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REFRIGERATION AND
AIRCONDITIONING
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 2
B.TECH. DEGREE COURSE
MAHATMA GANDHI UNIVERSITY
KOTTAYAM,KERALA
REFRIGERATION AND AIR CONDITIONING
M 704 2+1+0
Module 1
Principles of refrigeration: Thermodynamics of refrigeration - Carnot cycle,
reversed carnot cycle, heat pump, and refrigerating machine- coefficient of
performance - unit of refrigeration - refrigeration methods- conventional
refrigeration systems. Air refrigeration system- Bell Coleman cycle - C.O.P.
capacity work and refrigerant flow requirements in Bell - Coleman cycle.
Module 2
Vapour compression system: simple cycle -comparison with Carnot cycle -
theoretical, actual and reactive - COP effect of operating parameters on
COP - wet, dry and superheated compression - under cooling - actual cycle
representation on TS and PH diagrams simple problems. Advanced
vapour compression systems - multistage vapour compression systems -
flash chamber multiple compression and evaporation systems cascading -
simple problems.
Module 3
Vapour absorption systems: simple, cycles - actual cycle - ammonia water
and lithium bromide water systems - COP - electrolux system. Refrigerant
and their properties: Nomenclature - suitability of refrigerants for various
applications - unconventional refrigeration methods- Vortex tube, steam-
jet, magnetic (cryogenics) refrigeration and thermoelectric refrigeration -
applied refrigeration house hold refrigerators - unit air conditioners and
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 3
water coolers - ice plant cold storage.
Module 4
Refrigeration system components: condensers - water and air cooled
condensers - evaporative condensers - expansion devises - capillary tube-
constant pressure expansion valve - thermostatic expansion valve - float
valve and solenoid valve - evaporators - natural convection coils - flooded
evaporators - direct expansion coils. Reciprocating compressors: single
stage and multistage compressors - work done optimum pressure ratio-
effect of interfolding - volumetric efficiency -effect of clearance -
isothermal and adiabatic efficiency - compressed air motors. Rotodynamic
compressors: Screw and vane type compressors - principle of operation -
hermetic, semihermetic and open type refrigeration compressors.
Module 5
Principles of air conditioning: Psychrometry and psychrometric chart
thermodynamics of human comfort - effective temperature - comfort chart
applied psychrometry - sensible heat factor - psychometric process-
problems. Winter air conditioning: heating load calculations humidifiers
and humidistat. Summer air conditioning: cooling load calculations - year
round air conditioning - unitary and central systems - principles of air
distribution - design of air duct systems.
References
1. Refrigeration and air conditioning - Ballaney P. L.
2. Refrigeration and air conditioning - Stocker W. F.
3. Refrigeration and air conditioning - Jordan and Protester
4. Principles of Refrigeration - Roy J. Dossat
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 4
MODULE – I
Air Refrigeration Cycles
1.1 Introduction
In an air refrigeration cycle, the air is used as a refrigerant. In olden days,
air was widely used in commercial applications because of its availability at free
of cost. Since air does not change its phase i.e. remains gaseous throughout the
cycle, therefore the heat carrying capacity per kg of air is very small as compared
to vapour absorbing systems. The air-cycle refrigeration systems, as originally
designed and installed, are now practically obsolete because of their low
coefficient of performance and high power requirements. However, this system
continues to be favoured for air refrigeration because of the low weight and
volume of the equipment. The basic elements of an air cycle refrigeration system
are the compressor, the cooler or heat exchanger, the expander and the
refrigerator.
Before discussing the air refrigeration cycles, we should first know about
the unit of refrigeration, coefficient of performance of a refrigerator and the
difference between the heat engine, a refrigerator and a heat pump.
1.2 Units of Refrigeration
The practical unit of refrigeration is expressed in terms of ‘tonne of
refrigeration’ (briefly written as TR). A tonne of refrigeration is defined as the
amount of refrigeration effect produced by the uniform melting of one tonne
(1000 kg) of ice from and at 0
o
C in 24 hours.
Since the latent heat of ice is 335 kJ/kg, therefore one tonne of
refrigeration,
ITR = 1000 × 335 kJ in 24 hours
=
In actual practice, one tonne of refrigeration is taken as equivalent to 210
kJ/min or 3.5 kW (i.e. 3.5 kJ/s)
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 5
1.3 Coefficient of Performance of a Refrigerator
The coefficient of performance (briefly written as C.O.P.) is the ratio of
heat extracted in the refrigerator to the work done on the refrigerant. It is known
as theoretical coefficient of performance. Mathematically,
Theoretical C.O.P. =
where Q = Amount of heat extracted in the refrigerator (or the
amount of refrigeration produced, or the capacity of a refrigerator), and
W = Amount of work done.
Notes: 1. For per unit mass, C.O.P. =
2. The coefficient of performance is the reciprocal of the efficiency (i.e.
) of a heat engine. It is thus obvious, that the value of C.O.P. is always greater
than unity.
3. The ratio of the actual C.O.P. to the theoretical C.O.P. is known as
relative coefficient of performance, Mathematically.
Relative C.O.P =
Example 1.1 Find the C.O.P. of a refrigeration system if the work input is 80 kJ/kg
and refrigeration effect produced is 160 kJ/kg of refrigerant flowing.
Solution. Given : w = 80 kJ/kg ; q = 160 kJ/kg
We know that C.O.P. of a refrigeration system
= = = 2 Ans.
1.4 Difference Between a Heat Engine, Refrigerator and Heat Pump
In a heat engine, as shown in Fig. 2.1 (a), the heat supplied to the engine
is converted into useful work. If Q2 is the heat supplied to the engine and Q1 is the
heat rejected from the engine, then the net work done by the engine is given by
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 6
WE = Q2 - Q1
Fig. 1.1. Difference between a heat engine, refrigerator and heat pump.
The performance of a heat engine is expressed by its efficiency. We know
that the efficiency or coefficient of performance of an engine,
or (C.O.P.)E = = =
A refrigerator as shown in Fig.1.1 (b), is a reversed heat engine which
either cool or maintain the temperature of a body (T1) lower than the atmospheric
temperature (Ta). This is done by extracting the heat(Q1) from a cold body and
delivering it to a hot body (Q2). In doing so, work WR is required to be done on the
system. According to First Law of Thermodynamics,
WR = Q2 - Q1
The performance of a refrigerator is expressed by the ratio of amount of
heat taken from the cold body (Q1) to the amount of work required to be done on
the system (WR). This ratio is called coefficient of performance. Mathematically,
coefficient of performance of a refrigerator,
(C.O.P.)R = =
Any refrigerating system is a heat pump as shown in Fig 2.1 (c), which
extracts heat (Q1) from a cold body and delivers it to a hot body. Thus there is no
difference between the cycle of operations of a heat pump and a refrigerator. The
main difference between the two is in their operating temperatures. A refrigerator
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 7
works between the cold body temperature (T1) and the atmospheric temperature
(Ta)whereas the heat pump operates between the hot body temperature (T2) and
the atmospheric temperature (Ta). A refrigerator used for cooling in summer can
be used as a heat pump for heating in winter.
In the similar way, as discussed for refrigerator, we have
WP = Q2 - Q1
The performance of a heat pump is expressed by the ratio of the amount
of heat delivered to the hot body (Q2) to the amount of work required t o be done
on the system (WP). This ratio is called coefficient of performance or energy
performance ratio (E.P.R.) of a heat pump. Mathematically, coefficient of
performance or energy performance ratio of a heat pump.
(C.O.P.)P or E.P.R. = =
= = (C.O.P.)R + 1
From above we see that the C.O.P. may be less than one or greater than
one depending on the type of refrigeration system used. But the C.O.P. of a heat
pump is always greater than one.
1.5 Open Air Refrigeration Cycle
In an open air refrigeration cycle, the air is directly led to the space to be
cooled (i.e. a refrigerator), allowed to circulate through the cooler and then
returned to the compressor to start another cycle. Since the air is supplied to the
refrigerator at atmospheric pressure, therefore, volume of air handled by the
compressor and expander is large. Thus the size of compressor and expander
should be large. Another disadvantage of the open cycle system is that the
moisture is regularly carried away by the air circulated through the cooled space.
This leads to the formation of frost at the end of expansion process and clog the
line. Thus in an open cycle system, a drier should be used.
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 8
1.6 Closed or Dense Air Refrigeration Cycle
In a closed or dense air refrigeration cycle, the air is passed through the
pipes and component parts of the system at all times. The air, in this system, is
used for absorbing heat from the other fluid (say brine) and this cooled brine is
circulated into the space to be cooled. The air in the closed system does not
come in contact directly with the space to be cooled.
The closed air refrigeration cycle has the following thermodynamic
advantages :
1. Since it can work at a suction pressure higher than that of
atmospheric pressure, therefore the volume of air handled by the compressor
and expander are smaller as compared to an open air refrigeration cycle system.
2. The operating pressure ratio can be reduced, which results in higher
coefficient of performance.
1.7. Air Refrigerator Working on Reversed Carnot Cycle
In refrigerating systems, the Carnot cycle considered is the reversed
Carmot cycle. We know that a heat engine working on Carnot cycle has the
highest possible efficiency. Similarly, a refrigerating system working on the
reversed Carnot cycle, will have the maximum possible coefficient of
performance. We also know that it is not possible to make an engine working on
the Carnot cycle. However, it is used as the ultimate standard of comparison.
A reversed Carnot cycle, using air a working medium (or refrigerant) is
shown on p-v and T-s diagrams in Fig. 1.2 (a) and (b) respectively. At point 1, let
p1, v1, T1 be the pressure, volume and temperature of air respectively.
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 9
Fig. 1.2 Reversed Carnot cycle
The four processes of the cycle are as follows:
1. Isentropic compression process. The air is compressed isentropically
as shown by the curve 1-2 on p-v and T-s diagrams. During this process, the
pressure of air increases from p1 to p2, specific volume decreases from v1 to v2
and temperature increases from T1 to T2. We know that during isentropic
compression, no heat is absorbed or rejected by the air.
2. Isothermal compression process. The air is now compressed
isothermally (i.e. at constant temperature, T2 = T3) as shown by the curve 2-3 on p
-v and T-s diagrams. During this process, the pressure of air increases from p2 to
p3 and specific volume decreases from v2 to v3. We know that the heat rejected by
the air during isothermal compression per kg of air,
q2-3 = Area
=
3. Isentropic expansion process. The air is now expanded isentropically
as shown by the curve 3-4 on p-v and T-s diagrams. The pressure of air
decreases from p3 to p4, specific volume increases from v3 to v4 and the
temperature decreases from T3 to T4. We know that during isentropic expansion
no heat is absorbed or rejected by the air.
4. Isothermal expansion process. The air is now expanded isothermally
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 10
(i.e. at constant temperature, T4 = T1) as shown by the curve 4-1 on p-v and T-s
diagrams. The pressure of air decreases from p4 to p1, and specific volume
increases from v4 to v1. We know that the heat absorbed by the air (or heat
extracted from the cold body) during isothermal expansion per kg of air,
q4-1 = Area
=
We know that work done during the cycle per kg of air
= Heat rejected - Heat absorbed = q2-3-q4-1
= T2 (s2-s3)-T1(s2-s3) = (T2-T1) (s2-s3)
Coefficient of performance of the refrigeration system working on
reversed Carnot cycle,
(C.O.P.)R = =
= =
Though the reversed Carnot cycle is the most efficient between the fixed
temperature limits, yet no refrigerator has been made using this cycle. This is due
to the reason that the isentropic processes of the cycle require high speed while
the isothermal processes require an extremely low speed. This variation in speed
of air is not practicable.
Note : We have already discussed that C.O.P. of a heat pump,
(C.O.P.)P = (C.O.P.)R + 1 = +1 =
and C.O.P. or efficiency of a heat engine,
(C.O.P.)E = =
1.8 Temperature Limitations for Reversed Carnot Cycle
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 11
We have seen in the previous article that the C.O.P. of the refrigeration
system working on reversed Carnot cycle is given by
(C.O.P.)R =
where T1 = Lower temperature, and
T2 = Higher temperature.
The C.O.P. of the reversed Carnot cycle may be improved by
1. Decreasing the higher temperature (i.e. temperature of hot body, T2),
or
2. Increasing the lower temperature (i.e. temperature of cold body, T1).
This applies to all refrigerating machines, both theoretical and practical. It
may be noted that temperatures T1 and T2 cannot be varied at will, due to certain
functional limitations. It should be kept in mind that the higher temperature (T2) is
the temperature of cooling water or air available for rejection of heat and the
lower temperature (T1) is the temperature to be maintained in the refrigerator.
The heat transfer will take place in the right direction only when the higher
temperature is more than the temperature of cooling water or air to which heat is
to be rejected, while the lower temperature must be less than the temperature of
substance to be cooled.
Thus if the temperature of cooling water or air (i.e. T2) available for heat
rejection is low, the C.O.P. of the Carnot refrigerator will be high. Since T2 in
winter is less than T2 in summer, therefore, C.O.P. in winter will be higher than
C.O.P. in summer. In other words, the Carnot refrigerators work more efficiently
in winter than in summer. Similarly, if the lower temperature fixed by the
refrigeration application is high, the C.O.P. of the Carnot refrigerator will be high.
Thus a Carnot refrigerator used for making ice at 0o
C (273 K) will have less C.O.P.
than a Carnot refrigerator used for air-conditioned plant in summer at 20o
C when
the atmospheric temperature is 40o
C. In other words, we can say that the Carnot
C.O.P. of a domestic refrigerator is less than the Carnot C.O.P. of a domestic air-
conditioner.
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 12
Example 1.2. A machine working on a Carnot cycle operates between
305 K and 260 K. Determine the C.O.P. when it is operated as 1. a refrigerating
machine; 2. a heat pump; and 3. a heat engine.
Solution. Given : T2 = 305 K; T1 = 260 K
1. C.O.P. of refrigerating machine
We know that C.O.P. of a refrigerating machine,
(C.O.P.)R = = = 5.78 Ans.
2. C.O.P. of a heat pump
We know that C.O.P. of a heat pump,
(C.O.P.)P = = = 6.78 Ans.
3. C.O.P. of a heat engine
We know that C.O.P. of a eat engine,
(C.O.P.)E = = = 0.147 Ans.
Example 1.3. A Carnot refrigeration cycle absorbs heat at 270 K and
rejects it at 300 K.
1. Calculate the coefficient of performance of this refrigeration cycle.
2. If the cycle is absorbing 1130 kJ/min at 270 K, how many kJ of work is
required per second ?
3. If the Carnot heat pump operates between the same temperatures as
the above refrigeration cycle, what is the coefficient of performance ?
4. How many kJ/min will the heat pump deliver at 300 K if absorbs 1130
kJ/min at 270 K.
Solution. Given : T1 = 270 K; T2 = 300 K
1. Coefficient of performance of Carnot refrigeration cycle
We know that coefficient of performance of Carnot refrigeration cycle,
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 13
(C.O.P.)R = = = 9 Ans.
2. Work required per second
Let WR = Work required per second.
Heat absorbed at 270 K (i.e. T1),
Q1 = 1130 kJ/min = 18.83 kJ/s ... (Given)
We know that
(C.O.P.)R = or 9 =
WR = 2.1 kJ/s Ans.
3. Coefficient of performance of Carnot heat pump
We know that coefficient of performance of a Carnot heat pump,
(C.O.P.)P= = = 10 Ans.
4. Heat delivered by heat pump at 300 K
Let Q2 = Heat delivered by heat pump at 300 K.
Heat absorbed at 270 K (i.e. T1),
Q1 = 1130 kJ/min ... (Given)
We know that
(C.O.P.)P = or 10 =
10Q2 - 11300 = Q2 or Q2 = 1256 kJ/min Ans.
Example 1.4. A cold storage is to be maintained at -5oC while the
surroundings are at 35oC. The heat leakage from the surroundings into the cold
storage is estimated to be 29 kW. 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.
Solution. Given : T1 = -5o
C = -5 + 273 = 268 K;
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AKHIL K T, Assistant Professor, VJCET Page 14
T2 = 35
o
C = 35 + 273 = 308 K;
Q1 = 29 kW; (C.O.P.)actual = (C.O.P.)ideal
The refrigerating plant operating between the temperatures T1 and T2 is
shown in Fig. 1.3.
Let WR = Work or power required to drive the plant.
We know that the coefficient of performance of an ideal refrigeration
plant,
(C.O.P.)ideal = = = 6.7
Actual coefficient of performance,
(C.O.P.)actual = (C.O.P.)ideal = × 6.7 = 2.233
We also know that (C.O.P.)actual =
WR = = = 12.987 kW Ans.
Example 1.5. 1.5 kW per tonne of refregeration is required to maintain
the temperature of -40oC in the refrigerator. If the refrigeration cycle works on
Carnot cycle, determine the following:
1. C.O.P. of the cycle; 2. Temperature of the sink ; 3. Heat rejected to the
sink per tonne of refrigeration ; and 4. Heat supplied and E.P.R., if the cycle is
used as a heat pump.
Solution. Given : WR= 1.5 kW; Q1 = 1 TR; T1 = -40o
C = -40 + 273 = 233 K
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 15
1. C.O.P. of the cycle
Since 1.5 kW per tonne of refrigeration is required to maintain the
temperature in the refrigerator, therefore amount of work required to be done,
WR = 1.5 kW = 1.5 kJ/s = 1.5 × 60 = 90 kJ/min
and heat extracted from the cold body,
Q1 = 1 TR = 210 kJ/min
We know that (C.O.P.)R = = = 2.33 Ans.
Fig 1.4
2. Temperature of the sink
Let T2 = Temperature of the sink
We know that (C.O.P.)R = or 2.33=
T2 = + 233 = 333 K = 60o
C Ans.
3. Heat rejected to the sink per tonne of refrigeration
We know that heat rejected to the sink,
Q2 = Q1 + WR = 210 + 90 = 300 kJ/min Ans.
4. Heat supplied and E.P.R., if the cycle is used as a heat pump is
Q2 = 300 kJ/min Ans.
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 16
and E.P.R. = (C.O.P.)R + 1 = 2.33 + 1 = 3.33 Ans.
Example 2.6. The capacity of a refrigerator is 200 TR when working
between - 6o
C and 25o
C. Determine the mass of ice produced per day from water
at 25o
C. Also find the power required to drive the unit. Assume that the cycle
operates on reversed Carnot cycle and latent heat of ice is 335 kJ/kg.
Solution. Given : Q = 200 TR; T1 = -6o
C = - 6 + 273 = 267 K; T2 = 25o
C = 25
+ 273 = 298 K ; tw = 25o
C ; hfg(ice) = 335 kJ/kg
Mass of ice produced per day
We know that heat extraction capacity of the refrigerator
= 200 × 210 = 42 000 kJ/min ... ( 1TR = 210 kJ/min)
and heat removed from 1 kg of water at 25o
C to form ice at 0o
C
= Mass × Sp. heat × Rise in temperature + hfg(ice) =335 kJ/kg
= 1 × 4.187 (25 - 0) + 335 = 439.7 kJ/kg
Mass of ice produced per min
= = 95.52 kg/min
and mass of ice produced per day = 95.52 × 60 × 24 = 137 550 kg
= 137.55 tonnes Ans.
Power required to drive the unit
We know that C.O.P. of the reversed Carnot cycle
= = = 8.6
Also C.O.P. =
8.6 =
or Work done per min = 42 000/8.6 = 4884 kJ/min
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Power required to drive the unit
= 4884/60 = 81.4 kW Ans.
Example 1.7. Five hundred kgs of fruits are supplied to a cold storage at
20oC. The cold storage is maintained at -5oC and the fruits get cooled to the
storage temperature in 10 hours. The latent heat of freezing is 105 kJ/kg and
specific heat of fruit is 1.256 kJ/kg K. Find the refrigeration capacity of the plant.
Solution. Given : m = 500 kg ; T2 = 20o
C = 20 + 273 = 293 K ; T1 = -5o
C = -5
+ 273 = 268 K ; = 105 kJ/kg, cF = 1.256 kJ/kg K
We know that heat removed from the fruits in 10 hrs,
Q1 = m cF (T2 - T1)
= 500 × 1.256 (293 - 268) = 15 700 kJ
and total latent heat of freezing,
Q2 = m. = 500 × 105 = 52 500 kJ
Total heat removed in 10 hrs,
Q = Q1 + Q2 = 15 700 + 52 500 = 68 200 kJ
and total heat removed in one minute
= 68 200/10 × 60 = 1113.7 kJ/min
Refrigeration capacity of the plant
= 113.7/210 = 0.541 TR Ans. ...( 1 TR = 210 kJ/min)
Example 1.8. A cold storage plant is required to store 20 tonnes of fish.
The fish is supplied at a temperature of 30o
C. The specific heat of fish above
freezing point is 2.93 kJ/kg K. The specific heat of fish below freezing point is
1.26 kJ/kg K. The fish is stored in cold storage which is maintained at - 8o
C. the
freezing point of fish is - 4o
C. The latent heat of fish is 235 kJ/kg. If the plant
requires 75 kW to drive it, find :
1. The capacity of the plant, and 2. Time taken to achieve cooling.
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 18
Assume actual C.O.P. of the plant as 0.3 of the Carnot C.O.P.
Solution. Given : m = 20 t = 20 000 kg ; T2 = 30o
C = 30 + 273 = 303 K ; cAF
= 2.93 kJ/kg K; cBF = 1.26 kJ/kg K ; T1 = -8
o
C = -8 + 273 = 265 K ; T3 = -4
o
C = -4 +
273 = 269 K; hfg(Fish) = 235 kJ/kg ; P = 75 kW = 75 kJ/s
1. Capacity of the plant
We know that Carnot C.O.P.
