TABLE OF CONTENT
CONTENT PAGE
1.0 TITLE AND OBJECTIVE 2
2.0 INTRODUCTION 3
3.0 THEORETICAL BACKGROUND 4-7
4.0 APPARATUS 8-10
5.0 EXPERIMENTAL PROCEDURES 11
6.0 RESULT AND DATA ANALYSIS 12-17
7.0 DISCUSSION OF RESULT 18-23
8.0 CONCLUSION 24-27
9.0 REFERENCE 28
10.0 APPENDIX 29
TITLE: VARIATION IN REFRIGERATION COEFFICIENT OF PERFORMANCE
AT VARIOUS OPERATING CONDITIONS
1.0 OBJECTIVE
The objective of this experiment carried out is:
1. to investigate the variation in Coefficient of Performance (COP ref) of a vapor compression
system
2. to identify the refrigerant mass flow rate at different cooling load which is air and water.
3. to identify evaporator cooling load, Qevap at cooling load of water and air.
4. to identify Coefficient of Performance of the system.
LIST OF FIGURES
No Figure Page
1. Figure 1: Computer Controlled Refrigeration and Air Conditioning 9
Units.
2. Figure 2: System display 9
3. Figure 3: System schematic diagram 10
4. Figure 4: Collected data 10
2
2.0 INTRODUCTION
Heat flows in the direction of decreasing the temperature which is from high temperature
regions to low temperature regions. This heat transfer process occurs in nature without
requiring any devices. But the reverse process cannot occur by itself. The transfer of heat
from a low temperature region to a high temperature one requires special devices called
refrigerators. Refrigerators are cyclic devices and the working fluid used in the refrigeration
is called refrigerants.
Another device that transfers heat from a low temperature medium to a high
temperature one is the heat pump. Refrigerators and heat pumps are essentially the same
device but differ in terms of objectives. The objective of the refrigerator is to maintain the
refrigerated space at a low temperature by removing heat from it. Discharging this heat to
higher temperature medium is merely necessary part of the operation, not the purpose. The
objective of a heat pump is to maintain a heated space at a high temperature. This is
accomplished by absorbing heat from a low temperature source, such as well water or cold
outside air in winter, and supplying this heat to a warmer medium such as a house.
Refrigeration is used widely in various applications from industrial to domestic
situations, mainly for the storage and transport of perishable foodstuffs and chemical
substances. It has the prime function to remove heat from a low temperature region, and it
can also be applied as a heat pump for supplying heat to a region of high temperature.
3
3.0 THEORETICAL BACKGROUND
The Carnot cycle is a totally reversible cycle that consists of two reversible isothermal
and two isentropic processes. It has the maximum thermal efficiency for given temperature
limits and it serves as a standard against which actual power cycles can be compared. The
reversed Carnot cycle is the most efficient refrigeration cycle operating between two
specified temperature levels. But however, this cycle cannot be approximated in the actual
devices and is not a realistic model for refrigeration cycles due to difficulty in maintaining
isothermal condition during the heat absorption and heat rejection processes.
A refrigeration cycle works to lower and maintain the temperature of a controlled space by
heat transfer from a low to a high temperature region.
High Temperature Reservoir,
TH
QH
E Wnet
QL
.
Low Temperature Reservoir,
TL
Refrigeration duty is another term for the cooling effect of the refrigeration system,
which is the rate of heat being removed from the low temperature region with specified
evaporation and condensation temperatures. The unit for “duty” measurements is in Watts
(for 1 ton of refrigeration = 3517W).
4
3.1 The Vapor Compression Cycle
An ideal refrigeration system follows the theoretical Reversed Carnot Cycle process.
In practical refrigerators, compression and expansion of a gas and vapor mixture presents
practical problems in the compressor and expander. Therefore, in practical refrigeration,
compression usually takes place in the superheated condition and a throttling process is
substituted for the isentropic expansion.
5
In an ideal vapor-compression refrigeration cycle, the refrigerant enters the condenser
as superheated vapor at state 1 and leaves as saturated liquid at state 2 as a result of heat
rejection to the surroundings. The temperature of the refrigerant at this state is still above the
temperature of the surroundings. The saturated liquid refrigerant at state 2 is throttled to the
evaporator pressure by passing it through an expansion valve or capillary tube. The
temperature of the refrigerant drops below the temperature of the refrigerated space during
this process.
