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clc
clear
Ma=5500; //kg/hr
P=1; //bar
x=0.94; //Dryness Fraction
Tfw=40; //in C
Mf=600; //kg/hr
CV=32000; //kJ/kg
Hfw=Tfw*4.187;
//At 1 bar pressure
Hf=417.5; //kJ/kg
Hfg=2258; //kJ/kg
H=Hf+(x*Hfg);
Ms=Ma/Mf;
Me=(Ms*(H-Hfw))/2257;
printf('Equivalent Evaporation: %3.3f kg/kg of coal',Me);
printf('\n');
Eff=(Ms*100*(H-Hfw))/CV;
printf('Efficiency: %3.2f percent',Eff);
printf('\n');
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