1

CHAPTER 4REFRIGERATION &

HEAT PUMPS

THERMODYNAMIC II

2

TOPICS

3.1 Introduction3.2 Reversed Heat Engine Cycle3.3 Performance of Refrigeration Cycle and

Heat Pump3.4 The Ideal Vapor-Compression

Refrigeration Cycle3.5 The Practical Refrigeration Cycle3.6 Refrigeration Load

3.7 Flash Chamber3.8 Multistage Compression Refrigeration

System3.9 Cascade Refrigeration System3.10 Absorption Refrigeration Systems

3

INTRODUCTION

• A refrigerator is a heat engine in which work is done on a refrigerant substance in order to collect energy from a cold region and exhaust it in a higher temperature region, thereby further cooling the cold region.

• 2’nd Law of Thermodynamics (The Clausius statement) – “it is impossible for heat to flow from a colder body to a

warmer body without any work having been done to accomplish this flow”

4

REFRIGERATORS AND HEAT PUMPS

• Heat engines use heat to produce work.• Reversed heat engines use work to remove heat.• Refrigerators maintain cold space by removing heat

from it to a high-temperature region• Refrigerants are used as working fluid• Heat pumps maintain a hot space by absorbing heat

from a low temperature-region • Coefficient of Performance (COP) is the criteria used to

measure the performance of refrigerators & heat pumps

5

QL (Cooling Effect / Refrigerating Effect)

WARM ENVIRONMENT

MAINTAINED COLD ENVIRONMENT

REFRIGERATOR

Refrigerator

win

QH

in

L

in

L

WQ

WQ

&

&===

input workeffect cooling

COPR

6

MAINTAINED WARM

ENVIRONMENT

COLD ENVIRONMENT

HEAT PUMP

Heat Pump

win

QL

QH (Heating Effect)

in

H

in

H

WQ

WQ

&

&===

input workeffect heatingCOPHP

7

• From 1st Law of thermodynamics

• If COPR is positive, then COPHP > 1• The rate of heat removal from a system is called

cooling capacity.• Cooling capacity is normally measured in tons of

refrigeration• 1 ton = 211 kJ/min

1

1

+=

+=

+=

+=

RHP

in

LHP

in

inL

in

H

inLH

COPCOPWQCOP

WWQ

WQ

WQQ

8

CARNOT REVERSED HEAT ENGINE CYCLE

• The most efficient heat engine is represented by the Carnot cycle. (Remember that Carnot cycle is reversible)

• A reversed heat engine is represented by Carnot cycle which operates in a reversed direction

• This cycle is called a reversed Carnot cycle

• A refrigerator/heat pump using this cycle is called Carnot refrigerator/Carnot heat pump

• Its function is to remove heat from a low-temperature region to a high-temperature region.

9

Condenser

Evaporator

Turbine

4 1

23

QL

QH

Plant layout for Reversed Carnot Cycle

Pump

3

4 1

2

T

s

T-s Diagram with saturation line of Refrigerant

QH

QL

10

PROCESSES

• (1 – 2)– Wet vapor enters pump and is pumped

(Isentropic)– Temperature is increased

• (2 – 3)– Vapor is condensed at constant temperature– Heat rejected by refrigerant

• (3 – 4)– Isentropic expansion (Isentropic)– Temperature is reduced

• (4 – 1)– Heat for evaporation process is supplied from

cold source in evaporator at constant temperature.

11

PERFORMANCE OF REFRIGERATION CYCLE AND HEAT PUMP

• From T-s diagram,• TL= T1 = T4 and TH= T2 = T3

• s1 = s2 and s4 = s3

• QL= TL(s1 – s4) and QH= -TH(s3 – s2)

• Win = QH - QL

= -TH(s3 – s2)– T1(s1 – s4) = -T2(s4 – s1) – T1(s1 – s4) = (T2 – T1) (s1 – s4)

• So, COPR can be given as follows:

( )( )( )

1

1

21

1

4121

411

−=

−=

−=

−−−

=

=

L

HLH

LR

R

in

LR

TTTT

TTT

TCOP

ssTTssTCOP

WQCOP

12

•Similarly for COPHP, we get:

H

LLH

HHP

TTTT

TTT

TCOP−

=−

=−

=1

1

12

2

13

• EXAMPLE 4.1

• A refrigerator operates between evaporator temperature and condenser temperature of -30oC and 35oC respectively. Calculate the maximum possible COPR.

