Chapter 9. Refrigeration and Liquefaction
Introduction
Refrigeration
Maintain the temperature below that of the surrounding
- Absorption of heat at a low temperature level
- Accomplished by evaporation of a liquid in a steady-state flow process - Then, recycled via compression and condensation
Liquefaction
- A gas is liquefied
- Very similar to refrigeration process
Introduction
Refrigeration process
Air conditioning of building
Preservation of foods and beverages
Manufacture of ice
Dehydration of gases
Purifications, separations
Low temperature reactions
Liquefaction process
Propane gases in cylinders
Liquid oxygen for rockets
LNG (Liquid Natural Gas)
Separation of air
9.1 The Carnot Refrigeration
Reversed heat-engine cycle
Heat is absorbed at low TC
Heat is rejected at high TH
Requires external source of energy : W
Coefficient of Performance (COP), w
- A measure of the effectiveness of a refrigerator
C H
C C
C H
T T
T W
Q
Q Q
W
w
Coefficient of Performance(COP)
Hot reservoir at TH
Cold reservoir at TC
C W
Q
CQ
HCarnot refrigeration
9.2 Vapor-Compression Cycle
Liquid returns to its original P by
expansion (Throttling process)
Vapor is cooled and condensed
(Rejection of heat at high TH)
Vapor is compressed to high P
Liquid evaporates at const T & P (Heat absorption
at low TC)
9.2 Vapor-Compression Cycle
2 3
1 C 2
H H
H H
W Q
w
1 2
C
H H
m Q
Rate of circulation of refrigerant Coefficient of Performance
Throttling Process (Const. H)
1 2
C
H H
Q
2
3
H
H
W
4 3
H
H H
Q
Vapor-compression refrigeration cycle on P-H diagram
P-H diagrams are more commonly used in refrigeration cycle than TS diagrams.
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 and 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).
(a)
What is the value of COP for a Carnot Refrigerator ?
(b)
Calculate COP and m for the vapor-compression cycle of Fig.9.1 if the compressor
efficiency is 0.80.
Solution (a)
75 . ) 5
67 . 459 0
( ) 67 . 459 80
(
) 67 . 459 0
(
C H
C C
T T
T W
w Q
10
oF Minimum temperature difference : T
C= 0
oF
T
H= 80
oF
Solution (b)
Const. S
1 2
4 3’ 3
We should find the enthalpy at 1, 2, 3, and 4!
State 2
0oF, saturation condition (V) P = 21.162 psia
H2= 103.015 Btu/lbm S2= 0.22525 Btu/lbm.F State 4
80oF, saturation condition (L) P = 101.37 psia
H4= 37.987 Btu/lbm State 3’
Read H from Fig.G.2.
P = 101.37 psia, S=0.22525
H3’ = 117 Btu /lbm H1 = H4
ln P
H
Solution (b)
Compressor efficiency is 0.8
595 .
120 48
. 17 115
. 103 H
H H
, Then
48 . 8 17
. 0
98 . 13 H H
H 8 H
. 0
98 . 13 015
. 103 117
H H
H
2 3
S
S 3 2
S '
Solution (b)
COP
Refrigerant Circulation Rate
72 . 105 3
. 103 595
. 120
978 .
37 015
. 103 H
H
H H
2 3
1
2
w
hr / lb 845 ,
978 1 .
37 105
. 103
000 ,
120 H
H
m Q
m1 2
C
9.3 The Choice of Refrigerant
In principle, COP of Carnot refrigerator is independent of the refrigerant.
Many factors for irreversibility in the vapor-compression cycle of refrigerator
cause the COP to depend on the choice of refrigerant.
Two requirements for refrigerant
Air cannot leak into the refrigeration system vapor pressure at the evaporator
temperature should be greater than atmospheric pressure (1 atm)
The vapor pressure at the condenser temperature should not be too high cost for high pressure
Also, we should consider toxicity, flammability, cost, corrosion properties, and
vapor pressure, environmental influence, etc.
