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

+///Example 10.25 Page No:205

+// Find Initial specific volume of steam

+clc;

+clear;

+///Input data

+V=0.35;                     //Capacity of vessel in m^3

+P1=10*10^2;                //Absolute pressure in bar

+tsup1=250+273;              //Absolute temperature in degree celsius 

+P2=2.5*102;                 //Absolute pressure in the vessel fall in bar

+

+//From steam table (pressure basis at 10 bar)

+ts1=179.9+273;              //In degree celsius 

+Vg1=0.1943;                 //In m^3/Kg

+hf1=762.6;                  //In KJ/Kg

+hfg1=2013.6;                //In KJ/Kg

+hg1=2776.2;                 //In KJ/Kg

+

+//From steam table(pressure basis at 2.5 bar)

+V2=0.2247;                  //In m^3/Kg

+ts2=127.4;                  //In degree celsius

+Vg2=0.7184;                 //In m^3/Kg

+hf2=535.3;                  //In KJ/Kg

+hfg2=2181.0;                //In KJ/Kg

+hg2=2716.4;                 //In KJ/Kg

+Cps=2.3;

+///Calculation

+V1=Vg1*(tsup1/ts1);         //Initial specific volume of steam in m^3/Kg

+m=V/V1;                     //Initial mass of steam in Kg

+x2=V2/Vg2;                  //Final condition of wet steam

+h1=hg1+Cps*(tsup1-ts1);     //Initial enthalpy of steam in KJ/Kg

+u1=h1-P1*V1;                //Initial internal energy of steam in KJ/Kg

+h2=hf2+x2*hfg2;             //Final enthalpy of steam in KJ/Kg

+u2=h2-P2*V2;                //Final internal energy of steam in KJ/Kg

+deltaU=(u2-u1)*m;           //Change in internal energy in KJ

+

+//Output

+printf('Initial specific volume of steam=%f m^3/Kg \n ',V1);

+printf('Initial mass of steam=%fKg \n ',m);

+printf('Final condition of wet steam= %f \n ',x2);

+printf('Initial enthalpy of steam=%f KJ/Kg \n ',h1);

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

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

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

+printf('Change in internal energy= %f KJ/Kg \n',deltaU);