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//Chapter-18,Example 5,Page 406
clc();
close();
H2 =0.24 //composition of H2
CH4 =0.3 //composition of CH4
CO =0.06 //composition of CO
C2H6 =0.11 //composition of C2H6
C2H4 =0.045 //composition of C2H4
C4H8 =0.025 //composition of C4H8
N2=0.12 //composition of N2
CO2=0.08 //composition of CO2
O2=0.02 //composition of O2
//for reaction H2 + (1/2)O2 = H2O
V1=H2*(1/2) //volume of O2 required
//for reaction CH4 + 2O2 = CO2 + 2H2O
V2=CH4*2 //volume of O2 required
vCO2_1=CH4*1 //volume of CO2
//for reaction C2H6 + (7/2)O2 = 2CO2 +3H2O
V3=C2H6*(7/2) //volume of O2 required
vCO2_2=C2H6*2 //volume of CO2
//for reaction C2H4 + 3O2 = 2CO2 +2H2O
V4=C2H4*3 //volume of O2 required
vCO2_3=C2H4*2 //volume of CO2
//for reaction C4H8 + 6O2 = 4CO2 +4H2O
V5=C4H8*6 //volume of O2 required
vCO2_4=C4H8*4 //volume of CO2
//for reaction CO + (1/2)O2 = CO2
V6=CO*(1/2) //volume of O2 required
vCO2_5=CO*1 //volume of CO2
total_O2= V1+V2+V3+V4+V5+V6-O2 //total volume of oxygen
//as air contains 21% of O2 by volume
//when 40% excess
V_air = total_O2*(100/21)*(140/100) //volume of air
printf("the air to fuel ratio is %.3f",V_air)
total_CO2 = vCO2_1+vCO2_2+vCO2_3+vCO2_4+vCO2_5+CO2 //total volume of CO2
total_dry= total_CO2 +[N2+(79*V_air/100)]+[(V_air*21/100)-total_O2]
printf("\n the total volume of dry products is %.4f cubicmeter ",total_dry)
CO2_dry =total_CO2*100/total_dry
N2_dry =[N2+(79*V_air/100)]*100/total_dry
O2_dry =[(V_air*21/100)-total_O2]*100/total_dry
printf("\n Composition of products of combustion on dry basis")
printf("\n CO2 = %.3f",CO2_dry)
printf("\n N2 = %.3f",N2_dry)
printf("\n O2 = %.3f",O2_dry)
//calculation mistake in textbook
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