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+////Chapter No 14 Air Standard Cycles
+////Example No 14.9 Page No 311
+///Find Work done per Kg of air
+//Input data
+clc;
+clear;
+P1=0.1;                    //Beginning compression in MPa
+T1=40+273;                 //Beginning temp in degree celsius
+eta=0.55;                  //Standard effeciency in %
+QR=540;                    //Rejected heat in KJ/Kg
+r=7.36;                    //Compression ratio
+gamma1=1.4;
+Cv=0.718;
+
+//Calculation
+//eta=(1-(1/(r^(gamma-1))))
+QS=(-QR/(eta-1));          //Heat supplied/unit mass in KJ/Kg
+WD=QS-QR;                  //Work done per Kg of air in KJ/Kg
+T2=T1*(r^(gamma1-1));      //Temp at end of compression in K
+P2=P1*((r)^gamma1);        //pressure at point 2 in MPa
+T3=(QS/Cv)+T2;             //max temp of the cycle in K
+P3=(T3/T2)*P2;             //max pressure of the cycle in MPa
+
+//Output
+printf('Heat supplied/unit mass= %f KJ/Kg \n',QS);
+printf('Work done per Kg of air= %f KJ/Kg \n',WD);
+printf('Temp at end of compression= %f K \n ',T2);
+printf('pressure at point two= %f MPa \n',P2);
+printf('max temp of the cycle= %f K \n',T3);
+printf('max pressure of the cycle= %f MPa \n',P3);