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+// Example 15_8
+clc;funcprot(0);
+// Given data
+// For 100.% theoretical air, the combustion equation for methane is,CH_4+2.00[O_2+3.76 N_2]-->CO_2+2.00(H_2O)+7.52(N_2)
+// From Table 15.1, we find that
+h_f_CH4=-74.873;// MJ/kgmoleCH4
+h_R=-74.873;// MJ/kgmoleCH4
+h_f_N_2=0;// MJ/kgmole N2
+h_f_CO2=-393.522;// MJ/kgmole CO2
+h_f_H2O_g=-241.827;// MJ/kgmole H2O_g
+h_f_H2O_l=-285.838;// MJ/kgmole H2O_l
+
+// Calculation
+h_p_LHV=h_f_CO2+(2*h_f_H2O_g)+(7.52*h_f_N_2);// MJ/kgmole CH4
+h_p_HHV=h_f_CO2+(2*h_f_H2O_l)+(7.52*h_f_N_2);// MJ/kgmole CH4
+LHV=h_p_LHV-h_R;// MJ/kgmole CH4
+HHV=h_p_HHV-h_R;// MJ/kgmole CH4
+h_fg_H2O=44.00;// MJ/kgmole CH4
+n_H2O=2.00;// The stoichiometric coefficient for the reaction
+n_fuel=1.00;// The stoichiometric coefficient for the reaction
+HHV=LHV-((n_H2O/n_fuel)*h_fg_H2O);// MJ/kgmole CH4
+printf("\nThe higher heating value of methane,LHV=%3.2f MJ/kgmole CH4 \nThe lower heating value of methane,HHV=%3.2f MJ/kgmole CH4",HHV,LHV);