Choice of Refrigerant
Ammonia, Methyl chloride, Carbon Dioxide, Propane and
other hydrocarbons
Halogenated Hydrocarbons (CFC, HCFC)
Fully halogenated chlorofluorocarbons are most common
- CCl3F (CFC-11), CCl2F2 (CFC-12)
- Cause severe ozone depletion no more in use
Replacement
- CHCl2CF3 (HCFC-123), CF3CH2F (HCF-134a), CHF2CF3 (HCF-125)
Cascade refrigeration systems
The limit of operation
The operating pressures of evaporator and condenser also limit T
H– T
C T
H fixed by environments, T
C T level of refrigerants
To overcome this limit, two or more refrigeration cycle employing different
refrigerants can be used.
9.4 Absorption Refrigeration
Condenser
Evaporator
Compressor Throttling
Valve
This part can be replaced by heat engine – a work
producing device
W
Vapor-compression refrigerator
Work of compression is supplied by an electric motor mostly from heat engine
Absorption refrigeration
Direct use of heat as the energy source for refrigeration
Schematic Diagram of an Absorption-Refrigeration Unit
Throttling Valve
Same as vapor- compression
refrigerator Compression
by heat engine
Analysis
C C
C
S
Q
T T W T
S H
H H
H S
H
T T
W T Q
T and 1 T
Q W
Work required for the refrigeration cycle
Heat required for the production of the work.
C C S
S H
H C
H
T
T T
T T
Q T
Q
9.5 The Heat Pump
Dual-purpose, reversed heat engine
Winter : Heating
Summer : Cooling
Economic advantage depends on the cost of electricity comparing to the cost of fuels such as oil and natural gases.
1. Compressor: This increases the pressure of the refrigerant so that it will accept the maximum amount of heat from the air.
2. Condenser: Coils that move heat to or from the outside air.
3. Evaporator: Coils that move heat to or from the air inside the home.
4. Air handler: Fan that blows the air into the ducts of the home.
Components 1, 2, 3 and 4 are found in all standard air conditioners.
5. Reversing valve: Changes the heat pump from air conditioning to heating, and vice versa.
Example 9.2
A house has a winter heating requirement of 30 kJ/s and a summer cooling requirement of 60 kJ/s. Consider a heat-pump installation to maintain the house temperature at 20 oC in winter and 25 oC in summer. This requires circulation of the refrigerant through interior exchanger coils at 30 oC in winter and 5 oC in summer. Underground coils provide the heat source in winter and the heat sink in summer. For a year-round ground temperature of 15 oC, the heat-transfer characteristics of the coils necessitate refrigerant temperatures of 10 oC in winter and 25 oC in summer. What are the minimum power requirements for winter heating and summer cooling?
Example 9.2
Hot reservoir (House coil) TH= 30 oC
Cold reservoir (underground coil)
TC = 10 oC
C
W
Q
Cs / kJ 30 Q
H
In winter,
a house coil at 30 oC is a hot reservoir (heat sink) and
a underground coil at 10 oC is a cold reservoir (heat source)
Carnot heat pump
s / kJ 98 . 1 02 . 28 30 Q
Q W
s / kJ 02 . 28 15 30
. 303
15 . Q 283
T Q T
T T Q
Q
C H
H H C C
C H C
H
Example 9.2
Hot reservoir (underground coil)
TH= 25 oC
Cold reservoir (House coil) TC = 5 oC
s / kJ 60 Q
C
Q
HIn summer,
a house coil at 5 oC is a cold reservoir (heat source) and a underground coil at 25 oC is a heat reservoir (heat sink)
Carnot heat pump
s / kJ 31 . 4 60 31 . 64 Q
Q W
s / kJ 31 . 64 15 60
. 278
15 . Q 298
T Q T
T T Q
Q
C H
C C H H
C H C
H
C
W
9.6 Liquefaction Processes
Liquefaction processes
By heat exchange at constant pressure
By an expansion process from which work is obtained (Adiabatic expansion)
By a throttling process
Liquefaction Processes
For small-scale commercial liquefaction plant, throttling process is
commonly employed.
Sufficiently low T and high P desired.
compression Constant P
cooling