refrigeration.pptx

6. REFRIGERATION
Definition
Refrigeration is the process of providing and
maintaining temperature of the system below
that of the surrounding atmosphere.
The refrigeration effect can be accomplished by
non – cyclic processes, making use of substances
at temperature well below the temperature of the
surroundings – e.g., ice, snow, dry ice (solid CO2)
etc.
However, of greater importance are cyclic
refrigeration systems, wherein the cooling
substance (called refrigerant) is not consumed
and discarded, but used again and again in a
thermodynamic cycle.
2

A ton of refrigeration is defined as the quantity of
heat required to be removed to produce one ton
(1000kg) of ice within 24 hours when the initial
condition of water is 00C.
Consider a refrigerator of T tons capacity,
Refrigeration capacity = 3.5 kJ/s
Heat removed from refrigerator = Refrigeration
effect =R.E. kJ/s
Power of the compressor
=work/kg of refrigerant x mass flow rate
4
kJ/s
5
.
3
3600
24
335
1000
ion
refrigerat
of 

x
x
Ton

Reversed Heat Engine
Cycle
A reversed heat engine is a
potential refrigerating
machine. It receives heat
from a low temperature
region at T2, discharge heat
to a high temperature region
at T1, and requires a net
inflow of work. Removal of
heat from a low temperature
region reduces the
temperature of that region
below the temperature of
the surroundings, thus
producing refrigeration. 5

According to First Law
Q2 – Q1 = -W i.e., Q1 = Q2 + W
Such a device is called a Refrigerator or Heat Pump,
depending on whether the focus is on heat received
from the low temperature region Q2 or the heat
discharged to the high temperature region Q1.
Q2 is known as the refrigeration effect.
The performance of a refrigerator/heat pump is
measured by means of its coefficient of performance
(COP). COP of a refrigeration/heat pump is defined as
6
The working fluid in a refrigeration cycle is called a Refrigerant.

Important application of Refrigeration
1. Ice plants
2. Food processing units and transportation, including
dairies
3. Industrial air – conditioning
4. Comfort air – conditioning
5. Chemical and related industries.
6. Hospitals.
7. Laboratories.
8. Domestic applications
7

Basic processes (operations) in a Refrigeration
Cycle
Since a refrigeration cycle is essentially a reversed heat
engine cycle, the working substance (refrigerant) will
undergo the following basic operations.
1. Compression - resulting in increase in pressure and
temperature.
2. Heat rejection at high temperature.
3. Expansion – resulting in reduction in pressure and
temperature and
4. Heat addition at low temperature – during which heat
is transferred from the body to be cooled to the
refrigerant.
8

Vapour Compression Refrigeration Cycle
In this, the refrigerant used is a vapour (e.g., ammonia,
Freon-22, Freon-11, Freon -12 etc). The refrigerant
undergoes the following operations in a cyclic manner.
1. Compression in a compressor (Usually reciprocating),
with work input.
2. Condensation of the vapour into liquid in a condenser,
wherein heat is rejected to a cooling medium (air, water)
at high pressure and temperature.
3. Expansion of the liquid refrigerant in a suitable device
(engine, expansion valve, capillary etc). There may or
may not be work output. The liquid may evaporate
partially.
4. Evaporation of the mixture of liquid and vapour in an
evaporator where heat is added to the refrigerant from
the substance to be cooled, producing the necessary
refrigeration effect.
9

4-1: Reversible adiabatic (isentropic) compression, with work
input WC.
1-2: Condensation at constant pressure and temperature with
heat Q1 rejected to some cooling medium.
2-3: Reversible adiabatic expansion, with work output WE.
3-4: Evaporation at constant pressure and temperature wherein
heat Q2 is absorbed from the substance to be cooled.
10
Reversed Carnot Cycle as a Refrigeration Cycle

Q1 = area under 2 – 3 = Tmax (s2 – s3)
Q2 = area under 4 – 1 = Tmin (s1 – s4)
= Tmin (s2 – s3), s1 = s2 & s3 = s4
Wnet = WC – WE = Q1 – Q2
= (Tmax-Tmin) (s2-s3)
12

These are the maximum values for any refrigerator or
heat pump operating between two fixed temperatures
Tmax and Tmin. In other words, no refrigerator/ heat pump
has a COP greater than that of a Carnot refrigerator/heat
pump, operating between the same maximum and
minimum temperatures.
When the refrigerator/heat pump operates on a cycle
other than a Carnot cycle, the heat rejection
(condensation) and heat addition (evaporation) process
may not be isothermal. Then the COPs are given by
13
Where Tcond = average temperature during condensation.
Tevap = average temperature during evaporation.
It can be seen that the closer the temperatures Tcond and Tevap,
the higher the COP.

