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///Chapter No 11 Steam Boilers
////Example 11.1 Page No 228
///Find Mass of evaporation
//Input data
clc;
clear;
ms=5000; //Boiler produces wet steam in Kg/h
x=0.95; //Dryness function
P=10; //Operating pressure in bar
mf=5500; //Bour in the furnace in Kg
Tw=40; //Feed water temp in degree celsius
//Calculation
//from steam table
hfw=167.45; //In KJ/Kg
hf=762.61; //In KJ/Kg
hfg=2031.6; //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))*10; //Equivalent evaporation in Kg/Kg of coal
//Output
printf('Enthalpy of wet stream=%f KJ/Kg \n',hs);
printf('Mass of evaporation=%f KJ/Kg \n',me);
printf('Equivalent evaporation = %f Kg/Kg of coal \n',E);
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