= = = 6.97
Actual C.O.P. = 0.3 × Work required
= 2.091 × 75 = 156.8 kJ/s = 156.8 × 60 = 9408 kJ/min
Capacity of the plant
= 9408 / 210 = 44.8 TR Ans. ...( 1TR = 210 kJ/min)
2. Time taken to achieve cooling
We know that heat removed from the fish above freezing point,
Q1 = m × cAF (T2 - T3)
= 20 000 × 2.93 (303 - 269) = 1.992 × 106
kJ
Similarly, heat removed from the fish below freezing point,
Q2 = m × cBF (T3 - T1)
= 20 000 × 1.26 (269 - 265) = 0.101 × 106
kJ
and total latent heat of fish,
Q3 = m × = 20 000 × 235 = 4.7 × 106
kJ
Total heat removed by the plant
= Q1 + Q2 + Q3
= 1.992 × 106
+ 0.101 × 106
+ 4.7 × 106
= 6.793 × 106
kJ
and time taken to achieve cooling
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AKHIL K T, Assistant Professor, VJCET Page 19
= = = 722 min = 12.03 h Ans
1.9. Air Refrigerator Working on a Bell-Coleman Cycle (or Reversed Brayton or
Joule Cycle)
A Bell-Coleman air refrigeration machine was developed by Bell-Coleman
and Light Foot by reversing the Joule’s air cycle. It was one of the earliest types
of refrigerators used in ships carrying frozen meat. fig. 1.5 shows a schematic
diagram of such a machine which consists of a compressor, a cooler, an
expander and a refrigerator.
Fig 1.5 Open cycle air Bell- Coleman Fig 1.6 Closed cycle or dence air
bell-
Refrigerator Coleman refrigerator
The Bell-Coleman cycle (also known as reversed Brayton or Joule cycle)
is a modification of reversed Carnot cycle. The cycle is shown on p-v and T-s
diagrams in Fig. 1.7 (a) and (b). At point 1, p1, v1 and T1 be the pressure, volume
and temperature of air respectively. The four process of the cycle are as follows:
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 the curve 1-2 on p-v and T-s diagrams. During the
compression stroke, both the pressure and temperature increases and the
specific volume of air at delivery from compressor reduces from v1 to v2. We
know that during isentropic compression process, no heat is absorbed or
rejected by the air.
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Fig 1.7 Bell-Coleman cycle
2. Constant pressure cooling process. The warm air from the compressor
is now passed into the cooler where it is cooled at constant pressure p3 (equal to
p2), reducing the temperature from T2 to T3 (the temperature of cooling water) as
shown by the curve 2-3 on p-v and T-s diagrams. The specific volume also
reduces from v2 to v3. We know that heat rejected by the air during constant
pressure per kg of air,
Q2-3 = cp (T2-T3)
3. 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 air during expansion falls from T3 to T4 (i.e. the temperature much
below the temperature of cooling water, T3). The expansion process is shown by
the curve 3-4 on the p-v and T-s diagrams. The specific volume of air at entry to
the refrigerator increases from v3 to v4. We know that during isentropic expansion
of air, no heat is absorbed or rejected by the air.
4. Constant pressure expansion process. The cold air from the expander
is now passed to the refrigerator where it is expanded at constant pressure p4
(equal to p1). The temperature of air increases from T4 to T1. This process is
shown by the curve 4-1 on the p-v and T-s diagrams. Due to heat from the
refrigerator, the specific volume of the air changes from v4 to v1. We know that
the heat absorbed by the air (or heat extracted from the refrigerator) during
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AKHIL K T, Assistant Professor, VJCET Page 21
constant pressure expansion per kg of air is
q4 - 1 = cp (T1 - T4)
We know that work done during the cycle per kg of air
= Heat rejected - Heat absorbed
= Cp (T2 - T3) - cp (T1 - T4)
C.O.P. = =
=
= ... (i)
We know that for isentropic compression process 1-2,
= ... (ii)
Similarly, for isentropic expansion process 3-4,
= ... (iii)
Since p2 = p3 and p1 = p4, therefore from equations (ii) and (iii),
or ... (iv)
Now substituting these values in equation (i), we get
C.O.P. =
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= ... (v)
where rp = Compression or Expansion ratio = =
Sometimes, the compression and expansion process take place
according to the law pvn
= Constant. In such a case, the C.O.P. is obtained from
the fundamentals as discussed below:
We know that work done by the compressor during the process 1-2 per
kg of air,
... ( pv = RT)
and work done by the expander during the process 3-4 per kg of air,
Net work done during the cycle per kg of air,
We also know that heat absorbed during constant pressure process 4-1,
= cp (T1 - T4)
C.O.P. = = ... (vi)
We know that
Substituting the value of R in equation (vi),
C.O.P. =
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=
= ... (vii)
Notes : 1. In this case, the values of T2 and T4 are to be obtained from the
relations
and
2. For isentropic compression or expansion, n = . Therefore, the
equation (vii) may be written as
C.O.P. = ... (same as before)
3. We have already discussed that the main drawback of the open cycle
air refrigerator is freezing of the moisture in the air during expansion stroke which
is liable to choke up the valves. Due to this reason, a closed cycle or dense air
Bell-Coleman refrigerator as shown in Fig. 2.6 is preferred. In this case, the cold
air does not come in direct contact of the refrigerator. The cold air is passed
through the pipes and it is used for absorbing heat from the brine and this cooled
brine is circulated in the refrigerated space. The term dense air system is derived
from the fact that the suction to the compressor is at higher pressure than the
open cycle system (which is atmospheric).
Example 1.9. A refrigerator working on Bell-Coleman cycle operates
between pressure limits of 1.05 bar and 8.5 bar. Air is drawn from the cold
chamber at 10o
C, compressed and then it is cooled to 30o
C before entering the
expansion cylinder. The expansion and compression follows the law pv1.3
=
constant. Determine the theoretical C.O.P. of the system.
Solution: Given : p1 = p4 = 1.05 bar ; p2 = p3 = 8.5 bar; T1 = 10o
C = 10 + 273
= 283 K; T3 = 30o
C = 30 + 273 = 303 K; n = 1.3
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Fig 1.8
The p-v and T-s diagrams for a refrigerator working on the Bell-Coleman
cycle is shown in Fig. 1.8 (a) and (b) respectively.
Let T2 and T4 = temperatures of air at the end of compression and
expansion respectively.
Since the compression and expansion follows the law pv1.3
= C, therefore
Similarly
T4 = T3/1.62 = 303/1.62 = 187 K
Fig.
We know that theoretical coefficient of performance,
C.O.P. =
= ... (Taking = 1.4)
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= = 1.3 Ans.
MODULE– II
Vapour Compression Refrigeration Systems
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2.1 Introduction
A vapour compression refrigeration system is an improved type of air
refrigeration system in which a suitable working substance, termed as
refrigerant, is used. It condenses and evaporates at temperatures and pressures
close to the atmospheric conditions. The refrigerants, usually, used for this
purpose are ammonia (NH3), carbon dioxide (CO2) and sulphur dioxide (SO2).
The refrigerant used, does not leave the system, but is circulated throughout the
system alternately condensing and evaporating. In evaporating, the refrigerant
absorbs its latent heat from the brine (salt water) which is used for circulating it
around the cold chamber. While condensing, it gives out its latent heat to the
circulating water of the cooler. The vapour compression refrigeration system is,
therefore a latent heat pump, as it pumps its latent heat from the brine and
delivers it to the cooler.
The vapour compression refrigeration system is now-a-days used for all
purpose refrigeration. It is generally used for all industrial purposes from a small
domestic refrigerator to a big air conditioning plant.
2.2 Advantages and Disadvantages of vapour Compression Refrigeration
System over Air Refrigeration System
Following are the advantages and disadvantages of the vapour
compression refrigeration system over air refrigeration system:
Advantages
1. It has smaller size for the given capacity of refrigeration.
2. It has less running cost.
3. It can be employed over a large range of temperatures.
4. The coefficient of performance is quite high.
Disadvantages
1. The initial cost is high
2. The prevention of leakage of the refrigerant is the major problem in
vapour compression system
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2.3 Mechanism of a Simple Vapour Compression Refrigeration System
Fig. 2.1 shows the schematic diagram of a simple vapour compression
refrigeration system. It consists of the following five essential parts :
1. Compressor. The low pressure and temperature vapou refrigerant from
evaporator is drawn into the compressor through the inlet or suction valve A,
where it is compresses to a high pressure and temperature. This high pressure
and temperature vapour refrigerant is discharge the condenser through the
delivery or discharge valve B.
2. Condenser. The condenser or cooler consists of coils of pipe in which
the high pressure and temperature vapour refrigerant is cooled and condensed.
The refrigerant, while passing through the condenser, gives up its latent heat to
the surrounding condensing medium which is normally air or water.
3. Receiver. The condensed liquid refrigerant from the condenser is
stored in a vessel known as receiver from where it is supplied to the evaporator
through the expansion valve or refrigerant control valve.
4. Expansion valve. It is also called throttle valve or reftigerant control
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valve. The function of the expansion valve is to allow the liquid refrigerant under
high pressure and temperature to pass at a controlled rate after reducing its
pressure and temperature. Some of the liquid refrigerant evaporates as it passes
through the expansion valve, but the greater portion is vaporised in the
evaporator at the low pressure and temperature.
5. Evaporator. An evaporator consists of coils of pipe in which the liquid-
vapour refrigerant at low pressure and temperature is evaporated and changed
into vapour refrigerant at low pressure and temperature. In evaporating, the liquid
vapour refrigerant absorbs its latent heat of vaporisation from the medium (air,
water or brine) which is to be cooled.
Note : In any compression refrigeration system, there are two different
pressure conditions. One is called the high pressure side and other is known as
low pressure side. The high pressure side includes the discharge line (i.e. piping
from the evaporator to the suction valve A).
2.4 Pressure-Enthalpy (p-h) Chart
Fig 2.2 pressure enthalpy [p-h] chart
The most convenient chart for studying the behavior of a refrigerant is
the p-h chart, in which the vertical ordinates represent pressure and horizontal
ordinates represent enthalpy (i.e. total heat). A typical chart is shown in Fig. 2.2,
in which a few important lines of the complete chart are drawn. The saturated
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liquid line and the saturated vapour line merge into one another at the critical
point. A saturated liquid is one which has a temperature equal to the saturation
temperature corresponding to its pressure. The space to the left of the saturated
liquid line will, therefore, be sub-cooled liquid region. The space between the
liquid and the vapour lines is called wet vapour region and to the right of the
saturated vapour line is a superheated vapour region.
In the following pages, we shall drawn the p-h chart along with the T-s
diagram of the cycle.
2.5 Types of Vapour Compression Cycles
We have already discussed that vapour compression cycle essentially
consists of compression, condensation, throttling and evaporation. Many
scientists have focussed their attention to increase the coefficient of
performance of the cycle. Through there are many cycles, yet the following are
important from the subject point of view :
1. Cycle with dry saturated vapour after compression,
2. Cycle with wet vapour after compression,
3. Cycle with superheated vapour after compression,
4. Cycle with superheated vapour before compression, and
5. Cycle with undercooling or subcooling of refrigerant.
Now we shall discuss all the above mentioned cycles, one by one, in the
following pages.
2.6 Theoretical Vapour Compression Cycle with Dry Saturated Vapour
after Compression
A vapour compression cycle with dry saturated vapour after compression
is shown on T-s and p-h diagrams in Fig. 2.3 (a) and (b) respectively. At point 1,
let T1,p1 and s1 be the temperature, pressure and entropy of the vapour refrigerant
respectively. The four process of the cycle are as follows :
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Fig 2.3: theoretical vapour compression cycle with dry saturated vapour
after compression
1. Compression process. The vapour refrigerent at low pressure p1 and
temperature T1 is compressed isentropically to dry saturated vapour as shown by
the vertical line 1-2 on T-s diagram and by the curve 1-2 on p-h diagram. The
pressure and temperature rises from p1 to p2 and T1 to t2 respectively.
The work done during isentropic compression per kg of refrigerant is
given by
w = h2 - h1
where h1 = Enthalpy of vapour refrigerant at temperature T1, i.e. at suction
of the compressor, and
h2 = Enthalpy of the vapour refrigerant at temperature T2, i.e. at discharge
of the compressor.
2. Condensing process. The high pressure and temperature vapour
refrigerant from the compressor is passed through the condenser where it is
completely condensed at constant pressure p2 and temperature T2, as shown by
the horizontal line 2-3 on T-s and p-h diagrams. The vapour refrigerant is changed
into liquid refrigerant. The refrigerant, while passing through the condenser, gives
its latent heat to the surrounding condensing medium.
3. Expansion process. the liquid refrigerant at pressure p3 = p2 and
temperature T3 = T2 is expanded by *throttling process through the expansion
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valve to a low pressure p4 = p1 and temperature T4 = T1, as shown by the curve 3-4
on T-s diagram and by the vertical line 3-4 on p-h diagram. We have already
discussed that some of the liquid refrigerant evaporates as it passes through the
expansion valve, but the greater portion is vaporised in the evaporator. We know
that during the throttling process, no heat is absorbed or rejected by the liquid
refrigerant.
Notes : (a) In case an expansion cylinder is used in place of throttle or
expansion valve to expand the liquid refrigerant, then the refrigerant will expand
isentropically as shown by dotted vertical line on T-s diagram in Fig. 2.3 (a). The
isentropic expansion reduces the external work being expanded in running the
compressor and increases the refrigerating effect. Thus, the net result of using
the expansion cylinder is to increase the coefficient of performance.
Since the expansion cylinder system of expanding the liquid refrigerant is
quite complicated and involves greater initial cost, therefore its use is not
justified for small gain in cooling capacity. Moreover, the flow rate of the
refrigerant can be controlled with throttle valve which is not possible in case of
expansion cylinder which has a fixed cylinder volume.
(b) In modern domestic refrigerators, a capillary (small bore tube) is used
in place of an expansion valve.
4. Vaporising process. The liquid-vapour mixture of the refrigerant at
pressure p4 = p1 and temperature T4 = T1 is evaporated and changed into vapour
refrigerant at constant pressure and temperature, as shown by the horizontal line
4-1 on T-s and p-h diagrams. During evaporation, the liquid-vapour refrigerant
absorbs its latent heat of vaporisation from the medium (air, water or brine)
which is to be cooled. This heat which is absorbed by the refrigerant is called
refrigerating effect and it is briefly written as RE. The process of vaporisation
continues upto point 1 which is the starting point and thus the cycle is completed.
We know that the refrigerating effect or the heat absorbed or extracted
by the liquid-vapour refrigerant during evaporation per kg of refrigerant is given by
... ( = h4)
where = Sensible heat at temperature T3, i.e. enthalpy of liquid
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refrigerant leaving the condenser.
It may be noticed from the cycle that the liquid-vapour refrigerant has
extracted heat during evaporation and the work will be done by the compressor
for isentropic compression of the high pressure and temperature vapour
refrigerant.
Coefficient of performance,
C.O.P. = = =
Example 2.1 In an ammonia vapour compression system, the pressure in
the evaporator is 2 bar. Ammonia at exit is 0.85 dry and at entry its dryness
fraction is 0.19. During compression, the work done per kg of ammonia is 150 kJ.
Calculate the C.O.P. and the volume of vapour entering the compressor per
minute, if the rate of ammonia circulation is 4.5 kg/min. The latent heat and
specific volume at 2 bar are 1325 kJ/kg and 0.58 m3
/kg respectively.
Solution. Given : p1 = p4 = 2 bar ; x1 = 0.85 ; x4 = 0.19 ; w = 150 kJ/kg ; ma =
4.5 kg/min; hfg = 1325 kJ/kg ; vg = 0.58 m3
/kg
C.O.P.
The T-s and p-h diagrams are shown in Fig.4.3 (a) and (b) respectively.
Since the ammonia vapour at entry to the evaporator (i.e. at point 4) has
dryness fraction (x4) equal to 0.19, therefore enthalpy at point 4,
Similarly, enthaipy of ammonia vapour at exit i.e. at point 1,
Heat extracted from the evaporator or refrigerating effect,
We know that work done during compression,
w = 150 kJ/kg
C.O.P. = RE/w = 874.5/150 = 5.83 Ans.
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Volume of vapour entering the compressor per minute
We know that volume of vapour entering the compressor per minute
= Mass of refrigerant / min × Specific volume
= Ans.
Example 2.2. The temperature limits of an ammonia refrigerating system
are 25o
C and -10o
C. If the gas is dry at the end of compression, calculate the
coefficient of performance of the cycle assuming no undercooling of the liquid
ammonia. Use the following table for properties of ammonia:
Temperature
(o
C)
Liquid heat
(kJ/kg)
Latent heat
(kJ/kg)
Liquid entropy
(kJ/kg K)
25
-10
298.9
135.37
1166.94
1297.68
1.1242
0.5443
Solution. Given : ; T1 = T4 = -10
o
C = 263K ;
hf3 = h4 = 298.9 kJ/kg ; hfg2 = 1166.94 kJ/kg ; sf2 = 1.1242 kJ/kg K ; hf1 = 135.37
kJ/kg ; hfg1 = 1297.68 kJ/kg ; sf1 = 0.5443 kJ/kg K
The T-s and p-h diagrams are shown in Fig. 2. 4 (a) and (b) respectively.
Let x1 = Dryness fraction at point 1.
We know that entropy at point 1, Fig.
= 0.5443 + 4.934 x1
Similarly, entropy at point 2,
Since the entropy at point 1 is equal to entropy at point 2, therefore
equating equations (i) and (ii) 0.5443 + 4.934 x1 = 5.04 or x1 = 0.91
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We know that enthalpy at point 1,
h1 = hf1 + x1 hfg1 = 135.37 + 0.91 × 1297.68 = 1316.26 kJ/kg
and enthalpy at point 2,
h2 = hf2 + hfg2 = 298.9 + 1166.94 = 1465.84 kJ/kg
Coefficient of performance of the cycle
= Ans.
Example 2.3. A vapour compression refrigerator works between the
pressure limits of 60 bar and 25 bar. The working fluid is just dry at the end of
compression and there is no under-cooling of the liquid before the expansion
valve. Determine : 1. C.O.P. of the cycle ; and 2. Capacity of the refrigerator if the
fluid flow is at the rate of 5 kg/min.
Data :
Pressure
(bar)
Saturation
temperatur
e (K)
Enthalpy (kJ/kg) Entropy (kJ/kg K)
Liquid Vapour Liquid Vapour
60
25
295
261
151.96
56.32
293.29
322.58
0.554
0.226
1.0332
1.2464
Solution. Given : p2 = p= = 60 bar ; p1 = p4 = 25 bar ; T2 = T3 = 295 K ; T1 =
T4 = 261 K ; hf3 = h4 = 151.96 kJ/kg ; hf1 = 56.32 kJ/kg ; hg2 = h2 293.29 kJ/kg ; hg1 =
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322.58 kJ/kg ; sf2 = 0.554 kJ/kg K ; sf1 = 0.226 kJ/kg K ; sg2 = 1.0332 kJ/kg K; sg1 =
1.2464 kJ/kg K
Fig 2.5
1. C.O.P. of the cycle
The T-s and p-h diagrams are shown in Fig. 2.5 (a) and (b) respectively.
Let x1 = Dryness fraction of the vapour refrigerant entering the
compressor at point 1.
We know that entropy at point 1,
...( = sf1 + sfg1 )
= ... (i)
and entropy at point 2, s2 = 1.0332 kJ/kg K ... (Given) ... (ii)
Since the entropy at point 1 is equal to entropy at point 2, therefore
equating equations (i) and (ii),
0.226 + 1.0204 x1 = 1.0332 or x1 = 0.791
we know that enthalpy at point 1,
... ( = hf1 + hfg1)
= 56.32 + 0.791 (322.58 - 56.32) = 266.93 kJ/kg
C.O.P. of the cycle
= Ans.
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2. Capacity of the refrigerator
We know that the heat extracted or refrigerating effect produced per kg
of refrigerant
= h1 - hf3 = 266.93 - 151.93 = 114.97 kJ/kg
Since the fluid flow is at the rate of 5 kg/min, therefore total heat
extracted
= 5 × 114.97 = 574.85 kJ/min
Capacity of the refrigerator
= Ans. ...( 1 TR = 210 kJ/min)
2.7. Theoretical Vapour Compression Cycle with Wet Vapour after
Compression
A vapour compression cycle with wet vapour after compression is shown
on T-s and p-h diagrams in Fig. 2.6 (a) and (b) respectively. In this cycle, the
enthalpy at point 2 is found out with the help of dryness fraction at this point. The
dryness fraction at points 1 and 2 may be obtained by equating entropies at
points 1 and 2.
Now the coefficient of performance may be found out as usual from the
relation,
C.O.P. = =
Compound Vapour Compression Refrigeration Systems
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Introduction
In the previous chapter, we have discussed the simple vapour
compression refrigeration system in which the low pressure vapour refrigerant
from the evaporator is compressed in a single stage (or a single compressor) and
then delivered to a condenser at a high pressure. But sometimes, the vapour
refrigerant is required to be delivered at a very high pressure as in the case of low
temperature refrigerating systems. In such cases either we should compress the
vapour refrigerant by employing a single stage compressor with a very high
pressure ratio between the condenser and evaporator or compress it in two or
more compressors placed is series. The compression carried out in two or more
compressors is called compound or multistage compression.
In vapour compression refrigeration systems, the major operating cost is
the energy input to the system in the form of mechanical work. Thus any method
of increasing coefficient of performance is advantageous so long as it does not
involve too heavy an increase in other operating expenses, as well as initial plant
cost and consequent maintenance.
Since the coefficient of performance of a refrigeration system is the ratio
of refrigerating effect to the compression work, therefore the coefficient of
performance can be increased either by increasing the refrigerating effect or by
decreasing the compression work. A little consideration will show that in a
vapour compression system, the compression work is greatly reduced if the
refrigerant is compressed very close to the saturated vapour line. This can be
achieved by compressing the refrigerant in more stages with intermediate
intercooling. But it is economical only where the pressure ratio is considerable as
would be the case when very low evaporator temperatures are desired or when
high condenser temperature may be required. The compound compression is
generally economical in large plants.
