refrigeration unit
DESCRIPTION
lab experimentTRANSCRIPT
UNIVERSITI TEKNOLOGI MARAFAKULTI KEJURUTERAAN KIMIA
GEOLOGY AND DRILLING LABORATORY(CGE 558)
NAME : MUHAMMAD SHAFIE BIN HANAFIMUHAMMAD AIZUDDIN B ZAINAL ABIDIN SHAHNUR AMALINA EZRINA BINTI DZULKIFLI
STUDENT NO : 20144721582014489078
2014459386EXPERIMENT : REFRIGERATION UNITDATE PERFORMED : 3RD APRIL 2015SEMESTER : 3PROGRAMME/ CODE : EH 243GROUP : 4
No Title Allocated Marks % Marks1 Abstract/ Summary 52 Introduction 53 Aims/ Objectives 54 Theory 55 Apparatus 56 Procedure 107 Result 108 Calculations 109 Discussion 2010 Conclusions 1011 Recommendations 512 References 513 Appendices 5
TOTAL 100
Remarks:
Checked by:
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TABLE OF CONTENTS :
N
O
TOPICS PAGES
1 Abstract 2
2 Introduction 3
3 Objectives 3
4 Theory 4
5 Apparatus 8
6 Procedures 9
7 Results 11
8 Sample calculations 14
9 Discussion 23
10 Conclusions 25
11 Recommendations 26
12 References 27
13 Appendices 28
1
ABSTRACT
The experiment is done by using the SOLTEQ Refrigeration UnitSOLTEQ Refrigeration Unit
and is divided into 3 parts. There are 3 experiments that we conduct by using this refrigeration
unit that are experiment1, 3 and 5. The objectives of the first experiment is to determine the
power input, heat output and coefficient of performance of a vapour compression heat pump
system. The objectives of experiment 3 are to plot the vapour compression cycle on the p-h
diagram and compare with ideal cycle and to perform energy balances for the condenser and
compressor. The data obtained is then tabulated into table and is plotted into graph. The graph
is plotted in order to determine the heat pump performances against cooling water outlet
temperature and condensing temperature. The objective of experiment 5 is to determine the
compression ratio and volumetric efficiency. The data obtained was tabulated into table and
the compression ratio and volumetric efficiency are determine by calculation based on the
formula given.
2
INTRODUCTION
Refrigerant cycle is a thermodynamic process where heat in the cold body is
withdrawn and heat in the hot body is expelled from it. Generally, it is a process where it
removes the heat from the particular area and to lower the temperature. The device that
operates this system is called refrigerator and the substance that inside the refrigerator is
called refrigerant. There are four main components inside the refrigerant cycle which are
compressor, condenser, expansion valve and evaporator. Refrigerant and heat pumps apply
same concept which is vapor compression cycle and the operating principle is the same but
for refrigerant, it function is to remove heat while for heat pump is to give heat.
This experiment will demonstrate the flow rate of water will affect the refrigeration
unit to a certain extent. Knowing the cooling and refrigerant flow rate will help in determine
the power input, heat output and coefficient performance for the refrigerant unit. Besides, it
will also help to determine the energy balance of the refrigerant. Various physical refrigerant
units will be studied in this experiment and operating at different operating mode such as
different flow rate to acquaint this thermodynamic process. The apparatus is equipped with
control valve for the cooling water flow rate, temperature, pressure and compressor output
display to make it easier to run this experiment and get better understanding about this
process.
OBJECTIVES
To determine power output, heat output and coefficient of performance of refrigerant
unit
To construct or produce vapour compression cycle on p-h diagram
To study energy balance of refrigerant units
To estimate the effect of compressor pressure ration on volumetric efficiency
3
THEORY
Refrigeration cycle is a sequence of thermodynamic processes where heat is
withdrawn from a cold body and released it to surrounding. According to 2nd law of
thermodynamic, to transfer energy which is heat energy from a lower temperature to higher
temperature required an external source or external work done on the system. (Manohar,
2007). A schematically refrigerator process can be seen in figure 1. The heat from lower
temperature, QL, is the magnitude of the heat removed from the refrigerated space at lower
temperature, TL while the heat at higher temperature, QH, is the magnitude of the heat rejected
to the warm space where it has higher temperature, TH. Thus, work input, