IRJET- Enhancement of COP of Vapor Compression Refrigeration Cycle using CFD

International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 07 Issue: 01 | Jan 2020 www.irjet.net p-ISSN: 2395-0072
© 2020, IRJET | Impact Factor value: 7.34 | ISO 9001:2008 Certified Journal | Page 304
Enhancement of COP of Vapor Compression Refrigeration Cycle using CFD
Anjappa S B1, Y. Reddy prasad2
1PG Student, Department of Mechanical Engineering.
2Assistant professorDepartment of Mechanical Engineering.
Sir Vishveshwaraiah Institute of Science & Technology Madanapalle, Andhra Pradesh. India
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Abstract: - This Abstract presents the study on Vapor Compression Refrigeration system using diffuser to improve Coefficient of
Performance. To Improve the Coefficient of Performance, it is to require that Compressor work should decrease. The purpose of a
compressor in vapor compression system is to elevate the pressure of the refrigerant (refrigerant used is TETRA FLURO ETHANE),
but refrigerant leaves the compressor with comparatively high velocity which may cause splashing of liquid refrigerant in the
condenser, liquid hump and damage to condenser by erosion. It is needed to convert this kinetic energy to pressure energy, for
which diffuser can be used. The diffuser of increasing cross-sectional area profile was designed, fabricated and introduced
between Compressor and Condenser. By doing so, the power input to the compressor is reduced, thereby enhancing COP. The size
of diffuser selected was of 15 degree divergence angle. After result analysis, COP was enhanced from 3.83 to 5.55. With the
addition of diffuser Coefficient of Performance (COP) enhancement has been increased by 31% when compared without diffuser.
The Experimental results obtained are validated using CFD; Modeling and meshing will be done in ICEMCFD, analysis in CFX and
post results in CFD POST.
Keywords: ICEMCFD, Diffuser, COP, CFX, R134A,
I. INTRODUCTION
In thermodynamics Refrigeration is the major application area, in which the heat is transferred from a lower temperature part
to a higher temperature part. The devices which develop refrigeration are known as Refrigerators. The cycle on which it
operates are known as refrigeration cycles. There are many types of Refrigeration like Vapor compression refrigeration it is
the most commonly used refrigeration, cascade refrigeration and thermo electric refrigeration.
1.1. Refrigerators and Heat Pumps
It is known heat always flows from higher temperature medium to a lower temperature medium without any aid of devices
heat transfer occurs itself in nature. Reverse process, will not happen by itself. Special devices where the heat transfers from a
lower temperature medium to higher temperature medium are called Refrigerators.
The working fluids which are used in the cycles of refrigeration are called refrigerants, and the refrigerators are cyclic devices.
The schematic representation of a Refrigerator is displayed refer Fig 1.1a Where QL magnitude of heat removed at temperature
TL, QH is magnitude of heat rejected at temperature TH to the surrounding space, the refrigerator net work input is Wnet, in and
QH and QL represents magnitudes and they are positive quantities.
Heat pump transfers heat from a lower temperature medium to a higher temperature medium. Heat pumps, Refrigerators are
basically the same devices; but are dissimilar in aim. Main aim of refrigerator is to keep the refrigerated space at a very less
temperature by extracting the heat generated. Essential part of the operation, not the intent is discharging the heat to a higher-
temperature medium. Aim of the heat pump is maintaining the space that is heated at a high temperature. It’s capable of
riveting heat from small-temperature medium. (Fig1.1b)
Functioning of heat pumps, refrigerators is shown by the coefficient of performance (COP) -
COPR= = =
COPHP = =
Above relations can be shown by exchanging Qh, QL by ̇ ̇ ̇ . Both COPHP , COPR might be more than one. The
similarity of equations 1-1, 1-2 shows

