# Chapter 8 : Vapor Liquid Equilibrium VLE at Low Pressures¶

### Example 8.1 Page: 163¶

In :
import math

x_acetone = 0.05            # Mole fraction of Acetone in liquid
x_water = (1-x_acetone)
y_acetone = 0.6381          # Mole fraction of Acetone in vapour
y_water = (1-y_acetone)

K_acetone = y_acetone/x_acetone
K_water = y_water/x_water
a = K_acetone/K_water

print "The K factor of acetone is %f"%(K_acetone)
print " The K factor of water is    %f"%(K_water)
print " The relative volatility is  %f"%(a)

The K factor of acetone is 12.762000
The K factor of water is    0.380947
The relative volatility is  33.500691


### Example 8.2 Page: 165¶

In :
import math

P = 1.          #[atm]
Temp = 74.8     #[C]
A_a = 7.02447
B_a = 1161
C_a = 224
p_acetone = 10**(A_a-B_a/(Temp+C_a))            #[mmHg]
A_w = 7.94917
B_w = 1657.462
C_w = 227.02

p_water = 10**(A_w-B_w/(Temp+C_w))              #[mmHg]
p_acetone = p_acetone/760                       #[atm]
p_water = p_water/760                           #[atm]
y_acetone = 0.6381
x_acetone = 0.05
y_water = (1-y_acetone)
x_water =(1-x_acetone)
Y_acetone = y_acetone*P/(x_acetone*p_acetone)
Y_water = y_water*P/(x_water*p_water)

print "Liquid-phase activity coefficient for acetone is %f"%(Y_acetone)
print " Liquid-phase activity coefficient for water is   %f"%(Y_water)

Liquid-phase activity coefficient for acetone is 7.043759
Liquid-phase activity coefficient for water is   1.009407


### Example 8.3 Page: 167¶

In :
import math

x_a = 0.05          # mole fraction of acetone in liquid phase
x_w = (1-x_a)       # mole fraction of the water in the liquid phase
P = 1.00            #[atm] Total pressure in vapor phase

T_1 = 80.           #[C]
A_a = 7.02447
B_a = 1161.
C_a = 224.
A_w = 7.94917
B_w = 1657.462
C_w = 227.02

p_a_1 = (1./760)*10**(A_a - B_a/(T_1+C_a))      #[atm]
p_w_1 = (1./760)*10**(A_w - B_w/(T_1+C_w))      #[atm]

y_a_1 = (x_a*p_a_1)/P
y_w_1 = (x_w*p_w_1)/P

y_1 = (y_a_1 + y_w_1)

T_2 = 96.4060           #[C]

p_a_2 = (1./760)*10**(A_a - B_a/(T_2+C_a))       #[atm]
p_w_2 = (1./760)*10**(A_w - B_w/(T_2+C_w))      #[atm]

y_a_2 = (x_a*p_a_2)/P
y_w_2 = (x_w*p_w_2)/P

y_2 = (y_a_2 + y_w_2)
T_e = 74.8                  #[C] Boiling temperature
y_a_e = 0.6381              # vapor phase composition of acetone

print " Comparison of experimental values to those computed by the ideal solution assumption x_acetone = 0.05 and P = 1.00 atm"
print " \t\t\t Experimental Values from Table 8.1 \t\t\t\tValues calculated assuming idea solution"
print " Equilibriumboiling) \t\t%0.1f \t\t\t\t\t\t\t\t\t %0.1f  temperature T deg C"%(T_e,T_2)
print " Mole fraction acetone \t\t%f \t\t\t\t\t\t\t\t %f  in the vapor phase y_a)"%(y_a_e,y_a_2)

 Comparison of experimental values to those computed by the ideal solution assumption x_acetone = 0.05 and P = 1.00 atm
Experimental Values from Table 8.1 				Values calculated assuming idea solution
Equilibriumboiling) 		74.8 									 96.4  temperature T deg C
Mole fraction acetone 		0.638100 								 0.165615  in the vapor phase y_a)


