Introduction
How does
it work?
High Temperature
Reservoir
Heat Rejected
R
Work Input
Heat Absorbed
Low Temperature Reservoir
A N Khudaiwala,G.P.Porbandar
Refrigeration
It is defined as the process of providing and
maintaining a temperature well below that of
surrounding atmosphere.
In other words refrigeration is the process of cooling
substance.
A N Khudaiwala,G.P.Porbandar
Refrigerators and heat pumps
If the main purpose of the machine is to cool some
object the machine is named as refrigerator
If the main purpose of machine to heat a medium
warmer than the surroundings, the machine is
termed as heat pump.
A N Khudaiwala,G.P.Porbandar
Refrigerator and Heat pump
Warm Space
QR
Work Input
Heat
Pum
p
QR
Work Input
Refri
gerat
or
Cold Space
A N Khudaiwala,G.P.Porbandar
Refrigeration
REFRIGERATION – Science of producing and maintaining temperature below
that of
REFRIGERATION – Cooling of or removal of heat from a system.
surrounding / atmosphere.
Refrigerating System – Equipment employed to maintain the system at a
lowtemperature.
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.
ME0223 SEM-IV
Applied Thermodynamics & Heat Engines
A N Khudaiwala,G.P.Porbandar
Refrigerating EffectDefinitions
(N): It is defined as the
quantity of heat extracted from a cold body or
space to be cooled in a given time.
N= Heat extracted from the cold space
Time taken
Specific Heat of water and ice : It is the quantity
of heat required to raise or lower the
temperature of one kg of water (or ice),
through one kelvin or (10 c) in one second.
Specific heat of water, Cpw = 4.19 kJ/kg K
Specific heat of ice, Cpice = 2.1 kJ/kg K.
A N Khudaiwala,G.P.Porbandar
Definitions
Capacity of a Refrigeration
Unit :
 Capacity of a refrigerating machines are expressed by
their cooling capacity.
 The standard unit used for expressing the capacity of
refrigerating machine is ton of refrigeration.
 One ton of refrigeration is defined as, “the quantity of
heat abstracted (refrigerating effect) to freeze (into ice)
one ton of water in a duration of 24 hours at 0o c”.
Heat extracted from at oo c = latent heat of ice
Latent heat of ice = 336 kJ/kg
i.e., 336 kJ of heat should be extracted one kg of water at
0o C to convert it into ice.
A N Khudaiwala,G.P.Porbandar
One ton of Refrigeration
One ton of refrigeration= 336x1000 kJ/24 hrs.
= 336x1000 kJ/min
24x60
One ton of refrigeration = 233.333 kJ/min
= 3.8889 kJ/sec
For calculation purpose,
One ton of refrigeration = 12600 kJ/hr
= 210 kJ/min
Ton of refrigeration
= 3.5 kJ/s
A N Khudaiwala,G.P.Porbandar
- COP
Performance of Definitions
Refrigerators
(Co efficient of
Refrigerators) :
The performance of heat engine is expressed by its
thermal efficiency.
The performance of a refrigerator cannot be
expressed in terms of efficiency.
In case of a refrigerator the aim is to extract
maximum quantity of heat from the sink with
minimum of work input.
Hence a new term Co efficient of Performance is
brought into use to express the performance of
refrigerator.
A N Khudaiwala,G.P.Porbandar
Definitions -ItCOP
Co efficient of Performance:
is defined as the
ratio of heat extracted in a given time (refrigerating
effect) to the work input.
Co efficient of performance = Heat extracted in evaporator
Work Input
Co efficient of performance =
Refrigerating Effect
Work Input
Co efficient of performance =
N
W
The COP is always greater than 1 and known as theoretical
coefficient of performance.
A N Khudaiwala,G.P.Porbandar
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.
A N Khudaiwala,G.P.Porbandar
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
• Air refrigerant contained within
piping or components of
system.
• Pressures above atm. Pr.
Open System
• Refrigerator space is actual room to be
cooled.
• Air expansion to atm. Pr. And then
compressed to cooler pressure.
A N Khudaiwala,G.P.Porbandar
• Pressures limited to near atm. Pr. levels..
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.
A N Khudaiwala,G.P.Porbandar
Isotherms
3
T1
2
Adiabatic
Expansion
Temperature
Pressure
Reverse Carnot Cycle
Compression
4
T2
3
2
Expansion
T1
Compression
4
1
4’
1’
1
T2
Entropy
Volume
P –V Diagram
A N Khudaiwala,G.P.Porbandar
T –s Diagram
Reverse Carnot Cycle
Temperature
Operation :
T2
3
1 – 2 : Adiabatic Compression.
Requires external power.
2
Temp. rises from T1 to T2.
Expansion
T1
Cylinder in contact with Hot Body at T2
Compression
1
4
2 – 3 : Isothermal Compression.
Heat Rejection to Hot Body.
3 – 4 : Adiabatic Expansion.
4’
Temp. falls from T2 to T1.
1’
Cylinder in contact with Cold Body at T1.
Entropy
4 – 1 : Isothermal Expansion.
Heat Extraction from Cold Body.
A N Khudaiwala,G.P.Porbandar
Reverse Carnot Cycle
Heat extracted from cold Body : Area 1-1’-4’-4
Temperature
= T1 X 1-4
T2
3
2
Expansion
T1
Work done per cycle
= (T2 – T1) X 1-4
Compression
4
1
4’
1’
: Area 1-2-3-4
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
A N Khudaiwala,G.P.Porbandar
Example 1
A Carnot Refrigerator requires 1.3 kW per tonne of refrigeration to maintain a region at
low temperature of -38 ºC. Determine:
i)COP of Carnot Refrigerator.
ii)Higher temperature of the cycle.
iii)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 =
A N Khudaiwala,G.P.Porbandar
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 :
i)COP of Carnot Refrigerator.
ii)Heat rejected from the system per hour.
iii)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
A N Khudaiwala,G.P.Porbandar
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 =
A N Khudaiwala,G.P.Porbandar
Bell – Coleman / Reverse Bryaton Cycle
Elements of this system :
Cooling
Water
Heat Exchanger
Cold Air
1. Compressor.
Hot Air
2. Heat Exchanger.
3. Expander.
Expander
Compressor
Very Cold Air
4. Refrigerator.
Warm Air
Work gained from Expander is used
to drive Compressor.
Refrigerator
Hence, less external work is
required.