COPHP = COPR+1
Since COPR is a positive quantity. For set values of Ql, Qh, it gives us COPhp >1. That is, by supplying more energy to the house as
it consumes work of heat pump, least, like electric heater. In reality, however, because of piping and many more devices bit of
Qh will be lost to surroundings, COPHP may drop below unity. When it happens, normally system changes to fuel.
Rate of heat removal from the refrigerated space is shown by (TOR) Tons of refrigeration. Refrigeration system capacity will
freeze 1ton (2000lbm) of liquefied water at (32 ) into ice at 0 in 24h is told to be 1 ton. 1 ton of refrigeration is equal to
200 Btu/min or 211kj/min. The typical 200-m2 residence cooling load is 3-ton (120-kW) range.
Fig 1.1 a) Refrigerator Fig 1.1 b) Heat pump
II. OBJECTIVE OF THE PRESENT WORK
• To Increase the COP of Vapor compression refrigeration cycle
• To decrease the compressor work
• To increase the pressure of the refrigerant entering the condenser
III. METHODOLOGY
The schematic diagram of the vapor compression refrigeration system with diffuser at condenser inlet the system consists of
two flow lines one is simple VCR flow line without diffuser and other is flow line with diffuser. Thus we can calculate the
pressure with and without diffuser. P-h diagram has been shown in figure 3.1. Figure 3.1 shows the pressure enthalpy chart of
the system. The path 1-2-3-4-5-6-7 shows the VCR cycle with diffuser and path 1-2’-3’-4’-5-6-7 shows the VCR cycle without
diffuser at condenser inlet.
Fig.2. VCR system with diffuser
Fig 3. (P-H Diagram)
h4 hf = Enthalpy at condenser pressure
h5 h4 cp (T4 T5)
h7 hg = Enthalpyat evaporator pressure

h1 h7 cp (T1 T7)
h3 is hg at condenser temperature
h2 h3 cp (T2 T4)
COP = = (h7 h6)/ (h2 h1)
Mr Refrigeration capacity/refrigeration effect in kg/s
Refrigeration capacity=m (h1-h6)
Q Mr Cp dt Watts
Compressor power=m (h2-h1) kW
Result taken from refrigeration kit is validated through CFD
Further Diffuser has been used to decrease power input to the compressor which will enhance COP
DIFFUSER LENGTH = (D1-D2)/tanθ
Different D/L ratio of 0.5 and 0.6 for divergence angle of 150 has been carried so as to decrease power input of the compressor
which will result in increasing COP
Modeling and meshing done in ICEM-CFD, analysis in CFX and post result in CFD POST
The above chapter gives an overview on the methodology of the experiment.
IV. EXPERIMENTAL SETUP
In this chapter we discuss about the experimental setup and calculations of the experiment conducted.
Fig 4. Experimental set up (Refrigeration Test Rig)
4.1 Specifications
Refrigerator Capacity = 220 lts.
Pipe Diameter of the Evaporator =11 mm = 0.011 m
Length of the evaporator coil = 1539 mm = 1.539 m
4.2 Observations
T1 = Inlet Temperature at Compressor, (oC)
T2 = Outlet Temperature at Compressor, (oC)

T3 = Outlet Temperature at Condenser, (oC)
T4 = Inlet Temperature at Evaporator, (oC)
P1 = evaporator pressure (Kg/cm2)
P2 = condenser pressure (lb/in2)
Refrigerant used - R134A (Tetra Fluoro Ethane)
Compressor Suction pressure = Low Pressure
Compressor Discharge pressure = High Pressure
Compressor Suction temperature = Evaporator Outlet temperature
Compressor Discharge temperature = Compressor Outlet Temperature
Temperature Evaporator = Temperature at Evaporator Inlet
After condensation temperature = Temperature at Condenser Outlet
TABLE I
4.3Tabular Column:
Values of Enthalpy, Saturation temperature were taken from the table for R134A.
Fig.5. Actual Vapour Compression Cycle
4. 3.1. EXPERIMENT TRAIL 1. (Without Diffuser)
For P1 = 137.89 KPa saturation temp. is ts1 = -19.15oC. But observed temperature is 8oC. Therefore the condition of the
refrigerant before compression is Superheat.
H1 = hg1 + cp (tsup-ts1) = 238.98 + 0.958 (8-(-19.15))
= 264.99 KJ/Kg
Final pressure = P2 = 770KPa
Temperature is 75oC.
For P2 = 770 KPa saturation temp. is ts2 = 29.92oC.
Experiments
T1 T2 T3 T4 H.P (P2) L.P. (P1)
Kg/cm2 KPa lb/in2 KPa
1 8 75 29.92 -6.2 7.92 750 20 137.9
2 4.7 77.1 25.61 -12.6 7.6 720.5 12.5 86.18
3 3.5 77.9 25 -13.8 7.5 710.8 10 68.95

