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+// Example 15_12
+clc;funcprot(0);
+// Given data
+T=25.0+273;// K
+p_m=0.100;// MPa
+T_b=200+273;// K
+q_r=-134.158;// MJ
+R=8.3143;// kJ/(kgmole.K)
+
+// Calculation
+// The reaction equation for 100.% theoretical air is CH4+2O2+3.76N2-->CO2+2(H2O)+7.52(N2)
+n_CH4=1;// The stoichiometric coefficient of the reaction
+n_O2=2;// The stoichiometric coefficient of the reaction
+n_N2=7.52;// The stoichiometric coefficient of the reaction
+n_R=(n_CH4+n_O2+n_N2);// The stoichiometric coefficient of the reaction
+p_CH4=(n_CH4/n_R)*p_m;// kPa
+p_O2=(n_O2/n_R)*p_m;// kPa
+p_N2=(n_N2/n_R)*p_m;// kPa
+n_CO2=1;// The stoichiometric coefficient of the reaction
+n_H2O=2;// The stoichiometric coefficient of the reaction
+n_N2=7.52;// The stoichiometric coefficient of the reaction
+n_P=(n_CO2+n_H2O+n_N2);// The stoichiometric coefficient of the reaction
+p_CO2=(n_CO2/n_P)*p_m;// kPa
+p_H2O=(n_H2O/n_P)*p_m;// kPa
+p_N2=(n_N2/n_P)*p_m;// kPa
+s0_CH4=186.256;// kJ/(kgmole.K)
+s0_O2=205.138;// kJ/(kgmole.K)
+s0_N2=191.610;// kJ/(kgmole.K)
+sbar_CH4=s0_CH4-(R*log(p_CH4/p_m));// kJ/(kgmole.K)
+sbar_O2=s0_O2-(R*log(p_O2/p_m));// kJ/(kgmole.K)
+sbar_N2=s0_N2-(R*log(p_N2/p_m));// kJ/(kgmole.K)
+sbar_iR=(n_CH4*sbar_CH4)+(n_O2*sbar_O2)+(n_N2*sbar_N2);// kJ/(kgmole.K)
+s0_CO2=213.795;// kJ/(kgmole.K)
+s0_H2O=188.833;// kJ/(kgmole.K)
+s0_N2=191.610;// kJ/(kgmole.K)
+c_p_CO2=37.19;// kJ/(kgmole.K)
+c_p_H2O=33.64;// kJ/(kgmole.K)
+c_p_N2=29.08;// kJ/(kgmole.K)
+sbar_CO2=s0_CH4+(c_p_CO2*log(T_b/T))-(R*log(p_CO2/p_m));// kJ/(kgmole.K)
+sbar_H2O=s0_O2++(c_p_H2O*log(T_b/T))-(R*log(p_H2O/p_m));// kJ/(kgmole.K)
+sbar_N2=s0_N2+(c_p_N2*log(T_b/T))-(R*log(p_N2/p_m));// kJ/(kgmole.K)
+sbar_iP=(n_CO2*sbar_CO2)+(n_H2O*sbar_H2O)+(n_N2*sbar_N2);// kJ/(kgmole.K)
+sbar_p_r=sbar_iP-sbar_iR-((q_r*10^3)/T_b);// kJ/(kgmole.K)
+printf("\nThe entropy produced per mole of fuel,(sbar_p)_r=%3.0f kJ/(kgmole.K)",sbar_p_r);