Chapter 13 : Fugacity of a Component in a Mixture by Equations of State¶

Example 13.2 Page Number : 433¶

In [1]:

import math

# Variables
T = 310.93;			#[K] - Temperature
P = 2.76*10**(6);			#[Pa] - Pressure
y1 = 0.8942;			#[mol] - mole fraction of component 1
y2 = 1 - y1;			#[mol] - mole fraction of component 2
R=8.314;			#[J/mol*K] - Universal gas constant

#For component 1 (methane)
Tc_1 = 190.6;			#[K] - Critical temperature
Pc_1 = 45.99*10**(5);			#[N/m**(2)] - Critical pressure
Vc_1 = 98.6;			#[cm**(3)/mol] - Critical molar volume
Zc_1 = 0.286;			# - Critical compressibility factor
w_1 = 0.012;			# - Critical acentric factor
#Similarly for component 2 (n-Butane)
Tc_2 = 425.1;			#[K]
Pc_2 = 37.96*10**(5);			#[N/m**(2)]
Vc_2 = 255;			#[cm**(3)/mol]
Zc_2=0.274;
w_2=0.2;

# Calculations
#For component 1
Tr_1 = T/Tc_1;			#Reduced temperature
#At reduced temperature
B1_0 = 0.083-(0.422/(Tr_1)**(1.6));
B1_1 = 0.139-(0.172/(Tr_1)**(4.2));
#We know,(B*Pc)/(R*Tc) = B_0+(w*B_1)
B_11 = ((B1_0+(w_1*B1_1))*(R*Tc_1))/Pc_1;			#[m**(3)/mol]

#Similarly for component 2
Tr_2 = T/Tc_2;			#Reduced temperature
#At reduced temperature Tr_2,
B2_0 = 0.083-(0.422/(Tr_2)**(1.6));
B2_1 = 0.139-(0.172/(Tr_2)**(4.2));
B_22 = ((B2_0+(w_2*B2_1))*(R*Tc_2))/Pc_2;			#[m**(3)/mol]

#For cross coeffcient
Tc_12 = (Tc_1*Tc_2)**(1./2);			#[K]
w_12 = (w_1+w_2)/2;
Zc_12 = (Zc_1+Zc_2)/2;
Vc_12 = (((Vc_1)**(1./3)+(Vc_2)**(1./3))/2)**(3);			#[cm**(3)/mol]
Vc_12 = Vc_12*10**(-6);			#[m**(3)/mol]
Pc_12 = (Zc_12*R*Tc_12)/Vc_12;			#[N/m**(2)]

# Variables, Z = 1 + (B*P)/(R*T)
#Now we have,(B_12*Pc_12)/(R*Tc_12) = B_0 + (w_12*B_1)
#where B_0 and B_1 are to be evaluated at Tr_12
Tr_12 = T/Tc_12;
#At reduced temperature Tr_12
B_0 = 0.083-(0.422/(Tr_12)**(1.6));
B_1 = 0.139-(0.172/(Tr_12)**(4.2));
B_12 = ((B_0+(w_12*B_1))*R*Tc_12)/Pc_12;			#[m**(3)/mol]
#For the mixture
B = y1**(2)*B_11+2*y1*y2*B_12+y2**(2)*B_22;			#[m**(3)/mol]

#Now del_12 can be calculated as,
del_12 = 2*B_12 - B_11 - B_22;			#[m**(3)/mol]

#We have the relation, math.log(phi_1) = (P/(R*T))*(B_11 + y2**(2)*del_12), therefore
phi_1 = math.exp((P/(R*T))*(B_11 + y2**(2)*del_12));
#Similarly for component 2
phi_2 = math.exp((P/(R*T))*(B_22 + y1**(2)*del_12));

# Results
print " The value of fugacity coefficient of component 1 phi_1) is %f"%(phi_1);
print " The value of fugacity coefficient of component 2 phi_2) is %f"%(phi_2);

#Finally fugacity coefficient of the mixture is given by
#math.log(phi) = y1*math.log(phi_1) + y2*math.log(phi_2);
phi = math.exp(y1*math.log(phi_1) + y2*math.log(phi_2));

print " The value of fugacity coefficient of the mixture phi) is %f "%(phi);
#The fugacity coefficient of the mixture can also be obtained using
#math.log(phi) = (B*P)/(R*T)

 The value of fugacity coefficient of component 1 phi_1) is 0.965152
The value of fugacity coefficient of component 2 phi_2) is 0.675374
The value of fugacity coefficient of the mixture phi) is 0.929376


