Chapter 8 : Phase equilibria¶
Example 8.6¶
In [30]:
# Variables
P1 = 106.; #vapour pressure of n-heptane (kPa)
P2 = 74.; #vapour pressure of toluene (kPa)
P = 101.3; #total pressure (kPa)
# Calculations
x = (P-P2)/(P1-P2)
y = x*(P1/P)
# Results
print 'Composition of liquid heptane is %f mol percent'%(x*100)
print ' Composition of heptane in vapour form is %f mol percent'%(y*100)
Example 8.7¶
In [31]:
# Variables
P1 = 135.4; #vapour pressure of benzene (kPa)
P2 = 54.; #vapour pressure of toluene (kPa)
x = 0.5; #liquid phase composition
# Calculations
#Using eq. 8.51 (Page no. 332)
P_beg = P2 + (P1-P2)*x;
#At the end
y = 0.5; #vapour phase composition
#Using eq. 8.54 (Page no. 333) and rearranging
P_end = (P1*P2)/(P1-y*(P1-P2))
# Results
print 'Pressure at the beginning of the process is %f kPa'%(P_beg)
print ' Pressure at the end of the process is %f kPa'%(P_end)
Example 8.8¶
In [32]:
import math
def P1(T):
y1 = math.e**(14.5463 - 2940.46/(T-35.93)) #vapour pressure of acetone
return y1
def P2(T):
y2 = math.e**(14.2724 - 2945.47/(T-49.15)) #vapour pressure of acetonitrile
return y2
# Variables
T = 327.; #temperature in K
P = 65.; #pressure in kPa
# Calculations and Results
P1_s = P1(T)
P2_s = P2(T)
#Using eq. 8.51 (Page no. 332)
x1 = (P-P2_s)/(P1_s-P2_s)
#Using eq. 8.54 (Page no. 333)
y1 = x1*(P1_s/P)
print '(a)'
print ' x1 = %f'%x1
print ' y1 = %f'%y1
#(b). To calculate T and y1
P = 65.; #pressure in kPa
x1 = 0.4;
flag = 1.;
T2 = 340.; #temperatue (assumed)
while(flag==1):
P1_s = P1(T2)
P2_s = P2(T2)
P_calc = P2_s + x1*(P1_s-P2_s)
if((P_calc-P)<=1):
flag = 0;
else:
T2 = T2-0.8;
y1 = x1*(P1_s/P)
print ' (b)'
print ' Temperature is %f K'%T2
print ' y1 = %f'%y1
#(c). To calculate P and y1
T3 = 327.; #temperature in K
x1 = 0.4;
P1_s = P1(T3)
P2_s = P2(T3)
P = P2_s + x1*(P1_s-P2_s)
y1 = x1*(P1_s/P)
print ' (c)'
print ' Pressure is %f kPa'%P
print ' y1 = %f'%y1
#(d). To calculate T and x1
P = 65.; #pressure in kPa
y1 = 0.4;
flag = 1.;
T = 340.; #assumed temperature (K)
while(flag==1):
P1_s = P1(T)
P2_s = P2(T)
y1_calc = (P1_s*(P-P2_s))/(P*(P1_s-P2_s))
if((y1_calc-y1)>=0.001):
flag = 0;
else:
T = T-2;
x1 = y1*(P/P1_s)
print ' (d)'
print ' Temperature = %f K'%T
print ' x1 = %f'%x1
#(e). To calculate P and x1
T = 327.; #temperature (K)
y1 = 0.4;
P1_s = P1(T)
P2_s = P2(T)
#Using eq. 8.54 and 8.51
x1 = (y1*P2_s)/(P1_s-y1*(P1_s-P2_s))
P = x1*(P1_s/y1)
print ' (e)'
print ' Pressure = %f kPa'%P
print ' x1 = %f'%x1
#(f). To calculate fraction of the system is liquid and vapour in equilibrium
T = 327.; #temperature (K)
P = 65.; #pressure (kPa)
y1 = 0.7344;
P1_s = P1(T)
P2_s = P2(T)
x1 = (P-P2_s)/(P1_s-P2_s)
#Let f be the fraction of the mixture that is liquid
#Applying acetone balance
f = (0.7-y1)/(x1-y1)
print ' (f)'
print ' Fraction of mixture that is liquid is %f percent'%(f*100)
Example 8.9¶
In [6]:
%matplotlib inline
# Variables
P = 101.3; #total pressure over the system (kPa)
T = [371.4, 378, 383, 388, 393, 398.6];
Pa = [101.3 ,125.3, 140.0, 160.0 ,179.9, 205.3];
Pb = [44.4 ,55.6 ,64.5, 74.8, 86.6 ,101.3];
xa = [0,0,0,0,0,0]
ya = [0,0,0,0,0,0]
# Calculations
#To construct boiling point and equilibrium point diagram
for i in range(6):
xa[i] = (P-Pb[i])/(Pa[i]-Pb[i]) #Using eq. 8.51
ya[i] = xa[i]*(Pa[i]/P ) #Using eq. 8.54
from matplotlib.pyplot import *
#(a).
