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+// Example 8_12
+clc;funcprot(0);
+// Given data
+T_1=-20;// °C
+T_3=41.64;// °C
+m_L=0.6;// kg/s
+P_L=151;// kPa
+P_H=1000;// kPa
+
+// Calculation
+P_i=(P_L*P_H)^(1/2);// kPa
+// From appendix D we find,
+h_1=178.6;// kJ/kg
+s_1=0.7082;// kJ/kg.K
+s_2=s_1;// kJ/kg.K
+h_7=76.3;// kJ/kg
+h_8=h_7;// kJ/kg
+h_3=(((389-320)/(400-320))*(43.6-37.1))+37.1;// kJ/kg
+h_4=h_3;// kJ/kg
+s_6=(((389-320)/(400-320))*(0.6928-0.6960))+0.6960;// kJ/kg.K
+s_5=s_6;// kJ/kg.K
+h_5=(((389-320)/(400-320))*(190.97-188.0))+188.0;// kJ/kg
+// At P_i=389 kPa we interpolate and obtain
+// T=10°C s=0.6993 kJ/kg.K h=193.8 kJ/kg
+// T=20°C s=0.7226 kJ/kg.K h=200.3 kJ/kg
+// This gives
+h_2=(((0.7082-0.6993)/(0.7226-0.6993))*(200.3-193.8))+193.8;// kJ/kg
+// Also,extrapolating,we find
+h_6=(((0.6932-0.7021)/(0.7254-0.7021))*(217.8-210.2))+210.2;// kJ/kg
+Q_E=m_L*(h_1-h_4);// kW
+m_H=m_L*((h_2-h_3)/(h_5-h_8));// The mass flux in the high pressure stage in kg/s
+W_in=(m_L*(h_2-h_1))+(m_H*(h_6-h_5));// The power input to the compressors in kW
+COP=Q_E/W_in;// The coefficient of performance
+printf("\nThe rate of refrigeration,Q_E=%2.1f kW \nThe coefficient of performance,COP=%1.2f",Q_E,COP);