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/////////Chapter 10 Properties Of Steam
///Example 10.21 Page No:201
/// Find Final enthalpy of steam
//Input data
clc;
clear;
m=4; //Steam in Kg
P=13; //Absolute pressure in bar
tsup1=450; //Absolute temp in degree celsius
deltaH=2.8*10^3;
Cps=2.3; //loses in MJ
//from steam table at 13 bar
ts=191.6; //In degree celsius
Vg=0.1511; //In m^3/Kg
hf=814.7; //In m^3/Kg
hfg=1970.7; //In KJ/Kg
hg=2785.4; //In KJ/Kg
///Calculation
h1=hg+Cps*(tsup1-ts); //Initial enthalpy of steam in KJ/Kg
Deltah=deltaH/m; //Change in enthalpy/unit mass in KJ/Kg
h2=h1-Deltah; //Final enthalpy of steam in KJ/Kg
x2=(h2-hf)/hfg; //wet & dryness fraction
//Output
printf('Initial enthalpy of steam=%f Kj/Kg \n ',h1);
printf('Change in enthalpy/unit mass=%f Kj/Kg \n ',Deltah);
printf('Final enthalpy of steam= %f KJ/Kg \n',h2);
printf('wet & dryness fraction=%f \n',x2);
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