The refrigerant enters the evaporator at state 3 as a low-quality saturated mixture, and
it completely evaporates by absorbing heat from the refrigerated space. The refrigerant leaves
the evaporator as saturated vapor. Then the refrigerant enters the compressor at state 4 as
saturated vapor and is compressed isentropically back to the condenser pressure. The
temperature of the refrigerant increases during this isentropic compression process to well
above the temperature of the surrounding medium.
The cycle:
1–2 Condensation of the high-pressure vapour during which heat is transferred to the high
temperature region.
2–3 Adiabatic throttling of the condensed vapour from the condensing to the evaporating
pressure.
3–4 Evaporation of the low-pressure liquid during which heat is absorbed from the low
temperature source.
4–1 Isentropic compression of the vapour, from the evaporating to the condensing
pressures.
6
Energy Transfers Analysis
Compressor
q4-1 = h4 – h1 + w4-1
If compression is adiabatic, q4-1 = 0, and w4-1 = h1 – h4 = wcomp
Power requirement, P = ṁ (h1 – h4), where ṁ is the flow rate of working fluid per unit time.
Condenser
q1-2 = h2 – h1 + w
w = 0, therefore q1-2 = h2 – h1 and rate of heat rejection Q1-2 = ṁ (h2 – h1)
Expansion Valve
q2-3 = h3 – h2 + w
w = 0, therefore q 2-3 = h2 – h3 and process is assumed adiabatic ( q=0 )
therefore h2 = h3
Evaporator
q3–4 = h4 – h3 + w
w = 0 therefore q3–4 = h4 – h3 and rate of heat absorbed Q3–4 = ṁ (h4 – h3)
Coefficient of Performance, COPref:
q3−4 h4 −h 3
COPref = w = h1 −h4
7
4.0 APPARATUS
No Name of apparatus
1. Computer Controlled Air Conditioning Refrigeration and Heat Pump
Units, THAR22C
2. Water Condenser
3. Air Condenser
4. Thermocouple
5. Evaporator
6. Expansion Valve
7. Computer
8
Figure 1: The Computer Controlled Refrigeration and Air Conditioning Units.
Figure 2: The computer displays the information, and collect the reading of the system to be
analyse.
9
Figure 3: The schematic diagram of the overall system.
Figure 4: The collected data is transferred to spreadsheet to be recorded.
10
5.0 EXPERIMENTAL PROCEDURES
Condenser-water and evaporator-air
1. Select air as a heat source by opening valves AVS-3 and AVS-5. Then click “START”
2. Adjust the water flow rate at the condenser to 5 L/m and adjust the air flow of the
evaporator until 50% of the maximal flow (evaporator load).
3. Then click “COMPRESSOR”
4. When the system is stabilized, start recording the data by click “START SAVING”
5. Set the sampling rate at 180 second per sample.
6. Record the data for 15 minutes (5 samples @ 900 second). “STOP SAVING”
7. Then increase evaporator load to 100% and repeat step (c) to step (f).
Condenser-air and evaporator-air
1. Select air as a heat source by opening valves AVS-3 and AVS-6. Then click “START”
2. Adjust the air flow of the condenser to maximum flow (100%) and 50% of the maximal
flow at the evaporator (evaporator load).
3. Then click “COMPRESSOR”
4. When the system is stabilized, start recording the data by click “START SAVING”
5. Set the sampling rate at 180 second per sample.
6. Record the data for 15 minutes (3 samples @ 900 second). “STOP SAVING”
7. Then increase evaporator load to 100% and repeat step (c) to step (f).
11
6.0 RESULT AND DATA ANALYSIS
At T=1200 s, Cooling Load = Water.
ST-2 ST-3 ST-4 SP-1 SP-2 SC1 SW1
(°C) (°C) (°C) (bar) (bar) (L/h) (W)
29.07 33.99 36.10 2.21 0.04 27.99 355.54
a) Refrigerant mass flow rate (kg/s).
i. Find Vref.
Volume flow rate, Vref can be calculated by converting the volume flow rate obtained in
SC-1 which was unit in L/h to m3/kg.
SC-1 = 27.99 L/h
L 1 m3 1h
V ref =27.99 x x
h 1000 L 3600 s
¿ 7.77 x 10−6 m 3 / s
ii. Find v2
By referring Table A-11, find v2 by interpolation at the temperature obtained at ST-2.
Find vf as v2.
