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+/////////Chapter 10 Properties Of Steam
+////Example 10.20 Page No:200
+/// Find Initial enthalpy of steam
+//Input data
+clc;
+clear;
+P=10*10^2;                //Absolute pressure in bar
+x1=0.9;                    //Dryness enters
+tsup2=300+273;             //Temperature in degree celsius 
+//From steam table at 10 bar
+ts=179.9+273;              //In degree celsius
+Vg=0.1943;                 //In m^3/Kg
+hf=762.6;                  //In KJ/Kg
+hfg=2013.6;                //InK/Kg
+hg=2776.2;                 //In KJ/Kg
+Cps=2.3;
+
+//Calculation
+h1=hf+x1*hfg;              //Initial enthalpy of steam in KJ/Kg
+V1=x1*Vg;                  //Initial specific volume of steam
+u1=h1-P*V1;                //Initial internal energy of steam in KJ/Kg
+h2=hg+Cps*(tsup2-ts);      //Final enthalpy of steam in KJ/Kg
+V2=Vg*(tsup2/ts);          //Final specific volume of steam in m**3/Kg
+u2=h2-P*V2;                //Final internal energy of steam in KJ/K
+deltah=h2-h1;              //Heat gained by steam in KJ/Kg
+deltaU=(u2-u1);            //Change in internal energy in KJ/Kg
+
+//Output
+printf('Initial enthalpy of steam=%f KJ/Kg \n',h1);
+printf('Initial specific volume of steam=%f \n ',V1);
+printf('Initial internal energy of steam=%f KJ/Kg \n',u1);
+printf('Final enthalpy of steam= %f KJ/Kg \n ',h2);
+printf('Final specific volume of steam= %f m^3/kg \n',V2);
+printf('Final internal energy of steam=%f Kj/Kg \n ',u2);
+printf('Heat gained by steam= %f KJ/Kg \n ',deltah);
+printf('Change in internal energy=%f KJ/Kg \n ',deltaU);