Refrigeration and air conditioning

Introduction to Refrigeration and Air - Conditioning

Introduction to
Refrigeration & Air
Conditioning
Applied Thermodynamics & Heat Engines
S.Y. B. Tech.
ME0223 SEM - IV
Production Engineering
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Applied Thermodynamics & Heat Engines

S. Y. B. Tech. Prod Engg.


Introduction to Refrigeration and Air - Conditioning

Outline
•

Applications of Refrigeration.

•

Bell – Coleman Cycle.

•

COP and Power Calculations

•

Vapour – Compression Refrigeration System.

•

Presentation on T-S and P-h diagram.

•

Vapour – Absorption System.

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Introduction to Refrigeration and Air - Conditioning

Refrigeration
REFRIGERATION – Science of producing and maintaining temperature below that of
surrounding / atmosphere.
REFRIGERATION – Cooling of or removal of heat from a system.
Refrigerating System – Equipment employed to maintain the system at a low temperature.
Refrigerated System – System which is kept at lower temperature.
Refrigeration

– 1) By melting of a solid,

2) By sublimation of a solid,
3) By evaporation of a liquid.

Most of the commercial refrigeration production : Evaporation of liquid.
This liquid is known as Refrigerant.
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Introduction to Refrigeration and Air - Conditioning

Refrigeration Circuit

Evaporator

Compressor

Refrigeration Circuit
Expansion
Valve

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Condenser

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Introduction to Refrigeration and Air - Conditioning

Refrigeration - Elements
High Temp
Source

Surrounding Air
QH
Condenser

QH
Wnet, in

Wnet, in

Expansion
Valve
Compressor
Evaporator

QL

QL
Refrigerated Space

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Low Temp
Sink

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Introduction to Refrigeration and Air - Conditioning

Refrigeration - Applications
1. Ice making.
2. Transportation of food items above and below freezing.
2. Industrial Air – Conditioning.
4. Comfort Air – Conditioning.
5. Chemical and related industries.
6. Medical and Surgical instruments.

Applications :

7. Processing food products and beverages.
8. Oil Refining.
9. Synthetic Rubber Manufacturing.
10. Manufacture and treatment of metals.
11. Freezing food products.
12. Manufacturing Solid Carbon Dioxide.
13. Production of extremely low temperatures (Cryogenics)
14. Plumbing.
15. Building Construction.

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Introduction to Refrigeration and Air - Conditioning

Refrigeration Systems
1. Ice Refrigeration System.
2. Air Refrigeration System.
2. Vapour Compression Refrigeration System.

Refrigeration Systems :

4. Vapour Absorption Refrigeration System.
5. Adsorption Refrigeration System.
6. Cascade Refrigeration System.
7. Mixed Refrigeration System.
8. Thermoelectric Refrigeration System.
9. Steam Jet Refrigeration System.
10. Vortex Tube Refrigeration System.

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Introduction to Refrigeration and Air - Conditioning

Performance - COP
Performance of Refrigeration System :
- Measured in terms of COP (Coefficient of Performance).
COP – Ratio of Heat absorbed by the Refrigerant while passing through the Evaporator
to the Work Input required to compress the Refrigerant in the Compressor.
If;
Then;

Rn = Net Refrigerating Effect.

Rn
W

COP =

Re lative COP =
Actual COP

W = Work required by the machine.

Actual COP
Theoretical COP

= Ratio of Rn and W actually measured.

Theoretical COP = Ratio of Theoretical values of Rn and W obtained by applying
Laws of Thermodynamics to the Refrigerating Cycle.

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Introduction to Refrigeration and Air - Conditioning

Performance - Rating
Rating of Refrigeration System :
- Refrigeration Effect / Amount of Heat extracted from a body in a given time.
Definition :
- Refrigeration Effect produced by melting 1 tonne of ice from and at 0 ºC in 24 hours.
Unit :
- Standard commercial Tonne of Refrigeration / TR Capacity
Latent Heat of ice = 336 kJ/kg.

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Introduction to Refrigeration and Air - Conditioning

Air Refrigeration System
One of the earliest method.
Obsolete due to low COP and high operating cost.

Preferred in Aircraft Refrigeration due to its low weight.
Characteristic :
- Throughout the cycle, Refrigerant remains in gaseous state.

Air Refrigeration
Closed System

Open System

• Air refrigerant contained within
piping or components of system.
• Pressures above atm. Pr.

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• Refrigerator space is actual room to be cooled.
• Air expansion to atm. Pr. And then
compressed to cooler pressure.
• Pressures limited to near atm. Pr. levels..

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Introduction to Refrigeration and Air - Conditioning

Air Refrigeration System
Closed System Vs. Open System :
1. Suction to compressor in Closed System may be at high pressures. Hence, the

size of Expander and Compressor can be kept small.
2. In Open Systems, air picks up the moisture from refrigeration chamber. This
moisture freezes and chokes the valves.
3. Expansion in Open System is limited to atm. Pr. Level only. No such restriction to
Closed System.

