Question 9.5: Problem: A freezer working on a reverse Rankine cycle uses r...
Problem: A freezer working on a reverse Rankine cycle uses refrigerant 134a with a mass flow rate of 0.10 kg / s. The refrigerant leaves the evaporator as saturated vapour at a temperature of –8 °C. It leaves the condenser as saturated liquid at a pressure of 0.8 MPa. Find the power required to drive the compressor. Calculate the coefficient of performance ( COP_R ) of the freezer and compare it with the COP_R of a Carnot refrigerator operating between the same temperatures.
Find: Power \dot{W}_ c c required to drive the compressor, coefficient of performance of refrigeration COP_R of the Rankine cycle, the coefficient of performance of refrigeration COP_R of a Carnot cycle operating between the same temperatures.
Known: Refrigerant mass flow rate \dot{m} = 0.10 kg / s, evaporator temperature T_2 = T_1 = –8 °C, condenser pressure P_3 = P_4 = 0.8 MPa.
Process Diagram: The reverse Rankine cycle on a T‐s diagram (Figure E9.5).
Assumptions: Expansion through the compressor is isentropic so ∆S_{23} = 0.
Governing Equations:
Coefficient of performance Rankine COP_R=\frac{h_2 – h_1}{h_3 – h_2}
Power of compressor \dot{W} _c=\dot{m} (h_3-h_2)
We need to find the enthalpy at each of the four states marked in the process diagram.
For saturated refrigerant 134a at T_2 = –8 °C (Appendix 9a), specific enthalpy of vapour h_2 = h_g = 242.54 kJ / kg and specific entropy of vapour s_2 = s_g = 0.9239 kJ / kgK.
For saturated refrigerant 134a at P_4 = 0.8 MPa (Appendix 9b), specific enthalpy of liquid h_4 = h_f = 93.42 kJ / kg.
Expansion through the throttling valve is a constant enthalpy process, so that h_1 = h_4 = 93.42 kJ / kg.
Compression is isentropic, so s_3 = s_2 = 0.9239 kJ / kgK. At state 3, pressure P_3 = 0.8 MPa; with pressure and entropy at state 3, we can interpolate from the tables for superheated refrigerant 134a (Appendix 9c), giving h_3 = 269.49 kJ / kg.
The power required by the compressor is
\dot{W} _c=\dot{m} (h_3-h_2)=0.10 \ kg/s \times (269.49 \ kJ/kg – 242.54 \ kJ/kg)=2.6950 \ kW.
The coefficient of performance of the Rankine cycle is
COP_R=\frac{h_2 – h_1}{h_3 – h_2}=\frac{269.49 \ kJ/kg -93.42 \ kJ/kg}{269.49 \ kJ/kg – 242.54 \ kJ/kg} =6.5332.
Using the saturated refrigerant 134a tables (Appendix 9b), at 0.8 MPa T_H = T_{sat} = 31.33 °C = 304.48 K and T_C = –8 °C = 265.15 K. The coefficient of performance of the related Carnot cycle is
COP_{R,\text{Carnot}}=\frac{1}{T_H/T_C-1} =\frac{1}{(304.48 \ K/265.15 \ K)-1} =6.7417.
Answer: The compressor requires 2.70 kW of power. The COP_R of the freezer is 6.53, while the COP_R of a Carnot refrigerator operating between the same temperatures is 6.74.