1. Chapter 9. Refrigeration and Liquefaction (냉동과 액화)
고려대학교
화공생명공학과

2. Introduction Refrigeration Liquefaction Air conditioning of building
Preservation of foods and beverages
Manufacture of ice
Dehydration of gases
Purifications ,separations
Low temperature reactions
Liquefaction
Propane gases in cylinders
Liquid oxygen for rockets
LNG (Liquid Natural Gas)
Separation of air

3. 9.1 The Carnot Refrigeration
Reversed heat-engine cycle
Heat is absorbed at low T
Heat is rejected at high T
Requires external source of energy : W
Coefficient of Performance
(COP)

4. 9.2 Vapor-Compression Cycle4 3
Condenser
Const. H throttling process 3 3’
Throttling
Valve
Compressor 4 1 2 1
Evaporator 2
Coefficient of Performance
Rate of circulation of refrigerant

5. A Typical Compressor

6. Vapor-compression refrigeration cycle on P-H diagram
ln P
Const. S 4 3’ 3 1 2 H
P-H diagrams are more commonly used in refrigeration cycle than TS diagrams.

7. Example 9.1
A refrigerated space is maintained at 10 (oF) and cooling water is available at 70 (oF). Refrigeration capacity is 12,000 Btu/h. The evaporator condenser are of sufficient size that a 10 (oF) minimum temperature difference for heat transfer can be realized in each. The refrigerant is tetrafluoroethane (HCF-134a), for which data are given in Table 9.1 and Fig.G.2 (App.G).
Whate is the value of COP for a Carnot Refrigerator ?
Cacluate COP and m for the vapor-compression cycle of Fig.9.1 if the compressor efficiency is 0.80

8. Solution (a) 10 o F Minimum temperature difference : Tc = 0 oF
Th = 80 oF

9. Solution (b) H1,H2, H3, H4 값을 알면 모든 답을 구할 수 있다. State 2
0 oF, saturation condition (V)
P = psia
H2 = Btu/lbm
S2 = Btu/lbm.F
H1 = H4
ln P
Const. S
State 4
80 oF, saturation condition (L)
P = psia
H2 = Btu/lbm 4 3’ 3
State 3’
Read H from Fig.G.2.
P = psia, S=
 H3’ = 117 Btu /lbm 1 2 H

10. Solution (b)
Compressor efficiency is 0.8

11. Solution (b)
COP
Refrigerant Circulation Rate

12. 9.3 The Choice of Refrigerant
In principle, COP of Carnot refrigerator is independent of the refrigerant.
Irreversibility in the refrigerator cause the COP to depend on the choice of refrigerant .

13. The Choice of Refrigerant
Main characteristics
Vapor pressure of the refrigerant at evaporation T should be greater than 1 atm
Vapor pressure condenser T should not be high.
Additional characteristics
Toxicity
Flammability
Cost
Corrosion properties
Environmental consideration

14. Choice of Refrigerant
Ammonia, Methyl chloride, Carbon Dioxide, Propane and other hydrocarbons
Halogenated Hydrogcarons (CFC,HCFC)
CCl3F (CFC-11), CCl2F2 (CFC-12)
Replacement
CHCl2CF3 (HCF-123), CF3CH2F(HCF-134a), CHF2CF3 (HCF-125)

15. Cascade refrigeration systems
To overcome the limit of operation…
Tc fixed : environments
Two or more refrigeration cycle employing different refrigerant

16. 9.4 Absorption Refrigeration
This part can be replaced by heat engine – a work producing device
Condenser
Throttling
Valve
Compressor W
Evaporator

17. Schematic Diagram of an Absorption-Refrigeration Unit
Heat discarded at TS
QH at TH
Regenerator
Condenser
Throttling
Valve
Heat Exchanger
Absorber
Evaporator
QC at TC
Pump
LiBr Solution + Water
Water + Ammonia
Heat discarded at TS

18. Ammonia Absorption Refrigeration Unit

19. Analysis Work required for the refrigeration cycle Heat required for
the production of
the work.

20. 9.5 The Heat Pump Dual-purpose reversed heat engine
Winter : Heating
Summer : Cooling
If COP = 4, five times work has to be done to the compressor
Economic advantage depends on the cost of electricity vs. oil and natural gases.

21. 9.6 Liquefaction Processes
By heat exchange at const. P
By an expansion process from which work is obtained (Adiabatic expansion)
By a throttling process T S

22. Liquefaction Processes
For small-scale commercial liquefaction plant, throttling process is commonly employed.
Sufficiently low T and high P desired.

23. Linde Liquefaction Process4 7 3
Throttle
Cooler
Exchanger 8 10
Win 15 2 1 9
Liquid
Gas Feed

24. Linde Liquefaction Process

25. Congratulations !
This is the end !