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+/////////Chapter 10 Properties Of Steam

+///Example 10.23 Page No:203

+///Find Constant pressure process

+//Input data

+clc;

+clear;

+ms=1000;                      //Steam in Kg/h       

+P=16;                         //Absolute pressure in bar

+x2=0.9;                       //Steam is dry 

+t1=30+273;                    //temperature in degree celsius

+tsup=380;                     //tmperature rised in degree celsius                                                            

+                    

+//from steam table(pressure basis at 16 bar)

+h1=125.7;                     //in KJ/Kg

+ts=201.4;                     //In degree celsius

+hf=858.5;                     //in kJ/Kg

+hfg=1933.2;                   //in kJ/Kg

+hg=2791.7;                    //in kJ/Kg

+Cps=2.3;

+

+//Calculation       

+h2=hf+x2*hfg;                 //Final enthalpy of wet steam in KJ/Kg 

+Q1=(ms*(h2-h1))*(10^(-3));       //Constant pressure process in KJ/h 

+h3=hg+Cps*(tsup-ts);          //Final enthalpy of superheated steam in KJ/g

+Q2=(ms*(h3-h2))*(10^(-3));       //Suprheated steam in KJ/h

+

+//Output

+printf('Final enthalpy of wet steam= %f KJ/Kg \n ',h2);

+printf('Constant pressure process= %f KJ/h \n',Q1);

+printf('Final enthalpy of superheated steam= %f KJ/g \n',h3);

+printf('Suprheated steam= %f KJ/h \n',Q2);