The refrigerating effect can be increased by maintaining the condition of
the refrigerant in more liquid state at the entrance to the evaporator. This can be
achieved by expanding the refrigerant very close to the saturated liquid line. It
may be noted that by subcooling the refrigerant and by removing the flashed
vapour, as they are during multistage expansion, the expansion can be brought
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close to the liquid line.
2.2 Advantages of Compound (or Multi-stage) Vapour Compression with
Intercooler
Following are the main advantages of compound or multistage
compression over single stage compression:
I. The work done per kg of refrigerant is reduced in compound
compression with intercooler as compared to single stage compression for the
same delivery pressure.
2. It improves the volumetric efficiency for the given pressure ratio.
3. The sizes of the two cylinders (i.e., high pressure and low pressure)
may be adjusted to suit the volume and pressure of the refrigerant.
4. It reduces the leakage loss considerably.
5. It gives more uniform torque, and hence a smaller size flywheel is
needed.
6. It provides effective lubrication because of lower temperature range.
7. It reduces the cost of compressor.
2.3 Types of Compound Vapour Compression with Intercooler
In compound compression vapour refrigeration systems, the superheated
vapour refrigerant leaving the first stage of compression is cooled by suitable
method before being fed to the second stage of compression and so on. Such
type of cooling the refrigerant is called intercooling. Though there are many types
of compound compression with intercoolers, yet the following are important from
the subject point of view:
I. Two stage compression with liquid intercooler.
2. Two stage compression with water intercooler.
3. Two stage compression with water intercooler, liquid subcooler and
liquid flash chamber.
4. Two stage compression with water intercooler, liquid subcooler and
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flash intercooler.
5. Three stage compression with flash chambers.
6. Three stage compression with water intercoolers.
7. Three stage compression with flash intercoolers.
The above mentioned types are now discussed, in detail, one by one in
the following pages.
2.4 Two Stage Compression with Liquid Intercooler
The arrangement of a two stage compression with liquid intercooler is
shown in Fig. 2.7 (a). The corresponding p-h diagram is shown in Fig. 2.7 (b).
The various points on the p-h diagram are plotted as discussed below:
I. First of all, draw a horizontal pressure line representing the evaporator
pressure PE (or suction pressure of low pressure compressor) which intersects
the saturated vapour line at point 1. At this point, the saturated vapour is supplied
to the low pressure compressor. Let, at point I, the enthalpy of the saturated
vapour is h1 and entropy sv1
2. The saturated vapour refrigerant admitted at point 1 is compressed
isentropically in the low pressure compressor and delivers the refrigerant in a
superheated state. The pressure rises from PE to P2 The curve 1-2 represents the
isentropic compression in the low pressure compressor. In order to obtain point 2,
draw a line from point 1, with entropy equal to svl, along the constant entropy line
intersecting the intermediate pressure line P2 at point 2. Let enthalpy at this point
is h2.
3. The superheated vapour refrigerant leaving the low pressure
compressor at point 2 is cooled (or desuperheated) at constant pressure P2 = P3
in a liquid intercooler by the liquid refrigerant from the condenser. The refrigerant
leaving the liquid intercooler is in saturated vapour state. The line 2-3 represents
the cooling or desuperheating process. Let the enthalpy and entropy at point 3 is
h3 and sv3 respectively.
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Fig. 2.7
4. The dry saturated vapour refrigerant is now supplied to high pressure
compressor where it is compressed isentropically from intermediate or
interocooler pressure P2 to condensor pressure pc. The curve 3-4 represents the
isentropic compression in the high pressure compressor. The point 4 on the p-h
diagram is obtained by drawing a line of entropy equal to sv3 along the constant
entropy line as shown in Fig. 2.1 (b). Let the enthalpy of superheated vapour
refrigerant at point 4 is h4.
5. The superheated vapour refrigerant leaving the high pressure
compressor at point 4 is now passed through the condenser at constant pressure
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Pc as shown by a horizontal line 4-5. The condensing process 4-5 changes the
state of refrigerant from superheated vapour to saturated liquid.
6. The high pressure saturated liquid refrigerant from the condensor is
passed to the intercooler where some of liquid refrigerant evaporates in
desuperheating the superheated vapour refrigerant from the low pressure
compressor. In order to make up for the liquid evaporated, i.e. to maintain a
constant liquid level, an expansion valve E1 which acts as a float valve, is provided.
7. The liquid refrigerant from the intercooler is first expanded in an
expansion valve E2 and then evaporated in the evaporator to saturated vapour
condition, as shown in Fig. 2.1(b).
Let
ml = Mass of refrigerant passing through the evaporator (or low pressure
compressor) in kg/min, and
m2 = Mass of refrigerant passing through the condenser (or high
pressure compressor) in kg/min.
The high pressure compressor in a given system will compress the mass
of refrigerant from low pressure compressor (ml) and the mass of liquid
evaporated in the liquid intercooler during cooling or de superheating of
superheated vapour refrigerant from low pressure compressor. If m3 is the mass
of liquid evaporated in the intercooler, then
m3= m2 - ml
The value of m2 may be obtained by considering the thermal equilibrium
for the liquid intercooler as shown in Fig. 2.2, i.e .,
Heat taken by the liquid intercooler = Heat given by the liquid intercooler
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Fig2.8. Thermal equilibrium for liquid intercooler.
or m2 hf5 + m1 h2 = m1 h6 + m2 h3
... ( h6 = hf5)
and mass of liquid refrigerant evaporated in the intercooler,
We know that refrigerating effect,
RE = ml (hl - hf) = ml (hl - hf5) = 210 Q kJ/min
where Q is the load on the evaporator in tonne of refrigeration.
Total work done in both the compressors,
W = ml (h2 - h1) + m2 (h4 - h3)
Power required to drive the system,
and C.O.P. of the system
= =
Notes: 1. In case of ammonia, when liquid refrigerant is used for
intercooling, the total power requirement will decrease. It is due to the fact that
the mass of liquid evaporated during intercooling is extremely small because of
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its high latent heat of vaporisation and the constant entropy lines of ammonia
become very flat in the superheat region. Thus the intercooling by liquid
refrigerant is commonly used in multi-stage ammonia plants, because of less
power requirement.
2. In case of refrigerant R-12, when liquid refrigerant is used for
intercooling, the total power requirements may actually increase. It is due to the
fact that the latent heat of vaporisation is small and the constant entropy line of R
-12 does not change very much with the temperature. Thus in R-12 systems, the
saving in work by performing the compression close to the saturated vapour line
does not compensate for the increased mass flow rate through the high stage
compressor. Therefore, intercooling by liquid refrigerant in R-12 systems is never
employed.
Example 2.1. Calculate the power needed to compress 20 kg / min of
ammonia from saturated vapour at 1.4 bar to a condensing pressure of 10 bar by
two-stage compression with intercooling by liquid refrigerant at 4 bar. Assume
saturated liquid to leave the condenser and dry saturted vapours to leave the
evaporator. Use the p-h chart.
Determine, also, the power needed when intercooling is not employed.
Solution. Given : m1 = 20 kg/min ; PE = 1.4 bar ; pC = 10 bar ; p2 = p3 = 4 bar
Fig. 2.9
The p-h diagram for a two stage compression with intercooling by liquid
refrigerant is shown in Fig. 2.9. The various values for ammonia as read from the
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p-h diagram are as follows:
Enthalpy of saturated vapour refrigerant entering the low pressure
compressor at point 1, h1 = 1400 kJ/kg
Entropy of saturated vapour refrigerant entering the low pressure
compressor at point 1, Sl = 5.75 kJ/kg K
Enthalpy of superheated vapour refrigerant leaving the low pressure
compressor at point 2, h2 = 1527 kJ/kg
Enthalpy of saturated vapour refrigerant leaving the intercooler or
entering the high pressure compressor at point 3,
h3 = 1428 kJ/kg
Entropy of saturated vapour refrigerant leaving the intercooler or entering
the high pressure compressor at point 3,
s3 = 5.39 kJ/kg K
Enthalpy of superheated vapour refrigerant leaving the high pressure
compressor at point 4, h4 = 1550 kJ/kg
Enthalpy of saturated liquid refrigerant passing through the condenser at
point 5, hf5 = h6 = 284 kJ/kg
We know that mass of refrigerant passing through the condenser (or high
pressure compressor),
m2 = = = 21.73 kg/min
Work done in low pressure compressor,
WL = m1 (h2 - h1) = 20 (1527 - 1400) = 2540 kJ/min
Work done in high pressure compressor,
WH = m2 (h4 - h3) = 21.73 (1550 - 1428) = 2651 kJ/min
and total work done in both the compressors,
W = WL + WH = 2540 + 2651 = 5191 kJ/min
Power needed = 5191/60 = 86.5 kW Ans.
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Power needed when intercooling is not employed
When intercooling is not employed, the compression of refrigerant will
follow the path 1-2 in the low pressure compressor and 2-2' in the high pressure
compressor. In such a case,
Work done in the high pressure compressor,
WH = mi (h2'- h2) = 20 (1676- 1527) = 2980 kJ /min
... ( From p-h diagram, h2' = 1676 kJ/kg)
and total work done is both the compressors,
W = WL + WH = 2540 + 2980 = 5520 kJ/min
Power needed = 5520/60 = 92 kW Ans.
Example 2.2. Calculate the power needed to compress 20 kg/min of R-12
from saturated vapour at 1.4 bar to a condensing pressure of 10 bar by two-stage
compression with intercooling by liquid refrigerant at 4 bar. Assume saturated
liquid to leave the condenser and dry saturated vapours to leave the evaporator.
Use the p-h chart. Sketch the cycle on a skeleton p-h chart and label the
values of enthalpy at salient points.
Solution. Given: ml = 20 kg/min; PE = 1.4 bar; Pc = 10 bar; P2 = P3 = 4 bar
The p-h diagram for a two-stage compression with intercooling by liquid
refrigerant is shown in Fig. 5.4. The various values for R-12 as read from the p-h
diagram are as follows :..
Enthalpy of saturated vapour refrigerant entering the low pressure
compressor at point 1, h1= 178 kJ/kg
Entropy of saturated vapour refrigerant entering the low pressure
compressor at point 1, sl = 0.71 kJ/kg K
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Fig. 2.10
Enthalpy of superheated vapour refrigerant leaving the low pressure
compressor at point 2, h2 = 195 kJ/kg
Enthalpy of saturated vapour refrigerant leaving the intercooler or
entering the high pressure compressor at point 3,
h3=191 kJ/kg
Entropy of saturated vapour refrigerant entering the high pressure
compressor at point 3, s3 = 0.695 kJ/kg K
Enthalpy of superheated vapour refrigerant leaving the high pressure
compressor at point 4, h4 = 210 kJ/kg K
Enthalpy of saturated liquid refrigerant leaving the condenser at point 5,
hf5 = h6 = 77 kJ/kg
We know that mass of refrigerant passing through the condenser (or high
pressure compressor),
m2 = = = 20.7 kg/min
Work done in low pressure compressor,
WL = ml (h2 - h1) = 20 (195 - 178) = 340 kJ/min
Work done in high pressure compressor,
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WH= m2 (h4- h3) = 20.7 (210-191) = 393 kJ/min
We know that refrigerating effect,
RE = m (h1- hf6) = 210 Q kJ/min
C.O.P. of the system
= = =
Example 2.3. The following data refer to a two stage compression
ammonia refrigerating system with water intercooler.
Condenser pressure = 14 bar; Evaporator pressure = 2 bar; Intercooler
pressure = 5 bar; Load on the evaporator = 2 TR
If the temperature of the de-superheated vapour and sub-cooled liquid
refrigerant are limited to 30o
C, find (a) the power required to drive the system, and
(b) CO.P. of the system.
Solution. Given : Pc = 14 bar ; PE = 2 bar ; P2 = P3 = 5 bar ; Q = 10 TR ; t3 = t6
= 30° C
The p-h diagram for a two stage compression system with water
intercooler is shown in Fig. 2.11. The various values as read from the p-h diagram
for ammonia are as follows:
Enthalpy of saturated vapour refrigerant entering the low pressure
compressor at point 1. h1 = 1420 kJ/kg
Entropy of saturated vapour refrigerant at point 1, sl = 5.6244 kJ/kg K
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Fig. 2.11
Enthalpy of superheated vapour refrigerant leaving the water intercooler
at point 3. h3= 1510 kJ/kg
Entropy of superheated vapour refrigerant at point 3,
s3 = 5.424 kJ/kg K
Enthalpy of superheated vapour refrigerant leaving the high pressure
compressor at point 4.
h4 = 1672 kJ/kg
Enthalpy of liquid refrigerant leaving the liquid sub-cooler, hf6 = h7 = 323
kJlkg
The points 2 and 4 on the p-h diagram are obtained in the similar way as
discussed in Art. 5.3.
From the p-h diagram, we find that enthalpy of superheated vapour
refrigerant at point 2, h2 = 1550 kJ/kg.
(a) Power required to drive the system
We know that mass of refrigerant circulating through the system,
m = = = 1.91kg/min
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Total work done in both the compressors,
w= m [(h2 - h1) + (h4 - h3)]
= 1.91 [(1550- 1420) + (1672 - 1510) = 557.7 kJ/min
Power required to drive the system,
P = 557.7/60 = 9.3 kW Ans.
(b) CO.P. of system
We know that refrigerating effect of the system,
RE = 210Q = 2l0 × 10 = 2100 kJ/min
:. C.O.P. of the system
= = = 3.76 Ans.
2.6 Two Stage Compression with Water Intercooler, Liquid Sub-cooler and
Liquid Flash Chamber
The arrangement of a two stage compression with water intercooler,
liquid sub-cooler and liquid flash chamber is shown in Fig. 2.12 (a). The
corresponding p-h diagram is shown in Fig. 2.7 (b). The various processes, in this
system, are as follows:
I. The saturated vapour refrigerant at the evaporator pressure PE is
admitted to low pressure compressor at point I. In this compressor, the
refrigerant is compressed isentropically from evaporator pressure PE to water
intercooler (or flash chamber)
pressure PF as shown by the curve 1-2 in Fig. 2.12 (b).
2. The superheated vapour refrigerant leaving the low pressure
compressor at point 2 is now passed through the water intercooler at constant
pressure PF in order to reduce the degree of superheat (i.e., from temperature t2
to t3 ). The line 2-3 represents the water intercooling or de-superheating process.
3. The superheated vapour refrigerant leaving the water intercooler at
point 3 is mixed with the vapour refrigerant supplied by the flash chamber at
point 9. The condition of refrigerant after mixing is shown by point 4 which is in
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superheated state. Let the temperature at this point is t4.
4. The superheated vapour refrigerant admitted at point 4 to the high
pressure compressor is compressed isentropically from the intercooler or flash
chamber pressure PF to condenser pressure Pc as shown by the curve 4-5. The
temperature rises from t4 to t5.
5. The superheafed vapour leaving the high pressure compressor at
pressure Pc is passed through a condenser at constant pressure as shown by a
horizontal line 5-6. The condensing process 5-6 changes the state of refrigerant
from superheated vapour to saturated liquid.
Fig 2.12
6. The saturated liquid refrigerant from the condenser is now cooled in
liquid sub-cooler to a temperature, say t7. The line 6-7 represents a sub-cooling
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process.
7. The liquid refrigerant leaving the sub-cooler at pressure Pc is expanded
in an expansion valve E1 to a pressure equal to the flash chamber pressure PF, as
shown by vertical line 7-8. The expanded refrigerant which is a mixture of vapour
and liquid refrigerants is admitted to a flash chamber at point 8. The flash
chamber separates the vapour and liquid refrigerants at pressure PF. The vapour
refrigerant from the flash chamber at point 9 is mixed with the refrigerant from
the water intercooler. The liquid refrigerant from the flash chamber at Point 10 is
further expanded in an expansion valve E2 as shown by the vertical line 10-11.
8. The liquid refrigerant leaving the expansion valve E2 is evaporated in
the evaporator at the evaporator pressure PE (usually 2 bar) as shown by the
horizontal line 11-1 in Fig. 2.12 (b).
Let m2 = Mass of refrigerant passing through the condenser (or high
pressure compressor), and
m3 = Mass of vapour refrigerant formed in the flash chamber.
:. Mass of refrigerant passing through the evaporator (or low pressure
compressor),
ml = m2-m3
If Q tonne of refrigeration is the load on the evaporator, then the mass of
refrigerant passing through the evaporator,
m1= kg/min ...( h11 = hf10)
Now let us consider the thermal equilibrium of the flash chamber. Since
the flash chamber is an insulated vessel, therefore there is no heat exchange
between the flash chamber and atmosphere. In other words, the heat taken and
given by the flash chamber are same. Mathematically,
Heat taken by the flash chamber
= Heat given by the flash chamber
m2 h8 = m3 h9 + ml hf10
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= m3 h9 + (m2- m3) hf10 ...( m1 = m2 - m2)
m2 (h8 - hf10) = m3 (h9 - hf10)
m3 = m2 ...( h8 = hf7) ... (i)
The vapour refrigerant from the water intercooler (represented by point 3)
is mixed with vapour refrigerant m3 from the flash chamber (represented by point
9) at the same pressure before entering the high pressure compressor. The
enthalpy of the mixed refrigerant (represented by point 4) may be calculated by
using the equation,
m2 h4 = m3 h9 + ml h3
= m3 h9 + (m2 - m3) h3
We know that refrigerating effect of the system,
RE = ml (h1 - hll) = 210 Q kJ/min
Work done in low pressure compressor,
WL = ml (h2 - hl)
Work done in high pressure compressor,
WH = m2 (h5 - h4)
Total workdone in both the compressors,
W = WL + WH = m1 ( (h2 - hl) + m2 (h5 - h4)
Power required to drive the system,
p = kW
and C.O.P. of the system
= = =
Note: Since the mass of vapour refrigerant m, is cooled in the water
intercooler from condition 2 to 3, therefore cooling capacity of the intercooler
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= ml (h2 -h3)
Example 2.4. A two stage compression ammonia refrigeration system
operates between overall pressure limits of 14 bar and 2 bar. the temperature of
the desuperheated vapour and subcooled liquid refrigerant are limited to 30o
C.
The flash tank separates dry vapour at 5 bar pressure and the liquid refrigerant
then expands to 2 bar.
Estimate the C.O.P. of the machine and power required to drive the
compresor, if the mechanical effficiency of the drive is 80% and load on the
evaporator is 10 TR.
Solution. Given :Pc = 14 bar; PE = 2 bar; PF = 5 bar; t3 = t7 = 30o
C ; m =
80% = 0.8 ; Q = 10TR
The p-h diagram for a two-stage compression system with given
conditions is shown in Fig. 5.8. The values as read from p-h diagram for ammonia,
are as follows:
Fig 2.13
Enthalpy of saturated vapour refrigerant entering the low pressure
compressor at point 1. hi = 1420 kJ/kg
Entropy of saturated vapour refrigerant entering the low pressure
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compressor at point 1. s1 = 5.6244 kJ/kg K
Enthalpy of superheated vapour refrigerant leaving the low pressure
compressor at point 2, h2 =1550 kJ/kg
Enthalpy of superheated vapour refrigerant leaving the water intercooler
at point 3. h3 = 1510 kJ/kg
Enthalpy of saturated vapour refrigerant leaving the flash tank at point 9,
h9 = 1432 kJ/kg
Enthalpy of liquid refrigerant leaving the subcooler at point 7,
hf7 = h8 = 323 kJ/kg
Enthalpy of saturated liquid refrigerant leaving the second expansion
valve at point 10, hf10 = hll = 198 kJ/kg
Let m2 = Mass of refrigerant passing through the condenser.
We know that mass of the vapour refrigerant formed in the flash tank,
m3 = m2 = m2 = 0.l m2 ... (i)
and mass of refrigerant passing through the evaporator,
ml = m2 - m3 = = = 1.72 kg/min ... (ii)
From equations (i) and (ii),
m2 - 0.1 m2 = 1.72 or m2 = 1.9 kg/min
m3 = 0.l m2 = 0.l × 1.9 = 0.19 kg/min
The desuperheated vapour refrigerant (m2 - m3) as represented by point 3
is mixed with the vapour refrigerant from the flash tank as represented by point 9.
The enthalpy of the mixed refrigerant entering the high pressure compressor as
represented by point 4 is given by
m2 h4 = m3 h9 + (m2 - m3) h3
= 0.1 m2 h9 + (m2 - 0.1 m2) h3 ... [From equation (i)]
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or h4 = 0.1 × h9 + 0.9 × h3 = 0.1 × 1432+ 0.9 × 1510 = 1502 kJ/kg
We see from p-h diagram that at point 4 (intersection of pressure 5 bar
and enthalpy 1502 kJ/kg), the entropy is s4 = 5.51 kJ/kg K. Now from point 4,
draw a line of entropy equal to 5.51 kJ/kg K along the constant entropy line which
intersects the condenser pressure (14 bar) line at point 5. Thus, the point 5 is
located. From p-h diagram, we find that enthalpy of refrigerant leaving the high
pressure compressor at point 5 is
h5= 1650 kJ/kg
C.O.P. of the machine
We know that refrigerating effect
= ml (hl - hf10) = 210 Q = 210 × 10 = 2100 kJ/min
Work done in both the compressors
= ml (h2 - h1) + m2 (h5 - h4)
= 1.72 (1550 - 1420) + 1.9 (1650- 1502)
= 223.6 + 281.2 = 504.8 kJ/min
Since the mechanical efficiency of the drive is 80%, therefore actual work
done in both the compressors
= 504.8/0.8 = 631 kJ/min
Actual C.O.P = = = 3.32 Ans.
Power required to drive the compressors
We know that power required to drive the compressors
= = = 10.5 kW Ans.