A N Khudaiwala,G.P.Porbandar
Isobars
Isobars
2
2
3
Adiabatic
Expansion
Compression
1
4
Temperature
Pressure
Bell – Coleman / Reverse Bryaton Cycle
3
Compression
Expansion
4
Entropy
Volume
P –V Diagram
A N Khudaiwala,G.P.Porbandar
Adiabatic
T –s Diagram
1
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
1
Qexp n
Entropy
n
( P2 V2 − P1 V1 )
n −1
n
( P3 V3 − P4 V4 )
=
n −1
Qcomp =
We know,
A N Khudaiwala,G.P.Porbandar
 γ −1

R = C P 
 γ 
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 )
A N Khudaiwala,G.P.Porbandar
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 =
(T1 − T4 )
 n   γ −1 
 ( T4 − T3 + T2 − T1 )

 
 n −1  γ 
A N Khudaiwala,G.P.Porbandar
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
 P2 
T2 = T1  
 P1 
n −1
n
 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
n −1
n
 P3 
T3 = T4   ⇒ (302 K ) = T4
 P4 
⇒ T4 = 176.6 K
A N Khudaiwala,G.P.Porbandar
 8 bar 


 1 bar 
1.35 −1
1.35
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
A N Khudaiwala,G.P.Porbandar
Example 5
An air refrigeration open system operating between 1 MPa and 100 kPa is required to
produce a cooling effect of 2000 kJ/min. temperature of the air leaving the cold
chamber is -5 ºC, and at leaving the cooler is 30 ºC. Neglect losses and clearance in
the compressor and expander. Determine :
i)Mass of air circulated per min.
ii) Compressor Work, Expander Work, Cycle Work.
ii)COP and Power in kW required.
Polytropic Expansion 3-4 :
2
Pressure
303 K
P2
3
= 1 MPa
γ −1
γ
P1
= 100 kPa
PVγ=C
P 
T3 = T4  3  ⇒ (302 K ) = T4
 P4 
⇒ T4 = 156.9 K
 1 MPa 


 0.1 MPa 
Refrig. Effect per kg :
1
268 K
4
Volume
= C P (T1 − T4 )
= 1.003 (kJ / kg ) X (268 K − 156.9 K )
= 111.66 kJ / kg
A N Khudaiwala,G.P.Porbandar
1.4 −1
1 .4
Example 5….contd
Mass of air circulated per min :
Re frig . Effect
2000 kJ / min
=
=
= 17.91 kg / min ….ANS
Re frig . Effect per kg 111.66 kJ / kg
 P2 

Polytropic Compression 1-2 :T2 = T1 
 P1 
γ −1
γ
 1000 kPa 

= (268 K ) 
 100 kPa 
1.4 −1
1.4
= 517.4 K
Compressor Work :
 γ 
 m R ( T2 − T1 )
Wcomp = 
 γ −1 
 1.4 
Wcomp = 
 (17.91 kg / min) (0.287 kJ / kg ) ( 517.4 K − 268 K )
 1.4 − 1 
Wcomp = 4486.85 kJ / min ….ANS
A N Khudaiwala,G.P.Porbandar
….ANS
Example 5….contd
Expander Work :
 γ 
 m R ( T3 − T4 )
Wexp = 
 γ −1 
 1.4 
Wcomp = 
 (17.91 kg / min) (0.287 kJ / kg ) ( 303 K − 156.9 K )
 1.4 − 1 
Wcomp = 2628.42 kJ / min….ANS
cle Work = Wcycle = Wcomp – Wexp
= 4486.85 kJ/min – 2628.42 kJ/min
COPrefrig
= 1858.43 kJ/min…ANS
Re frig . Effect
2000 kJ / min
=
=
= 1.076 ….ANS
Work required 1858.43 kJ / min
Wcycle
1858.43 kJ / min
=
= 30.97 kW….ANS
Power required : P =
time
60 sec/ min
A N Khudaiwala,G.P.Porbandar
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).
A N Khudaiwala,G.P.Porbandar
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.
A N Khudaiwala,G.P.Porbandar
Applications
of Refrigeration
In chemical industries, for separating and liquefying
the gases.
In manufacturing and storing ice.
For the preservation of perishable food items in cold
storages.
For cooling water.
For controlling humidity of air manufacture and heat
treatment of steels.
For chilling the oil to remove wax in oil refineries.
For the preservation of tablets and medicines in
pharmaceutical industries.
For the preservation of blood tissues etc.,
For comfort air conditioning the hospitals, theatres,
etc.,
A N Khudaiwala,G.P.Porbandar
Types of Refrigerators
Ice Refrigerators : Ice is kept in the cabinet of
refrigerators and this acts as the refrigerating means.
Air Refrigerators : Air is used as working agent in
these types of refrigerators.
E.g., Bell Coleman Cycle.
Vapour Refrigerators: The working agents
employed in this type of refrigerators are ammonia,
CO2, SO2, freons etc.,
A N Khudaiwala,G.P.Porbandar
Vapour Compression
Refrigeration System
A N Khudaiwala,G.P.Porbandar
Refrigeration - Elements
High Temp
Source
Surrounding Air
QH
Condenser
QH
Wnet, in
Wnet, in
Expansion
Valve
Compressor
Evaporator
QL
QL
Refrigerated Space
A N Khudaiwala,G.P.Porbandar
Low Temp
Sink
Vapour Compression Refrigeration System Construction
This system consists of a compressor, condenser, a
receiver tank, an expansion valve and an evaporator.
Compressor : Reciprocating
compressors generally used.
For very big plants centrifugal
compressors directly coupled
with high speed rotating
engines (gas turbine) are used.
A N Khudaiwala,G.P.Porbandar
Vapour Compression Refrigeration System Construction
Compressor: For very big plants
Centrifugal compressors
directly coupled with high
speed rotating engines
(gas turbine) are used
A N Khudaiwala,G.P.Porbandar
Vapour Compression Refrigeration System Construction
Condenser : It is a coil of tubes made of copper.
Receiver tank: It is the reservoir of liquid
refrigerant.
Expansion Valve: This is a throttle valve. High
pressure refrigerant is made to flow at a controlled
rate through this valve.
Evaporator : It is the actual cooler and kept in the
space to be cooled. The evaporator is a coil of
tubes made of copper
A N Khudaiwala,G.P.Porbandar
Vapour Compression Refrigeration System Working
Working :
1. The low pressure refrigerant vapour coming out of the
evaporator flows into the compressor.