ST-2 = 29.27°C
TEMPERATURE (oC) SPECIFIC VOLUME, vf (m3/kg)
28 0.0008362
29.27 v2
30 0.0008421
29.27−28 v 2−0.0008362
=
30−28 0.0008421−0.0008362
v 2=8.399 x 10−4 m3 /kg
iii. Find mref
Calculate the mass flow rate mref by using formula
V ref
m ref =
v2
12
7.77 x 10−6 −3
¿ −4
=9.312 x 10 kg /s
8.399 x 10
b) Evaporator cooling load, Qevap. (kW).
i. By referring Table A-10, find h3 at ST-3. Use hf as h3
ST-3 =33.99°C
TEMPERATURE (oC) ENTHALPY hf (kJ/kg)
34.00 97.31
h3 =97.31 kJ /kg
ii. By referring Table A-10, find h4 by interpolation at the temperature obtained at ST-
4. Use hg as h4
ST-4 = 36.10°C
TEMPERATURE (oC) ENTALPHY hf (kJ/kg)
36 100.25
36.10 h4
38 103.21
36.10−36 h 4−100.25
=
38−36 103.21−100.25
h 4=100.39 kJ /kg
iii. Find Qevap
Calculate evaporator cooling load, Qevap by using formula
Q evap =m ref ( h4 −h3 )
¿ 9.312−3 (100.39−97.31)
¿ 0.028 kW
c) Coefficient of performance, COP.
13
Q
COP= ¿ evap
¿¿
SW-1 = 355.24 W
0.028
COP=
¿¿¿
¿ 0.079
14
At T=1200 s, Cooling load =Air.
ST-2 ST-3 ST-4 SP-1 SP-2 SC1 SW1
(°C) (°C) (°C) (bar) (bar) (L/h) (W)
27.63 24.97 26.95 2.38 0.04 28.49 363.09
d) Refrigerant mass flow rate (kg/s).
iv. Find Vref.
Volume flow rate, Vref can be calculated by converting the volume flow rate obtained in
SC-1 which was unit in L/h to m3/kg.
SC-1 = 27.99 L/h
L 1 m3 1h
V ref =28.49 x x
h 1000 L 3600 s
¿ 7.914 x 10−6 m3 / s
v. Find v2
By referring Table A-11, find v2 by interpolation at the temperature obtained at ST-2.
Find vf as v2.
ST-2 = 27.63°C
TEMPERATURE (oC) SPECIFIC VOLUME, vf (m3/kg)
26 0.0008309
27.63 v2
28 0.0008362
27.63−26 v 2−0.0008309
=
28−26 0.0008362−0.0008309
v 2=8.352 x 10−4 m3 /kg
vi. Find mref
15
Calculate the mass flow rate mref by using formula
V ref
m ref =
v2
7.914 x 10−6 −3
¿ −4
=9.475 x 10 kg /s
8.352 x 10
e) Evaporator cooling load, Qevap. (kW).
vii. By referring Table A-10, find h3 at ST-3. Use hf as h3
ST-3 =24.97°C
TEMPERATURE (oC) ENTHALPY hf (kJ/kg)
24 82.9
24.97 x
26 85.75
h3 =84.28 kJ /kg
viii. By referring Table A-10, find h4 by interpolation at the temperature obtained at ST-
4. Use hg as h4
ST-4 = 26.95°C
TEMPERATURE (oC) ENTALPHY hf (kJ/kg)
26 85.75
26.95 h4
28 88.61
.
26.95−26 h 4−85.75
=
28−26 88.61−85.75
h 4=87.11 kJ / kg
ix. Find Qevap
Calculate evaporator cooling load, Qevap by using formula
16
Q evap =m ref ( h4 −h3 )
¿ 9.475 x 10−3 (87.11−84.28)
¿ 0.026 kW
f) Coefficient of performance, COP.
Q
COP= ¿ evap
¿¿
SW-1 = 363.09 W
0.026
COP=
¿¿¿
¿ 0.072
17
7.0 DISCUSSION OF RESULT
MUHAMAD IZZAT IHSAN BIN PUZER (2016229376)
DISCUSSION
From the result obtained, as cooling load increase, the COP ref increases. This is the
effect from the mass flow rate is increase or the value of h3 decrease. When we increase the
amount of cooling load definitely the mass flow rate is increase. Furthermore, according from
reference, a rule of thumb is that the COP improves by 2 to 4 percent for each C the
evaporating temperature is raised or condensing temperature is lowered. But however, when
we change the working fluid of condenser with air and the working fluid of condenser with
water (experiment d) the COPref is constant. If we look closely at this state, the mass flow rate
and h3 is remain constant at any time and so does h4, that is why the COPref also does not
change and remain constant. Along the experiment was done, there might have some error
such as before saving, the system maybe does not 100% stabilized. The graph shown on the
screen which shows the stabilization cannot be 100% accurate as same as the actual system.
