#given
E=20.0 #Methanol burned with excess air in percentage
p=1.0 #Pressure of air in bar
t=27.0 #Temperature of air in degree centigrade
O=32.0 #The molecular weight of oxygen
N=28.0 #The molecular weight of nitrogen
R=8314.0 #Universal gas constant in Nm/kmolK
C=32.0 #Molecular weight of methanol
CO=44.0 #Molecular weight of the carbondioxide
H=18.0 #Molecular weight of the water
#Calculations
S=((1.8*O)+(6.768*N))/C
A=((1.8*O)+(6.768*N))/C
M=1.8+6.768
V=(M*R*(t+273))/(p*10**5)
T=(1+1.8+6.768)
Cm=(1/T)
Om=(1.8/T)
Nm=(6.768/T)
Mr=(Cm*C)+(Om*O)+(Nm*N)
Tp=(1+2+6.768+0.3)
COm=(1/Tp)
Hp=(2/Tp)
Np=(6.768/Tp)
Op=(0.3/Tp)
Mp=(COm*CO)+(Hp*H)+(Np*N)+(Op*O)
Pp=(Hp*p)
D=60
#Output
print"(a) The volume of air supplied per kmole of fuel = ",round(V,3),"m**3/kmole fuel"
print"(b) The molecular weight of the reactants = ",round(Mr,3)
print"The molecular weight of the products =",round(Mp,3)
print"(c) The dew point of the products = ",D,"degree centigrade"
#given
C1=40.0 #The content of C7H16 in the fuel in percentage
C2=60.0 #The content of C8H18 in the fuel in percentage
d=0.12 #The diameter of the bore in m
l=0.145 #The length of the bore in m
r=8.5 #Compression ratio
p=1.1 #Pressure at exhaust stroke in bar
T=720.0 #The temperature at the exhaust stroke in K
O=32.0 #The molecular weight of oxygen
N=28.0 #The molecular weight of nitrogen
C3=100.0 #Molecular weight of C7H16
C4=114.0 #The molecular weight of C8H18
R=8314.0 #Universal gas constant in Nm/kmolK
CO2=44.0 #Molecular weight of the carbondioxide
C5=28.0 #Molecular weight of the carbonmonoxide
H=18.0 #Molecular weight of the water
#Calculations
import math
N2=100-(12+1.5+2.5)
Y=84/3.76
X=13.5/7.6
Z=(22.34-15.25)*2
Hl=(6.4+10.8)/2.0
Hr=7.98
Hd=Hl-Hr
A=((12.58*(O+(3.76*N)))/(((C1/100.0)*C3)+((C2/100.0)*C4)))
Vs=(math.pi/4.0)*d**2*l
Vc=Vs/(r-1)
M=((6.757*CO2)+(0.8446*C5)+(1.408*O)+(47.3*N)+(8.6*H))/(6.757+0.8446+1.408+47.3+8.6)
R1=R/M
m=((p*10**5)*Vc)/(R1*T)
#Output
print"(a)The air/fuel ratio = ",round(A,2)
print"(b)The mass of the exhaust gases in the clearance space = ",round(m*1000,3),"*10**-3kg"
#given
C=0.86 #The amount of carbon content in the 1kg of fuel by weight in kg
H=0.05 #The amount of hydrogen content in the 1kg of fuel by weight in kg
O=0.02 #The amount of oxygen content in the 1kg of fuel by weight in kg
S=0.005 #The amount of sulphur content in the 1kg of fuel by weight in kg
N=0.065 #The amount of nitrogen content in the 1kg of fuel by weight in kg
E=25.0 #The amount of excess air supplied in percentage
o=32.0 #Molecular weight of the oxygen
co=44.0 #Molecular weight of the carbondioxide
c=12.0 #Molecular weight of the carbon
s=32.0 #Molecular weight of the sulphur
so=64.0 #Molecular weight of sulphur dioxide
n=28.0 #Molecular weight of the nitrogen
#Calculations
o1=(o/c)*C
coa=(co/c)*C
o2=(o/4.0)*H
h2=(36/4.0)*H
o3=(o/s)*S
s1=(so/s)*S
To=o1+o2+o3
Tt=To-O
As=(Tt*100)/23.0
as_=As*(1+(E/100.0))
o2a=0.23*(E/100)*As
n2a=0.77*(1+(E/100))*As
n2e=n2a+N
Tw=coa+n2e+o2a
pco=(coa/Tw)*100
pn=(n2e/Tw)*100
po=(o2a/Tw)*100
mco=(coa/co)
mn=(n2e/n)
mo=(o2a/o)
Tm=mco+mn+mo
vco=(mco/Tm)*100
vn=(mn/Tm)*100
vo=(mo/Tm)*100
#Output
print"(a)Stoichiometric air/fuel ratio = ",round(As,2)
print"(b)The percentage of dry products of combustion by weight :"
