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+///Chapter No 11 Steam Boilers
+////Example 11.2 Page No 229
+///Find Enthalpy of wet stream
+///Input data
+clc;
+clear;
+p=14; //Boiler pressure in bar
+me=9; //Evaporates of water in Kg
+Tw=35; //Feed water entering in degree celsius
+x=0.9; //Steam stop value
+CV=35000; //Calorific value of the coal
+
+///Calculation
+//From Steam Table
+hfw=146.56; //In KJ/Kg
+hf=830.07; //In KJ/Kg
+hfg=1957.7; //In KJ/Kg
+hs=hf+x*hfg; //Enthalpy of wet stream in KJ/Kg
+E=((me*(hs-hfw))/2257); //Equivalent evaporation in Kg/Kg of coal
+etaboiler=((me*(hs-hfw))/CV)*100;//Boiler efficiency in %
+
+///Output
+printf('Enthalpy of wet stream=%f KJ/Kg \n',hs);
+printf('Equivalent evaporation=%f Kg/Kg of coal \n',E);
+printf('Boiler efficiency=%f percent \n',etaboiler);