Chapter 8 : Vapor Liquid Equilibrium VLE at Low Pressures

Example 8.1 Page: 163

In [18]:
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 [19]:
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 [20]:
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 [22]:
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 [23]:
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 [4]:
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 [25]:
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 [26]:
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 [27]:
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 [28]:
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 [29]:
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 [30]:
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 [31]:
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 [32]:
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 [6]:
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 [34]:
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 [ ]: