“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 gianluigi.bovesecchi@gmail.com 06-7259-7127 (7249) Accademic Year 2018-2019

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

Refrigeration Cycle C A 5 6 7 8 R134a R22

Refrigeration Cycle

Refrigeration Cycle R22

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:

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:

Refrigeration Cycle R22

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:

Refrigeration Cycle

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:

Refrigeration Cycle

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:

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.

Refrigeration Cycle 1 682

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.

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

Refrigeration Cycle 2 1 727

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.

Refrigeration Cycle 3 2 4 1 490

Refrigeration Cycle 3 2 4 1

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:

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

Refrigeration Cycle 5 387

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

Refrigeration Cycle 6 5 427

Refrigeration Cycle State 6: State 7:

Refrigeration Cycle 7 6 5 250

Refrigeration Cycle State 6: State 7: State 8:

Refrigeration Cycle 7 6 8 5 250

Refrigeration Cycle 7 6 8 5

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

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