In practice, an expansion engine is not used in a vapour
compression refrigeration unit.
This is because; the power output of such an engine is too
small to justify its cost. Instead, some kind of expansion
device – like a throttling valve or a capillary tube – is used
to reduce the pressure and temperature of the refrigerant.
14

• The most convenient property diagram.
15

Process 1-2 or 1’–2’:
Reversible adiabatic
compression. Process 1–2,
starting with saturated
vapour (state 1) and
ending in the superheated
region (state 2) is called
Dry compression. Process
1’-2’, starting with wet
vapour (state 1’) and
ending as saturated
vapour (state 2’) is called
wet compression. Dry
compression is always
preferred to wet
compression.
. 17

Process 2-3 (or 2’–3):
Reversible constant
pressure heat rejection,
at the end of which the
refrigerant is in
saturated liquid state.
2–2’ is
desuperheating, and
2’-3 is condensation.
18

Process 3-4: Adiabatic
throttling process, for
which enthalpy before
is equal to enthalpy
after throttling. This
process is adiabatic but
not isentropic. Since it
is irreversible, it cannot
be shown on a property
diagram. States 3 and 4
are equilibrium points
and are simply joined
by a dotted line
following a constant
enthalpy line. 19

Process 3-4: Adiabatic
throttling process, for which
enthalpy before is equal to
enthalpy after throttling.
This process is adiabatic but
not isentropic. Since it is
irreversible, it cannot be
shown on a property
diagram. States 3 and 4 are
equilibrium points and are
simply joined by a dotted
line following a constant
enthalpy line.
20

Analysis:
The compressor, the condenser and the evaporator can be
treated as steady–flow devices, governed by the Steady
Flow Energy Equation. Application of S.F.E.E. to these
devices results in:
Compressor: Process 1–2 isentropic
Q1-2 = 0 W1-2 = - ∆h
W1-2 = - (h2 - h1)
Compressor work WC = (h2-h1) kJ/kg, on a unit mass basis.
If mr is the mass flow rate of the refrigerant in kg/sec, then
the power input in the compressor is given by
Power input = mr (h2 – h1) kW
21

Condenser:
Process 2–3: reversible constant pressure process
W2-3 = 0
Q2-3 = ∆h = (h3 – h2) kJ/kg
This is negative i.e., heat rejected.
Heat rejected per unit mass of the refrigerant is
Q1 = (h2 - h3) kJ/kg.
Rate of heat rejection Q1 = mr (h2 - h3) kJ/sec
22

Evaporator: Process 4 – 1: reversible constant pressure
process , W4-1 = 0
Q4-1 = ∆h = (h1 - h4) kJ/kg
Heat received by unit mass of the refrigerant
= heat received from the substance being cooled
= Q2 = (h1 – h4) kJ/kg of refrigerant.
Rate of heat removed = Refrigerating effect
= Q2 = m r(h1 - h4) kJ/sec
Refrigerating Effect in terms of refrigeration
23

Expansion: for process 3 – 4,
h3 = h4
but, it is not a constant enthalpy process.
Note: Values of enthalpy h1, h2, h3 & h4 can be obtained
from property Tables or Property Charts (Diagrams).
24

Actual Vapour Compression
Refrigeration Cycle:
A constant amount of
superheating of the vapour
before it enters the
compressor is recommended.
This is to ensure that no liquid
refrigerant droplets enter the
compressor. Further, a small
degree of sub cooling (under
cooling) of the liquid
refrigerant at the condenser
exit is desirable, in order to
reduce the mass of vapour
formed during expansion.
Excessive formation of vapour
bubbles may obstruct the
flow of liquid refrigerant
through the expansion valve. 25

Both the superheating at the evaporator outlet
and the subcooling at the condenser outlet
contribute to an increase in the refrigerating
effect. However, the load on the condenser also
increases. There will be an increase in the
compressor discharge temperature. Since the
compressor input more or less remains
unchanged, the COP of the cycle appears to
increase due to this superheating/subcooling.
However, for a fixed temperature of the
refrigerated space, the evaporation temperature
must be lowered (i.e., Tevap is reduced). Further,
for a fixed temperature of the cooling medium,
the condensation temperature must be raised
(i.e., Tcond will be higher). Hence COP will reduce. 26

