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///Chapter No 11 Steam Boilers
////Example 11.7 Page No 234
///Find Boiler efficiency
///Input data
clc;
clear;
E=12; //Boiler found steam in Kg/Kg
CV=35000; //Calorific value in KJ/Kg
ms=15000; //Boiler produces in Kg/h
P=20; //Boiler pressure in bar
Tw=40; //Feed water in degree celsius
mf=1800; //Fuel consumption
//Calculation
//R=me(hs-hfw)
hfw=167.45; //In KJ/Kg
hg=2797.2; //In KJ/Kg
Ts=211.37; //In degree celsius
Cps=2.3;
R=E*2257; //Equivalent evaporation in KJ/Kg of coal
etaboiler=(R/CV)*100; //Boiler efficiency in %
me=ms/mf; //Equivalent mass evaporation in KJ/Kg of coal
hs=(R/me)+hfw; //In KJ/Kg
Tsup=((hs-hg)/Cps)+Ts; //Enthalpy of superheated steam in degree celsius
//Output
printf('Equivalent evaporation=%f KJ/Kg of coal \n',R);
printf('Boiler efficiency=%f percent \n',etaboiler);
printf('Equivalent mass evaporation= %f KJ/Kg of coal \n',me);
printf('hs=%f KJ/Kg \n',hs);
printf('Enthalpy of superheated steam=%f degree celsius \n',Tsup);
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