Heat Engine, Pump, Refrigerator
Heat Engine, Pump, Refrigerator
Heat Engine, Pump, Refrigerator
While the name “HEAT PUMP” is the thermodynamic term used to describe a
cyclic device that allows the transfer of heat energy from a low temperature to a
higher temperature, we use the terms “REFREGRATOR” and “heat pump” to
apply to particular devices. Here a refrigerator is a device that operates on a
thermodynamic cycle and extracts heat from a low-temperature medium. The heat
pump also operates on a thermodynamic cycle but rejects heat to the high-
temperature medium.
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Coefficient of Performance, COP
For the refrigerator the desired result is the heat supplied at the low temperature and
the input is the net work into the device to make the cycle operate.
QL
COPR
Wnet , in
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Now apply the first law to the cyclic refrigerator.
( QL QH ) ( 0 Win ) U cycle 0
Win Wnet , in QH QL
and the coefficient of performance becomes
QL
COPR
QH Q L
For the device acting like a “heat pump,” the primary function of the device is the
transfer of heat to the high-temperature system. The coefficient of performance for a
heat pump is
QH QH
COPHP
Wnet , in QH Q L
Noteunder the same operating conditions the COPHP and COPR are related by
COPHP COPR 1
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Heat Pump and Air Conditioner Ratings
Heat pumps and air conditioners are rated using the SEER system. SEER is the
seasonal adjusted energy efficiency (bad term for HP and A/C devices) rating. The
SEER rating is the amount of heating (cooling) on a seasonal basis in Btu/hr per unit
rate of power expended in watts, W.
The heat transfer rate is often given in terms of tons of heating or cooling. One ton
equals 12,000 Btu/hr = 211 kJ/min.
The following two statements of the second law of thermodynamics are based on the
definitions of the heat engines and heat pumps.
It is impossible for any device that operates on a cycle to receive heat from a single
reservoir and produce a net amount of work.
Heat engine that violates the Kelvin-Planck statement of the second law
The Clausius statement of the second law states that it is impossible to construct a
device that operates in a cycle and produces no effect other than the transfer of heat
from a lower-temperature body to a higher-temperature body.
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Heat pump that violates the Clausius statement of the second law
Or energy from the surroundings in the form of work or heat has to be expended to
force heat to flow from a low-temperature medium to a high-temperature medium.
Thus, the COP of a refrigerator or heat pump must be less than infinity.
COP
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A violation of either the Kelvin-Planck or Clausius statements of the
second law implies a violation of the other. Assume that the heat engine
shown below is violating the Kelvin-Planck statement by absorbing heat
from a single reservoir and producing an equal amount of work W. The
output of the engine drives a heat pump that transfers an amount of heat QL
from the low-temperature thermal reservoir and an amount of heat QH + QL
to the high-temperature thermal reservoir. The combination of the heat
engine and refrigerator in the left figure acts like a heat pump that transfers
heat QL from the low-temperature reservoir without any external energy
input. This is a violation of the Clausius statement of the second law.
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The Carnot Cycle
French military engineer Nicolas Sadi Carnot (1769-1832) was among the first to
study the principles of the second law of thermodynamics. Carnot was the first to
introduce the concept of cyclic operation and devised a reversible cycle that is
composed of four reversible processes, two isothermal and two adiabatic.
The Carnot Cycle
Process 2-3:Reversible adiabatic expansion during which the system does work
as the working fluid temperature decreases from TH to TL.
Process 3-4:The system is brought in contact with a heat reservoir at TL < TH and
a reversible isothermal heat exchange takes place while work of compression is
done on the system.
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You may have observed that power cycles operate in the clockwise direction when
plotted on a process diagram. The Carnot cycle may be reversed, in which it
operates as a refrigerator. The refrigeration cycle operates in the counterclockwise
direction.
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Carnot Principles
The second law of thermodynamics puts limits on the operation of cyclic devices as
expressed by the Kelvin-Planck and Clausius statements. A heat engine cannot
operate by exchanging heat with a single heat reservoir, and a refrigerator cannot
operate without net work input from an external source.
Consider heat engines operating between two fixed temperature reservoirs at TH > TL.
We draw two conclusions about the thermal efficiency of reversible and irreversible
heat engines, known as the Carnot principles.
As the result of the above, Lord Kelvin in 1848 used energy as a thermodynamic
property to define temperature and devised a temperature scale that is independent
of the thermodynamic substance.
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The following is Lord Kelvin's Carnot heat engine arrangement.
The thermal efficiencies of actual and reversible heat engines operating between the
same temperature limits compare as follows:
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REFRIGERATORS
AND HEAT PUMPS
The transfer of heat from a low-temperature
region to a high-temperature one requires
special devices called refrigerators.
Refrigerators and heat pumps are essentially
the same devices; they differ in their
objectives only.
