Chapter7 A

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( ):=


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( ):=


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


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⋅=


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,

⎛⎜⎝

⎞


⎛⎜⎝

⎞

⎠⋅+

...⎡⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎦

⋅=

Given

(guess)τ 1.4:=


∆H 5673.2J

mol=∆H

∆H'η

:=η 0.82:=


∆H' 4652J

mol=


∆Hig 4.826kJ

mol=

∆Hig R ICPH T0 T, 3.578, 3.020 10 3−⋅, 0.0, 0.186− 105⋅,( )⋅:=

250

Pr 0.386=



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=


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=


η 0.80:= ∆H∆H'η

:= ∆H 1205.2J

mol=

ndot 1000molsec

⋅:= Wdot ndot ∆H⋅:= Wdot 1205.2kW= Ans.


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,

⎛⎜⎝

⎞


⎛⎜⎝

⎞

⎠⋅+

...⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦

⋅=

τ Find τ( ):= τ 1.079= T τ T0⋅:= T 327.15K= Ans.

252

(guess)τ 1.1:=


Ans.Wdot 2228.4kW=Wdot ndot ∆H⋅:=ndot 1500molsec

⋅:=

∆H 1485.6J

mol=∆H

∆H'η

:=η 0.78:=


∆H' 1158.8J

mol=


∆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:=



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,

⎛⎜⎝

⎞


⎛⎜⎝

⎞

⎠⋅+

...⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦

⋅=

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

Chapter7 A

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