Heat Engine ,Pump,Refrigerator

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Heat Pump

A heat pump is a thermodynamic system operating in a thermodynamic cycle that removes heat from a low-temperature body and delivers heat to a high-temperature body. To accomplish this energy transfer, the heat pump receives external energy in the form of work or heat from the surroundings.

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.

The following figure illustrates a refrigerator as a heat pump operating in a thermodynamic cycle.

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Coefficient of Performance, COP

The index of performance of a refrigerator or heat pump is expressed in terms of the coefficient of performance, COP, the ratio of desired result to input. This measure of performance may be larger than 1, and we want the COP to be as large as possible.

COP Desired Result

Required Input

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For the heat pump acting like a refrigerator or an air conditioner, the primary function of the device is the transfer of heat from the low- temperature system.

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.

COPQ

WRL

net in

,

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Now apply the first law to the cyclic refrigerator.

( ) ( )

,

Q Q W U

W W Q QL H in cycle

in net in H L

0 0

and the coefficient of performance becomes

COPQ

Q QRL

H 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

COPQ

W

Q

Q QHPH

net in

H

H L

,

Noteunder the same operating conditions the COPHP and COPR are related by

COP COPHP R 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.

Second Law Statements

The following two statements of the second law of thermodynamics are based on the definitions of the heat engines and heat pumps.

Kelvin-Planck statement of the second law

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.

The Kelvin-Planck statement of the second law of thermodynamics states that no heat engine can produce a net amount of work while exchanging heat with a single reservoir only. In other words, the maximum possible efficiency is less than 100 percent.

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th < 100%

Heat engine that violates the Kelvin-Planck statement of the second law

Clausius 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 1-2:Reversible isothermal heat addition at high temperature, TH > TL, to the working fluid in a piston-cylinder device that does some boundary work.

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.

Process 4-1:A reversible adiabatic compression process increases the working fluid temperature from TL to TH

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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.

(a)The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs.

th th Carnot ,

(b) The efficiencies of all reversible heat engines operating between the same two constant-temperature heat reservoirs have the same efficiency.

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.

Since the thermal efficiency in general is

th L

H

Q

Q 1

For the Carnot engine, this can be written as

th L H L Hg T T f T T ( , ) ( , )1

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Considering engines A, B, and C

Q

Q

Q

Q

Q

Q1

3

1

2

2

3

This looks like

f T T f T T f T T( , ) ( , ) ( , )1 3 1 2 2 3

One way to define the f function is

f T TT

T

T

T

T

T( , )

( )

( )

( )

( )

( )

( )1 32

1

3

2

3

1

The simplest form of is the absolute temperature itself.

f T TT

T( , )1 3

3

1

The Carnot thermal efficiency becomes

th rev L

H

T

T, 1

This is the maximum possible efficiency of a heat engine operating between two heat reservoirs at temperatures TH and TL. Note that the temperatures are absolute temperatures.

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These statements form the basis for establishing an absolute temperature scale, also called the Kelvin scale, related to the heat transfers between a reversible device and the high- and low-temperature heat reservoirs by

Q

Q

T

TL

H

L

H

Then the QH/QL ratio can be replaced by TH/TL for reversible devices, where TH and TL are the absolute temperatures of the high- and low-temperature heat reservoirs, respectively. This result is only valid for heat exchange across a heat engine operating between two constant temperature heat reservoirs. These results do not apply when the heat exchange is occurring with heat sources and sinks that do not have constant temperature.

The thermal efficiencies of actual and reversible heat engines operating between the same temperature limits compare as follows:

REFRIGERATORS AND HEAT PUMPS

The objective of a refrigerator is to remove heat (QL) from the cold medium; the objective of a heat pump is to supply heat (QH) to a warm medium.

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.

for fixed values of QL and QH

THE REVERSED CARNOT CYCLE

Schematic of a Carnot refrigerator and T-s diagram of the reversed Carnot cycle.

Both COPs increase as the difference between the two temperaturesdecreases, that is, as TL rises or TH falls.

