# Chapter 10 - Vapour liquid equillibrium¶

## Example: 10.1 Page: 390¶

In :
print "Example: 10.1 - Page: 390\n\n"

# Mathematics is involved in proving but just that no numerical computations are involved.
# For prove refer to this example 10.1 on page number 390 of the book.

print " Mathematics is involved in proving but just that no numerical computations are involved.\n\n"
print " For prove refer to this example 10.1 on page 390 of the book."

Example: 10.1 - Page: 390

Mathematics is involved in proving but just that no numerical computations are involved.

For prove refer to this example 10.1 on page 390 of the book.


## Example: 10.2 Page: 399¶

In :
print "Example: 10.2 - Page: 399\n\n"

# Mathematics is involved in proving but just that no numerical computations are involved.
# For prove refer to this example 10.2 on page number 399 of the book.

print " Mathematics is involved in proving but just that no numerical computations are involved.\n\n"
print " For prove refer to this example 10.2 on page 399 of the book."

Example: 10.2 - Page: 399

Mathematics is involved in proving but just that no numerical computations are involved.

For prove refer to this example 10.2 on page 399 of the book.


## Example: 10.3 Page: 400¶

In :
from __future__ import division
print "Example: 10.3 - Page: 400\n\n"

# Solution

#*****Data******#
x1 = 0.6## [mole fraction of ethylene]
x2 = 0.4## [mole fraction of propylene]
T = 423## [K]
P1_sat = 15.2## [vapour pressure of ethylene, atm]
P2_sat = 9.8## [vapour pressure of propylene, atm]
#**************#

P = x1*P1_sat + x2*P2_sat## [atm]
print "The total pressure is %.2f atm\n"%(P)#
# In vapour phase:
y1 = x1*P1_sat/P## [mole fraction of ethylene]
y2 = x2*P2_sat/P## [mole fraction of propylene]
print "Mole fraction of ethylene in vapour phase is %.1f\n"%(y1)#
print "Mole fraction of propylene in the vapour phase is %.1f\n"%(y2)#

Example: 10.3 - Page: 400

The total pressure is 13.04 atm

Mole fraction of ethylene in vapour phase is 0.7

Mole fraction of propylene in the vapour phase is 0.3



## Example: 10.4 Page: 400¶

In :
from __future__ import division
from scipy.optimize import fsolve
from math import exp
print "Example: 10.4 - Page: 400\n\n"

# Solution

#*****Data******#
Temp = 77## [OC]
P = 75## [kPa]
#deff('[P1] = f1(T)','P1 = exp(14.35 - 2942/(T + 220))')#
#deff('[P2] = f2(T)','P2 = exp(14.25 - 2960/(T + 210))')#
def f1(T):
P1 = exp(14.35 - 2942/(T + 220))
return P1

def f2(T):
P2 = exp(14.25 - 2960/(T + 210))
return P2

#*************#

P1sat = f1(Temp)## [kPa]
P2sat = f2(Temp)## [kPa]
#deff('[y] = f3(x1)','y = P - (x1*P1sat) - (1 - x1)*P2sat')#
def f3(x1):
y = P - (x1*P1sat) - (1 - x1)*P2sat
return y
x1 = fsolve(f3,7)#
x2 = 1 - x1#
print "In Liquid phase\n"
print "The mole fraction of X is %.3f\n"%(x1)#
print "The mole fraction of Y is %.3f\n"%(x2)#

y1 = x1*P1sat/P#
y2 = 1 - y1#
print "In Vapour phase\n"
print "The mole fraction of X is %.3f\n"%(y1)#
print "The mole fraction of Y is %.3f\n"%(y2)#

Example: 10.4 - Page: 400

In Liquid phase

The mole fraction of X is 0.701

The mole fraction of Y is 0.299

In Vapour phase

The mole fraction of X is 0.796

The mole fraction of Y is 0.204



## Example: 10.5 Page: 401¶

In :
from __future__ import division
from math import exp
print "Example: 10.5 - Page: 401\n\n"

# Solution

#*****Data******#
#deff('[P1] = f1(T)','P1 = exp(14.3916 - 2795/(T + 230))')#
def f1(T):
P1 = exp(14.3916 - 2795/(T + 230))
return P1

