This laboratory report describes an experiment that studied the performance of a vapor compression refrigeration cycle in an air conditioning unit. Temperature and humidity readings of the air were taken at different points as it was heated, humidified, cooled, dehumidified, and reheated. Readings of the refrigerant temperatures and pressures were also recorded. The air conditions were plotted on a psychrometric chart, and the refrigerant cycle was plotted on a p-h diagram. The cooling loads of the air and refrigerant were calculated, and the coefficient of performance (COP) was determined based on each. The COP value based on the refrigerant was considered more accurate due to inaccuracies in the wet bulb temperature readings for the air.

1. Page 1 of 13
Short Laboratory Report 2015
MP2.2
Vapour Compression Refrigeration Cycle
Chloé Marie Taylor
1322748
Lab Group: Mech 18
Date of Experiment: 10th
November 2015
Date of Lab Report: 24th
December 2015

2. Page 2 of 13
Contents
1 – Summary ------------------------------------------------------------------------------------------------------------ 2
2 – Introduction -------------------------------------------------------------------------------------------------------- 3
3 – Aims of Experiment ---------------------------------------------------------------------------------------------- 3
4 – Theory of Vapour Compression Refrigeration Cycles --------------------------------------------------. 4
5 – Set-up and Procedure ------------------------------------------------------------------------------------------- 6
6 – Observations and Results--------------------------------------------------------------------------------------- 6
7 – Discussion and Analysis of Results --------------------------------------------------------------------------- 8
8 – Sources and Discussion of Error ------------------------------------------------------------------------------ 11
9 – Conclusion ---------------------------------------------------------------------------------------------------------- 12
10 – References --------------------------------------------------------------------------------------------------------- 12
11 – Appendix ------------------------------------------------------------------------------------------------------------13
1 – Summary
An air conditioning unit was used for the experiment. The purpose of the experiment was to
study the performance of a vapour compression refrigeration cycle by calculating the unit’s coefficient
of performance (COPR). Air was heated, humidified, cooled and dehumidified and then reheated
during the process. Temperature readings for the air and the refrigerant and working parameters of
the unit were recorded during the experiment. The state of the air at each point was plotted on a
psychometric chart, which showed that the unit did not reduce the humidity of the air as much as
expected. This was suggested to be due to the refrigerant not removing as much heat from the air as
expected. The refrigerant cycle was plotted on a p-h diagram, where the ideal and actual cycles were
shown along with irreversible processes. It was shown that the actual cycle moved more into the
super-heated region meaning more heat was rejected into the environment than the ideal cycle.
The cooling loads from the air and into the refrigerant were calculated, giving values of
2.55kW and 2.78kW respectively. The difference between these values was said to be due to the unit
being badly insulated hence the refrigerant absorbed heat from the environment as well as the air.
COPR values for the refrigeration cycle were calculated based on the air and the refrigerant cooling
loads, yielding values of 1.19 and 1.3 respectively. The COPR based on the refrigerant was said to be
more accurate as it did not have the inaccuracies of the wet bulb temperature readings that the air
based COPR had. Sources of error, how to minimise them and improvements for the experiment and
the air conditioning unit are discussed.

3. Page 3 of 13
2 - Introduction
Introduction
Heat pumps are very important in today’s society, being the basis for refrigerators and
freezers as well as some heating methods. Heat pumps are machines which move heat from a lower
temperature reservoir (called the source) to a higher temperature reservoir (called the heat sink)
driven by a work input. Depending on the aim, they can be heaters by rejecting heat into the heat
sink, or refrigerators by absorbing heat from the source. Their performance is measured by the ratio
of cooling obtained to work required, called the coefficient of performance (COP) shown in equation
1 below.
!"#$ =
'(
)
, +,. 1
Where:
'( = cooling load (heat removed from source) (W)
) = Work input (W)
COP values will generally be above 1 as more heat will be removed from the source than work
is put in due to the nature of the cycle (discussed in 3 – Theory of Vapour Compression Refrigeration
Cycles).
This experiment aims to find the COP of a vapour compression refrigeration cycle as it is used
to cool and dehumidify air. Irreversibilities of the process and other points of interest will be discussed.
3 - Aims of Experiment
An air conditioning unit operating a vapour compression refrigeration cycle will be used to heat
air to a warm and humid state, then dehumidify and cool air. Temperatures of the air and the
refrigerant working fluid will be recorded from the air conditioning unit. Other working parameters
including differential pressures, mass flow rates and voltage/current values will be recorded. These
set of results will allow:
• The path of the air conditioning process to be plotted on a psychometric chart
• The state of the air at each point to be found and comments to be made on the final quality
of the air
• The cooling load from the air during the air conditioning cycle to be calculated
• The ideal and real cycle for the refrigerant on a p-h diagram to be plotted
• The cooling load for the refrigerant during the cycle to be calculated
• The coefficient of performance of the refrigerator (COP) to be calculated

