Chapter 3 Homework Limited distribution permitted Below is a plot of

Pr1

P1

Pc

:= Tr2

T2

Tc

:= Pr2

P2

Pc

:=

Tr1 0.62=Pr1 0.03=Tr2 0.88=Pr2 3.561=

From Fig. (3.16): ρr1 2.69:= ρr2 2.27:=

3.54 For ethanol: Tc513.9 K⋅:= T 453.15 K⋅:= Tr

T

Tc

:= Tr0.882=

Pc61.48 bar⋅:= P 200 bar⋅:= Pr

P

Pc

:= Pr3.253=

Vc167 cm3

mol

⋅:= molwt 46.069 gm

mol

⋅:=

From Fig. (3.17): ρr1 2.45:=

The final T > Tc, and Fig. 3.16 probably should not be used. One can easily

show that

with Z from Eq. (3.57) and

Tables E.3 and E.4. Thus

ρr

PV

c

⋅

ZR⋅T⋅

=

Vc262.7 cm3

mol

⋅:= ω 0.181:= Z00.3356:= Z10.0756−:=

3.53 For n-pentane: Tc469.7 K⋅:= Pc33.7 bar⋅:= ρ10.63 gm

cm3

⋅:=

T1291.15 K⋅:= P11 bar⋅:= T2413.15 K⋅:= P2120 bar⋅:=

Tr1

T1

Tc

:=

59

Vvapor 2616 cm3

mol

=Vvapor

RT⋅

PB0ωB1

⋅+

()

R⋅Tc

Pc

⋅+:=

B10.534−=B10.139 0.172

Tr4.2

−:=

B00.627−=

B00.083 0.422

Tr1.6

−:=

Vliquid 27.11 cm3

mol

=Vliquid VcZc

1T

r

−

()

0.2857

⎡

⎣

⎤

⎦

⋅:=

Eq. (3.72):

ω0.253:=Zc0.242:=Vc72.5 cm3

mol

⋅:=

From Fig. 3.16: ρr2.28:= ρρ

rρc

⋅=

ρr

Vc

=

3.55 For ammonia:

Tc405.7 K⋅:= T 293.15 K⋅:= Tr

T

Tc

:= Tr0.723=

Pc112.8 bar⋅:= P 857 kPa⋅:= Pr

P

Pc

:= Pr0.076=

60

For methane at 3000 psi and 60 degF:

Tc190.6 1.8⋅rankine⋅:= T 519.67 rankine⋅:= Tr

T

Tc

:= Tr1.515=

Pc45.99 bar⋅:= P 3000 psi⋅:= Pr

P

Pc

:= Pr4.498=

ω0.012:=

From Tables E.3 & E.4:

Alternatively, use Tables E.1 & E.2 to get the vapor volume:

Z00.929:= Z10.071−:= ZZ

0ωZ1

⋅+:= Z 0.911=

Vvapor

ZR⋅T⋅

P

:= Vvapor 2591 cm3

mol

=

3.58 10 gal. of gasoline is equivalent to 1400 cu ft. of methane at 60 degF and 1

atm. Assume at these conditions that methane is an ideal gas:

R 0.7302 ft3atm⋅

lbmol rankine⋅

=T 519.67 rankine⋅:= P 1 atm⋅:=

V 1400 ft3

⋅:= nPV⋅

RT⋅

:= n 3.689 lbmol=

61

For Redlich/Kwong EOS:

σ1:= ε 0:= Ω 0.08664:= Ψ 0.42748:= Table 3.1

αTr

()

Tr0.5−

:= Table 3.1 qT

r

()

Ψα Tr

()

⋅

ΩTr

⋅

:= Eq. (3.54)

βTrPr

,

()

ΩPr

⋅

Tr

:= Eq. (3.53)

3.59 T 25K:= P 3.213bar:=

Calculate the effective critical parameters for hydrogen by equations (3.58)

and (3.56)

Tc

43.6

121.8K

2.016T

+

K⋅:= Tc30.435 K=

Pc

20.5

144.2K

2.016T

+

bar⋅:= Pc10.922 bar=

ω0:=

Pr

P

Pc

:= Pr0.294=Tr

T

Tc

:= Tr0.821=

Initial guess of volume: VRT⋅

P

:= V 646.903 cm3

mol

=

Use the generalized Pitzer correlation

62

B00.134−= B10.139 0.172

Tr4.2

−:= B10.109=

Z01B

0

Pr

Tr

⋅+:= Z00.998=Z1B1

Pr

Tr

⋅:= Z10.00158=

ZZ

0ωZ1

⋅+:= Z 0.998=V1

ZR⋅T⋅

P

:= V10.024 m3

mol

=

(a) At actual condition: T5032−()

5

9

⋅273.15+

⎡

⎢

⎣

⎤

⎥

⎦K:= P 300psi:=

Pitzer correlations: T 283.15 K=

Tr

T

Tc

:= Tr1.486=Pr

P

Pc

:= Pr0.45=

B00.083 0.422

Tr1.6

−:= B00.141−=

B10.139 0.172

Tr4.2

−:= B10.106=

Calculate Z Guess: Z 0.9:=

Given Eq. (3.52)

3.61 For methane: ω0.012:= Tc190.6K:= Pc45.99bar:=

At standard condition: T6032−()

5

9

⋅273.15+

⎡

⎢

⎣

⎤

⎥

⎦K:= T 288.706 K=

Pitzer correlations: P 1atm:=

Tr

T

Tc

:= Tr1.515=Pr

P

Pc

:= Pr0.022=

B00.083 0.422

Tr1.6

−:=

63

V20.00109 m3

mol

=V2

ZR⋅T⋅

P

:=Z 0.958=ZZ

0ωZ1

⋅+:=

Z10.0322=Z1B1

Pr

Tr

⋅:=Z00.957=Z01B

0

Pr

Tr

⋅+:=

64

3.62 Use the first 29 components in Table B.1

slope and intercept functions.

0.012

0.1

0.152

0.187

0.191

0.196

0.205

0.21

0.212

0.235

0.262

⎛

⎜

⎞

⎟

0.286

0.279

0.276

0.271

0.277

0.273

0.272

0.269

0.264

⎛

⎜

⎞

⎟

()

0 0.1 0.2 0.3 0.4

0.27

0.28

0.29

ZC

Step 2->3 Isobaric cooling

Step 3->1 Isothermal expansion

3.65 Cp

7

2R:= Cv

5

2R:= γ Cp

Cv

:= γ 1.4=

Step 1->2 Adiabatic compression

T2T1

P2

⎛

⎜

⎞

γ1−

γ

⋅:= T2472.216 K=

66

Step 2->3 Isobaric cooling

Step 3->1 Isothermal expansion

For the cycle

Step 1->2 Adiabatic compression

67

If a linear equation is fit to the points then the value of B is the y-intercept.

Use the Mathcad intercept function to find the y-intercept and hence, the

value of B

→

⎯

→

⎯

M 18.01 gm

:=

1757.0

875.29

⎛

⎜

⎝

⎞

⎟

⎠

150

300

⎛

⎜

⎝

⎞

⎟

⎠

a) PV data are taken from Table F.2 at pressures above 1atm.3.67

68

If a linear equation is fit to the points then the value of B is the y-intercept.

Use the Mathcad intercept function to find the y-intercept and hence, the

value of B

→

⎯

→

⎯

mol

2295.6

1638.3

1145.2

1040.7

953.52

⎛

⎜

⎝

⎞

⎟

⎠

125

175

250

275

300

⎛

⎜

⎝

⎞

⎟

⎠

PV data are taken from Table F.2 at pressures above 1atm.

Repeat part a) for T = 350 Cb)

125

115

69

Below is a plot of the data along with the linear fit and the extrapolation to

the y-intercept.

02

.10 54.10 56.10 58.10 5

110

105

95

90

(Z-1)/p

Linear fit

p

c) Repeat part a) for T = 400 C

PV data are taken from Table F.2 at pressures above 1atm.

150

225

300

⎛

⎜

⎝

⎞

⎟

⎠

2066.9

1376.6

1031.4

⎛

⎜

⎝

⎞

⎟

⎠

gm

M 18.01 gm

:=

70

ZPV⋅M⋅

RT⋅

→

⎯

:= ρ 1

VM⋅

→

⎯

:= i07..:=

If a linear equation is fit to the points then the value of B is the

y-intercept.

Use the Mathcad intercept function to find the y-intercept and hence,

the value of B

Below is a plot of the data along with the linear fit and the extrapolation to

the y-intercept.

75

70

71

Ans.

Bhat 0.332−=Bhat Intercept:=

0.3

Y

Slope 0.033=Slope slope X Y,():=

Create a linear fit of Y vs X

Tr1:=Z

0.9832

0.9300

0.7574

0.6355

⎛

⎜

⎝

⎞

⎟

⎠

:=

Pr

0.05

0.20

0.60

0.80

⎛

⎜

⎝

⎞

⎟

⎠

:=

Data from Appendix E at Tr = 1

Create a plot of Z1−()Z⋅Tr

⋅

Pr

vs Pr

ZT

r

⋅

3.70

72

Eq. (3.53)

Calculate Z Guess: Z 0.9:=

Given Eq. (3.52)

Z1βTrPr

,

()

+qT

r

()

βTrPr

,

()

⋅ZβTrPr

,

()

−

Zεβ TrPr

,

()

⋅+

()

Zσβ TrPr

,

()

⋅+

()

⋅

⋅−=

3.72 After the reaction is complete, there will be 5 moles of C2H2 and 5 moles of

Ca(OH)2.

First calculate the volume available for the gas.

3.71 Use the SRK equation to calculate Z

Table 3.1

αTrω,

()

1 0.480 1.574ω+ 0.176ω2

−

()

1T

r

1

2

−

⎛

⎜

⎝

⎞

⎠

⋅+

⎡

⎢

⎣

⎤

⎥

⎦

2

:=

qT

r

()

Ψα Trω,

()

⋅

ΩTr

⋅

:= Eq. (3.54) βTrPr

,

()

ΩPr

⋅

Tr

:=

73

3.73 mass 35000kg:= T 10 273.15+()K:=

ω0.152:= Tc369.8K:= Pc42.48bar:= M 44.097 gm

mol

:=

Zc0.276:= Vc200.0 cm3

mol

:= nmass

M

:= n 7.937 105

×mol=

a) Estimate the volume of gas using the truncated virial equation

Tr

T

Tc

:= Tr0.766=P 1atm:= Pr

P

Pc

:=

Use SRK equation to calculate pressure.

Table 3.1

αTrω,

()

1 0.480 1.574ω+ 0.176ω2

−

()

1T

r

1

2

−

⎛

⎜

⎝

⎞

⎠

⋅+

⎡

⎢

⎣

⎤

⎥

⎦

2

:=

qT

r

()

Ψα Trω,

()

⋅

ΩTr

⋅

:= Eq. (3.54)

aΨαTrω,

()

R2

⋅Tc2

⋅

Pc

⋅:= Eq. (3.45) bΩRT

c

⋅

Pc

⋅:= Eq. (3.46)

74

Pr0.15=Pr

P

Pc

:=P 6.294atm:=

Vliq 85.444 cm3

mol

=

Vliq VcZc

1T

r

−

()

0.2857

⋅:=

Calculate the molar volume of the liquid with the Rackett equation(3.72)b)

75