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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+///Chapter No 11 Steam Boilers
+////Example 11.7 Page No 234
+///Find Boiler efficiency
+///Input data
+clc;
+clear;
+E=12; //Boiler found steam in Kg/Kg
+CV=35000; //Calorific value in KJ/Kg
+ms=15000; //Boiler produces in Kg/h
+P=20; //Boiler pressure in bar
+Tw=40; //Feed water in degree celsius
+mf=1800; //Fuel consumption
+
+
+//Calculation
+//R=me(hs-hfw)
+hfw=167.45; //In KJ/Kg
+hg=2797.2; //In KJ/Kg
+Ts=211.37; //In degree celsius
+Cps=2.3;
+R=E*2257; //Equivalent evaporation in KJ/Kg of coal
+etaboiler=(R/CV)*100; //Boiler efficiency in %
+me=ms/mf; //Equivalent mass evaporation in KJ/Kg of coal
+hs=(R/me)+hfw; //In KJ/Kg
+Tsup=((hs-hg)/Cps)+Ts; //Enthalpy of superheated steam in degree celsius
+
+
+
+//Output
+printf('Equivalent evaporation=%f KJ/Kg of coal \n',R);
+printf('Boiler efficiency=%f percent \n',etaboiler);
+printf('Equivalent mass evaporation= %f KJ/Kg of coal \n',me);
+printf('hs=%f KJ/Kg \n',hs);
+printf('Enthalpy of superheated steam=%f degree celsius \n',Tsup);