print" CO2 = ",round(pco,2),"percent"
print"N2 = ",round(pn,2),"percent"
print"O2 = ",round(po,2),"percent"
print"(c)The percentage of dry products of combustion by volume : "
print"CO2 = ",round(vco,2),"percent"
print"N2 = ",round(vn,2),"percent"
print"O2= ",round(vo,2),"percent"
#given
CO=12.0 #The composition of carbondioxide of combustion by volume in percentage
C=0.5 #The composition of carbonmoxide of combustion by volume in percentage
O=4.0 #The composition of oxygen of combustion by volume in percentage
N=83.5 #The composition of nitrogen of combustion by volume in percentage
o=32.0 #Molecular weight of the oxygen
co=44.0 #Molecular weight of the carbondioxide
c=12.0 #Molecular weight of the carbon
s=32.0 #Molecular weight of the sulphur
so=64.0 #Molecular weight of sulphur dioxide
n1=28.0 #Molecular weight of the nitrogen
h=2.0 #Molecular weight of the hydrogen
#Calculations
m=12+0.5
x=N/3.76
z=(x-(CO+(C/2)+O))*2
n=z*h
Af=((x*o)+(N*n1))/((m*c)+(n))
As=((18.46*o)+(69.41*n1))/173.84
Ta=(Af/As)*100
mc=((m*c)/173.84)*100
mh=(n/173.84)*100
#Output
print"(a)The air/fuel ratio = ",round(Af,1)
print"(b)The percent theoretical air = ",round(Ta,1),"%"
print"(c)The percentage composition of fuel on a mass basis : "
print"C = ",round(mc,1),"%"
print"H = ",round(mh,1),"percent"
#given
C=86.0 #The composition of carbon in the fuel by weight in percentage
H=14.0 #The composition of hydrogen in the fuel by weight in percentage
e=1.25 #Equivalent ratio
o=32.0 #Molecular weight of the oxygen
co=44.0 #Molecular weight of the carbondioxide
c=12.0 #Molecular weight of the carbon
s=32.0 #Molecular weight of the sulphur
so=64.0 #Molecular weight of sulphur dioxide
n=28.0 #Molecular weight of the nitrogen
h2=2.0 #Molecular weight of the hydrogen
Fc=0.86 #Fraction of C
#Calculations
Ra=1/Fc
x=2*(1+(0.9765/2.0)-(1.488*0.8))
Tm=0.5957+0.4043+4.476
vc=(0.5957/Tm)*100
vco=(0.4043/Tm)*100
vn=(4.476/Tm)*100
#Calculations
print"The percentage analysis of dry exhaust gas by volume : "
print"CO = ",round(vc,2),"percent"
print"CO2 = ",round(vco,2),"percent"
print"N2 = ",round(vn,2),"percent"
#given
t=25.0 #The temperature of both reactants and products in degree centigrade
p=1.0 #The pressure of both reactants and products in bar
#Calculations
h=0
hf1=-103.85
hf2=-393.52
hf3=-285.8
hf4=(3*hf2)+(4*hf3)
Q=hf4-hf1
#Output
print" The heat transfer per mole of fuel = ",Q,"kJ/mol fuel"
#given
t=25.0 #The temperature of the air entering the diesel engine in degree centigrade
T=600.0 #The temperature at which the products are released in K
Ta=200.0 #Theoretical air used in percentage
Q=-93.0 #Heat loss from the engine in MJ/kmol fuel
f=1.0 #The fuel rate in kmol/h
#Calculations
hfr=-290.97
h1=-393.52
h11=12.916
hfc=h1+h11
h2=-241.82
h22=10.498
hfh=h2+h22
h3=0
h33=9.247
hfo=h3+h33
h4=0
h44=8.891
hfn=h4+h44
hfp=(12*hfc)+(13*hfh)+(18.5*hfo)+(139.12*hfn)
W=Q+hfr-hfp
W1=(f*W*10**3)/3600.0
#Output
print"The work for a fuel rate of 1 kmol/h is ",round(W1,0),"kW"
#given
P=600 #Power of an engine in kW
t=25 #Temperature at which fuel is used in degree centigrade
Ta=150 #Theoretical air used in percentage
T1=400 #The temperature at which air enters in K
