Transcript Document 7199112

Refrigeration Cycles
‫ محمود عبدالوهاب‬/‫د‬
The vapor compression refrigeration
cycle is a common method for
transferring heat from a low
temperature to a high temperature.
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The above figure shows the objectives of refrigerators
and heat pumps. The purpose of a refrigerator is the
removal of heat, called the cooling load, from a lowtemperature medium. The purpose of a heat pump is
the transfer of heat to a high-temperature medium,
called the heating load. When we are interested in the
heat energy removed from a low-temperature space,
the device is called a refrigerator. When we are
interested in the heat energy supplied to the hightemperature space, the device is called a heat pump.
In general, the term heat pump is used to describe the
cycle as heat energy is removed from the lowtemperature space and rejected to the hightemperature space.
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The performance of refrigerators and heat pumps is
expressed in terms of coefficient of performance
(COP), defined as
Desired output Cooling effect
QL
COPR 


Required input
Work input
Wnet ,in
Desired output Heating effect
QH
COPHP 


Required input
Work input
Wnet ,in
Both COPR and COPHP can be larger than 1. Under the
same operating conditions, the COPs are related by
COPHP  COPR  1
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Can you show this to be true?
Refrigerators, air conditioners, and heat pumps are rated
with a SEER number or seasonal adjusted energy
efficiency ratio. The SEER is defined as the Btu/hr of
heat transferred per watt of work energy input. The Btu
is the British thermal unit and is equivalent to 778 ft-lbf of
work (1 W = 3.4122 Btu/hr). An SEER of 10 yields a
COP of 2.9.
SEER = COP * 3.4122
Refrigeration systems are also rated in terms of tons of
refrigeration. One ton of refrigeration is equivalent to
12,000 Btu/hr or 211 kJ/min. How did the term “ton of
cooling” originate?
1 TR = 12000 Btu/hr = 3.517 kW
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Reversed Carnot Refrigerator and Heat Pump
Shown below are the cyclic refrigeration device
operating between two constant temperature
reservoirs and the T-s diagram for the working
fluid when the reversed Carnot cycle is used.
Recall that in the Carnot cycle heat transfers
take place at constant temperature. If our
interest is the cooling load, the cycle is called
the Carnot refrigerator. If our interest is the
heat load, the cycle is called the Carnot heat
pump.
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The standard of comparison for refrigeration
cycles is the reversed Carnot cycle. A
refrigerator or heat pump that operates on the
reversed Carnot cycle is called a Carnot
refrigerator or a Carnot heat pump, and their
COPs are
COPR , Carnot
1
TL


TH / TL  1 TH  TL
COPHP , Carnot
1
TH


1  TL / TH TH  TL
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Notice that a turbine is used for the expansion process
between the high and low-temperatures. While the
work interactions for the cycle are not indicated on the
figure, the work produced by the turbine helps supply
some of the work required by the compressor from
external sources.
Why not use the reversed Carnot refrigeration cycle?
•Easier to compress vapor only and not liquid-vapor
mixture.
•Cheaper to have irreversible expansion through an
expansion valve.
What problems result from using the turbine instead of
the expansion valve?
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The Vapor-Compression Refrigeration Cycle
The vapor-compression refrigeration cycle has
four components: evaporator, compressor,
condenser, and expansion (or throttle) valve.
The most widely used refrigeration cycle is the
vapor-compression refrigeration cycle. In an
ideal vapor-compression refrigeration cycle, the
refrigerant enters the compressor as a
saturated vapor and is cooled to the saturated
liquid state in the condenser. It is then throttled
to the evaporator pressure and vaporizes as it
absorbs heat from the refrigerated space.
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The Vapor-Compression Refrigeration Cycle
The ideal vapor-compression cycle consists of four
processes.
Ideal Vapor-Compression Refrigeration Cycle
Process
Description
1-2
Isentropic compression
2-3
Constant pressure heat rejection in the
condenser
3-4
Throttling in an expansion valve
4-1
Constant pressure heat addition in the
evaporator
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The following diagrams, show the flow and T-s
diagram illustrates the refrigeration cycle.
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The P-h diagram is another convenient diagram
often used to illustrate the refrigeration cycle.
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The ordinary household refrigerator is a good example
of the application of this cycle.
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Thermodynamic analysis of ideal vapor-compression
cycle “ SSVCC” “ simple saturation vapor
compression refrigeration cycle
Q L
h h
COPR 
 1 4
Wnet ,in h2  h1
COPHP
Q H
h2  h3