The condition of the refrigerant after compression is Superheat.
H2 = hg2 + cp (tsup-ts2) = 266.64 + 0.958 (75- 29.92)
= 309.83 KJ/Kg
H6 = h5 = 93.155 KJ/Kg
Coefficient of Performance
C.O.P. = Refrigeration Effect / Work Done
= (H1 – H4) / (H2 – H1)
= (264.99 – 93.155) / (309.83 – 264.99)
= 3.83
4.3.2. TRAIL 2. (Without Diffuser)
For P1 = 86.18 KPa saturation temp. is ts1 = -29.69oC. But observed temperature is 4.7oC. Therefore the condition of the
H1 = hg1 + cp (tsup-ts1) = 232.4 + 0.958 (4.7-(-29.69))
= 265.35 KJ/Kg
For P2 = 720.5 KPa saturation temp. is ts2 = 28oC. But observed temperature is 61.2oC. Therefore the condition of the
refrigerant after compression is Superheat.
H2 = hg2 + cp (tsup-ts2) = 265.232 + 0.958 (77.1- 28) = 312.26 KJ/Kg
h3 = h4 = 89.202 KJ/Kg
Coefficient of Performance = C.O.P. = Refrigeration Effect / Work done= (H1 – h4) / (H2 – H1)
= (265.35 – 89.202) / (312.26 – 265.35)
= 3.75
4.3.3. TRAIL 3. (Without Diffuser)
For P1 = 68.95 KPa saturation temp. is ts1 = -34.29oC. But observed temperature is 3.5oC. Therefore the condition of the
H1 = hg1 + cp (tsup-ts1) = 229.465 + 0.958 (3.5-(-34.29)) = 265.67 KJ/Kg
For P2 = 700.8 KPa saturation temp. is ts2 = 26.318oC. But observed temperature is 77.9oC. Therefore the condition of the
H2 = hg2 + cp (tsup-ts2) = 264.722 + 0.958 (77.9- 26.318) = 314.13 KJ/Kg
h3 = h4 = 89.202 KJ/Kg
Coefficient of Performance = C.O.P. = Refrigeration Effect / Work Done
= (H1 – h4) / (H2 – H1)
= (265.67 – 89.202) / (314.13 – 265.67)

= 3.64
CFD VALIDATION OF THE ABOVE RESULTS
4.5. Evaporator Flow Analysis
Fig 6. Modeling and meshing
4.5.1. Analysis Carried Through CFX
Fig 7. Model Imported From ICEM CFD
V. RESULTS AND DISCUSSIONS
We discuss about the results obtained from the experiment.
5.1 GEOMETRY OF DIFFUSER (D/L=0.5)
INNER DIA= 15 mm
ANGLE OF DIVERGENCE=150
LENGTH OF DIFFUSER= 30 mm
DIFFUSER ANGLE SELECTED ON THE BASIS OF REFERENCE PAPER
Fig. 8. Diffuser Model
Fig 9. Meshing of Diffuser

Fig 10. Imported from ICEMCFD to CFX for Analysis
Fig 11. Inlet Pressure of the diffuser is validated
Fig 12. Outlet Pressure of the diffuser is validated
Fig 13. Inlet Temperature of the diffuser is validated
Fig 14. Outlet Temperature of the diffuser is validated
Fig 16. Actual Vapour Compression Cycle (T-S diagram)