2. The compressor is driven by a prime mover.
3. In the compressor the refrigerant vapour is compressed.
4. The high pressure refrigerant vapour from the compressor
is then passed through the condenser.
5. The refrigerant gives out the heat it had taken in the
evaporator (N)
A N Khudaiwala,G.P.Porbandar
Vapour Compression Refrigeration System Working
Working :
6. The heat equivalent of work done on it (w) on the
compressor.
7. This heat is carried by condenser medium which may be
air or water.
8. The high pressure liquid refrigerant then enters the
expansion valve.
9. This valve allows the high pressure liquid refrigerant to
flow at a controlled rate into the evaporator.
10. While passing though this valve the liquid partially
evaporates. A N Khudaiwala,G.P.Porbandar
Vapour Compression Refrigeration System Working
Working :
11.Most of the refrigerant is vapourised only in the
evaporator, at a low pressure.
12. In the evaporator the liquid refrigerant absorbs
its latent heat of vapourisation from the material
which is to be cooled.
13. Thus the refrigerating effect (N) is obtained.
14. Then the low pressure refrigerant enters the
Khudaiwala,G.P.Porbandar
compressorA Nand
the cycle is repeated.
Refrigeration Circuit
Evaporator
Compressor
Refrigeration Circuit
Expansion
Valve
ME0223 SEM-IV
Condenser
A N Khudaiwala,G.P.Porbandar
Applied Thermodynamics & Heat Engines
Vapour Compression System
Elements of this system :
3
1. Compressor.
4
2. Condenser.
2
1
3
2
3. Expansion Valve.
4. Evaporator.
Vapour @ ↓ Pr. and ↓ Temp. (State 1)
Isentropic Compression :
↑ Pr. and ↑ Temp. (State 2)
Condenser : ↑ Pr. Liquid (State 3)
4
1
Throttling : ↓ Pr.
↓ Temp. (State 4)
Evaporator : Heat Extraction from surrounding;
↓ Pr. vapour (State 1).
A N Khudaiwala,G.P.Porbandar
Vapour Compression System
Merits :
1. High COP; as very close to Reverse Carnot Cycle.
2. Running Cost is 1/5th of that of Air Refrigeration Cycle.
3. Size of Evaporator is small; for same Refrigeration Effect.
4. Evaporator temperature adjustment is simple; by adjusting Throttle Valve.
Demerits :
1. Initial cost is high.
2. Inflammability.
3. Leakage.
4. Toxicity.
A N Khudaiwala,G.P.Porbandar
Vapour Compression System : T-s Diagram
Case A. Dry and Saturated Vapour after Compression :
Temperature, T
Work done by Compressor
= W = Area 1-2-3-4-1
T2
Compressor Work,
Condensation2 (W)
3
Heat Absorbed
= W = Area 1-4-g-f-1
Sat. Vapour Line
Expansion
T1
4
g
Sat. Liq. Line
Evaporation
Heat Absorbed
COP =
Compression
Work Done
1
Area 1 − 4 − g − f − 1
=
Net Refrig. Effect,
Area 1 − 2 − 3 − 4 − 1
(R )
f
n
Entropy, s
A N Khudaiwala,G.P.Porbandar
=
h1 − h4
h2 − h1
Vapour Compression System : T-s Diagram
Case B. Superheated Vapour after Compression :
Temperature, T
2
T2
3
Condensation
2’
Compressor Work,
(W)
Sat. Vapour Line
Expansion
T1
4
g
Sat. Liq. Line
Compression
1
Evaporation
f
Net Refrig. Effect,
(Rn)
Entropy, s
Work done by Compressor
= W = Area 1-2-2’-3-4-1
Heat Absorbed
= W = Area 1-4-g-f-1
Heat Absorbed
Work Done
Area 1 − 4 − g − f − 1
=
Area 1 − 2 − 2'−3 − 4 − 1
COP =
h1 − h4
=
h2 − h1
NOTE : h2 = h2’ + Cp (Tsup – Tsat)
A N Khudaiwala,G.P.Porbandar
Vapour Compression System : T-s Diagram
Case C. Wet Vapour after Compression :
Temperature, T
Work done by Compressor
= W = Area 1-2-3-4-1
3 Condensation
T2
Compressor Work,
(W)
2
Sat. Vapour Line
Expansion
T1
4
g
Sat. Liq. Line
Compression
Evaporation
f
1
Net Refrig. Effect,
(Rn)
Entropy, s
Heat Absorbed
= W = Area 1-4-g-f-1
Heat Absorbed
Work Done
Area 1 − 4 − g − f − 1
=
Area 1 − 2 − 3 − 4 − 1
COP =
h1 − h4
=
h2 − h1
NOTE : h2 = (hf + x.hfg)2
A N Khudaiwala,G.P.Porbandar
Vapour Compression System : P-h Diagram
Isobaric,
P = Const
S ent
= rop
C o ic
ns ,
t.
. Line
Is
2 – phase
region
Sat. Va
p
Isenthalpic,
h = Const.
Const. D
ryness
Fraction
Sub-cooled
Liq. region
Sat.
Liq. L
ine
Pressure, Pr
Isothermal,
T = Const
,
ric .
