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Chapter 9. Refrigeration and Liquefaction

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(1)

Chapter 9. Refrigeration and Liquefaction

(2)

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

(3)

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

(4)

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

C

Q

H

Carnot refrigeration

(5)

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)

(6)

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  

(7)

Vapor-compression refrigeration cycle on P-H diagram

P-H diagrams are more commonly used in refrigeration cycle than TS diagrams.

(8)

Example 9.1

A refrigerated space is maintained at 10 (

o

F) and cooling water is available at 70 (

o

F).

Refrigeration capacity is 12,000 Btu/h. The evaporator and condenser are of sufficient size that a 10 (

o

F) 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.

(9)

Solution (a)

75 . ) 5

67 . 459 0

( ) 67 . 459 80

(

) 67 . 459 0

( 

 

 

C H

C C

T T

T W

w Q

10

o

F Minimum temperature difference : T

C

= 0

o

F

T

H

= 80

o

F

(10)

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

(11)
(12)
(13)

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 '

 

 

 

(14)

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

m

1 2

C

 

 

(15)

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.

(16)

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)

(17)

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.

(18)

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

(19)

Schematic Diagram of an Absorption-Refrigeration Unit

Throttling Valve

Same as vapor- compression

refrigerator Compression

by heat engine

(20)

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 

 

(21)

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.

(22)

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?

(23)

Example 9.2

Hot reservoir (House coil) TH= 30 oC

Cold reservoir (underground coil)

TC = 10 oC

C

W 

Q 

C

s / 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

 

(24)

Example 9.2

Hot reservoir (underground coil)

TH= 25 oC

Cold reservoir (House coil) TC = 5 oC

s / kJ 60 Q 

C

Q 

H

In 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 

(25)

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

(26)

Liquefaction Processes

For small-scale commercial liquefaction plant, throttling process is

commonly employed.

Sufficiently low T and high P desired.

compression Constant P

cooling

(27)

Linde Liquefaction Process

Linde liquefaction process

Only depends on throttling expansion

Compression  precooling to ambient T  cooling by refrigeration

 throttling

(28)

Final Exam

Dec 13 (월요일), 7:00 – 9:00 PM

창의관 B113호 (08~09학번), 창의관 117호 (03~07학번)

CH.6 – CH.9

Recommend Problems, Quiz, Lecture Note, etc

(29)

Congratulations !

This is the end !

Merry Christmas & Happy New Year !

(30)

Homework

Problems

9.4, 9.10, 9.14

Due:

Recommend Problems

9.2, 9.7, 9.8, 9.13, 9.18, 9.22

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