#deff('[P2] = f2(T)','P2 = exp(14.2724 - 2945.47/(T + 224))')#
def f2(T):
P2 = exp(14.2724 - 2945.47/(T + 224))
return P2
#deff('[P3] = f3(T)','P3 = exp(14.2043 - 2972.64/(T + 209))')#
def f3(T):
P3 = exp(14.2043 - 2972.64/(T + 209))
return P3

#*************#

# Solution (i)

#*****Data******#
Temp = 75## [OC]
P = 75## [kPa]
x1 = 0.30#
x2 = 0.40#
#*************#

x3 = 1 - (x1 + x2)#
P1sat = f1(Temp)## [kPa]
P2sat = f2(Temp)## [kPa]
P3sat = f3(Temp)## [kPa]
P = x1*P1sat + x2*P2sat + x3*P3sat## [kPa]
y1 = x1*P1sat/P#
y2 = x2*P2sat/P#
y3 = x3*P3sat/P#
print "Solution (i)\n"
print "The mole fraction of acetone is %.3f\n"%(y1)#
print "The mole fraction of acetonitrile is %.3f\n"%(y2)#
print "The mole fraction of nitromethane is %.3f\n"%(y3)#

# Solution (ii)

#*****Data*****#
Temp = 80## [OC]
y1 = 0.45#
y2 = 0.35#
#**************#

y3 = 1 - (y1 + y2)#
P1sat = f1(Temp)## [kPa]
P2sat = f2(Temp)## [kPa]
P3sat = f3(Temp)## [kPa]
P = 1/((y1/P1sat) + (y2/P2sat) + (y3/P3sat))## [kPa]
x1 = y1*P/P1sat#
x2 = y2*P/P2sat#
x3 = y3*P/P3sat#
print "Solution (ii)\n"
print "The mole fraction of acetone is %.3f\n"%(x1)#
print "The mole fraction of acetonitrile is %.3f\n"%(x2)#
print "The mole fraction of nitromethane is %.3f\n"%(x3)#

Example: 10.5 - Page: 401

Solution (i)

The mole fraction of acetone is 0.549

The mole fraction of acetonitrile is 0.327

The mole fraction of nitromethane is 0.124

Solution (ii)

The mole fraction of acetone is 0.216

The mole fraction of acetonitrile is 0.371

The mole fraction of nitromethane is 0.413



## Example: 10.6 Page: 403¶

In :
from __future__ import division
from math import exp
print "Example: 10.6 - Page: 403\n\n"

# Solution

#*****Data******#
#deff('[P1] = f1(T)','P1 = exp(16.5915 - 3643.31/(T - 33.424))')#
def f1(T):
P1 = exp(16.5915 - 3643.31/(T - 33.424))
return P1
#deff('[P2] = f2(T)','P2 = exp(14.2532 - 2665.54/(T - 53.424))')#
def f2(T):
P2 = exp(14.2532 - 2665.54/(T - 53.424))
return P2
#deff('[A] = f3(T)','A = 2.771 - 0.00523*T')#
def f3(T):
A = 2.771 - 0.00523*T
return A

Temp = 318.15## [K]
x1 = 0.25#
#**************#

P1sat = f1(Temp)## [kPa]
P2sat = f2(Temp)## [kPa]
A = f3(Temp)#
x2 = 1 - x1#
gama1 = exp(A*x2**2)#
gama2 = exp(A*x1**2)#
P = x1*gama1*P1sat + x2*gama2*P2sat#
y1 = x1*gama1*P1sat/P#
y2 = x2*gama2*P2sat/P#
print "In Vapour phase\n"
print "The mole fraction of methanol is %.3f\n"%(y1)#
print "The mole fraction of methyl acetate is %.3f\n"%(y2)#

Example: 10.6 - Page: 403

In Vapour phase

The mole fraction of methanol is 0.282

The mole fraction of methyl acetate is 0.718



## Example: 10.7 Page: 408¶

In :
from __future__ import division
from math import exp,log
print "Example: 10.7 - Page: 408\n\n"