If the COPR for actual refrigerator is 80% of ideal refrigerator, calculate the power required for a cooling effect of 5kW.

14

• The reversed Carnot is the most efficient refrigeration cycle operating between two fixed temperatures

• This cycle is impractical because,– In process (1 – 2) it is difficult to compress

liquid-vapor mixture– In process (3 – 4) it is difficult to expand high-

moisture-content refrigerant.• Reversed Carnot cycle is only for comparison with

the actual refrigeration cycles

15

THE IDEAL VAPOR-COMPRESSION CYCLE

• To make the cycle practical;– the refrigerant is vaporized completely before

compression– The expansion engine (turbine) is replaced by a throttle

valve (expansion with no enthalpy change)• Since compression process is carried out in vapor state, the

cycle is then called the Vapor-Compression Cycle• This type of cycle is commonly used in domestic

refrigerators and air conditioning systems.

Condenser

Evaporator

WinExpansion Valve

1

2

4

3

16

PROCESS OF VAPOR-COMPRESSION CYCLE• (1 – 2)-Isentropic compression until vapor is superheated

• (2 – 3)-Constant pressure heat rejection in condenser

• (3 – 4)-Throttling in an expansion device

• (4 – 1)-Constant pressure heat absorption in an evaporator

T (K)

s(kJ/kgK

3 2

14

17

Condenser

Evaporator

WinExpansion Valve

1

2

4

3

T

s

1

2

4

Win

QL

QH

18

P-h DIAGRAM

• 2 methods can be used for cycle analysis.– Using property table for refrigerants– Using the P-h diagram

P

h

1

23

4s

cons

tant

v constant

x con

stant

q2 = h2 – h3

q2 = h1 – h4

win = h2 – h1

19

P-h D

iagram for R

efrigerant 134a

20

T

s

1

2

4

3’Cooling water temperature

4’

3

UNDERCOOLING (SUBCOOLING) AND ITS EFFECTS

• In the condenser, the vapor can be further cooled at constant pressure to a temperature that is lower than temperature in condenser

21

• Undercooling (subcooling) increases the refrigerating effect (h1 – h4) > (h1 – h4’) where h4 is enthalpy with undercooling(subcooling) and h4’ is initial enthalpy

• Undercooling (subcooling) is limited by temperature of cooling water and temperature difference of cycle

T

s

1

2

4

3’Cooling water temperature

4’

3

22

EXAMPLE 4.2

• In a refrigeration cycle, pressure of ammonia refrigerant is 1.902 bar and condenser pressure of 12.37 bar. Calculate refrigerating effect (QL) per unit mass and COPR for:-

i) Ideal reverse Carnot cycleii) Ideal vapor compression cycle without

superheating and undercoolingiii) Ideal vapor compression cycle with

superheatingiv) Ideal vapor compression cycle with

superheating and undercooling (subcooling) by 10K

23

THE PRACTICAL VAPOR-COMPRESSION CYCLES

• Because of the irreversible nature of most processes of the cycle, the actual cycle deviates from actual cycle.

• Source of irreversibilities– Pressure drop in fluid– Heat transfer with surroundings

• It is difficult to get saturated vapor at compressor inlet. So in practice the refrigerant is slightly superheated at compressor inlet.

• It is also difficult to get saturated liquid at condenser exit. So in practice undercooling (subcooling) (3 – 3a)is used.

24

T

s

Pressure drop

Undercooling

1a

2s

3

4

3a

4a 1

2a

2bPressure drop occurs in:

(1 – 1a) :line connecting evaporator and compressor(2 – 3):within condenser(4 – 4a):line between expansion valve and evaporator(4a – 1):within evaporator

• During actual compression, entropy might increase or decrease. (Point 2 might be somewhere between 2a and 2b)

25

EXAMPLE 4.3

• R12 enters compressor as saturated vapor at 1.509 bar and -20oC at a rate of 0.05 kg/s and leaves compressor at 8.477 bar. After cooling in condenser, the temperature is 26oC and 7.449 bar Then the refrigerant is throttled to 1.509 bar.

