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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);