2.7 Two Stage Compression with Water Intercooler, Liquid Sub-cooler and
Flash Intercooler
A two stage compression with water intercooler, liquid sub-cooler and
flash intercooler is shown in Fig. 2.9 (a). The corresponding p-h diagram is shown
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in Fig. 5.9 (b).
Fig 2.14
We have seen in the previous article that when the vapour refrigerant
from the low pressure compressor is passed through the water intercooler, its
temperature does not reduce to the saturated vapour line or even very near to it,
before admitting it to the high pressure compressor [ Refer point 4 of Fig. 2.12
(b)]. In fact, with water cooling there may be no saving of work in compression.
But the improvement in performance and the reduction in compression work may
be achieved by using a flash chamber as an intercooler as well as flash separator,
as shown in Fig. 2.14 (a). The corresponding p-h diagram is shown in Fig. 2.14 (b).
The various processes, in this system, are as follows:
1. The saturated vapour refrigerant at the evaporator pressure PE is
admitted to the low pressure compressor at point 1. In this compressor, the
refrigerant is compressed isentropically from evaporator pressure PE to the flash
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intercooler pressure PF,as shown by the curve 1-2 in Fig2.14 (b).
2. The superheated vapour refrigerant leaving the low pressure
compressor at point 2 is now passed through the water intercooler at constant
pressure PF, in order to reduce the degree of superheat ( i.e. from temperature t2
to t3). The line 2-3 represents the water intercooling or desuperheating process.
3. The superheated vapour refrigerant leaving the water intercooler at
point 3 is passed through a flash intercooler which cools the superheated vapour
refrigerant to saturated vapour refrigerant as shown by the line 3-4. The cooling
of superheated vapour refrigerant is done by the evaporation of a part of the
liquid refrigerant from the flash intercooler placed at point 8.
4. The saturated vapour refrigerant leaving the flash intercooler enters
the high pressure compressor at point 4 where it is compressed isentropically
from flash intercooler pressure PF to condenser pressure PC, as shown by the
curve 4-5.
5. The superheated vapour refrigerant leaving the high pressure
compressor at pressure Pc is passed through a condenser at constant pressure.
The condensing process as shown by line 5-6 changes the state of refrigerant
from superheated vapour to saturated liquid.
6. The saturated liquid refrigerant leaving the condenser at point 6 is now
cooled at constant pressure Pc in the liquid sub-cooler to a temperature t7 as
shown in Fig. 2.9 (b). The line 6-7 shows the sub-cooling process.
7. The liquid refrigerant leaving the sub-cooler at point 7 is expanded in
an expansion valve E1 to a pressure equal to the flash intercooler pressure PF , as
shown by the vertical line 7-8. The expanded refrigerant (which is a mixture of
vapour and liquid refrigerant) is admitted to flash intercooler at point 8 which also
acts as a flash separator.
8. The liquid refrigerant leaving the flash intercooler at point 9 is passed
through the second expansion valve E2 (process 9-10) and then evaporated in the
evaporator as shown by the horizontal line 10-1.
Let m1 = Mass of the refrigerant passing through the evaporator or low
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pressure compressor), and
m2 = Mass of the refrigerant passing through the condenser (or high
pressure compressor).
If Q tonne of refrigeration is the load on the evaporator, then the mass of
refrigerant passing through the evaporator is given by,
m1 = = kg/min ... ( h10 = hf9)
Now for the thermal equilibrium of the flash intercooler, Heat taken by the
flash intercooler
= Heat given by the flash intercooler
m2 h8 + m1 h3 = m2 h4 + m1 hf9
m1(h3 - hf9) = m2 (h4 -h8)
m2 = m1 = m1 kg/min
We know that refrigerating effect,
RE = ml (hl - h10) = ml (hl - hf9) = 210 Q kJ/min
and work done in both the compressors,
W = Work done in L.P. compressor + Work done in H.P. compressor
= ml (h2 - hl) + m2 (h5- h4)
Power required to drive the system,
kW
and coefficient of performance of the system,
C.O.P. = = =
MODULE – III
Vapour Absorption Refrigeration Systems
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3.1 Introduction
The vapour absorption refrigeration system is one of the oldest method
of producing refrigerating effect. The principle of vapour absorption was first
discovered by Michael Faraday in 1824 while performing a set of experiments to
liquify certain gases. The first vapour absorption refrigeration machine was
developed by a French scientist Ferdinand Carre in 1860. This system may be
used in both the domestic and large industrial refrigerating plants. The refrigerant,
commonly used in a vapour absorption system, is ammonia.
The vapour absorption system uses heat energy, instead of mechanical
energy as in vapour compression systems, in order to change the conditions of
the refrigerant required for the operation of the refrigeration cycle. We have
discussed in the previous chapters that the function of a compressor, in a vapour
compression system, is to withdraw the vapour refrigerant from the evaporator. It
then raises its temperature and pressure higher than the cooling agent in the
condenser so that the higher pressure vapours can reject heat in the condenser.
The liquid refrigerant leaving the condenser is now ready to expand to the
evaporator conditions again.
In the vapour absorption system, the compressor is replaced by an
absorber, a pump, a generator and a pressure reducing valve. These components
in vapour absorption system perform the same function as that of a compressor
in vapour compression system. In this system, the vapour refrigerant from the
evaporator is drawn into an absorber where it is absorbed by the weak solution of
the refrigerant forming a strong solution. This strong solution is pumped to the
generator where it is heated by some external source. During the heating process,
the vapour refrigerant is driven off by the solution and enters into the condenser
where it is liquefied. The liquid refrigerant then flows into the evaporator and thus
the cycle is completed.
3.2 Simple Vapour Absorption System -
The simple vapour absorption system, as shown in Fig. 3.1, consists of
an absorber, a pump, a generator and a pressure reducing valve to replace the
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compressor of vapour compression system. The other components of the
system are condenser, receiver, expansion valve and evaporator as in the vapour
compression system.
Fig 3.1 Simple vapour absorbtion system
. In this system, the low pressure ammonia vapour leaving the evaporator
enters the absorber where it is absorbed by the cold water in the absorber. The
water has the ability to absorb very large quantities of ammonia vapour and the
solution thus formed, is known as aqua-ammonia. The absorption of ammonia
vapour in water lowers the pressure in the absorber which in turn draws more
ammonia vapour from the evaporator and thus raises the temperature of solution.
Some form of cooling arrangement (usually water cooling) is employed in the
absorber to remove the heat of solution evolved there. This is necessary in order
to increase the absorption capacity of water, because at higher temperature
water absorbs less ammonia vapour. The strong solution thus formed in the
absorber is pumped to the generator by the liquid pump. The pump increases the
pressure of the solution upto 10 bar.
The *strong solution of ammonia in the generator is heated by some
external source such as gas or steam. During the heating process, the ammonia
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vapour is driven off the solution at high pressure leaving behind the hot weak
ammonia solution in the generator. This weak ammonia solution flows back to
the absorber at low pressure after passing through the pressure reducing valve.
The high pressure ammonia vapour from the generator is condensed in the
condenser to a high pressure liquid ammonia. This liquid ammonia is passed to
the expansion valve through the receiver and then to the evaporator. This
completes the simple vapour absorption cycle.
3.3 actical Vapour Absorption System
The simple absorption system as discussed in the previous article is not
very economical. In order to make the system more practical, it is fitted with an
analyser, a rectifier and two heat exchangers as shown in Fig. 3.2. These
accessories help to improve the performance and working of the plant, as
discussed below :-
Fig. 3.2. Practical vapour absorption system.
1. Analyser. When ammonia is vaporised in the generator, some water is
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also vaporised and will flow into the condenser along with the ammonia vapours
in the simple system. If these unwanted water particles are not removed before
entering into the condenser, they will enter into the expansion valve where they
freeze and choke the pipe line. In order to remove these unwanted particles
flowing to the condenser, an analyser is used. The analyser may be built as an
integral part of the generator or made as a separate piece of equipment. It
consists of a series of trays mounted above the generator. The strong solution
from the absorber and the aqua from the rectifier are introduced at the top of the
analyser and flow downward over the trays and into the generator. In this way,
considerable liquid surface area is exposed to the vapour rising from the
generator. The vapour is cooled and most of the water vapour condenses, so that
mainly ammonia vapour leaves the top of the analyser. Since the aqua is heated
by the vapour, less external heat is required in the generator.
2. Rectifier. In case the water vapours are not completely removed in the
analyser, a closed type vapour cooler called rectifier (also known as dehydrator)
is used. It is generally water cooled and may be of the double pipe, shell and coil
or shell and tube type. Its function is to cool further the ammonia vapours leaving
the analyser so that the remaining water vapours are condensed. Thus, only dry
or anhydrous ammonia vapours flow to the condenser. The condensate from the
rectifier is returned to the top of the analyser by a drip return pipe.
3. Heat exchangers. The heat exchanger provided between the pump and
the generator is used to cool the weak hot solution returning from the generator
to the absorber. The heat removed from the weak solution raises the temperature
of the strong solution leaving the pump and going to analyser and generator. This
operation reduces the heat supplied to the generator and the amount of cooling
required for the absorber. Thus the economy of the plant increases.
The heat exchanger provided between the condenser and the evaporator
may also be called liquid sub-cooler. In this heat exchanger, the liquid refrigerant
leaving the condenser is sub- cooled by the low temperature ammonia vapour
from the evaporator as shown in Fig. 7.2. This sub-cooled liquid is now passed to
the expansion valve and then to the evaporator.
In this system, the net refrigerating effect is the heat absorbed by the
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 63
refrigerant in the evaporator. The total energy supplied to the system is the sum
of work done by the pump and the heat supplied in the generator. Therefore, the
coefficient of performance of the system is given by
C.O.P. =
3.4 Advantages of Vapour Absorption Refrigeration System over Vapour
Compression Refrigeration System
Following are the advantages of vapour absorption system over vapour
compression system:
1. In the vapour absorption system, the only moving part of the entire
system is a pump which has a small motor. Thus, the operation of this system is
essentially quiet and is subjected to little wear.
The vapour compression system of the same capacity has more wear,
tear and noise due to moving parts of the compressor.
2. The vapour absorption system uses heat energy to change the
condition of the refrigerant from the evaporator. The vapour compression system
uses mechanical energy to change the condition of the refrigerant from the
evaporator.
3. The vapour absorption systems are usually designed to use steam,
either at high pressure or low pressure. The exhaust steam from furnaces and
solar energy may also be used. Thus this system can be used where the electric
power is difficult to obtain or is very expensive.
4. The vapour absorption systems can operate at reduced evaporator
pressure and temperature by increasing the steam pressure to the generator,
with little decrease in capacity. But the capacity of vapour compression system
drops rapidly with lowered evaporator pressure.
5. The load variations does not effect the performance of a vapour
absorption system. The load variations are met by controlling the quantity of
aqua circulated and the quantity of steam supplied to the generator.
The performance of a vapour compression system at partial loads is
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 64
poor.
6. In the vapour absorption system, the liquid refrigerant leaving the
evaporator has no bad effect on the system except that of reducing the
refrigerating effect. In the vapour compression system, it is essential to
superheat the vapour refrigerant leaving the evaporator so that no liquid may
enter the compressor.
7. The vapour absorption systems can be built in capacities well above
1000 tonnes of refrigeration each which is the largest size for single compressor
units.
8. The space requirements and automatic control requirements favour
the absorption system more and more as the desired evaporator temperature
drops.
3.5 Coefficient of Performance of an Ideal Vapour Absorption
Refrigeration System
We have discussed earlier that in an ideal vapour absorption refrigeration
system,
(a) the heat (QG) is given to the refrigerant in the generator,
(b) the heat (Qc) is discharged to the atmosphere or cooling water from
the condenser and absorber.
(c) the heat (QE) is absorbed by the refrigerant in the evaporator, and
(d) the heat (Qp) is added to the refrigerant due to pump work.
Neglecting the heat due to pump work (Qp), we have according to First
Law of Thermodynamics,
... (i)
Let TG = Temperature at which heat (QG) is given to the generator,
Tc = Temperature at which heat (QC) is discharged to atmosphere or
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 65
cooling water from the condenser and absorber, and
TE = Temperature at which heat (QE) is absorbed in the evaporator.
Since the vapour absorption system can be considered as a perfectly
reversible system, therefore the initial entropy of the system must be equal to the
entropy of the system after the change in its condition.
...(ii)
= ... [ From equation (i) ]
or
QG =
=
= ... (iii)
Maximum coefficient of performance of the system is given by
(C.O.P.)max = =
= ...(iv)
It may noted that,
I. The expression is the C.O.P. of a Carnot refrigerator working
between the temperature limits of T E and T C.
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 66
2. The expression is the efficiency of a Carnot engine working
between the temperature limits of TG and TC.
Thus an ideal vapour absorption refrigeration system may be regarded as
a combination of a Carnot engine and a Carnot refrigerator. The maximum C.O.P.
may be written as
(C.O.P.)max = (C.O.P)carnot × carnot
In case the heat is discharged at different temperatures in condenser and
absorber, then
(C.O.P.)max =
where TA= Temperature at which heat is discharged in the absorber.
Example 3.1. In a vapour absorption refrigeration system, heating,
cooling and refrigeration takes place at the temperatures of 100o
C, 20o
C and _5°
C respectively. Find the maximum C.O.P. of the system.
Solution. Given: TG = 100o
C = 100 + 273 = 373 K ; Tc = 20o
C = 20 + 273 =
293 K ; T E = - 5° C = - 5 + 273 = 268 K
We know that maximum C.O.P. of the system
= = 2.3 Ans.
Example 3.2. In an absorption type refrigerator, the heat is supplied to
NH3 generator by condensing steam at 2 bar and 90% dry. The temperature in the
refrigerator is to be maintained at - 5° C. Find the maximum C.O.P. possible.
If the refrigeration load is 20 tonnes and actual C.O.P. is 10% of the
maximum C.O.P ., find the mass of steam required per hour. Take temperature of
the atmosphere as 30o
C.
Solution. Given: p = 2 bar; x = 90% = 0.9; TE = -5°C = -5 + 273 = 268 K ; Q =
20 TR; Actual C.O.P. = 70% of maximum C.O.P. ; TC = 30° C = 30 + 273 = 303 K
Maximum C.O.P.
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 67
From steam tables, we find that the saturation temperature of steam at a
pressure of 2 bar is
TG = 120.2° C = 120.2 + 273 = 393.2 K
We know that maximum C.O.P.
= = = 1.756 Ans.
Mass of steam required per hour
We know that actual C.O.P.
= 70% of maximum C.O.P. = 0.7 × 1.756 = 1.229
Actual heat supplied
= = = 3417.4 kJ/min
Assuming that only latent heat of steam is used for heating purposes,
therefore from steam tables, the latent heat of steam at 2 bar is
hfg = 2201.6 kJ/kg
Mass of steam required per hour
= = = 1.552 kg/min= 93.12 kg/h Ans.
3.6 Domestic Electrolux (Ammonia Hydrogen) Refrigerator
The domestic absorption type refrigerator was invented by two Swedish
engineers Carl Munters and Baltzer Von Platan in 1925 while they were studying
for their under-graduate course of Royal Institute of Technology in Stockholm.
The idea was first developed by the ‘Electrolux Company’ of Luton, England.
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 68
Fig. 3.3. Domestic electrolux type refrigerator.
This type of refrigerator is also called three-fluids absorption system. The
main purpose of this system is to eliminate the pump so that in the absence of
moving parts. the machine becomes noise-less. The three fluids used in this
sytem are ammonia, hydrogen and water. The ammonia is used as a refrigerant
because it possesses most of the desirable properties. It is toxic. but due to
absence of moving parts, there is very little changes for the leakage and the total
amount of refrigeration used is small. The hydrogen being the lightest gas. is
used to increase the rate of evaporation of the liquid ammonia passing through
the evaporator. The hydrogen is also non-corrosive and insoluble in water. This is
used in the low-pressure side of the system. The water is used as a solvent
because it has the ability to absorb ammonia readily. The principle of operation of
a domestic electolux type refrigerator. as shown in Fig. 3.3. is discussed below:
The strong ammonia solution from the absorber through heat exchanger
is heated in the generator by applying heat from an external source usually a gas
burner. During this heating process, ammonia vapours are removed from the
solution and passed to the condenser. A rectifirer or a water separator fitted
before the condenser removes water vapour carried with the ammonia vapours.
so that dry ammonia vapours are supplied to the condenser. These water
vapours,.
f not removed, they will enter into the evaporator causing freezing and
choking of the machine . The hot weak solution left behind in the generator flow
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 69
to the absorber through the heat excl anger. This hot weak solution while passing
through the exchanger is cooled. The heat removed by the weak solution is
utilised in raising the temperature of strong solution passing through the heat
exchanger. In this way, the absorption is accelerated and the improvement in the
performance of a plant is achieved.
The ammonia vapours in the condenser are condensed by using external
cooling source. The liquid refrigerant leaving the condenser flows under gravity to
the evaporator where it meets the hydrogen gas. The hydrogen gas which is
being fed to the evaporator permit the liquid ammonia to evaporate at a low
pressure and temperature according to Dalton's principle. During the process of
evaporation, the ammonia absorbs latent heat from the refrigerated space and
thus produces cooling effect.
The mixture of ammonia vapour and hydrogen is passed to the absorber
where ammonia is absorbed in water while the hydrogen rises to the top and
flows hack to the evaporator. This completes the cycle. The coefficient of
performance of this refrigerator is given by :
C.O.P. =
Notes: 1. The hydrogen gas only circulates from the absorber to the
evaporator and back.
2. The whole cycle is carried out entirely by gravity flow of the refrigerant.
3. It can not be used for industrial purposes as the C.O.P. of the system
is very low.
3.7 Lithium Bromide Absorption Refrigeration System
The lithium-bromide absorption refrigeration system uses a solution of
lithium bromide in water. In this system, the *water is being used as a refrigerant
whereas lithium bromide, which is a highly hydroscopic salt, as an absorbent. The
lithium bromide solution has a strong affinity for water vapour because of its very
low vapour pressure. Since lithium bromide solution is corrosive, therefore
inhibitors should be added in order to protect the metal parts of the system
against corrosion. Lithium chromate is often uged as a corrosion inhibitor. This
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 70
system is very popular for air conditioning in which low refrigeration
temperatures (not below 0° C)** are required.
Fig. 3.4 shows a lithium bromide vapour absorption system. In this
system. the absorber and the evaporator are placed in one shell which operates
at the same low pressure of the system. The generator and condenser are placed
in another shell which operates at the same high pressure of the system. The
principle of operation of this system is discussed below :
The water for air-conditioning coils or process requirements is chilled as
it is pumped through the chilled-water tubes in.the evaporator by giving up heat to
the refrigerant water sprayed over the tubes. Since the pressure inside the
evaporator is maintained very low, therefore, the refrigerant waterevaporates. The
water vapours thus formed will be absorbed by the strong lithium-bromide
solution which is sprayed in the absorber. In absorbing the water vapour, the
lithium bromide solution helps in maintaining very low pressure (high vacuum)
needed in the evaporator, and the solution becomes weak. This weak solution is
pumped by a pump to the generator where it is heated up by using steam or hot
water in the heating coils. A portion of water is evaporated by the heat and the
solution now becomes more strong. This strong solution is passed through the
heat exchanger and then sprayed in the absorber as discussed above. The weak
solution of lithium bromide from the absorber to the generator is also passed
through the heat exchanger. This weak solution gets heat from the strong
solution in the heat exchanger, thus reducing the quantity of steam required to
heat the weak solution in the generator.
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 71
Fig. 3.4. Lithium-Bromide absorption refrigeration system.
The refrigerant water vapours formed in the generator due to heating of
solution are passed to the condenser where they are cooled and condensed by
the cooling water flowing through the condenser water tubes. The cooling water
for condensing is pumped from the cooling water pond or tower. This cooling
water first enters the absorber where it takes away the heat of condensation and
dilution. The condensate from the condenser is supplied to the evaporator to
compensate the water vapour formed in the evaporator. The pressure reducing
valve reduces the pressure of condensate from the condenser pressure to the
evaporator pressure. The cooled water from the evaporator is pumped and
sprayed in the evaporator in order to cool the water for air conditioning flowing
through the chilled tubes. This completes the cycle.
Note: The pressure difference between the generator and the absorber
and the gravity due to the height difference of the two shells is utilised to create
the pressure for the spray.
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 72
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 73
3.8 Refrigerats
The refrigerant is a heat carrying medium which during their cycle (i.e.
compression, condensation, expansion and evaporation) in the refrigeration
system absorb heat from a low temperature system and discard the heat so
absorbed to a higher temperature system.
The natural ice and a mixture of ice and salt were the first refrigerants. In
1834, ether, ammonia, sulphur dioxide, methyl chloride and carbon dioxide came
into use as refrigerants in compression cycle refrigeration machines. Most of the
early refrigerant materials have been discarded for safety reasons or for lack of
chemical or thermal stability. In the present days, many new refrigerants
including halo-carbon compounds, hydro-carbon compounds are used for air
conditioning and refrigeration applications.
The suitability of a refrigerant for a certain application is determined by
its physical, thermodynamic, chemical properties and by various practical factors.
There is no one refrigerant which can be used for all types of applications i.e.,
there is no ideal refrigerant. If one refrigerant has certain good advantages, it will
have some disadvantages also. Hence, a refrigerant is chosen which has greater
advantages and less disadvantages.
3.9 Desirable Properties of an Ideal Refrigerant
We have discussed above that there is no ideal refrigerant. A refrigerant
is said to be ideal if it has all of the following properties:
I. Low boiling point,
2. High critical temperature,
3. High latent heat of vaporisation.
4. Low specific heat of liquid.
5. Low specific volume of vapour,
6. Non-corrosive to metal,
7. Non-flammable and non-explosive,
Viswajyothi College Of Engineering And Technology, Vazhakulam
AKHIL K T, Assistant Professor, VJCET Page 74
8. Non-toxic,
9. Low cost,
10. Easy to liquify at moderate pressure and temperature,
11. Easy of locating leaks by odour or suitable indicator, and
12. Mixes well with oil.
The standard comparison of refrigerants, as used in the refrigeration
industry, is based on an evaporating temperature of - 15° C and a condensing
temperature of + 30° C.