4. Page 4 of 13
4 - Theory of Vapour Compression Refrigeration Cycles
Carnot Cycle
This experiment uses a vapour compression refrigeration cycle which cools air down. Though
the air is heated and cooled in the air conditioning unit it is the refrigeration cycle (using R12
refrigerant as the working fluid) which is the main focus of the experiment. The refrigeration cycle
removes heat from the air and rejects it to the environment, acting as a reverse heat engine –
commonly called a heat pump. Heat pumps are based around reverse Carnot cycles, which are exactly
the same as Carnot cycle but the process directions and hence work and heat inputs and outputs are
reversed. Figure 1 and 2 below are the pressure-volume (p-v) and temperature-entropy (T-s) diagrams
for a reverse Carnot cycle.
Figure 1: Pressure Volume diagram showing Figure 2: Temperature-Entropy graph with
Carnot cycle. ssssssssssss sssssssssssddfffffffffffhsslkcsd a saturation curve showing Carnot cycle.
The processes relating to the points in figure 1 and 2 are detailed below:
1-2: Isentropic compression – work is done on the working fluid at a constant entropy, resulting in
an increase in temperature
2-3: Isothermal heat rejection – heat is rejected into the high temperature reservoir
3-4: Isentropic expansion – work is done by the working fluid at a constant entropy, resulting in a
decrease in temperature
4-1: Isothermal heat absorption – heat is absorbed ('() from a low temperature reservoir
Note that figure 2 shows the saturation curve for the cycle. Enclosed in the saturation curve
the refrigerant is a mixed state of vapour and liquid. To the left of the saturated liquid line the
refrigerant is a sub-cooled liquid, to the right of the saturated vapour line the refrigerant is
superheated vapour. The importance of this is discussed below.
To base a refrigeration cycle around the reverse Carnot cycle, a working fluid must be found
that is capable of isothermal heat rejection and absorption at the temperatures of the high and low
temperature reservoirs (TH and TL). This can be obtained by using fluids that condense and evaporate
around these temperatures as phase changes are processes in which heat is input or output with no
temperature change. This can be understood more by looking at the Carnot cycle encased in a
saturation curve (figure 2). By choosing a refrigerant which has saturation temperatures (TH and TL)
at the required pressures, the phase changes of the refrigerant can be used to facilitate the isothermal
processes 2-3 and 4-1. The refrigerant will have a low boiling point allowing it to change phase (and
absorb heat) from an already cool temperature.
P
V
11
3
4
2
T
s
1
3
4
2TH
TL