Schematic of a
Carnot refrigerator
and T-s diagram
of the reversed
Carnot cycle.
THE IDEAL VAPOR-COMPRESSION REFRIGERATION CYCLE
The vapor-compression refrigeration cycle is the ideal model for refrigeration
systems. Unlike the reversed Carnot cycle, the refrigerant is vaporized completely
before it is compressed and the turbine is replaced with a throttling device.
This is the
most widely
used cycle for
refrigerators,
A-C systems,
and heat
pumps.
An ordinary
household
refrigerator.
The P-h diagram of an ideal vapor-
compression refrigeration cycle.
Reversed Carnot Device Coefficient of Performance
If the Carnot device is caused to operate in the reversed cycle, the reversible heat
pump is created.
• The COP of reversible refrigerators and heat pumps are given in a similar manner
to that of the Carnot heat engine as
QH
QL 1
COPR COPHP
QH
QL
QH Q L QH 1 QH Q L QH 1
QL QL
TL 1 TH
TH TL TH 1 TH TL
TL TH TL TH 1
TL
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Again, these are the maximum possible COPs for a refrigerator or a heat pump
operating between the temperature limits of TH and TL.
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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.
DIFFERENCES
Non-isentropic
compression
Superheated vapor
at evaporator exit
Subcooled liquid at
condenser exit
Pressure drops in
condenser and
evaporator
The claim is false since no refrigerator may have a COP larger than the COP for the
reversed Carnot device.
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Q1. A machine working on a Carnot cycle operates
between 305 k and 260 k. Determine the C.O.P. when it is
operated as: 1. a refrigerating machine; 2. A heat pump; 3. A
heat engine.
Solution. Given : T2 =305k ; T1= 260k
1.C.O.P. of a refrigerating machine
We know that C.O.P. of a refrigerating machine
(C.O.P.)R= {T1/(T2-T1)} = {260/(305-260)} = 5.78
2.C.O.P. of a heat pump
We know that C.O.P. of a heat pump
(C.O.P.)P= {T2/(T2-T1)} = {305/(305-260)} = 6.78
3.C.O.P. of a heat engine
We know that C.O.P. of a heat engine
(C.O.P.)E= {(T2-T1)/T2} = {(305-260)/305} = 0.147
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Q2. A carnot refrigeration cycle absorbs heat at 270k & reject it
at 300k.
1.calculate the cofficient of performance of this refrigeration
cycle.
2. if the cycle absorbing 1130kj/min at 270k, how many kj of
work is required per sec.?
3. if the carnot heat pumpoperates between the same
temperature as the above refrigeration cycle, what is the
cofficient of performance?
4. How many kj/min will the heat pump deliver at 300k if it
absorbs1130kj/min at 270k.
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Q3. The capacity of a refrigerator is 200 TR when working between
-60C & 250C. determine the mass of ice produced per day from water
at 250C. Also find the power required to drive the unit.Assume the
cycle operates on reversed carnot cycle and latent heat of ice is 335
kj/kg.
Solution: Given Q =200 TR ; T1= -60C = 267 k ; T2=298 k ; TW= 250C ;
hfg(ice)= 335 kj/kg
Mass of ice produced per day
Heat extraction capacity of refrigerator
= 200* 210 = 42000 kj/min
Heat removed from 1kg of water at 250C to form ice at 00C
=1*4.187(25-0)+335 = 439.7 kj/kg
Mass of ice produced per min
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=42000/439.7 = 95.52 kg/min
Mass of ice produced per day
=95.52*24 = 137550 kg
Power required to drive the unit
C.O.P. of reversed carnot cycle
(C.O.P.)R= {T1/(T2-T1)}
= {267/(298-267)} =8.6
Also C.O.P. =heat extraction capacity/ W.D per min
Work done per min = 42000/8.6 = 48.84 kj/min
Power required to drive the unit = 42000 / 60 = 81.4 kw.
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Q4. Five hundred kgs of fruits are supplied to a cold
storage at 200C. the cold storage is maintained at -50C
and fruits get cooled to the storage temp in 10 hours. The
latent heat of freezing is 105 kj/kg. and specific heat of
fruit is 1.256kj/kg k. find the refrigeration capacity of
plant.
Solution: Given m = 500 kg; T2= 293k ; T2 = 268 k; hfg=
105 kj/kg; cf=1.256kj/kg k.
Heat removed from fruit in 10 hours
Q1= m cf (T2-T1)
= 500* 1.256* (293-268) = 15700 kj
Total latent heat of freezing
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Q2= m* hfg = 500 * 105 = 52500 kj
Total heat removed in 10 hours,
Q = Q1+ Q2 =15700+52500 =68200 kj
And total heat removed in one min.
= 68200 / 10 *60 =113.7 kj/min
Refrigeration capacity of plant
= 113.7/210 = 0.541 TR.
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