Example 13.7 Page Number : 447¶

In [2]:

from scipy.optimize import fsolve
import math

# Variables
T = 460.;			#[K] - Temperature
P = 40.*10**(5);			#[Pa] - Pressure
R=8.314;			#[J/mol*K] - Universal gas constant
# component 1 = nitrogen
# component 2 = n-Butane
y1 = 0.4974;			# Mole percent of nitrogen
y2 = 0.5026;			# Mole percent of n-Butane
Tc_nit = 126.2;			#[K]
Pc_nit = 34.00*10**(5);			#[Pa]
Tc_but = 425.1;			#[K]
Pc_but = 37.96*10**(5);			#[Pa]

# (1). van der Walls equation of state

# The fugacity coefficient of component 1 in a binary mixture following van der Walls equation of state is given by,
# math.log(phi_1) = b_1/(V-b) - math.log(Z-B) -2*(y1*a_11 + y2*a_12)/(R*T*V)
# and for component 2 is given by,
# math.log(phi_2) = b_2/(V-b) - math.log(Z-B) -2*(y1*a_12 + y2*a_22)/(R*T*V)
# Where B = (P*b)/(R*T)

# Calculations
# For componenet 1 (nitrogen)
a_1 = (27*R**(2)*Tc_nit**(2))/(64*Pc_nit);			#[Pa-m**(6)/mol**(2)]
b_1 = (R*Tc_nit)/(8*Pc_nit);			#[m**(3)/mol]

# Similarly for componenet 2 (n-Butane)
a_2 = (27*R**(2)*Tc_but**(2))/(64*Pc_but);			#[Pa-m**(6)/mol**(2)]
b_2 = (R*Tc_but)/(8*Pc_but);			#[m**(3)/mol]

# Here
a_11 = a_1;
a_22 = a_2;
# For cross coefficient
a_12 = (a_1*a_2)**(1./2);			#[Pa-m**(6)/mol**(2)]

# For the mixture
a = y1**(2)*a_11 + y2**(2)*a_22 + 2*y1*y2*a_12;			#[Pa-m**(6)/mol**(2)]
b = y1*b_1 + y2*b_2;			#[m**(3)/mol]

# The cubic form of the van der Walls equation of state is given by,
# V**(3) - (b+(R*T)/P)*V**(2) + (a/P)*V - (a*b)/P = 0
# Substituting the value and solving for V, we get
# Solving the cubic equation
def f(V):
return V**(3)-(b+(R*T)/P)*V**(2)+(a/P)*V-(a*b)/P
V_1=fsolve(f,-1)
V_2=fsolve(f,0)
V_3=fsolve(f,1)
# The molar volume V = V_3, the other two roots are imaginary
V = V_3;			#[m**(3)/mol]

# The comprssibility factor of the mixture is
Z = (P*V)/(R*T);
# And B can also be calculated as
B = (P*b)/(R*T);

# The fugacity coefficient of component 1 in the mixture is
phi_1 = math.exp(b_1/(V-b) - math.log(Z-B) -2*(y1*a_11 + y2*a_12)/(R*T*V));
# Similarly fugacity coefficient of component 2 in the mixture is
phi_2 = math.exp(b_2/(V-b) - math.log(Z-B) -2*(y1*a_12 + y2*a_22)/(R*T*V));

# The fugacity coefficient of the mixture is given by,
# math.log(phi) = y1*math.log(phi_1) + y2*math.log(phi_2)
phi = math.exp(y1*math.log(phi_1) + y2*math.log(phi_2));

# Also the fugacity coefficient of the mixture following van der Walls equation of state is given by,
# math.log(phi) = b/(V-b) - math.log(Z-B) -2*a/(R*T*V)
phi_dash = math.exp(b/(V-b) - math.log(Z-B) -2*a/(R*T*V));
# The result is same as obtained above

# Results
print " 1van der Walls equation of state";
print "  The value of fugacity coefficient of component 1 nitrogen) is %f"%(phi_1);
print "  The value of fugacity coefficient of component 2 n-butane) is %f"%(phi_2);
print "  The value of fugacity coefficient of the mixture is %f"%(phi);
print "  Also the fugacity coefficient of the mixture from van der Walls equation of state is %f which is same as above)"%(phi_dash);

# (2). Redlich-Kwong equation of state

# For component 1,
a_1_prime = (0.42748*R**(2)*Tc_nit**(2.5))/Pc_nit;			#[Pa-m**(6)/mol**(2)]
b_1_prime = (0.08664*R*Tc_nit)/Pc_nit;			#[m**(3)/mol]

#similarly for component 2,
a_2_prime = (0.42748*R**(2)*Tc_but**(2.5))/Pc_but;			#[Pa-m**(6)/mol**(2)]
b_2_prime = (0.08664*R*Tc_but)/Pc_but;			#[m**(3)/mol]

