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+clear;
+clc;
+disp('Example 5.19')
+
+//  aim : To determine the 
+//  (a) Gamma,
+//  (b) del_U
+
+//  Given Values
+P1 = 1400;//  [kN/m^2]
+P2 = 100;//  [kN/m^2]
+P3 = 220;//  [kN/m^2]
+T1 = 273+360;//  [K]
+m = .23;// [kg]
+cp = 1.005;//  [kJ/kg*K]
+
+//  Solution
+T3 = T1;//  since process 1-3 is isothermal
+
+//  (a)
+//  for process 1-3, P1*V1=P3*V3,so
+V3_by_V1 = P1/P3;
+//  also process 1-2 is adiabatic,so P1*V1^(Gamma)=P2*V2^(Gamma),hence
+//  and process process 2-3 is iso-choric so,V3=V2 and
+V2_by_V1 = V3_by_V1;
+//  hence,
+Gamma = log(P1/P2)/log(P1/P3); //  heat capacity ratio
+
+mprintf('\n (a) The value of adiabatic index Gamma is  =  %f\n',Gamma);
+
+//  (b)
+cv = cp/Gamma;//  [kJ/kg K]
+//  for process 2-3,P3/T3=P2/T2,so
+T2 = P2*T3/P3;//  [K]
+
+//   now
+del_U = m*cv*(T2-T1);//  [kJ]
+mprintf('\n (b) The change in internal energy during the adiabatic expansion is U2-U1 =  %f kJ (This is loss of internal energy)\n',del_U);
+//   End