#To construct boiling point diagram
plot(xa,T)
plot(ya,T)
#title("Boiling Point diagram xa and ya Temperature"
#(b).
#To construct the equilibrium diagram
plot(ya,xa)
#title("Equilibrium Diagram","xa","ya"
show()
#(c).
# Results
print '(c). The given subpart is theoretical and does not involve any numerical computation'
Example 8.11¶
In [34]:
# Variables
x1 = 46.1/100; #mole percent of A
P = 101.3; #total pressure of system (kPa)
P1_s = 84.8; #vapour pressure of component A (kPa)
P2_s = 78.2; #vapour pressure of component B (kPa)
# Calculations
#To calculate van Laar constants
gama1 = P/P1_s;
gama2 = P/P2_s;
x2 = 1-x1;
import math
#van Laar constants:
#Using eq. 8.69 (Page no. 348)
A = math.log (gama1)*(1 + (x2*math.log(gama2))/(x1*math.log(gama1)))**2;
B = math.log (gama2)*(1 + (x1*math.log(gama1))/(x2*math.log(gama2)))**2;
# Results
print 'van Laar constants are:'
print ' A = %f'%A
print ' B = %f'%B
Example 8.12¶
In [35]:
# Variables
x2 = 0.448; #mole fraction of ethanol
P = 101.3; #total pressure (kPa)
P1_s = 68.9; #Vapour pressure of benzene (kPa)
P2_s = 67.4; #vapour pressure of ethanol (kPa)
# Calculations
#To calculate activity coeffecients in a solution containing 10% alcohol
x1 = 1-x2;
gama1 = P/P1_s;
gama2 = P/P2_s;
import math
#Using eq. 8.69 (Page no. 348)
#van Laar constants:
A = math.log(gama1)*(1 + (x2*math.log(gama2))/(x1*math.log(gama1)))**2;
B = math.log(gama2)*(1 + (x1*math.log(gama1))/(x2*math.log(gama2)))**2;
#For solution containing 10% alcohol
x2 = 0.1;
x1 = 1-x2;
ln_g1 = (A*x2**2)/(((A/B)*x1+x2)**2)
ln_g2 = (B*x1**2)/((x1+(B/A)*x2)**2)
gama1 = math.e**ln_g1;
gama2 = math.e**ln_g2;
# Results
print 'Activity coeffecients:'
print ' For component 1: %f'%gama1
print ' For component 2: %f'%gama2
Example 8.13¶
In [36]:
# Variables
x2 = 0.585; #mol fraction of hydrazine
P = 101.3; #total pressure of system (kPa)
P2_s = 124.76; #vapour pressure of hydrazine (kPa)
# Calculations
#To calculate equilibrium vapour composition for solution containing 20% (mol) hydrazine
x1 = 1-x2;
P1_s = 1.6*P2_s; #vapour pressure of water (kPa)
gama1 = P/P1_s;
gama2 = P/P2_s;
import math
#Using eq. 8.69 (Page no. 348)
#van Laar constants:
A = math.log(gama1)*(1 + (x2*math.log(gama2))/(x1*math.log(gama1)))**2;
B = math.log(gama2)*(1 + (x1*math.log(gama1))/(x2*math.log(gama2)))**2;
#For solution containing 20% hydrazine
x2 = 0.2;
x1 = 1-x2;
ln_g1 = (A*x2**2)/(((A/B)*x1+x2)**2)
ln_g2 = (B*x1**2)/((x1+(B/A)*x2)**2)
gama1 = math.e**ln_g1;
gama2 = math.e**ln_g2;
#Using eq. 8.47 (Page no. 325) for components 1 and 2 and rearranging
alpha = 1.6; #alpha = P1_s/P2_s
y1 = 1./(1 + (gama2*x2)/(gama1*x1*alpha))
y2 = 1-y1;
