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clear;
clc;
disp('Example 4.18');
// aim : To determine
// the heat received by the steam per kilogram
// Given values
// initial
P1 = 4;// pressure, [MN/m^2]
x1 = .95; // dryness fraction
// final
t2 = 350;// temperature,[C]
// solution
// from steam table, at 4 MN/m^2 and x1=.95
hf = 1087.4;// [kJ/kg]
hfg = 1712.9;// [kJ/kg]
// hence
h1 = hf+x1*hfg;// [kJ/kg]
// since pressure is kept constant ant temperature is raised so at this condition
h2 = 3095;// [kJ/kg]
// so by energy balance
Q = h2-h1;// Heat received,[kJ/kg]
mprintf('\n The heat received by the steam is = %f kJ/kg \n',Q);
// End
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