chapter7 a
DESCRIPTION
Introduction to chemical engineering thermodynamicsTRANSCRIPT
Interpolation in Table F.2 at P = 525 kPa and S = 7.1595 kJ/(kg*K) yields:
H2 2855.2kJkg⋅:= V2 531.21
cm3
gm⋅:= mdot 0.75
kgsec⋅:=
With the heat, work, and potential-energy terms set equal to zero andwith the initial velocity equal to zero, Eq. (2.32a) reduces to:
∆Hu2
2
2+ 0= Whence u2 2− H2 H1−( )⋅:=
u2 565.2msec
= Ans.
By Eq. (2.27), A2mdot V2⋅
u2:= A2 7.05cm2= Ans.
7.5 The calculations of the preceding problem may be carried out for aseries of exit pressures until a minimum cross-sectional area is found.The corresponding pressure is the minimum obtainable in the convergingnozzle. Initial property values are as in the preceding problem.
Chapter 7 - Section A - Mathcad Solutions
7.1 u2 325msec⋅:= R 8.314
Jmol K⋅⋅:= molwt 28.9
gmmol
:= CP72
Rmolwt⋅:=
With the heat, work, and potential-energy terms set equal to zero andwith the initial velocity equal to zero, Eq. (2.32a) reduces to
∆Hu2
2
2+ 0= But ∆H CP ∆T⋅=
Whence ∆Tu2
2−
2 CP⋅:= ∆T 52.45− K= Ans.
7.4 From Table F.2 at 800 kPa and 280 degC:
H1 3014.9kJkg⋅:= S1 7.1595
kJkg K⋅⋅:=
220
Ans.A pmin( ) 7.021cm2=Ans.pmin 431.78kPa=
pmin Find pmin( ):=pmin
A pmin( )dd
0cm2
kPa⋅=Given
(guess)pmin 400 kPa⋅:=
A P( ) interp s p, a2, P,( ):=s cspline P A2,( ):=
a2iA2i
:=pi Pi:=i 1 5..:=
Fit the P vs. A2 data with cubic spline and findthe minimum P at the point where the first derivative of the spline is zero.
A2
7.05
7.022
7.028
7.059
7.127
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
cm2=u2
565.2
541.7
518.1
494.8
471.2
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
msec
=
A2mdot V2⋅
u2
→⎯⎯⎯
:=u2 2− H2 H1−( )⋅→⎯⎯⎯⎯⎯⎯
:=mdot 0.75kgsec⋅:=
V2
531.21
507.12
485.45
465.69
447.72
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
cm3
gm⋅:=H2
2855.2
2868.2
2880.7
2892.5
2903.9
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
kJkg⋅:=P
400
425
450
475
500
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
kPa⋅:=
Interpolations in Table F.2 at several pressures and at the givenentropy yield the following values:
S2 S1=S1 7.1595kJ
kg K⋅⋅:=H1 3014.9
kJkg⋅:=
221
Show spline fit graphically: p 400 kPa⋅ 401 kPa⋅, 500 kPa⋅..:=
400 420 440 460 480 5007.01
7.03
7.05
7.07
7.09
7.11
7.13
A2i
cm2
A p( )
cm2
PikPa
pkPa
,
7.9 From Table F.2 at 1400 kPa and 325 degC:
H1 3096.5kJkg⋅:= S1 7.0499
kJkg K⋅⋅:= S2 S1:=
Interpolate in Table F.2 at a series of downstream pressures and at S =7.0499 kJ/(kg*K) to find the minimum cross-sectional area.
P
800
775
750
725
700
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
kPa⋅:= H2
2956.0
2948.5
2940.8
2932.8
2924.9
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
kJkg⋅:= V2
294.81
302.12
309.82
317.97
326.69
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
cm3
gm⋅:=
u2 2− H2 H1−( )⋅→⎯⎯⎯⎯⎯⎯
:= A2V2
u2
⎛⎜⎝
⎞
⎠mdot⋅=
222
Svap 1.6872Btu
lbm rankine⋅⋅:=Sliq 0.3809
Btulbm rankine⋅⋅:=
Hvap 1167.1Btulbm⋅:=Hliq 228.03
Btulbm⋅:=
From Table F.4 at 35(psi), we see that the final state is wet steam:
H2 1154.8Btulbm
=H2 H1 ∆H+:=
∆H 78.8−Btulbm
=∆Hu1
2 u22−
2:=By Eq. (2.32a),
S1 1.6310Btu
lbm rankine⋅⋅:=H1 1233.6
Btulbm⋅:=
From Table F.4 at 130(psi) and 420 degF:
u2 2000ft
sec⋅:=u1 230
ftsec⋅:=7.10
x 0.966=xS1 Sliq−
Svap Sliq−:=
Svap 7.2479kJ
kg K⋅⋅:=Sliq 1.4098
kJkg K⋅⋅:=
At the nozzle exit, P = 140 kPa and S = S1, the initial value. FromTable F.2 we see that steam at these conditions is wet. Byinterpolation,
Ans.mdot 1.081kgsec
=mdotA2 u23
⋅
V23
:=
A2 6 cm2⋅:=At the throat,
V2
u2
⎛⎜⎝
⎞
⎠
→⎯⎯
5.561
5.553
5.552
5.557
5.577
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
cm2 sec⋅kg
=
Since mdot is constant,the quotient V2/u2 is ameasure of the area. Itsminimum value occurs very close to the value atvector index i = 3.
223
∆Tu2
2−
2 CP⋅:= ∆T 167.05− K= Ans.
Initial t = 15 + 167.05 = 182.05 degC Ans.
7.12 Values from the steam tables for saturated-liquid water:
At 15 degC: V 1.001cm3
gm⋅:= T 288.15 K⋅:=
Enthalpy difference for saturated liquid for a temperature change from14 to 15 degC:
∆H 67.13 58.75−( )J
gm⋅:= ∆t 2 K⋅:= Cp
∆H∆t
:=
Cp 4.19J
gm K⋅=β
1.5 10 4−⋅K
:= ∆P 4− atm⋅:=
Apply Eq. (7.25) to the constant-enthalpy throttling process. Assumesvery small temperature change and property values independent of P.
xH2 Hliq−
Hvap Hliq−:= x 0.987= (quality)
S2 Sliq x Svap Sliq−( )⋅+:= S2 1.67BTU
lbm rankine⋅=
SdotG S2 S1−:= SdotG 0.039Btu
lbm rankine⋅= Ans.
7.11 u2 580msec⋅:= T2 273.15 15+( ) K⋅:= molwt 28.9
gmmol
:= CP72
Rmolwt⋅:=
By Eq. (2.32a), ∆Hu1
2 u22−
2=
u22−
2=
But ∆H CP ∆T⋅= Whence
224
D
1.157−
0.0
0.040
0.0
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
105⋅ K2⋅:=C
0.0
4.392−
0.0
8.824−
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
10 6−
K2⋅:=
B
1.045
14.394
.593
28.785
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
10 3−
K⋅:=A
5.457
1.424
3.280
1.213
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
:=
ω
.224
.087
.038
.152
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
:=Pc
73.83
50.40
34.00
42.48
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
bar:=Tc
304.2
282.3
126.2
369.8
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
K:=
P1
80
60
60
20
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
bar:=T1
350
350
250
400
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
K:=
P2 1.2bar:=7.13--7.15
Ans.Wlost 0.413kJkg
=orWlost 0.413J
gm=Wlost Tσ ∆S⋅:=
Tσ 293.15 K⋅:=Apply Eq. (5.36) with Q=0:
∆S 1.408 10 3−×J
gm K⋅=∆S Cp ln
T ∆T+T
⎛⎜⎝
⎞⎠
⋅ β V⋅ ∆P⋅−:=
The entropy change for this process is given by Eq. (7.26):
∆T 0.093K=∆TV− 1 β T⋅−( )⋅ ∆P⋅
Cp1
9.86923joule
cm3 atm⋅⋅⎛
⎜⎝
⎞
⎠⋅:=
225
T2
280
302
232
385
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
K:=Guesses
The simplest procedure here is to iterate by guessing T2, and thencalculating it.
Eq. (6.68)SRi R ln Z β i qi,( )( ) β i− 0.5 qi⋅ Ii⋅−( )⋅:=
The derivative in these
equations equals -0.5
Eq. (6.67)HRi R T1i⋅ Z β i qi,( ) 1−( ) 1.5 qi⋅ Ii⋅−⎡⎣ ⎤⎦⋅:=
Eq. (6.65b)Ii lnZ β i qi,( ) β i+
Z β i qi,( )⎛⎜⎝
⎞
⎠:=i 1 4..:=
Z β q,( ) Find z( ):=
Eq. (3.52)z 1 β+ q β⋅z β−
z z β+( )⋅⋅−=
As in Example 7.4, Eq. (6.93) is applied to this constant-enthalpyprocess. If the final state at 1.2 bar is assumed an ideal gas, then Eq.(A) of Example 7.4 (pg. 265) applies. Its use requires expressions for HRand Cp at the initial conditions.
