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/////////Chapter 10 Properties Of Steam
///Example 10.24 Page No:204
///Find Enthalpy of steam of first boiler
clc;
clear;
//Input data;
FB=15; //First boiler in bar
SB=15; //Second boiler in bar
tsup1=300; //Temperature of the steam in degree celsius
tsup2=200; //Temperature of the steam in degree celsius
//From steam table (pressure basis at 15 bar )
ts=198.3; //In degree celsius
hf=844.7; //In KJ/Kg
hfg=1945.2; //In KJ/Kg
hg=2789.9; //In KJ/I
Cps=2.3;
//Calculation
h1=hg+Cps*(tsup1-ts); //Enthalpy of steam of first boiler in KJ/Kg
h3=hg+Cps*(tsup2-ts); //Enthalpy of steam in steam main in KJ/Kg
h2=2*h3-h1; //Energy balance in KJ/Kg
x2=(h2-hf)/hfg; //Enthalpy of wet steam
//OUTPUT
printf('Enthalpy of steam of first boiler= %f KJ/Kg\n',h1);
printf('Enthalpy of steam in steam main=%f KJ/Kg \n ',h3);
printf('Energy balance=%f KJ/Kg \n ',h2);
printf('Enthalpy of wet steam= %f \n ',x2);
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