Refrigerants and desirable properties:
The most commonly used refrigerants are a group of
halogenated hydrocarbons, marketed under various
proprietary names of freon, genetron, arcton etc.
Among them Freon–22 (Mono-chloro Difluoro Methane),
Freon–11 (Tri-chloro – mono-fluoro methane) & Freon–12
(Dichloro Difluoro methane) are extensively used.
Ammonia is another commonly used refrigerant. Other
refrigerants include CO2, SO2, Methyl chloride, Methylene
chloride, Ethyl chloride etc.
27

Desirable properties of a good refrigerant:
Thermodynamic properties
1. Low boiling point
2. Low freezing point
3. Positive gauge pressure in condenser and evaporator,
but not very high
4. High latent heat of vaporization
Chemical properties
1. Non–toxic
2. Non–inflammable & non–explosive
3. Non–corrosive
4. Chemically stable
5. No effect on quality of stored products 28

Desirable properties of a good refrigerant:
Physical properties.
1. Low specific volume of vapour
2. Low specific heat
3. High thermal conductivity
4. Low viscosity
Other properties
1. Ease of leakage detection
2. Cost
3. Ease of handling
29

Ammonia is a good refrigerant with the highest
refrigerating effect per unit mass.
It is relatively cheap. But it is toxic and corrosive.
Leakage can be easily detected because if its pungent
odour.
Freons are Non–toxic & non–inflammable. Leakage
cannot be detected easily as they are odour less and
colour less.
Some coloured additives are sometimes mixed with
Freons to facilitate detection of leakage.
30

Gas (Air) Cycle Refrigeration:
Refrigeration can also be accomplished by means of
a gas cycle, the most common being the one using air as a
refrigerant.
In such a cycle, a throttle valve cannot be used for
expansion of the working fluid.
During the throttling process, enthalpy at the beginning is
equal to enthalpy at the end. For an ideal gas, (all gases
including air are assumed to be ideal), enthalpy is a
function of temperature only.
Hence, during throttling temperature at the beginning will
be equal to temperature at the end.
31

Gas (Air) Cycle Refrigeration contin…..:
Since there is no cooling of air during expansion,
refrigeration is not possible. In place of a throttle valve, an
expander is used.
Work output obtained from the expander can be utilized
for compression, thus decreasing the net work input.
In a gas refrigeration cycle, the refrigerant (gas/air)
remains in a gaseous state throughout the cycle.
Since there is no phase change, the terms ‘condenser’ and
‘evaporator’ are not appropriate.
The device in which heat is rejected at a higher
temperature can be called a cooler, while the device in
which heat is absorbed at a lower temperature is called the
‘refrigerator’.
32

Reversed Carnot Cycle
A reversed Carnot cycle
using air as the working
substance can be a
Refrigeration cycle, through it is
not practicable.
1 – 2: isentropic compression.
2 – 3: heat rejection at constant
temperature.
3 – 4: Expansion
4 – 1: heat addition at constant
temperature (refrigeration)
33

Heat rejected during process 2 – 3
= Q1 = Tmax (s2-s3)
= Tmax (s1-s4)
Heat received during process 4 – 1
= Q2 = Tmin (s1-s4)
Wnet = WC - WE = Q1 - Q2 (First Law)
= (Tmax - Tmin) (s1-s4)
36
These COPs are the maximum possible COPs for given maximum
and minimum temperatures.

Reversed Brayton Cycle.
A reversed Brayton cycle with air as the working substance is a
more practical refrigeration cycle.
1 – 2: isentropic compression
2 – 3: constant pressure heat rejection
3 – 4: isentropic expansion
4 - 1: constant pressure heat addition.
37

On a unit mass basis,
Compressor work input = WC = h2 - h1 = Cp (T2 - T1)
Expansion work output = WE = h3 - h4 = Cp (T3 - T4)
Heat rejected at constant pressure = Q1 = h2 - h3 = Cp (T2 - T3)
Heat received at constant pressure = Q2 = h1 - h4 = Cp (T1 - T4)
39

On a unit mass basis,
Compressor work input = WC = h2 - h1 = Cp (T2 - T1)
Expansion work output = WE = h3 - h4 = Cp (T3 - T4)
Heat rejected at constant pressure = Q1 = h2 - h3 = Cp (T2 - T3)
Heat received at constant pressure = Q2 = h1 - h4 = Cp (T1 - T4)
40

For the isentropic process 1 – 2,
For the isentropic process 3-4,
41

The COP of a gas cycle refrigeration system is low.
The power required per unit capacity is high. Its main application is
in aircrafts and missiles, where a vapour compression refrigeration
system becomes heavy and bulky. Another application of gas cycle
refrigeration is in the liquefaction of gases.
Shown below is a schematic flow diagram of an open cycle
air refrigeration system.
42