# Solution

#*****Data******#
Temp = 30## [OC]
A = 0.625#
#**************#

P1sat = exp(13.71 - 3800/Temp)## [kPa]
P2sat = exp(14.01 - 3800/Temp)## [kPa]
# At azeotropic point:
# P = gama1*P1sat + gama2*P2sat
# gama1/gama2 = P2sat/P1sat
# log(gama1) - log(gama2) = log(P2sat) - log(P1sat)
# Val = log(gama1) - gama2
Val = log(P2sat) - log(P1sat)#
# log(gama1) = (A*x2**2)
# log(gama2) = (A*x1**2)
# A(x2**2 - x1**2) = 0.625*(x2**2 - x1**2)..................... (1)
# x1 + x2 = 1............................................. (2)
# On simplifying, we get:
# A*(1 - (2*x1)) = Val
x1 = (1/2)*(1 - Val/A)#
x2 = 1 - x1#
print "Azeotropic Composition\n"
print "The mole fraction of component 1 is %.3f\n"%(x1)#
print "The mole fraction of component 2 is %.3f\n"%(x2)#

Example: 10.7 - Page: 408

Azeotropic Composition

The mole fraction of component 1 is 0.260

The mole fraction of component 2 is 0.740



## Example: 10.8 Page: 410¶

In :
print "Example: 10.8 - Page: 410\n\n"

# This problem involves proving a relation in which no mathematics and no calculations are involved.
# For prove refer to this example 10.8 on page number 410 of the book.

print " This problem involves proving a relation in which no mathematics and no calculations are involved.\n\n"
print " For prove refer to this example 10.8 on page 410 of the book."

Example: 10.8 - Page: 410

This problem involves proving a relation in which no mathematics and no calculations are involved.

For prove refer to this example 10.8 on page 410 of the book.


## Example: 10.9 Page: 412¶

In :
from __future__ import division
from math import exp,log

print "Example: 10.9 - Page: 412\n\n"

# Solution

#*****Data******#
x1 = 0.42#
x2 = 0.58#
P = 760## [mm of Hg]
P1sat = 786## [mm of Hg]
P2sat = 551## [mm of Hg]
#***************#

gama1 = P/P1sat#
gama2 = P/P2sat#
A = log(gama1)*(1 + (x2*log(gama2))/(x1*log(gama1)))**2#
B = log(gama2)*(1 + (x1*log(gama1))/(x2*log(gama2)))**2#
print "Van Laar Constants\n"
print "A = %.3f\n"%(A)#
print "B = %.3f\n"%(B)#

Example: 10.9 - Page: 412

Van Laar Constants

A = -5.008

B = 0.275



## Example: 10.10 Page: 412¶

In :
from __future__ import division
from math import exp,log
print "Example: 10.10 - Page: 412\n\n"

# Solution

#*****Data******#
P = 101.3## [kPa]
P1sat = 100.59## [kPa]
P2sat = 99.27## [kPa]
x1 = 0.532#
#****************#

x2 = 1 - x1#
gama1 = P/P1sat#
gama2 = P/P2sat#
A = log(gama1)*(1 + (x2*log(gama2))/(x1*log(gama1)))**2#
B = log(gama2)*(1 + (x1*log(gama1))/(x2*log(gama2)))**2#

# For solution containing 10 mol percent benzene:
x1 = 0.10#
x2 = 1 - x1#
gama1 = exp(A/(1 + (A*x1/(B*x2))**2))#
gama2 = exp(B/(1 + (B*x2/(A*x1))**2))#
print "Activity Coeffecient\n"
print "gama1 = %.3f\n"%(gama1)#
print "gama2 = %.3f\n"%(gama2)#

Example: 10.10 - Page: 412

Activity Coeffecient

gama1 = 1.086

gama2 = 1.002



## Example: 10.11 Page: 413¶

In :
from __future__ import division
from math import exp,log
print "Example: 10.11 - Page: 413\n\n"

# Solution

#*****Data******#
P = 760## [mm of Hg]
P1sat = 995## [mm of Hg]
P2sat = 885## [mm of Hg]
x1 = 0.335#
T = 64.6## [OC]
#****************#

x2 = 1 - x1#
gama1 = P/P1sat#
gama2 = P/P2sat#
A = log(gama1)*(1 + (x2*log(gama2))/(x1*log(gama1)))**2#
B = log(gama2)*(1 + (x1*log(gama1))/(x2*log(gama2)))**2#

