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+
+clc
+clear
+//Input data
+C=84;//The mass of carbon present in the fuel in %
+H=10;//The mass of hydrogen present in the fuel in %
+S=3.2;//The mass of sulphur present in the fuel in %
+O=1.6;//The mass of oxygen present in the fuel in %
+I=1.2;//The mass of incombustible in the fuel in %
+X=15.72;//The flue gas of combined CO2 and SO2 by volume in %
+Og=1;//The flue gas of O2 by volume in %
+Y=100;//Let us consider the fuel oil in kg
+C1=12;//Molecular weight of Carbon 
+H1=2;//Molecular weight of hydrogen
+S1=32;//Molecular weight of sulphur
+O1=32;//Molecular weight of oxygen
+Co2=44;//Molecular weight of carbondioxide
+So2=64;//Molecular weight of sulphurdioxide
+N1=28;//Molecular weight of nitrogen 
+H2O=18;//Molecular weight of water
+
+//Calculations
+b=C/C1;//Equating coefficients of the carbon from equation
+g=H/H1;//Equating coefficients of the hydrogen from equation
+d=S/S1;//Equating coefficients of the sulphur from the equation
+e=(b+d)/(X/Og);//By volumetric analysis
+x=b+d+e+(g/2)-(O/O1);//Moles of oxygen are supplied for combustion
+f=3.76*x;//Equating coefficients of the nitrogen from equation
+Mo=x*O1;//Mass of oxygen supplied in kg
+Ma=Mo/0.232;//Mass of air supplied for 100 kg of fuel in kg
+Wa=Ma/100;//Mass of air supplied for 1 kg fuel in kg
+Wrh={(11.5*C)+(34.5*[(H)-(O/8)])+(4.3*S)}/100;//Theoretical air required per kg of fuel in kg
+E=[(Wa-Wrh)/Wrh]*100;//Percentage of excess air in %
+D=(b*Co2)+(d*So2)+(e*O1)+(f*N1);//Mass of dry flue gas formed for 100 kg fuel in kg
+dfg=D/100;//Mass of dry flue gas formed per kg of fuel in kg
+Mw=(g*H2O)/100;//Mass of water vapour formed per kg of fuel
+
+//Output
+printf('(a) Mass of air supplied WA = %3.2f kg \n (b)Percentage excess air supplied = %3.2f percentage \n (c)mass of dry flue gas formed = %3.2f kg \n (d) Mass of water vapour formed = %3.2f kg',Wa,E,dfg,Mw)