The reversed Carnot cycle is the most efficient refrigeration cycle operating between TL and TH.

However, it is not a suitable model for refrigeration cycles since processes 2-3 and 4-1 are not practical because

Process 2-3 involves the compression of a liquid–vapor mixture, which requires a compressor that will handle two phases, and process 4-1 involves the expansion of high-moisture-content refrigerant in a turbine.

THE IDEAL VAPOR-COMPRESSION REFRIGERATION CYCLEThe 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.

Schematic and T-s diagram for the ideal vapor-compression refrigeration cycle.

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.

The ideal vapor-compression refrigeration cycle involves an irreversible (throttling) process to make it a more realistic model for the actual systems. Replacing the expansion valve by a turbine is not practical since the added benefits cannot justify the added cost and complexity.

Steady-flow energy balance

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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

COPQ

Q Q QQ

T

T T TT

RL

H L H

L

L

H L H

L

1

1

1

1

COPQ

Q Q

QQQQ

T

T T

TTTT

HPH

H L

H

L

H

L

H

H L

H

L

H

L

1

1

Reversed Carnot Device Coefficient of Performance

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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.

The coefficients of performance of actual and reversible (such as Carnot) refrigerators operating between the same temperature limits compare as follows:

ACTUAL VAPOR-COMPRESSION REFRIGERATION CYCLE

Schematic and T-s diagram for the 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.

DIFFERENCESNon-isentropic compressionSuperheated vapor at evaporator exitSubcooled liquid at condenser exitPressure drops in condenser and evaporator

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Example 6-4

An inventor claims to have developed a refrigerator that maintains the refrigerated space at 2oC while operating in a room where the temperature is 25oC and has a COP of 13.5. Is there any truth to his claim?

QL

Win

QH

TH = 25oC

TL = 2oC

R

COPQ

Q Q

T

T T

K

K

RL

H L

L

H L

( )

( )

.

2 273

25 2

1196

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= 260k1.C.O.P. of a refrigerating machineWe know that C.O.P. of a refrigerating machine

(C.O.P.)R= {T1/(T2-T1)} = {260/(305-260)} = 5.782.C.O.P. of a heat pumpWe know that C.O.P. of a heat pump(C.O.P.)P= {T2/(T2-T1)} = {305/(305-260)} = 6.783.C.O.P. of a heat engineWe 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.

Solution: Given : T2 =300k ; T1= 270k1.C.O.P. of a carnot refrigeration cycleWe know that C.O.P. of a carnot refrigeration cycle,

(C.O.P.)R= {T1/(T2-T1)} = {270/(300-270)} = 9

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2. Work required per secLet WR = work required per sec.Heat absorbed at 270k,

Q1= 1130 kj/min = 18.83 kj/sec.We know that (C.O.P.)R = Q1/WR

WR= 18.83/9 = 2.1 kj/s3.C.O.P. of a carnot heat pumpWe know that C.O.P. of a heat pump(C.O.P.)P= {T2/(T2-T1)} = {300/(300-270)} = 104.Heat delivered by heat pump at 300kLet Q2 = heat delivered by pump at

300kHeat absorbed by 270k

Q1= 1130 kj/min

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We know that(C.O.P.)P= {Q2/(Q2-Q1)} ; 10 = Q2/( Q2-1130)10 Q2-11300 = Q2 ; Q2=1256 kj/min

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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/kgMass 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/kgMass of ice produced per min

=42000/439.7 = 95.52 kg/min

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Mass of ice produced per day=95.52*24 = 137550 kg

Power required to drive the unitC.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 minWork done per min = 42000/8.6 = 48.84 kj/minPower 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 kjTotal latent heat of freezing

Q2= m* hfg = 500 * 105 = 52500 kj

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Total heat removed in 10 hours,Q = Q1+ Q2 =15700+52500

=68200 kjAnd total heat removed in one min.

= 68200 / 10 *60 =113.7 kj/minRefrigeration capacity of plant

= 113.7/210 = 0.541 TR.