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+/////////Chapter 10 Properties Of Steam
+////Example 10.13 Page No:196
+////Find Volume occupied by water
+///Input data
+clc;
+clear;
+V=0.04;                  //Volume of vessel in m^3 
+x=1;
+t=250+273;               //Saturated steam temp in degree celsius
+mw=9;                    //Mass of liquid in Kg
+//From steam table(temp basis,at t=250)
+P=39.78*10^5;               //in bar
+Vf=0.001251;             //In m^3/kg
+Vg=0.05004;              //In m^3/Kg
+hf=1085.7;               //KJ/Kg
+hfg=2800.4;              //KJ/Kg
+hg=1714.7;               //KJ/Kg
+
+//Calculation
+Vw=mw*Vf;                //Volume occupied by water in m^3
+Vs=V-Vw;                 //Volume of waterin m^3
+ms=Vs/Vg;                //Volume of dry and saturated steam in Kg 
+m=mw+ms;                 //Total mass of steam in Kg
+x=ms/(ms+mw);            //Dryness fraction of steam 
+Vwet=(1-x)*Vf+x*Vg;      //Specific volume of steam in m^3/Kg
+h=hf+x*hfg;              //Enthalpy of wet steam in KJ/Kg
+EOWS=h*m;                //Enthalpy of 9.574 Kg of wet steam KJ
+u=h-((P*Vwet)/10^3);    //Internal Energy in KJ/Kg
+IEOS=u*m;                //Internal energy of 9.574 Kg of steam in KJ
+
+
+///Output
+printf('Volume occupied by water=%f m^3 \n ',Vw);
+printf('Volume of water=%f m^3 \n ',Vs);
+printf('Volume of dry and saturated steam=%f Kg \n',ms);
+printf('Total mass of steam= %f Kg \n ',m);
+printf('Dryness fraction of steam= %f  \n',x);
+printf('Specific volume of steam=%f m^3/Kg \n ',Vwet);
+printf('Enthalpy of wet steam=%f KJ/Kg \n ',h);
+printf('Enthalpy of 9.574 Kg of wet steam=%f KJ \n ',EOWS);
+printf('Internal Energy= %f KJ/Kg \n',u);
+printf('Internal energy of 9.574 Kg of steam=%f KJ \n ',IEOS);