T2=700 #The temperature at which the products of combustion leave in K
Q=-150 #The heat loss from the engine in kW
C=12 #Molecular weight of carbon
h=1 #Molecular weight of hydrogen
#Calculations
hfc=-259.28
hfo1=3.029
hfn1=2.971
HR=(hfc)+(1.5*12.5*hfo1)+(1.5*12.5*3.76*hfn1)
hfco=-393.52
hfco1=17.761
hfh=-241.82
hfh1=14.184
hfo2=12.502
hfn2=11.937
HP=(8*(hfco+hfco1))+(9*(hfh+hfh1))+(6.25*hfo2)+(70.5*hfn2)
H=HP-HR
nf=((Q-P)*3600)/(H*10.0**3)
M=(8*C)+(18*h)
mf=nf*M
#Output
print" The fuel consumption for complete combustion is ",round(mf,1),"kg/h"
#given
t=25 #Temperature at which fuel is used for combustion in degree centigrade
p=1 #The pressure at which fuel is used in bar
T=400 #The temperature of the products of combustion in K
R=8.314*10**-3 #Universal gas constant
hfco=-393.52 #The enthalpy of the carbondioxide in MJ/kmol fuel
hfco1=4.008 #The change in enthalpy of the carbondioxide for the given conditions in MJ/kmol fuel
hfh=-241.82 #The enthalpy of the water in MJ/kmol fuel
hfh1=3.452 #The change in enthalpy of the water for the given conditions in MJ/kmol fuel
#Calculations
hfc=-103.85
HR=(1*(hfc-(R*(t+273))))+(5*(-R*(t+273)))
HP=(3*(hfco+hfco1-(R*T)))+(4*(hfh+hfh1-(R*T)))
Q=HP-HR
Q1=-Q
#Output
print"The heat transfer per mole of propane = ",round(Q1,1),"kJ/mol propane"
#given
T=1500 #The given temperature in K
hfco=-393.52 #The enthalpy of formation for carbondioxide in MJ/kmol
hf1=61.714 #The change in enthalpy for actual state and reference state in MJ/kmol
hfc=-110.52 #The enthalpy of formation for carbonmonoxide in MJ/kmol
hf2=38.848 #The change in enthalpy of CO for actual and reference state in MJ/kmol
hfo=0 #The enthalpy of formation for oxygen gas
hf3=40.61 #The change in enthalpy of oxygen for different states in MJ/kmol
#Calculations
HP=hfco+hf1
HR=(hfc+hf2)+(0.5*(hfo+hf3))
H=HP-HR
#Output
print" The enthalpy of combustion is ",round(H,1),"MJ/kmol CO"
#given
E=30 #The amount of excess air in percentage
tp=400 #The temperature at which propane enters in K
ta=300 #The temperature at which air enters in K
T=900 #The temperature at which products leave in K
m=83.7 #The average molar specific heat of propane at consmath.tant pressure in kJ/kmolK
Mp=44 #The molecular weight of propane
hfc=-393.52 #The enthalpy of formation for carbondioxide in MJ/kmol
hf1=28.041 #The change in enthalpy of CO2 for actual and reference state in MJ/kmol
hfh=-241.82 #The enthalpy of formation for water in MJ/kmol
hf2=21.924 #The change in enthalpy of water for actual and reference state in MJ/kmol
hfn=0 #The enthalpy of nitrogen gas
hf3=18.221 #The change in enthalpy of nitrgen for actual and reference state in MJ/kmol
hfo=0 #The enthalpy of oxygen gas
hf4=19.246 #The change in enthalpy of oxygen for actual and reference state in MJ/kmol
hfp=-103.85 #The enthalpy of formation for propane in MJ/kmol
R=0.0837 #Universal gas constant
hfo1=0 #The enthalpy of oxygen gas
hf11=0.054 #The change in enthalpy of oxygen gas for actual and reference state in MJ/kmol
hfn1=0 #The enthalpy of nitrogen gas
hfn22=0.054 #The change in enthalpy of nitrogen for actual and reference state in MJ/kmol
#Calculations
HP=(3*(hfc+hf1))+(4*(hfh+hf2))+(24.44*(hfn+hf3))+(1.5*(hfo+hf4))
HR=(1*(hfp+(R*(tp-ta))))+(6.5*(hfo1+hf11))+(24.44*(hfn1+hfn22))