Calculate:-

i) Rate of heat removal from the refrigerated space

ii) power input to the compressor

iii) COPref

26

4.6 REFRIGERATION LOAD

massunitpereffectingrefrigeratcapacityorrefrigeratm

=&

• Refrigeration Capacity,

– defined as the amount of heat that has to be transferred from a cold space per unit time

– determines the mass flow rate of refrigerant

• 1 ton = 200Btu/min = 211kJ/min = 3.516kW

• ton : “the rate of heat transfer to produce 2000 lb of ice at 0oC (32o)F from liquid water at 0oC (32oF) in 24 hours”

• Mass flow rate of refrigerant

LQ&

27

EXAMPLE 4.4

• Calculate the refrigerating capacity of the refrigerator in unit ton if given the enthalpy at the outlet and inlet of the evaporator, and the mass flow rate are 179.01 kJ/kg, 60.58 kJ/kg and 0.05 kg/s respectively.

( )( )

ton.

ton.

...hhmQ

So, ..

.mGiven,

kJkJ

kgkJkg

kg

skg

6841211

129355

58600117903

0360050050

41

=

×=

−=

−=

=×=

=

minmin

min

min

&&

&

28

FLASH CHAMBER

• Flash chamber is used in a multi-staging refrigeration system

• It separates vapor and liquid refrigerant during the throttling process

• The purpose is to avoid vapor refrigerants from entering evaporator

• The vapor developed during throttling (flash vapor) is bled out of the throttling device and fed back to the compressor

29

MULTISTAGE COMPRESSION REFRIGERATION SYSTEM

• A multistage compression refrigeration system is one example of a system that uses a flash chamber

• It can be carried out with the use of one or more compressors

Condenser

Evaporator

WinExpansion Valve

1

4

4

5

Flash Chamber

Expansion Valve

8

7

6 9

2

3

Win

QH

QL

Cycle Layout of a Two-Stage Compression Refrigeration System

30

T-s DIAGRAM

• The T-s diagram representing the cycle of a two-stage vapor-compression cycle

1

2

39

4

5

67

8

s

T

31

TWO-STAGE REFRIGERATION CYCLE REPRESENTED BY THE P-h DIAGRAM

• The P-h diagram is a more convenient representation of the cycle because it can easily be compared to the plant layout

P

h

1

29

45

67

8

3

Condenser

Evaporator

Flash Chamber

32

• 1kg refrigerant starts his journey through condenser

• 1kg liquid enters 1st throttle valve• 1kg (mostly liquid) enters flash

chamber starts to evaporate and becomes mixture of gas (x)kg and liquid (1–x)kg

• (x) leaves early and is ready to enter 2nd stage compressor at Pi

• (1–x)kg liquid make its way through the 2nd throttle valve into the evaporator

• (1–x)kg vapor enters the 1st stage compressor where it is compressed to Pi

• At Pi (state 3) (1-x)kg vapor mixes with (x)kg vapor adiabatically and becomes 1kg vapor

• 1kg vapor is compressed in 2nd stage compressor

• 1kg vapor enters condenser to be condensed and becomes 1kg liquid

P

h

1

29

45

67

8

3

Condenser

Evaporator

33

ANALYSIS

• Fraction of refrigerant which evaporates in the flash chamber can be given as follows.

• Refrigerating Effect, QL= (1 – x)(h1 – h8)

• Total work input, ∑Win = W12 + W54

= (1 – x)(h2 – h1) + (h4 – h9)

• Heat rejected in condenserQH = (h4 – h5)

i

i

fg

f

hhh

x−

= 6

34

• Coefficient of Performance

( )( )( )( ) ( )9412

81

11

hhhhxhhx

WQCOP

in

LR

−+−−−−

=

=∑

35

EXAMPLE 4.5

• A vapor compression of a two-stage compression plant uses R-134a and has an evaporator temperature of -5oC and condenser temperature of 45oC. The vapor is dry saturated on entering the compressor. A flash chamber is employed at an inter stage saturation temperature of 15oC.Calculate:

i. The amount of vapor bled off at the flash chamberii. The state of vapor at the inlet to the 2nd stage of

compressioniii. The refrigerating effect per unit mass of refrigerant in

the evaporatoriv. The work done per unit mass of refrigerant in the

compressors.v. COPR

36

CASCADE REFRIGERATION SYSTEM

• A refrigeration process can be carried out in stages• We call refrigeration cycles that operate in series as

cascade refrigeration cycles• We will look at a two-stage cascade refrigeration

system• A heat exchanger will serve as an evaporator for

one cycle and a condenser for another

37

Cycle Layout of a Two-Stage Cascade System

WinExpansion

Valve

1

6

4

7

Expansion Valve

8

5

2

3Win

QH

QL

Heat Exchanger

Compressor

Compressor

Condenser

Evaporator

A

B

38

• The heat exchanger connects cycle A with cycle B• For cycle A, the heat exchanger acts as condenser• For cycle B, the heat exchanger acts as the evaporator• Assumptions