A small amount of compressed air is blend from the
main compressor of a turbojet or a supercharged
aircraft engine, and is cooled by rejecting heat to large
amounts of cooler ambient air.
The cooled compressed air expands in an expander,
and as a result cools further.
The cool air enters the cabin. The output of the
expander is used to run a blower which sucks the
ambient air in.
In addition to cooling, replacement of stale air in the
cabin is possible. At high altitudes the pressurization of
the cabin is also possible. Because of this
consideration air cycle refrigeration is extensively used
in aircrafts.
43

Vapour Absorption Refrigeration
A vapour absorption refrigeration system uses a
refrigerant as well as an absorbent which can be a
liquid or solid.
Possibly the best known combination is ammonia as
the refrigerant and water as the absorbent.
A vapour absorption refrigeration system does not
have a compressor.
The compressor is replaced by a combination of
generator, an absorber and a pump.
44

Working:
The generator acts as a reservoir for the solution of
ammonia in water.
Heat from an external source QG is supplied to the solution,
leading to evaporation of ammonia and water.
The mixture of ammonia vapour and water vapour rises
through the analyzer, where most of the water vapour
condenses, gets separated from NH3 and drops back into
the generator.
The analyzer is a direct–contact heat exchanger consisting
of a series of trays mounted above the generator.
The strong solution of NH3 from the absorber flows down
over the trays, comes into contact with and cools the rising
vapours. Since the saturation temperature of water is
higher than that of NH3 at a given pressure, water vapour
will condense first.
46

As the vapour passes upward through the analyzer, it is
cooled and enriched by ammonia.
The ammonia vapour leaving the analyzer may still
contain traces of water vapour.
If allowed to flow through the condenser and expansion
valve, the water vapour will freeze and block the
expansion valve.
Traces of water vapour are separated from ammonia
vapour in the rectifier.
The rectifier is a water cooled heat exchanger, wherein all
of the remaining water vapour and some ammonia
vapour condense and return to the generator through the
drip line.
47

The net result is that pure ammonia vapour flows into the
condenser and condenses to form saturated or slightly
under cooled liquid.
The refrigerant then expands through the valve, resulting
in a drop in its pressure and temperature.
The cold refrigerant then flows through the evaporator,
extracting heat from the substance to be cooled.
Saturated or slightly saturated ammonia vapour from
evaporator flows into the absorber.
The weak solution of ammonia (with low concentration of
ammonia in water) coming from the generator is sprayed
into the absorber.
The ammonia vapour comes into contact with the weak
solution, and gets readily absorbed, releasing the latent
heat of condensation.
48

This heat QA taken away by cooling water, thereby
maintaining the temperature in the absorber
constant.
The resulting strong NH3 solution is pumped to the
generator, where heat Qa is supplied to it from an
external source.
The weak solution leaving the generator and the
pressurized strong solution going to the generator
flow through a heat exchanger.
In this heat exchanger, the strong solution is
preheated while the weak solution is pre-cooled,
reducing both Qa, the heat to be supplied in the
generator and QA the heat to be removed in the
absorber.
49

The combination of the generator and absorber is
equivalent to a heat engine, which does the job of the
compressor, namely, receiving from the evaporator at
low pressure, comparatively low temperature ammonia
vapour and delivering high pressure, higher temperature
ammonia vapour to the condenser. This is shown in the
diagram.
50

Advantages of vapour absorption Refrigeration System:
1. No moving parts (in some cases, there is a small pump) Less
wear and tear. Less maintenance cost.
2. Low grade fuels can be used. Waste heat can be used.
3. System not affected by variation of loads.
4. No electricity required.
5. No chance of leakage.
Disadvantages
1. Low COP
2. Bulky
3. Higher cost initially
Vapour absorption refrigeration systems using solar energy as
the heat source to the generator, hold a lot of promise for the
future, in the areas of food preservation and comfort cooling.
51

Steam jet refrigeration:
It operates on the principle of reducing the boiling point
of water below 100OC by reducing the pressure on the
surface of water below the atmospheric pressure.
This low pressure or high vacuum is maintained by
throttling the steam through jet or nozzles.
52

Working: The system consists of an evaporator, one or
more booster ejectors, a surface type of barometric
condenser and two stage ejector air pump.
53

Boiler supplies high pressure steam to the nozzle inlet
where steam expands.
The warm water returning from the A/C plant is
sprayed through the nozzle in the flash chamber to
ensure maximum surface area for evaporation.
The water vapor leaving the flash chamber mixes with
the high velocity steam from the nozzle and gets
further compressed in thermo compressor.
The K.E is converted into pressure energy and the mass
discharged into the condenser returns as condensate
to the boiler.
About 1% evaporation of water in evaporator is
sufficient to reduce the temperature of water to 6OC.
the chilled water is circulated to the A/C plant, which
returns as warm water into the flash chamber.
54