3.10 Classification of Refrigerants
The refrigerants may, broadly, be classified into the following two groups:
1. Primary refrigerants, and 2. Secondary refrigerants.
The refrigerants which directly take part in the refrigeration system are
called primary refrigerants whereas the refrigerants which are first cooled by
primary refrigerants and then used for cooling purposes are known as secondary
refrigerants.
The primary refrigerants are further classified into the following four
groups:
I. Halo-carbon refrigerants,
2. Azeotrope refrigerants,
3. Inorganic refrigerants. and
4. Hydro-carbon refrigerants.
These above mentioned refrigerants are discussed, in detail, in the following
pages.
8.4 Halo-carbon Refrigerants
The American Society of Heating, Refrigeration and Air-conditioning
Engineers (ASHARAE) identifies 42 halo-carbon compounds as refrigerants, but
only a few of them are commonly used. The following table gives some of the
commonly used halo-carbon refrigerants:
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Refrigeration and air conditioning (full note)

  • 2. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 2 B.TECH. DEGREE COURSE MAHATMA GANDHI UNIVERSITY KOTTAYAM,KERALA REFRIGERATION AND AIR CONDITIONING M 704 2+1+0 Module 1 Principles of refrigeration: Thermodynamics of refrigeration - Carnot cycle, reversed carnot cycle, heat pump, and refrigerating machine- coefficient of performance - unit of refrigeration - refrigeration methods- conventional refrigeration systems. Air refrigeration system- Bell Coleman cycle - C.O.P. capacity work and refrigerant flow requirements in Bell - Coleman cycle. Module 2 Vapour compression system: simple cycle -comparison with Carnot cycle - theoretical, actual and reactive - COP effect of operating parameters on COP - wet, dry and superheated compression - under cooling - actual cycle representation on TS and PH diagrams simple problems. Advanced vapour compression systems - multistage vapour compression systems - flash chamber multiple compression and evaporation systems cascading - simple problems. Module 3 Vapour absorption systems: simple, cycles - actual cycle - ammonia water and lithium bromide water systems - COP - electrolux system. Refrigerant and their properties: Nomenclature - suitability of refrigerants for various applications - unconventional refrigeration methods- Vortex tube, steam- jet, magnetic (cryogenics) refrigeration and thermoelectric refrigeration - applied refrigeration house hold refrigerators - unit air conditioners and
  • 3. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 3 water coolers - ice plant cold storage. Module 4 Refrigeration system components: condensers - water and air cooled condensers - evaporative condensers - expansion devises - capillary tube- constant pressure expansion valve - thermostatic expansion valve - float valve and solenoid valve - evaporators - natural convection coils - flooded evaporators - direct expansion coils. Reciprocating compressors: single stage and multistage compressors - work done optimum pressure ratio- effect of interfolding - volumetric efficiency -effect of clearance - isothermal and adiabatic efficiency - compressed air motors. Rotodynamic compressors: Screw and vane type compressors - principle of operation - hermetic, semihermetic and open type refrigeration compressors. Module 5 Principles of air conditioning: Psychrometry and psychrometric chart thermodynamics of human comfort - effective temperature - comfort chart applied psychrometry - sensible heat factor - psychometric process- problems. Winter air conditioning: heating load calculations humidifiers and humidistat. Summer air conditioning: cooling load calculations - year round air conditioning - unitary and central systems - principles of air distribution - design of air duct systems. References 1. Refrigeration and air conditioning - Ballaney P. L. 2. Refrigeration and air conditioning - Stocker W. F. 3. Refrigeration and air conditioning - Jordan and Protester 4. Principles of Refrigeration - Roy J. Dossat
  • 4. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 4 MODULE – I Air Refrigeration Cycles 1.1 Introduction In an air refrigeration cycle, the air is used as a refrigerant. In olden days, air was widely used in commercial applications because of its availability at free of cost. Since air does not change its phase i.e. remains gaseous throughout the cycle, therefore the heat carrying capacity per kg of air is very small as compared to vapour absorbing systems. The air-cycle refrigeration systems, as originally designed and installed, are now practically obsolete because of their low coefficient of performance and high power requirements. However, this system continues to be favoured for air refrigeration because of the low weight and volume of the equipment. The basic elements of an air cycle refrigeration system are the compressor, the cooler or heat exchanger, the expander and the refrigerator. Before discussing the air refrigeration cycles, we should first know about the unit of refrigeration, coefficient of performance of a refrigerator and the difference between the heat engine, a refrigerator and a heat pump. 1.2 Units of Refrigeration The practical unit of refrigeration is expressed in terms of ‘tonne of refrigeration’ (briefly written as TR). A tonne of refrigeration is defined as the amount of refrigeration effect produced by the uniform melting of one tonne (1000 kg) of ice from and at 0 o C in 24 hours. Since the latent heat of ice is 335 kJ/kg, therefore one tonne of refrigeration, ITR = 1000 × 335 kJ in 24 hours = In actual practice, one tonne of refrigeration is taken as equivalent to 210 kJ/min or 3.5 kW (i.e. 3.5 kJ/s)
  • 5. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 5 1.3 Coefficient of Performance of a Refrigerator The coefficient of performance (briefly written as C.O.P.) is the ratio of heat extracted in the refrigerator to the work done on the refrigerant. It is known as theoretical coefficient of performance. Mathematically, Theoretical C.O.P. = where Q = Amount of heat extracted in the refrigerator (or the amount of refrigeration produced, or the capacity of a refrigerator), and W = Amount of work done. Notes: 1. For per unit mass, C.O.P. = 2. The coefficient of performance is the reciprocal of the efficiency (i.e. ) of a heat engine. It is thus obvious, that the value of C.O.P. is always greater than unity. 3. The ratio of the actual C.O.P. to the theoretical C.O.P. is known as relative coefficient of performance, Mathematically. Relative C.O.P = Example 1.1 Find the C.O.P. of a refrigeration system if the work input is 80 kJ/kg and refrigeration effect produced is 160 kJ/kg of refrigerant flowing. Solution. Given : w = 80 kJ/kg ; q = 160 kJ/kg We know that C.O.P. of a refrigeration system = = = 2 Ans. 1.4 Difference Between a Heat Engine, Refrigerator and Heat Pump In a heat engine, as shown in Fig. 2.1 (a), the heat supplied to the engine is converted into useful work. If Q2 is the heat supplied to the engine and Q1 is the heat rejected from the engine, then the net work done by the engine is given by
  • 6. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 6 WE = Q2 - Q1 Fig. 1.1. Difference between a heat engine, refrigerator and heat pump. The performance of a heat engine is expressed by its efficiency. We know that the efficiency or coefficient of performance of an engine, or (C.O.P.)E = = = A refrigerator as shown in Fig.1.1 (b), is a reversed heat engine which either cool or maintain the temperature of a body (T1) lower than the atmospheric temperature (Ta). This is done by extracting the heat(Q1) from a cold body and delivering it to a hot body (Q2). In doing so, work WR is required to be done on the system. According to First Law of Thermodynamics, WR = Q2 - Q1 The performance of a refrigerator is expressed by the ratio of amount of heat taken from the cold body (Q1) to the amount of work required to be done on the system (WR). This ratio is called coefficient of performance. Mathematically, coefficient of performance of a refrigerator, (C.O.P.)R = = Any refrigerating system is a heat pump as shown in Fig 2.1 (c), which extracts heat (Q1) from a cold body and delivers it to a hot body. Thus there is no difference between the cycle of operations of a heat pump and a refrigerator. The main difference between the two is in their operating temperatures. A refrigerator
  • 7. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 7 works between the cold body temperature (T1) and the atmospheric temperature (Ta)whereas the heat pump operates between the hot body temperature (T2) and the atmospheric temperature (Ta). A refrigerator used for cooling in summer can be used as a heat pump for heating in winter. In the similar way, as discussed for refrigerator, we have WP = Q2 - Q1 The performance of a heat pump is expressed by the ratio of the amount of heat delivered to the hot body (Q2) to the amount of work required t o be done on the system (WP). This ratio is called coefficient of performance or energy performance ratio (E.P.R.) of a heat pump. Mathematically, coefficient of performance or energy performance ratio of a heat pump. (C.O.P.)P or E.P.R. = = = = (C.O.P.)R + 1 From above we see that the C.O.P. may be less than one or greater than one depending on the type of refrigeration system used. But the C.O.P. of a heat pump is always greater than one. 1.5 Open Air Refrigeration Cycle In an open air refrigeration cycle, the air is directly led to the space to be cooled (i.e. a refrigerator), allowed to circulate through the cooler and then returned to the compressor to start another cycle. Since the air is supplied to the refrigerator at atmospheric pressure, therefore, volume of air handled by the compressor and expander is large. Thus the size of compressor and expander should be large. Another disadvantage of the open cycle system is that the moisture is regularly carried away by the air circulated through the cooled space. This leads to the formation of frost at the end of expansion process and clog the line. Thus in an open cycle system, a drier should be used.
  • 8. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 8 1.6 Closed or Dense Air Refrigeration Cycle In a closed or dense air refrigeration cycle, the air is passed through the pipes and component parts of the system at all times. The air, in this system, is used for absorbing heat from the other fluid (say brine) and this cooled brine is circulated into the space to be cooled. The air in the closed system does not come in contact directly with the space to be cooled. The closed air refrigeration cycle has the following thermodynamic advantages : 1. Since it can work at a suction pressure higher than that of atmospheric pressure, therefore the volume of air handled by the compressor and expander are smaller as compared to an open air refrigeration cycle system. 2. The operating pressure ratio can be reduced, which results in higher coefficient of performance. 1.7. Air Refrigerator Working on Reversed Carnot Cycle In refrigerating systems, the Carnot cycle considered is the reversed Carmot cycle. We know that a heat engine working on Carnot cycle has the highest possible efficiency. Similarly, a refrigerating system working on the reversed Carnot cycle, will have the maximum possible coefficient of performance. We also know that it is not possible to make an engine working on the Carnot cycle. However, it is used as the ultimate standard of comparison. A reversed Carnot cycle, using air a working medium (or refrigerant) is shown on p-v and T-s diagrams in Fig. 1.2 (a) and (b) respectively. At point 1, let p1, v1, T1 be the pressure, volume and temperature of air respectively.
  • 9. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 9 Fig. 1.2 Reversed Carnot cycle The four processes of the cycle are as follows: 1. Isentropic compression process. The air is compressed isentropically as shown by the curve 1-2 on p-v and T-s diagrams. During this process, the pressure of air increases from p1 to p2, specific volume decreases from v1 to v2 and temperature increases from T1 to T2. We know that during isentropic compression, no heat is absorbed or rejected by the air. 2. Isothermal compression process. The air is now compressed isothermally (i.e. at constant temperature, T2 = T3) as shown by the curve 2-3 on p -v and T-s diagrams. During this process, the pressure of air increases from p2 to p3 and specific volume decreases from v2 to v3. We know that the heat rejected by the air during isothermal compression per kg of air, q2-3 = Area = 3. Isentropic expansion process. The air is now expanded isentropically as shown by the curve 3-4 on p-v and T-s diagrams. The pressure of air decreases from p3 to p4, specific volume increases from v3 to v4 and the temperature decreases from T3 to T4. We know that during isentropic expansion no heat is absorbed or rejected by the air. 4. Isothermal expansion process. The air is now expanded isothermally
  • 10. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 10 (i.e. at constant temperature, T4 = T1) as shown by the curve 4-1 on p-v and T-s diagrams. The pressure of air decreases from p4 to p1, and specific volume increases from v4 to v1. We know that the heat absorbed by the air (or heat extracted from the cold body) during isothermal expansion per kg of air, q4-1 = Area = We know that work done during the cycle per kg of air = Heat rejected - Heat absorbed = q2-3-q4-1 = T2 (s2-s3)-T1(s2-s3) = (T2-T1) (s2-s3) Coefficient of performance of the refrigeration system working on reversed Carnot cycle, (C.O.P.)R = = = = Though the reversed Carnot cycle is the most efficient between the fixed temperature limits, yet no refrigerator has been made using this cycle. This is due to the reason that the isentropic processes of the cycle require high speed while the isothermal processes require an extremely low speed. This variation in speed of air is not practicable. Note : We have already discussed that C.O.P. of a heat pump, (C.O.P.)P = (C.O.P.)R + 1 = +1 = and C.O.P. or efficiency of a heat engine, (C.O.P.)E = = 1.8 Temperature Limitations for Reversed Carnot Cycle
  • 11. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 11 We have seen in the previous article that the C.O.P. of the refrigeration system working on reversed Carnot cycle is given by (C.O.P.)R = where T1 = Lower temperature, and T2 = Higher temperature. The C.O.P. of the reversed Carnot cycle may be improved by 1. Decreasing the higher temperature (i.e. temperature of hot body, T2), or 2. Increasing the lower temperature (i.e. temperature of cold body, T1). This applies to all refrigerating machines, both theoretical and practical. It may be noted that temperatures T1 and T2 cannot be varied at will, due to certain functional limitations. It should be kept in mind that the higher temperature (T2) is the temperature of cooling water or air available for rejection of heat and the lower temperature (T1) is the temperature to be maintained in the refrigerator. The heat transfer will take place in the right direction only when the higher temperature is more than the temperature of cooling water or air to which heat is to be rejected, while the lower temperature must be less than the temperature of substance to be cooled. Thus if the temperature of cooling water or air (i.e. T2) available for heat rejection is low, the C.O.P. of the Carnot refrigerator will be high. Since T2 in winter is less than T2 in summer, therefore, C.O.P. in winter will be higher than C.O.P. in summer. In other words, the Carnot refrigerators work more efficiently in winter than in summer. Similarly, if the lower temperature fixed by the refrigeration application is high, the C.O.P. of the Carnot refrigerator will be high. Thus a Carnot refrigerator used for making ice at 0o C (273 K) will have less C.O.P. than a Carnot refrigerator used for air-conditioned plant in summer at 20o C when the atmospheric temperature is 40o C. In other words, we can say that the Carnot C.O.P. of a domestic refrigerator is less than the Carnot C.O.P. of a domestic air- conditioner.
  • 12. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 12 Example 1.2. A machine working on a Carnot cycle operates between 305 K and 260 K. Determine the C.O.P. when it is operated as 1. a refrigerating machine; 2. a heat pump; and 3. a heat engine. Solution. Given : T2 = 305 K; T1 = 260 K 1. C.O.P. of refrigerating machine We know that C.O.P. of a refrigerating machine, (C.O.P.)R = = = 5.78 Ans. 2. C.O.P. of a heat pump We know that C.O.P. of a heat pump, (C.O.P.)P = = = 6.78 Ans. 3. C.O.P. of a heat engine We know that C.O.P. of a eat engine, (C.O.P.)E = = = 0.147 Ans. Example 1.3. A Carnot refrigeration cycle absorbs heat at 270 K and rejects it at 300 K. 1. Calculate the coefficient of performance of this refrigeration cycle. 2. If the cycle is absorbing 1130 kJ/min at 270 K, how many kJ of work is required per second ? 3. If the Carnot heat pump operates between the same temperatures as the above refrigeration cycle, what is the coefficient of performance ? 4. How many kJ/min will the heat pump deliver at 300 K if absorbs 1130 kJ/min at 270 K. Solution. Given : T1 = 270 K; T2 = 300 K 1. Coefficient of performance of Carnot refrigeration cycle We know that coefficient of performance of Carnot refrigeration cycle,
  • 13. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 13 (C.O.P.)R = = = 9 Ans. 2. Work required per second Let WR = Work required per second. Heat absorbed at 270 K (i.e. T1), Q1 = 1130 kJ/min = 18.83 kJ/s ... (Given) We know that (C.O.P.)R = or 9 = WR = 2.1 kJ/s Ans. 3. Coefficient of performance of Carnot heat pump We know that coefficient of performance of a Carnot heat pump, (C.O.P.)P= = = 10 Ans. 4. Heat delivered by heat pump at 300 K Let Q2 = Heat delivered by heat pump at 300 K. Heat absorbed at 270 K (i.e. T1), Q1 = 1130 kJ/min ... (Given) We know that (C.O.P.)P = or 10 = 10Q2 - 11300 = Q2 or Q2 = 1256 kJ/min Ans. Example 1.4. A cold storage is to be maintained at -5oC while the surroundings are at 35oC. The heat leakage from the surroundings into the cold storage is estimated to be 29 kW. 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. Solution. Given : T1 = -5o C = -5 + 273 = 268 K;
  • 14. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 14 T2 = 35 o C = 35 + 273 = 308 K; Q1 = 29 kW; (C.O.P.)actual = (C.O.P.)ideal The refrigerating plant operating between the temperatures T1 and T2 is shown in Fig. 1.3. Let WR = Work or power required to drive the plant. We know that the coefficient of performance of an ideal refrigeration plant, (C.O.P.)ideal = = = 6.7 Actual coefficient of performance, (C.O.P.)actual = (C.O.P.)ideal = × 6.7 = 2.233 We also know that (C.O.P.)actual = WR = = = 12.987 kW Ans. Example 1.5. 1.5 kW per tonne of refregeration is required to maintain the temperature of -40oC in the refrigerator. If the refrigeration cycle works on Carnot cycle, determine the following: 1. C.O.P. of the cycle; 2. Temperature of the sink ; 3. Heat rejected to the sink per tonne of refrigeration ; and 4. Heat supplied and E.P.R., if the cycle is used as a heat pump. Solution. Given : WR= 1.5 kW; Q1 = 1 TR; T1 = -40o C = -40 + 273 = 233 K
  • 15. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 15 1. C.O.P. of the cycle Since 1.5 kW per tonne of refrigeration is required to maintain the temperature in the refrigerator, therefore amount of work required to be done, WR = 1.5 kW = 1.5 kJ/s = 1.5 × 60 = 90 kJ/min and heat extracted from the cold body, Q1 = 1 TR = 210 kJ/min We know that (C.O.P.)R = = = 2.33 Ans. Fig 1.4 2. Temperature of the sink Let T2 = Temperature of the sink We know that (C.O.P.)R = or 2.33= T2 = + 233 = 333 K = 60o C Ans. 3. Heat rejected to the sink per tonne of refrigeration We know that heat rejected to the sink, Q2 = Q1 + WR = 210 + 90 = 300 kJ/min Ans. 4. Heat supplied and E.P.R., if the cycle is used as a heat pump is Q2 = 300 kJ/min Ans.
  • 16. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 16 and E.P.R. = (C.O.P.)R + 1 = 2.33 + 1 = 3.33 Ans. Example 2.6. The capacity of a refrigerator is 200 TR when working between - 6o C and 25o C. Determine the mass of ice produced per day from water at 25o C. Also find the power required to drive the unit. Assume that the cycle operates on reversed Carnot cycle and latent heat of ice is 335 kJ/kg. Solution. Given : Q = 200 TR; T1 = -6o C = - 6 + 273 = 267 K; T2 = 25o C = 25 + 273 = 298 K ; tw = 25o C ; hfg(ice) = 335 kJ/kg Mass of ice produced per day We know that heat extraction capacity of the refrigerator = 200 × 210 = 42 000 kJ/min ... ( 1TR = 210 kJ/min) and heat removed from 1 kg of water at 25o C to form ice at 0o C = Mass × Sp. heat × Rise in temperature + hfg(ice) =335 kJ/kg = 1 × 4.187 (25 - 0) + 335 = 439.7 kJ/kg Mass of ice produced per min = = 95.52 kg/min and mass of ice produced per day = 95.52 × 60 × 24 = 137 550 kg = 137.55 tonnes Ans. Power required to drive the unit We know that C.O.P. of the reversed Carnot cycle = = = 8.6 Also C.O.P. = 8.6 = or Work done per min = 42 000/8.6 = 4884 kJ/min
  • 17. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 17 Power required to drive the unit = 4884/60 = 81.4 kW Ans. Example 1.7. Five hundred kgs of fruits are supplied to a cold storage at 20oC. The cold storage is maintained at -5oC and the fruits get cooled to the storage temperature in 10 hours. The latent heat of freezing is 105 kJ/kg and specific heat of fruit is 1.256 kJ/kg K. Find the refrigeration capacity of the plant. Solution. Given : m = 500 kg ; T2 = 20o C = 20 + 273 = 293 K ; T1 = -5o C = -5 + 273 = 268 K ; = 105 kJ/kg, cF = 1.256 kJ/kg K We know that heat removed from the fruits in 10 hrs, Q1 = m cF (T2 - T1) = 500 × 1.256 (293 - 268) = 15 700 kJ and total latent heat of freezing, Q2 = m. = 500 × 105 = 52 500 kJ Total heat removed in 10 hrs, Q = Q1 + Q2 = 15 700 + 52 500 = 68 200 kJ and total heat removed in one minute = 68 200/10 × 60 = 1113.7 kJ/min Refrigeration capacity of the plant = 113.7/210 = 0.541 TR Ans. ...( 1 TR = 210 kJ/min) Example 1.8. A cold storage plant is required to store 20 tonnes of fish. The fish is supplied at a temperature of 30o C. The specific heat of fish above freezing point is 2.93 kJ/kg K. The specific heat of fish below freezing point is 1.26 kJ/kg K. The fish is stored in cold storage which is maintained at - 8o C. the freezing point of fish is - 4o C. The latent heat of fish is 235 kJ/kg. If the plant requires 75 kW to drive it, find : 1. The capacity of the plant, and 2. Time taken to achieve cooling.
