Chapter 10 - Vapour liquid equillibrium¶
Example: 10.1 Page: 390¶
In [1]:
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: 10.2 Page: 399¶
In [2]:
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: 10.3 Page: 400¶
In [3]:
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: 10.4 Page: 400¶
In [1]:
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: 10.5 Page: 401¶
In [2]:
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: 10.6 Page: 403¶
In [9]:
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: 10.7 Page: 408¶
In [10]:
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: 10.8 Page: 410¶
In [11]:
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: 10.9 Page: 412¶
In [3]:
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: 10.10 Page: 412¶
In [4]:
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: 10.11 Page: 413¶
In [5]:
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: 10.12 Page: 414¶
In [15]:
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: 10.13 Page: 418¶
In [6]:
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: 10.14 Page: 426¶
In [7]:
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: 10.15 Page: 428¶
In [18]:
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: 10.16 Page: 429¶
In [19]:
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: 10.17 Page: 432¶
In [8]:
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: 10.18 Page: 433¶
In [10]:
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: 10.19 Page: 413¶
In [11]:
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"