Fig 17. Pressure Enthalpy diagram
When diffuser added pressure from 2 to 21 i.e. from bar to bar.
But, if diffuser is not there compressor take additional power input to reach bar.
WITH DIFFUSER
For P1 = 137.89 KPa saturation temperature is
ts1 = -19.15oC. But observed temperature is 8oC. Therefore the condition of the refrigerant before compression is Superheat.
H1 = hg1 + cp (tsup-ts1) = 238.98 + 0.958 (8-(-19.15)) = 264.99 KJ/Kg
For P2 = 764.9 KPa saturation temp. is ts2 = 29.69oC. But observed temperature is 60.4oC. Therefore the condition of the
H2 = hg2 + cp (tsup-ts2) = 266.53 + 0.958 (60.4- 29.69) = 295.95 KJ/Kg
H5 = h6 = 93.155 KJ/Kg
= (H1 – h6) / (H2 – H1)
= (264.99 – 93.155) / (295.95 – 264.99 )
= 5.55
WITHOUT DIFFUSER
Therefore Compressor input = h2
1- h1
Refrigeration effect = h1 - h6
For P1 = 137.89 KPa saturation temp. is ts1 = -19.15oC. But observed temperature is 8oC. Therefore the condition of the
H1 = hg1 + cp (tsup-ts1) = 238.98 + 0.958 (8-(-19.15)) = 264.99 KJ/Kg
Final pressure = P2
1 = 770KPa
Final Temperature = 74.09oC.
For P2
1 = 770 KPa saturation temp. is t3
1 = 29.92oC.
The condition of the refrigerant after compression is Superheat.
H2
1 = h3
1 + cp (tsup-t3
1) = 266.64 + 0.958 (75- 29.92) = 309.83 KJ/Kg

H6 = h5 = 93.155 KJ/Kg
= (H1 – h6) / (H2
1 – H1)
= (264.99 – 93.155) / (309.83 – 264.99)
= 3.83
Enhancement of COP =
= 0.31 = 31%
Table5.1 Results
SL
NO
REFRIGERATION EFFECT
WITHOUT DIFFUSER KJ/Kg
COMPRESSOR INPUT
WITHOUT DIFFUSER
KJ/Kg
REFRIGERATION
EFFECT WITH
DIFFUSER KJ/Kg
COMPRESSOR
INPUT WITH
DIFFUSER KJ/Kg
1 171.84 44.84 171.84 30.96
Table 5.2 Results
SL NO POWER INPUT TO
COMPRESSOR
WITHOUT DIFFUSER
KW
POWER INPUT TO
COMPRESSOR WITH
DIFFUSER
KW
COP
WITHOUT
DIFFUSER
COP WITH
DIFFUSER
% OF
ENHANCEMENT OF
COP
1 0.2 0.14 3.83 5.55 31%
VI. CONCLUSIONS
From above analysis following conclusion has been arrived.
1. With the addition of diffuser pressure inlet of refrigerant to the Condenser is increased.
2. Compressor Work input is decreased
3. Since compressor work input decreased COP has been increased.
4. With the addition of diffuser Coefficient of Performance (COP) Enhancement has been increased by 31% when
compared without diffuser.
5. Compressor Power input obtained without diffuser is 0.2 KW.
6. Compressor Power input with diffuser obtained is 0.14KW
7. Finally conclusion drawn is Coefficient of Performance (COP) can be enhanced by placing diffuser.
FUTURE SCOPE OF WORK
1. With varying divergence angle to the diffuser analysis can be carried out
2. By adopting passive method for diffuser enhancement of pressure can be done
3. By providing heat exchanger in between evaporator and compressor temperature outlet can be raised.
4. Bleed vapour to the condenser can be sent directly from evaporator COP can be raised.

REFERENCES
[1]. Amit prakash “to improve the performance of vapour compression refrigeration system by using sub cooling and diffuser”
IJEBEA 13-129 2013
[2]. R. Rejikumar et.al “enhancement of heat transfer in domestic refrigerator using r600a/mineral oil/ nano-al2o3 as working
fluid” IJCETR APRIL 2013 VOL 03, ISSUE 04.
[3]. M. Krishna Prasanna M.E. student, and P.S Kishore “enhancement of coefficient of performance (cop) in vapour
compression refrigeration system”
[4]. Lucia vilceanu and Mihaela flori “performance characteristics of vapour- compression systems”
[5]. R.T.Saudagar and Dr .U.S. Wankhede “introducing diffuser at compressor inlet in a vapour compression refrigeration
system”
[6]. A. Baskaran and P.Koshy Mathews “thermal analysis of vapour compression refrigeration system with R152a and its
blends R429a, R430a, R431a, and R435a”

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