Wnet ,in h2  h1
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Relative efficiency
Refrigerator relative efficiency is defined as the
ratio between the actual refrigeration cycle
coefficient of performance and that for Carnot
cycle at same heat source and heat sink
temperatures.
hR =
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Definitions
Refrigeration capacity: is the amount of heat transfer to the
evaporator in kW; for steady operations the refrigeration
capacity equal to the refrigeration load
Refrigeration effect: the amount of heat transfer to the
evaporator per one kg of refrigerant mass flow rate through it "
kJ/kg"
Condenser duty: is the amount of heat transfer from
condenser to surroundings in kW
Compressor indicated power: is power added to the
refrigerant during compression process ;kW
Compressor brake power: the power required to brake the
compressor kW; compressor brake power greater than the
compressor indicate power" defined the compressor
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mechanical efficiency"
Example 1
Refrigerant-134a is the working fluid in an
ideal compression refrigeration cycle. The
o
refrigerant leaves the evaporator at -20 C
and has a condenser pressure of 0.9 MPa.
The mass flow rate is 3 kg/min. Find
COPR and COPR, Carnot for the same Tmax
and Tmin , and the refrigeration capacity in
tons of refrigeration.
Using the Refrigerant-134a Tables, we
have
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
kJ
h1  238.41


Compressor inlet 
kg

o
T1  20 C
 s  0.9456 kJ
1

kg  K
x1  1.0



kJ
Compressor exit
 h2 s  278.23

kg
P2 s  P2  900 kPa 
o
kJ T2 s  43.79 C

s2 s  s1  0.9456
kg  K 
State 3

kJ
h3  101.61


Condenser exit 
kg

P3  900 kPa 
kJ
s3  0.3738

kg  K
x3  0.0


 x4  0.358

Throttle exit 
kJ
o 
T4  T1  20 C  s4  0.4053
kg  K


h4  h3
State1
State 2
State 4
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COPR 
QL
m(h1  h4 ) h1  h4


Wnet , in m(h2  h1 ) h2  h1
kJ
kg

kJ
(278.23  238.41)
kg
 3.44
(238.41  101.61)
The tons of refrigeration, often called the cooling load or refrigeration effect, are
QL  m(h1  h4 )
kg
kJ 1Ton
(238.41  101.61)
min
kg 211 kJ
min
 1.94 Ton
3
COPR , Carnot
TL

TH  TL
(20  273) K

(43.79  ( 20)) K
 3.97
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Another measure of the effectiveness of the
refrigeration cycle is how much input power to
the compressor, in horsepower, is required for
each ton of cooling.
The unit conversion is 4.715 hp per ton of
cooling.
Find the system SEER?
Wnet , in
QL
4.715

COPR
4.715 hp

3.44 Ton
hp
 1.37
Ton
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ACTUAL VAPOR-COMPRESSION REFRIGERATION CYCLE
An actual vapor-compression
refrigeration cycle differs from the ideal
one in several ways, owing mostly to the
irreversibilities that occur in various
components, mainly due to fluid friction
(causes pressure drops) and
heat transfer to or from the
surroundings.
The COP decreases as a result of
irreversibilities.
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ACTUAL VAPOR-COMPRESSION REFRIGERATION CYCLE
Schematic and T-s diagram for the actual
vapor-compression refrigeration cycle.
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ACTUAL VAPOR-COMPRESSION REFRIGERATION CYCLE
DIFFERENCE:
The difference between actual and simple
refrigeration cycles are:
1- Non-isentropic compression
2- Superheated vapor at evaporator exit
3- Sub-cooled liquid at condenser exit
4- Pressure drops in condenser and evaporator
5- pressure drop in suction and discharge valves.
6- refrigerant heating during suction stroke.
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ACTUAL VAPOR-COMPRESSION REFRIGERATION CYCLE
Compressor isentropic efficiency
Polytrophic power
Pressure drop in suction and discharge valves
Refrigerant heating during suction process.
Actual volumetric efficiency
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Reciprocating compressors
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Single acting
Clearance volumetric efficiency=
Apparent “clearance” volumetric
efficiency
• Clearance ratio C = clearance volume/
swept volume
• Clearance volume = volume above the top
dead center = Vc
• Swept volume = volume of the cylinder
between top and bottom dead centers =
volume above the bottom dead center volume above the top dead center
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• P
Pd
PS
VC
V1
V2
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• Actual volume = volume above bottom dead
center – volume of residual gages after
expansion from discharge to suction
pressure = V2 – V1
• Theoretical volume = swept volume =
volume above bottom dead center - volume
above top dead center = V2 - Vc
• Clearance ratio C = clearance volume /
swept volume = Vc / (V2 – V1)
•
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Volumetric efficiency =
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Compressor power
Compressors are classified into the following
categories from Th.D. point of view
1- isentropic compressors “ Adiabatic + reversible”
Q =0 and no internal and external
irreversibleness
S = constant IW isent = m ref Dh isent
2- Adiabatic compressors, Q= 0 and DS
0
IW adiab = m ref Dh act.
Compressor efficiency = compressor isentropic
efficiency = IW isent / IW adiab = Dh isent / Dh act
Compressor power
• 3- poly-tropic compressors
• During the compression process the
refrigerant follows the relation
Pv n = constant = C
Since the indicated work is given as
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Compressor power
• Since,
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Heat transfer to poly-tropic
compressor
• Find the rate of heat transfer to the polytropic compressor?
•
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Heat Pump Systems
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