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diff --git a/3821/CH10/EX10.26/Example10_26.sce b/3821/CH10/EX10.26/Example10_26.sce new file mode 100644 index 000000000..02474da0f --- /dev/null +++ b/3821/CH10/EX10.26/Example10_26.sce @@ -0,0 +1,34 @@ +/////////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);
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