TrT1Tc
→⎯
:= Tr
1.151
1.24
1.981
1.082
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
= PrP1Pc
→⎯
:= Pr
1.084
1.19
1.765
0.471
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
=
7.13 Redlich/Kwong equation: Ω 0.08664:= Ψ 0.42748:=
β ΩPrTr⋅⎛⎜
⎝⎞⎠
→⎯⎯⎯
:= Eq. (3.53) qΨ
Ω Tr1.5⋅
⎛⎜⎝
⎞
⎠
→⎯⎯⎯⎯
:= Eq. (3.54)
Guess: z 1:=
Given
226
α 1 c 1 Tr0.5−( )⋅+⎡⎣ ⎤⎦2
→⎯⎯⎯⎯⎯⎯⎯⎯
:=
β ΩPrTr⋅⎛⎜
⎝⎞⎠
→⎯⎯⎯
:= Eq. (3.53) qΨ α⋅
Ω Tr⋅⎛⎜⎝
⎞⎠
→⎯⎯⎯
:= Eq. (3.54)
Guess: z 1:=
Given z 1 β+ q β⋅z β−
z z β+( )⋅⋅−= Eq. (3.52) Z β q,( ) Find z( ):=
i 1 4..:= Ii lnZ β i qi,( ) β i+
Z β i qi,( )⎛⎜⎝
⎞
⎠:= Eq. (6.65b)
Eq. (6.67)HRi R T1i⋅ Z β i qi,( ) 1− ciTriαi
⎛⎜⎝
⎞⎠
0.5⋅ 1+
⎡⎢⎣
⎤⎥⎦
qi⋅ Ii⋅−⎡⎢⎣
⎤⎥⎦
⋅:=
Z β i qi,( )0.7210.773
0.956
0.862
=
HR
2.681−
2.253−
0.521−
1.396−
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
kJmol
= SR
5.177−
4.346−
1.59−
2.33−
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
Jmol K⋅
=
τT2T1
→⎯
:= Cp R AB2
T1⋅ τ 1+( )⋅+C3
T12⋅ τ2
τ+ 1+( )⋅+D
τ T12⋅+⎡
⎢⎣
⎤⎥⎦
⋅⎡⎢⎣
⎤⎥⎦
→⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
:=
T2HRCp
T1+⎛⎜⎝
⎞⎠
→⎯⎯⎯⎯
:= ∆S Cp lnT2T1⎛⎜⎝
⎞⎠
⋅ R lnP2P1⎛⎜⎝
⎞⎠
⋅− SR−⎛⎜⎝
⎞⎠
→⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
:=
∆S
31.545
29.947
31.953
22.163
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
Jmol K⋅
= Ans.T2
279.971
302.026
232.062
384.941
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
K= Ans.
7.14 Soave/Redlich/Kwong equation: Ω 0.08664:= Ψ 0.42748:=
c 0.480 1.574 ω⋅+ 0.176 ω2
⋅−( )→⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
:=
227
Ans.∆S
31.565
30.028
32.128
22.18
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
Jmol K⋅
=∆S Cp lnT2T1⎛⎜⎝
⎞⎠
⋅ R lnP2P1⎛⎜⎝
⎞⎠
⋅− SR−⎛⎜⎝
⎞⎠
→⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
:=
Ans.T2
272.757
299.741
231.873
383.554
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
K=T2HRCp
T1+⎛⎜⎝
⎞⎠
→⎯⎯⎯⎯
:=
Cp R AB2
T1⋅ τ 1+( )⋅+C3
T12⋅ τ2
τ+ 1+( )⋅+D
τ T12⋅+⎡
⎢⎣
⎤⎥⎦
⋅⎡⎢⎣
⎤⎥⎦
→⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
:=τT2T1
→⎯
:=
SR
6.126−
4.769−
1.789−
2.679−
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
Jmol K⋅
=HR
2.936−
2.356−
0.526−
1.523−
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
kJmol
=
Z β i qi,( )0.750.79
0.975
0.866
=
T2
273
300
232
384
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
K:=Guesses
Now iterate for T2:
ci−Triαi
⎛⎜⎝
⎞⎠
0.5⋅The derivative in these equations equals:
Eq. (6.68)SRi R ln Z β i qi,( )( ) β i− ciTriαi
⎛⎜⎝
⎞⎠
0.5⋅ qi⋅ Ii⋅−
⎡⎢⎣
⎤⎥⎦
⋅:=
228
T2
270
297
229
383
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
K:=Guesses
Now iterate for T2:
ci−Triαi
⎛⎜⎝
⎞⎠
0.5⋅The derivative in these equations equals:
Eq. (6.68)SRi R ln Z β i qi,( )( ) β i− ciTriαi
⎛⎜⎝
⎞⎠
0.5⋅ qi⋅ Ii⋅−
⎡⎢⎣
⎤⎥⎦
⋅:=
Eq. (6.67)HRi R T1i⋅ Z β i qi,( ) 1− ciTriαi
⎛⎜⎝
⎞⎠
0.5⋅ 1+
⎡⎢⎣
⎤⎥⎦
qi⋅ Ii⋅−⎡⎢⎣
⎤⎥⎦
⋅:=
Eq. (6.65b)Ii1
2 2⋅ln
Z β i qi,( ) σ β i⋅+
Z β i qi,( ) ε β i⋅+
⎛⎜⎝
⎞
⎠⋅:=i 1 4..:=
Z β q,( ) Find z( ):=
Eq. (3.52)z 1 β+ q β⋅z β−
z ε β⋅+( ) z σ β⋅+( )⋅⋅−=Given
z 1:=Guess:
Eq. (3.54)qΨ α⋅
Ω Tr⋅⎛⎜⎝
⎞⎠
→⎯⎯⎯
:=Eq. (3.53)β ΩPrTr⋅⎛⎜
⎝⎞⎠
→⎯⎯⎯
:=
α 1 c 1 Tr0.5−( )⋅+⎡⎣ ⎤⎦2
→⎯⎯⎯⎯⎯⎯⎯⎯
:=c 0.37464 1.54226 ω⋅+ 0.26992 ω2
⋅−( )→⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
:=
Ψ 0.45724:=Ω 0.07779:=ε 1 2−:=σ 1 2+:=
Peng/Robinson equation:7.15
229
(quality)x 0.92=xS2 Sliq−
Svap Sliq−:=
S2 S1:=Svap 7.9094kJ
kg K⋅⋅:=Sliq 0.8321
kJkg K⋅⋅:=
For isentropic expansion, exhaust is wet steam:
Ans.mdot 4.103kgsec
=mdotWdot
H2 H1−:=
By Eq. (7.13),
S1 7.3439kJ
kg K⋅⋅:=H2 2609.9
kJkg⋅:=H1 3462.9
kJkg⋅:=
Data from Table F.2:Wdot 3500− kW⋅:=7.18
∆S
31.2
29.694
31.865
22.04
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
Jmol K⋅
= Ans.∆S Cp lnT2T1⎛⎜⎝
⎞⎠
⋅ R lnP2P1⎛⎜⎝
⎞⎠
⋅− SR−⎛⎜⎝
⎞⎠
→⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
:=
Ans.T2
269.735
297.366
229.32
382.911
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
K=T2HRCp
T1+⎛⎜⎝
⎞⎠
→⎯⎯⎯⎯
:=
Cp R AB2
T1⋅ τ 1+( )⋅+C3
T12⋅ τ2
τ+ 1+( )⋅+D
τ T12⋅+⎡
⎢⎣
⎤⎥⎦
⋅⎡⎢⎣
⎤⎥⎦
→⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
:=τT2T1
→⎯
:=
SR
6.152−
4.784−
1.847−
2.689−
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
Jmol K⋅
=HR
3.041−
2.459−
0.6−
1.581−
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
kJmol
=
Z β i qi,( )0.7220.76
0.95
0.85
=
230
Hliq 251.453kJkg⋅:= Hvap 2609.9
kJkg⋅:=
H'2 Hliq x Hvap Hliq−( )⋅+:= H'2 2.421 103×kJkg
=
ηH2 H1−
H'2 H1−:= η 0.819= Ans.
