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+//Engineering and Chemical Thermodynamics
+// Example 3.16
+//Page no:144
+
+clear ; clc ;
+//Given
+P_1 = 120 * 10^3 ; //[N]
+P_2 = 900 * 10^3 ; //[N]
+h_4 = 25.486 ; //[kJ/mol], From table
+h_1 = h_4 ;
+h_2 = 39.295 ; //[kJ/mol], From table
+S_2 = 177.89 ; //[kJ/molK], From table
+S_3 = S_2 ; //[kJ/mol]
+h_3 = 43.578 ; //[kJ/mol] , Enthalpy corresponding to S3 value which equales to S2
+Q_dot_c_des = 10 ; //[kW]
+
+q_c = h_2 - h_1 ;
+Q_dot_c = h_2 - h_1 ;
+W_dot_c = h_3 - h_2 ;
+
+COP = Q_dot_c / W_dot_c ;
+n_dot = Q_dot_c_des / q_c ;
+disp(" Example: 3.16 Page no : 144") ;
+printf("\n COP of the refrigerator is = %.2f \n\n Mass flow rate needed = %.3f mol/s",COP,n_dot) \ No newline at end of file