The water lost due to evaporation in the flash
chamber and any loss of cold water is
compensated by make-up water.
Air dissolved in feed water is released in the
condenser of the system and covers the
condensing surfaces along with other gases and
increases the condenser pressure.
This air is removed by using small two stage air
ejector to maintain high vacuum.
In order to maintain temperature in the
evaporator below 0OC it is necessary to add
antifreeze or brine.
55

Limitations:
1. It requires very high vacuum and cannot be used if
temperature in the evaporator is below 0OC.
2. Size of the compressor should be larger to handle
larger volume.
3. The compression ratio used in the thermo compressor
is limited to 8 only.
4. Heat removed in the condenser is almost double as
compared to vapor compression system.
56

Advantages:
1. Due to no moving parts it is noiseless.
2. Flexible in operation as cooling capacity can be quickly
changed.
3. Weight/Ton of Ref is low and plant life is more.
4. Used in cold water processing of rubber mills, chemical
and food processing plants, breweries, refineries etc.
5. Safer in operation and absolutely no hazard from
leakage.
6. Cheaper in operation and easy to maintain.
7. Useful in comfort air conditioning, but not suitable if
water temperature is below 4OC.
57

Prob 1. A cold storage is to be maintained at -5°C (268k) while
the surroundings are at 35°C. the heat leakage from the
surroundings into the cold storage is estimated to be 29kW.
The actual C.O.P of the refrigeration plant is one third of an
ideal plant working between the same temperatures. Find
the power required to drive the plant. VTU Jan 2007.
58
7
.
6
68
2
308
268
T
T
T
ideal
C.O.P
cycle.
carnot
on
based
C.O.P
but
nothing
is
plant
ideal
the
of
C.O.P
268
5
T
308
35
T
-
:
Solution
2
1
2
1 2











 k
C
k
C -

59
233
.
2
7
.
6
3
1
.
.
3
1
.
.



x
P
O
idealC
P
O
C
Actual
Q2 = The heat removed from low temperature reservoir
(cold storage) must be equal to heat leakage from surroundings
to the cold storage(which is 29kw)
kW
12.98
required
233
.
2
29
.
.
W
.
.
29
2
2
2





Power
P
O
C
Actual
Q
W
Q
P
O
C
Actual
kW
Q

2. A refrigeration machine of 6 tones capacity working on
Bell coleman cycle has an upper limit pressure of 5.2
bar. The pressure and temperature at the start of the
compression are 1 bar and 18°C respectively. The
cooled compressed air enters the expander at 41°C,
assuming both expansion and compression to be
adiabatic with an index of 1.4.
Calculate:-
Co-efficient of performance.
Quantity of air circulated per minute.
Piston displacement of compressor and expander
Bore of compression and expansion cylinder when the
unit runs at 240 rpm and is double acting with stroke
length =200 mm , Power required to drive the unit
60

61
bar
2
.
5
P
41
T
1bar
P
18
T
-
:
Solution
2
3
1
1






C
C
   
 
   
  kg
kJ
T
T
T
T
Work
/
57
196
314
291
466
005
.
1
C
input 4
3
1
2
p









21kJ/s
6x3.5
tons
6
capacity
Re
67
.
1
57
95.42
input
Work
Re
.
.






n
frigeratio
effect
n
griferatio
P
O
C

62
13.2kg/min
0.22x60
air/min
of
Mass
12.54kW
0.22
x
57
air/sec
of
Mass
air x
of
/kg
required
/
22
.
0
95.42
21
R.E
apacity
Re
air/sec
of








workdone
Power
s
kg
c
n
griferatio
Mass
min
/
42
.
7
V
expander
of
nt
displaceme
min
/
42
.
7
10
1
196
287
.
0
2
.
13
min
/
11
V
compressor
of
nt
displaceme
min
/
11
10
1
291
287
.
0
2
.
13
3
4
3
2
4
4
4
3
1
3
2
1
1
1
m
Piston
m
x
x
x
P
mRT
V
m
Piston
m
x
x
x
P
mRT
V









63
31.3cm
0.313m
cylinder
expander
of
240
2
.
0
4
2
42
.
7
4
2
38cm
0.38m
cylinder
compressor
of
240
2
.
0
4
2
11
4
2
1
2
1
2
2
4
1
2
1
2
1
1










diameter
d
x
x
d
LN
d
V
diameter
d
x
x
d
LN
d
V
But





Problem3 An air refrigerator system operating on
Bell Column cycle, takes in air from cold room at
268 K and compresses it from 1 bar to 5.5 bar
the index of compression being 1.25. the
compressed air is cooled to 300 K. the ambient
temperature is 200C. Air expands in expander
where the index of expansion is 1.35.
Calculate:
C.O.P of the system
Quantity of air circulated per minute for
production of 1500 kg of ice per day at 0°C from
water at 20ºC.
Capacity of the plant.
64

65
 
K
8
.
376
5
.
5
268 25
.
1
1
25
.
1
1
1
2
1
2















P
P
T
T
Solution
K
83
.
192
5
.
5
1
300
35
.
1
1
35
.
1
1
3
4
3
4 




















P
P
T
T
 
  kg
kJ
T
T
C
n
n
p
/
2
.
156
268
8
.
376
005
.
1
4
.
1
1
4
.
1
1
25
.
1
25
.
1
1
1
W 1
2
C







 
















 











66
 
  kg
kJ
T
T
C
n
n
p
/
69
.
118
83
.
192
300
005
.
1
4
.
1
1
4
.
1
1
35
.
1
35
.
1
1
1
W 4
3
E







 
















 










2
5
.
37
54
.
75
.
.
/
54
.
75
)
83
.
192
268
(
005
.
1
)
(
.
/
5
.
37
69
.
118
2
.
156
4
1













work
RE
P
O
C
s
kJ
T
T
C
E
R
kg
kJ
W
W
Network
p
E
C
s
kg
heat
Latent
Heat
/
0173
.
0
24x3600
1500
ec
produced/s
ice
of
Mass
g
418.74kJ/k
335
4.187(20)
_
)
0
20
(
C
ice
of
kg
extracted/ pw









67
s
kg
Mass
tons
heat
Actual
/
096
.
0
54
.
75
26
.
7
efect
ion
Refrigerat
Capacity
ion
Refrigerat
rate
flow
02
.
2
5
.
3
7.26
7.26kJ/s
capacity
ion
Refrigerat
or
173
418.74x0.0
sec
extracted/








Problem 4
An air refrigeration system is to be designed according to
the following specifications
Pressure of air at compressor inlet=101kPa
Pressure of work at compressor outlet=404kPa
Pressure loss in the inter cooler=12kPa
Pressure loss in the cold chamber=3kPa
Temperature of air at compressor inlet=7°
Temperature of air at turbine inlet=27°
Isentropic efficiency of compressor =85%
Isentropic efficiency of turbine =85%
Determine
C.O.P of cycle
Power required to produce 1 ton of refrigeration
Mass flow rate of air required for 1 ton of refrigeration
68

69
85
.
0
;
85
.
0
27
T
101kPa
P
7
T
-
:
Solution
C
T
3
1
1










C
C
K
4
.
395
101
404
266
'
Hence
,
isentropic
is
2
-
1
Pr
4
.
1
1
4
.
1
1
1
2
1
2





















P
P
T
T
ocess
kPa
x
P
P
P
P
P
kPa
x
P
P
P
P
P
k
T
T
T
T
T
T
T
C
392
404
97
.
0
97
.
0
03
.
0
104
101
03
.
1
03
.
1
03
.
0
2
.
418
'
88
.
0
266
4
.
395
'
or
'
2
3
2
3
2
1
4
1
1
4
2
1
2
1
2
1
2





















70
202.3K
392
104
300
P
P
T
T
,
isentropic
is
4
-
3
Process
1.4
1
1.4
γ
1
γ
3
4
3
4




















 
 
 
g
152.96kJ/k
266]
-
.2
1.005x[418
'
C
air
of
work/kg
ompressor
kJ/kg
47
.
0
5
216.53]
-
1.005x[266
C
air
of
effect/kg
Re
53
.
216
]
3
.
205
300
[
85
.
0
300
'
'
'
1
2
p
4
1
p
4
4
3
3
4
4
3
4
3



















T
T
C
T
T
n
frigeratio
k
x
T
T
T
T
T
T
T
T
T
T
E 


71
 
kg
kJ
T
T
T
/
06
.
72
9
.
80
96
.
152
W
W
air W
of
Input/kg
Net work
kJ/kg
9
.
84
216.53]
-
1.005x[300
'
C
air W
of
work/kg
urbine
T
C
net
4
3
p
T









73
.
0
06
.
72
73
.
46
.
. 


Work
RE
P
O
C
C.O.P
capacity
ion
Refrigerat
ion
refergerat
of
per tons
required

Power

72
kW
x
xmassofair
W
Power
s
kg
net 42
.
5
075
.
0
06
.
72
sec
/
/
075
.
0
47
.
50
5
.
3
RE
capacity
ion
Refrigerat
air
of
Mass
3.5kJ/s
ton
1
capacity
ion
Refrigerat









Problem5:
20 tons of ice is produced from water at 200C
to ice at -60C in a day of 24 hours, when the
temperature range in the compressor is from
-150C to 250C. The condition of the vapour is
dry at the end of compression.
Assuming relative C.O.P as 80%, calculate the
power required to drive the compressor.
Take Cpice=2.1kJ/kg, Latent heat of ice=335k/kg
73

74
Temp
ºC
Liquid
Vapour
Enthalpy hf Entropy Sf Enthalpy hg Entropy Sg
25 100.04 0.347 1319.2 4.4852
-15 -54.55 -2.1338 1304.99 5.0585

75
K
T
T
Ln
T
T
Ln
C
To
b
b
b
b
p
15
.
312
7019
.
0
303
56
.
0
6853
.
0
s
s
a
at
Entropy
b
at
Entropy
b'.
'
point
at
vapour
of
condition
the
find
ga
'
gb'









kg
kJ
H
ork
kg
kJ
H
T
T
C
h
H
kg
kJ
h
T
T
C
h
H
a
c
c
c
PL
fc
c
ga
b
b
p
gb
b
/
55
.
21
19
.
183
74
.
204
H
w
/
61
.
123
575
.
59
19
.
183
H
R.E
kJ/kg
59.575
25)
-
1.003(30
-
64.59
)
(
/
19
.
183
H
/kg
204.74kJ
303)
-
5
0.56(312.1
100.62
)
(
'
b
a
'
'
'
a
'






















 

76
431.34
6
x
2.1
335
20
x
4.187
)]
6
(
0
[
C
heat
)
0
20
(
C
ice
of
kg
extracted/
Heat
7.13
0.8x8.913
C.O.P
ctual
0.8
C.O.P
Re
913
.
8
98
.
122
18
.
1096
.
.
pice
pw
















Latent
A
lative
work
RE
P
O
C
kW
Power
Actual
Actual
Actual
s
kg
x
Mass
14
7.13
99.84
work/sec
work/sec
ec
extrated/s
heat
Actual
C.O.P
Actual
99.84kJ/s
31
431.34x0.2
sec
extracted/
heat
/
231
.
0
3600
24
20x1000
ec
produced/s
ice
of










77
s
kJ
capacity
f
n
frigeratio
work
RE
P
O
C
/
424
.
0
123.61
52.5
RE
.
Re
feron
of
Mass
52.5kJ/kg
15x3.5
15tons
capacity
Re
73
.
5
55
.
21
61
.
123
.
.









9.152kW
4
21.55x0.42
freon/s
of
xMass
work/kg
required



Power

Problem6: A food storage locker requires a refrigeration
system of 12 tons capacity at an evaporator temperature
of -80C and a condenser temperature of 300C. The
refrigerant freon-12 is sub cooled to 250C before
entering the expansion valve and the vapour is
superheated to -20C before entering the compressor. The
compression of the refrigerant is reversible adiabatic. A
double action compressor with stroke equal to 1.5 times
the bore is to be used operating at 900 rpm.
Determine
COP
Theoretical piston displacement/min
Mass of refrigerant to be circulated/min
Theoretical bore and stroke of the compressor.
Take liquid specific heat of refrigerant as 1.23 kJ/kg K and
the specific heat of vapour refrigerant is 0.732 kJ/kg K.
78

79
Temp
ºC
Enthalpy
Entropy
hf hg Sf Sg
30 64.59 199.62 0.24 0.6853
-8 25.75 184.2 0.1142 0.7002
Solution:
From tables the properties of Freon 12 are

80
265
271
Ln
733
.
0
7002
.
0
303
T
Ln
732
.
0
6853
.
0
S
S
a
at
entropy
b
at
/
235
.
1
,
/
732
.
0
b
'
ga'
'
gb'









a
a
p
b
b
p
PL
p
T
T
Ln
C
T
T
Ln
C
Entropy
kgK
kJ
C
kgK
kJ
C



K
Tb 22
.
317

kJ/kg
02
.
210
)
303
22
.
318
(
732
.
0
62
.
99
1
)
(
kJ/kg
59
.
188
)
265
271
(
732
.
0
2
.
84
1
)
(
'
'
'
'














b
b
p
gb
b
a
a
p
ga
a
T
T
C
h
H
T
T
C
h
H



81
kg
kJ
H
ork
kg
kJ
H
T
T
C
h
H
a
c
c
c
PL
gc
c
/
43
.
21
59
.
188
02
.
210
H
w
/
181
.
130
41
.
58
59
.
188
H
R.E
kJ/kg
41
.
8
5
298)
-
1.235(303
-
64.59
)
(
b
a
'
'