7.19 The following vectors contain values for Parts (a) through (g). For intakeconditions:
H1
3274.3kJkg⋅
3509.8kJkg⋅
3634.5kJkg⋅
3161.2kJkg⋅
2801.4kJkg⋅
1444.7Btulbm⋅
1389.6Btulbm⋅
⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠
:= S1
6.5597kJ
kg K⋅⋅
6.8143kJ
kg K⋅⋅
6.9813kJ
kg K⋅⋅
6.4536kJ
kg K⋅⋅
6.4941kJ
kg K⋅⋅
1.6000Btu
lbm rankine⋅⋅
1.5677Btu
lbm rankine⋅⋅
⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠
:= η
0.80
0.77
0.82
0.75
0.75
0.80
0.75
⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟⎟
⎠
:=
231
For discharge conditions:
Sliq
0.9441kJ
kg K⋅⋅
0.8321kJ
kg K⋅⋅
0.6493kJ
kg K⋅⋅
1.0912kJ
kg K⋅⋅
1.5301kJ
kg K⋅⋅
0.1750Btu
lbm rankine⋅⋅
0.2200Btu
lbm rankine⋅⋅
⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠
:= Svap
7.7695kJ
kg K⋅⋅
7.9094kJ
kg K⋅⋅
8.1511kJ
kg K⋅⋅
7.5947kJ
kg K⋅⋅
7.1268kJ
kg K⋅⋅
1.9200Btu
lbm rankine⋅⋅
1.8625Btu
lbm rankine⋅⋅
⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠
:= S'2 S1=
Hliq
289.302kJkg⋅
251.453kJkg⋅
191.832kJkg⋅
340.564kJkg⋅
504.701kJkg⋅
94.03Btulbm⋅
120.99Btulbm⋅
⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠
:= Hvap
2625.4kJkg⋅
2609.9kJkg⋅
2584.8kJkg⋅
2646.0kJkg⋅
2706.3kJkg⋅
1116.1Btulbm⋅
1127.3Btulbm⋅
⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠
:= mdot
80kgsec⋅
90kgsec⋅
70kgsec⋅
65kgsec⋅
50kgsec⋅
150lbm
sec⋅
100lbm
sec⋅
⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠
:=
232
Ans.Wdot
91230−
117544−
109523−
60126−
17299−
87613−
46999−
⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟⎟
⎠
hp=Wdot
68030−
87653−
81672−
44836−
12900−
65333−
35048−
⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟⎟
⎠
kW=
S26
S27
⎛⎜⎜⎝
⎞
⎠
1.7762
1.7484⎛⎜⎝
⎞⎠
Btulbm rankine⋅
=H26
H27
⎛⎜⎜⎝
⎞
⎠
1031.9
1057.4⎛⎜⎝
⎞⎠
Btulbm
=
Ans.
S21
S22
S23
S24
S25
⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟⎟
⎠
7.1808
7.6873
7.7842
7.1022
6.7127
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
kJkg K⋅
=
H21
H22
H23
H24
H25
⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟⎟
⎠
2423.9
2535.9
2467.8
2471.4
2543.4
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
kJkg
=
S2 Sliq x2 Svap Sliq−( )⋅+⎡⎣ ⎤⎦→⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
:=x2H2 Hliq−
Hvap Hliq−
→⎯⎯⎯⎯⎯
:=
Wdot ∆H mdot⋅( )→⎯⎯⎯⎯
:=H2 H1 ∆H+:=∆H η H'2 H1−( )⋅⎡⎣ ⎤⎦→⎯⎯⎯⎯⎯⎯
:=
H'2 Hliq x'2 Hvap Hliq−( )⋅+⎡⎣ ⎤⎦→⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
:=x'2S1 Sliq−
Svap Sliq−
→⎯⎯⎯⎯
:=
233
T0 762.42K= Ans.
Thus the initial temperature is 489.27 degC
7.21 T1 1223.15 K⋅:= P1 10 bar⋅:= P2 1.5 bar⋅:=
CP 32J
mol K⋅⋅:= η 0.77:=
Eqs. (7.18) and (7.19) derived for isentropic compression apply equally wellfor isentropic expansion. They combine to give:
W's CP T1⋅P2
P1
⎛⎜⎝
⎞
⎠
RCP
1−
⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
⋅:= W's 15231−J
mol=
Ws η W's⋅:= ∆H Ws:= Ws 11728−J
mol= Ans.
7.20 T 423.15 K⋅:= P0 8.5 bar⋅:= P 1 bar⋅:=
For isentropic expansion, ∆S 0J
mol K⋅⋅:=
For the heat capacity of nitrogen:
A 3.280:= B0.593 10 3−⋅
K:= D 0.040 105⋅ K2⋅:=
For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15)with C = 0. Substitute:
τ 0.5:= (guess)
Given
∆S R A ln τ( )⋅ BTτ⋅
D
T2τ 1+
2⎛⎜⎝
⎞⎠
⋅+⎡⎢⎣
⎤⎥⎦
τ 1−( )⋅+ lnPP0
⎛⎜⎝
⎞⎠
−⎡⎢⎣
⎤⎥⎦
⋅=
τ Find τ( ):= T0Tτ
:=
234
Tr0T0
Tc:= Tr0 1.282= Pr0
P0
Pc:= Pr0 1.3706=
PrPPc
:= Pr 0.137=
The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0:
τ 0.5:= (guess)
Given
∆S R A ln τ( )⋅ B T0⋅ C T02⋅
τ 1+2
⎛⎜⎝
⎞⎠
⋅+⎡⎢⎣
⎤⎥⎦
τ 1−( )⋅+ lnPP0
⎛⎜⎝
⎞⎠
−
SRBτ T0⋅
TcPr, ω,
⎛⎜⎝
⎞
⎠SRB Tr0 Pr0, ω,( )−+
...⎡⎢⎢⎢⎢⎣
⎤⎥⎥⎥⎥⎦
⋅=
τ Find τ( ):= T τ T0⋅:= T 445.71K=
TrTTc
:= Tr 1.092=
Eq. (7.21) also applies to expansion:
T2 T1∆HCP
+:= T2 856.64K= Ans.
7.22 Isobutane: Tc 408.1 K⋅:= Pc 36.48 bar⋅:= ω 0.181:=
T0 523.15 K⋅:= P0 5000 kPa⋅:= P 500 kPa⋅:=
∆S 0J
mol K⋅⋅:= For the heat capacity of isobutane:
A 1.677:= B37.853 10 3−⋅
K:= C
11.945− 10 6−⋅
K2:=
235
Sliq 0.6493kJ
kg K⋅⋅:=x2 0.95:=At 10 kPa:
S1 6.5138kJ
kg K⋅⋅:=H1 2851.0
kJkg⋅:=
From Table F.2 @ 1700 kPa & 225 degC:7.23
Ans. T 457.8K=T τ T0⋅:=τ 0.875=τ Find τ( ):=
∆H R A T0⋅ τ 1−( )⋅B2
T02⋅ τ
2 1−( )⋅+C3
T03⋅ τ
3 1−( )⋅+
Tc HRBτ T0⋅
TcPr, ω,
⎛⎜⎝
⎞
⎠HRB Tr0 Pr0, ω,( )−
⎛⎜⎝
⎞
⎠⋅+
...⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
⋅=
Given
(guess)τ 0.7:=
The actual final temperature is now found from Eq. (6.91) combined with Eq(4.7), written:
Ans.Wdot 4665.6− kW=Wdot ndot ∆H⋅:=
∆H 6665.1−J
mol=∆H η ∆H'⋅:=ndot 700
molsec
⋅:=η 0.8:=
The actual enthalpy change from Eq. (7.16):
∆H' 8331.4−J
mol=
∆H' ∆Hig R Tc⋅ HRB Tr Pr, ω,( ) HRB Tr0 Pr0, ω,( )−( )⋅+:=
∆Hig 11.078−kJ
mol=
∆Hig R ICPH T0 T, 1.677, 37.853 10 3−⋅, 11.945− 10 6−⋅, 0.0,( )⋅:=
The enthalpy change is given by Eq. (6.91):
236
Ans.
7.24 T0 673.15 K⋅:= P0 8 bar⋅:= P 1 bar⋅:=
For isentropic expansion, ∆S 0J
mol K⋅⋅:=
For the heat capacity of carbon dioxide:
A 5.457:= B1.045 10 3−⋅
K:= D 1.157− 105⋅ K2⋅:=
For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with C = 0:τ 0.5:= (guess)
Given
∆S R A ln τ( )⋅ B T0⋅D
T0 τ⋅( )2τ 1+
2⎛⎜⎝
⎞⎠
⋅+⎡⎢⎣
⎤⎥⎦
τ 1−( )⋅+ lnPP0
⎛⎜⎝
⎞⎠
−⎡⎢⎣
⎤⎥⎦
⋅=
τ Find τ( ):= τ 0.693= T' τ T0⋅:= T' 466.46K=
Hliq 191.832kJkg⋅:= Hvap 2584.8
kJkg⋅:= Svap 8.1511
kJkg K⋅⋅:=
mdot 0.5kgsec⋅:= Wdot 180− kW⋅:=
H2 Hliq x2 Hvap Hliq−( )⋅+:= ∆H H2 H1−:=
H2 2.465 103×kJkg
= ∆H 385.848−kJkg
=
(a) Qdot mdot ∆H⋅ Wdot−:= Qdot 12.92−kJsec
= Ans.
