MEngg 6122 Advanced Refrigeration & Air conditioning
Lectured by
Dr. P. Kumaran
Professor(Assistant)
Department of Mechanical Engineering
College of Engineering
Wolaita Sodo University
Refrigeration and Air conditioning
Introduction
• Refrigeration: The process of removing heat from a substance
under controlled conditions.
• It also includes the process of reducing and maintaining the
temperature of a body below the atmospheric conditions
• Also, refrigeration means a continued extraction of heat from
a body whose temperature is already below the temperature
of its surroundings.
• For example
The objective of a refrigerator is to remove QL
from the cooled space.
Coefficient of Performance:
• The efficiency of a refrigerator is expressed in terms of the coefficient of
performance (COP), denoted by COPR.
• The objective of a refrigerator is to remove heat (QL) from the refrigerated
space. To accomplish this objective requires a work input of (Wnet,in). Then
the COP of a refrigerator can be expressed as
The conservation of energy principle for a cyclic device requires that
➢ Notice that the value of COPR can be greater than unity.
➢ That is, the amount of heat removed from the refrigerated space can be
greater than the amount of work input.
➢ This is in contrast to the thermal efficiency, which can never be greater
than 1.
➢ In fact, one reason for expressing the efficiency of a
refrigerator by another term—the coefficient of performance—
is the desire to avoid the oddity of having efficiencies greater
than unity.
Air Refrigeration cycles
- In Air refrigeration cycles, the air is used as a refrigerant.
- Air was used in commercial applications because of its
availability at free cost
- Air cycle refrigeration systems as originally designed and
installed are now practically outdated because of their low
coefficient of performance and high power requirements.
- The basic elements of an air cycle refrigeration system are
1. compressor
2. cooler or heat exchanger
3. the expander
4. refrigerator
Unit of Refrigeration
• The practical unit of refrigeration is expressed in terms of “ tonne of
refrigeration (TR)”
• A tonne of refrigeration is defined as the amount of refrigeration
effect produced by the uniform melting of one tonne (1000kg) of ice
from and at 0oC in 24 hours.
• Since the latent heat of ice is 335kJ/kg, therefore one tonne of
refrigeration.
Since the latent heat of ice is 335kJ/kg, therefore one tonne of refrigeration,
1TR = 1000 335 kJ in 24 hours
1000 335
1TR = = 232.6 kJ/min
24 60
In actual practice, one tonne of refrigeration is taken as
equivalent to 210kJ/min or 3.5kJ/s (3.5kW).
Air Refrigerator Working on a Bell-Coleman cycle ( or Reversed Brayton or joule
cycle)
➢ It was one of the earliest types of refrigerators used in ships
➢ Figure shows a schematic diagram that consists of a compressor, a cooler, an
expander, and a refrigerator.
1-2 Isentropic Compression process
• The cold air from the refrigerator is drawn into the compressor cylinder
where it is compressed isentropically in the compressor as shown by
curves 1-2 in p-v and T-s diagrams.
• During the compression stroke, both the pressure and temperature
increase and
• the specific volume of air at delivery from the compressor reduces from
v1 and v2.
2-3 Constant pressure cooling process
• The warm air from the compressor is now passed into the cooler where it
is cooled at constant pressure.
• Temperature T2 is reduced to T3 and sp. Volume also reduced from v2
also v3
Therefore the heat rejected by the air during constant pressure per kg of air
qR = Q2-3 = cp(T2-T3)
3-4 Isentropic expansion process
• The air from the cooler is now drawn into the expander cylinder
where it is expanded isentropically from pressure p3 to the
refrigerator pressure p4 which is equal to the atmospheric pressure.
• The temperature of the air during expansion falls from T3 to T4
• Specific volume of air increases from v3 to v4.
4-1 Constant pressure expansion process
• The cold air from the expander is now passed to the refrigerator
where it is expanded at constant pressure p1.
