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clc
clear
Ma=18000; //in kg/hr
P=10; //in bar
x=0.97; //Dryness Fraction
Tfw=40; //in C
Mf=2050; //in kg/hr
CV=28000; //kJ/kg
//At 10 bar
Hf1=762.8;
Hfg1=2015.3;
H=Hf1+(x*Hfg1);
Hfw=167.6;
Eff=(Ma*100*(H-Hfw))/(Mf*CV);
printf('Boiler efficiency: %3.2f Percent',Eff);
printf('\n');
EA=((Ma/Mf)*(H-Hfw))/2257;
printf('Equivalent Evaporation: %3.2f kg/kg of coal',EA);
printf('\n');
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