(b) For isentropic expansion to 10 kPa, producing wet steam:
x'2S1 Sliq−
Svap Sliq−:= H'2 Hliq x'2 Hvap Hliq−( )⋅+:=
x'2 0.782= H'2 2.063 103×kJkg
=
Wdot' mdot H'2 H1−( )⋅:= Wdot' 394.2− kW=
237
∆HS Cp T1⋅P2P1⎛⎜⎝
⎞⎠
RCp
1−
⎡⎢⎢⎣
⎤⎥⎥⎦
⋅
⎡⎢⎢⎣
⎤⎥⎥⎦
→⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
:=
Eq. (7.22) Applies to expanders aswell as to compressors
Ideal gases with constant heat capacities∆H Cp T2 T1−( )⋅[ ]→⎯⎯⎯⎯⎯⎯
:=
Cp
3.5
4.0
5.5
4.5
2.5
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
R⋅:=P2
1.2
2.0
3.0
1.5
1.2
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
:=T2
371
376
458
372
403
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
:=P1
6
5
10
7
4
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
:=T1
500
450
525
475
550
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
:=
Vectors containing data for Parts (a) through (e):7.25
Thus the final temperature is 246.75 degC
Ans.T 519.9K=T τ T0⋅:=τ 0.772=τ Find τ( ):=
∆H R A T0⋅ τ 1−( )⋅B2
T02⋅ τ
2 1−( )⋅+DT0
τ 1−
τ⎛⎜⎝
⎞⎠
⋅+⎡⎢⎣
⎤⎥⎦
⋅=
Given
For the enthalpy change of an ideal gas, combine Eqs. (4.2) and (4.7)with C = 0:
∆H 7.326−kJ
mol=∆H Work:=
Ans.Work 7.326−kJ
mol=Work η ∆H'⋅:=η 0.75:=
∆H' 9.768−kJ
mol=
∆H' R ICPH T0 T', 5.457, 1.045 10 3−⋅, 0.0, 1.157− 105⋅,( )⋅:=
238
Ans.SdotG 1.126 103×J
K sec⋅=SdotG ndot ∆S⋅:=
By Eq. (5.37), for adiabatic operation :
∆S 6.435J
mol K⋅=∆S R
CpR
lnT2T1⎛⎜⎝
⎞⎠
⋅ lnP2P1⎛⎜⎝
⎞⎠
−⎛⎜⎝
⎞⎠
⋅:=
By Eq. (5.14):
T2 433.213K=T2 T1 1 ηP2P1⎛⎜⎝
⎞⎠
RCp
1−
⎡⎢⎢⎣
⎤⎥⎥⎦
⋅+
⎡⎢⎢⎣
⎤⎥⎥⎦
⋅:=
For an expander operating with an ideal gas with constant Cp, one canshow that:
Ans.η 0.576=η 0.065 0.08 lnWdotkW
⎛⎜⎝
⎞⎠
⋅+⎛⎜⎝
⎞⎠
:=
Ans.Wdot 594.716kW=Wdot Find Wdot( ):=
Wdot 0.065 .08 lnWdotkW
⎛⎜⎝
⎞⎠
⋅+⎛⎜⎝
⎞⎠
− ndot⋅ Cp⋅ T1⋅P2P1⎛⎜⎝
⎞⎠
RCp
1−
⎡⎢⎢⎣
⎤⎥⎥⎦
⋅=
Given
Wdot 600kW:=η 0.75:=Guesses:
P2 1.2bar:=P1 6bar:=T1 550K:=ndot 175molsec
:=Cp72
R⋅:=7.26
η
0.7
0.803
0.649
0.748
0.699
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
=η∆H∆HS
→⎯⎯
:=
239
If η were 0.8, the pressure would be higher, because a smaller pressuredrop would be required to produce the same work and ∆H.
t=120 degC; P=198.54 kPa
These are sufficiently close, and we conclude that:
xS 0.925=xH 0.924=The trial values given produce:
xS6.7093 Sl−
Sv Sl−:=xH
Hv 801.7− .75 Hl⋅−.75 Hv Hl−( )⋅
:=
The two equations for x are:
Sv 7.1293:=Sl 1.5276:=
Hv 2706.0:=Hl 503.7:=
If the exhaust steam (Point 2, Fig. 7.4) is "dry," i.e., saturated vapor, thenisentropicexpansion to the same pressure (Point 2', Fig. 7.4) must produce"wet" steam, withentropy:
S2 = S1 = 6.7093 = (x)(Svap) + (1-x)(Sliq) [x is quality]
A second relation follows from Eq. (7.16), written:
∆H = Hvap - 3207.1 = (η)(∆HS) = (0.75)[ (x)(Hvap) + (1-x)(Hliq) - 3207.1]
Each of these equations may be solved for x. Given a final temperatureand the corresponding vapor pressure, values for Svap, Sliq, Hvap, andHliq are found from the table for saturated steam, and substitution into theequations for x produces two values. The required pressure is the one forwhich the two values of x agree. This is clearly a trial process. For a finaltrial temperature of 120 degC, the following values of H and S forsaturated liquid and saturated vapor are found in the steam table:
S1 6.7093:=H1 3207.1:=
Properties of superheated steam at 4500 kPa and 400 C from Table F.2,p. 742.
7.27
240
i 1 5..:=η
0.80
0.75
0.78
0.85
0.80
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
:=ndot
200
150
175
100
0.5 453.59⋅
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
molsec
⋅:=
∆S 0J
mol K⋅⋅:=
P
1 bar⋅
1 bar⋅
1 bar⋅
2 bar⋅
15 psi⋅
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
:=P0
6 bar⋅
5 bar⋅
7 bar⋅
8 bar⋅
95 psi⋅
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
:=T0
753.15
673.15
773.15
723.15
755.37
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
K⋅:=
Assume nitrogen an ideal gas. First find the temperature after isentropicexpansion from a combination of Eqs. (5.14) & (5.15) with C = 0. Thenfind the work (enthalpy change) of isentropic expansion by a combinationof Eqs. (4.2) and (4.7) with C = 0. The actual work (enthalpy change) isfound from Eq. (7.20). From this value, the actual temperature is found bya second application of the preceding equation, this time solving it for thetemperature. The following vectors contain values for Parts (a) through(e):
7.30
Ans.∆T 0.044degC=
∆T∆H V P2 P1−( )⋅−
Cp:=Eq. (7.25) with β=0 is solved for ∆T:
Ws 0.223−kJkg
=(7.14)Ws ∆H:=
∆H η V⋅ P2 P1−( )⋅:=Eqs. (7.16) and (7.24) combine to give:
Cp 4.190kJ
kg degC⋅⋅:=V 1001
cm3
kg⋅:=
Data in Table F.1 for saturated liquid water at 15 degC give:
η 0.55:=T1 15 degC⋅:=P2 1 atm⋅:=P1 5 atm⋅:=7.29
241
Ti T0iτi⋅:=τi Tau T0i
∆Hi,( ):=Tau T0 ∆H,( ) Find τ( ):=
∆H R A T0⋅ τ 1−( )⋅B2
T02⋅ τ
2 1−( )⋅+DT0
τ 1−
τ⎛⎜⎝
⎞⎠
⋅+⎡⎢⎣
⎤⎥⎦
⋅=
Given
(guess)τ 0.5:=
∆H
7103.4−
5459.8−
7577.2−
5900.5−
7289.7−
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
Jmol
=∆H ∆H' η⋅( )→⎯⎯⎯
:=∆H'
8879.2−
7279.8−
9714.4−
6941.7−
9112.1−
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
Jmol
=
∆H'i R ICPH T0iTi, 3.280, 0.593 10 3−⋅, 0.0, 0.040 105⋅,⎛
⎝⎞⎠
⋅:=
T
460.67
431.36
453.48
494.54
455.14
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
K=Ti T0iτi⋅:=
τi Tau T0iP0i, Pi,( ):=Tau T0 P0, P,( ) Find τ( ):=
∆S R A ln τ( )⋅ B T0⋅D
T02 τ
2⋅
τ 1+2
⎛⎜⎝
⎞⎠
⋅+⎡⎢⎣
⎤⎥⎦
τ 1−( )⋅+ lnPP0
⎛⎜⎝
⎞⎠
−⎡⎢⎣
⎤⎥⎦
⋅=
Given
(guess)τ 0.5:=
D 0.040 105⋅ K2⋅:=B0.593 10 3−⋅
K:=A 3.280:=
For the heat capacity of nitrogen:
242
η t 0.761= Ans.
The process is adiabatic; Eq. (5.33) becomes:
SdotG mdot S2 S1−( )⋅:= SdotG 58.949kWK
= Ans.