# Results
print 'Equilibrium vapour composition for solution containing 20 mol percent hydrazine'
print ' Hydrazine is %f percent'%(y2*100)
print ' Water is %f percent'%(y1*100)
Example 8.15¶
In [37]:
#Given:
x1 = 0.047; #mol fraction of isopropanol
P1 = 91.11; #vapour pessure of pure propanol (kPa)
P = 91.2; #toatl pressure of system (kPa)
P2 = 47.36; #vapour pressure of water (kPa)
#van Laar consatnts:
A = 2.470;
B = 1.094;
#To determine the total pressure:
x2 = 1-x1;
#Using eq. 8.68 (Page no. 348)
ln_g1 = (A*x2**2)/(((A/B)*x1 + x2)**2);
ln_g2 = (B*x1**2)/((x1 + (B/A)*x2)**2);
gama1 = math.e**ln_g1;
gama2 = math.e**ln_g2;
#Total pressure:
P_tot = (gama1*x1*P1) + (gama2*x2*P2);
# Results
if(P==P_tot):
print 'This is equal to total pressure'
else:
print 'This is less than the total pressure. This error must have been caused by air leak'
Example 8.16¶
In [7]:
# Variables
P1 = 24.62; #vapour pressure of cyclohexane (kPa)
P2 = 24.41; #vapour pressure of benzene (kPa)
from numpy import array
import math
x1 = array([0, 0.2, 0.4, 0.6, 0.8, 1.0])
x2 = 1-x1;
g1 = [0,0,0,0,0,0]
g2 = [0,0,0,0,0,0]
P = [0,0,0,0,0,0]
y1 = [0,0,0,0,0,0]
# Calculations
for i in range(6):
g1[i] = math.e**(0.458*x2[i]**2) #activity coeffecient for component 1
g2[i] = math.e**(0.458*x1[i]**2 ) #activity coeffecient for component 2
P[i] = (g1[i]*x1[i]*P1) + (g2[i]*x2[i]*P2) #total pressure (kPa)
y1[i] = (g1[i]*x1[i]*P1)/P[i];
from matplotlib.pyplot import *
# Results
#To construct P-x-y diagram
plot(x1,P)
plot(y1,P)
#title("P-x-y Diagram","x1 and y1","Pressure"
show()
Example 8.17¶
In [39]:
# Variables
P = 40.25; #total pressure (kPa)
y1 = 0.566; #mol fraction of benzene in vapour phase
x1 = 0.384; #mol fraction of benzene in liquid state
P1 = 29.6; #vapour pressure of benzene (kPa)
P2 = 22.9; #vapour pressure of ethanol (kPa)
#To determine the composition and total pressure of azeotrope
x2 = 1-x1;
y2 = 1-y1;
# Calculations
#Using eq. 8.47 (Page no. 325)
#Activity coeffecients:
g1 = (y1*P)/(x1*P1)
g2 = (y2*P)/(x2*P2)
import math
#Using eq. 8.69 (Page no. 348)
#van Laar constants:
A = math.log(g1)*((1 + (x2*math.log(g2))/(x1*math.log(g1)))**2)
B = math.log(g2)*((1 + (x1*math.log(g1))/(x2*math.log(g2)))**2)
#Assuming azeotropic comp. (for hit and trial method)
x1 = 0.4;
flag = 1.;
while(flag==1):
x2 =1-x1;
ln_g1 = (A*x2**2)/(((A/B)*x1 + x2)**2)
ln_g2 = (B*x1**2)/((x1 + (B/A)*x2)**2)
g1 = math.e**ln_g1;
g2 = math.e**ln_g2;
P_1 = g1*P1;
P_2 = g2*P2;
if((P_1-P_2)<=1) and ((P_1-P_2)>=-1):
flag = 0;
else:
x1 = x1+0.1;
# Results
print 'Azeotropic compositon of benzene is %i percent'%(x1*100)
print ' Total pressure of azeotrope is %f kPa'%((P_1+P_2)/2)
Example 8.19¶
In [40]:
# Variables
a12 = 1225.31; #(J/mol)
a21 = 6051.01; #(J/mol)