We look forward to the cooling medium effect, there is two types of medium used
which is water (experiment a) and air (experiment b). When water is used as the cooling
medium, the COP is much higher compare to air. The higher the COP equate to lower
operating cost. This shows that liquid/water cooling is a better cooling medium. This is
because air is not an outstanding thermal conductor (air has a thermal conductivity of 0.026
W/mK). A more effective method of cooling than air cooling is liquid/water cooling as the
heat transferring medium. One way of increasing the thermal efficiency of a water-cooling
system is by placing coils inside the cooling channel to induce a turbulent flow of the cooling
liquid.
Load is the amount of heat energy to be removed from refrigerators by the HVAC
equipment to maintain the design temperature. In domestic fridge, the refrigeration cycle for
the fresh food compartment could be used directly subcool the condensate for the freezer
cycle, thereby shifting some of the cooling load from the freezer to the fresh food cycle.
Comparing the COP of a dual-cycle system to single-cycle system, in a related study found a
23% improvement in overall COP for a dual cycle system using refrigerants R-142 and R-152
18
when the total cabinet loads are evenly distributed between the freezer and the fresh food
compartment.
Figure 1: Refrigeration System Operating Characteristics
The refrigeration practice (with actual loads) in a factory. The refrigeration process
begins with the compressor. The process starts when Ammonia gas is compressed until it
becomes very hot from the increased pressure. This heated gas flows through the coils behind
the refrigerator, which allow excess heat to be released into the surrounding air. This is why
users sometimes feel warm air circulating around the fridge. Eventually the ammonia cools
down to the point where it becomes a liquid. This liquid form of ammonia is then forced
through a device called an expansion valve. Essentially, the expansion valve has such a small
opening that the liquid ammonia is turned into a very cold, fast-moving mist, evaporating as
it travels through the coils in the freezer. Since this evaporation occurs at -27 degrees F (-32
degrees Celsius), the ammonia draws heat from the surrounding area. This is the Second Law
of Thermodynamics in effect. Cold material, such as the evaporating ammonia gas, tends to
take heat from warmer materials, such as the water in the ice cube tray.
As the evaporating ammonia gas absorbs more heat, its temperature rises. Coils
surrounding the lower refrigerator compartment are not as compact. The cool ammonia still
draws heat from the warmer objects in the fridge, but not as much as the freezer section. The
19
ammonia gas is drawn back into the compressor, where the entire cycle of pressurization,
cooling and evaporation begins anew
Figure 2: Batch-continuous air blast freezer with counter flow air circulation
Figure 3: Batch continuous air blast freezer with cross flow
20
MUHAMMAD AMIRUL ADLI BIN KAMARUDIN (2016238344)
DISCUSSION
Refrigerant cycle is a compulsory laboratory experiment for EM220 students. This is
because it provides student knowledge and understanding for the process of controlling the
temperature.
In the experiment, firstly we use water as the load. This means refrigerant system
needs to cool down the water. In order to obtain successful results, the system need to achieve
stable state before recording the data. The interval for each data is taken for every 2 minutes.
The data that being obtained are ST-2, ST-3, ST-4, SP-1, SP-2, SC-1, Sc-2 and Sw-1.
Q
To calculate the COP, we use the formula COP= evap
¿¿¿
from the table of result. We can compare the COP between the water load and the air load.
We found that the COP of 0.079 and 0.072. We can see that COP of air as load is lower than
water.
21
MUHAMAD HAZIQ BIN HASSAN NORDIN (2016218312)
DISCUSSION
Based on the result obtained, the evaporator cooling load calculated and from there,
the coefficient of performance was calculated for both different cooling loads which were
water and air. For water, the coefficient of performance calculated was 0.079. For air, the
coefficient of performance calculated was 0.072. Comparing both coefficient of
performances, water is a better coolant for a system compared to air. The low coefficient of
performance indicates that it performs better at cooling a system. Water is a better heat
conductor compared to air. So, the transfer of heat produced by the system to the cooling load
is at a faster rate thus controlling the temperature of the system efficiently.
In this experiment, precaution steps have to be taken in order to obtain an accurate
data. Firstly, the system needs to achieve a stable state before the time of 1200 seconds start.
This will give a more accurate data. Next, let the timer go a bit further than 1200 seconds.