Wdotlost Tσ SdotG⋅:= Wdotlost 17685kW= Ans.
7.32 For sat. vapor steam at 1200 kPa, Table F.2:
H2 2782.7kJkg⋅:= S2 6.5194
kJkg K⋅⋅:=
The saturation temperature is 187.96 degC.The exit temperature of the exhaust gas is therefore 197.96 degC, andthe temperature CHANGE of the exhaust gas is -202.04 K. For the water at 20 degC from Table F.1,
H1 83.86kJkg⋅:= S1 0.2963
kJkg K⋅⋅:=
T
520.2
492.62
525.14
529.34
516.28
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
K= Ans. Wdot ndot ∆H⋅( )→⎯⎯⎯⎯
:= Wdot
1421−
819−
1326−
590−
1653−
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
kW= Ans.
7.31 Property values and data from Example 7.6:
H1 3391.6kJkg⋅:= S1 6.6858
kJkg K⋅⋅:= mdot 59.02
kgsec⋅:=
H2 2436.0kJkg⋅:= S2 7.6846
kJkg K⋅⋅:= Wdot 56400− kW⋅:=
Tσ 300 K⋅:= By Eq. (5.26)
Wdotideal mdot H2 H1− Tσ S2 S1−( )⋅−⎡⎣ ⎤⎦⋅:= Wdotideal 74084− kW=
η tWdot
Wdotideal:=
243
∆Sgas R MCPS T1 T2, 3.34, 1.12 10 3−⋅, 0.0, 0.0,( )⋅ lnT2
T1
⎛⎜⎝
⎞
⎠⋅:=
∆Hgas R MCPH T1 T2, 3.34, 1.12 10 3−⋅, 0.0, 0.0,( )⋅ T2 T1−( )⋅:=
molwt 18gmmol
:=
T2 471.11K=T1 673.15K=
T2 273.15 197.96+( ) K⋅:=T1 273.15 400+( ) K⋅:=
ndot 125molsec
⋅:=For the exhaust gases:
x3 0.883= S3 7.023kJ
kg K⋅=
S3 Sliq x3 Slv⋅+:=x3H3 Hliq−
Hlv:=
H3 2.345 103×kJkg
=∆H23 437.996−kJkg
=
H3 H2 ∆H23+:=∆H23 η H'3 H2−( )⋅:=
H'3 2.174 103×kJkg
=x'3 0.811=S'3 6.519kJ
kg K⋅=
H'3 Hliq x'3 Hlv⋅+:=x'3S'3 Sliq−
Slv:=S'3 S2:=
For isentropic expansion of steam in the turbine:
η 0.72:=Slv 6.9391kJ
kg K⋅⋅:=Sliq 0.8932
kJkg K⋅⋅:=
Hlv 2346.3kJkg⋅:=Hliq 272.0
kJkg⋅:=
The turbine exhaust will be wet vapor steam.For sat. liquid and sat. vapor at the turbine exhaust pressure of 25 kPa, thebest property values are found from Table F.1 by interpolation between 64and 65 degC:
244
For both the boiler and the turbine, Eq. (5.33) applies with Q = 0. For the boiler:
SdotG ndot ∆Sgas⋅ mdot S2 S1−( )⋅+:=
Boiler: SdotG 0.4534kWK
= Ans.
For the turbine: SdotG mdot S3 S2−( )⋅:=
Turbine: SdotG 0.156kWK
= Ans.
(d) Wdotlost.boiler 0.4534kWK
⋅ Tσ⋅:= Wdotlost.boiler 132.914kW=
Wdotlost.turbine 0.1560kWK
⋅ Tσ⋅:= Wdotlost.turbine 45.731kW=
FractionboilerWdotlost.boiler
Wdotideal:= Fractionboiler 0.4229= Ans.
∆Hgas 6.687− 103×kJ
kmol= ∆Sgas 11.791−
kJkmol K⋅
=
Energy balance on boiler:
mdotndot− ∆Hgas⋅
H2 H1−:= mdot 0.30971
kgsec
=
(a) Wdot mdot H3 H2−( )⋅:= Wdot 135.65− kW= Ans.
(b) By Eq. (5.25): Tσ 293.15 K⋅:=
Wdotideal ndot ∆Hgas⋅ mdot H3 H1−( )⋅+Tσ− ndot ∆Sgas⋅ mdot S3 S1−( )⋅+⎡⎣ ⎤⎦⋅+
...:=
Wdotideal 314.302− kW= η tWdot
Wdotideal:= η t 0.4316= Ans.
(c)
245
Ans.Wdot 1173.4kW=Wdot mdot ∆H⋅:=mdot 2.5kgsec⋅:=
Ans.S2 7.4586kJ
kg K⋅⋅:=
Interpolation in Table F.2 at 700 kPa for the entropy of steam with thisenthalpy gives
Ans.H2 3154.6kJkg
=H2 H1 ∆H+:=
∆H 469.359kJkg
=∆HH'2 H1−
η:=η 0.78:=H'2 3051.3
kJkg⋅:=
Interpolation in Table F.2 at 700 kPa for the enthalpy of steam with thisentropy gives
S'2 S1= 7.2847kJ
kg K⋅⋅=For isentropic expansion,
S1 7.2847kJ
kg K⋅⋅:=H1 2685.2
kJkg⋅:=
From Table F.2 for sat. vap. at 125 kPa:7.34
η t Fractionboiler+ Fractionturbine+ 1=Note that:
Ans.Fractionturbine 0.1455=FractionturbineWdotlost.turbine
Wdotideal:=
246
τi Tau T0iP0i, Pi,( ):=Tau T0 P0, P,( ) Find τ( ):=
∆S R A ln τ( )⋅ B T0⋅D
T02 τ
2⋅
τ 1+2
⎛⎜⎝
⎞⎠
⋅+⎡⎢⎣
⎤⎥⎦
τ 1−( )⋅+ lnPP0
⎛⎜⎝
⎞⎠
−⎡⎢⎣
⎤⎥⎦
⋅=
Given
(guess)τ 0.5:=
D 0.016− 105⋅ K2⋅:=B0.575 10 3−⋅
K:=A 3.355:=
For the heat capacity of air:
i 1 6..:=η
0.75
0.70
0.80
0.75
0.75
0.70
⎛⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟
⎠
:=ndot
100
100
150
50
0.5 453.59⋅
0.5 453.59⋅
⎛⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟
⎠
molsec
⋅:=∆S 0
Jmol K⋅⋅:=
P
375 kPa⋅
1000 kPa⋅
500 kPa⋅
1300 kPa⋅
55 psi⋅
135 psi⋅
⎛⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟
⎠
:=P0
101.33 kPa⋅
375 kPa⋅
100 kPa⋅
500 kPa⋅
14.7 psi⋅
55 psi⋅
⎛⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟
⎠
:=T0
298.15
353.15
303.15
373.15
299.82
338.71
⎛⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟
⎠
K⋅:=
Assume air an ideal gas. First find the temperature after isentropiccompression from a combination of Eqs. (5.14) & (5.15) with C = 0. Thenfind the work (enthalpy change) of isentropic compression by acombination of Eqs. (4.2) and (4.7) with C = 0. The actual work (enthalpychange) is found from Eq. (7.20). From this value, the actual temperatureis found by a second application of the preceding equation, this timesolving it for the temperature. The following vectors contain values forParts (a) through (f):
7.35
247
Ti T0iτi⋅:= T
431.06
464.5
476.19
486.87
434.74
435.71
⎛⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟
⎠
K=
∆H'i R ICPH T0iTi, 3.355, 0.575 10 3−⋅, 0.0, 0.016− 105⋅,⎛
⎝⎞⎠
⋅:=
∆H'
3925.2
3314.6
5133.2
3397.5
3986.4
2876.6
⎛⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟
⎠
Jmol
=
∆H∆H'η
⎛⎜⎝
⎞⎠
→⎯⎯
:= ∆H
5233.6
4735.1
6416.5
4530
5315.2
4109.4
⎛⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟
⎠
Jmol
=
τ 1.5:= (guess)
Given ∆H R A T0⋅ τ 1−( )⋅B2
T02⋅ τ
2 1−( )⋅+DT0
τ 1−
τ⎛⎜⎝
⎞⎠
⋅+⎡⎢⎣
⎤⎥⎦
⋅=
Tau T0 ∆H,( ) Find τ( ):= τi Tau T0i∆Hi,( ):= Ti T0i
τi⋅:=
Wdot ndot ∆H⋅( )→⎯⎯⎯⎯
:=
248
Tr0 0.725= Pr0P0
Pc:= Pr0 0.0177=
PrPPc
:= Pr 0.089=
Use generalized second-virial correlation:
The entropy change is given by Eq. (6.92) combined with Eq. (5.15); C = 0:
τ 1.4:= (guess)
Given
∆S R A ln τ( )⋅ B T0⋅D
τ T0⋅( )2τ 1+
2⎛⎜⎝
⎞⎠
⋅+⎡⎢⎣
⎤⎥⎦
τ 1−( )⋅+ lnPP0
⎛⎜⎝
⎞⎠
−
SRBτ T0⋅
TcPr, ω,
⎛⎜⎝
⎞
⎠SRB Tr0 Pr0, ω,( )−+
...⎡⎢⎢⎢⎢⎣
⎤⎥⎥⎥⎥⎦
⋅=
τ Find τ( ):= τ 1.437= T τ T0⋅:= T 422.818K=
TrTTc
:= Tr 1.042=
T
474.68
511.58
518.66
524.3
479.01
476.79
⎛⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟
⎠
K= Wdot
702
635
1291
304
1617
1250
⎛⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟
⎠
hp= Wdot
523
474
962
227
1205
932
⎛⎜⎜⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟⎟⎟
⎠
kW= Ans.
7.36 Ammonia: Tc 405.7 K⋅:= Pc 112.8 bar⋅:= ω 0.253:=
T0 294.15 K⋅:= P0 200 kPa⋅:= P 1000 kPa⋅:=
∆S 0J
mol K⋅⋅:= For the heat capacity of ammonia:
A 3.578:= B3.020 10 3−⋅
K:= D 0.186− 105⋅ K2⋅:=
Tr0T0
Tc:=
249
Ans.∆S 2.347J
mol K⋅=
∆S R A ln τ( )⋅ B T0⋅D
τ T0⋅( )2τ 1+
2⎛⎜⎝
⎞⎠
⋅+⎡⎢⎣
⎤⎥⎦
τ 1−( )⋅+ lnPP0
⎛⎜⎝
⎞⎠
−
SRB Tr Pr, ω,( ) SRB Tr0 Pr0, ω,( )−+
...⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
⋅:=
Tr 1.103=TrTTc
:=
Ans.T 447.47K=T τ T0⋅:=τ 1.521=τ Find τ( ):=
∆H R A T0⋅ τ 1−( )⋅B2
T02⋅ τ
2 1−( )⋅+DT0
τ 1−
τ⎛⎜⎝
⎞⎠
⋅+
Tc HRBτ T0⋅
TcPr, ω,
⎛⎜⎝
⎞
⎠HRB Tr0 Pr0, ω,( )−
⎛⎜⎝
⎞
⎠⋅+
...⎡⎢⎢⎢⎢⎣
⎤⎥⎥⎥⎥⎦
⋅=
Given
(guess)τ 1.4:=
The actual final temperature is now found from Eq. (6.91) combined with Eq(4.7), written:
∆H 5673.2J
mol=∆H
∆H'η
:=η 0.82:=
The actual enthalpy change from Eq. (7.17):
∆H' 4652J
mol=
∆H' ∆Hig R Tc⋅ HRB Tr Pr, ω,( ) HRB Tr0 Pr0, ω,( )−( )⋅+:=
∆Hig 4.826kJ
mol=
∆Hig R ICPH T0 T, 3.578, 3.020 10 3−⋅, 0.0, 0.186− 105⋅,( )⋅:=
250
Pr 0.386=
Use generalized second-virial correlation:
The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0:
τ 1.1:= (guess)
Given
∆S R A ln τ( )⋅ B T0⋅ C T02⋅
τ 1+2
⎛⎜⎝
⎞⎠
⋅+⎡⎢⎣
⎤⎥⎦
τ 1−( )⋅+ lnPP0
⎛⎜⎝
⎞⎠
−
SRBτ T0⋅
TcPr, ω,
⎛⎜⎝
⎞
⎠SRB Tr0 Pr0, ω,( )−+
...⎡⎢⎢⎢⎢⎣
⎤⎥⎥⎥⎥⎦
⋅=
τ Find τ( ):= τ 1.069= T τ T0⋅:= T 324.128K=
TrTTc
:= Tr 0.887=
The enthalpy change for the final T is given by Eq. (6.91), with HRB forthis T:
∆Hig R ICPH T0 T, 1.637, 22.706 10 3−⋅, 6.915− 10 6−⋅, 0.0,( )⋅:=
∆Hig 1.409 103×J
mol=
∆H' ∆Hig R Tc⋅ HRB Tr Pr, ω,( ) HRB Tr0 Pr0, ω,( )−( )⋅+:=
7.37 Propylene: Tc 365.6 K⋅:= Pc 46.65 bar⋅:= ω 0.140:=
T0 303.15 K⋅:= P0 11.5 bar⋅:= P 18 bar⋅:=
∆S 0J
mol K⋅⋅:= For the heat capacity of propylene:
A 1.637:= B22.706 10 3−⋅
K:= C
6.915− 10 6−⋅
K2:=
Tr0T0
Tc:= Tr0 0.8292= Pr0
P0
Pc:= Pr0 0.2465=
PrPPc
:=
251
7.38 Methane: Tc 190.6 K⋅:= Pc 45.99 bar⋅:= ω 0.012:=
T0 308.15 K⋅:= P0 3500 kPa⋅:= P 5500 kPa⋅:=
∆S 0J
mol K⋅⋅:= For the heat capacity of methane:
A 1.702:= B9.081 10 3−⋅
K:= C
2.164− 10 6−⋅
K2:=
Tr0T0
Tc:= Tr0 1.6167= Pr0
P0
Pc:= Pr0 0.761=
PrPPc
:= Pr 1.196=
∆H' 964.1J
mol=
The actual enthalpy change from Eq. (7.17):
η 0.80:= ∆H∆H'η
:= ∆H 1205.2J
mol=
ndot 1000molsec
⋅:= Wdot ndot ∆H⋅:= Wdot 1205.2kW= Ans.
The actual final temperature is now found from Eq. (6.91) combined with Eq(4.7), written:
τ 1.1:= (guess)
Given
∆H R A T0⋅ τ 1−( )⋅B2
T02⋅ τ
2 1−( )⋅+C3
T03⋅ τ
3 1−( )⋅+
Tc HRBτ T0⋅
TcPr, ω,
⎛⎜⎝
⎞
⎠HRB Tr0 Pr0, ω,( )−
⎛⎜⎝
⎞
⎠⋅+
...⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
⋅=
τ Find τ( ):= τ 1.079= T τ T0⋅:= T 327.15K= Ans.
252
(guess)τ 1.1:=
The actual final temperature is now found from Eq. (6.91) combined with Eq(4.7), written:
Ans.Wdot 2228.4kW=Wdot ndot ∆H⋅:=ndot 1500molsec
⋅:=
∆H 1485.6J
mol=∆H
∆H'η
:=η 0.78:=
The actual enthalpy change from Eq. (7.17):
∆H' 1158.8J
mol=
∆H' ∆Hig R Tc⋅ HRB Tr Pr, ω,( ) HRB Tr0 Pr0, ω,( )−( )⋅+:=
∆Hig 1.298 103×J
mol=
∆Hig R ICPH T0 T, 1.702, 9.081 10 3−⋅, 2.164− 10 6−⋅, 0.0,( )⋅:=
The enthalpy change for the final T is given by Eq. (6.91), with HRB forthis T:
Tr 1.802=TrTTc
:=
T 343.379K=T τ T0⋅:=τ 1.114=τ Find τ( ):=
∆S R A ln τ( )⋅ B T0⋅ C T02⋅
τ 1+2
⎛⎜⎝
⎞⎠
⋅+⎡⎢⎣
⎤⎥⎦
τ 1−( )⋅+ lnPP0
⎛⎜⎝
⎞⎠
−
SRBτ T0⋅
TcPr, ω,
⎛⎜⎝
⎞
⎠SRB Tr0 Pr0, ω,( )−+
...⎡⎢⎢⎢⎢⎣
⎤⎥⎥⎥⎥⎦
⋅=
Given
(guess)τ 1.1:=
The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0:
Use generalized second-virial correlation:
253
∆H 5288.2J
mol=
∆S R ICPS T1 T2, 1.702, 9.081 10 3−⋅, 2.164− 10 6−⋅, 0.0,( ) lnP2
P1
⎛⎜⎝
⎞
⎠−
⎛⎜⎝
⎞
⎠⋅:=
∆S 3.201J
mol K⋅=
Since the process is adiabatic: SG ∆S:= SG 3.2012J
mol K⋅= Ans.
Wideal ∆H Tσ ∆S⋅−:= Wideal 4349.8J
mol= Ans.
Wlost Tσ ∆S⋅:= Wlost 938.4J
mol= Ans.
η tWideal
Work:= η t 0.823= Ans.