V1 = 74.05*10**-6; #(m**3/mol)
V2 = 18.07*10**-6; #(m**3/mol)
R = 8.314; #ideal gas constant
T = 349; #temperature in K
# Calculations
#Antoine Equation:
#Vapour pressure of 1st element
def P1(T):
y1 = math.e**(14.39155-(2795.817/(T-43.198)))
return y1
#Vapour pressure of 2nd element
def P2(T):
y2 = math.e**(16.26205-(3799.887/(T-46.854)))
return y2
#To calculate equilibrium pressure and composition
#Using eq. 8.73 (Page no. 350)
#Wilson Parameters:
W12 = (V2/V1)*math.e**(-a12/(R*T));
W21 = (V1/V2)*math.e**(-a21/(R*T));
import math
#Using Antoine equation
P1_s = P1(T);
P2_s = P2(T);
#(a). Composition of vapour in equilibrium
x1 = 0.43;
x2 = 1-x1;
#Using eq. 8.72 (Page no. 350)
#Wilson equations:
#Activity coeffecient of 1st component
def g_1(n1,n2): #n1 is mol fraction of 1 and n2 is for 2
y3 = math.e**(-math.log(n1 + W12*n2) + n2*((W12/(n1+W12*n2))-(W21/(W21*n1+n2))));
return y3
#Activity coeffecint of 2nd component
def g_2(n1,n2):
y4 = math.e**(-math.log(n2 + W21*n1) - n1*((W12/(n1+W12*n2))-(W21/(W21*n1+n2))));
return y4
#Activity coeffecients:
g1 = g_1(x1,x2);
g2 = g_2(x1,x2);
P = (g1*x1*P1_s) + (g2*x2*P2_s);
y1 = (g1*x1*P1_s)/P;
# Results
print '(a).'
print ' Equilibrium pressure is %f kPa'%P
print ' Composition of acetone vapour in equilibrium is %f'%y1
#(b). Composition of liquid in equilibrium
y1 = 0.8;
y2 = 1-y1;
g1 = 1; g2 = 1; #assumed activity coeffecients
P_as = 1/((y1/(g1*P1_s)) + (y2/(g2*P2_s)));
#Hit and trial method:
flag = 1;
while(flag==1):
x1 = (y1*P_as)/(g1*P1_s);
x2 = 1-x1;
g1 = g_1(x1,x2);
g2 = g_2(x1,x2);
P_calc = 1/((y1/(g1*P1_s)) + (y2/(g2*P2_s)));
if((P_calc-P_as)<=1) and ((P_calc-P_as)>=-1):
flag = 0;
else:
P_as = P_calc;
print ' (b).'
print ' Equilibrium Pressure is %f kPa'%P_calc
print ' Composition of acetone in liquid in equilibrium is %f'%x1
Example 8.20¶
In [41]:
# Variables
P = 101.3; #total pressure of system (kPa)
T = 337.5; #temperature in K
x1 = 0.842;
#Antoine constants
#For methanol(1)
A1 = 16.12609;
B1 = 3394.286;
C1 = 43.2;
#For methyl ethyl ketone (2)
A2 = 14.04357;
B2 = 2785.225;
C2 = 57.2;
import math
# Calculations
#To determine parameters in Wilson's equation
P1_s = math.e**(A1-(B1/(T-C1)))
P2_s = math.e**(A2-(B2/(T-C2)))
x2 = 1-x1;
g1 = P/P1_s;
g2 = P/P2_s;
#Using eq. 8.72 and rearranging:
def Wils(n): #n is the Wilson's parameter W12
y1 = (((g1*x2)/(1-(n*x1/(x1+n*x2))+(x1/x2)*math.log(g1*(x1+n*x2))))**(x2/x1))*(g1*(x1+n*x2))
return y1
flag = 1;
W12 = 0.5; #assumed value
while(flag==1):
res = Wils(W12)
if ((res-1)>=-0.09):
flag = 0;
else:
W12 = W12+0.087;
#For 2nd Wilson parameter:
#Using eq. 8.72 and rearranging:
k = math.log(g1*(x1+W12*x2))/x2 - (W12/(x1+W12*x2))
W21 = (-k*x2)/(1+k*x1)
