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clc
disp("(i) The stoichiometric A/F ratio")
// 1 kg of coal contains 0.82 kg C and 0.10 kg H2.
// Let the oxygen required for complete combustion = x moles
// the nitrogen supplied with the oxygen = x*79/21=3.76*x
// 0.82/12*C+0.10/2*H2 + x CO2 + 3.76x N2 → a CO2 + b H2O + 3.76 x N2
a=0.82/12; // Carbon balance
b=0.10/2; //Hydrogen balance
x=(2*a+b)/2; // Oxygen balance
Stoichiometric_AF_ratio=2.976/0.233;
disp("Stoichiometric AF ratio =")
disp(Stoichiometric_AF_ratio)
n=a+b+3.76*x;
CO2=0.068/n*100;
H2=0.05/n*100;
N2=3.76*0.093/n*100;
disp("the analysis of the products is")
disp("CO2 =")
disp(CO2)
disp("%")
disp("H2 =")
disp(H2)
disp("%")
disp("N2 =")
disp(N2)
disp("%")
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