Given
∆H R A T0⋅ τ 1−( )⋅B2
T02⋅ τ
2 1−( )⋅+C3
T03⋅ τ
3 1−( )⋅+
Tc HRBτ T0⋅
TcPr, ω,
⎛⎜⎝
⎞
⎠HRB Tr0 Pr0, ω,( )−
⎛⎜⎝
⎞
⎠⋅+
...⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
⋅=
τ Find τ( ):= τ 1.14= T τ T0⋅:= T 351.18K= Ans.
7.39 From the data and results of Example 7.9,
T1 293.15 K⋅:= T2 428.65 K⋅:= P1 140 kPa⋅:= P2 560 kPa⋅:=
Work 5288.3J
mol⋅:= Tσ 293.15 K⋅:=
∆H R ICPH T1 T2, 1.702, 9.081 10 3−⋅, 2.164− 10 6−⋅, 0.0,( )⋅:=
254
T'2 T2 T1−( ) η⋅ T1+⎡⎣ ⎤⎦:=
T'2 415.4K= Eq. (7.18) written for a single stage is:
T'2 T1P2P1⎛⎜⎝
⎞⎠
R1N Cp⋅
⋅= Put in logarithmic form and solve for N:
(a) Although any number ofstages greater than thiswould serve, design for 4stages.
NRCp
lnP2P1⎛⎜⎝
⎞⎠
lnT'2T1
⎛⎜⎝
⎞⎠
⋅:= N 3.743=
(b) Calculate r for 4 stages: N 4:= rP2P1⎛⎜⎝
⎞⎠
1N
:= r 2.659=
Power requirement per stage follows from Eq. (7.22). In kW/stage:
Wdotrndot Cp⋅ T1⋅ r
RCp 1−
⎛⎜⎝
⎞⎠⋅
η:= Wdotr 87.944kW= Ans.
7.42 P1 1atm:= T1 35 273.15+( )K:= T1 308.15K=
P2 50atm:= T2 200 273.15+( )K:= T2 473.15K=
η 0.65:= Vdot 0.5m3
sec:= Cp 3.5 R⋅:=
VR T1⋅P1
:= ndotVdot
V:= ndot 19.775
molsec
=
With compression from the same initial conditions (P1,T1) to the samefinal conditions (P2,T2) in each stage, the same efficiency in each stage,and the same power delivered to each stage, the applicable equations are:
(where r is the pressure ratio in each stage and N isthe number of stages.)r
P2P1⎛⎜⎝
⎞⎠
1N
=
Eq. (7.23) may be solved for T2prime:
255
(7.22) ∆HS Cp T1⋅P2P1⎛⎜⎝
⎞⎠
RCp
1−
⎡⎢⎢⎣
⎤⎥⎥⎦
⋅
⎡⎢⎢⎣
⎤⎥⎥⎦
→⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
:=
Ideal gases with constant heat capacities∆H Cp T2 T1−( )⋅[ ]→⎯⎯⎯⎯⎯⎯
:=
Cp
3.5
2.5
4.5
5.5
4.0
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
R⋅:=P2
6
5
6
8
7
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
bar:=T2
464
547
455
505
496
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
K:=
P1
2.0
1.5
1.2
1.1
1.5
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
bar:=T1
300
290
295
300
305
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
K:=
7.44
(in each interchanger)Ans.mdotw 1.052kgsec
=mdotwQdotr∆Hw
:=
∆Hw 83.6kJkg
=∆Hw 188.4 104.8−( )kJkg
:=
With data for saturated liquid water from the steam tables:
(d) Energy balance on each interchanger (subscript w denotes water):
Heat duty = 87.94 kW/interchanger
Ans.Qdotr 87.944− kW=Qdotr Wdotr−:=
(c) Because the gas (ideal) leaving the intercooler and the gas enteringthe compressor are at the same temperature (308.15 K), there is noenthalpy change for the compressor/interchanger system, and the first lawyields:
256
∆H∆HS
η
→⎯⎯
:=∆HS V P2 P1−( )⋅⎡⎣ ⎤⎦→⎯⎯⎯⎯⎯
:=By Eq. (7.24)
CP
4.15
4.20
4.20
4.185
4.20
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
kJkg K⋅⋅:=V
1.003
1.036
1.017
1.002
1.038
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
cm3
gm⋅:=
From the steam tables for sat.liq. water at the initial temperature (heatcapacity calculated from enthalpy values):
β
257.2
696.2
523.1
217.3
714.3
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
10 6−
K⋅:=η
0.75
0.70
0.75
0.70
0.75
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
:=P2
2000 kPa⋅
5000 kPa⋅
5000 kPa⋅
20 atm⋅
1500 psi⋅
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
:=
mdot
20 kg⋅
30 kg⋅
15 kg⋅
50 lb⋅
80 lb⋅
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
1sec⋅:=P1
100 kPa⋅
200 kPa⋅
20 kPa⋅
1 atm⋅
15 psi⋅
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
:=T1
298.15
363.15
333.15
294.26
366.48
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
K⋅:=
The following vectors contain values for Parts (a) through (e). Intakeconditions first:
7.47
Ans.η
0.675
0.698
0.793
0.636
0.75
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
=η∆HS
∆H
→⎯⎯
:=∆HS
3.219
3.729
4.745
5.959
4.765
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
kJmol
=
257
degFt24
t25
⎛⎜⎜⎝
⎞
⎠
70.41
202.7⎛⎜⎝
⎞⎠
=t2T2
K1.8⋅ 459.67−
⎛⎜⎝
⎞⎠
→⎯⎯⎯⎯⎯⎯⎯
:=
degC
t21
t22
t23
⎛⎜⎜⎜⎜⎝
⎞
⎟⎟
⎠
25.19
90.81
60.61
⎛⎜⎜⎜⎝
⎞⎟⎠
=t2T2
K273.15−
⎛⎜⎝
⎞⎠
→⎯⎯⎯⎯⎯⎯⎛⎜⎜⎝
⎞
⎠:=
T2
298.338
363.957
333.762
294.487
367.986
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
K=T2 T1 ∆T+( )→⎯⎯⎯⎯
:=
Ans.Wdot
68.15
285.8
135.84
83.81
689.56
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
hp=Wdot
50.82
213.12
101.29
62.5
514.21
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
kW=Wdot ∆H mdot⋅( )→⎯⎯⎯⎯
:=
∆T
0.188
0.807
0.612
0.227
1.506
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
K=∆T∆H V 1 β T1⋅−( )⋅ P2 P1−( )⋅−
CP
→⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
:=By Eq. (7.25)
∆H
2.541
7.104
6.753
2.756
14.17
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
kJkg
=∆HS
1.906
4.973
5.065
1.929
10.628
⎛⎜⎜⎜⎜⎜⎝
⎞
⎟⎟⎟
⎠
kJkg
=
258
P2 5bar:=
T3 200 273.15+( )K:= P3 5bar:=
Cpv 105J
mol K⋅:= ∆Hlv 30.72
kJmol
:= η 0.7:=
Estimate the specific molar volume of liquid benzene using the Rackett equation (3.72).
From Table B.1 for benzene: Tc 562.2K:= Zc 0.271:= Vc 259cm3
mol:=
From Table B.2 for benzene: Tn 80.0 273.15+( )K:= TrnTn
Tc:=
Assume Vliq = Vsat: V Vc Zc1 Trn−( )
27
⋅:= Eq. (3.72) V 96.802cm3
mol=
Calculate pump power
WsV P2 P1−( )⋅
η:= Ws 0.053
kJmol
= Ans.
7.48 Results from Example 7.10:
∆H 11.57kJkg⋅:= W 11.57
kJkg⋅:= ∆S 0.0090
kJkg K⋅⋅:=
Tσ 300 K⋅:= Wideal ∆H Tσ ∆S⋅−:= η tWideal
W:=
Wideal 8.87kJkg
= Ans. η t 0.767= Ans.
Since the process is adiabatic.
SG ∆S:= SG 9 10 3−×kJ
kg K⋅= Ans.
Wlost Tσ ∆S⋅:= Wlost 2.7kJkg
= Ans.
7.53 T1 25 273.15+( )K:= P1 1.2bar:=
259
Ans.Q 51.1kJ
mol=
Q R ICPH T2 Tsat, 0.747−, 67.96 10 3−⋅, 37.78− 10 6−⋅, 0,( )⋅∆Hlv2 Cpv T3 Tsat−( )⋅++
...:=
Calculate the heat exchanger heat duty.