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Introduction to Refrigeration and Air - Conditioning

Isotherms

3

T1

T2

2
Adiabatic

Expansion
Compression

4

3

Temperature

Pressure

Reverse Carnot Cycle

Expansion

T1

Compression

4

1

4’

1’

1

T2

Entropy

Volume

P –V Diagram
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2

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T –s Diagram
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Introduction to Refrigeration and Air - Conditioning

Reverse Carnot Cycle
Operation :

Temperature

T2

3

1 – 2 : Adiabatic Compression.
Requires external power.

2

Temp. rises from T1 to T2.
Expansion

T1

Compression

1

4

Cylinder in contact with Hot Body at T2
2 – 3 : Isothermal Compression.
Heat Rejection to Hot Body.
3 – 4 : Adiabatic Expansion.

1’

4’
Entropy

Temp. falls from T2 to T1.
Cylinder in contact with Cold Body at T1.
4 – 1 : Isothermal Expansion.
Heat Extraction from Cold Body.

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Introduction to Refrigeration and Air - Conditioning

Reverse Carnot Cycle
Heat extracted from cold Body : Area 1-1’-4’-4
= T1 X 1-4

Temperature

T2

3

2

Expansion

T1

Work done per cycle

= (T2 – T1) X 1-4
Compression

4

1

4’

1’

Heat Extracted
Work Done
Area 1 − 1'−4'−4
=
Area 1 − 2 − 3 − 4

COP =

=

T1 X (1 − 4)
(T2 − T1 ) X (1 − 4)

=

T1
T2 − T1

Entropy

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: Area 1-2-3-4

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Introduction to Refrigeration and Air - Conditioning

Example 1
A Carnot Refrigerator requires 1.3 kW per tonne of refrigeration to maintain a region at low
temperature of -38 ºC. Determine:
ii)COP of Carnot Refrigerator.
iii)Higher temperature of the cycle.
iv)Heat delivered and COP, if the same device is used Heat Pump.

Heat absorbed 1 tonne
14,000 kJ / hr
COPrefrig =
=
=
= 2.99….ANS
Work done
1.3 kW (1.3 kW ) (3600 sec/ hr )
T1
235 K
COPrefrig =
⇒ 2.99 =
⇒ T1 = 313.6 K ….ANS
T2 − T1
T2 − 235 K
Heat Delivered as Heat Pump ;

= Heat absorbed + Work done
14,000 kJ / hr
+ 1.3 = 5.189 kJ / sec ….ANS
3600

Heat delivered 5.189 kJ / sec
COPHP =
=
= 3.99 ….ANS
Work done
1.3 kW
= 1 tonne + 1.3 kW =

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Introduction to Refrigeration and Air - Conditioning

Example 2
A refrigerating system works on reverse Carnot cycle. The higher temperature in the system is
35 ºC and the lower temperature is -15 ºC. The capacity is to be 12 tonnes. Determine :
ii)COP of Carnot Refrigerator.
iii)Heat rejected from the system per hour.
iv)Power required.

COPrefrig =

T1
258 K
=
= 5.18 ….ANS

T2 − T1 308 K − 258 K

Re frig . Effect
12 tonne
12 X 14,000 kJ / hr
⇒ 5.16 =
=
Work Input
Work Input
Work Input
⇒ Work Input = 32558 kJ / hr

COPrefrig =

Heat Rejected / hr = Refrig. Effect / hr + Work Input / hr
= 12 x 14,000 (kJ/hr) + 32,558 (kJ/hr) = 2,00,558 kJ/hr. ….ANS

Power =

Work Input / hr
32558 kJ / hr
=
= 9.04 kW ….ANS
3600
3600

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Introduction to Refrigeration and Air - Conditioning

Example 3
Ice is formed at 0 ºC from water at 20 ºC. The temperature of the brine is -8 ºC. Find out the kg
of ice per kWh. Assume that the system operates on reversed Carnot cycle. Take latent heat of
ice as 335 kJ/kg.

COPrefrig

T1
265 K
=
=
= 9.46
T2 − T1 293 K − 265 K

Heat to be extracted per kg of water ( to from ice at 0 ºC)
Rn = 1 (kg) x Cpw (kJ/kg.K) x (293– 273) (K) + Latent Heat (kJ/kg) of ice
= 1 (kg) x 4.18 (kJ/kg.K) x 20 (K) + 335 (kJ/kg)
= 418.6 kJ/kg.
Also,

1 kWh = 1 (kJ) x 3600 (sec/hr) = 3600 kJ.