o
ch nst
I so Co
Superheated
v=
region
Enthalpy, h
A N Khudaiwala,G.P.Porbandar
Condensation
s io
n
2
Evaporation
Co
mp
res
3
Compression
Pressure, Pr
Vapour Compression System : P-h Diagram
1
4
Enthalpy, h
Rn = h1 − h4
W = h2 − h1
}
Rn h1 − h4
COP =
=
W h2 − h1
A N Khudaiwala,G.P.Porbandar
Factors Affecting Vapour Compression System
A. Effect of Suction Pressure :
Pressure, Pr
COP of Original Cycle :
P2
P1
2 2’
3
4
4’
Rn h1 − h4
COP =
=
W h2 − h1
COP when Suction Pr. decreased :
Rn h1' − h4 '
COP =
=
W h2 ' − h1'
1
=
1’
( h1 − h4 ) − ( h1 − h1' )
( h2 − h1 ) + ( h1 − h1' ) + ( h2' − h2 )
Thus,
Refrig. Effect ↓
Enthalpy, h
Work Input ↑
⇒ COP ↓
A N Khudaiwala,G.P.Porbandar
Factors Affecting Vapour Compression System
B. Effect of Delivery Pressure :
Pressure, Pr
COP of Original Cycle :
P2
3
3’
2’
2
P1
4
Rn h1 − h4
COP =
=
W h2 − h1
COP when Delivery Pr. increased :
Rn h1 − h4 '
COP =
=
W h2 ' − h1
1
=
4’
( h1 − h4 ) − ( h4' − h4 )
( h2 − h1 ) + ( h2' − h2 )
Thus,
Refrig. Effect ↓
Enthalpy, h
Work Input ↑
⇒ COP ↓
A N Khudaiwala,G.P.Porbandar
Factors Affecting Vapour Compression System
C. Effect of Superheating :
Pressure, Pr
COP of Original Cycle :
P2
P1
2
3
4
2’
Rn h1 − h4
COP =
=
W h2 − h1
COP when Delivery Pr. increased :
Rn h1' − h4
COP =
=
W h2 ' − h1'
1
=
1’
( h1 − h4 ) + ( h1' − h1 )
( h2 − h1 ) + ( h2' − h2 ) + ( h1' − h1 )
Thus,
Refrig. Effect ↑
Work Input ↑
Enthalpy, h
⇒
A N Khudaiwala,G.P.Porbandar
or ↓
COP ↓ or ↑
Factors Affecting Vapour Compression System
E. Effect of Suction & Condenser Temperatures :
Temperature, T
COP of Original Cycle :
Heat Absorbed
COP =
Work Done
T2
3
2’
3’
4
Compression
Evaporation
g
1
COP of Modified Cycle :
Heat Absorbed ( ↑ )
Work Done ( ↓ )
Area 1 − 1'−4 − 4'− g − f − 1
=
Area 1'−2'−3'−4'−1'
COP =
f
Entropy, s
⇒
A N Khudaiwala,G.P.Porbandar
h1 − h4
h2 − h1
Evaporator Temp. ↑
1’
4’
=
Condenser Temp. ↓
Now,
Expansion
T1
Area 1 − 4 − g − f − 1
Area 1 − 2 − 3 − 4 − 1
=
Condensation2
COP ↑
>
h1 − h4
h2 − h1
Vapour Compression System – Mathematical
Analysis
A. Refrigerating Effect :
= Amount of Heat absorbed in Evaporator.
Qevap = ( h1 − h4 ) + Latent Heat + Superheated Heat (kJ / kg )
B. Mass of Refrigerant :
= Amount of Heat absorbed / Refrigerating Effect.
m=
14,000
3600 ( h1 − h4 )
( kg / sec− tonne)
1 TR = 14,000 kJ/hr
C. Theoretical Piston Displacement :
= Mass of Refrigerant X Sp. Vol. of Refrigerant Gas (vg)1.
14,000
Th. Piston Displ. =
∗ ( v g )1 (m 3 / sec− tonne)
3600 ( h1 − h4 )
ME0223 SEM-IV
A N Khudaiwala,G.P.Porbandar
Applied Thermodynamics & Heat Engines
Vapour Compression System – Mathematical
Analysis
D. Theoretical Power Required :
a) Isentropic Compression :
Wcomp = ( h2 − h1 )
Ptheor = m ( h2 − h1 )
(kJ / kg )
(kW )
a) Polytropic Compression :
n
( P2V2 − P1V1 ) (kJ / kg )
Wcomp =
n −1
n
( P2V2 − P1V1 ) (kW )
Ptheor = m
n −1
E. Heat removed through Condenser :
Qcond = m ( h2 − h3 )
A N Khudaiwala,G.P.Porbandar
( kJ / kg )
Example 6
7
A refrigeration machine is required to produce ice at 0º C from water at 20 ºC. The
machine has a condenser temperature of 298 K while the evaporator temperature is
268 K. The relative efficiency of the machine is 50 % and 6 kg of Freon-12 refrigerant
is circulated through the system per minute. The refrigerant enters the compressor with
a dryness fraction of 0.6. Specific heat of water is 4.187 kJ/kg.K and the latent heat of
ice is 335 kJ/kg. Calculate the amount of ice produced on 24 hours. The table of
properties if Freon-12 is given below:
Given :
}
Temperature
(K)
Liquid Heat
(kJ/kg)
Latent Heat
(kJ/kg)
Entropy of Liquid
(kJ/kg)
298
59.7
138.0
0.2232
268
31.4
154.0
0.1251
m = 6 kg/min
hf1 = 31.4 kJ/kg
ηrel = 50 %
hfg1 = 154.0 kJ/kg
x2 = 0.6
hf2 = 59.7 kJ/kg
Cpw = 4.187 kJ/kg.K
hfg2 = 138.0 kJ/kg
Latent Heat of ice = 335.7 kJ/kg
hf3 = h4 = 59.7 kJ/kg
A N Khudaiwala,G.P.Porbandar
Example 6….contd
h1 = h f1 + x h fg1
= 31.4 + (0.6)154.0 = 123.8 kJ / kg
Temperature, T
Isentropic Compression : 1-2
298 K 3
s2 = s1
2
s f 2 + x2 ∗ s fg 2 = s f 1 + x1 ∗ s fg1
h2 = h f 2 + x2 h fg 2
= 59.7 + (0.5325)138.0 = 133.2 kJ / kg
 h fg 2 
 h fg1 
 = s f 1 + x1 

s f 2 + x2 
 T2 
 T1 
 138.0 
 154.0 
(
)
0.2232 + x2 
=
0
.
1251
+
0
.
6



 298 
 268 
⇒ x2 = 0.5325
Sat. Vapour Line
268
K
4
g
Sat. Liq. Line
1
f
Entropy, s
h4 = h f 3
= 59.7 kJ / kg
(
Rn h1 − h4
123.8 − 59.7 ) kJ / kg
=
=
= 6.82
COP of Original Cycle : COP =
W h2 − h1 (133.2 − 123.8) kJ / kg
ME0223 SEM-IV
A N Khudaiwala,G.P.Porbandar
Applied Thermodynamics & Heat Engines
Example 6….contd
Actual COP = ηrel X COPtheor = 0.5 X 6.82 = 3.41
Temperature, T
Heat extracted from 1 kg of water at 20 ºC to form 1 kg of ice at 0 ºC
:
298 K 3
= 1 (kg ) X 4.187 (kJ / kg.K ) X (20 − 0) (°C )
+ 335 (kJ / kg )
= 418.74 kJ / kg
2
Sat. Vapour Line
268
K
4
g
Sat. Liq. Line
1
f
Entropy, s
A N Khudaiwala,G.P.Porbandar
Now;
COPactual = 3.41 =
Rn ( actual )
W
mice X 418.74
=
m ( h2 − h1 )
6 (kg ) X (133.2 − 123.8) (kJ / kg )
⇒ mice =
∗ 3.41
418.74 kJ / kg
= 0.459 kg / min
0.459 X 60 X 24
=
= 0.661 tonne in 24 hrs
1000
….ANS
Example 7
28 tonnes of ice from and at 0 ºC is produced per day in an ammonia refrigerator.