# For solution containing 11.1 mol percent acetone:
x1 = 0.111#
x2 = 1 - x1#
gama1 = exp((A*x2**2)/(x2 + (A*x1/(B))**2))#
gama2 = exp((B*x1**2)/(x1 + (B*x2/(A))**2))#
y1 = 1/(1 + (gama2*x2*P2sat/(gama1*x1*P1sat)))#
y2 = 1 - y1#
print "Equilibrium Composition\n"
print "Acetone Composition = %.2f %%\n"%(y1*100)#
print "Chloform composition = %.2f %%\n"%(y2*100)#

Example: 10.11 - Page: 413

Equilibrium Composition

Acetone Composition = 4.98 %

Chloform composition = 95.02 %



## Example: 10.12 Page: 414¶

In :
print "Example: 10.12 - Page: 414\n\n"

# Mathematics is involved in proving but just that no numerical computations are involved.
# For prove refer to this example 10.12 on page number 414 of the book.

print " Mathematics is involved in proving but just that no numerical computations are involved.\n\n"
print " For prove refer to this example 10.12 on page 414 of the book."

Example: 10.12 - Page: 414

Mathematics is involved in proving but just that no numerical computations are involved.

For prove refer to this example 10.12 on page 414 of the book.


## Example: 10.13 Page: 418¶

In :
from __future__ import division
from math import exp,log
print "Example: 10.13 - Page: 418\n\n"

# Solution

#*****Data******#
# 1: iso - butanol
# 2: iso - propanol
T = 50 + 273## [K]
x1 = 0.3#
V1 = 65.2## [cubic cm/mol]
V2 = 15.34## [cubic cm/mol]
# For Wilson equation:
a12 = 300.55## [cal/mol]
a21 = 1520.32## [cal/mol]
# For NTRL equation:
b12 = 685.21## [cal/mol]
b21 = 1210.21## [cal/mol]
alpha = 0.552#
R = 2## [cal/mol K]
#******************#

x2 = 1 - x1#
# A: Estimation of activity coeffecient using Wilson equation:
# From Eqn. 10.65:
A12 = (V2/V1)*exp(-a12/(R*T))#
# From Eqn. 10.66:
A21 = (V1/V2)*exp(-a21/(R*T))#
# From Eqn. 10.67:
gama1 = exp(-log(x1 + A12*x2) + x2*((A12/(x1 + A12*x2)) - (A21/(A21*x1 + x2))))#
# From Eqn. 10.68:
gama2 = exp(-log(x2 + A21*x1) - x1*((A12/(x1 + A12*x2)) - (A21/(A21*x1 + x2))))#
print "Wilson equation\n"
print "Activity Coeffecient of iso - butanol is %.3f\n"%(gama1)#
print "Activity Coeffecient of iso - propanol is %.3f\n"%(gama2)#
print "\n"

# A: Estimation of activity coeffecient using NTRL equation:
t12 = b12/(R*T)#
t21 = b21/(R*T)#
G12 = exp(-alpha*t12)#
G21 = exp(-alpha*t21)#
# From Eqn. 10.70:
gama1 = exp((x2**2)*(t21*(G21/(x1 + x2*G21))**2 + (t12*G12/(x2 + x1*G12)**2)))#
# From Eqn. 10.71:
gama2 = exp((x1**2)*(t12*(G12/(x2 + x1*G12))**2 + (t21*G21/(x1 + x2*G21)**2)))#
print "NTRL equation\n"
print "Activity Coeffecient of iso - butanol is %.3f\n"%(gama1)#
print "Activity Coeffecient of iso - propanol is %.3f\n"%(gama2)#

Example: 10.13 - Page: 418

Wilson equation

Activity Coeffecient of iso - butanol is 2.270

Activity Coeffecient of iso - propanol is 1.265

NTRL equation

Activity Coeffecient of iso - butanol is 2.160

Activity Coeffecient of iso - propanol is 1.269



## Example: 10.14 Page: 426¶

In :
print "Example: 10.14 - Page: 426\n\n"

# Solution

#*****Data******#
x1 = 0.4## [mole fraction of ethane in vapour phase]
x2 = 0.6## [mole fraction of propane in vapour phase]
P = 1.5## [MPa]
#***************#