Rn Re frig . Effect (kJ )
=
W

Work done (kJ )
m (kg ) X 418.6 (kJ / kg )
⇒ 9.46 = ice
⇒ mice = 81.35 kg….ANS
3600 kJ
COPrefrig =

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Introduction to Refrigeration and Air - Conditioning

Bell – Coleman / Reverse Bryaton Cycle
Elements of this system :
Cooling
Water

Heat Exchanger

Cold Air

Hot Air

Expander

Compressor

Very Cold Air

•

Compressor.

•

Heat Exchanger.

•

Expander.

•

Refrigerator.

Warm Air

Work gained from Expander is used
to drive Compressor.

Refrigerator

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Hence, less external work is required.

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Introduction to Refrigeration and Air - Conditioning

Isobars

Isobars

2

2

3

Adiabatic
Expansion

Temperature

Pressure

Bell – Coleman / Reverse Bryaton Cycle

3

Compression

1

4

Compression
Expansion

1

4

Entropy

Volume
P –V Diagram
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Adiabatic

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T –s Diagram
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Introduction to Refrigeration and Air - Conditioning

Bell – Coleman / Reverse Bryaton Cycle
Heat Absorbed in Refrigerator :

Qadded = m C P (T1 − T4 )
Heat Rejected in Heat Exchanger :

Isobars

Temperature

2
3

Adiabatic

Qrejected = m C P (T2 − T3 )
If process changes from Adiabatic to Polytropic;

Compression
Expansion

4

Qexp n
Entropy

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1

n
( P2 V2 − P1 V1 )
n −1

n
( P3 V3 − P4 V4 )
=
n −1

Qcomp =

We know,

 γ −1

R = C P 
 γ 

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Introduction to Refrigeration and Air - Conditioning

Bell – Coleman / Reverse Bryaton Cycle
Net Work Done :

W = Wcomp − Wexp n
n
( P2 V2 − P1 V1 − P3 V3 + P4 V4 )
=
n −1
n

=
m R ( T2 − T1 − T3 + T4 )
n −1
n  γ −1 

 m C P ( T4 − T3 + T2 − T1 )
=
n −1  γ 
For Isentropic Process :

W = Wcomp − Wexp n

= m C P ( T4 − T3 + T2 − T1 )

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Introduction to Refrigeration and Air - Conditioning

Bell – Coleman / Reverse Bryaton Cycle
COP :

Qadded
Work Added
COP =
=

Qrejected − Qadded
Wnet
=

m C P (T1 − T4 )
 n   γ −1 
 m C P ( T4 − T3 + T2 − T1 )

 
 n −1  γ 

COP =

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(T1 − T4 )
 n   γ −1 
 ( T4 − T3 + T2 − T1 )

 
 n −1  γ 

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Introduction to Refrigeration and Air - Conditioning

Air Refrigeration Cycle - Merits / Demerits

Merits :
1. No risk of fire (as in case of NH3); as air is non – flammable.
2. Cheaper (than other systems); as air is easily available.
3. Weight per tonne of refrigeration is quite low (compared to other systems).

Demerits :
1. Low COP (compared with other systems).
2. Weight of air (as Refrigerant) is more (compared to other systems).

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Introduction to Refrigeration and Air - Conditioning

Example 4
A Bell – Coleman refrigerator operates between pressure limits of 1 bar and 8 bar. Air is drawn
from the cold chamber at 9 ºC, compressed and then cooled to 29 ºC before entering the
expansion cylinder. Expansion and compression follow the law PV1.35 = Const. Calculate the
theoretical COP.
For air, take γ = 1.4 and Cp = 1.003 kJ/kg.
Polytropic Compression 1-2 :

2

Pressure

302 K
P2
= 8 bar 3

P1
= 1 bar

 8 bar 

= (282 K ) 
 1 bar 

1.35−1
1.35

= 482.2 K

PV1.35=C

Polytropic Expansion 3-4 :
1
282 K

4
Volume

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 P2 
T2 = T1  

 P1 

n −1
n

n −1
n

 P3 
T3 = T4   ⇒ (302 K ) = T4
 P4 
⇒ T4 = 176.6 K

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 8 bar 


 1 bar 

1.35 −1
1.35

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Introduction to Refrigeration and Air - Conditioning

Example 4….contd
Heat Extracted from Cold Chamber :

= C P (T1 − T4 ) = 1.003 (kJ / kg ) X (282 K − 176.6 K ) = 105.7 kJ / kg
Heat Rejected to Heat Exchanger :

= C P (T2 − T3 ) = 1.003 (kJ / kg ) X (482.2 K − 302 K ) = 180.7 kJ / kg
Net Work Done :

Wnet
Wnet
Wnet

n  γ −1 

 m C P ( T4 − T3 + T2 − T1 )
=
n −1  γ 
1.35  1.4 − 1 
=

 (1.003 kJ / kg ) (176.6 K − 302 K + 482.2 K − 282 K )
1.35 − 1  1.4 
= 82.8 kJ / kg

COPrefrig

Heat absorbed 105.7 kJ / kg
=
=
= 1.27….ANS
Work done

82.8 kJ / kg

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