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+//clc()
+Nflue = 100;//kmol
+NCO2 = 9;
+NCO = 2;
+NO2 = 3;
+NN2 = 86;
+NCflue = NCO2 + NCO ;
+MC = 12;
+mC = MC * NCflue ;
+//let A kmol air supplied, taking N2 balance,
+Nair = NN2 * 100/79;
+NO2supplied = Nair - NN2;
+// if CO in the flue gas was to be completely converted to CO2, then, the moles of oxygen present in the flue gas would be 3-1 =2kmol
+Noexcess = NO2 - NCO/2;
+Pexcess =  Noexcess * 100 / ( NO2supplied - Noexcess);
+disp("%",Pexcess,"(a)Percentage excess air = ")
+NwaterO = NO2supplied - NCO2 - NCO/2 - NO2;
+NH2 = NwaterO*2;
+mH2 = NH2 * 2;
+xCF = 0.7
+R = mC / mH2;
+disp(R,"(b)Ratio of carbon to hydrogen in the fuel = ")
+//let x be the amount of moisture in the feed, n it is given that 70% is carbon,therefore,
+//0.7 = 3.32 / ( 1 + 3.32 + x )
+x = R / xCF - 1 - R;
+mH = x * 2.016 / 18.016;
+mHtotal = mH + mH2;
+Rtotal = mC / mHtotal;
+disp(Rtotal,"(c)Ratio of carbon to total hydrogen in the fuel = ")
+ntotal = R + 1  +x;
+PH2 = 1*100/ntotal;
+PH2O = x * 100 / ntotal;
+disp("%",PH2,"(d)percentage of combustible hydrogen in the fuel = ")
+disp("%",PH2O,"percentage of moisture in the fuel = ")
+nH2Ototal = (PH2O + PH2 * 18.016 / 2.016)/100;
+disp("kg",nH2Ototal,"(e)The mass of moisture in the flue gas per kg of fuel burned = ")