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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);