∆Hlv2 26.822kJ
mol=Eq. (4.13)∆Hlv2 ∆Hlv
1 Tr2−
1 Tr1−⎛⎜⎝
⎞
⎠
0.38
⋅:=
Tr2 0.74=Tr2Tsat
Tc:=Tr1 0.628=Tr1
80 273.15+( )KTc
:=
∆Hlv 30.72kJ
mol:=From Table B.2
At 80 C:
Estimate the heat of vaporization at Tsat using Watson's method
Tsat 415.9K=Tsat Tsat 273.15K+:=
Tsat 142.77degC=TsatB
A lnP2
kPa⎛⎜⎝
⎞⎠
−
C−⎛⎜⎜⎝
⎞
⎠
degC:=
C 217.572:=B 2726.81:=A 13.7819:=For benzene fromTable B.2:
Estimate the saturation temperature at P = 5 bar using the AntoineEquation and values from Table B.2
T2 T1:=Therefore:
Assume that no temperature change occurs during the liquid compression.
260
Ans.
Calculate the heat exchanger duty. Note that the exchanger outlettemperature, T2, is equal to the compressor inlet temperature. Thebenzene enters the exchanger as a subcooled liquid. In the exchanger theliquid is first heated to the saturation temperature at P1, vaporized andfinally the vapor is superheated to temperature T2. Estimate the saturation temperature at P = 1.2 bar using theAntoine Equation and values from Table B.2
For benzene fromTable B.2: A 13.7819:= B 2726.81:= C 217.572:=
TsatB
A lnP1
kPa⎛⎜⎝
⎞⎠
−
C−⎛⎜⎜⎝
⎞
⎠
degC:= Tsat 85.595degC=
Tsat Tsat 273.15K+:= Tsat 358.7K=
Estimate the heat of vaporization at Tsat using Watson's method
From Table B.2At 25 C:
From Table B.1 for benzene:
Tc 562.2K:=∆Hlv 30.72kJ
mol:=
7.54 T1 25 273.15+( )K:= P1 1.2bar:= P2 1.2bar:=
T3 200 273.15+( )K:= P3 5bar:=
Cpv 105J
mol K⋅:= η 0.75:=
Calculate the compressor inlet temperature.
Combining equations (7.17), (7.21) and (7.22) yields:
T2T3
11η
P3
P2
⎛⎜⎝
⎞
⎠
RCpv
1−
⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
⋅+
:= T2 408.06K=
T2 273.15K− 134.91degC=
Calculate the compressor power
Ws Cpv T3 T2−( )⋅:= Ws 6.834kJ
mol=
261
Ans.C_motor 32572dollars=C_motor 380dollarsWdote
kW⎛⎜⎝
⎞⎠
0.855⋅:=
Ans.C_compressor 307452dollars=C_compressor 3040dollarsWdotskW
⎛⎜⎝
⎞⎠
0.952⋅:=
Wdote 182.345kW=WdoteWdotsη
:=
Wdots 127.641kW=Wdots ndot Cp⋅ T2 T1−( )⋅:=
T2 390.812K=(Pg. 77)T2P2
P1
⎛⎜⎝
⎞
⎠
RCp
T1⋅:=
Assume the compressor is adaiabatic.
η 0.70:=
Tr180 273.15+( )K
Tc:= Tr1 0.628= Tr2
Tsat
Tc:= Tr2 0.638=
∆Hlv2 ∆Hlv1 Tr2−
1 Tr1−⎛⎜⎝
⎞
⎠
0.38⋅:= Eq. (4.13) ∆Hlv2 30.405
kJmol
=
Q R ICPH T1 Tsat, 0.747−, 67.96 10 3−⋅, 37.78− 10 6−⋅, 0,( )⋅∆Hlv2 Cpv T2 Tsat−( )⋅++
...:=
Q 44.393kJ
mol= Ans.
7.57 ndot 100kmol
hr:= P1 1.2bar:= T1 300K:= P2 6bar:=
Cp 50.6J
mol K⋅:=
262
For throttling process, assume the process is adiabatic. Find T2 such that∆H = 0.
∆H Cpmig T2 T1−( )⋅ HR2+ HR1−= Eq. (6-93)
Use the MCPH function to calculate the mean heat capacity and the HRBfunction for the residual enthalpy.
Guess: T2 T1:=Given
0J
mol⋅ MCPH T1 T2, A, B, C, D,( ) R⋅ T2 T1−( )⋅
R Tc⋅ HRBT2
TcPr2, ω,
⎛⎜⎝
⎞
⎠⋅ R Tc⋅ HRB Tr1 Pr1, ω,( )⋅−+
...=
T2 Find T2( ):= T2 365.474K= Ans. Tr2T2
Tc:= Tr2 1.295=
Calculate change in entropy using Eq. (6-94) along with MCPS function forthe mean heat capacity and SRB function for the residual entropy.
∆S R MCPS T1 T2, A, B, C, D,( )⋅ lnT2
T1
⎛⎜⎝
⎞
⎠⋅ R ln
P2
P1
⎛⎜⎝
⎞
⎠⋅−
⎛⎜⎝
⎞
⎠R SRB Tr2 Pr2, ω,( )⋅ R SRB Tr1 Pr1, ω,( )⋅−+
...:= Eq. (6-94)
∆S 22.128J
mol K⋅= Ans.
7.59 T1 375K:= P1 18bar:= P2 1.2bar:=
For ethylene: ω 0.087:= Tc 282.3K:= Pc 50.40bar:=
Tr1T1
Tc:= Tr1 1.328= Pr1
P1
Pc:= Pr1 0.357=
Pr2P2
Pc:= Pr2 0.024=
A 1.424:= B 14.394 10 3−⋅:= C 4.392− 10 6−⋅:= D 0:=
a)
263
Ans.T2 268.536K=T2 Find T2( ):=
η ∆HS⋅ MCPH T1 T2, A, B, C, D,( ) R⋅ T2 T1−( )⋅
R Tc⋅ HRBT2
TcPr2, ω,
⎛⎜⎝
⎞
⎠⋅ R Tc⋅ HRB Tr1 Pr1, ω,( )⋅−+
...=
Given
Find T2 such that ∆H matches the value above.
∆H 4.496− 103×J
mol=∆H η ∆HS⋅:=
Calculate actual enthalpy change using the expander efficiency.
∆HS 6.423− 103×J
mol=
∆HS R MCPH T1 T2, A, B, C, D,( )⋅ T2 T1−( )⋅⎡⎣ ⎤⎦HRB Tr2 Pr2, ω,( ) R⋅ Tc⋅ HRB Tr1 Pr1, ω,( ) R⋅ Tc⋅−+
...:=
HR2 HRB Tr2 Pr2, ω,( ) R⋅ Tc⋅:=
Now calculate the isentropic enthalpy change, ∆HS.
Tr2 0.779=Tr2T2
Tc:=T2 219.793K=T2 Find T2( ):=
Eq. (6-94)0
Jmol K⋅
R MCPS T1 T2, A, B, C, D,( )⋅ lnT2
T1
⎛⎜⎝
⎞
⎠⋅ R ln
P2
P1
⎛⎜⎝
⎞
⎠⋅−
SRBT2
TcPr2, ω,
⎛⎜⎝
⎞
⎠R⋅ SRB Tr1 Pr1, ω,( ) R⋅−+
...=
Given
T2 T1:=Guess:
First find T2 for isentropic expansion. Solve Eq. (6-94) with ∆S = 0.
η 70%:=For expansion process. b)
264
Using liquid oil to quench the gas stream requires a smaller oil flow rate.This is because a significant portion of the energy lost by the gas is usedto vaporize the oil.
c)
Ans.DF 0.643=DFCpgas T3 T1−( )⋅⎡⎣ ⎤⎦−
∆Hlv Cpoil T3 T2−( )⋅+⎡⎣ ⎤⎦:=
Solving for D/F gives:
F Cpgas⋅ T3 T1−( )⋅ D ∆Hlv Coilp T3 T2−( )⋅+⎡⎣ ⎤⎦⋅+ 0=
Assume that the oil vaporizes at 25 C. For an adiabatic column, the overallenergy balance is as follows.
b)
T3 200degC:=Exit stream:
∆Hlv 35000J
mol:=Cpoil 200
Jmol K⋅
:=T2 25degC:=Light oil:
Cpgas 150J
mol K⋅:=T1 500degC:=Hydrocarbon gas:7.60
The advantage of the expander is that power can be produced in theexpander which can be used in the plant. The disadvantages are the extracapital and operating cost of the expander and the low temperature of thegas leaving the expander compared to the gas leaving the throttle valve.
Ans.P 3.147−kJ
mol=P η ∆H⋅:=
Calculate power produced by expander
Ans.∆S 7.77J
mol K⋅=
Eq. (6-94)∆S R MCPS T1 T2, A, B, C, D,( )⋅ lnT2
T1
⎛⎜⎝
⎞
⎠⋅ R ln
P2
P1
⎛⎜⎝
⎞
⎠⋅−
⎛⎜⎝
⎞
⎠R SRB Tr2 Pr2, ω,( )⋅ R SRB Tr1 Pr1, ω,( )⋅−+
...:=
Now recalculate ∆S at calculated T2
265