5. Page 5 of 13
Practical vapour compression refrigeration cycle
In practice the reverse Carnot cycle is adjusted to the vapour compression refrigeration cycle
shown below in figure 3. The working fluid is a refrigerant that suits the high and low temperature
reservoirs temperatures.
Figure 3: Temperature Enthalpy diagram showing
vapour compression refrigeration cycle.
The processes in figure 3 are as follows:
1-2: Isentropic compression – the refrigerant is compressed to a super-heated vapour
2-3: Heat rejection – the superheated vapour is cooled to the saturation temperature (TH) then rejects
heat at constant temperature until it reaches the saturated liquid point
3-4: Expansion – the refrigerant is cooled quickly by use of a throttle valve to TL, forming a liquid
vapour mixture
4-1: Heat absorption – the refrigerant absorbs heat ('() from low temperature reservoir until it
reaches the saturated vapour point
The differences in figure 3 are mainly due to the practicalities of running a reverse Carnot
cycle. Stage 1-2 is the most notably different as it is shifted completely out of the saturation curve.
Compressors do not have high efficiency and require more maintenance when they operate with a
mixed state medium, hence the compression stage is shifted into the superheated vapour region. This
means that process 2-3 can no longer be isothermal as the super-heated vapour must first be cooled
to its saturation temperature (TH) before it can undergo the phase transition to the saturated liquid.
Process 3-4 is an irreversible expansion process using a throttling valve. Throttling valves are
used rather than isentropic expansion engines (which expand saturated liquids) as there is only a small
amount of work output to be gained – the costs for an engine would not be justifiable. The throttle
valve reduces the pressure of the saturated liquid abruptly, causing flash evaporation (partial
evaporation of liquid due to sudden drop in pressure) of the liquid and reducing its temperature. It is
ideally a constant enthalpy process.
Figure 3 shows an ideal refrigeration cycle. In reality the compression process (1-2) is likely to
not be isentropic hence an increase in entropy will be seen. This will also mean that the super-heated
vapour will also reach a higher temperature meaning the heat rejection process will have to reduce
the temperature more before reaching the saturation temperature (TH). There will be other generally
frictional losses during the process which will can cause pressure drops within the processes.
T
s
1
TH
TL
4
2
3

6. Page 6 of 13
5 - Set-up and Procedure
Equipment
The apparatus used is an air conditioning unit. The unit is divided into stations where various
readings can be taken. At station A air is taken in from the room and enters the unit. Station B is for
mixing re-circulating air that has already passed through the unit in with the fresh air from the room.
Air will not be re-circulated during the experiment so any readings from station A and B should be
identical. Between station B and C electric pre-heaters and steam injection occurs to bring the air to
the conditions of a warm humid climate. Between station C and D the air passes though an
evaporator where it is cooled and excess moisture is condensed out. The evaporator uses
dichlorodifluoromethane (R12) refrigerant – the cycle of the refrigerant is of key importance to the
experiment and will be discussed in more detail in Section 7 – Discussion and Analysis of Results.
Between station D and E there is re-heating of the air to increase its temperature and reduce its
humidity. Each station has two thermocoupled thermometers – a wet and a dry bulb. By taking these
two readings we can determine the state of the air (its relative humidity). Other readings about the
unit such as mass flow rates, current in compressors, relative pressures and temperatures in the
refrigerant cycle will be recorded.
Set-up and Procedure
Hot and humid air must be produced for the air conditioning cycle. This air will then be
cooled and dehumidified and then reheated to a required temperature of 20°C.
The unit is turned on and air flow is set to a minimum of 0.07kgs-1
(any lower and the reliability
of the wet bulb thermocouples will be poor). The boilers and refrigerator units are turned on and
allowed to stabilise for about 5 minutes. Once steam is being produced by the boilers, two of the three
boilers are turned off and the re-heater between station D and E is turned on, adjusting it so that the
air before the cooling is heated to 25°C and a relative humidity of 90% (a warm humid climate). The
unit is then left again to stabilise for 10 minutes, after which the re-heater is adjusted to give a
temperature of about 20°C and relative humidity of 50-60% after the cooling as it exits the unit. After
allowing the unit to stabilise for 10 minutes once more, thermocouple readings and other parameters
at each station can be taken. The results can be seen in the section below.
6 - Observations and Results
The temperature readings from the thermocouples (wet and dry bulbs) were recorded and
shown in table 1. The corrected values for the wet bulb temperature are also shown – the correction
value was found using figure 6 which shows the relationship between ‘Screen’ and ‘Sling’ wet bulb
temperatures when the dry bulb temperature is known. Figure 6 and a brief explanation of how
correction values are obtained is shown in the Appendix.
Station Dry Bulb Temperature
(°C)
Wet Bulb
Temperature
(observed) (°C)
Wet Bulb
Temperature
(corrected) (°C)
A – Intake 20.0 16.0 16.4
B – After Mixing 21.0 17.0 17.4
C – After Pre-heating
and Steam Injection
27.5 27.5 27.5
D – After Cooling and
Dehumidification
18.3 18.3 18.3
E – After Re-heating 24.0 22.0 22.2
Table 1: Dry, Wet bulb and corrected wet bulb temperature readings from each station.