# Results
print "wilson parameters are: %f, %f"%(W12,W21)
Example 8.21¶
In [8]:
from numpy import *
# Variables
P = 101.3; #total pressure in kPa
T = [333, 343, 353, 363]; #temperatures(K)
Pa = [81.97 ,133.29 ,186.61, 266.58]; #Partial pressure of component A (kPa)
Pb = [49.32 ,73.31, 106.63, 166.61]; #Partial pressure of component B (kPa)
Pc = [39.32 ,62.65, 93.30, 133.29]; #Partial pressure of component C (kPa)
xa = 0.45; #mole fraction of methanol
xb = 0.3; #mole fraction of ethanol
# Calculations and Results
xc = 1-xa-xb; #mole fraction of propanol
#To calculate bubble and dew point and the composition
#(a). To calculate bubble point and vapour composition
from matplotlib.pyplot import *
plot(T,Pa);
plot(T,Pb);
plot(T,Pc);
#title(" ","Temperature","Vapour pressures");
#legend("Pa","Pb","Pc");
#Using eq. 8.84 (Page no. 362)
#At bubble temperature, sum(yi) = sum((xi*Pi)/P) = 1
sum_y = [0,0,0,0]
for i in range(4):
sum_y[i] = (xa*Pa[i])/P + (xb*Pb[i])/P + (xc*Pc[i])/P;
Tb = interp(1,sum_y,T); #obtaining temperature at which sum (yi) = 1
#Obtaining vapour pressures at bubble temperature
Pb1 = interp(Tb,T,Pa);
Pb2 = interp(Tb,T,Pb);
Pb3 = interp(Tb,T,Pc);
#Calculating equilibrium vapour composition
ya = (xa*Pb1*100)/P;
yb = (xb*Pb2*100)/P;
yc = (xc*Pb3*100)/P;
print '(a).'
print ' The bubble temperature is %f K'%Tb
print ' The equilibrium vapour contains %f methanol, %f ethanol and %f propanol'%(ya,yb,yc)
#(b). The dew point and liquid composition
#Vapour phase compositions at dew point
ya = 0.45; #methanol
yb = 0.30; #ethanol
yc = 0.25; #propanol
sum_x = zeros(4)
#At dew point, sum(xi) = sum ((yi*P)/Pi) = 1
for i in range(4):
sum_x[i] = (ya*P)/Pa[i] + (yb*P)/Pb[i] + (yc*P)/Pc[i];
Td = interp(1,sum_x,T); #obtaining temperature at which sum (xi) = 1
#Obtaining vapour pressures at dew temperature
Pd1 = interp(Td,T,Pa);
Pd2 = interp(Td,T,Pb);
Pd3 = interp(Td,T,Pc);
#Calculating liquid composition
xa = (ya*P*100)/Pd1;
xb = (yb*P*100)/Pd2;
xc = (yc*P*100)/Pd3;
print ' (c).'
print ' The dew point is %f K'%Td
print ' At dew point liquid contains %f methanol, %f ethanol and %f propanol'%(xa,xb,xc)
Example 8.22¶
In [42]:
#Given:
P = 1447.14; #pressure of the system (kPa)
x = [0.25 ,0.4, 0.35]; #composition of the components
T = [355.4 ,366.5]; #assumed temperatures (K)
K1 = [2.00, 0.78 ,0.33]; #value of Ki at 355.4 K
K2 = [2.30, 0.90 ,0.40]; #value of Ki at 366.5 K
# Calculation and Result
Kx = [0, 0];
for i in range(3):
Kx[0] = Kx[0]+K1[i]*x[i];
Kx[1] = Kx[1]+K2[i]*x[i];
Tb = interp(1,Kx,T);
#At Tb K, from Fig. 13.6 of Chemical Engineer's Handbook
Kb = [2.12 ,0.85 ,0.37]
#Calculation of vapour composition
y1 = Kb[0]*x[0]*100;
y2 = Kb[1]*x[1]*100;
y3 = Kb[2]*x[2]*100;
print '(a).'
