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clc
clear
//DATA GIVEN
V=0.15; //volume of wet steam in m^3
p=4; //pressure of wet steam in bar
x=0.8; //dryness fraction
//At 4 bar, from steam tables
vg=0.462; //m^3/kg
hf=604.7; //kJ/kg
hfg=2133; //kJ/kg
rho=1/(x*vg); //density in kg/m^3
m=rho*V; //mass of 0.15 m^3 of steam
Htotal=(rho*1)*(hf+x*hfg); //total heat of 1 m^3 of steam which has a mass of rho(2.7056) kg
printf('(i)The Mass of 0.15 m^3 of steam is: %1.4f kg. \n',m);
printf('(ii)The Total heat of 1 m^3 of steam which has a mass of 2.7056 kg is: %4.2f kJ. \n',Htotal);
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