The
temperature range in the compressor is from 25 ºC to -15oC. The vapour is dry and
saturated at the end of compression and an expansion valve is used. Assuming a
co-efficient of performance of 62% of the theoretical, calculate the power required
Temp
Enthalpy (kJ/kg)
Entropy of
Entropy of Vapour
to
(ºC)
Liquid
(kJ/kg.K)
Liquid
Vapour
drive the compressor.
Take latent
heat of ice
=
335
kJ/kg.
(kJ/kg.K)
Given :
}
25
100.04
1319.22
0.3473
4.4852
-15
-54.56
1304.99
-2.1338
5.0585
Tcond = 25 ºC
hf1 = -54.56 kJ/kg
Tevap = -15 ºC
hg1 = 1304.99kJ/kg
x2 = 1….dry saturated vapour
hf2 = 100.04 kJ/kg
COPactual = 0.62 (COPtheor)
hg2 = 1319.22 kJ/kg
Latent Heat
A N Khudaiwala,G.P.Porbandar
of ice = 335.7 kJ/kg
hf3 = h4 = 100.04 kJ/kg
Example 7….contd
h2 = hg 2 = 1319.22 kJ / kg
h3 = h4 = 100.04 kJ / kg.....Isenthalpic process
Isentropic Compression : 1-2
Temperature, T
s2 = s1
298 K 3
s g 2 = s f 1 + x1 ∗ s fg1
2
Sat. Vapour Line
258
K
⇒
4
g
Sat. Liq. Line
4.4852 = (−2.1338) + ( x1 ) [ 5.0585 − ( − 2.1338) ]
1
x2 = 0.92
h1 = h f 1 + x1 (h fg1 )
= (−54.56) + (0.92) [1304.99 − (−54.56)]
f
= 1196.23 kJ / kg
Entropy, s
COP of the Cycle : COPtheoretical =
(1196.23 − 100.04) = 8.91
h1 − h4
=
h2 − h1 (1319.22 − 1196.23)
A N Khudaiwala,G.P.Porbandar
Example 7….contd
Actual Rn = COPactual X Work done
Temperature, T
Actual COP = ηrel X COPtheor
= 0.62 X 8.91
= 5.52 X (h2 – h1)
= 5.52
= 5.52 X (1319.22 – 1196.23)
298 K 3
2
= 678.9 kJ/kg
Heat extracted from 28 tonnes of water at 0 ºC to form ice at 0 ºC
:
Sat. Vapour Line
258
K
4
g
Sat. Liq. Line
1
28 (kg ) X 1000 (kg / tonne) X 335 (kJ / kg )
=
24 (hr ) X 3600 (sec/ hr )
= 108.56 kJ / sec (kW )
Mass of refrigerant : =
f
Entropy, s
Total Work done by
Compressor :
108.56 (kJ / sec)
= 0.1599 kg
678.9 (kJ / kg )
= mrefrig X ( h2 − h1 ) = 0.1599 (kg ) X (1319.22 − 1196.23) kJ / kg
= 19.67 kJ / sec (kW ) ….ANS
A N Khudaiwala,G.P.Porbandar
Example 8
In a standard vapour compression refrigeration cycle, operating between an
evaporator
temperature of -10 ºC and a condenser temperature of 40 ºC, the enthalpy of the
refrigerant, Freon-12, at the end of compression is 220 kJ/kg. Show the cycle diagram
on
T-s plane. Calculate:
1. The C.O.P. of the cycle.
2. The refrigerating capacity and the compressor power assuming a refrigerant
flow rate Temp
of 1 kg/min.
(ºC)
Pr (MPa)
hf (kJ/kg)
hg (kJ/kg)
You may use the extract of Freon-12 property table given below:
Given :
}
-10
0.2191
26.85
183.1
40
0.9607
74.53
203.1
Tcond = 40 ºC
Tevap = -10 ºC
x1 = 1….dry saturated vapour
h2 = 220 kJ/kg
A N Khudaiwala,G.P.Porbandar
hf1 = 26.85 kJ/kg
hg1 = h1 = 183.1 kJ/kg
hf2 = 74.53 kJ/kg
hg2 = 203.1 kJ/kg
hf3 = h4 = 74.53 kJ/kg
Example 8….contd
COP of Original Cycle : COP =
Temperature, T
=
2
40 ºC
3
Rn h1 − h4
=
W h2 − h1
(183.1 − 74.53) kJ / kg = 2.94
….ANS
( 220.0 − 183.1) kJ / kg
2’
Refrigerating Capacity :
= m ( h1 − h4 ) = 1 (kg ) X (183.1 − 74.53) kJ / kg
-10 ºC
1
4
Sat. Vapour Line
g
Sat. Liq. Line
= 108.57 kJ / min ….ANS
f
Entropy, s
Compressor Power :
= m ( h2 − h1 ) = 1 (kg ) X ( 220.0 − 183.1) kJ / kg
= 36.9 kJ / min
= 0.615 kW ….ANS
A N Khudaiwala,G.P.Porbandar
Example 9
A Freon-12 refrigerator producing a cooling effect of 20 kJ/sec operates on a simple
cycle with pressure limits of 1.509 bar and 9.607 bar. The vapour leaves the
evaporator
dry saturated and there is no undercooling. Determine the power required by the
machine. If the compressor operates at 300 rpm and has a clearance volume of 3%
of
strokeTemp
volume,Pdetermine
piston displacement
of the compressor.
