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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+// Exa 3.25
+clc;
+clear;
+close;
+// Given data
+P1 = 14;// in bar
+P3 = 2.222;// in bar
+V3byV1 = P1/P3;
+P2 = 1.05;// in bar
+Gamma = log(P1/P2)/log(V3byV1);
+C_P = 1.005;// in kJ/kg-K
+C_V = C_P/Gamma;// in kJ/kg-K
+T3 = 343;// in degree C
+T3 = T3 + 273;// in K
+T2 = ceil(T3*P2)/P3;// in K
+m = 0.5;// in kg
+del_U = m*C_V*(T2-T3);// in kJ
+disp(del_U,"Change in internal energy in kJ is");
+disp("i.e. there is a loss of "+string(abs(del_U))+" kJ of internal energy")