# For cross coefficient
a_12_prime = (a_1_prime*a_2_prime)**(1./2);			#[Pa-m**(6)/mol**(2)]
# For the mixture
a_prime = y1**(2)*a_1_prime + y2**(2)*a_2_prime +2*y1*y2*a_12_prime;			#[Pa-m**(6)/mol**(2)]
b_prime = y1*b_1_prime +y2*b_2_prime;			#[m**(3)/mol]

#The cubic form of Redlich Kwong equation of state is given by,
#  V**(3)-((R*T)/P)*V**(2)-((b**(2))+((b*R*T)/P)-(a/(T**(1/2)*P))*V-(a*b)/(T**(1/2)*P)=0
# Solving the cubic equation
def f1(V):
return V**(3)-((R*T)/P)*V**(2)-((b_prime**(2))+((b_prime*R*T)/P)-(a_prime/(T**(1./2)*P)))*V-(a_prime*b_prime)/(T**(1./2)*P)
V_4=fsolve(f1,1)
V_5=fsolve(f1,0)
V_6=fsolve(f1,-1)
# The molar volume V = V_4, the other two roots are imaginary
V_prime = V_4;			#[m**(3)/mol]

# The compressibility factor of the mixture is
Z_prime = (P*V_prime)/(R*T);
# And B can also be calculated as
B_prime = (P*b_prime)/(R*T);

# The fugacity coefficient of component 1 in the binary mixture is given by
# math.log(phi_1) = b_1/(V-b) - math.log(Z-B) + ((a*b_1)/((b**(2)*R*T**(3/2))))*(math.log((V+b)/V)-(b/(V+b)))-(2*(y1*a_1+y2*a_12)/(R*T**(3/2)b))*(math.log(V+b)/b)

phi_1_prime = math.exp((b_1_prime/(V_prime-b_prime))-math.log(Z_prime-B_prime)+((a_prime*b_1_prime)/((b_prime**(2))*R*(T**(3./2))))*(math.log((V_prime+b_prime)/V_prime)-(b_prime/(V_prime+b_prime)))-(2*(y1*a_1_prime+y2*a_12_prime)/(R*(T**(3./2))*b_prime))*(math.log((V_prime+b_prime)/V_prime)));

# Similarly fugacity coefficient of component 2 in the mixture is
phi_2_prime = math.exp((b_2_prime/(V_prime-b_prime))-math.log(Z_prime-B_prime)+((a_prime*b_2_prime)/((b_prime**(2))*R*(T**(3./2))))*(math.log((V_prime+b_prime)/V_prime)-(b_prime/(V_prime+b_prime)))-(2*(y1*a_12_prime+y2*a_2_prime)/(R*(T**(3./2))*b_prime))*(math.log((V_prime+b_prime)/V_prime)));

# The fugacity coefficient of the mixture is given by,
# math.log(phi) = y1*math.log(phi_1) + y2*math.log(phi_2)
phi_prime = math.exp(y1*math.log(phi_1_prime) + y2*math.log(phi_2_prime));

# Also the fugacity coefficient for the mixture following Redlich-kwong equation of state is also given by
# math.log(phi) = b/(V-b) - math.log(Z-B) - (a/(R*T**(3/2)))*(1/(V+b)+(1/b)*math.log((V+b)/V))
phi_prime_dash = math.exp(b_prime/(V_prime-b_prime) - math.log(Z_prime-B_prime) - (a_prime/(R*T**(3./2)))*(1/(V_prime+b_prime)+(1./b_prime)*math.log((V_prime+b_prime)/                   V_prime)));

print "  \nRedlich-Kwong equation of state";
print "  The value of fugacity coefficient of component 1 nitrogen) is %f"%(phi_1_prime);
print "  The value of fugacity coefficient of component 2 n-butane) is %f"%(phi_2_prime);
print "  The value of fugacity coefficient of the mixture is %f"%(phi_prime);
print "  Also the fugacity coefficient for the mixture from Redlich-kwong equation of state is %f which is same as above)"%(phi_prime_dash);

 1van der Walls equation of state
The value of fugacity coefficient of component 1 nitrogen) is 1.057500
The value of fugacity coefficient of component 2 n-butane) is 0.801865
The value of fugacity coefficient of the mixture is 0.920192
Also the fugacity coefficient of the mixture from van der Walls equation of state is 0.920192 which is same as above)

Redlich-Kwong equation of state
The value of fugacity coefficient of component 1 nitrogen) is 1.071129
The value of fugacity coefficient of component 2 n-butane) is 0.793063
The value of fugacity coefficient of the mixture is 0.920948
Also the fugacity coefficient for the mixture from Redlich-kwong equation of state is 0.920948 which is same as above)