7. Page 7 of 13
Note that when the readings for station C and D were taken the wet bulb temperatures were
higher than the dry bulb temperatures. The wet bulb is meant to be less than or equal to the dry bulb
so the wet bulbs were given the same temperature as the dry bulb. No correction values were
obtained for these temperatures. This is further discussed in the section 8 - Sources of Error.
The air at A and B is expected to be identical as there is no change in the air between these
two points. However at station B there is an increase in temperature of 1°C for both the wet and the
dry bulb. This could be due to the air conditioning unit acting as an insulation to the outside
environment so air further in the unit (station B) will be warmer whereas station A is not well enclosed.
It could also be due to a too short stabilisation period.
The temperature readings from the refrigerant cycle are shown in table 2. These will be used
in Section 7 – Discussion and Analysis of Results to plot a more realistic cycle on a p-h diagram.
R12 Temperature: Temperature (°C)
Before Expansion valve 33
After Expansion valve 0
After Evaporator 21.5
After Compressor 92
Table 2: Temperature of R12 refrigerant at various stages
through cycle.
The voltage supply, and currents through the relevant equipment are recorded, shown in table 3.
Reading
Voltage (V) 233
Pre-heater current (A) 2
Boiler current (A) 8
Compressor and cooling fan current (A) 9.2
Re-heater current (A) 2
Circulating fan current (A) 0.8
Table 3: Voltage and current readings from the unit.
Using the value of voltage and compressor and cooling fan current we can calculate the
amount of power the compressor and cooling fan require. This is calculated using equation 2 below
and we obtain a power of 2.144kW.
#/0+1 = 23 = 9.2×233 = 2143.6), +,. 2
This value of power is the work input to the refrigerant cycle. It will be used later to find a
value for the COPR of the refrigerant cycle.
Other variables of the unit which were recorded are shown in table 4. They will be used in
later calculations finding the COP and mass flow rates of the air and refrigerant.
Reading Units
Orifice Differential (intake) mmH20 0.7
Orifice Differential (outlet) mmH20 0.7
Evaporator Pressure (gauge) kPa 200
Condenser Pressure (gauge) kPa 750
Refrigerant mass flow rate kgs-1
0.024
Table 4: Orifice differential intake and outlet, evaporator and condenser pressure and refrigerant mass
flow rate as recorded from the unit.

8. Page 8 of 13
7 - Discussion and analysis of Results
Air Calculations
Psychometric chart
The temperature readings (dry and wet corrected) from table 1 are plotted on a psychometric
chart to enable the state of the air at each station to be found. This is shown in figure 4 below [1].
Each point has been labelled and from the graph we can find the specific enthalpy, % saturation and
specific volume of the air at each station. The data obtained from the graph is shown in table 5 below.
Figure 4: Psychometric chart for air at 101.325kPa with points of air at each station
plotted on. [1]
Station A Station B Station C Station D Station E
Specific Enthalpy
(kJ/kg)
45 49 87 50 66
% Saturation 72 70 100 100 88
Specific Volume
(m3
/kg)
0.843 0.848 0.884 0.842 0.863
Table 5: Specific enthalpy, % saturation and specific volume for air at each station read from
figure 4.
We can see that the states of A and B are very similar – this is expected and was discussed
earlier in section 5 - Set-up and Procedure. Station C was aimed to have a % saturation of 90 and a
temperature of around 25°C, table 5 and table 1 show that actual values were 100% saturation and
27.5°C. Station E was aimed to be at 50-60% saturation and around 20°C, table 5 and 1 show that
actual values were 88% and 22.2°C. The actual % saturation at E is considerably higher than expected,
showing the re-heating process between D and E did not reduce the relative humidity as much as
expected. The final temperature of the air leaving the conditioning unit (station E) was 2.2°C above
the aimed value, suggesting that the refrigeration process may not have reduced the temperature as
much as was required. Had the refrigeration cycle cooled the air more, more heat could have been