print ' The bubble point temperature is %f K'%Tb
print ' At bubble point vapour contains %f percent propane, %f percent butane and %f percent pentane'%(y1,y2,y3)
#(b). The dew point temperature and composition of the liquid
T = [377.6 ,388.8]; #assumed temperatures (K)
y = [0.25, 0.40, 0.35]; #vapour composition at dew point
K1 = [2.6, 1.1, 0.5]; #at 377.6 K
K2 = [2.9, 1.3, 0.61]; #at 388.8 K
#At dew point, sum(yi/Ki) = 1
Ky = [0, 0];
for i in range(3):
Ky[0] = Ky[0] + y[i]/K1[i];
Ky[1] = Ky[1] + y[i]/K2[i];
Td = interp(1,Ky,T);
#At Td K,
Kd = [2.85, 1.25, 0.59];
#Calculation of liquid composition
x1 = y[0]*100/Kd[0];
x2 = y[1]*100/Kd[1];
x3 = y[2]*100/Kd[2];
print ' (b).'
print ' The dew point temperature is %f K'%Td
print ' Liquid at dew point contains %f percent propane, %f percent butane and %f percent pentane'%(x1,x2,x3)
#(c). Temperature and composition when 45% of initial mixture is vaporised
#Basis:
F = 100;
V = 45;
L = 55;
#For the given condition eq. 8.91 (Page no. 364) is to be satisfied
#sum(zi/(1+ L/(VKi))) = 0.45
z = [0.25, 0.4, 0.35];
T = [366.5 ,377.6]; #assumed temperatures
K1 = [2.3 ,0.9 ,0.4]; #at 366.5 K
K2 = [2.6 ,1.1 ,0.5]; #at 377.6 K
Kz = [0 ,0];
for i in range(3):
Kz[0] = Kz[0] + z[i]/(1 + L/(V*K1[i]));
Kz[1] = Kz[1] + z[i]/(1 + L/(V*K2[i]));
#The required temperature is T3
T3 = interp(.45,Kz,T);
#At T3 K
K3 = [2.5, 1.08, 0.48];
#Calculating liquid and vapour compositions
for i in range(3):
y[i] = (z[i]/(1 + L/(V*K3[i])))/0.45;
x[i] = ((F*z[i]) - (V*y[i]))/L;
print (x[i]);
print ' (c).'
print ' The equilibrium temperature is %f K'%T3
print ' Liquid composition in equilibrium is %f percent propane, %f percent butane \
and %f percent pentane'%(x[0]*100,x[1]*100,x[2]*100)
print ' Vapour composition in equilibrium is %f percent propane, %f percent butane \
and %f percent pentane'%(y[0]*100,y[1]*100,y[2]*100);
Example 8.23¶
In [4]:
from numpy import array
# Variables
P = 101.3; #total pressure (kPa)
x1 = array([0.003, 0.449, 0.700, 0.900])
y1 = array([0.432, 0.449, 0.520, 0.719])
P1 = [65.31 ,63.98, 66.64, 81.31]; #(kPa)
P2 = [68.64 ,68.64, 69.31, 72.24]; #(kPa)
import math
# Calculations
#To test whether the given data are thermodynamically consistent or not
x2 = 1-x1;
y2 = 1-y1;
g1= [0,0,0,0]
g2 = [0,0,0,0]
c=[0,0,0,0]
import math
for i in range(4):
g1[i] = (y1[i]*P)/(x1[i]*P1[i])
g2[i] = (y2[i]*P)/(x2[i]*P2[i])
c[i] = math.log(g1[i]/g2[i]) #k = ln (g1/g2)
from matplotlib.pyplot import *
plot(x1,c)
#a = get("current_axes"
#set(a,"x_location","origin"
# Results
show()
#As seen from the graph net area is not zero
print 'The given math.experimental data do not satisfy the Redlich-Kistern criterion'
Example 8.25¶
In [44]:
# Variables
x1 = [0.0331, 0.9652]; #composition of chloroform
P = [40.84 ,84.88]; #total pressure for system (kPa)
P1 = 82.35; #vapour pressure of chloroform at 328 K (kPa)
P2 = 37.30; #vapour pressure of acetone at 328 K (kPa)
# Calculations
#To estimate the constants in Margules equation
#Using eq. 8.103 and 8.104 (Page no. 375)
g1_inf = (P[0]-(1-x1[0])*P2)/(x1[0]*P1)
g2_inf = (P[1]-(x1[1]*P1))/((1-x1[1])*P2)
import math
A = math.log(g1_inf)
B = math.log(g2_inf)
# Results
print 'Margules constants are:'
