b)For water as solvent: Ms 18.015gmmol

:=

For CO2 in H2O: ki 0.034mol

kg bar⋅:=

By Eq. (5): Hi1

Ms ki⋅:= Hi 1633bar= Ans.

The value is Table 10.1 is 1670 bar. The values agree within about 2%.

10.36

Acetone: Psat1 T( ) e

14.31452756.22

TdegC

228.060+−

kPa⋅:=

Acetonitrile Psat2 T( ) e

14.89503413.10

TdegC

250.523+−

kPa⋅:=

a) Find BUBL P and DEW P values

T 50degC:= x1 0.5:= y1 0.5:=

10.35 a) The equation from NIST is: Mi ki yi⋅ P⋅= Eq. (1)

The equation for Henry's Law is:xi Hi⋅ yi P⋅= Eq. (2)

Solving to eliminate P gives: HiMi

ki xi⋅= Eq. (3)

By definition: Mini

ns Ms⋅= where M is the molar mass and the

subscript s refers to the solvent.

Dividing by the toal number of moles gives: Mixi

xs Ms⋅= Eq. (4)

Combining Eqs. (3) and (4) gives: Hi1

xs Ms⋅ ki⋅=

If xi is small, then xs is approximately equal to 1 and: Hi1

Ms ki⋅= Eq. (5)

333

x1 Psat1 T( )⋅ y1 P⋅= 1 x1−( ) Psat2 T( )⋅ 1 y1−( ) P⋅=

x1

DEWT

⎛⎜⎝

⎞

⎠Find x1 T,( ):= DEWT 51.238degC= Ans.

At P = 0.5 atm, two phases will form between T = 46.3 C and 51.2 C

10.37 Calculate x and y at T = 90 C and P = 75 kPa

Benzene: Psat1 T( ) e

13.78192726.81

TdegC

217.572+−

kPa⋅:=

Toluene: Psat2 T( ) e

13.93203056.96

TdegC

217.625+−

kPa⋅:=

a) Calculate the equilibrium composition of the liquid and vapor at the flash T and P

T 90degC:= P 75kPa:= Guess: x1 0.5:= y1 0.5:=

BUBLP x1 Psat1 T( )⋅ 1 x1−( ) Psat2 T( )⋅+:= BUBLP 0.573atm= Ans.

DEWP1

y1

Psat1 T( )

1 y1−( )Psat2 T( )

+

:= DEWP 0.478atm= Ans.

At T = 50 C two phases will form between P = 0.478 atm and 0.573 atm

b)Find BUBL T and DEW T values

P 0.5atm:= x1 0.5:= y1 0.5:= Guess: T 50degC:=

Given x1 Psat1 T( )⋅ 1 x1−( ) Psat2 T( )⋅+ P=

BUBLT Find T( ):= BUBLT 46.316degC= Ans.

Given

334

1 x1−( ) Psat2 T( )⋅ 1 y1− y3−( ) P⋅= y1 y2+ y3+ 1=

y2

y3

⎛⎜⎝

⎞

⎠Find y2 y3,( ):= y2 0.608=

y3 0.1= Ans.

Conclusion: An air leak is consistent with the measured compositions.

10.38 yO21 0.0387:= yN21 0.7288:= yCO21 0.0775:= yH2O1 0.1550:=

ndot 10kmol

hr:= T1 100degC:= T2 25degC:= P 1atm:=

PsatH2O T( ) e

16.38723885.70

TdegC

230.170+−

kPa⋅:=

Given x1 Psat1 T( )⋅ y1 P⋅= 1 x1−( ) Psat2 T( )⋅ 1 y1−( ) P⋅=

x1

y1

⎛⎜⎝

⎞

⎠Find x1 y1,( ):= x1 0.252= y1 0.458=

The equilibrium compositions do not agree with the measured values.

b) Assume that the measured values are correct. Since air will not dissolvein the liquid to any significant extent, the mole fractions of toluene in theliquid can be calculated.

x1 0.1604:= y1 0.2919:= x2 1 x1−:= x2 0.8396=

Now calculate the composition of the vapor. y3 represents the molefraction of air in the vapor.

Guess: y2 0.5:= y3 1 y2− y1−:=

Given

335

yH2O2 0.031=yCO22 0.089=yN22 0.835=yO22 0.044=

ndotvap 8.724kmol

hr=ndotliq 1.276

kmolhr

=

ndotliq

ndotvap

yO22

yN22

yCO22

⎛⎜⎜⎜⎜⎜⎜⎝

⎞

⎟⎟⎟⎟

⎠

Find ndotliq ndotvap, yO22, yN22, yCO22,( ):=

Summation equationyO22 yN22+ yCO22+ yH2O2+ 1=

CO2 balancendot yCO21⋅ ndotvap yCO22⋅=

N2 balancendot yN21⋅ ndotvap yN22⋅=

O2 balancendot yO21⋅ ndotvap yO22⋅=

Overall balancendot ndotliq ndotvap+=Given

yCO22 0.0775:=yN22 0.7288:=yO22 0.0387:=

ndotliqndot

2:=ndotvap

ndot2

:=Guess:

Assume that two streams leave the process: a liquid water stream at ratendotliq and a vapor stream at rate ndotvap. Apply mole balances aroundthe cooler to calculate the exit composition of the vapor phase.