This is a precaution step to prevent the timer from being stopped before 1200 seconds and
disturb the data accuracy.
22
AINOL HIJRAHTUL HAKIM BIN ARBAEN (2016218402)
DISCUSSION
Based on the outcome and the information that we have obtained, the COP for both
examinations demonstrates a slight distinction. The results show that the value of coefficient
of performance (COP) for the first experiment is 0.079 but in the second experiment is 0.072.
The difference is because of the different cooling load which are water and air. After
conducting the experiment, experimentally the uses of water as the cooling medium resulting
in higher performance than the air. However, theoretically, air as cooling medium is much
more efficient because of its properties that have lower molecular weight and less viscose
than water.
The coefficient of performance (COP) is a proportion of the useful heating or cooling
gave to work required. The COP for both examinations demonstrates a slight distinction. The
COP of the framework are likewise related with the measure of load entering at the
condenser. By investigating the outcome, we can state that the bigger the measure of
refrigerant enters the condenser, the lower the temperature of refrigerant exits from the
condenser.
The terms load is referred to the excessive temperature over the required temperature.
Thus, it must be removed in order to maintain the desired temperature. As the application, for
example is the domestic fridge. The temperature of the leftovers, groceries is considered as
the excessive temperature. Therefore, those temperature is needed to be remove to maintain
cold temperature in the fridge. Besides, other example like in the room, the excessive heat is
the temperature of human body and at the factory which is the temperature of the workers.
23
8.0 CONCLUSION
MUHAMAD IZZAT IHSAN BIN PUZER (2016229376)
CONCLUSION
As the conclusion, refrigerator consists with two compartments - one for frozen items
and the other for items requiring refrigeration but not freezing. The objective of this
experiment has been completely achieved by understanding the relation of various condition
with Coefficient of Performance, all the parameters required to be solved have been
calculated and solved accordingly. In addition, all of the experiments have eventually been
done according to the procedures given systematically and appropriately.
24
MUHAMMAD AMIRUL ADLI BIN KAMARUDIN (2016238344)
CONCLUSION
In a nutshell, it is concluded that the experiment is a success and a report is produced. We
have achieved our main objective to investigate the variation in Coefficient of Performance
(COPref) of a vapour compression refrigeration system. We are also able to find out that the
higher the value of COPref, the better the refrigeration cycle.
25
MUHAMAD HAZIQ BIN HASSAN NORDIN (2016218312)
CONCLUSION
In conclusion, the experiment was deemed successful because the comparison of the
coefficients of performance seemed logical. All the precaution steps were also followed in
order to obtain accurate result. All the objectives have been fulfilled and we, the student also
learned something new thus expanding our knowledge. From the experiment, we can
conclude that water is a better cooling load compared to air due to its nature of heat
conductivity being better than air.
26
AINOL HIJRAHTUL HAKIM BIN ARBAEN (2016218402)
CONCLUSION
As for conclusion, the experiment is success. We have achieved the objective by
following the procedures accordingly, collected all the required data and make the calculation
using the suitable and correct approach. However, the precaution is still not to be overlooked.
To avoid any error, make sure that the system that being analysed is in a good condition and
the team members performed they ask well by avoiding misreading during data collection
and do the calculation correctly.
27
9.0 REFERENCE
Bailes, A. (March 28, 2018). Converting Heating and Cooling Loads to Air Flow - The
Physics. Retrieved October 19, 2018 from
https://www.energyvanguard.com/blog/converting-heating-and-cooling-loads-air-
flow-physics
Coefficient of Performance for a Refrigeration Cycle. Retrieved October 19, 2018 from
https://www.learnthermo.com/T1-tutorial/ch04/lesson-F/pg09.php
Why is COP calculated for refrigerator and heat pump but not efficiency as in case of heat
engine? (2014). Retrieved October 19, 2018 from
https://www.researchgate.net/post/Why_is_COP_calculated_for_refrigerator_and_hea
t_pump_but_not_efficiency_as_in_case_of_heat_engine
COPs, EERs, and SEERs (March 1, 2011). Retrieved October 19 2018 from
http://www.powerknot.com/2011/03/01/cops-eers-and-seers/
Coefficient of Performance – COP – Refrigerator, Air Conditioner. Retrieved October 19,
2018 from
https://www.nuclear-power.net/nuclear-engineering/thermodynamics/thermodynamic-
cycles/heating-and-air-conditioning/coefficient-of-performance-cop-refrigerator-air-
conditioner/
28
10.0 APPENDIX
29
