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///Chapter No 11 Steam Boilers
////Example 11.8 Page No 236
///Find Equivalent mass evaporation
//Input data
clc;
clear;
ms=6000; //Steam generated in Kg/h
mf=700; //Coal burnt in Kg/h
CV=31500; //Cv of coal in KJ/Kg
x=0.92; //Dryness in fraction of steam
P=12; //Boiler pressure in bar
Tsup=259; //Temperature of steam in degree celsius
Tw=45; //Hot well temperature in degree celsius
//Calculation
hfw=188.35; //In KJ/Kg
Ts=187.96; //In degree celsius
hf=798.43; //In KJ/Kg
hfg=1984.3; //In KJ/Kg
hg=2782.7; //In KJ/Kg
Cps=2.3;
me=ms/mf; //Equivalent mass evaporation
hs=hf+x*hfg; //Enthalpy of wet steam in KJ/Kg
E=((me*(hs-hfw))/2257); //Equivalent evaporation in Kg/Kg of coal
hs1=(hg+Cps*(Tsup-Ts)); //Enthalpy of superheated steam in KJ/Kg
E1=((me*(hs1-hfw))/2257); //Equivalent evaporation(with superheater) in Kg/Kg of coal
etaboiler=((me*(hs-hfw))/CV)*100; //Boiler efficiency without superheater in %
etaboiler1=((me*(hs1-hfw))/CV)*100;//Boiler efficiency with superheater in %
//Output
printf('Equivalent mass evaporation=%f \n',me);
printf('Enthalpy of wet steam=%f KJ/Kg \n',hs);
printf('Equivalent evaporation=%f Kg/Kg of coal\n',E);
printf('Enthalpy of superheated steam=%f KJ/Kg \n',hs1);
printf('Equivalent evaporation(with superheater)=%f Kg/Kg of coal\n',E1);
printf('Boiler efficiency without superheater=%f percent \n',etaboiler);
printf('Boiler efficiency without superheater=%f percent \n',etaboiler1);
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