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

+///Example 10.26 Page No:207

+//Find Constant volume process

+clc;

+clear;

+//Input data

+m=1.5;                          //Saturated steam in Kg

+x1=1;                           

+x2=0.6;  

+P1=5*10^5;                     //Absolute  pressure in bar

+//From steam table at pressure basis 5 bar

+hg1=2747.5;                     //In KJ/Kg

+Vg1=0.3747;                     //In m^3/Kg

+V1=0.3747;                      //In m^3/Kg

+V2=0.3747;                      //In m^3/Kg

+//From steam table at Vg2 is 2.9 bar

+P2=2.9*10^5;                   //Absolute pressure in bar 

+t2=132.4;                       //In degree celsius 

+hf2=556.5;                      //In KJ/Kg

+hfg2=2166.6;                    //In KJ/Kg

+

+

+    

+//Calculation

+Vg2=V2/x2;                       //Constant volume process in m^3/Kg

+u1=hg1-((P1*Vg1)/1000);          //Initial internal energy in KJ/Kg

+u2=(hf2+x2*hfg2)-((P2*V2)/1000); //Final internal energy in KJ

+deltaU=(u1-u2)*m;                //Heat supplied in KJ

+

+//Output

+printf('Constant volume process=%f m^3/Kg \n ',Vg2);

+printf('Initial internal energy=%f KJ/Kg \n ',u1);

+printf('Final internal energy= %f KJ \n',u2);

+printf('Heat supplied=%f KJ \n ',deltaU);