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clc
clear
//At 10 bar pressure
Tsat=179.9;
Tsup=250;
Cps=2.1; //in kJ/kg K
Hg=2778.1; //in kJ/kg
Ms=10; //in kg/kg of coal
Hsup=Hg+(Cps*(Tsup-Tsat));
Hfw=155;
Me=(Ms*(Hsup-Hfw))/2257;
FOE=Me/Ms; //Factor of Evaporation
BP=(Me*370)/21.296;
printf('Equivalent Evaporation: %3.1f kg/kg of coal',Me);
printf('\n');
printf('Boiler Power: %3.1f kW',BP);
printf('\n');
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