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+/////////Chapter 10 Properties Of Steam
+///Example 10.24 Page No:204
+///Find Enthalpy of steam of first boiler
+clc;
+clear;
+//Input data;
+FB=15;                    //First boiler in bar
+SB=15;                    //Second boiler in bar
+tsup1=300;                //Temperature of the steam in degree celsius
+tsup2=200;                //Temperature of the steam in degree celsius
+//From steam table (pressure basis at 15 bar )
+ts=198.3;                 //In degree celsius                
+hf=844.7;                 //In KJ/Kg
+hfg=1945.2;               //In KJ/Kg
+hg=2789.9;                //In KJ/I
+Cps=2.3;
+
+//Calculation
+h1=hg+Cps*(tsup1-ts);     //Enthalpy of steam of first boiler in KJ/Kg 
+h3=hg+Cps*(tsup2-ts);     //Enthalpy of steam in steam main in KJ/Kg
+h2=2*h3-h1;               //Energy balance in KJ/Kg
+x2=(h2-hf)/hfg;           //Enthalpy of wet steam
+
+//OUTPUT
+printf('Enthalpy of steam of first boiler= %f KJ/Kg\n',h1);
+printf('Enthalpy of steam in steam main=%f KJ/Kg \n ',h3);
+printf('Energy balance=%f KJ/Kg \n ',h2);
+printf('Enthalpy of wet steam= %f \n ',x2);