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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
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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⋅:=
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(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⋅:=
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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⎛⎜⎝
⎞⎠
⋅:=
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The Carnot refrigerator is a reverse Carnot engine. Combine Eqs. (5.8) & (5.7) to get:
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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=
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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
⎛⎜⎝
⎞
⎠⋅−:=
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PROPRIETARY MATERIALonly to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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⋅
=
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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.
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
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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⋅:=
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(a) Assuming an isenthalpic process: T2 T1:= T2 298.15K= Ans.
(b)
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PROPRIETARY MATERIALonly to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
PROPRIETARY MATERIALonly to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
PROPRIETARY MATERIALonly to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
PROPRIETARY MATERIALonly to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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:
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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
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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.
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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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
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