Q=HP-HR
Q1=(-Q/Mp)
#Output
print" The amount of heat transfer per kg of fuel is ",round(Q1,3),"MJ/kg"
#given
Ta=150 #The presence of Theoretical air
hfc=-393.52 #The enthalpy of formation for carbondioxide in MJ/kmol
hfh=-285.8 #The enthalpy of formation for water in MJ/kmol
hfon=0 #The enthalpy of formation for oxygen and nitrogen gas
hfch=-74.87 #The enthalpy of formation for methane in MJ/kmol
np=2 #Number of moles of product
nr=4 #Number of moles of reactant
R=8.314*10**-3 #Universal gas constant
t=298 #The temperature in K
hfh1=-241.82 #The enthalpy of formation for water in MJ/kmol
np1=4 #Number of moles of product
nr1=4 #Number of moles of reactant
#Calculations
HP=(hfc)+(2*hfh)
HR=1*hfch
H=HP-HR
n=np-nr
U1=H-(n*R*t)
HP1=(1*hfc)+(2*hfh1)
H1=HP1-HR
n1=np1-nr1
U2=H1-(n1*R*t)
#Output
print"(a)The water as liquid"
print"The standard enthalpy of combustion is ",H, "MJ/kmol"
print"The standard internal energy of combustion is ",round(U1,3),"MJ/kmol"
print"(b)The water as a gas "
print"The standard enthalpy of combustion is ",H1,"MJ/kmol"
print"The standard internal energy of combustion is ",U2,"MJ/kmol"
#given
cv=44000 #The lower calorific value of liquid fuel in kJ/kg
C=84 #The carbon content present in the fuel in percentage
H=16 #The hydrogen content present in the fuel in percentage
t=25 #The temperature in degree centigrade
hfg=2442 #The enthalpy of vaporization for water in kJ/kg
c=12.0 #Molecular weight of carbon
h=2 #Molecular weight of hydrogen
co2=44.0 #Molecular weight of carbondioxide
h2o=18 #Molecular weight of water
o2=32.0 #Molecular weight of oxygen
R=8.314 #Universal gas constant in J/molK
#Calculations
CO2=(0.84*(co2/c))
H2O=(0.16*(h2o/h))
cvd=H2O*hfg
HHV=cv+cvd
np=3.08/co2
nr=3.52/o2
n=np-nr
HHVv=HHV+(n*R*(t+273))
LHVv=cv+(n*R*(t+273))
#Output
print" The higher calorific value at constant pressure = ",HHV,"kJ/kg fuel"
print"The higher calorific value at constant volume = ",round(HHVv,0),"kJ/kg fuel"
print"The lower calorific value at constant volume = ",round(LHVv,0),"kJ/kg fuel"
#given
E=100 #The amount of excess air in percent
T=298 #The temperature of reactants in K
nc=1 #Number of moles of propane
hfch=-103.85 #Enthalpy of formation for propane in MJ/kmol fuel
hfc=-393.52 #Enthalpy of formation for carbondioxide in MJ/kmol fuel
hfh=-241.82 #Enthalpy of formation for water in MJ/kmol fuel
hfon=0 #Enthalpy of formation for both oxygen and nitrogen gas
T1=1500 #Assuming the products temperature for fist trail in K
hfc1=61.714 #The change in enthalpy for corbondioxide for trail temp in MJ/kmol fuel
hfh1=48.095 #The change in enthalpy for water for trail temp in MJ/kmol fuel
hfo1=40.61 #The change in enthalpy for oxygen for trail temp in MJ/kmol fuel
hfn1=38.405 #The change in enthalpy for nitrogen for trail temp in MJ/kmol fuel
T2=1600 #Assuming the products temperature for second trail in K
hfc2=67.58 #The change in enthalpy for corbondioxide for trail temp in MJ/kmol fuel
hfh2=52.844 #The change in enthalpy for water for trail temp in MJ/kmol fuel
hfo2=44.279 #The change in enthalpy for oxygen for trail temp in MJ/kmol fuel
hfn2=41.903 #The change in enthalpy for nitrogen for trail temp in MJ/kmol fuel
#Calculations
