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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);
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