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+// Example 14_14

+clc;funcprot(0);

+// Given data

+m_ref=0.500;// kg/s

+T_0=25.0;// °C

+n_c=70.0;// The isentropic efficiency of compressor

+// Using Figure 14.36 as the illustration for this example, the properties at the four stations can be found in Tables C.7e, C.7f, and C.8d as

+// Station 1

+// Compressor inlet

+x_1=1.00;// The quality of steam

+T_1=-20.0;// °C

+h_1=235.31;// kJ/kg

+s_1=0.9332;// kJ/kg.K

+p_1=132.99;// kPa

+// Station 2

+// Compressor outlet

+p_2s=800;// kPa

+s_2=s_1;// kJ/kg.K

+h_2s=271.10;// kJ/kg

+T_2s=39.8;// °C

+// Station 3

+// Condenser outlet

+x_3=0.00;// The quality of steam

+p_3=725;// kPa

+h_3=87.46;// kJ/kg

+s_3=0.3257;// kJ/kg.K

+T_3=27.9;// °C

+// Station 4h

+// Expansion valve outlet

+h_4h=h_3;// kJ/kg

+p_4h=160;// kPa

+h_4h=87.46;// kJ/kg

+x_4h=0.280;// The quality of steam

+s_4h=0.3449;// kJ/kg.K

+T_4h=-15.6;// °C

+T_e=-15.6;// °C

+

+// Calculation

+// (a)

+h_2=((h_2s-h_1)/(n_c/100))+h_1;// kJ/kg

+p_2=p_2s;// kPa

+//Interpolation in Table C.7f in Thermodynamic Tables to accompany Modern Engineering Thermodynamics (or through the use of an appropriate computer program) gives the following additional properties at this state:

+s_2=0.9814;// kJ/kg.K

+T_2=54.97;// °C

+Q_condenser=m_ref*(h_3-h_2);// kJ/s

+Q_evaporator=m_ref*(h_1-h_4h);// kJ/s

+Q_compressor=m_ref*(h_2-h_1);// kJ/s

+I_ac=m_ref*(T_0+273.15)*(s_2-s_1);// kW

+I_con=(T_0+273.15)*((m_ref*(s_3-s_2))-(Q_condenser/(T_0+273.15)));// kW

+I_ev=m_ref*(T_0+273.15)*(s_4h-s_3);// kW

+I_e=(T_0+273.15)*((m_ref*(s_1-s_4h))-(Q_evaporator/(T_e+273.15)));// kW

+I_total=I_ac+I_con+I_ev+I_e;// kW

+W_compressor=Q_compressor;// kW

+// (b)

+COP=Q_evaporator/W_compressor;// The system coefficient of performance

+T_L=T_e;// °C

+COP_act=2.85;// The second law efficiency for a refrigeration system

+E_RAC=(abs(1-((T_0+273.15)/(T_e+273.15)))*COP_act)*100;// %

+printf("\n(a)The irreversibility rate of each component in the system are given below: \n   I_adiabatic compressor=%1.2f kW \n   I_condenser=%1.2f kW \n   I_expansion valve=%1.2f kW \n   I_evaporator=%1.2f kW \n   The total irreversibility rate of the system,I_total=%2.0f kW \n(b)The system coefficient of performance,COP=%1.2f \n   The second law efficiency for a refrigeration system,E_RAC=%2.1f percentage",I_ac,I_con,I_ev,I_e,I_total,COP,E_RAC);

+// The answer provided in the text book is wrong(The value of h_2 changed little bit)