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+///Chapter No 11 Steam Boilers
+////Example 11.5 Page No 232
+//Find Enthalpy of wet stream
+//Input data
+clc;
+clear;
+ms=5000; //Steam generted in Kg/h
+mf=700; //Coal burnt in Kg/h
+CV=31402; //Cv of coal in KJ/Kg
+x=0.92; //quality of steam
+P=1.2; //Boiler pressure in MPa
+Tw=45; //Feed water temperature in degree celsius
+
+
+//Calculation
+hfw=188.35; //In KJ/Kg
+hf=798.43; //In KJ/Kg
+hfg=1984.3; //In KJ/Kg
+hs=hf+x*hfg; //Enthalpy of wet stream in KJ/Kg
+me=ms/mf; //mass of evaporation
+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('mass of evaporation=%f \n',me);
+printf('Equivalent evaporation=%f Kg/Kg of coal \n',E);
+printf('Boiler efficiency=%f percent \n',etaboiler);