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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);