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diff --git a/3821/CH10/EX10.25/Example10_25.sce b/3821/CH10/EX10.25/Example10_25.sce new file mode 100644 index 000000000..c78ea5cb6 --- /dev/null +++ b/3821/CH10/EX10.25/Example10_25.sce @@ -0,0 +1,45 @@ +/////////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);
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