# Assume T = 10 OC
T = 10## [OC]
# From Fig. 10.14 (Pg 426):
K1 = 1.8#
K2 = 0.5#
# From Eqn. 10.83:
y1 = K1*x1#
y2 = K2*x2#
# Since y1 + y2 > 1, so we assume another value of T = 9 OC.
T = 9## [OC]
# From Fig. 10.14 (Pg 426):
K1 = 1.75#
K2 = 0.5#
# From Eqn. 10.83:
y1 = K1*x1#
y2 = K2*x2#
# Since y1 + y2 = 1. Therefore:
print "Bubble Temperature is %d OC\n"%(T)#
print "Composition of the vapour bubble:\n y1 = %.2f\n y2 = %.2f"%(y1,y2)#

Example: 10.14 - Page: 426

Bubble Temperature is 9 OC

Composition of the vapour bubble:
y1 = 0.70
y2 = 0.30


## Example: 10.15 Page: 428¶

In :
print "Example: 10.15 - Page: 428\n\n"

# Solution

#*****Data******#
y1 = 0.20## [mole fraction of methane in vapour phase]
y2 = 0.30## [mole fraction of ethane in vapour phase]
y3 = 0.50## [mole fraction of propane in vapour phase]
T = 30## [OC]
#*************#

# Assume P = 2.0 MPa
P = 2.0## [MPa]
# From Fig. 10.14 (Pg 426):
K1 = 8.5#
K2 = 2.0#
K3 = 0.68#
# From Eqn. 10.83:
x1 = y1/K1#
x2 = y2/K2#
x3 = y3/K3#
# Since x1 + x2 +x3 < 1, so we assume another value of P = 2.15 MPa at 30 OC.
P = 2.15## [MPa]
# From Fig. 10.14 (Pg 426):
K1 = 8.1#
K2 = 1.82#
K3 = 0.62#
# From Eqn. 10.83:
x1 = y1/K1#
x2 = y2/K2#
x3 = y3/K3#
# Since x1 + x2 +x3 = 1. Therefore:
print "Dew Pressure is %.2f MPa\n"%(P)#
print "Composition of the liquid drop:\n x1 = %.4f\n x2 = %.4f\n x3 = %.4f"%(x1,x2,x3)#

Example: 10.15 - Page: 428

Dew Pressure is 2.15 MPa

Composition of the liquid drop:
x1 = 0.0247
x2 = 0.1648
x3 = 0.8065


## Example: 10.16 Page: 429¶

In :
print "Example: 10.16 - Page: 429\n\n"

# Solution

# Dew point Pressure
#*****Data******#
y1 = 0.10## [mole fraction of methane in vapour phase]
y2 = 0.20## [mole fraction of ethane in vapour phase]
y3 = 0.70## [mole fraction of propane in vapour phase]
T = 10## [OC]
#*************#

# Assume P = 690 kPa
P = 690## [kPa]
# From Fig. 10.14 (Pg 426):
K1 = 20.0#
K2 = 3.25#
K3 = 0.92#
# From Eqn. 10.83:
x1 = y1/K1#
x2 = y2/K2#
x3 = y3/K3#
# Since x1 + x2 +x3 < 1, so we assume another value of P = 10135 kPa at 10 OC.
P = 10135## [kPa]
# From Fig. 10.14 (Pg 426):
K1 = 13.20#
K2 = 2.25#
K3 = 0.65#
# From Eqn. 10.83:
x1 = y1/K1#
x2 = y2/K2#
x3 = y3/K3#
# Since x1 + x2 +x3 > 1, so we assume another value of P = 870 kPa at 10 OC.
P = 870## [kPa]
# From Fig. 10.14 (Pg 426):
K1 = 16.0#
K2 = 2.65#
K3 = 0.762#
# From Eqn. 10.83:
x1 = y1/K1#
x2 = y2/K2#
x3 = y3/K3#
# Since x1 + x2 +x3 = 1. Therefore:
print "Dew Pressure is %d kPa\n"%(P)#
print "Composition of the liquid drop:\n x1 = %.4f\n x2 = %.4f\n x3 = %.4f\n"%(x1,x2,x3)#
print "\n"

# Bubble point Pressure
#*****Data******#
x1 = 0.10## [mole fraction of methane in vapour phase]
x2 = 0.20## [mole fraction of ethane in vapour phase]
x3 = 0.70## [mole fraction of propane in vapour phase]
T = 10## [OC]
#*************#