For
compressor
vg the
Enthalpy
Enthalpy Entropy
Entropy
Specific
s
assume
that (bar)
the expansion
the law
(oC)
hf
hg PV1.3 = Constant.
sf
sg
heat
(m3/kg) following
(kJ/kg)
(kJ/kg)
(kJ/kg)
(kJ/kg)
(kJ/kg.K)
-20
1.509
0.1088
17.8
178.61
0.073
0.7082
---
}
9.607
---
74.53
203.05
0.2716
0.682
0.747
40
Given :
Tcond = 40 ºC
Tevap = -20 ºC
x1 = 1….dry saturated vapour
h2 = 220 kJ/kg
A N Khudaiwala,G.P.Porbandar
hf1 = 17.8 kJ/kg
hg1 = h1 = 178.61 kJ/kg
hf2 = 74.53 kJ/kg
hg2’ = 203.05 kJ/kg
hf3 = h4 = 74.53 kJ/kg
Example 9….contd
•
Refrigerating Capacity : = m ( h1 − h4 )
•
⇒ 20 kW = m X (178.61 − 74.53) kJ / kg
•
Temperature, T
⇒ m = 0.192 kg / sec
2
313 K
3
Isentropic Compression : 1-2
s1 = s2
2’
 T2 
s1 = s2 ' + C P ln 
 T2' 
253 K
1
4
g
Sat. Liq. Line
f
Entropy, s
 T 
0.7082 = 0.682 + ( 0.747 ) ln 2 
 313 
Sat. Vapour Line
⇒ T2 = 324.2 K
h2 = h2' + C P ( T2 − T2 ' )
= 203.05 (kJ / kg ) + ( 0.747 kJ / kg.K ) ( 324.2 − 313.0 ) K
= 211.4 kJ / kg
A N Khudaiwala,G.P.Porbandar
Example 9
•
Power Required : = m ( h2 − h1 ) = 0.192 ( kg / sec) X ( 211.4 − 178.61) kJ / kg
Temperature, T
= 6.29 kW ….ANS
Vol. Efficiency :
2
313 K
3
2’
Sat. Vapour Line
1
4
Sat. Liq. Line



1 / 1.13
253 K
g
η vol
 Pd
= 1 + k − k 
 PS
1/ n
f
Entropy, s
 9.607 bar 

= 1 + 0.03 − 0.03 
 1.509 bar 
= 87.6 %
•
Vol of Refrigerant = m ∗ v g
at Intake :
= 0.192 (kg / sec) X 0.1088 (m 3 / kg )
= 0.02089 m 3 / sec
(
Actual Vol.) 0.02089 (m 3 / sec) ∗ 60 (sec/ min)
=
=
Piston Displ. Vol. :
η vol ∗ (rpm)
A N Khudaiwala,G.P.Porbandar
0.876 ∗ 300 (rpm)
= 0.00477 m 3
….ANS
Example 10
A food storage locker requires a refrigeration capacity of 50 kW. It works between a
condenser temperature of 35 ºC and an evaporator temperature of -10 ºC. The
refrigerator
is ammonia. It is sub-cooled by 5 ºC before entering the expansion valve. By the dry
saturated vapour leaving the evaporator. Assuming a single-cylinder, single-acting
compressor operating at 1000 rpm with stroke equal to 1.2 times the bore, determine :
1. The power required.
2. The cylinder dimensions.
Properties
of ammonia
are :
Sat.
Pr.
Enthalpy
Entropy
Sp. Vol.
Sp. Heat
Temp.
(oC)
(bar)
(kJ/kg)
(kJ/kg)
(m3/kg)
(kJ/kg.K)
Liquid
Vapour
Liquid
Vapou
r
Liquid
Vapour
Liquid
Vapour
-10
2.9157
154.056
1450.22
0.82965
5.7550
---
0.417477
---
2.492
35
13.522
366.072
1488.57
1.56605
5.2086
1.7023
0.095629
4.556
2.903
Given :
}
Tcond = 35 ºC
Tevap = -10 ºC
x1 = 1….dry saturated vapour
State 3 = Sub-cooled by 5 ºC
A N Khudaiwala,G.P.Porbandar
h1 = 1450.22 kJ/kg
h2’ = 1488.57 kJ/kg
hf3 = 366.072 kJ/kg
Temperature, T
Example 10….contd
2
3
308 K
303 K
Isentropic Compression : 1-2
2’
3’
263 K
 T2 
s1 = s2 ⇒ s1 = s2 ' + C P ln 
 T2 ' 
 T2 
h = h = h − C ( T 5−.T755) = 5.2086 + ( 2.903) ln 

1= 366.07 (kJ / kg ) − 405.56 ( 308 − 303) (kJ / kg )
308


Sat. Vapour Line
= 343.29 kJ / kg
⇒ T2 = 371.8 K
3'
4
g
Sat. Liq. Line
f
4
f3
P liq
sat
subcool
Entropy, s
h2 = h2' + C P ( T2 − T2' )
= 1488.57 ( kJ / kg ) + ( 2.903 kJ / kg.K ) ( 371.8 − 308.0 ) K
= 1673.8 kJ / kg
ME0223 SEM-IV
A N Khudaiwala,G.P.Porbandar
Applied Thermodynamics & Heat Engines
Example 10….contd
Temperature, T
Mass of Refrigerant :
303 K
m=
50 (kW )
50 (kW )
=
( h1 − h4 ) kJ / kg (1450.22 − 343.29) kJ / kg
= 0.04517 kg / sec
2
308 K 3
•
2’
Compressor Power :
•
= m ( h2 − h1 )
3’
= 0.04517 (kg ) X (1673.8 − 1450.22 ) kJ / kg
263 K
1
4
g
Sat. Liq. Line
= 10.1 kW ….ANS
Sat. Vapour Line
f
Cylinder Dimensions :
1000 (rpm) 
π 2 N  π 2
 D L   D (1.2 D)

Entropy,
•
s
4
60   4
60

m = 0.04517 (kg / sec) = 
=
vg
0.417477 m 3 / kg
(
⇒
⇒
ME0223 SEM-IV
D = 0.19 m ….ANS
L =1.2 ∗ (0.19 m) = 0.228 m ….ANS
A N Khudaiwala,G.P.Porbandar
Applied Thermodynamics & Heat Engines
)
Vapour Absorption Refrigeration system
In this system compression process of vapour
compression cycle is eliminated. Instead of that
the following three processes are carried out.
1. Absorbing ammonia vapour into water.
2. Pumping this solution to a high pressure cycle
3. Producing ammonia vapours from ammonia
solution by heating.
