# Chapter6-Combustion Chambers and Afterburners¶

## Ex1-pg309¶

In [2]:
#determine number of mole of hydrogen and oxygen
nH2=12/2. ##molecular mass og hydrogen =2kg/kmol
nO2=8/32. ##Molecular mass of O2=32kg/kmol
print'%s %.f %s'%("No. of kilomoles of H2",nH2,"")
print'%s %.2f %s'%("No. of kilomoles of O2",nO2,"")

No. of kilomoles of H2 6
No. of kilomoles of O2 0.25


# Ex3-pg317¶

In [3]:
import math
#calculate lower and higher heating values of hydrogen
T=298.16 ##in K
dhf=-241827. ##heat of formation of H2O(g in kJ.
n=1 ##kmol
Qr=n*dhf ##kJ/kmol
LHV=(-1.)*Qr/2.
print'%s %.1f %s'%("LHV in",LHV,"kJ/kg")
HHV=LHV+9*2443
print'%s %.1f %s'%("HHV in ",HHV,"kJ/kg")

LHV in 120913.5 kJ/kg
HHV in  142900.5 kJ/kg


# Ex5-pg320¶

In [4]:
import math
#calcualte the ratio Nh2/no2 of the reactants and fuel oxdizer and adiabatic flame temperature
##from equation CH4+2.4(O2+3.76N2)-->CO2+2H2O+0.4O2+9.02N2
f=(12+4.)/(2.4*(32.+3.76*28.)) ##fuel to air ratio based on mass.
fs=(12+4.)/(2.*(32.+3.76*28.)) ##fuel to air ratio based on stoichometric condition.
feq=f/fs
print'%s %.7f %s'%("fuel to air ratio based on mass",f,"")
print'%s %.7f %s'%("fuel to air ratio based on stoichometric condition",fs,"")
print'%s %.7f %s'%("Equivalent ratio",feq,"")
dH=-802303 ##kJ
dC=484.7 ##kJ
Dt=(-1)*dH/dC ##Dt=T2-Tf
Tf=25+273
T2=Dt+Tf
print'%s %.4f %s'%("The diabatic flame temperature in",T2," K")

fuel to air ratio based on mass 0.0485625
fuel to air ratio based on stoichometric condition 0.0582751
Equivalent ratio 0.8333333
The diabatic flame temperature in 1953.2569  K


## Ex6-pg323¶

In [26]:
import math
import numpy
#calculate mole fraction of N2 at equlibrium when pm is 1 atm and 10 atms
print("Example 6.6")
Kp=0.1

pm=1.
y=2
d=numpy.roots(y)
x=0.1561738
print'%s %.2f %s '%("(a)Mole fraction of N2 at equilibrium when pm is 1 atm:",x,"")
#part (b)
Kp=0.1

pm=10.
x=0.0499376
y=- 0.1 + 40.1*x
d=numpy.roots(y)
i=numpy.linspace(1,2,num=1);
print'%s %.2f %s '%("(b)Mole fraction of N2 at equilibrium when pm is 10 atm:",x,"")


Example 6.6
(a)Mole fraction of N2 at equilibrium when pm is 1 atm: 0.16
(b)Mole fraction of N2 at equilibrium when pm is 10 atm: 0.05