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/////////Chapter 10 Properties Of Steam
////Example 10.20 Page No:200
/// Find Initial enthalpy of steam
//Input data
clc;
clear;
P=10*10^2; //Absolute pressure in bar
x1=0.9; //Dryness enters
tsup2=300+273; //Temperature in degree celsius
//From steam table at 10 bar
ts=179.9+273; //In degree celsius
Vg=0.1943; //In m^3/Kg
hf=762.6; //In KJ/Kg
hfg=2013.6; //InK/Kg
hg=2776.2; //In KJ/Kg
Cps=2.3;
//Calculation
h1=hf+x1*hfg; //Initial enthalpy of steam in KJ/Kg
V1=x1*Vg; //Initial specific volume of steam
u1=h1-P*V1; //Initial internal energy of steam in KJ/Kg
h2=hg+Cps*(tsup2-ts); //Final enthalpy of steam in KJ/Kg
V2=Vg*(tsup2/ts); //Final specific volume of steam in m**3/Kg
u2=h2-P*V2; //Final internal energy of steam in KJ/K
deltah=h2-h1; //Heat gained by steam in KJ/Kg
deltaU=(u2-u1); //Change in internal energy in KJ/Kg
//Output
printf('Initial enthalpy of steam=%f KJ/Kg \n',h1);
printf('Initial specific volume of steam=%f \n ',V1);
printf('Initial internal energy of steam=%f KJ/Kg \n',u1);
printf('Final enthalpy of steam= %f KJ/Kg \n ',h2);
printf('Final specific volume of steam= %f m^3/kg \n',V2);
printf('Final internal energy of steam=%f Kj/Kg \n ',u2);
printf('Heat gained by steam= %f KJ/Kg \n ',deltah);
printf('Change in internal energy=%f KJ/Kg \n ',deltaU);
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