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+///Chapter No 11 Steam Boilers
+////Example 11.1 Page No 228
+///Find Mass of evaporation
+//Input data
+clc;
+clear;
+ms=5000;                 //Boiler produces wet steam in Kg/h
+x=0.95;                  //Dryness function
+P=10;                    //Operating pressure in bar
+mf=5500;                 //Bour in the furnace in Kg
+Tw=40;                   //Feed water temp in degree celsius
+
+//Calculation
+//from steam table
+hfw=167.45;              //In KJ/Kg
+hf=762.61;               //In KJ/Kg
+hfg=2031.6;              //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))*10; //Equivalent evaporation in Kg/Kg of coal
+
+//Output
+printf('Enthalpy of wet stream=%f KJ/Kg \n',hs);
+printf('Mass of evaporation=%f KJ/Kg \n',me);
+printf('Equivalent evaporation = %f Kg/Kg of coal \n',E);