print ' A = %f'%A
print ' B = %f'%B
Example 8.26¶
In [5]:
# Variables
x1 = [0., 0.033, 0.117, 0.318, 0.554, 0.736, 1.000]; #liquid composition of acetone
pp1 = [0.,25.33, 59.05, 78.37, 89.58,94.77, 114.63]; #partial pressure of acetone (kPa)
Pw = 19.91; #vapour pressure of water at 333 K (kPa)
k = [0,0,0,0,0,0,0]
# Calculations
for i in range(1,6):
k[i] = x1[i]/((1.-x1[i])*pp1[i])
k[6] = 0.1; #k(7) should tend to infinity
from matplotlib.pyplot import *
plot(pp1,k)
show()
#From graph% area gives the integration and hence partiaal pressure of water is calculated
pp2 = [19.91, 19.31, 18.27, 16.99, 15.42, 13.90, 0];
# Results
print "The results are:"
print ' Acetone composition Partial pressure of water'
for i in range(7):
print ' %f %f'%(x1[i],pp2[i])
Example 8.27¶
In [46]:
# Variables
P = 93.30; #total pressure in kPa
T1 = 353.; #(K)
T2 = 373.; #(K)
Pa1 = 47.98; #Vapour pressure of water at 353 K (kPa)
Pb1 = 2.67; #Vapour pressure of liquid at 353 K (kPa)
Pa2 = 101.3; #Vapour pressure of water at 373 K (kPa)
Pb2 = 5.33; #Vapour pressure of liquid at 373 K (kPa)
# Calculations and Results
#To calculate under three phase equilibrium:
#(a). The equilibrium temperature
P1 = Pa1+Pb1; #sum of vapour pressures at 353 K
P2 = Pa2+Pb2; #at 373 K
#Since vapour pressure vary linearly with temperature% so T at which P = 93.30 kPa
T = T1 + ((T2-T1)/(P2-P1))*(P-P1)
print '(a). The equilibrium temperature is %f K'%T
#(b). The composition of resulting vapour
#At equilibrium temp:
Pa = 88.5; #vapour pressure of water (kPa)
Pb = 4.80; #vapour pressure of liquid (kPa)
#At 3-phase equilibrium% ratio of mol fractions of components is same as the ratio of vapour pressures
P = Pa+Pb; #sum of vapour pressures
y = Pa/P; #mole fraction of water
print ' The vapour contains %f mol percent water vapour'%(y*100)
Example 8.28¶
In [47]:
# Variables
T = [323 ,333, 343, 348, 353, 363, 373]; #temperatures (K)
P2 = [12.40 ,19.86, 31.06, 37.99, 47.32, 70.11, 101.3]; #vapour pressure for benzene (kPa)
P1 = [35.85 ,51.85, 72.91, 85.31, 100.50, 135.42, 179.14]; #vapour pressure for water (kPa)
Tb = 353.1; #boiling temperature (K)
Pb = 101.3; #boiling pressure (kPa)
# Calculations and Results
#To prepare temperature composition diagram
#To find three phase temperature
P = [0,0,0,0,0,0,0]
for i in range(7):
P[i] = P1[i] + P2[i];
from matplotlib.pyplot import *
plot(P,T)
#From graph, at P = 101.3 kPa..
T_ = 340.; #three phase temperature
#At three phase temperature
P1_ = 71.18; #(kPa)
P2_ = 30.12; #(kPa)
xb_ = P1_/Pb; #mol fraction of benzene at triple point
#For the dew point curve
#For curve BE in temp range from 342 to 373 K
y1 = [0,0,0,0,0,0,0]
for i in range(2,7):
y1[i] = 1-(P2[i]/Pb )
#xset('window',1
T1 = [0,0,0,0,0,0,0]
y1_ = [0,0,0,0,0,0,0]
T1[0] = 342
y1_[0] = 0.7;
for i in range(1,6):
T1[i] = T[i+1];
y1_[i] = y1[i+1];
plot(y1_,T1)
y2 = [0,0,0,0,0,0,0]
#For the curve Ae in the temp range of 342 K to 353.1 K
for i in range(2,5):
y2[i] = P1[i]/Pb;
T2 = [0,0,0,0,0,0,0]
y2_ = [0,0,0,0,0,0,0]
T2[0] = 342.;
y2_[0] = 0.7;
for i in range(1,4):
T2[i] = T[i+1]
y2_[i] = y2[i+1]
plot(y2_,T2)
#axhspan(0.25,0.75,facecolor='0.5'
#title("Temperature Composition diagram","xa,ya","Temperature"
show()