– Heat exchanger is insulated– Kinetic & potential energy is negligible– Same refrigerant is used in both cycles

• So heat leaving condenser in A is equal to heat entering absorbed by evaporator in B

THE PROCESS

( ) ( )( )( )85

32

3285

hhhh

mm

hhmhhmQQ

B

A

BA

BA

−−

=

−=−

=

&

&

&&

&&

39

• Refrigeration Coefficient of Performance for the system:

( )( ) ( )1256

41

hhmhhmhhm

WQCOP

A

B

in

LR

−+−−

=

=∑

&&

&

40

T-s AND P-h DIAGRAM OF THE CASCADE REFRIGERATION SYSTEM

1

2

5

6

7

83

4

s

T

B

A

P

h

1

2

67

83

4

5

B

A

41

CHARACTERISTICS OF CASCADE SYSTEM

• In a cascade system, no mixing of refrigerant takes place in the heat exchanger, – so no mixing of refrigerant between cycles, – so different refrigerants can be used

• Using a cascade system– Increases the refrigeration capacity– Decreases compressor work– So COPR increases

• In practice the working fluid of the lower cycle will be at a higher pressure and temperature in the heat exchanger for effective heat transfer

42

• A two stage cascade refrigeration system operates between pressure limits of 0.8 MPa and 0.14 MPa. Each stage operates on an ideal vapor-compression refrigeration cycle with R-134a as working fluid. Heat rejection from lower cycle to the upper cycle takes place in an adiabatic counterflow heat exchanger where both streams enter at about 0.4 MPa. If mass flow rate of refrigerant through the upper cycle is 0.24 kg/s. Determine

i. Mass flow rate through lower cycleii. Rate of heat removal from refrigerated spaceiii. Power input of compressoriv. COPR of this Cascade

EXAMPLE 4.6

43

ABSORPTION & GAS REFRIGERATION SYSTEM

Absorption Refrigeration System

• Introduced because:– saves energy– uses environmental friendly refrigerant

• Types of absorption refrigeration systems– Ammonia-Water– Water-Lithium bromide

• The system is similar to a vapor-compression system except that the compressor is replaced by an absorption system

• Refrigerant is absorbed by a transport medium through the absorbing system

44

SYSTEM LAYOUT OF AN AMMONIA-WATER ABSORPTION SYSTEM

Condenser

Evaporator

Win, Pump

Expansion Valve

QL

RectifierGeneratorNH3 + H2O

AbsorberNH3 + H2O

Pump

Expansion Valve

QGeneratorNH3

H2O

QH

NH3

Absorption system

Cooling water

Note:An identical layout is used for the water-ammonia absorption system

45

PROCESS IN THE ABSORPTION SYSTEM

• Take an Ammonia-Water system as an example. Here, Ammonia is the refrigerant and water is the absorber.

• In the case of water-lithium bromide system, water is the refrigerant and lithium bromide is the absorber.

• NH3 from evaporator enters the absorber. It reacts with cooling water and releases heat to form NH3 + H2O solution

• The NH3 + H2O is pumped to generator. Heat is transferred to NH3 + H2O to vaporize it

• Then it is passed to the rectifier to separate NH3 and H2O

• High pressure NH3 in rectifier goes to condenser• H2O & the rest of NH3 is passed through a regenerator.

It transfers heat to the solution rich with NH3 that is on its way to the generator

• Then it is throttled back to the absorber

46

ADVANTAGE & DISADVANTAGE

• Advantage – Pump work is relatively small compared to the

heat supplied to the generator• Disadvantage

– Rather complex• Coefficient of performance

generator

LR

pump

pumpgenerator

LR

QQCOP

WWQ

QCOP

=

+=

small is because

47

GAS REFRIGERATION CYCLE

• Can be represented by a reversed Brayton cycle

Turbine Compressor

Heat Exchanger

1 4

23

QH

Win

Heat Exchanger

QL

Figure: A Reversed Brayton Cycle Refrigeration system

48

T-S DIAGRAM AND COPR

T

s

4

2

1

3

QH

QL

( )( ) ( )4312

41

hhhhhh

WWQ

WQCOP

turcomp

L

in

LR

−−−−

=

−=

=