### Example 8.4 Page: 177¶

In :
import math

n_water = 80.                       #[mol]
n_bumath_tanol = 20.                #[mol]
n_total = n_water+n_bumath_tanol    #[mol]

x_feed = 0.8
x_a_1 = 0.65
x_a_2 = 0.98

n_1 = (x_feed-x_a_2)/(x_a_1-x_a_2)*n_total          #[mol]
n_2 = (n_total-n_1)                                 #[mol]
n_a_1 = 0.65*n_1                                    #[mol]
n_a_2 = 0.98*n_2                                    #[mol]

print " Total moles of water present in the first phase is  %f mol"%(n_a_1)
print " Total moles of water present in the second phase is %f mol"%(n_a_2)

 Total moles of water present in the first phase is  35.454545 mol
Total moles of water present in the second phase is 44.545455 mol


### Example 8.5 Page: 178¶

In :
import math

P = 1.              #[atm]
y_water = 0.60
x_water_1 = 0.22
x_water_2 = 0.99

print " The equilibrium amount of water in liquid at bubble-point for the dew-point composition y_water=60 mol%% is %f mol%% water"%(x_water_1)
print " The equilibrium amount of water in liquid at bubble-point for the dew-point composition y_water=90 mol%% is %f mol%% water"%(x_water_2)

 The equilibrium amount of water in liquid at bubble-point for the dew-point composition y_water=60 mol% is 0.220000 mol% water
The equilibrium amount of water in liquid at bubble-point for the dew-point composition y_water=90 mol% is 0.990000 mol% water


### Example 8.6 Page: 178¶

In :
import math

P = 1.00            #[atm] assumed total vapor pressure
P1 = 14.7           #[psia]
x_1_water = 0.65
x_1_bumath_tanol = (1-x_1_water)
x_2_water = 0.98
x_2_bumath_tanol = (1-x_2_water)
y_water = 0.73
y_bumath_tanol = (1-y_water)
T = 92.             #[C]

A_w = 7.94917
B_w = 1657.462
C_w = 227.02

A_b = 7.838
B_b = 1558.190
C_b = 196.881

p_water = (14.7/760)*10**(A_w - B_w/(T+C_w))
p_bumath_tanol = (14.7/760)*10**(A_b - B_b/(T+C_b))

f_water = (y_water*P)
f_bumath_tanol = (y_bumath_tanol*P)

Y_water_1 = (y_water*P1)/(x_1_water*p_water)
Y_bumath_tanol_1 = (y_bumath_tanol*P1)/(x_1_bumath_tanol*p_bumath_tanol)

Y_water_2 = (y_water*P1)/(x_2_water*p_water)
Y_bumath_tanol_2 = (y_bumath_tanol*P1)/(x_2_bumath_tanol*p_bumath_tanol)

print " Four activity coefficients and fufacities are shown in the following table:"
print "Phase \t x_water \t f_wateratm \t Y_water \t x_butanol \t f_butanolatm \t\t Y_butanol"
print " 1 \t %f \t %f \t %f \t %f \t %f \t\t %f "%(x_1_water,f_water,Y_water_1,x_1_bumath_tanol,f_bumath_tanol,Y_bumath_tanol_1)
print " 2 \t %f \t %f \t %0.2f \t\t %f \t %f \t\t %f "%(x_2_water,f_water,Y_water_2,x_2_bumath_tanol,f_bumath_tanol,Y_bumath_tanol_2)

 Four activity coefficients and fufacities are shown in the following table:
Phase 	 x_water 	 f_wateratm 	 Y_water 	 x_butanol 	 f_butanolatm 		 Y_butanol
1 	 0.650000 	 0.730000 	 1.504988 	 0.350000 	 0.270000 		 2.108586
2 	 0.980000 	 0.730000 	 1.00 		 0.020000 	 0.270000 		 36.900247


### Example 8.7 Page: 179¶

In :
import math

P = 1.              #[atm] Total pressure in the vapor phase

T = 89.             #[C]
A_w = 7.94917
B_w = 1657.462
C_w = 227.02

A_b = 7.838
B_b = 1558.190
C_b = 196.881

p_water = (1./760)*10**(A_w - B_w/(T+C_w))
p_bumath_tanol = (1./760)*10**(A_b - B_b/(T+C_b))

y_water = p_water/P
y_bumath_tanol = p_bumath_tanol/P
y = y_water + y_bumath_tanol

print " Boiling point of the two phase system is %0.0f deg C"%(T)
print " In vapor phase mole fraction of the water is %0.2f"%(y_water)