This is less than the mole fraction of water in the feed. Therefore, someof the water will condense.

yH2O2 0.0315=yH2O2PsatH2O T2( )

P:=

Calculate the mole fraction of water in the exit gas if the exit gas issaturated with water.

336

xC3 KC3⋅ xC4 KC4⋅+ xC5 KC5⋅+ 1.004=

The vapor mole fractions must sum to 1.

KC5 0.23:=xC5 0.10:=

KC4 0.925:=xC4 0.85:=

KC3 3.9:=xC3 0.05:=

Ans.P 18psia:=Taking values from Fig 10.14 at pressure:

Assume the liquid is stored at the bubble point at T = 40 F10.39

Ans.Qdot 19.895− kW=

Qdot ndotvap yO22⋅ R⋅ ICPH T1 T2, 3.639, 0.506 10 3−⋅, 0, 0.227− 105⋅,( )⋅

ndotvap yN22⋅ R⋅ ICPH T1 T2, 3.280, 0.539 10 3−⋅, 0, 0.040 105⋅,( )⋅+

...

ndotvap yCO22⋅ R⋅ ICPH T1 T2, 5.457, 1.045 10 3−⋅, 0, 1.157− 105⋅,( )⋅+

...

ndotvap yH2O2⋅ R⋅ ICPH T1 T2, 3.470, 1.450 10 3−⋅, 0, 0.121 105⋅,( )⋅+

...

∆HlvH2O ndotliq⋅( )−+...

:=

T2 T2 273.15K+:=T1 T1 273.15K+:=∆HlvH2O 40.66kJ

mol:=

Apply an energy balance around the cooler to calculate heat transfer rate.

337

yH2Ovap 0.308= Ans.

ySO2 1 yH2Ovap−:= ySO2 0.692= Ans.

b)Calculate the vapor stream molar flow rate using balance on SO2

ndotvapndotSO2

ySO2:= ndotvap 14.461

kmolhr

= Ans.

Calculate the liquid H2O flow rate using balance on H2O

ndotH2Ovap ndotvap yH2Ovap⋅:= ndotH2Ovap 4.461kmol

hr=

ndotH2Oliq ndotH2O ndotH2Ovap−:= ndotH2Oliq 5.539kmol

hr= Ans.

10.40 H2S + 3/2 O2 -> H2O + SO2

By a stoichiometric balance, calculate the following total molar flow rates

ndotH2S 10kmol

hr:= ndotO2

32

ndotH2S:=Feed:

Products ndotSO2 ndotH2S:= ndotH2O ndotH2S:=

Exit conditions:

P 1atm:= T2 70degC:= PsatH2O T( ) e

16.38723885.70

TdegC

230.170+−

kPa⋅:=

a) Calculate the mole fraction of H2O and SO2 in the exiting vapor streamassuming vapor is saturated with H2O

yH2OvapPsatH2O T2( )

P:=

338

Tdp Find T( ):=

Tdp 14.004degC= Tdp Tdp 32degF+:= Tdp 57.207degF= Ans.

10.42 ndot1 50kmol

hr:= Tdp1 20degC:= Tdp2 10degC:= P 1atm:=

MH2O 18.01gmmol

:=PsatH2O T( ) e

16.38723885.70

TdegC

230.170+−

kPa⋅:=

y1PsatH2O Tdp1( )

P:= y1 0.023= y2

PsatH2O Tdp2( )P

:= y2 0.012=

By a mole balances on the process

Guess: ndot2liq ndot1:= ndot2vap ndot1:=

10.41 NCL 0.01kgkg

:= MH2O 18.01gmmol

:= Mair 29gmmol

:=

a) YH2O NCLMair

MH2O⋅:= YH2O 0.0161=

yH2OYH2O

1 YH2O+:= yH2O 0.0158= Ans.

b) P 1atm:= ppH2O yH2O P⋅:= ppH2O 1.606kPa= Ans.

Guess: T 20degC:=c) PsatH2O T( ) e

16.38723885.70

TdegC

230.170+−

kPa⋅:=

Given yH2O P⋅ PsatH2O T( )=

339

Cyclohexane: A2 13.6568:= B2 2723.44:= C2 220.618:=

Psat1 T( ) exp A1B1

TdegC

C1+−⎛

⎜⎜⎝

⎞

⎠

kPa:=

Psat2 T( ) exp A2B2

TdegC

C2+−⎛

⎜⎜⎝

⎞

⎠

kPa:=

Guess: T 66degC:=

Given Psat1 T( ) Psat2 T( )= T Find T( ):=

The Bancroft point for this system is:

Psat1 T( ) 39.591kPa= T 52.321degC= Ans.

Component 1 Component 2 T ( C) P (kPa)Benzene Cyclohexane 52.3 39.62-Butanol Water 87.7 64.2

Acetonitrile Ethanol 65.8 60.6

Given ndot1 y1⋅ ndot2vap y2⋅ ndot2liq+= H2O balance

ndot1 ndot2vap ndot2liq+= Overall balance

ndot2liq

ndot2vap

⎛⎜⎝

⎞

⎠Find ndot2liq ndot2vap,( ):=

ndot2vap 49.441kmol

hr= ndot2liq 0.559

kmolhr

=

mdot2liq ndot2liq MH2O⋅:= mdot2liq 10.074kghr

= Ans.

10.43 Benzene: A1 13.7819:= B1 2726.81:= C1 217.572:=

340