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clear;
clc;
disp('Example 10.1');
// aim : To determine
// the equivalent evaporation
// Given
P = 1.4;// [MN/m^2]
m = 8;// mass of water,[kg]
T1 = 39;// entering temperature,[C]
T2 = 100;// [C]
x = .95;//dryness fraction
// solution
hf = 830.1;// [kJ/kg]
hfg = 1957.7;// [kJ/kg]
// steam is wet so specific enthalpy of steam is
h = hf+x*hfg;// [kJ/kg]
// at 39 C
h1 = 163.4;// [kJ/kg]
// hence
q = h-h1;// [kJ/kg]
Q = m*q;// [kJ]
evap = Q/2256.9;// equivalent evaporation[kg steam/(kg coal)]
mprintf('\n The equivalent evaporation, from and at 100 C is = %f kg steam/kg coal\n ',evap);
// End
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