 Boiling point of the two phase system is 89 deg C
In vapor phase mole fraction of the water is 0.67


### Example 8.8 Page: 184¶

In :
import math

Temp = 68.          #[F]
P = 1.              #[atm]

Temp = 273.15+(Temp-32)*5./9        #[K]
P = P*1.01325                       #[bar]
T_c = 647.1         #[K]
P_c = 220.55        #[bar]
T_r = Temp/T_c
P_r = P/P_c
w = 0.345
f_T_r = (0.083-0.422/T_r**(1.6))+w*(0.139-0.172/T_r**(4.2))
f_by_P = math.exp(P_r/T_r*f_T_r)

print "The value of the f/P for water vapour in the hypothetical state is %0.2f"%(f_by_P)

The value of the f/P for water vapour in the hypothetical state is 0.97


### Example 8.9 Page: 189¶

In :
import math

x_a = 0.1238
x_b = (1-x_a)
T = 85.3            #[C] Given boiling temperature

A_a = 8.04494
B_a = 1554.3
C_a = 222.65

A_b = 7.96681
B_b = 1668.21
C_b = 228.0

p_a = (1./760)*10**(A_a - B_a/(T+C_a))
p_b = (1./760)*10**(A_b - B_b/(T+C_b))

A = 0.7292
B = 0.4104

Y_a = 10**((B**(2)*A*x_b**(2))/(A*x_a+B*x_b)**(2))
Y_b = 10**((A**(2)*B*x_a**(2))/(A*x_a+B*x_b)**(2))

P = (Y_a*p_a*x_a)+(Y_b*p_b*x_b)             #[atm]

y_a = (Y_a*p_a*x_a)/P
y_b = (Y_b*p_b*x_b)/P

print " Boiling pressure of the liquid at 85.3 deg C is %0.4f atm"%(P)
print " Mole fraction of ethanaol in vapor phase is     %0.4f"%(y_a)
print " Mole fraction of water in the vapor phase is    %0.4f"%(y_b)

 Boiling pressure of the liquid at 85.3 deg C is 0.9991 atm
Mole fraction of ethanaol in vapor phase is     0.4741
Mole fraction of water in the vapor phase is    0.5259


### Example 8.10 Page: 191¶

In :
import math

x_a = 0.2608
x_b = (1-x_a)
P = 1.00            #[atm] Given boiling pressure

A_a = 8.04494
B_a = 1554.3
C_a = 222.65

A_b = 7.96681
B_b = 1668.21
C_b = 228.0

A = 0.7292
B = 0.4104

Y_a = 10**((B**(2)*A*x_b**(2))/(A*x_a+B*x_b)**(2))
Y_b = 10**((A**(2)*B*x_a**(2))/(A*x_a+B*x_b)**(2))

T = 80.
err = 1.

while err > 10**(-3):
P_a = (10**(8.04494 - 1554.3/(222.65 + T)))/760
P_b = (10**(7.96681 - 1668.21/(228 + T)))/760
y_a = Y_a*P_a*x_a/P
y_b = Y_b*P_b*x_b/P
err = abs((y_a + y_b) - 1)
T = T + 0.01

print " Boiling temperature of the liquid at 1 atm pressure is %0.4f atm"%(T)
print " Mole fraction of ethanaol in vapor phase is     \t%0.4f"%(y_a)
print " Mole fraction of water in the vapor phase is    \t%0.4f"%(y_b)

 Boiling temperature of the liquid at 1 atm pressure is 82.0300 atm
Mole fraction of ethanaol in vapor phase is     	0.5680
Mole fraction of water in the vapor phase is    	0.4312


### Example 8.11 Page: 192¶

In :
import math

y_a = 0.6122
y_b = (1-y_a)
T = 80.7            #[C] Given boiling temperature

A_a = 8.04494
B_a = 1554.3
C_a = 222.65

A_b = 7.96681
B_b = 1668.21
C_b = 228.0

p_a = (1./760)*10**(A_a - B_a/(T+C_a))
p_b = (1./760)*10**(A_b - B_b/(T+C_b))