9. Page 9 of 13
added back during the re-heating process which may have decreased the % saturation more, giving a
closer value of both % saturation and temperature to the aim.
Other differences between the aimed and actual values are commented on in the section 8 -
Sources of Error.
Cooling Load
The cooling load on the air is the amount of heat removed from it as it passes through the
heat exchanger where the cooled refrigerant flows. It can be calculated using equation 3 below.
'( = :(ℎ= − ℎ?), +,. 3
Where:
: = mass flow rate (kgs-1
) (air)
ℎ= = specific enthalpy of air before cooling (station C) (kJ/kg)
ℎ? = specific enthalpy of air after cooling (station D) (kJ/kg)
The values of ℎ= and ℎ? from table 5 are 87 kJ/kg and 50 kJ/kg respectively. The mass flow
rate of air can be calculated using equation 4 shown below.
: = 0.0757
∆E
FG
, +,. 4
Where:
∆E = intake orifice differential pressure (mmH20)
FG = specific volume at intake (station A) (m3
/kg)
The value for ∆E was recorded to be 0.7 mmH20 (table 4) and the value of FG was found to be
0.843 m3
/kg (graph 1 and table 5). These two values can be substituted into equation 4 and give a
value of 0.0690kgs-1
for the air mass flow rate. Substituting the mass flow rate and both specific
enthalpies for station D and C (table 5) into equation 3 a cooling load on the air of 2.55kW is obtained.
Refrigerant Calculations
Refrigeration Cycle on pressure-enthalpy Diagram
The readings that were taken during the experiment can be used to plot a p-h diagram for the
refrigerant cycle. A pressure enthalpy (p-h) diagram shown in figure 5 [2] can then be used to find the
enthalpy of the refrigerant at each stage and hence the cooling load can be calculated. This can then
be used to calculate the COPR of the refrigeration cycle.
A pressure enthalpy diagram for dichlorodifluoromethane (R12) was used, by plotting the
points the changes in phase of the refrigerant can clearly be seen. Note that on this graph there are
multiple axes. In order to plot the points the absolute pressures at which the condenser and
evaporator are running on must be calculated. The gauge running pressures of the evaporator and
condenser were recorded to be 200kPa and 750kPa respectively (table 4). To convert the values to
absolute pressure the atmospheric pressure (taken as 100 kPa) is added to each value. From this the
evaporator pressure is found to be 300kPa and the condenser pressure is found to be 850kPa.
Each point is plotted on the diagram (figure 5), note that the same nomenclature for the points
is used as in the theory section (4) above. Point one represents the refrigerant as a superheated
vapour as it exits the evaporator. It is plotted at 0.3 bar (evaporator pressure) and on the saturated
vapour curve. Process 1-2 is a constant entropy compression of the refrigerant to the condenser
pressure (0.85 bar), so point 2 is plotted at 0.85 bar (condenser pressure) and at entropy as point 1
(0.7 kJ/kg/K). Process 2-3 is a condensing process where the refrigerant is cooled and phase changes
into a liquid. Point 3 can be plotted at 0.85bar (condenser pressure) and on the saturated liquid curve.
Process 3-4 is a constant enthalpy expansion process where the refrigerant is bought to the
evaporator pressure (0.3 bar) so point 4 is plotted at the same enthalpy as point 3 and on the
evaporator pressure line.

