Chapter 11 : Chemical Thermodynamics and Equilibrium

Example 11.1 pg : 287

In [4]:
			
# Variables
x = 1.5
P = 14.696 	    		#psia
m = 28.96
			
# Calculations
mf = 114.    			# lbm/mol fuel
ma = x*12.5*(1+3.76)*m
AF = ma/mf
n1 = 8.
n2 = 9.
n3 = (x-1)*12.5 
n4 =  x*3.76*12.5
np = n1+n2+n3+n4
x1 = n1/np
x2 = n2/np
x3 = n3/np
x4 = n4/np
ph = x2*P
Td = 113.5 			#F
			
# Results
print "Air fuel ratio  =  %.1f lbm air/lbm fuel"%(AF)
print " Mole fraction of CO2  =  %.2f percent"%(x1*100)
print " Mole fraction of H2O  =  %.2f percent"%(x2*100)
print " Mole fraction of O2  =  %.2f percent"%(x3*100)
print " Mole fraction of N2  =  %.2f percent"%(x4*100)
print ("From tables of saturation pressure")
print "Dew point   =  %.1f F"%(Td)

Air fuel ratio  =  22.7 lbm air/lbm fuel
 Mole fraction of CO2  =  8.53 percent
 Mole fraction of H2O  =  9.60 percent
 Mole fraction of O2  =  6.67 percent
 Mole fraction of N2  =  75.20 percent
From tables of saturation pressure
Dew point   =  113.5 F

Example 11.2 pg : 290

In [5]:
			
# Variables
x1 = 9.
x2 = 1.2
x3 = 1.5
x4 = 88.3
			
# Calculations
a = x1+x2
b = 2*a
xO = (2*x1 + x2+ 2*x3 + b)/2
xN = x4/3.76
ratio = xO/a
percent = ratio/2 *100
			
# Results
print "Percent theoretical air  =  %.1f percent"%(percent)

Percent theoretical air  =  104.4 percent

Example 11.3 pg : 291

In [6]:
			
# Variables
T = 440. 			#F
			
# Calculations
h1 = -169290
h2 = 7597.6
h3 = 4030.2
ht = h1+h2-h3
			
# Results
print "Molal enthalpy of CO2  =  %d Btu/lbm mole"%(ht)

Molal enthalpy of CO2  =  -165722 Btu/lbm mole

Example 11.4 pg : 291

In [7]:
			
# Variables
T = 77. 	    		#F
			
# Calculations
Hr = -36420. 			#B
hc = -169290. 			#B/lb mol
hh = -122970. 			#B/lb mol
Hp = 2*hc+3*hh
Q = Hp-Hr
			
# Results
print "Heat transfer  =  %d B/mol fuel"%(Q)

Heat transfer  =  -671070 B/mol fuel

Example 11.5 pg : 294

In [9]:
			
# Variables
T2 = 440. 			#F
T1 = 77. 			#F
Mch4 = 16.
Mw = 18.
			
# Calculations
h77 = 3725.1
ht = 6337.9
ht2 = 7597.6
h772 = 4030.2
hwt = 1260.3
h77w = 45.02
hr77 = -383040. 			#B/lbm mol
dHR = 1*Mch4*0.532*(T1-T2) + 2*(h77-ht)
dHp = 1*(ht2-h772) + 2*Mw*(hwt - h77w)
hrp = dHp+hr77+dHR
			
# Results
print "Enthalpy of combustion of gaseous methane  =  %d B/lbm mol fuel"%(hrp)

#The calculation in textbook is wrong Please check it using a calculator.

Enthalpy of combustion of gaseous methane  =  -344037 B/lbm mol fuel

Example 11.6 pg : 295

In [10]:
			
# Variables
Hr = -107530. 			#B/mol fuel
print ("By iteration of temperatures, T = 2700 R")
T = 2700. 			#R
			
# Results
print "Adiabatic flame temperature  =  %d R"%(T)

By iteration of temperatures, T = 2700 R
Adiabatic flame temperature  =  2700 R

Example 11.7 pg : 306

In [1]:
from numpy import poly1d,roots

# Variables
import math 
Kp = 0.668
y = Kp**2
			
# Calculations
x = poly1d(0)
vec = roots([y,2,-y,-2,0])        #x**3 + y*x**3 + 2*y*x**2 -y*x -2*y)

eps = vec[0]
x1 = (1-eps)/(1+ eps/2)
x2 = eps/(1+eps/2)
x3 = eps/2/(1+ eps/2)
			
# Results
print "degree of reaction  =  %.3f "%(eps)
print " Equilibrium concentration of CO2  =  %.3f "%(x1)
print " Equilibrium concentration of CO  =  %.3f "%(x2)
print " Equilibrium concentration of O2  =  %.3f "%(x3)

#the answers are different due to approximation in textbook

degree of reaction  =  -4.482 
 Equilibrium concentration of CO2  =  -4.417 
 Equilibrium concentration of CO  =  3.612 
 Equilibrium concentration of O2  =  1.806

Example 11.8 pg : 307

In [2]:
from numpy import roots

# Variables
Kp = 15.63
y = Kp
			
# Calculations
vec = roots([y+1,0,-y])#x**2 + y*x**2 - y)
eps = vec[0]
x1 = (1-eps)/(1+eps)
x2 = eps/(1+eps)
x3 = eps/(1+eps)
			
# Results
print " Equilibrium concentration of Cs  =  %.4f "%(x1)
print " Equilibrium concentration of Cs+  =  %.4f "%(x2)
print " Equilibrium concentration of e-  =  %.4f "%(x3)

#the answers are a bit different due to approximation in textbook

 Equilibrium concentration of Cs  =  0.0155 
 Equilibrium concentration of Cs+  =  0.4922 
 Equilibrium concentration of e-  =  0.4922