• the temperature of air increases from T4 to T1. also sp. Volume
increases as shown in the figure.
• The heat absorbed by the air during Constant pressure expansion
per kg of air.
Therefore
q A = q4−1 = c p (T1 − T4 )
work done during the cycle per kg of air
W = Heat Rejected - Heat absorbed = q R − q A
= c p (T2 − T3 ) − c p (T1 − T4 )
Heat absorbed
coefficient of performance (COP) =
work done
qA c p (T1 − T4 )
COP = =
qR − q A c p (T2 − T3 ) − c p (T1 − T4 )
T
T4 1 − 1
(T1 − T4 ) T4
COP = =
(T2 − T3 ) − (T1 − T4 ) T2 T1
T3 − 1 − T4 − 1
T3 T4
Therefore
T4 1 1 1
COP = = = −1
= −1
T3 − T4 T3 − 1
p3 − 1 p2 − 1
T4 p p
4 1
1
COP = −1
(r )
p
−1
Sometimes, the compression and expansion process take place according to the
law pvn=C
work done by the compressor during the process 1-2 per kg of air
n nR
wc = ( p2v2 − p1v1 ) = (T2 − T1 )
n −1 n −1
work done by expander during the process 3-4 per kg of air
n nR
wE = ( 3 3 4 4)
p v − p v = (T3 − T4 )
n −1 n −1
Net work done during the cyce per kg of air
nR
w = wc − wE = (T2 − T1 ) − (T3 − T4 )
n −1
WkT
Heat absorbed during constant pressure process 4-1
q A = c p (T1 − T4 )
Heat absorbed
COP =
Workdone
After substituting the COP is
T1 − T4
COP =
n −1
(T2 − T3 ) − (T1 − T4 )
n −1
In this case, values of T2 and T4 are obtained from the following relations:
n −1 n −1
T2 p2 n T3 p3 n
= and =
T1 p1 T4 p4
For Isentropic compression or expansion n = , Therefore
T1 − T4
COP =
(T2 − T3 ) − (T1 − T4 )
Problem 1: In a refrigeration plant working on bell coleman cycle, air is
compressed to 5 bar from 1 bar. Its initial temperature is 10oC. After compression,
the air is cooled up to 20oC in a cooler before expanding back to a pressure of 1
bar. Determine the theoretical COP of the plant and net refrigerating effect. Take cp
= 1.005kJ/kgK and cv = 0.718kJ/kgK.
Given data:
Intake pressure of the compressor p1 = 1 bar = 1*105N/m2 = p4
Compressed pressure p2 = 5 bar = 5*105N/m2 = p3
Initial temperature of the air T1 = 10oC = 10+273=283K
After compression, air is cooled T3 = 20oC = 20+273 = 293K
Specific heat cp = 1.005kJ/kgK and cv = 0.718kJ/kgK
cp 1.005
= = = 1.4
cv 0.718
Formula Used
T4
COP =
T3 − T4
Isentropic Compression 1-2
−1
T2 p2
=
T1 p1
Isentropic Expansion 3-4
−1
T3 p3
=
T4 p4
Solution
T3 = 293K, p3 =p2 = 5bar and p4=p1 = 1 bar
Theoretical COP of the plant
T4 185
COP = = = 1.713
T3 − T4 293 − 185
Net Refrigerating effect q A
q A = c p (T1 − T4 ) = 1.005(283 − 185) = 98.5kJ / kg
A refrigerator working on the Bell-Coleman cycle operates between pressure
limits of 1.05bar and 8.5bar. Air is drawn from the cold chamber at 10oC,
Compressed, and then it is cooled to 30oC before entering the expansion
cylinder. The expansion and compression follow the law pv1.3 = constant.
Determine the theoretical COP of the system
Answer: COP = 1.3
T1 − T4
COP =
n −1
(T2 − T3 ) − (T1 − T4 )
n −1