10. Page 10 of 13
Point 1 has been plotted assuming that the evaporator only heats the refrigerant to the
saturated vapour point. The actual point 1 (1a) can be plotted on the graph using the temperature of
the refrigerant as it exits the evaporator which from table 2 is found to be 21.5°C. Note that this is
much higher than the ideal temperature value of 0°C. It is seen that point 1a is in the superheated
region in the p-h diagram, meaning the evaporator has heated the refrigerant more than is expected.
The importance of this will be discussed when calculating the cooling load on the refrigerant (later in
section).
Point 2 has been plotted assuming the process 1-2 is isentropic (constant entropy). In reality
it will not have been an isentropic process, there will have been a small amount of heat transfer into
the refrigerant resulting in a higher final temperature. A more realistic point 2 can be plotted by using
the temperature of the refrigerant at the end of the compression process. From table 2 the value is
found to be 92°C and can be plotted (as point 2a) on the condenser pressure line (0.85 bar) using a
constant temperature line. It is seen that the actual point (2a) is further to the right than the ideal
isentropic point (2), as the ideal point is at a temperature of 40°C. This means that more cooling is
needed to cool the superheated refrigerant to a saturated vapour, and also represents an
irreversibility in the cycle.
Point 1a and 2a have been connected and drawn on figure 5, to allow comparison between the
actual and ideal cycle.
Figure 5: Pressure-enthalpy diagram showing the refrigerant cycle. Note that point 1
and 1a and the isentropic (2) and actual (2a) points of the compression are shown.
Dashed lines represent irreversible processes. [2]
Irreversibilities in the vapour compression refrigeration cycle have been shown on figure 5 by
dashed lines. Process 1a to 2a is irreversible as it does is not an isentropic process (due to there being
small amounts of heat transfer) hence it cannot be reversed. Process 3 to 4 is irreversible due to the
nature of the air conditioning unit, this was discussed in Section 4 – Theory of Vapour Compression
Refrigeration Cycles.
Specic Enthalpy (kJkg)
AbsolutePressure(MPa)
Constant Specic
Entropy (kJkgK)
4
3 2 2
11 a
a

11. Page 11 of 13
Cooling Load
The cooling load on the air can be calculated using equation 3 from above, note that ℎ= and ℎ?
correspond to the specific enthalpies of the refrigerant at points 1 and 4 on figure 5 – 186 kJ/kg and
70 kJkg respectively. These values and the mass flow rate of the refrigerant from table 4 of
0.024 kgs-1
can be substituted into equation 3, giving a cooling load of 2.78kW. Note that this is larger
than the cooling load calculated for the air (2.55kW), showing that the refrigerant gained more energy
than the air in the unit lost. This difference is further discussed in the section 8 - Sources of Error.
Coefficient of Performance
The COPR can be calculated for the refrigeration unit using equation 1 shown below.
!#$ =
'(
)
, +,. 1
Where:
'( = cooling load (W)
) = Work input (power of the compressor) (W)
The work input into this cycle is the compressor, its power was calculated in section 6 -
Observations and Results and was found to be 2.144kW. This and the value for cooling load from
above (2.78kW) can be substituted into equation 1 yielding a COPR of 1.30. This COPR value means that
for every 1kW of work input, 1.3kW of heat will be removed from the air.
The COPR can also be calculated using the cooling load based on the air’s calculations. The
compressor power and the cooling load based on the air (2.55kW) can be substituted into equation 4
to give a COPR of 1.19. This is smaller than the COPR based around the refrigerant, and is due to the
cooling load of the air being lower than the cooling load for the refrigerant. This COPR value means
that for every 1kW of work input, 1.19kW of heat will be removed from the air. The significance of this
is discussed in section 8 – Sources and Discussion of Error.
8 - Sources and Discussion of Error
As this experiment was mainly descriptive, it is difficult to quantify any errors. No fits or
expected trends can be compared to the results, instead the systematic and experimental errors and
improvements for the experiment and the air conditioning unit are discussed.
The air conditioning unit was used in a large room with other large pieces of equipment running
nearby. Intermittent construction work was going on in the room adjacent, meaning that the quality
of the air may have fluctuated during the experiment. It is also possible that the air had a higher dust
and particulate content than normal. The experiment was conducted within a 2 hour session, this time
constraint limited the time that could be used for stabilisation of the unit. If the experiment were to
be done again, an isolated, air conditioned and filtered room should be used and longer times should
be allowed for the stabilisation of the unit.
All readings were taken from needle scales which were fluctuating, increasing the error in the
results. The unit was not calibrated before usage and hence the results obtained may have systematic
error. A digital scale may have increased the precision of the results obtained though this would
require adjustment on the unit.
In section 6 the results obtained showed that for two of the recorded temperatures the wet
bulb was higher than the dry bulb temperature - this may have happened due to the unit not having
enough time to fully stabilise.
In section 7 it was shown that the aimed temperature and saturations at stations C and E were
different to the values aimed for. This was suggested to be due to the refrigeration cycle not removing
as much heat from the air as expected. This is supported by the fact that the cooling load on the air
was found to be less than the cooling load gained by the refrigerant (2.55kW to 2.78kW, section 7),
showing the refrigerant was removing more heat than was lost from the air. The extra heat gained by
the refrigerant could be due to it also absorbing heat from the surrounding environment as the unit

