) R⋅+

...:=

First check the First Law using Eqn. (2.33) neglect changes in kinetic andpotential energy.

P2 1atm:=P1 5bar:=

T3 22− 273.15+( )K:=T2 27 273.15+( ) K⋅:=T1 20 273.15+( )K:=5.46

138



SdotG SdotGsteam SdotGgas+=

Calculate the rate of entropy generation in the boiler. This is the sum of theentropy generation of the steam and the gas.

mdotndot 15.043lb

lbmol=mdotndot

T1

T2

TCp T( )⌠⎮⌡

d−

∆Hv:=

ndotgasT1

T2

TCp T( )⌠⎮⌡

d⋅ mdotsteam ∆Hv⋅+ 0=

First apply an energy balance on the boiler to get the ratio of steam flowrate to gas flow rate.:

a)

Tsteam 212 459.67+( )rankine:=Tσ 70 459.67+( )rankine:=

M 29gmmol

:=∆Hv 970BTUlbm

:=Cp T( ) 3.83 0.000306T

rankine⋅+⎛⎜

⎝⎞⎠

R⋅:=

T2 300 459.67+( )rankine:=T1 2000 459.67+( )rankine:=5.48

Wideal 1.952kW=

Wideal ndot R ICPH T1 T2, 3.355, 0.575 10 3−⋅, 0, 0.016− 105⋅,( )⋅

Tσ− R ICPS T1 T2, 3.355, 0.575 10 3−⋅, 0, 0.016− 105⋅,( )⋅( )⋅+

...⎡⎢⎢⎣

⎤⎥⎥⎦

⋅:=

Calculate ideal work using Eqn. (5.26)

ndot 34.064mol

s=ndot

P Vdot⋅R T1⋅

:=

Assume air is an Ideal Gas

Tσ 25 273.15+( )K:=P 1atm:=

T2 8− 273.15+( )K:=T1 25 273.15+( )K:=Vdot 3000m3

hr:=b)

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Ans.Wideal 9.312− 103×BTUlbmol

=Wideal ∆Hgas Tσ ∆Sgas⋅−:=

∆HgasT1

T2

TCp T( )⌠⎮⌡

d:=c)

Ans.Wideal mn⋅ 3.085− 103×BTUlbmol

=

Use ratio to calculate ideal work of steam per lbmol of gas

mn 15.043lb

lbmol=mn

T1

T2

TCp T( )⌠⎮⌡

d

∆Hv−:=

Calculate lbs of steam generated per lbmol of gas cooled.

Wideal 205.071−BTU

lb=Wideal ∆Hsteam Tσ ∆Ssteam⋅−( ):=

∆Ssteam 1.444−BTU

lb rankine⋅=∆Ssteam

∆Hv−

Tsteam:=∆Hsteam ∆Hv−:=b)

Ans.Wlost 6227BTUlbmol

=Wlost SdotG Tσ⋅:=

Calculate lost work by Eq. (5.34)

SdotG 11.756BTU

lbmol rankine⋅=SdotG mdotndot ∆Ssteam⋅ ∆Sgas+:=

∆Sgas 9.969− 10 3−×kgmol

BTUlb rankine⋅

=∆Sgas

T1

T2

TCp T( )

T

⌠⎮⎮⌡

d:=

∆Ssteam 1.444BTU

lb rankine⋅=∆Ssteam

∆Hv

Tsteam:=

SdotGndotgas

mdotsteam

ndotgas∆Ssteam⋅ ∆Sgas+=

Calculate entropy generation per lbmol of gas:

140



Ans.Wlost 14.8kJ

mol=Wlost SdotG Tσ⋅:=

Calculate lost work by Eq. (5.34)

SdotG 49.708J

mol K⋅=SdotG mdotndot ∆Ssteam⋅ ∆Sgas+:=

∆Sgas 41.835−J

mol K⋅=∆Sgas

T1

T2

TCp T( )

T

⌠⎮⎮⌡

d:=

∆Ssteam 6.048 103×J

kg K⋅=∆Ssteam

∆Hv

Tsteam:=

SdotGndotgas

mdotsteam

ndotgas∆Ssteam⋅ ∆Sgas+=

Calculate entropy generation per lbmol of gas:

SdotG SdotGsteam SdotGgas+=

Calculate the rate of entropy generation in the boiler. This is the sum of theentropy generation of the steam and the gas.

mdotndot 15.135gmmol

=mdotndotT1

T2

TCp T( )⌠⎮⌡

d−

∆Hv:=

ndotgasT1

T2

TCp T( )⌠⎮⌡

d⋅ mdotsteam ∆Hv⋅+ 0=

First apply an energy balance on the boiler to get the ratio of steam flow rate togas flow rate.:

a)

Tsteam 100 273.15+( )K:=Tσ 25 273.15+( )K:=

M 29gmmol

:=∆Hv 2256.9kJkg

:=Cp T( ) 3.83 0.000551TK⋅+⎛⎜

⎝⎞⎠

R⋅:=

T2 150 273.15+( )K:=T1 1100 273.15+( )K:=5.49

141



Now place a heat engine between the ethylene and the surroundings. Thiswould constitute a reversible process, therefore, the total entropy generatedmust be zero. calculate the heat released to the surroundings for ∆Stotal = 0.

Wlost 33.803kJ

mol=Wlost Tσ ∆Sethylene⋅ Qethylene−:=

Qethylene 60.563−kJ

mol=

Qethylene R ICPH T1 T2, 1.424, 14.394 10 3−⋅, 4.392− 10 6−⋅, 0,( )⋅:=

∆Sethylene 0.09−kJ

mol K⋅=

∆Sethylene R ICPS T1 T2, 1.424, 14.394 10 3−⋅, 4.392− 10 6−⋅, 0,( )⋅:=a)

Tσ 25 273.15+( )K:=T2 35 273.15+( )K:=T1 830 273.15+( )K:=5.50

Ans.Wideal 21.686−kJ

mol=Wideal ∆Hgas Tσ ∆Sgas⋅−:=

∆HgasT1

T2

TCp T( )⌠⎮⌡

d:=c)

Ans.Wideal mn⋅ 6.866−kJ

mol=

Use ratio to calculate ideal work of steam per lbmol of gas

mn 15.135gmmol

=mnT1

T2

TCp T( )⌠⎮⌡

d

∆Hv−:=

Calculate lbs of steam generated per lbmol of gas cooled.

Wideal 453.618−kJkg

=Wideal ∆Hsteam Tσ ∆Ssteam⋅−( ):=

∆Ssteam 6.048− 103×J

kg K⋅=∆Ssteam

∆Hv−

Tsteam:=∆Hsteam ∆Hv−:=b)

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∆SethyleneQC

Tσ− 0= Solving for QC gives: QC Tσ ∆Sethylene⋅:=

QC 26.76−kJ

mol=

Now apply an energy balance around the heat engine to find the workproduced. Note that the heat gained by the heat engine is the heat lost bythe ethylene.

QH Qethylene−:= WHE QH QC+:= WHE 33.803kJ

mol=

The lost work is exactly equal to the work that could be produced by the heatengine

143



Chapter5 A

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Transcript of Chapter5 A