A N Khudaiwala,G.P.Porbandar
Vapour Absorption Refrigeration system Construction:
The vapour absorption
system consists of a
condenser, an expansion
valve and an evaporator.
They perform the same as
they do in vapour
compression method.
In addition to these, this
system has an absorber, a
heat exchanger, an
analyser and a rectifier.
A N Khudaiwala,G.P.Porbandar
Vapour Absorption Refrigeration system
Dry ammonia vapour at low
pressure passes in to the
absorber from the evaporator.
2. In the absorber the dry
ammonia vapour is dissolved
in cold water and strong
solution of ammonia is
formed.
3. Heat evolved during the
absorption of ammonia is
removed by circulating cold
water through the coils kept
in the absorber.
4. The highly concentrated
ammonia (known as Aqua
Ammonia) is then pumped
by a pump to generator
through a heat exchanger.
1.
A N Khudaiwala,G.P.Porbandar
Vapour Absorption Refrigeration system
6. In the heat exchanger the
strong ammonia solution is
heated by the hot weak
solution returning from the
generator to the absorber.
7. In the generator the warm
solution is further heated by
steam coils, gas or electricity
and the ammonia vapour is
driven out of solution.
8. The boiling point of
ammonia is less than that of
water.
9. Hence the vapours leaving
the generator are mainly of
ammonia. A N Khudaiwala,G.P.Porbandar
Vapour Absorption Refrigeration system
9. The weak ammonia solution is
left in the generator is called
weak aqua.
10. This weak solution is returned
to the absorber through the heat
exchanger.
11. Ammonia vapours leaving the
generator may contain some
water vapour.
12. If this water vapour is allowed
to the condenser and expansion
valve, it may freeze resulting in
chocked flow.
13. Analyser and rectifiers are
incorporated in the system
before condenser.
A N Khudaiwala,G.P.Porbandar
Vapour Absorption Refrigeration system
14. The ammonia vapour from the
generator passes through a series
of trays in the analyser and
ammonia is separated from water
vapour.
15. The separated water vapour
returned to generator.
16. Then the ammonia vapour passes
through a rectifier.
17. The rectifier resembles a
condenser and water vapour still
present in ammonia vapour
condenses and the condensate is
returned to analyser.
18. The virtually pure ammonia
vapour then passes through the
A N Khudaiwala,G.P.Porbandar
condenser.
Vapour Absorption Refrigeration system
19. The latent heat of ammonia
vapour is rejected to the cooling
water circulated through the
condenser and the ammonia
vapour is condensed to liquid
ammonia.
20. The high pressure liquid
ammonia is throttled by an
expansion valve or throttle valve.
21. This reduces the high
temperature of the liquid
ammonia to a low value and
liquid ammonia partly evaporates.
22. Then this is led to the
evaporator.
23. In the evaporator
the liquid
A N Khudaiwala,G.P.Porbandar
Vapour Absorption Refrigeration system
24. The latent heat of
evaporation is obtained
from the brine or other
body which is being
cooled.
25. The low pressure
ammonia vapour leaving
the evaporator again
enters the absorber and
the cycle is completed.
26. This cycle is repeated
again to provide the
refrigerating effect.
A N Khudaiwala,G.P.Porbandar
Comparison between Vapour compression & Vapour
Absorption refrigeration systems
S.No. Vapour Compression System
Vapour Absorption System
1
This system has more wear and
Only moving part in this system is
tear and produces more noise due an aqua pump. Hence the quieter
to the moving parts of the
in operation and less wear and tear
compressor.
2.
Electric power is needed to drive
the system
Waste of exhaust steam may be
used. No need of electric power
3.
Capacity of the system drops
rapidly with lowered evaporator
pressure
Capacity of the system decreases
with the lowered evaporative
pressure, by increasing the steam
pressure in generator.
4.
At partial loads performance is
poor.
At partial loads performance is not
affected.
5.
Mechanical energy is supplied
through compressor
Heat energy is utilised
6.
Energy supplied is ¼ to ½ of the
Energy supplied is about one and
A N Khudaiwala,G.P.Porbandar
refrigerating effect
half times the refrigerating effect
Comparison between Vapour compression & Vapour
Absorption refrigeration systems
S.No. Vapour Compression System
Vapour Absorption System
7.
Charging of the refrigerating to
the system is easy
Charging of refrigerant is difficult
8.
Preventive measure is needed,
since liquid refrigerant
accumulated in the cylinder may
damage to the cylinder
Liquid refrigerant has no bad effect
on the system.
A N Khudaiwala,G.P.Porbandar
Layout of Domestic Refrigerator
A N Khudaiwala,G.P.Porbandar
Window Type Air Conditioner
A N Khudaiwala,G.P.Porbandar
Window Type Air Conditioner  This is also called room air
conditioner.
 This unit consists of the
following.
1. A cooling system to cool and
dehumidify the air involves a
condenser, a compressor and a
refrigerant coil.
2. A filter to any impurities in
the air. The filter is made of
mesh, glass wool or fiber.
3. A fan and adjustable grills to
circulate the air.
4. Controls to regulate the
equipment operation.
A N Khudaiwala,G.P.Porbandar
Window Type Air Conditioner
 The low pressure refrigerant
vapour is drawn from the
evaporator to the hermetic
compressor through suction
pipe.
 It is compressed from low
pressure to the high pressure
and supplied to the condenser.
 It is condensed in the
condenser by passing the
outdoor air over the condenser
coil by a fan.
 The liquid refrigerant is passed
through the capillary into the
evaporator. A N Khudaiwala,G.P.Porbandar
Window Type Air Conditioner
5. In the evaporator the liquid
refrigerant picks up the heat from
the refrigerator surface and gets
vaporized.
6. A motor driven fan draws air from
the room through the air filter and
this air is cooled by losing its heat
to the low temperature refrigerant
and cold air is circulated back into
the room.
7. The vapour refrigerant from the
evaporator goes to the compressor
from evaporator and the cycle is
repeated.
8. Thus the room is air conditioned
A N Khudaiwala,G.P.Porbandar
Window Type Air Conditioner
9. The quantity of air circulated
can be controlled by the
dampers.
10. The moisture in the air passing
over the evaporator coil is
dehumidified and drips into
the trays.
11. This water evaporator to certain
extent and thus helps in
cooling the compressor and
condenser.
12. The unit automatically stops
when the required temperature
is reached in the room. This is
accomplished by the
A N Khudaiwala,G.P.Porbandar
thermostat and control panel.
