1. “THERMODYNAMIC AND HEAT TRANSFER”
University of Rome – Tor Vergata
Faculty of Engineering – Department of Industrial Engineering
“THERMODYNAMIC AND HEAT TRANSFER”
REFRIGERATION CYCLE
dr. G. Bovesecchi
(7249)
Accademic Year

2. Refrigeration Cycle Ex. 16
A refrigeration room cools 4 tons of meat (cp=3.5 kJ/kg K) from -20 to -40°C in 2 hours. The refrigerator is made with a cascade of two refrigeration cycle, one working with R22 and one with R134a.
The ambient temperature is 25°C. The condenser of the R22 cycle is an heat exchanger which also is the evaporator of the R134a cycle. Draw the 2 cycle on the proper p-h diagram and then calculate:
the thermal power absorbed from the room;
the R22 and R134a mass flow rate;
the R22 and R134a C.O.P.;
the C.O.P. of the whole cycle.
Assume a 10°C ∆T in the R22 evaporator, in the R134a condenser and between R22 and R134a in the heat exchanger.
The R22 condenser pressure is 3.5bars

3. Refrigeration Cycle C A 5 6 7 8
R134a
R22

4. Refrigeration Cycle

5. Refrigeration Cycle
R22

6. Refrigeration Cycle Solution
We begin by calculating the thermal power absorbed from the room, which is equal to the one taken from the meat.
Then fixing each of the principal states of the two cycles on the respective p-h diagram.
The thermal power absorbed from the room is given by the relation:

7. Refrigeration Cycle
Fixing ∆T=10°C in the R22 evaporator, in the R134a condenser and between R22 and R134a in the heat exchanger we have:

8. Refrigeration Cycle
R22

9. Refrigeration Cycle
Fixing ∆T=10°C in the R22 evaporator, in the R134a condenser and between R22 and R134a in the heat exchanger we have:

10. Refrigeration Cycle

11. Refrigeration Cycle
Fixing ∆T=10°C in the R22 evaporator, in the R134a condenser and between R22 and R134a in the heat exchanger we have:

12. Refrigeration Cycle

13. Refrigeration Cycle
Fixing ∆T=10°C in the R22 evaporator, in the R134a condenser and between R22 and R134a in the heat exchanger we have:

14. Refrigeration Cycle
Now we can fix the principal states located of the two cycles on the two p–h diagram.
R22 Cycle:
At the inlet to the compressor (state 1), the refrigerant is a saturated vapor at -50°C.
From the diagram pressure, enthalpy and entropy can be read.

15. Refrigeration Cycle 1
682

16. Refrigeration Cycle
Now we can fix the principal states located of the two cycles on the two p–h diagram.
R22 Cycle:
At the inlet to the compressor (state 1), the refrigerant is a saturated vapor at -50°C.
From the diagram pressure, enthalpy and entropy are:
At the outlet of the compressor (state 2), the vapor is superheated. We use entropy of state 1 and pressure of state 3 to fix this state.

17. Refrigeration Cycle We consider ideal compression, so:
The evaporation is a T=const process, so:

18. Refrigeration Cycle 2 1
727

19. Refrigeration Cycle We consider ideal compression, so:
The evaporation is a T=const process, so:
At the exit of the evaporator, R22 is liquid (state 3) at 3.5 bar.

20. Refrigeration Cycle 3 2 4 1
490

21. Refrigeration Cycle 3 2 4 1

22. Refrigeration Cycle We consider ideal compression, so:
The evaporation is a T=const process, so:
At the exit of the evaporator, R22 is liquid (state 3) at 3.5 bar.
State 4 represent the exit from the valve, this is an isenthalpic process, so:

23. Refrigeration Cycle The Coefficient of Performance (COP) is:
In the same way the R134a:
State 5:

24. Refrigeration Cycle 5
387

25. Refrigeration Cycle
The Coefficient of Performance (COP) for R22 cycle is:
In the same way the R134a:
State 5:
State 6:

26. Refrigeration Cycle 6 5
427

27. Refrigeration Cycle
State 6:
State 7:

28. Refrigeration Cycle 7 6 5
250

29. Refrigeration Cycle
State 6:
State 7:
State 8:

30. Refrigeration Cycle 7 6 8 5
250

31. Refrigeration Cycle 7 6 8 5

32. Refrigeration Cycle State 6: State 7: State 8:
The Coefficient of Performance (COP) for R134a cycle is:

33. Refrigeration Cycle The R22 refrigerant mass flow rate is:
The Coefficient of Performance (COP) for the whole cycle is: