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+// Example 8_11
+clc;funcprot(0);
+// Given data
+T_1=-10;// °C
+T_3=40;// °C
+P_1=0.15;// MPa
+n_c=0.80;// The efficiency of the compressor
+mdot=0.6;// kg/s
+
+// Calculation
+// From appendix D we find,using T_3=40°C
+h_4=74.5;// kJ/kg
+h_3=h_4;// kJ/kg
+// From table D-3 at P_1=0.15 MPa and T_1=10°C
+h_1=185;// kJ/kg
+s_1=0.732;// kJ/kg.K
+s_2a=s_1;// kJ/kg.K
+P_2=1.0;// MPa
+h_2a=(((0.732-0.7254)/(0.7476-0.7254))*(225.3-217.8))+218;// kJ/kg
+h_2=((h_2a-h_1)/n_c)+h_1;// kJ/kg
+Q_E=mdot*(h_1-h_4);// The rate of refrigeration in kW
+COP=Q_E/(mdot*(h_2-h_1));// The coefficient of performance
+printf("\nThe rate of refrigeration,Q_E=%2.1f kW \nThe coefficient of performance,COP=%1.2f",Q_E,COP);