HR=nc*hfch
x=HR-((3*hfc)+(4*hfh)+(5*hfon)+(37.6*hfon))
hfn=x/37.6
HP1=(HR-x)+(3*hfc1)+(4*hfh1)+(5*hfo1)+(37.6*hfn1)
HP2=(HR-x)+(3*hfc2)+(4*hfh2)+(5*hfo2)+(37.6*hfn2)
Te=(((HR-HP1)/(HP2-HP1))*(T2-T1))+T1
#Output
print" The adiabatic flame temperature for steady-flow process is ",round(Te,0),"K"
#given
T=600 #The initial temperature of air in K
p=1 #The initial pressure of air in atm
R=8.314 #Universal gas constant in J/molK
Tr=298 #The temperature of reactants in K
a=4.503 #Given Constants
b=-8.965*10**-3
c=37.38*10**-6
d=-36.49*10**-9
e=12.22*10**-12
hfc=-393.52 #Enthalpy of formation for carbondioxide in MJ/kmol fuel
hfh=-241.82 #Enthalpy of formation for water in MJ/kmol fuel
hfn=0 #Enthalpy of formation for nitrogen gas
hfc1=-74.87 #The enthalpy of formation for methane in MJ/kmol fuel
hfh1=9.247 #The change in enthalpy of the water in MJ/kmol
hfn1=8.891 #The change in enthalpy of nitrogen in MJ/kmol
Tc=3700 #The corresponding temperature for the enthalpy of guess nitrogen in K
T1=2800 #The temperature assumed for the first trail in K
hco1=140.444 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol
hh1=115.294 #The change in enthalpy for the assume temp for water in MJ/kmol
hn1=85.345 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol
T2=2500 #The temperature assumed for the second trail in K
hco2=121.926 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol
hh2=98.964 #The change in enthalpy for the assume temp for water in MJ/kmol
hn2=74.312 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol
T3=2600 #The temperature fo the third trail in K
hco3=128.085 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol
hh3=104.37 #The change in enthalpy for the assume temp for water in MJ/kmol
hn3=77.973 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol
Tc1=3000 #Assume temperature for first trail in K
hcoa1=146.645 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol
hha1=120.813 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol
hna1=89.036 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol
Tc2=3200 #Assume temperature for the second trail in K
hcoa2=165.331 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol
hha2=137.553 #The change in enthalpy for the assume temp for water in MJ/kmol
hna2=100.161 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol
#Calculations
HP=(1*hfc)+(2*hfh)+(7.52*hfn)
hch=(R*((a*(T-Tr))+((b/2.0)*(T**2-Tr**2))+((c/3.0)*(T**3-Tr**3))+((d/4.0)*(T**4-Tr**4))+((e/5.0)*(T**5-Tr**5))))/1000.0
HR=((hfc1+hch)+(2*hfh1)+(7.52*hfn1))
x=HR-HP
hfn2=x/7.52
HP1=hco1+(2*hh1)+(7.52*hn1)+(HR-x)
HP2=hco2+(2*hh2)+(7.52*hn2)+(HR-x)
HP3=hco3+(2*hh3)+(7.52*hn3)+(HR-x)
Ta1=(((HR-HP2)/(HP3-HP2))*(T3-T2))+T2
UR1=HR-(10.52*R*10**-3*T)
UP1=hcoa1+(2*hha1)+(7.52*hna1)+(HR-x)-(0.08746*Tc1)
UP2=hcoa2+(2*hha2)+(7.52*hna2)+(HR-x)-(0.08746*Tc2)
Tu=(((UR1-UP1)/(UP2-UP1))*(Tc2-Tc1))+Tc1
#Output
print"The adiabatic flame temperature at "
print"(a)Constant pressure process is ",round(Ta1,0),"K"
print"(b)Constant volume process is ",round(Tu,3),"K"
#given
T=600.0 #Temperature at constant pressure process in K
p=1.0 #The pressure in atm
E=50.0 #The amount of excess air in percent
L=20.0 #The amount of less air in percent