  • 18. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 18 Assume actual C.O.P. of the plant as 0.3 of the Carnot C.O.P. Solution. Given : m = 20 t = 20 000 kg ; T2 = 30o C = 30 + 273 = 303 K ; cAF = 2.93 kJ/kg K; cBF = 1.26 kJ/kg K ; T1 = -8 o C = -8 + 273 = 265 K ; T3 = -4 o C = -4 + 273 = 269 K; hfg(Fish) = 235 kJ/kg ; P = 75 kW = 75 kJ/s 1. Capacity of the plant We know that Carnot C.O.P. = = = 6.97 Actual C.O.P. = 0.3 × Work required = 2.091 × 75 = 156.8 kJ/s = 156.8 × 60 = 9408 kJ/min Capacity of the plant = 9408 / 210 = 44.8 TR Ans. ...( 1TR = 210 kJ/min) 2. Time taken to achieve cooling We know that heat removed from the fish above freezing point, Q1 = m × cAF (T2 - T3) = 20 000 × 2.93 (303 - 269) = 1.992 × 106 kJ Similarly, heat removed from the fish below freezing point, Q2 = m × cBF (T3 - T1) = 20 000 × 1.26 (269 - 265) = 0.101 × 106 kJ and total latent heat of fish, Q3 = m × = 20 000 × 235 = 4.7 × 106 kJ Total heat removed by the plant = Q1 + Q2 + Q3 = 1.992 × 106 + 0.101 × 106 + 4.7 × 106 = 6.793 × 106 kJ and time taken to achieve cooling
  • 19. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 19 = = = 722 min = 12.03 h Ans 1.9. Air Refrigerator Working on a Bell-Coleman Cycle (or Reversed Brayton or Joule Cycle) A Bell-Coleman air refrigeration machine was developed by Bell-Coleman and Light Foot by reversing the Joule’s air cycle. It was one of the earliest types of refrigerators used in ships carrying frozen meat. fig. 1.5 shows a schematic diagram of such a machine which consists of a compressor, a cooler, an expander and a refrigerator. Fig 1.5 Open cycle air Bell- Coleman Fig 1.6 Closed cycle or dence air bell- Refrigerator Coleman refrigerator The Bell-Coleman cycle (also known as reversed Brayton or Joule cycle) is a modification of reversed Carnot cycle. The cycle is shown on p-v and T-s diagrams in Fig. 1.7 (a) and (b). At point 1, p1, v1 and T1 be the pressure, volume and temperature of air respectively. The four process of the cycle are as follows: 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 the curve 1-2 on p-v and T-s diagrams. During the compression stroke, both the pressure and temperature increases and the specific volume of air at delivery from compressor reduces from v1 to v2. We know that during isentropic compression process, no heat is absorbed or rejected by the air.
  • 20. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 20 Fig 1.7 Bell-Coleman cycle 2. Constant pressure cooling process. The warm air from the compressor is now passed into the cooler where it is cooled at constant pressure p3 (equal to p2), reducing the temperature from T2 to T3 (the temperature of cooling water) as shown by the curve 2-3 on p-v and T-s diagrams. The specific volume also reduces from v2 to v3. We know that heat rejected by the air during constant pressure per kg of air, Q2-3 = cp (T2-T3) 3. 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 air during expansion falls from T3 to T4 (i.e. the temperature much below the temperature of cooling water, T3). The expansion process is shown by the curve 3-4 on the p-v and T-s diagrams. The specific volume of air at entry to the refrigerator increases from v3 to v4. We know that during isentropic expansion of air, no heat is absorbed or rejected by the air. 4. Constant pressure expansion process. The cold air from the expander is now passed to the refrigerator where it is expanded at constant pressure p4 (equal to p1). The temperature of air increases from T4 to T1. This process is shown by the curve 4-1 on the p-v and T-s diagrams. Due to heat from the refrigerator, the specific volume of the air changes from v4 to v1. We know that the heat absorbed by the air (or heat extracted from the refrigerator) during
  • 21. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 21 constant pressure expansion per kg of air is q4 - 1 = cp (T1 - T4) We know that work done during the cycle per kg of air = Heat rejected - Heat absorbed = Cp (T2 - T3) - cp (T1 - T4) C.O.P. = = = = ... (i) We know that for isentropic compression process 1-2, = ... (ii) Similarly, for isentropic expansion process 3-4, = ... (iii) Since p2 = p3 and p1 = p4, therefore from equations (ii) and (iii), or ... (iv) Now substituting these values in equation (i), we get C.O.P. =
  • 22. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 22 = ... (v) where rp = Compression or Expansion ratio = = Sometimes, the compression and expansion process take place according to the law pvn = Constant. In such a case, the C.O.P. is obtained from the fundamentals as discussed below: We know that work done by the compressor during the process 1-2 per kg of air, ... ( pv = RT) and work done by the expander during the process 3-4 per kg of air, Net work done during the cycle per kg of air, We also know that heat absorbed during constant pressure process 4-1, = cp (T1 - T4) C.O.P. = = ... (vi) We know that Substituting the value of R in equation (vi), C.O.P. =
  • 23. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 23 = = ... (vii) Notes : 1. In this case, the values of T2 and T4 are to be obtained from the relations and 2. For isentropic compression or expansion, n = . Therefore, the equation (vii) may be written as C.O.P. = ... (same as before) 3. We have already discussed that the main drawback of the open cycle air refrigerator is freezing of the moisture in the air during expansion stroke which is liable to choke up the valves. Due to this reason, a closed cycle or dense air Bell-Coleman refrigerator as shown in Fig. 2.6 is preferred. In this case, the cold air does not come in direct contact of the refrigerator. The cold air is passed through the pipes and it is used for absorbing heat from the brine and this cooled brine is circulated in the refrigerated space. The term dense air system is derived from the fact that the suction to the compressor is at higher pressure than the open cycle system (which is atmospheric). Example 1.9. A refrigerator working on Bell-Coleman cycle operates between pressure limits of 1.05 bar and 8.5 bar. Air is drawn from the cold chamber at 10o C, compressed and then it is cooled to 30o C before entering the expansion cylinder. The expansion and compression follows the law pv1.3 = constant. Determine the theoretical C.O.P. of the system. Solution: Given : p1 = p4 = 1.05 bar ; p2 = p3 = 8.5 bar; T1 = 10o C = 10 + 273 = 283 K; T3 = 30o C = 30 + 273 = 303 K; n = 1.3
  • 24. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 24 Fig 1.8 The p-v and T-s diagrams for a refrigerator working on the Bell-Coleman cycle is shown in Fig. 1.8 (a) and (b) respectively. Let T2 and T4 = temperatures of air at the end of compression and expansion respectively. Since the compression and expansion follows the law pv1.3 = C, therefore Similarly T4 = T3/1.62 = 303/1.62 = 187 K Fig. We know that theoretical coefficient of performance, C.O.P. = = ... (Taking = 1.4)
  • 25. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 25 = = 1.3 Ans. MODULE– II Vapour Compression Refrigeration Systems
  • 26. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 26 2.1 Introduction A vapour compression refrigeration system is an improved type of air refrigeration system in which a suitable working substance, termed as refrigerant, is used. It condenses and evaporates at temperatures and pressures close to the atmospheric conditions. The refrigerants, usually, used for this purpose are ammonia (NH3), carbon dioxide (CO2) and sulphur dioxide (SO2). The refrigerant used, does not leave the system, but is circulated throughout the system alternately condensing and evaporating. In evaporating, the refrigerant absorbs its latent heat from the brine (salt water) which is used for circulating it around the cold chamber. While condensing, it gives out its latent heat to the circulating water of the cooler. The vapour compression refrigeration system is, therefore a latent heat pump, as it pumps its latent heat from the brine and delivers it to the cooler. The vapour compression refrigeration system is now-a-days used for all purpose refrigeration. It is generally used for all industrial purposes from a small domestic refrigerator to a big air conditioning plant. 2.2 Advantages and Disadvantages of vapour Compression Refrigeration System over Air Refrigeration System Following are the advantages and disadvantages of the vapour compression refrigeration system over air refrigeration system: Advantages 1. It has smaller size for the given capacity of refrigeration. 2. It has less running cost. 3. It can be employed over a large range of temperatures. 4. The coefficient of performance is quite high. Disadvantages 1. The initial cost is high 2. The prevention of leakage of the refrigerant is the major problem in vapour compression system
  • 27. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 27 2.3 Mechanism of a Simple Vapour Compression Refrigeration System Fig. 2.1 shows the schematic diagram of a simple vapour compression refrigeration system. It consists of the following five essential parts : 1. Compressor. The low pressure and temperature vapou refrigerant from evaporator is drawn into the compressor through the inlet or suction valve A, where it is compresses to a high pressure and temperature. This high pressure and temperature vapour refrigerant is discharge the condenser through the delivery or discharge valve B. 2. Condenser. The condenser or cooler consists of coils of pipe in which the high pressure and temperature vapour refrigerant is cooled and condensed. The refrigerant, while passing through the condenser, gives up its latent heat to the surrounding condensing medium which is normally air or water. 3. Receiver. The condensed liquid refrigerant from the condenser is stored in a vessel known as receiver from where it is supplied to the evaporator through the expansion valve or refrigerant control valve. 4. Expansion valve. It is also called throttle valve or reftigerant control
  • 28. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 28 valve. The function of the expansion valve is to allow the liquid refrigerant under high pressure and temperature to pass at a controlled rate after reducing its pressure and temperature. Some of the liquid refrigerant evaporates as it passes through the expansion valve, but the greater portion is vaporised in the evaporator at the low pressure and temperature. 5. Evaporator. An evaporator consists of coils of pipe in which the liquid- vapour refrigerant at low pressure and temperature is evaporated and changed into vapour refrigerant at low pressure and temperature. In evaporating, the liquid vapour refrigerant absorbs its latent heat of vaporisation from the medium (air, water or brine) which is to be cooled. Note : In any compression refrigeration system, there are two different pressure conditions. One is called the high pressure side and other is known as low pressure side. The high pressure side includes the discharge line (i.e. piping from the evaporator to the suction valve A). 2.4 Pressure-Enthalpy (p-h) Chart Fig 2.2 pressure enthalpy [p-h] chart The most convenient chart for studying the behavior of a refrigerant is the p-h chart, in which the vertical ordinates represent pressure and horizontal ordinates represent enthalpy (i.e. total heat). A typical chart is shown in Fig. 2.2, in which a few important lines of the complete chart are drawn. The saturated
  • 29. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 29 liquid line and the saturated vapour line merge into one another at the critical point. A saturated liquid is one which has a temperature equal to the saturation temperature corresponding to its pressure. The space to the left of the saturated liquid line will, therefore, be sub-cooled liquid region. The space between the liquid and the vapour lines is called wet vapour region and to the right of the saturated vapour line is a superheated vapour region. In the following pages, we shall drawn the p-h chart along with the T-s diagram of the cycle. 2.5 Types of Vapour Compression Cycles We have already discussed that vapour compression cycle essentially consists of compression, condensation, throttling and evaporation. Many scientists have focussed their attention to increase the coefficient of performance of the cycle. Through there are many cycles, yet the following are important from the subject point of view : 1. Cycle with dry saturated vapour after compression, 2. Cycle with wet vapour after compression, 3. Cycle with superheated vapour after compression, 4. Cycle with superheated vapour before compression, and 5. Cycle with undercooling or subcooling of refrigerant. Now we shall discuss all the above mentioned cycles, one by one, in the following pages. 2.6 Theoretical Vapour Compression Cycle with Dry Saturated Vapour after Compression A vapour compression cycle with dry saturated vapour after compression is shown on T-s and p-h diagrams in Fig. 2.3 (a) and (b) respectively. At point 1, let T1,p1 and s1 be the temperature, pressure and entropy of the vapour refrigerant respectively. The four process of the cycle are as follows :
  • 30. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 30 Fig 2.3: theoretical vapour compression cycle with dry saturated vapour after compression 1. Compression process. The vapour refrigerent at low pressure p1 and temperature T1 is compressed isentropically to dry saturated vapour as shown by the vertical line 1-2 on T-s diagram and by the curve 1-2 on p-h diagram. The pressure and temperature rises from p1 to p2 and T1 to t2 respectively. The work done during isentropic compression per kg of refrigerant is given by w = h2 - h1 where h1 = Enthalpy of vapour refrigerant at temperature T1, i.e. at suction of the compressor, and h2 = Enthalpy of the vapour refrigerant at temperature T2, i.e. at discharge of the compressor. 2. Condensing process. The high pressure and temperature vapour refrigerant from the compressor is passed through the condenser where it is completely condensed at constant pressure p2 and temperature T2, as shown by the horizontal line 2-3 on T-s and p-h diagrams. The vapour refrigerant is changed into liquid refrigerant. The refrigerant, while passing through the condenser, gives its latent heat to the surrounding condensing medium. 3. Expansion process. the liquid refrigerant at pressure p3 = p2 and temperature T3 = T2 is expanded by *throttling process through the expansion
  • 31. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 31 valve to a low pressure p4 = p1 and temperature T4 = T1, as shown by the curve 3-4 on T-s diagram and by the vertical line 3-4 on p-h diagram. We have already discussed that some of the liquid refrigerant evaporates as it passes through the expansion valve, but the greater portion is vaporised in the evaporator. We know that during the throttling process, no heat is absorbed or rejected by the liquid refrigerant. Notes : (a) In case an expansion cylinder is used in place of throttle or expansion valve to expand the liquid refrigerant, then the refrigerant will expand isentropically as shown by dotted vertical line on T-s diagram in Fig. 2.3 (a). The isentropic expansion reduces the external work being expanded in running the compressor and increases the refrigerating effect. Thus, the net result of using the expansion cylinder is to increase the coefficient of performance. Since the expansion cylinder system of expanding the liquid refrigerant is quite complicated and involves greater initial cost, therefore its use is not justified for small gain in cooling capacity. Moreover, the flow rate of the refrigerant can be controlled with throttle valve which is not possible in case of expansion cylinder which has a fixed cylinder volume. (b) In modern domestic refrigerators, a capillary (small bore tube) is used in place of an expansion valve. 4. Vaporising process. The liquid-vapour mixture of the refrigerant at pressure p4 = p1 and temperature T4 = T1 is evaporated and changed into vapour refrigerant at constant pressure and temperature, as shown by the horizontal line 4-1 on T-s and p-h diagrams. During evaporation, the liquid-vapour refrigerant absorbs its latent heat of vaporisation from the medium (air, water or brine) which is to be cooled. This heat which is absorbed by the refrigerant is called refrigerating effect and it is briefly written as RE. The process of vaporisation continues upto point 1 which is the starting point and thus the cycle is completed. We know that the refrigerating effect or the heat absorbed or extracted by the liquid-vapour refrigerant during evaporation per kg of refrigerant is given by ... ( = h4) where = Sensible heat at temperature T3, i.e. enthalpy of liquid
  • 32. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 32 refrigerant leaving the condenser. It may be noticed from the cycle that the liquid-vapour refrigerant has extracted heat during evaporation and the work will be done by the compressor for isentropic compression of the high pressure and temperature vapour refrigerant. Coefficient of performance, C.O.P. = = = Example 2.1 In an ammonia vapour compression system, the pressure in the evaporator is 2 bar. Ammonia at exit is 0.85 dry and at entry its dryness fraction is 0.19. During compression, the work done per kg of ammonia is 150 kJ. Calculate the C.O.P. and the volume of vapour entering the compressor per minute, if the rate of ammonia circulation is 4.5 kg/min. The latent heat and specific volume at 2 bar are 1325 kJ/kg and 0.58 m3 /kg respectively. Solution. Given : p1 = p4 = 2 bar ; x1 = 0.85 ; x4 = 0.19 ; w = 150 kJ/kg ; ma = 4.5 kg/min; hfg = 1325 kJ/kg ; vg = 0.58 m3 /kg C.O.P. The T-s and p-h diagrams are shown in Fig.4.3 (a) and (b) respectively. Since the ammonia vapour at entry to the evaporator (i.e. at point 4) has dryness fraction (x4) equal to 0.19, therefore enthalpy at point 4, Similarly, enthaipy of ammonia vapour at exit i.e. at point 1, Heat extracted from the evaporator or refrigerating effect, We know that work done during compression, w = 150 kJ/kg C.O.P. = RE/w = 874.5/150 = 5.83 Ans.
  • 33. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 33 Volume of vapour entering the compressor per minute We know that volume of vapour entering the compressor per minute = Mass of refrigerant / min × Specific volume = Ans. Example 2.2. The temperature limits of an ammonia refrigerating system are 25o C and -10o C. If the gas is dry at the end of compression, calculate the coefficient of performance of the cycle assuming no undercooling of the liquid ammonia. Use the following table for properties of ammonia: Temperature (o C) Liquid heat (kJ/kg) Latent heat (kJ/kg) Liquid entropy (kJ/kg K) 25 -10 298.9 135.37 1166.94 1297.68 1.1242 0.5443 Solution. Given : ; T1 = T4 = -10 o C = 263K ; hf3 = h4 = 298.9 kJ/kg ; hfg2 = 1166.94 kJ/kg ; sf2 = 1.1242 kJ/kg K ; hf1 = 135.37 kJ/kg ; hfg1 = 1297.68 kJ/kg ; sf1 = 0.5443 kJ/kg K The T-s and p-h diagrams are shown in Fig. 2. 4 (a) and (b) respectively. Let x1 = Dryness fraction at point 1. We know that entropy at point 1, Fig. = 0.5443 + 4.934 x1 Similarly, entropy at point 2, Since the entropy at point 1 is equal to entropy at point 2, therefore equating equations (i) and (ii) 0.5443 + 4.934 x1 = 5.04 or x1 = 0.91
  • 34. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 34 We know that enthalpy at point 1, h1 = hf1 + x1 hfg1 = 135.37 + 0.91 × 1297.68 = 1316.26 kJ/kg and enthalpy at point 2, h2 = hf2 + hfg2 = 298.9 + 1166.94 = 1465.84 kJ/kg Coefficient of performance of the cycle = Ans. Example 2.3. A vapour compression refrigerator works between the pressure limits of 60 bar and 25 bar. The working fluid is just dry at the end of compression and there is no under-cooling of the liquid before the expansion valve. Determine : 1. C.O.P. of the cycle ; and 2. Capacity of the refrigerator if the fluid flow is at the rate of 5 kg/min. Data : Pressure (bar) Saturation temperatur e (K) Enthalpy (kJ/kg) Entropy (kJ/kg K) Liquid Vapour Liquid Vapour 60 25 295 261 151.96 56.32 293.29 322.58 0.554 0.226 1.0332 1.2464 Solution. Given : p2 = p= = 60 bar ; p1 = p4 = 25 bar ; T2 = T3 = 295 K ; T1 = T4 = 261 K ; hf3 = h4 = 151.96 kJ/kg ; hf1 = 56.32 kJ/kg ; hg2 = h2 293.29 kJ/kg ; hg1 =
  • 35. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 35 322.58 kJ/kg ; sf2 = 0.554 kJ/kg K ; sf1 = 0.226 kJ/kg K ; sg2 = 1.0332 kJ/kg K; sg1 = 1.2464 kJ/kg K Fig 2.5 1. C.O.P. of the cycle The T-s and p-h diagrams are shown in Fig. 2.5 (a) and (b) respectively. Let x1 = Dryness fraction of the vapour refrigerant entering the compressor at point 1. We know that entropy at point 1, ...( = sf1 + sfg1 ) = ... (i) and entropy at point 2, s2 = 1.0332 kJ/kg K ... (Given) ... (ii) Since the entropy at point 1 is equal to entropy at point 2, therefore equating equations (i) and (ii), 0.226 + 1.0204 x1 = 1.0332 or x1 = 0.791 we know that enthalpy at point 1, ... ( = hf1 + hfg1) = 56.32 + 0.791 (322.58 - 56.32) = 266.93 kJ/kg C.O.P. of the cycle = Ans.