12. Page 12 of 13
may not be well insulated on the section where the heat exchange occurs. The extra heat absorption
is also supported by the actual cycle plotted on the p-h diagram where the evaporator increases the
temperature of the refrigerant as well as changes its phase (pushing it into the superheated region).
Two COPR values were calculated, one using the cooling load to the refrigerant (1.3) and one
from the cooling load from the air (1.19). The COPR based on the air is more conservative as it uses
the cooling load that was directly measured from the air and is the desired outcome of the air-
conditioning process. However the refrigerant based COPR is more accurate as it does not have the
inaccuracies in measurement of the wet bulb readings as the air COPR does.
The COPR values obtained for the unit could be improved in a number of ways. From equation
1 it is clear that the COP value can be increased by making the cooling load larger, or the work input
smaller. The cooling load could be made larger by insulating the heat exchange section to ensure that
all heat absorbed by the refrigerant is from the air. It could also be increased by using a contra-current
flow of the refrigerant and air when the exchange takes place as this maintains a higher temperature
gradient increasing the amount of heat exchange. Work input could be reduced by using a more
efficient compressor.
9 – Conclusion
The results obtained allowed us to plot the state of the air on a psychometric chart. It was shown
that the % saturations of the air were different to the values that were aimed for, this was attributed
to the refrigerant removing less heat from the air than expected hence removing less humidity. The
temperatures of the air were relatively close to the aimed values, though the final temperature of the
air was 2.2°C above the aimed value, again suggesting that the refrigeration cycle did not reduce the
air’s temperature enough.
The refrigerant cycle was plotted on a p-h diagram, showing the ideal and actual cycle along
with irreversibilities. The actual cycle moved into the super-heated vapour region of the graph much
more than the actual cycle, meaning that more heat was rejected to the environment than expected.
The cooling loads on the air and to the refrigerant were calculated, giving 2.55kW and 2.78kW
respectively. This higher refrigerant cooling load was said to be due to the air conditioning unit not
being well insulated so heat was removed from the surrounding environment as well as the air.
COPR values were calculated based on the cooling loads for the air and the refrigerant, yielding
values of 1.3 and 1.19 respectively. The COPR based on the air’s cooling load was said to be more
conservative as it used a smaller cooling load and the desired outcome of the refrigeration, however
it is more inaccurate as it includes the measurements of the wet bulb temperatures which are
inherently inaccurate. Improvements for the air conditioning unit were discussed including using a
more efficient compressor and a counter-current flow in the heat exchanger.
10 – References
[1] - Heikal Morgan R., Miller A. J. (2011). AIR CONDITIONING. Available:
http://www.thermopedia.com/content/550. Last accessed 24th Dec 2015. Edited.
[2] - University of Birmingham (2015). Laboratory Experiment MP2.2. Appendix 5 – Enthalpy diagram
for R12. p10. Edited.
[3] - University of Birmingham (2015). Laboratory Experiment MP2.2. Appendix 6: Relationship
between “Screen” and “Sling” wet bulb temperatures. p11. Edited.

13. Page 13 of 13
11 – Appendix
Relationship between “Screen” and “Sling” wet bulb temperatures
Figure 6: Relationship between ‘Screen’ and ‘Sling’ wet bulb temperatures. [3]
For the readings from station A values of 20°C and 16°C were obtained for the dry and the
wet bulb respectively. To obtain the correction value for the wet bulb temperature a line was drawn
up from 16°C on the x axis until the dry bulb temperature of 20°C was reached. A line was then drawn
horizontally and we obtain a correction value of 0.4°C. The same process was used for each set of
thermocoupled readings and the corrected values are shown in Table 1.