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+// Example 12_11
+clc;funcprot(0);
+// Given data
+T=25;// °C
+P=1;// atm
+// The combustion equation C_3H_8+5(O_2+3.76N_2)--->3CO_2+4H_2O+18.8N_2
+N_p=1;// mol
+N_CO2=3;// mol
+N_H2O=4;// mol
+N_N2=18.8;// mol
+hbar0_fp=-103850;// kJ/kmol (C_3H_8)
+hbar0_fCO2=-393520;// kJ/kmol
+hbar0_CO2=9360;// kJ/kmol
+hbar0_fH2O=-241810;// kJ/kmol
+hbar0_H2O=9900;// kJ/kmol
+hbar0_fN2=0;// kJ/kmol
+hbar0_N2=8670;// kJ/kmol
+Q=0;// kJ/kmol
+H_R=-103850;// kJ/kmol fuel
+
+// Calculation
+H_P=H_R;// kJ/kmol fuel
+hbar_p=((H_R-((N_CO2*(hbar0_fCO2))+(N_H2O*(hbar0_fH2O))))/25.8)+hbar0_N2;// kJ/kmol
+// Suggests T_P=1380 K
+T_P1=2600;// K
+hbar_CO2=137400;// kJ/kmol
+hbar_H2O=114300;// kJ/kmol
+hbar_N2=86600;// kJ/kmol 
+H_P1=(N_CO2*(hbar0_fCO2+hbar_CO2-hbar0_CO2))+(N_H2O*(hbar0_fH2O+hbar_H2O-hbar0_H2O))+(N_N2*(hbar0_fN2+hbar_N2-hbar0_N2));// The enthalpy of the products of combustion in kJ/kmol fuel
+// The temperature is obiviously too high.We select aa lower value,T_p=1300 K
+T_P2=2400;// K
+hbar_CO2=125200;// kJ/kmol
+hbar_H2O=103500;// kJ/kmol
+hbar_N2=79320;// kJ/kmol 
+H_P2=(N_CO2*(hbar0_fCO2+hbar_CO2-hbar0_CO2))+(N_H2O*(hbar0_fH2O+hbar_H2O-hbar0_H2O))+(N_N2*(hbar0_fN2+hbar_N2-hbar0_N2));// The enthalpy of the products of combustion in kJ/kmol fuel
+T_P=T_P2-([(-H_P+H_P2)/(H_P1-H_P2)]*(T_P1-T_P2));// K
+printf("\nThe adiabatic flame temperature,T_P=%4.0f K",T_P);