  • 36. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 36 2. Capacity of the refrigerator We know that the heat extracted or refrigerating effect produced per kg of refrigerant = h1 - hf3 = 266.93 - 151.93 = 114.97 kJ/kg Since the fluid flow is at the rate of 5 kg/min, therefore total heat extracted = 5 × 114.97 = 574.85 kJ/min Capacity of the refrigerator = Ans. ...( 1 TR = 210 kJ/min) 2.7. Theoretical Vapour Compression Cycle with Wet Vapour after Compression A vapour compression cycle with wet vapour after compression is shown on T-s and p-h diagrams in Fig. 2.6 (a) and (b) respectively. In this cycle, the enthalpy at point 2 is found out with the help of dryness fraction at this point. The dryness fraction at points 1 and 2 may be obtained by equating entropies at points 1 and 2. Now the coefficient of performance may be found out as usual from the relation, C.O.P. = = Compound Vapour Compression Refrigeration Systems
  • 37. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 37 Introduction In the previous chapter, we have discussed the simple vapour compression refrigeration system in which the low pressure vapour refrigerant from the evaporator is compressed in a single stage (or a single compressor) and then delivered to a condenser at a high pressure. But sometimes, the vapour refrigerant is required to be delivered at a very high pressure as in the case of low temperature refrigerating systems. In such cases either we should compress the vapour refrigerant by employing a single stage compressor with a very high pressure ratio between the condenser and evaporator or compress it in two or more compressors placed is series. The compression carried out in two or more compressors is called compound or multistage compression. In vapour compression refrigeration systems, the major operating cost is the energy input to the system in the form of mechanical work. Thus any method of increasing coefficient of performance is advantageous so long as it does not involve too heavy an increase in other operating expenses, as well as initial plant cost and consequent maintenance. Since the coefficient of performance of a refrigeration system is the ratio of refrigerating effect to the compression work, therefore the coefficient of performance can be increased either by increasing the refrigerating effect or by decreasing the compression work. A little consideration will show that in a vapour compression system, the compression work is greatly reduced if the refrigerant is compressed very close to the saturated vapour line. This can be achieved by compressing the refrigerant in more stages with intermediate intercooling. But it is economical only where the pressure ratio is considerable as would be the case when very low evaporator temperatures are desired or when high condenser temperature may be required. The compound compression is generally economical in large plants. The refrigerating effect can be increased by maintaining the condition of the refrigerant in more liquid state at the entrance to the evaporator. This can be achieved by expanding the refrigerant very close to the saturated liquid line. It may be noted that by subcooling the refrigerant and by removing the flashed vapour, as they are during multistage expansion, the expansion can be brought
  • 38. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 38 close to the liquid line. 2.2 Advantages of Compound (or Multi-stage) Vapour Compression with Intercooler Following are the main advantages of compound or multistage compression over single stage compression: I. The work done per kg of refrigerant is reduced in compound compression with intercooler as compared to single stage compression for the same delivery pressure. 2. It improves the volumetric efficiency for the given pressure ratio. 3. The sizes of the two cylinders (i.e., high pressure and low pressure) may be adjusted to suit the volume and pressure of the refrigerant. 4. It reduces the leakage loss considerably. 5. It gives more uniform torque, and hence a smaller size flywheel is needed. 6. It provides effective lubrication because of lower temperature range. 7. It reduces the cost of compressor. 2.3 Types of Compound Vapour Compression with Intercooler In compound compression vapour refrigeration systems, the superheated vapour refrigerant leaving the first stage of compression is cooled by suitable method before being fed to the second stage of compression and so on. Such type of cooling the refrigerant is called intercooling. Though there are many types of compound compression with intercoolers, yet the following are important from the subject point of view: I. Two stage compression with liquid intercooler. 2. Two stage compression with water intercooler. 3. Two stage compression with water intercooler, liquid subcooler and liquid flash chamber. 4. Two stage compression with water intercooler, liquid subcooler and
  • 39. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 39 flash intercooler. 5. Three stage compression with flash chambers. 6. Three stage compression with water intercoolers. 7. Three stage compression with flash intercoolers. The above mentioned types are now discussed, in detail, one by one in the following pages. 2.4 Two Stage Compression with Liquid Intercooler The arrangement of a two stage compression with liquid intercooler is shown in Fig. 2.7 (a). The corresponding p-h diagram is shown in Fig. 2.7 (b). The various points on the p-h diagram are plotted as discussed below: I. First of all, draw a horizontal pressure line representing the evaporator pressure PE (or suction pressure of low pressure compressor) which intersects the saturated vapour line at point 1. At this point, the saturated vapour is supplied to the low pressure compressor. Let, at point I, the enthalpy of the saturated vapour is h1 and entropy sv1 2. The saturated vapour refrigerant admitted at point 1 is compressed isentropically in the low pressure compressor and delivers the refrigerant in a superheated state. The pressure rises from PE to P2 The curve 1-2 represents the isentropic compression in the low pressure compressor. In order to obtain point 2, draw a line from point 1, with entropy equal to svl, along the constant entropy line intersecting the intermediate pressure line P2 at point 2. Let enthalpy at this point is h2. 3. The superheated vapour refrigerant leaving the low pressure compressor at point 2 is cooled (or desuperheated) at constant pressure P2 = P3 in a liquid intercooler by the liquid refrigerant from the condenser. The refrigerant leaving the liquid intercooler is in saturated vapour state. The line 2-3 represents the cooling or desuperheating process. Let the enthalpy and entropy at point 3 is h3 and sv3 respectively.
  • 40. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 40 Fig. 2.7 4. The dry saturated vapour refrigerant is now supplied to high pressure compressor where it is compressed isentropically from intermediate or interocooler pressure P2 to condensor pressure pc. The curve 3-4 represents the isentropic compression in the high pressure compressor. The point 4 on the p-h diagram is obtained by drawing a line of entropy equal to sv3 along the constant entropy line as shown in Fig. 2.1 (b). Let the enthalpy of superheated vapour refrigerant at point 4 is h4. 5. The superheated vapour refrigerant leaving the high pressure compressor at point 4 is now passed through the condenser at constant pressure
  • 41. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 41 Pc as shown by a horizontal line 4-5. The condensing process 4-5 changes the state of refrigerant from superheated vapour to saturated liquid. 6. The high pressure saturated liquid refrigerant from the condensor is passed to the intercooler where some of liquid refrigerant evaporates in desuperheating the superheated vapour refrigerant from the low pressure compressor. In order to make up for the liquid evaporated, i.e. to maintain a constant liquid level, an expansion valve E1 which acts as a float valve, is provided. 7. The liquid refrigerant from the intercooler is first expanded in an expansion valve E2 and then evaporated in the evaporator to saturated vapour condition, as shown in Fig. 2.1(b). Let ml = Mass of refrigerant passing through the evaporator (or low pressure compressor) in kg/min, and m2 = Mass of refrigerant passing through the condenser (or high pressure compressor) in kg/min. The high pressure compressor in a given system will compress the mass of refrigerant from low pressure compressor (ml) and the mass of liquid evaporated in the liquid intercooler during cooling or de superheating of superheated vapour refrigerant from low pressure compressor. If m3 is the mass of liquid evaporated in the intercooler, then m3= m2 - ml The value of m2 may be obtained by considering the thermal equilibrium for the liquid intercooler as shown in Fig. 2.2, i.e ., Heat taken by the liquid intercooler = Heat given by the liquid intercooler
  • 42. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 42 Fig2.8. Thermal equilibrium for liquid intercooler. or m2 hf5 + m1 h2 = m1 h6 + m2 h3 ... ( h6 = hf5) and mass of liquid refrigerant evaporated in the intercooler, We know that refrigerating effect, RE = ml (hl - hf) = ml (hl - hf5) = 210 Q kJ/min where Q is the load on the evaporator in tonne of refrigeration. Total work done in both the compressors, W = ml (h2 - h1) + m2 (h4 - h3) Power required to drive the system, and C.O.P. of the system = = Notes: 1. In case of ammonia, when liquid refrigerant is used for intercooling, the total power requirement will decrease. It is due to the fact that the mass of liquid evaporated during intercooling is extremely small because of
  • 43. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 43 its high latent heat of vaporisation and the constant entropy lines of ammonia become very flat in the superheat region. Thus the intercooling by liquid refrigerant is commonly used in multi-stage ammonia plants, because of less power requirement. 2. In case of refrigerant R-12, when liquid refrigerant is used for intercooling, the total power requirements may actually increase. It is due to the fact that the latent heat of vaporisation is small and the constant entropy line of R -12 does not change very much with the temperature. Thus in R-12 systems, the saving in work by performing the compression close to the saturated vapour line does not compensate for the increased mass flow rate through the high stage compressor. Therefore, intercooling by liquid refrigerant in R-12 systems is never employed. Example 2.1. Calculate the power needed to compress 20 kg / min of ammonia from saturated vapour at 1.4 bar to a condensing pressure of 10 bar by two-stage compression with intercooling by liquid refrigerant at 4 bar. Assume saturated liquid to leave the condenser and dry saturted vapours to leave the evaporator. Use the p-h chart. Determine, also, the power needed when intercooling is not employed. Solution. Given : m1 = 20 kg/min ; PE = 1.4 bar ; pC = 10 bar ; p2 = p3 = 4 bar Fig. 2.9 The p-h diagram for a two stage compression with intercooling by liquid refrigerant is shown in Fig. 2.9. The various values for ammonia as read from the
  • 44. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 44 p-h diagram are as follows: Enthalpy of saturated vapour refrigerant entering the low pressure compressor at point 1, h1 = 1400 kJ/kg Entropy of saturated vapour refrigerant entering the low pressure compressor at point 1, Sl = 5.75 kJ/kg K Enthalpy of superheated vapour refrigerant leaving the low pressure compressor at point 2, h2 = 1527 kJ/kg Enthalpy of saturated vapour refrigerant leaving the intercooler or entering the high pressure compressor at point 3, h3 = 1428 kJ/kg Entropy of saturated vapour refrigerant leaving the intercooler or entering the high pressure compressor at point 3, s3 = 5.39 kJ/kg K Enthalpy of superheated vapour refrigerant leaving the high pressure compressor at point 4, h4 = 1550 kJ/kg Enthalpy of saturated liquid refrigerant passing through the condenser at point 5, hf5 = h6 = 284 kJ/kg We know that mass of refrigerant passing through the condenser (or high pressure compressor), m2 = = = 21.73 kg/min Work done in low pressure compressor, WL = m1 (h2 - h1) = 20 (1527 - 1400) = 2540 kJ/min Work done in high pressure compressor, WH = m2 (h4 - h3) = 21.73 (1550 - 1428) = 2651 kJ/min and total work done in both the compressors, W = WL + WH = 2540 + 2651 = 5191 kJ/min Power needed = 5191/60 = 86.5 kW Ans.
  • 45. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 45 Power needed when intercooling is not employed When intercooling is not employed, the compression of refrigerant will follow the path 1-2 in the low pressure compressor and 2-2' in the high pressure compressor. In such a case, Work done in the high pressure compressor, WH = mi (h2'- h2) = 20 (1676- 1527) = 2980 kJ /min ... ( From p-h diagram, h2' = 1676 kJ/kg) and total work done is both the compressors, W = WL + WH = 2540 + 2980 = 5520 kJ/min Power needed = 5520/60 = 92 kW Ans. Example 2.2. Calculate the power needed to compress 20 kg/min of R-12 from saturated vapour at 1.4 bar to a condensing pressure of 10 bar by two-stage compression with intercooling by liquid refrigerant at 4 bar. Assume saturated liquid to leave the condenser and dry saturated vapours to leave the evaporator. Use the p-h chart. Sketch the cycle on a skeleton p-h chart and label the values of enthalpy at salient points. Solution. Given: ml = 20 kg/min; PE = 1.4 bar; Pc = 10 bar; P2 = P3 = 4 bar The p-h diagram for a two-stage compression with intercooling by liquid refrigerant is shown in Fig. 5.4. The various values for R-12 as read from the p-h diagram are as follows :.. Enthalpy of saturated vapour refrigerant entering the low pressure compressor at point 1, h1= 178 kJ/kg Entropy of saturated vapour refrigerant entering the low pressure compressor at point 1, sl = 0.71 kJ/kg K
  • 46. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 46 Fig. 2.10 Enthalpy of superheated vapour refrigerant leaving the low pressure compressor at point 2, h2 = 195 kJ/kg Enthalpy of saturated vapour refrigerant leaving the intercooler or entering the high pressure compressor at point 3, h3=191 kJ/kg Entropy of saturated vapour refrigerant entering the high pressure compressor at point 3, s3 = 0.695 kJ/kg K Enthalpy of superheated vapour refrigerant leaving the high pressure compressor at point 4, h4 = 210 kJ/kg K Enthalpy of saturated liquid refrigerant leaving the condenser at point 5, hf5 = h6 = 77 kJ/kg We know that mass of refrigerant passing through the condenser (or high pressure compressor), m2 = = = 20.7 kg/min Work done in low pressure compressor, WL = ml (h2 - h1) = 20 (195 - 178) = 340 kJ/min Work done in high pressure compressor,
  • 47. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 47 WH= m2 (h4- h3) = 20.7 (210-191) = 393 kJ/min We know that refrigerating effect, RE = m (h1- hf6) = 210 Q kJ/min C.O.P. of the system = = = Example 2.3. The following data refer to a two stage compression ammonia refrigerating system with water intercooler. Condenser pressure = 14 bar; Evaporator pressure = 2 bar; Intercooler pressure = 5 bar; Load on the evaporator = 2 TR If the temperature of the de-superheated vapour and sub-cooled liquid refrigerant are limited to 30o C, find (a) the power required to drive the system, and (b) CO.P. of the system. Solution. Given : Pc = 14 bar ; PE = 2 bar ; P2 = P3 = 5 bar ; Q = 10 TR ; t3 = t6 = 30° C The p-h diagram for a two stage compression system with water intercooler is shown in Fig. 2.11. The various values as read from the p-h diagram for ammonia are as follows: Enthalpy of saturated vapour refrigerant entering the low pressure compressor at point 1. h1 = 1420 kJ/kg Entropy of saturated vapour refrigerant at point 1, sl = 5.6244 kJ/kg K
  • 48. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 48 Fig. 2.11 Enthalpy of superheated vapour refrigerant leaving the water intercooler at point 3. h3= 1510 kJ/kg Entropy of superheated vapour refrigerant at point 3, s3 = 5.424 kJ/kg K Enthalpy of superheated vapour refrigerant leaving the high pressure compressor at point 4. h4 = 1672 kJ/kg Enthalpy of liquid refrigerant leaving the liquid sub-cooler, hf6 = h7 = 323 kJlkg The points 2 and 4 on the p-h diagram are obtained in the similar way as discussed in Art. 5.3. From the p-h diagram, we find that enthalpy of superheated vapour refrigerant at point 2, h2 = 1550 kJ/kg. (a) Power required to drive the system We know that mass of refrigerant circulating through the system, m = = = 1.91kg/min
  • 49. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 49 Total work done in both the compressors, w= m [(h2 - h1) + (h4 - h3)] = 1.91 [(1550- 1420) + (1672 - 1510) = 557.7 kJ/min Power required to drive the system, P = 557.7/60 = 9.3 kW Ans. (b) CO.P. of system We know that refrigerating effect of the system, RE = 210Q = 2l0 × 10 = 2100 kJ/min :. C.O.P. of the system = = = 3.76 Ans. 2.6 Two Stage Compression with Water Intercooler, Liquid Sub-cooler and Liquid Flash Chamber The arrangement of a two stage compression with water intercooler, liquid sub-cooler and liquid flash chamber is shown in Fig. 2.12 (a). The corresponding p-h diagram is shown in Fig. 2.7 (b). The various processes, in this system, are as follows: I. The saturated vapour refrigerant at the evaporator pressure PE is admitted to low pressure compressor at point I. In this compressor, the refrigerant is compressed isentropically from evaporator pressure PE to water intercooler (or flash chamber) pressure PF as shown by the curve 1-2 in Fig. 2.12 (b). 2. The superheated vapour refrigerant leaving the low pressure compressor at point 2 is now passed through the water intercooler at constant pressure PF in order to reduce the degree of superheat (i.e., from temperature t2 to t3 ). The line 2-3 represents the water intercooling or de-superheating process. 3. The superheated vapour refrigerant leaving the water intercooler at point 3 is mixed with the vapour refrigerant supplied by the flash chamber at point 9. The condition of refrigerant after mixing is shown by point 4 which is in
  • 50. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 50 superheated state. Let the temperature at this point is t4. 4. The superheated vapour refrigerant admitted at point 4 to the high pressure compressor is compressed isentropically from the intercooler or flash chamber pressure PF to condenser pressure Pc as shown by the curve 4-5. The temperature rises from t4 to t5. 5. The superheafed vapour leaving the high pressure compressor at pressure Pc is passed through a condenser at constant pressure as shown by a horizontal line 5-6. The condensing process 5-6 changes the state of refrigerant from superheated vapour to saturated liquid. Fig 2.12 6. The saturated liquid refrigerant from the condenser is now cooled in liquid sub-cooler to a temperature, say t7. The line 6-7 represents a sub-cooling
  • 51. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 51 process. 7. The liquid refrigerant leaving the sub-cooler at pressure Pc is expanded in an expansion valve E1 to a pressure equal to the flash chamber pressure PF, as shown by vertical line 7-8. The expanded refrigerant which is a mixture of vapour and liquid refrigerants is admitted to a flash chamber at point 8. The flash chamber separates the vapour and liquid refrigerants at pressure PF. The vapour refrigerant from the flash chamber at point 9 is mixed with the refrigerant from the water intercooler. The liquid refrigerant from the flash chamber at Point 10 is further expanded in an expansion valve E2 as shown by the vertical line 10-11. 8. The liquid refrigerant leaving the expansion valve E2 is evaporated in the evaporator at the evaporator pressure PE (usually 2 bar) as shown by the horizontal line 11-1 in Fig. 2.12 (b). Let m2 = Mass of refrigerant passing through the condenser (or high pressure compressor), and m3 = Mass of vapour refrigerant formed in the flash chamber. :. Mass of refrigerant passing through the evaporator (or low pressure compressor), ml = m2-m3 If Q tonne of refrigeration is the load on the evaporator, then the mass of refrigerant passing through the evaporator, m1= kg/min ...( h11 = hf10) Now let us consider the thermal equilibrium of the flash chamber. Since the flash chamber is an insulated vessel, therefore there is no heat exchange between the flash chamber and atmosphere. In other words, the heat taken and given by the flash chamber are same. Mathematically, Heat taken by the flash chamber = Heat given by the flash chamber m2 h8 = m3 h9 + ml hf10
  • 52. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 52 = m3 h9 + (m2- m3) hf10 ...( m1 = m2 - m2) m2 (h8 - hf10) = m3 (h9 - hf10) m3 = m2 ...( h8 = hf7) ... (i) The vapour refrigerant from the water intercooler (represented by point 3) is mixed with vapour refrigerant m3 from the flash chamber (represented by point 9) at the same pressure before entering the high pressure compressor. The enthalpy of the mixed refrigerant (represented by point 4) may be calculated by using the equation, m2 h4 = m3 h9 + ml h3 = m3 h9 + (m2 - m3) h3 We know that refrigerating effect of the system, RE = ml (h1 - hll) = 210 Q kJ/min Work done in low pressure compressor, WL = ml (h2 - hl) Work done in high pressure compressor, WH = m2 (h5 - h4) Total workdone in both the compressors, W = WL + WH = m1 ( (h2 - hl) + m2 (h5 - h4) Power required to drive the system, p = kW and C.O.P. of the system = = = Note: Since the mass of vapour refrigerant m, is cooled in the water intercooler from condition 2 to 3, therefore cooling capacity of the intercooler
  • 53. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 53 = ml (h2 -h3) Example 2.4. A two stage compression ammonia refrigeration system operates between overall pressure limits of 14 bar and 2 bar. the temperature of the desuperheated vapour and subcooled liquid refrigerant are limited to 30o C. The flash tank separates dry vapour at 5 bar pressure and the liquid refrigerant then expands to 2 bar. Estimate the C.O.P. of the machine and power required to drive the compresor, if the mechanical effficiency of the drive is 80% and load on the evaporator is 10 TR. Solution. Given :Pc = 14 bar; PE = 2 bar; PF = 5 bar; t3 = t7 = 30o C ; m = 80% = 0.8 ; Q = 10TR The p-h diagram for a two-stage compression system with given conditions is shown in Fig. 5.8. The values as read from p-h diagram for ammonia, are as follows: Fig 2.13 Enthalpy of saturated vapour refrigerant entering the low pressure compressor at point 1. hi = 1420 kJ/kg Entropy of saturated vapour refrigerant entering the low pressure
  • 54. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 54 compressor at point 1. s1 = 5.6244 kJ/kg K Enthalpy of superheated vapour refrigerant leaving the low pressure compressor at point 2, h2 =1550 kJ/kg Enthalpy of superheated vapour refrigerant leaving the water intercooler at point 3. h3 = 1510 kJ/kg Enthalpy of saturated vapour refrigerant leaving the flash tank at point 9, h9 = 1432 kJ/kg Enthalpy of liquid refrigerant leaving the subcooler at point 7, hf7 = h8 = 323 kJ/kg Enthalpy of saturated liquid refrigerant leaving the second expansion valve at point 10, hf10 = hll = 198 kJ/kg Let m2 = Mass of refrigerant passing through the condenser. We know that mass of the vapour refrigerant formed in the flash tank, m3 = m2 = m2 = 0.l m2 ... (i) and mass of refrigerant passing through the evaporator, ml = m2 - m3 = = = 1.72 kg/min ... (ii) From equations (i) and (ii), m2 - 0.1 m2 = 1.72 or m2 = 1.9 kg/min m3 = 0.l m2 = 0.l × 1.9 = 0.19 kg/min The desuperheated vapour refrigerant (m2 - m3) as represented by point 3 is mixed with the vapour refrigerant from the flash tank as represented by point 9. The enthalpy of the mixed refrigerant entering the high pressure compressor as represented by point 4 is given by m2 h4 = m3 h9 + (m2 - m3) h3 = 0.1 m2 h9 + (m2 - 0.1 m2) h3 ... [From equation (i)]
  • 55. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 55 or h4 = 0.1 × h9 + 0.9 × h3 = 0.1 × 1432+ 0.9 × 1510 = 1502 kJ/kg We see from p-h diagram that at point 4 (intersection of pressure 5 bar and enthalpy 1502 kJ/kg), the entropy is s4 = 5.51 kJ/kg K. Now from point 4, draw a line of entropy equal to 5.51 kJ/kg K along the constant entropy line which intersects the condenser pressure (14 bar) line at point 5. Thus, the point 5 is located. From p-h diagram, we find that enthalpy of refrigerant leaving the high pressure compressor at point 5 is h5= 1650 kJ/kg C.O.P. of the machine We know that refrigerating effect = ml (hl - hf10) = 210 Q = 210 × 10 = 2100 kJ/min Work done in both the compressors = ml (h2 - h1) + m2 (h5 - h4) = 1.72 (1550 - 1420) + 1.9 (1650- 1502) = 223.6 + 281.2 = 504.8 kJ/min Since the mechanical efficiency of the drive is 80%, therefore actual work done in both the compressors = 504.8/0.8 = 631 kJ/min Actual C.O.P = = = 3.32 Ans. Power required to drive the compressors We know that power required to drive the compressors = = = 10.5 kW Ans. 2.7 Two Stage Compression with Water Intercooler, Liquid Sub-cooler and Flash Intercooler A two stage compression with water intercooler, liquid sub-cooler and flash intercooler is shown in Fig. 2.9 (a). The corresponding p-h diagram is shown
  • 56. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 56 in Fig. 5.9 (b). Fig 2.14 We have seen in the previous article that when the vapour refrigerant from the low pressure compressor is passed through the water intercooler, its temperature does not reduce to the saturated vapour line or even very near to it, before admitting it to the high pressure compressor [ Refer point 4 of Fig. 2.12 (b)]. In fact, with water cooling there may be no saving of work in compression. But the improvement in performance and the reduction in compression work may be achieved by using a flash chamber as an intercooler as well as flash separator, as shown in Fig. 2.14 (a). The corresponding p-h diagram is shown in Fig. 2.14 (b). The various processes, in this system, are as follows: 1. The saturated vapour refrigerant at the evaporator pressure PE is admitted to the low pressure compressor at point 1. In this compressor, the refrigerant is compressed isentropically from evaporator pressure PE to the flash
