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+/////////Chapter 10 Properties Of Steam
+///Example 10.9 Page No:190
+////Find Volume of wet steam
+
+//Input data
+clc;
+clear;
+P=12*10^5;             //Absolute pressure
+//From steam table (pressure basis at 12 bar)
+ts=188+273;             //In degree celsius
+vf=0.001139;            //In m**3/Kg 
+vg=0.1632;              //In m**3/Kg 
+hf=798.4;               //In KJ/Kg
+hfg=1984.3;             //In KJ/Kg
+hg=2782.7;              //In KJ/Kg
+x=0.94;
+Cps=2.3;
+tsup=350+273;           //In degree celsius
+
+//Calcuation
+h=hf+x*hfg;             //Enthalpy of wet steam in KJ/Kg
+v=(1-x)*vf+x*vg;        //Volume of wet steam m**3/Kg
+u=h-((P*v)/10^3);      //Internal Energy in KJ/Kg
+hg=2782.7;              //Enthalpy of dry & saturated steam in KJ/Kg
+v1=vg;                  //Volume of dry & saturated steam  m**3/Kg
+u1=hg-((P*vg)/10^3);   //Internal Energy in KJ/Kg       
+h1=hg+Cps*(tsup-ts);    //Enthalpy of superheated steam in KJ/Kg
+vsup=vg*(tsup/ts);      //Volume of superheated steam in m**3/Kg
+u2=h1-((P*v)/10**3);    //Internal Energy in KJ/Kg
+
+
+///Output
+printf('Enthalpy of wet steam=%f KJ/Kg \n ',h);
+printf('Volume of wet steam= %f m^3/Kg \n ',v);
+printf('Internal Energy= %f KJ/Kg \n ',u);
+printf('Enthalpy of dry & saturated steam=%f KJ/Kg \n',hg);
+printf('Volume of dry & saturated steam=%f m^3/Kg \n ',v1);
+printf('Internal Energy= %f KJ/Kg \n',u1);
+printf('Enthalpy of superheated steam=%f KJ/Kg \n ',h1);
+printf('Volume of superheated steam= %f m^3/kg \n',vsup);
+printf('Internal Energy= %f KJ/Kg \n ',u2);