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///Chapter No 11 Steam Boilers
////Example 11.2 Page No 229
///Find Enthalpy of wet stream
///Input data
clc;
clear;
p=14; //Boiler pressure in bar
me=9; //Evaporates of water in Kg
Tw=35; //Feed water entering in degree celsius
x=0.9; //Steam stop value
CV=35000; //Calorific value of the coal
///Calculation
//From Steam Table
hfw=146.56; //In KJ/Kg
hf=830.07; //In KJ/Kg
hfg=1957.7; //In KJ/Kg
hs=hf+x*hfg; //Enthalpy of wet stream in KJ/Kg
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('Equivalent evaporation=%f Kg/Kg of coal \n',E);
printf('Boiler efficiency=%f percent \n',etaboiler);
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