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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);