Split Type Air Conditioner - Construction
A N Khudaiwala,G.P.Porbandar
Thermodynamic and thermo-physical properties of
Refrigerants
a) Suction pressure: Higher suction pressure is better as it
leads to smaller compressor displacement.
b) Discharge pressure: At a given condenser temperature,
the discharge pressure should be as small as possible to
allow light-weight construction of compressor, condenser
etc.
c) Pressure ratio: Should be as small as possible for high
volumetric efficiency and low power consumption.
d) Latent heat of vaporization: Should be as large as
possible so that the required mass flow rate per unit
cooling capacity will be small.
A N Khudaiwala,G.P.Porbandar
Thermodynamic and thermo-physical properties of
Refrigerants
e) Isentropic index of compression: Should be as small as
possible so that the temperature rise during compression will
be small.
f) Liquid specific heat: Should be small so that degree of
subcooling will be large leading to smaller amount of flash gas
at evaporator inlet.
g) Vapour specific heat: Should be large so that the degree of
superheating will be small.
h) Thermal conductivity: Thermal conductivity in both liquid
as well as vapour phase should be high for higher heat transfer
coefficients.
i) Viscosity: Viscosity should be small in both liquid and vapour
phases for smaller frictional pressure drops
A N Khudaiwala,G.P.Porbandar
Environmental and safety properties
a) Ozone Depletion Potential (ODP): According to
the Montreal protocol, the ODP of refrigerants
should be zero, i.e., they should be non-ozone
depleting substances. Refrigerants having non-zero
ODP have either already been phased-out (e.g. R 11, R
12) or will be phased-out in near-future(e.g. R22).
Since ODP depends mainly on the presence of
chlorine or bromine in the molecules, refrigerants
having either chlorine (i.e., CFCs and HCFCs) or
bromine cannot be used under the new regulations
b) Global Warming Potential (GWP): Refrigerants
should have as low a GWP value as possible to
minimize the problem of global warming.
Refrigerants with zero ODP but a high value of GWP
(e.g. R134a) are
likely to be regulated in future.
A N Khudaiwala,G.P.Porbandar
Environmental and safety properties
c) Total Equivalent Warming Index (TEWI): The
factor TEWI considers both direct (due to release
into atmosphere) and indirect (through energy
consumption) contributions of refrigerants to global
warming. Naturally, refrigerants with as a low a value
of TEWI are preferable from global warming point of
view.
d) Toxicity: Ideally, refrigerants used in a
refrigeration system should be non-toxic.Some fluids
are toxic even in small concentrations. Some fluids
are mildly toxic, i.e., they are dangerous only when
the concentration is large and duration of exposure is
long.
A N Khudaiwala,G.P.Porbandar
Environmental and safety properties
e) Flammability: The refrigerants should preferably be
non-flammable and non-explosive. For flammable
refrigerants special precautions should be taken to
avoid accidents.
f) Chemical stability: The refrigerants should be
chemically stable as long as they are inside the
refrigeration system.
g) Compatibility :with common materials of
construction (both metals and non-metals)
h) Miscibility with lubricating oils: Oil separators
have to be used if the refrigerant is not miscible with
lubricating oil (e.g. ammonia). Refrigerants that are
completely miscible with oils are easier to
handle(R12).
A N Khudaiwala,G.P.Porbandar
Halocarbon Refrigerants
Halocarbon Refrigerants are all synthetically
produced and were developed as the Freon family of
refrigerants.
Examples :
CFC’s : R11, R12, R113, R114, R115
HCFC’s : R22, R123
HFC’s : R134a, R404a, R407C, R410a
A N Khudaiwala,G.P.Porbandar
Inorganic Refrigerants
Carbon Dioxide
Water
Ammonia
Air
Sulphur dioxide
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R22
ODP-0.05, GWP-1700
R22 has 40% more refrigerating capacity
Higher pressure and discharge temp and not
suitable for low temp application
Extensively used in commercial air-
conditioning and frozen food storage and
display cases
A N Khudaiwala,G.P.Porbandar
R123
ODP-0.02,GWP-90
As a replacement for R11 as similar thermodynamic
properties.
Very short atmospheric life but classified as
carcinogen
Retrofit alternative to R11
A N Khudaiwala,G.P.Porbandar
HFC
Zero ODP as no chlorine atom contains only
Hydrogen and Flurodine
Very small GWP values
No phase out date in Montreal Protocol
R134a and R152 a – Very popular refrigerants
HFC refrigerants are costly refrigerants
A N Khudaiwala,G.P.Porbandar
R134a
ODP-0, GWP-1300
Used as a substitute for R12 and to a limited range
for R22
Good performance in medium and high temp
application
Toxicity is very low
Not miscible with mineral oil
A N Khudaiwala,G.P.Porbandar
Ammonia
ODP = 0 and GWP = 0
Excellent thermodynamic characteristics: small
molecular mass, large latent heat, large vapour density
and excellent heat transfer characteristics
High critical temperature (132C) : highly efficient
cycles at high condensing temperatures
Its smell causes leaks to be detected and fixed before
reaching dangerous concentration
Relatively Low price
A N Khudaiwala,G.P.Porbandar
Drawbacks of Ammonia
Toxic
Flammable ( 16 – 28% concentration )
Not compatible with copper
Temperature on discharge side of compressor is
higher compared to other refrigerants
A N Khudaiwala,G.P.Porbandar
Water
Zero ODP & GWP
Water as refrigerant is used in absorption
system .New developing technology has created
space for it for use in compression cycles also.
But higher than normal working pressure in the
system can be a factor in restricted use of water as
refrigerant.
A N Khudaiwala,G.P.Porbandar
Application of New Eco-friendly Refrigerants
Application of Refrigerants
HFCs used
Eco-friendly
refrigerant
Domestic refrigeration
R134a,R152a
HC600a and blends
Commercial refrigeration ,CO2
R134a,R404A,R407C HC blends,NH3 ,CO2 **
Cold storage ,food processing
And industrial refrigeration
R134a,R404A,R507A NH3 ,HCs,CO2 **
Unitary air conditioners
R410A,R407C
CO2 , HC s
Centralized AC (chillers)
R134a,R410A,R407C
NH3 ,HCs,CO2, water **
Transport refrigeration
R134a,R404A
CO 2,
Mobile air conditioner
R134a
CO2 ,HCs
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A N Khudaiwala,G.P.Porbandar