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/////////Chapter 10 Properties Of Steam
///Example 10.23 Page No:203
///Find Constant pressure process
//Input data
clc;
clear;
ms=1000; //Steam in Kg/h
P=16; //Absolute pressure in bar
x2=0.9; //Steam is dry
t1=30+273; //temperature in degree celsius
tsup=380; //tmperature rised in degree celsius
//from steam table(pressure basis at 16 bar)
h1=125.7; //in KJ/Kg
ts=201.4; //In degree celsius
hf=858.5; //in kJ/Kg
hfg=1933.2; //in kJ/Kg
hg=2791.7; //in kJ/Kg
Cps=2.3;
//Calculation
h2=hf+x2*hfg; //Final enthalpy of wet steam in KJ/Kg
Q1=(ms*(h2-h1))*(10^(-3)); //Constant pressure process in KJ/h
h3=hg+Cps*(tsup-ts); //Final enthalpy of superheated steam in KJ/g
Q2=(ms*(h3-h2))*(10^(-3)); //Suprheated steam in KJ/h
//Output
printf('Final enthalpy of wet steam= %f KJ/Kg \n ',h2);
printf('Constant pressure process= %f KJ/h \n',Q1);
printf('Final enthalpy of superheated steam= %f KJ/g \n',h3);
printf('Suprheated steam= %f KJ/h \n',Q2);
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