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clc
clear
Ms=7.5; //kg/kg of coal
P=11; //in bar
Tf=70; //in C
Eff=0.75; //Efficiency
FOE=1.15; //Factor of Evaporation
Cps=2.1; //in kJ/kg K
Hfw=293; //in kJ/kg
H=(FOE*2257)+Hfw;
//At 11 bar
Hg=2781.7; //in kJ/kg
Tsat=184.1; //in C
Tsup=((H-Hg)/Cps)+Tsat;
DOS=Tsup-Tsat; //Degree of Superheat
printf('Degree of Superheat: %3.1f C',DOS);
printf('\n');
Me=(Ms*(H-Hfw))/2257;
printf('Equivalent evaporation: %3.2f kg/kg of coal',Me);
printf('\n');
CV=(Ms*(H-Hfw))/Eff;
printf('Calorific value of Boiler: %3.2f kJ/kg ',CV);
printf('\n');
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