# Assume P = 2622 kPa
P = 2622## [kPa]
# From Fig. 10.14 (Pg 426):
K1 = 5.60#
K2 = 1.11#
K3 = 0.335#
# From Eqn. 10.83:
y1 = K1*x1#
y2 = K2*x2#
y3 = K3*x3#
# Since x1 + x2 +x3 > 1, so we assume another value of P = 2760 kPa at 10 OC.
P = 2760## [kPa]
# From Fig. 10.14 (Pg 426):
K1 = 5.25#
K2 = 1.07#
K3 = 0.32#
# From Eqn. 10.83:
y1 = K1*x1#
y2 = K2*x2#
y3 = K3*x3#
# Since x1 + x2 +x3 < 1, so we assume another value of P = 2656 kPa at 10 OC.
P = 2656## [kPa]
# From Fig. 10.14 (Pg 426):
K1 = 5.49#
K2 = 1.10#
K3 = 0.33#
# From Eqn. 10.83:
y1 = K1*x1#
y2 = K2*x2#
y3 = K3*x3#
# Since x1 + x2 +x3 = 1. Therefore:
print "Bubble Pressure is %d kPa\n"%(P)#
print "Composition of the vapour bubble:\n y1 = %.4f\n y2 = %.4f\n y3 = %.4f"%(y1,y2,y3)#

Example: 10.16 - Page: 429

Dew Pressure is 870 kPa

Composition of the liquid drop:
x1 = 0.0063
x2 = 0.0755
x3 = 0.9186

Bubble Pressure is 2656 kPa

Composition of the vapour bubble:
y1 = 0.5490
y2 = 0.2200
y3 = 0.2310


## Example: 10.17 Page: 432¶

In :
from __future__ import division
from scipy.optimize import fsolve
from math import exp,log
print "Example: 10.17 - Page: 432\n\n"

# Solution

#*****Data******#
# 1: acetone 2: acetonitrile 3: nitromethane
z1 = 0.45#
z2 = 0.35#
z3 = 0.20#
P1sat = 195.75## [kPa]
P2sat = 97.84## [kPa]
P3sat = 50.32## [kPa]
#***************#

# Bubble Point Calculation:
Pbubble = z1*+P1sat + z2*P2sat +z3*P3sat## [kPa]

# Dew Point Calculation:
Pdew = 1/((z1/P1sat) + (z2/P2sat) + (z3/P3sat))## [kPa]
K1 = P1sat/Pdew#
K2 = P2sat/Pdew#
K3 = P3sat/Pdew#
# Overall Material balance:
# For 1 mol of the feed.
# L + V = 1......................................... (1)
# F*zi = L*xi + V*yi ............................... (2)
# zi = (1 - V)*xi + V*yi ........................... (3)
# Substituting xi = yi/K in eqn. (3)
# yi = zi*Ki/(1 + V*(Ki - 1))
# Since, Sum(yi) = 1.
#deff('[y] = f(V)','y = (z1*K1/(1 + V*(K1 - 1))) + (z2*K2/(1 + V*(K2 - 1))) + (z3*K3/(1 + V*(K3 - 1))) - 1')#
def f(V):
y = (z1*K1/(1 + V*(K1 - 1))) + (z2*K2/(1 + V*(K2 - 1))) + (z3*K3/(1 + V*(K3 - 1))) - 1
return y

V = fsolve(f,0.8)#
L = 1 - V#
y1 = z1*K1/(1 + V*(K1 - 1))#
y2 = z2*K2/(1 + V*(K2 - 1))#
y3 = z3*K3/(1 + V*(K3 - 1))#
# From Eqn. 10.83:
x1 = y1/K1#
x2 = y2/K2#
x3 = y3/K3#
print " L = %e mol\n"%(L)#
print " V = %e mol\n"%(V)#
print " y1 = %.4f\n y2 = %.4f\n y3 = %.4f\n"%(y1,y2,y3)#
print " x1 = %.4f\n x2 = %.4f\n x3 = %.4f\n"%(x1,x2,x3)#

Example: 10.17 - Page: 432

L = 3.154304e-08 mol

V = 1.000000e+00 mol

y1 = 0.4500
y2 = 0.3500
y3 = 0.2000

x1 = 0.2334
x2 = 0.3631
x3 = 0.4035



## Example: 10.18 Page: 433¶

In :
from __future__ import division
from scipy.optimize import fsolve
from math import exp
print "Example: 10.18 - Page: 433\n\n"