  • 57. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 57 intercooler pressure PF,as shown by the curve 1-2 in Fig2.14 (b). 2. The superheated vapour refrigerant leaving the low pressure compressor at point 2 is now passed through the water intercooler at constant pressure PF, in order to reduce the degree of superheat ( i.e. from temperature t2 to t3). The line 2-3 represents the water intercooling or desuperheating process. 3. The superheated vapour refrigerant leaving the water intercooler at point 3 is passed through a flash intercooler which cools the superheated vapour refrigerant to saturated vapour refrigerant as shown by the line 3-4. The cooling of superheated vapour refrigerant is done by the evaporation of a part of the liquid refrigerant from the flash intercooler placed at point 8. 4. The saturated vapour refrigerant leaving the flash intercooler enters the high pressure compressor at point 4 where it is compressed isentropically from flash intercooler pressure PF to condenser pressure PC, as shown by the curve 4-5. 5. The superheated vapour refrigerant leaving the high pressure compressor at pressure Pc is passed through a condenser at constant pressure. The condensing process as shown by line 5-6 changes the state of refrigerant from superheated vapour to saturated liquid. 6. The saturated liquid refrigerant leaving the condenser at point 6 is now cooled at constant pressure Pc in the liquid sub-cooler to a temperature t7 as shown in Fig. 2.9 (b). The line 6-7 shows the sub-cooling process. 7. The liquid refrigerant leaving the sub-cooler at point 7 is expanded in an expansion valve E1 to a pressure equal to the flash intercooler pressure PF , as shown by the vertical line 7-8. The expanded refrigerant (which is a mixture of vapour and liquid refrigerant) is admitted to flash intercooler at point 8 which also acts as a flash separator. 8. The liquid refrigerant leaving the flash intercooler at point 9 is passed through the second expansion valve E2 (process 9-10) and then evaporated in the evaporator as shown by the horizontal line 10-1. Let m1 = Mass of the refrigerant passing through the evaporator or low
  • 58. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 58 pressure compressor), and m2 = Mass of the refrigerant passing through the condenser (or high pressure compressor). If Q tonne of refrigeration is the load on the evaporator, then the mass of refrigerant passing through the evaporator is given by, m1 = = kg/min ... ( h10 = hf9) Now for the thermal equilibrium of the flash intercooler, Heat taken by the flash intercooler = Heat given by the flash intercooler m2 h8 + m1 h3 = m2 h4 + m1 hf9 m1(h3 - hf9) = m2 (h4 -h8) m2 = m1 = m1 kg/min We know that refrigerating effect, RE = ml (hl - h10) = ml (hl - hf9) = 210 Q kJ/min and work done in both the compressors, W = Work done in L.P. compressor + Work done in H.P. compressor = ml (h2 - hl) + m2 (h5- h4) Power required to drive the system, kW and coefficient of performance of the system, C.O.P. = = = MODULE – III Vapour Absorption Refrigeration Systems
  • 59. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 59 3.1 Introduction The vapour absorption refrigeration system is one of the oldest method of producing refrigerating effect. The principle of vapour absorption was first discovered by Michael Faraday in 1824 while performing a set of experiments to liquify certain gases. The first vapour absorption refrigeration machine was developed by a French scientist Ferdinand Carre in 1860. This system may be used in both the domestic and large industrial refrigerating plants. The refrigerant, commonly used in a vapour absorption system, is ammonia. The vapour absorption system uses heat energy, instead of mechanical energy as in vapour compression systems, in order to change the conditions of the refrigerant required for the operation of the refrigeration cycle. We have discussed in the previous chapters that the function of a compressor, in a vapour compression system, is to withdraw the vapour refrigerant from the evaporator. It then raises its temperature and pressure higher than the cooling agent in the condenser so that the higher pressure vapours can reject heat in the condenser. The liquid refrigerant leaving the condenser is now ready to expand to the evaporator conditions again. In the vapour absorption system, the compressor is replaced by an absorber, a pump, a generator and a pressure reducing valve. These components in vapour absorption system perform the same function as that of a compressor in vapour compression system. In this system, the vapour refrigerant from the evaporator is drawn into an absorber where it is absorbed by the weak solution of the refrigerant forming a strong solution. This strong solution is pumped to the generator where it is heated by some external source. During the heating process, the vapour refrigerant is driven off by the solution and enters into the condenser where it is liquefied. The liquid refrigerant then flows into the evaporator and thus the cycle is completed. 3.2 Simple Vapour Absorption System - The simple vapour absorption system, as shown in Fig. 3.1, consists of an absorber, a pump, a generator and a pressure reducing valve to replace the
  • 60. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 60 compressor of vapour compression system. The other components of the system are condenser, receiver, expansion valve and evaporator as in the vapour compression system. Fig 3.1 Simple vapour absorbtion system . In this system, the low pressure ammonia vapour leaving the evaporator enters the absorber where it is absorbed by the cold water in the absorber. The water has the ability to absorb very large quantities of ammonia vapour and the solution thus formed, is known as aqua-ammonia. The absorption of ammonia vapour in water lowers the pressure in the absorber which in turn draws more ammonia vapour from the evaporator and thus raises the temperature of solution. Some form of cooling arrangement (usually water cooling) is employed in the absorber to remove the heat of solution evolved there. This is necessary in order to increase the absorption capacity of water, because at higher temperature water absorbs less ammonia vapour. The strong solution thus formed in the absorber is pumped to the generator by the liquid pump. The pump increases the pressure of the solution upto 10 bar. The *strong solution of ammonia in the generator is heated by some external source such as gas or steam. During the heating process, the ammonia
  • 61. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 61 vapour is driven off the solution at high pressure leaving behind the hot weak ammonia solution in the generator. This weak ammonia solution flows back to the absorber at low pressure after passing through the pressure reducing valve. The high pressure ammonia vapour from the generator is condensed in the condenser to a high pressure liquid ammonia. This liquid ammonia is passed to the expansion valve through the receiver and then to the evaporator. This completes the simple vapour absorption cycle. 3.3 actical Vapour Absorption System The simple absorption system as discussed in the previous article is not very economical. In order to make the system more practical, it is fitted with an analyser, a rectifier and two heat exchangers as shown in Fig. 3.2. These accessories help to improve the performance and working of the plant, as discussed below :- Fig. 3.2. Practical vapour absorption system. 1. Analyser. When ammonia is vaporised in the generator, some water is
  • 62. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 62 also vaporised and will flow into the condenser along with the ammonia vapours in the simple system. If these unwanted water particles are not removed before entering into the condenser, they will enter into the expansion valve where they freeze and choke the pipe line. In order to remove these unwanted particles flowing to the condenser, an analyser is used. The analyser may be built as an integral part of the generator or made as a separate piece of equipment. It consists of a series of trays mounted above the generator. The strong solution from the absorber and the aqua from the rectifier are introduced at the top of the analyser and flow downward over the trays and into the generator. In this way, considerable liquid surface area is exposed to the vapour rising from the generator. The vapour is cooled and most of the water vapour condenses, so that mainly ammonia vapour leaves the top of the analyser. Since the aqua is heated by the vapour, less external heat is required in the generator. 2. Rectifier. In case the water vapours are not completely removed in the analyser, a closed type vapour cooler called rectifier (also known as dehydrator) is used. It is generally water cooled and may be of the double pipe, shell and coil or shell and tube type. Its function is to cool further the ammonia vapours leaving the analyser so that the remaining water vapours are condensed. Thus, only dry or anhydrous ammonia vapours flow to the condenser. The condensate from the rectifier is returned to the top of the analyser by a drip return pipe. 3. Heat exchangers. The heat exchanger provided between the pump and the generator is used to cool the weak hot solution returning from the generator to the absorber. The heat removed from the weak solution raises the temperature of the strong solution leaving the pump and going to analyser and generator. This operation reduces the heat supplied to the generator and the amount of cooling required for the absorber. Thus the economy of the plant increases. The heat exchanger provided between the condenser and the evaporator may also be called liquid sub-cooler. In this heat exchanger, the liquid refrigerant leaving the condenser is sub- cooled by the low temperature ammonia vapour from the evaporator as shown in Fig. 7.2. This sub-cooled liquid is now passed to the expansion valve and then to the evaporator. In this system, the net refrigerating effect is the heat absorbed by the
  • 63. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 63 refrigerant in the evaporator. The total energy supplied to the system is the sum of work done by the pump and the heat supplied in the generator. Therefore, the coefficient of performance of the system is given by C.O.P. = 3.4 Advantages of Vapour Absorption Refrigeration System over Vapour Compression Refrigeration System Following are the advantages of vapour absorption system over vapour compression system: 1. In the vapour absorption system, the only moving part of the entire system is a pump which has a small motor. Thus, the operation of this system is essentially quiet and is subjected to little wear. The vapour compression system of the same capacity has more wear, tear and noise due to moving parts of the compressor. 2. The vapour absorption system uses heat energy to change the condition of the refrigerant from the evaporator. The vapour compression system uses mechanical energy to change the condition of the refrigerant from the evaporator. 3. The vapour absorption systems are usually designed to use steam, either at high pressure or low pressure. The exhaust steam from furnaces and solar energy may also be used. Thus this system can be used where the electric power is difficult to obtain or is very expensive. 4. The vapour absorption systems can operate at reduced evaporator pressure and temperature by increasing the steam pressure to the generator, with little decrease in capacity. But the capacity of vapour compression system drops rapidly with lowered evaporator pressure. 5. The load variations does not effect the performance of a vapour absorption system. The load variations are met by controlling the quantity of aqua circulated and the quantity of steam supplied to the generator. The performance of a vapour compression system at partial loads is
  • 64. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 64 poor. 6. In the vapour absorption system, the liquid refrigerant leaving the evaporator has no bad effect on the system except that of reducing the refrigerating effect. In the vapour compression system, it is essential to superheat the vapour refrigerant leaving the evaporator so that no liquid may enter the compressor. 7. The vapour absorption systems can be built in capacities well above 1000 tonnes of refrigeration each which is the largest size for single compressor units. 8. The space requirements and automatic control requirements favour the absorption system more and more as the desired evaporator temperature drops. 3.5 Coefficient of Performance of an Ideal Vapour Absorption Refrigeration System We have discussed earlier that in an ideal vapour absorption refrigeration system, (a) the heat (QG) is given to the refrigerant in the generator, (b) the heat (Qc) is discharged to the atmosphere or cooling water from the condenser and absorber. (c) the heat (QE) is absorbed by the refrigerant in the evaporator, and (d) the heat (Qp) is added to the refrigerant due to pump work. Neglecting the heat due to pump work (Qp), we have according to First Law of Thermodynamics, ... (i) Let TG = Temperature at which heat (QG) is given to the generator, Tc = Temperature at which heat (QC) is discharged to atmosphere or
  • 65. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 65 cooling water from the condenser and absorber, and TE = Temperature at which heat (QE) is absorbed in the evaporator. Since the vapour absorption system can be considered as a perfectly reversible system, therefore the initial entropy of the system must be equal to the entropy of the system after the change in its condition. ...(ii) = ... [ From equation (i) ] or QG = = = ... (iii) Maximum coefficient of performance of the system is given by (C.O.P.)max = = = ...(iv) It may noted that, I. The expression is the C.O.P. of a Carnot refrigerator working between the temperature limits of T E and T C.
  • 66. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 66 2. The expression is the efficiency of a Carnot engine working between the temperature limits of TG and TC. Thus an ideal vapour absorption refrigeration system may be regarded as a combination of a Carnot engine and a Carnot refrigerator. The maximum C.O.P. may be written as (C.O.P.)max = (C.O.P)carnot × carnot In case the heat is discharged at different temperatures in condenser and absorber, then (C.O.P.)max = where TA= Temperature at which heat is discharged in the absorber. Example 3.1. In a vapour absorption refrigeration system, heating, cooling and refrigeration takes place at the temperatures of 100o C, 20o C and _5° C respectively. Find the maximum C.O.P. of the system. Solution. Given: TG = 100o C = 100 + 273 = 373 K ; Tc = 20o C = 20 + 273 = 293 K ; T E = - 5° C = - 5 + 273 = 268 K We know that maximum C.O.P. of the system = = 2.3 Ans. Example 3.2. In an absorption type refrigerator, the heat is supplied to NH3 generator by condensing steam at 2 bar and 90% dry. The temperature in the refrigerator is to be maintained at - 5° C. Find the maximum C.O.P. possible. If the refrigeration load is 20 tonnes and actual C.O.P. is 10% of the maximum C.O.P ., find the mass of steam required per hour. Take temperature of the atmosphere as 30o C. Solution. Given: p = 2 bar; x = 90% = 0.9; TE = -5°C = -5 + 273 = 268 K ; Q = 20 TR; Actual C.O.P. = 70% of maximum C.O.P. ; TC = 30° C = 30 + 273 = 303 K Maximum C.O.P.
  • 67. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 67 From steam tables, we find that the saturation temperature of steam at a pressure of 2 bar is TG = 120.2° C = 120.2 + 273 = 393.2 K We know that maximum C.O.P. = = = 1.756 Ans. Mass of steam required per hour We know that actual C.O.P. = 70% of maximum C.O.P. = 0.7 × 1.756 = 1.229 Actual heat supplied = = = 3417.4 kJ/min Assuming that only latent heat of steam is used for heating purposes, therefore from steam tables, the latent heat of steam at 2 bar is hfg = 2201.6 kJ/kg Mass of steam required per hour = = = 1.552 kg/min= 93.12 kg/h Ans. 3.6 Domestic Electrolux (Ammonia Hydrogen) Refrigerator The domestic absorption type refrigerator was invented by two Swedish engineers Carl Munters and Baltzer Von Platan in 1925 while they were studying for their under-graduate course of Royal Institute of Technology in Stockholm. The idea was first developed by the ‘Electrolux Company’ of Luton, England.
  • 68. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 68 Fig. 3.3. Domestic electrolux type refrigerator. This type of refrigerator is also called three-fluids absorption system. The main purpose of this system is to eliminate the pump so that in the absence of moving parts. the machine becomes noise-less. The three fluids used in this sytem are ammonia, hydrogen and water. The ammonia is used as a refrigerant because it possesses most of the desirable properties. It is toxic. but due to absence of moving parts, there is very little changes for the leakage and the total amount of refrigeration used is small. The hydrogen being the lightest gas. is used to increase the rate of evaporation of the liquid ammonia passing through the evaporator. The hydrogen is also non-corrosive and insoluble in water. This is used in the low-pressure side of the system. The water is used as a solvent because it has the ability to absorb ammonia readily. The principle of operation of a domestic electolux type refrigerator. as shown in Fig. 3.3. is discussed below: The strong ammonia solution from the absorber through heat exchanger is heated in the generator by applying heat from an external source usually a gas burner. During this heating process, ammonia vapours are removed from the solution and passed to the condenser. A rectifirer or a water separator fitted before the condenser removes water vapour carried with the ammonia vapours. so that dry ammonia vapours are supplied to the condenser. These water vapours,. f not removed, they will enter into the evaporator causing freezing and choking of the machine . The hot weak solution left behind in the generator flow
  • 69. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 69 to the absorber through the heat excl anger. This hot weak solution while passing through the exchanger is cooled. The heat removed by the weak solution is utilised in raising the temperature of strong solution passing through the heat exchanger. In this way, the absorption is accelerated and the improvement in the performance of a plant is achieved. The ammonia vapours in the condenser are condensed by using external cooling source. The liquid refrigerant leaving the condenser flows under gravity to the evaporator where it meets the hydrogen gas. The hydrogen gas which is being fed to the evaporator permit the liquid ammonia to evaporate at a low pressure and temperature according to Dalton's principle. During the process of evaporation, the ammonia absorbs latent heat from the refrigerated space and thus produces cooling effect. The mixture of ammonia vapour and hydrogen is passed to the absorber where ammonia is absorbed in water while the hydrogen rises to the top and flows hack to the evaporator. This completes the cycle. The coefficient of performance of this refrigerator is given by : C.O.P. = Notes: 1. The hydrogen gas only circulates from the absorber to the evaporator and back. 2. The whole cycle is carried out entirely by gravity flow of the refrigerant. 3. It can not be used for industrial purposes as the C.O.P. of the system is very low. 3.7 Lithium Bromide Absorption Refrigeration System The lithium-bromide absorption refrigeration system uses a solution of lithium bromide in water. In this system, the *water is being used as a refrigerant whereas lithium bromide, which is a highly hydroscopic salt, as an absorbent. The lithium bromide solution has a strong affinity for water vapour because of its very low vapour pressure. Since lithium bromide solution is corrosive, therefore inhibitors should be added in order to protect the metal parts of the system against corrosion. Lithium chromate is often uged as a corrosion inhibitor. This
  • 70. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 70 system is very popular for air conditioning in which low refrigeration temperatures (not below 0° C)** are required. Fig. 3.4 shows a lithium bromide vapour absorption system. In this system. the absorber and the evaporator are placed in one shell which operates at the same low pressure of the system. The generator and condenser are placed in another shell which operates at the same high pressure of the system. The principle of operation of this system is discussed below : The water for air-conditioning coils or process requirements is chilled as it is pumped through the chilled-water tubes in.the evaporator by giving up heat to the refrigerant water sprayed over the tubes. Since the pressure inside the evaporator is maintained very low, therefore, the refrigerant waterevaporates. The water vapours thus formed will be absorbed by the strong lithium-bromide solution which is sprayed in the absorber. In absorbing the water vapour, the lithium bromide solution helps in maintaining very low pressure (high vacuum) needed in the evaporator, and the solution becomes weak. This weak solution is pumped by a pump to the generator where it is heated up by using steam or hot water in the heating coils. A portion of water is evaporated by the heat and the solution now becomes more strong. This strong solution is passed through the heat exchanger and then sprayed in the absorber as discussed above. The weak solution of lithium bromide from the absorber to the generator is also passed through the heat exchanger. This weak solution gets heat from the strong solution in the heat exchanger, thus reducing the quantity of steam required to heat the weak solution in the generator.
  • 71. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 71 Fig. 3.4. Lithium-Bromide absorption refrigeration system. The refrigerant water vapours formed in the generator due to heating of solution are passed to the condenser where they are cooled and condensed by the cooling water flowing through the condenser water tubes. The cooling water for condensing is pumped from the cooling water pond or tower. This cooling water first enters the absorber where it takes away the heat of condensation and dilution. The condensate from the condenser is supplied to the evaporator to compensate the water vapour formed in the evaporator. The pressure reducing valve reduces the pressure of condensate from the condenser pressure to the evaporator pressure. The cooled water from the evaporator is pumped and sprayed in the evaporator in order to cool the water for air conditioning flowing through the chilled tubes. This completes the cycle. Note: The pressure difference between the generator and the absorber and the gravity due to the height difference of the two shells is utilised to create the pressure for the spray.
  • 72. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 72
  • 73. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 73 3.8 Refrigerats The refrigerant is a heat carrying medium which during their cycle (i.e. compression, condensation, expansion and evaporation) in the refrigeration system absorb heat from a low temperature system and discard the heat so absorbed to a higher temperature system. The natural ice and a mixture of ice and salt were the first refrigerants. In 1834, ether, ammonia, sulphur dioxide, methyl chloride and carbon dioxide came into use as refrigerants in compression cycle refrigeration machines. Most of the early refrigerant materials have been discarded for safety reasons or for lack of chemical or thermal stability. In the present days, many new refrigerants including halo-carbon compounds, hydro-carbon compounds are used for air conditioning and refrigeration applications. The suitability of a refrigerant for a certain application is determined by its physical, thermodynamic, chemical properties and by various practical factors. There is no one refrigerant which can be used for all types of applications i.e., there is no ideal refrigerant. If one refrigerant has certain good advantages, it will have some disadvantages also. Hence, a refrigerant is chosen which has greater advantages and less disadvantages. 3.9 Desirable Properties of an Ideal Refrigerant We have discussed above that there is no ideal refrigerant. A refrigerant is said to be ideal if it has all of the following properties: I. Low boiling point, 2. High critical temperature, 3. High latent heat of vaporisation. 4. Low specific heat of liquid. 5. Low specific volume of vapour, 6. Non-corrosive to metal, 7. Non-flammable and non-explosive,
  • 74. Viswajyothi College Of Engineering And Technology, Vazhakulam AKHIL K T, Assistant Professor, VJCET Page 74 8. Non-toxic, 9. Low cost, 10. Easy to liquify at moderate pressure and temperature, 11. Easy of locating leaks by odour or suitable indicator, and 12. Mixes well with oil. The standard comparison of refrigerants, as used in the refrigeration industry, is based on an evaporating temperature of - 15° C and a condensing temperature of + 30° C. 3.10 Classification of Refrigerants The refrigerants may, broadly, be classified into the following two groups: 1. Primary refrigerants, and 2. Secondary refrigerants. The refrigerants which directly take part in the refrigeration system are called primary refrigerants whereas the refrigerants which are first cooled by primary refrigerants and then used for cooling purposes are known as secondary refrigerants. The primary refrigerants are further classified into the following four groups: I. Halo-carbon refrigerants, 2. Azeotrope refrigerants, 3. Inorganic refrigerants. and 4. Hydro-carbon refrigerants. These above mentioned refrigerants are discussed, in detail, in the following pages. 8.4 Halo-carbon Refrigerants The American Society of Heating, Refrigeration and Air-conditioning Engineers (ASHARAE) identifies 42 halo-carbon compounds as refrigerants, but only a few of them are commonly used. The following table gives some of the commonly used halo-carbon refrigerants: