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///Chapter No 11 Steam Boilers
////Example 11.5 Page No 232
//Find Enthalpy of wet stream
//Input data
clc;
clear;
ms=5000; //Steam generted in Kg/h
mf=700; //Coal burnt in Kg/h
CV=31402; //Cv of coal in KJ/Kg
x=0.92; //quality of steam
P=1.2; //Boiler pressure in MPa
Tw=45; //Feed water temperature in degree celsius
//Calculation
hfw=188.35; //In KJ/Kg
hf=798.43; //In KJ/Kg
hfg=1984.3; //In KJ/Kg
hs=hf+x*hfg; //Enthalpy of wet stream in KJ/Kg
me=ms/mf; //mass of evaporation
E=((me*(hs-hfw))/2257); //Equivalent evaporation in Kg/Kg of coal
etaboiler=((me*(hs-hfw))/CV)*100; //Boiler efficiency in %
//Output
printf('Enthalpy of wet stream= %f KJ/Kg \n',hs);
printf('mass of evaporation=%f \n',me);
printf('Equivalent evaporation=%f Kg/Kg of coal \n',E);
printf('Boiler efficiency=%f percent \n',etaboiler);
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