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+///Chapter No 11 Steam Boilers

+////Example 11.3 Page No 230

+///Find mass of evaporation 

+//Input data

+clc;

+clear;

+ms=2500;                          //Saturated steam per bour in Kg

+x=1;         

+P=15;                             //Boiler pressure in bar

+Tw=25;                            //Feed water entering in degree celsius 

+mf=350;                           //Coal burnt in Kg/bour

+CV=32000;                         //Calorific value in Kj/Kg 

+

+//Calculation

+//steam table

+hfw=104.77;                       //In KJ/Kg

+hf=844.66;                        //In KJ/Kg

+hfg=1945.2;                       //In KJ/Kg

+hg=2789.9;                        //In KJ/Kg

+hs=2789.9;                        //Enthalpy of dry steam in KJ/Kg

+me=ms/mf;                         //mass of evaporation 

+E=((me*(hs-hfw))/2257);           //Equivalent evaporation in Kg/Kg ofcoal

+etaboiler=((me*(hs-hfw))/CV)*100; //Boiler efficiency in %

+

+//Output

+printf('mass of evaporation= %f \n',me);

+printf('Equivalent evaporation= %f Kg/Kg of coal\n',E);

+printf('Boiler efficiency= %f percent \n',etaboiler);