# Solution

#*****Data******#
# 1: Benzene 2: Toulene
z1 = 0.81#
Temp = 60## [OC]
P = 70## [kPa]
# Antonine Constants:
A1 = 14.2321#
B1 = 2773.61#
C1 = 220.13#
A2 = 15.0198#
B2 = 3102.64#
C2 = 220.02#
#******************#

#deff('[P1] = f1(T)','P1 = exp(A1 - B1/(T + C1))')#
def f1(T):
P1 = exp(A1 - B1/(T + C1))
return P1

P1sat = f1(Temp)## [kPa]
#deff('[P2] = f2(T)','P2 = exp(A2 - B2/(T + C2))')#
def f2(T):
P2 = exp(A2 - B2/(T + C2))
return P2

P2sat = f2(Temp)## [kPa]
# P = x1*P1sat + x2*P2sat#
# x2 = 1 - x1#
#deff('[y] = f3(x1)','[y] = P - (x1*P1sat + (1 - x1)*P2sat)')#
def f3(x1):
y= P - (x1*P1sat + (1 - x1)*P2sat)
return y

x1 = fsolve(f3,7)#
y1 = x1*P1sat/P#
x2 = 1 - x1#
y2 = 1 - y1#

# Basis: 1 mol of feed stream.
F = 1## [mol]
# F*zi = L*xi + V*yi = L*xi + (1 - L)*yi
#deff('[y] = f4(L)','[y] = F*z1 - (L*x1 + (1 - L)*y1)')#
def f4(L):
y = F*z1 - (L*x1 + (1 - L)*y1)
return y

L = fsolve(f4,7)## [mol]
V = 1 - L## [mol]
print " L = %.4f mol\n"%(L)#
print " V = %.4f mol\n"%(V)#
print " y1 = %.4f\n y2 = %.4f\n"%(y1,y2)#
print " x1 = %.4f\n x2 = %.4f\n"%(x1,x2)#

Example: 10.18 - Page: 433

L = 0.1615 mol

V = 0.8385 mol

y1 = 0.8205
y2 = 0.1795

x1 = 0.7555
x2 = 0.2445



## Example: 10.19 Page: 413¶

In :
from math import exp, log
from numpy import nditer, shape, zeros,mat
%matplotlib inline
from matplotlib.pyplot import plot, xlabel, ylabel, show, grid

print "Example: 10.19 - Page: 436\n\n"

# Solution

#*****Data******#
# (1): acetone (2): carbon tetrachloride
T = 45## [OC]
# Data = [P (torr), x1, y1]

Data = mat([[315.32, 0.0556, 0.2165],[339.70, 0.0903, 0.2910],[397.77, 0.2152, 0.4495],[422.46, 0.2929, 0.5137],[448.88, 0.3970, 0.5832],[463.92, 0.4769, 0.6309],[472.84, 0.5300, 0.6621],[485.16, 0.6047, 0.7081],[498.07, 0.7128, 0.7718],[513.20, 0.9636, 0.9636]])
#*************#

# From the standard data (Pg 531):
# For Acetone:
A1 = 14.2342#
B1 = 2691.46#
C1 = 230.00#
# For carbon tetrachloride:
A2 = 13.6816#
B2 = 2355.82#
C2 = 220.58#
P1sat = exp(A1 - B1/(T + C1))## [kPa]
P2sat = exp(A2 - B2/(T + C2))## [kPa]
P1sat = P1sat*760/101.325## [torr]
P2sat = P2sat*760/101.325## [torr]
P = Data[:,0]#
x1 = Data[:,1]#
y1 = Data[:,2]#
x2 = 1 - x1#
y2 = 1 - y1#

gama1 = zeros(x1.shape)
i=0
for a,b,c in nditer([y1,P,x1]):
gama1[i]=(a*b/c)
i+=1

gama2 = zeros(x2.shape)
i=0
for a,b,c in nditer([y2,P,x2]):
gama2[i]=(a*b/c)
i+=1

Value = zeros(gama1.shape)
i=0
for x,y in nditer([gama1, gama2]):
Value[i]=(log(x/y))
i+=1
plot(x1,Value)
grid()#
xlabel("x1")
ylabel("ln(y1/y2)")
# Since the whole area is above X - axis:
print "The data is not consistent thermodynamically\n"

Example: 10.19 - Page: 436

The data is not consistent thermodynamically 