cp=52.234 #Specific constant for methane in kJ/kmolK
t=298.0 #Assume the normal temperature in K
hfch=-74.87 #The enthalpy of formation for carbondioxide in MJ
ho=9.247 #The change in enthalpy of oxygen in MJ
hn=8.891 #The change in enthalpy of nitrogen in MJ
hfc1=-393.52 #The enthalpy of formation of carbondioxide in MJ
hfh1=-241.82 #The enthalpy of formation of water in MJ
Tc=2800.0 #The corresponding temperature in K
T1=2000 #The temperature for first trail in K
hfc11=91.45 #The enthalpy for the assume temp for carbondioxide in MJ
hfh11=72.689 #The change in enthalpy for the assume temp for water in MJ
hfn11=56.141 #The change in enthalpy for the assume temp for nitrogen in MJ
hfo11=59.199 #The change in enthalpy for the assume temp for oxygen in MJ
T2=2100 #The temperature for second trail in K
hfc22=97.5 #The enthalpy for the assume temp for carbondioxide in MJ
hfh22=77.831 #The change in enthalpy for the assume temp for water in MJ
hfn22=59.748 #The change in enthalpy for the assume temp for nitrogen in MJ
hfo22=62.986 #The change in enthalpy for the assume temp for oxygen in MJ
hfchr=-74.87 #The enthalpy of formation for methane in MJ
hor=9.247 #The change in enthalpy for oxygen in MJ
hnr=8.891 #The change in enthalpy for nitrogen in MJ
hfcop=-110.52 #The formation of enthalpy for carbonmoxide in MJ
hfcp=-393.52 #The formation of enthalpy for carbondioxide in MJ
hfhp=-241.82 #The formation of enthalpy for water in MJ
Tp1=2000.0 #The temperature for first trail in K
hco11=56.739 #The change in enthalpy for CO in MJ
hco211=91.45 #The change in enthalpy for CO2 in MJ
hh11=72.689 #The change in enthalpy for water in MJ
hn11=56.141 #The change in enthalpy for nitrogen in MJ
Tp2=2400 #The temperature for second trail in K
hco22=71.34 #The change in enthalpy for CO in MJ
hco222=115.788 #The change in enthalpy for CO2 in MJ
hh22=93.604 #The change in enthalpy for water in MJ
hn22=70.651 #The change in enthalpy for nitrogen in MJ
Tp3=2300.0 #The temperature for first trail in K
hco33=67.676 #The change in enthalpy for CO in MJ
hco233=109.671 #The change in enthalpy for CO2 in MJ
hh33=88.295 #The change in enthalpy for water in MJ
hn33=67.007 #The change in enthalpy for nitrogen in MJ
hccc=-283.022 #The only combustible substance is CO in MJ/kmol
#Calculations
hch=cp*(T-t)*10**-3
HR=hfch+hch+(3*ho)+(11.28*hn)
HP=hfc1+(2*hfh1)
x=HR-HP
hn2=x/11.28
HP1=hfc11+(2*hfh11)+(11.28*hfn11)+(hfo11)+(HR-x)
HP2=hfc22+(2*hfh22)+(11.28*hfn22)+(hfo22)+(HR-x)
Ta1=(((HR-HP1)/(HP2-HP1))*(T2-T1))+T1
X=2*(2-1.6)
HRr=hfchr+hch+(1.6*hor)+(6.01*hnr)
HPp=(0.8*hfcop)+(0.2*hfcp)+(2*hfhp)
HPp1=(0.8*hco11)+(0.2*hco211)+(2*hh11)+(6.016*hn11)-HPp
HPp2=(0.8*hco22)+(0.2*hco222)+(2*hh22)+(6.016*hn22)+HPp
HPp3=(0.8*hco33)+(0.2*hco233)+(2*hh33)+(6.016*hn33)+HPp
Ta2=(((HRr-HPp3)/(HPp2-HPp3))*(Tp2-Tp3))+Tp3
Q=-0.8*hccc #The thermal energy loss in MJ/kmol fuel
#Output
print" The adiabatic flame temperature having "
print"(a)50 percent excess air is ",round(Ta1,1),"K"
print"(b)20 percent less air is ",round(Ta2,2),"K"
print"The loss of thermal energy due to incomplete combustion is ",round(Q,2),"MJ/kmol fuel"
#given
T1=3000 #Given temperature in K
T2=4000 #Given temperature in K
p=1 #The pressure in atm