07
.
6
43
.
21
18
.
130
.
. 


work
RE
P
O
C
a
a
a
ga
T
PV
T
PV
kg
m
s
kg
capacity
f








'
'
3
ga' /
0441995
.
0
V
C,
8
-
at
tables
From
n
19.35kg/mi
0.322x60
/
322
.
0
130.18
12x3.5
RE
.
Re
t
refrigeran
of
Mass

82
0.0203m
900
4
5
.
1
2
87462
.
0
)
5
.
1
(
4
2
min
/
0.87462m
52
19.35x0.04
xV
mass
V
nt
displaceme
piston
0452
.
0
0441995
.
0
265
271
2
2
3
a
'
'










x
d
x
xd
d
L
LN
d
V
l
Theoretica
x
T
xV
T
V
a
ga
a
a


11.08cm
1.5x7.38
1.5d
L
7.38cm
0.0738m
d






83
Temp Pressure Vf Vg hf hg Sf Sg
°C bar m3/kg kJ/kg kJ/kgK
-10 226 0.7x10-3 0.08 190 345 0.95 1.5
30 7.5 0.77x10-3 0.02 220 220 1.10 1.45
Problem7:
A vapour compression refrigeration system of 5kW cooling
capacity operates between -10ºC and 30ºC. The enthalpy of
refrigerant vapour after compression is 370kJ/kg. Find the COP,
refrigerating effect, mass flow rate of the refrigerant and the
compressor power. The extract of the refrigerant property table is
given below

84
kg
kJ
ork
kg
kJ
given
kg
kJ
H
kg
kJ
h
kg
kJ
h
H
b
ga
fc
c
/
25
345
370
H
H
w
/
125
220
345
H
H
R.E
)
(
/
370
/
345
H
/
220
a
b
c
a
'
a















Solution:
Assume the condition before compression as dry saturated vapour
5
25
125
.
. 


work
RE
P
O
C

85
1kW
25x0.04
-
t
refrigeran
of
mass
x
.
/
04
.
0
125
5
RE
.
Re
t
refrigeran
of
Mass
kJ/s
or
5kW
capacity
ion
Refrigerat






kg
work
work
Compressor
s
kg
capacity
f

86
Problem8: A vapour compression refrigerator uses
methyl chloride and works in the pressure rang of
1.19 bar and 5.67 bar. At the beginning of
compression, the refrigerant is 0.96 dry and at the
end of isentropic compression, its temperature is
55ºC. The refrigerant liquid leaving the condenser is
saturated.
If the mass flow of refrigerant is 1.8kg/min,
Determine
COP
The rise in temperature of cooling water if the water
flow rate is16 kg/min. the properties of methyl
chloride is given below

87
Temp
ºC
Pressure
bar
Enthalpy
Entropy
hf Hfg hg Sf Sg
30 1.19 64.59 135.03 199.62 0.24 0.6853
-10 5.67 26.87 156.31 183.19 0.108 0.7019
kg
kJ
h
h
x
h
h
x
h
H
C
T
x
Solution
Take
fa
ga
a
fa
fga
a
fa
a
b
a
/
196
.
438
30.1)
-
0.96(455.2
430.1
)
(
)
(
55
96
.
0
K
0.75kJ/kg
as
chloride
methyl
heat
super
of
heat
specific












88
kg
kJ
H
ork
kg
kJ
H
kg
kJ
h
T
T
C
h
H
a
c
fc
b
b
p
gb
b
/
8
.
60
196
.
438
499
H
w
/
669
.
337
5
.
100
196
.
438
H
R.E
/
5
.
100
H
kJ/kg
499
25)
-
(55
75
.
0
5
.
76
4
)
(
b
a
c
'
'


















55
.
5
8
.
60
669
.
337
.
. 


work
RE
P
O
C
water
cooling
by
gain
heat
condenser
in the
vapour
by the
lost

Heat
C
rise
e
Temperatur
rise
e
temperatur
x
rise
e
temperatur
x
C
m
h
m
T
T
C
m p
w
fgb
r
b
b
p
r








7
.
10
x
187
.
4
16
100.5)
-
1.8(476.5
25)
-
0.75(55
x
8
.
1
)
( '
' 

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