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///Chapter No 11 Steam Boilers
////Example 11.3 Page No 230
///Find mass of evaporation
//Input data
clc;
clear;
ms=2500; //Saturated steam per bour in Kg
x=1;
P=15; //Boiler pressure in bar
Tw=25; //Feed water entering in degree celsius
mf=350; //Coal burnt in Kg/bour
CV=32000; //Calorific value in Kj/Kg
//Calculation
//steam table
hfw=104.77; //In KJ/Kg
hf=844.66; //In KJ/Kg
hfg=1945.2; //In KJ/Kg
hg=2789.9; //In KJ/Kg
hs=2789.9; //Enthalpy of dry steam in KJ/Kg
me=ms/mf; //mass of evaporation
E=((me*(hs-hfw))/2257); //Equivalent evaporation in Kg/Kg ofcoal
etaboiler=((me*(hs-hfw))/CV)*100; //Boiler efficiency in %
//Output
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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