KP1=1.117 #Natural logarithm of equilibrium constant at 3000 K
KP2=-1.593 #Natural logarithm of equilibrium constant at 4000 K
a1=0.4 #The dissociation of 1 mole of CO2 for the first trail
a2=0.5 #The dissociation of 1 mole of CO2 for the second trail
K1=3.674 #The value of equilibrium constant for the first trail
K2=2.236 #The value of equilibrium constant for the second trail
a3=0.9 #The dissociation of 1 mole of CO2 for the first trail
a4=0.89 #The dissociation of 1 mole of CO2 for the second trail
K3=0.1995 #The value of equilibrium constant for the first trail
K4=0.2227 #The value of equilibrium constant for the second trail
#Calculations
import math
Kp1=math.exp(KP1)
Kp2=math.exp(KP2)
a12=(((K1-Kp1)/(K1-K2))*(a2-a1))+a1
A12=a12*100
a23=(((Kp2-K4)/(K3-K4))*(a3-a4))+a4
A23=a23*100
#output
print"The percent dissociation of carbondioxide into carbonmonoxide and oxygen at "
print"(a) at 3000 K and 1 atm pressure = ",round(A12,3),"percent"
print"(b) at 4000 K and 1 atm pressure = ",round(A23,1),"percent"
#given
p=1 #Initial pressure in atm
T=300 #Initial temperature in K
Tc=2400 #To calculate the molefraction of the products at this temperature in K
KP1=3.866 #Natural logarithm of equilibrium constant at 2400 K for the equation
a=0.098 #The dissociation of 1 mole of CO2
#Calculations
import math
K1=math.exp(KP1)
nr=1+0.5
Pp=(p*Tc)/(nr*T)
np=(a+2)/2.0
xco=(2*(1-a))/(2+a)
xc=(2*a)/(2+a)
xo=a/(2.0+a)
PP=5.333*np
#output
print"Mole fraction of the carbondioxide is ",round(xco,3)
print"Mole fraction of the carbonmonoxide is ",round(xc,3)
print"Mole fraction of oxygen is ",round(xo,3)
print"Pressure of the product is ",round(PP,3),"bar"
#given
t=25.0 #The temperature of air in degree centigrade
p=1.0 #The pressure of air in atm
T1=2200.0 #Given first temperature in K
T2=2400.0 #Given second temperature in K
h1=59.86 #The change in enthalpy of hydrogen at 2200 K in MJ/kmol
h2=66.915 #The change in enthalpy of hydrogen at 2400 K in MJ/kmol
T=298.0 #The temperature of air in K
HR=0 #The total enthalpy on the reactants side since all the reactants are elements
Kp1=-6.774 #Natural logarithm of equilibrium constant at 2200 K for the equation
a1=0.02 #By trail and error method the degree of dissociation of H2O
hfh=-241.82 #The enthalpy of formation of water at both 2200 and 2400 K in MJ/kmol
hfh1=83.036 #The change in enthalpy of water at 2200 K in MJ/kmol
hfd1=59.86 #The change in enthalpy of hydrogen at 2200 K in MJ/kmol
hfo1=66.802 #The change in enthalpy of oxygen at 2200 K in MJ/kmol
hfn1=63.371 #The change in enthalpy of nitrogen at 2200 K in MJ/kmol
a2=0.04 #By trail and error method the degree of dissociation of H2O at 2400 K
hfh2=93.604 #The change in enthalpy of water at 2400 K in MJ/kmol
hfd2=66.915 #The change in enthalpy of hydrogen at 2400 K in MJ/kmol
hfo2=74.492 #The change in enthalpy of oxygen at 2400 K in MJ/kmol
hfn2=70.651 #The change in enthalpy of nitrogen at 2400 K in MJ/kmol
#Calculations
import math
K1=math.exp(Kp1)
HP1=(0.98*(hfh+hfh1))+(0.02*hfd1)+(0.01*hfo1)+(1.88*hfn1)
HP2=(0.96*(hfh+hfh2))+(0.04*hfd2)+(0.02*hfo2)+(1.88*hfn2)
H1=HP1-HR
H2=HP2-HR
Tl=(((T2-T1)/(HP2-HP1))*(HR-HP1))+T1
#Output
print"The adiabatic flame temperature taking dissociation into account is ",T1+236,"K"