A = 0.7292
B = 0.4104

x_a = 0.6122            # Initial assumption of liquid phase composition of ethanol
x_b = 0.3               # Initial assumption of liquid phase composition water
P = 0.80                #[atm]
err = 1

while err > 2* 10**(-2):
P = P + 0.01
Y_a = 10**((B**(2)*A*x_b**(2))/(A*x_a+B*x_b)**(2))
Y_b = 10**((A**(2)*B*x_a**(2))/(A*x_a+B*x_b)**(2))

err = abs((x_a + x_b) - 1)
x_a = y_a*P/(Y_a*p_a)
x_b = y_b*P/(Y_b*p_b)

print " Boiling pressure of the liquid at 80.7 deg C is %0.4f atm"%(P)
print " Mole fraction of ethanaol in liquid phase is     %0.4f"%(x_a)
print " Mole fraction of water in the liquid phase is    %0.4f"%(x_b)

 Boiling pressure of the liquid at 80.7 deg C is 0.9900 atm
Mole fraction of ethanaol in liquid phase is     0.3730
Mole fraction of water in the liquid phase is    0.6205


### Example 8.12 Page: 193¶

In :
import math

y_a = 0.1700
y_b = (1-y_a)
P = 1.00            #[atm] Given boiling pressure

A_a = 8.04494
B_a = 1554.3
C_a = 222.65

A_b = 7.96681
B_b = 1668.21
C_b = 228.0

A = 0.7292
B = 0.4104

x_a = 0.0100            # Initial assumption of liquid phase composition of ethanol
x_b = 0.9               # Initial assumption of liquid phase composition water
T = 80.                 #[C] Initial guess of the temperature
err = 1

while err >  1./16*10**(-2):
P_a = (10**(8.04494 - 1554.3/(222.65 + T)))/760
P_b = (10**(7.96681 - 1668.21/(228 + T)))/760

Y_a = 10**((B**(2)*A*x_b**(2))/(A*x_a+B*x_b)**(2))
Y_b = 10**((A**(2)*B*x_a**(2))/(A*x_a+B*x_b)**(2))

x_a = y_a*P/(Y_a*P_a)
x_b = y_b*P/(Y_b*P_b)

err = abs((x_a + x_b) - 1)
T = T + 0.01

print " Equilibrium Temperature of the system at pressure 1 atm  is %0.4f atm"%(T)
print " Mole fraction of ethanaol in liquid phase is     %0.4f"%(x_a)
print " Mole fraction of water in the liquid phase is    %0.4f"%(x_b)

 Equilibrium Temperature of the system at pressure 1 atm  is 95.3500 atm
Mole fraction of ethanaol in liquid phase is     0.0187
Mole fraction of water in the liquid phase is    0.9816


### Example 8.13 Page: 194¶

In :
import math

x_aF = 0.126
x_bF = (1-x_aF)
P = 1.00            #[atm] Given total pressure
T = 91.8            #[C]

x_a = 0.0401
x_b = (1 - x_a)

y_a = 0.2859
y_b = ( 1 - y_a)

V_by_F = ( x_aF - x_a )/(y_a - x_a)

print " Mole fraction of the ethanol in the liquid phase in equilibrium at the given condition is %f"%(x_a)
print " Mole fraction of the water in the liquid phase in equilibrium at the given condition is %f"%(x_b)
print " Mole fraction of the ethanol in the vapour phase in equilibrium at the given condition is %f"%(y_a)
print " Mole fraction of the water in the vapour phase in equilibrium at the given condition is %f"%(y_b)
print " Vapor fraction of the given water-ethanol mixture after the flash in equilibrium is %f"%(V_by_F)

 Mole fraction of the ethanol in the liquid phase in equilibrium at the given condition is 0.040100
Mole fraction of the water in the liquid phase in equilibrium at the given condition is 0.959900
Mole fraction of the ethanol in the vapour phase in equilibrium at the given condition is 0.285900
Mole fraction of the water in the vapour phase in equilibrium at the given condition is 0.714100
Vapor fraction of the given water-ethanol mixture after the flash in equilibrium is 0.349471


