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+clc
+clear
+//Input data
+t=25;//The temperature of air in degree centigrade
+p=1;//The pressure of air in atm
+T1=2200;//Given first temperature in K
+T2=2400;//Given second temperature in K
+h1=59.86;//The change in enthalpy of hydrogen at 2200 K in MJ/kmol
+h2=66.915;//The change in enthalpy of hydrogen at 2400 K in MJ/kmol
+T=298;//The temperature of air in K
+
+//Calculations
+HR=0;//The total enthalpy on the reactants side since all the reactants are elements
+Kp1=-6.774;//Natural logarithm of equilibrium constant at 2200 K for the equation 
+K1=exp(Kp1);//The value of equilibrium constant at 2200 K
+a1=0.02;//By trail and error method the degree of dissociation of H2O
+hfh=-241.82;//The enthalpy of formation of water at both 2200 and 2400 K in MJ/kmol
+hfh1=83.036;//The change in enthalpy of water at 2200 K in MJ/kmol
+hfd1=59.86;//The change in enthalpy of hydrogen at 2200 K in MJ/kmol
+hfo1=66.802;//The change in enthalpy of oxygen at 2200 K in MJ/kmol
+hfn1=63.371;//The change in enthalpy of nitrogen at 2200 K in MJ/kmol
+HP1=(0.98*(hfh+hfh1))+(0.02*hfd1)+(0.01*hfo1)+(1.88*hfn1);//The enthalpy on the products side in MJ/kmol 
+a2=0.04;//By trail and error method the degree of dissociation of H2O at 2400 K
+hfh2=93.604;//The change in enthalpy of water at 2400 K in MJ/kmol
+hfd2=66.915;//The change in enthalpy of hydrogen at 2400 K in MJ/kmol
+hfo2=74.492;//The change in enthalpy of oxygen at 2400 K in MJ/kmol
+hfn2=70.651;//The change in enthalpy of nitrogen at 2400 K in MJ/kmol
+HP2=(0.96*(hfh+hfh2))+(0.04*hfd2)+(0.02*hfo2)+(1.88*hfn2);//The enthalpy on the products side in MJ/kmol 
+H1=HP1-HR;//The total change in enthalpy at 2200 K in MJ/kmol
+H2=HP2-HR;//The total change in enthalpy at 2400 K in MJ/kmol
+Tl=[[[T2-T1]/[HP2-HP1]]*[HR-HP1]]+T1;//The required temperature in K
+
+//Output
+printf('The adiabatic flame temperature taking dissociation into account is %3.0f K',Tl)