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+clear;
+clc;
+disp('Example 5.13');
+
+//  aim : To determine the
+//  final volume, work done and the change in internal energy
+
+//  Given values
+P1 = 700;//  initial pressure,[kN/m^2]
+V1 = .015;// initial volume, [m^3]
+P2 = 140;//  final pressure, [kN/m^2]
+cp = 1.046;//  [kJ/kg K]
+cv = .752; //  [kJ/kg K]
+
+//  solution
+
+Gamma = cp/cv;
+//  for adiabatic expansion, P*V^gamma=constant, so
+V2 = V1*(P1/P2)^(1/Gamma);//  final volume, [m^3]
+mprintf('\n The final volume of the gas is V2  =  %f  m^3\n',V2);
+
+//  work done
+W = (P1*V1-P2*V2)/(Gamma-1);//  [kJ]
+mprintf('\n The work done by the gas is  =  %f  kJ\n',W);
+
+//  for adiabatic process
+del_U = -W;//  [kJ]
+mprintf('\n The change of internal energy is  =  %f kJ',del_U);
+if(del_U>0)
+    disp('since del_U>0, so the the gain in internal energy of the gas ')
+else
+    disp('since del_U<0, so this is a loss of internal energy from the gas')
+end
+
+//  End