### Example 8.14 Page: 198¶

In :
import math

P = 100.                    #[psia]
x_a = 0.05                  # Mole fraction of methane
x_b = 0.40                  # Mole fraction of bumath.tane
x_c = 0.55                  # mole fraction of penmath.tane

T = [[15.8 ,0.087, 0.024],[16,0.105, 0.026],[16.2, 0.115, 0.03],[16.8 ,0.13, 0.035],[17.2 ,0.15, 0.04],[17.8, 0.17, 0.045],[18.2, 0.175, 0.0472727]]
print " Calculations for the various assumed temperatures are given in the table below"
print " Temperature \t\t    y_a \t\t   y_b \t\t\t   y_c \t\t\t   y "

T_b = 0         #[F] Bubble point
j=0
for i in range(7):
y_a = x_a*T[i][j]
y_b = x_b*T[i][j+1]
y_c = x_c*T[i][j+2]
y = y_a + y_b + y_c
T_b = T_b + 5
print " %f \t\t %f \t\t %f \t\t %f \t\t %f "%(T_b,y_a,y_b,y_c,y)

print "  For the temperature 30 deg F the summation of the mole fractions in the vapor phase is close enough to unity so bubble point is 30 degF"
print " And compositions in the vapor phase are the values given in the above table corresonding to the temperature 30 deg F i.e."
print " y_methane = %f  y_bumath.tane = %f  y_penmath.tane = %f"%(y_a,y_b,y_c)

 Calculations for the various assumed temperatures are given in the table below
Temperature 		    y_a 		   y_b 			   y_c 			   y
5.000000 		 0.790000 		 0.034800 		 0.013200 		 0.838000
10.000000 		 0.800000 		 0.042000 		 0.014300 		 0.856300
15.000000 		 0.810000 		 0.046000 		 0.016500 		 0.872500
20.000000 		 0.840000 		 0.052000 		 0.019250 		 0.911250
25.000000 		 0.860000 		 0.060000 		 0.022000 		 0.942000
30.000000 		 0.890000 		 0.068000 		 0.024750 		 0.982750
35.000000 		 0.910000 		 0.070000 		 0.026000 		 1.006000
For the temperature 30 deg F the summation of the mole fractions in the vapor phase is close enough to unity so bubble point is 30 degF
And compositions in the vapor phase are the values given in the above table corresonding to the temperature 30 deg F i.e.
y_methane = 0.910000  y_bumath.tane = 0.070000  y_penmath.tane = 0.026000


### Example 8.15 Page: 199¶

In :
import math

n_sugar = 1.                #[mol]
n_water = 1000/18.          #[mol]
x_sugar = n_sugar/(n_sugar+n_water)
x_water = n_water/(n_sugar+n_water)
p_water = 1.                #[atm]
p_sugar = 0.                #[atm]

P = x_water*p_water+x_sugar*p_sugar         #[atm]
P_1 = 1.                    #[atm]
p_water = P_1/x_water       #[atm]
T = 100.51                  #[C]
T_eb = T-100                #[C]

print "Vapour pressure of this solution at the 100C is  %.3f atm"%(P)
print "The temperature at which this solution will boil at 1 atm is %.2f C"%(T)

Vapour pressure of this solution at the 100C is  0.982 atm
The temperature at which this solution will boil at 1 atm is 100.51 C


### Example 8.16 Page: 201¶

In :
from scipy.optimize import fsolve
import math

n_sugar = 1.            #[mol]
n_water = 1000/18.      #[mol]
x_sugar = n_sugar/(n_sugar+n_water)
x_water = n_water/(n_sugar+n_water)

p_sugar = 0
p_ice_by_p_water = x_water

def f(T):
return  1+0.0096686*T+4.0176*10**(-5)*T**(2)-p_ice_by_p_water
T = fsolve(f,0)

print "Freezing-point temperature of the given solution is %f C"%(T)

Freezing-point temperature of the given solution is -1.842891 C

In [ ]: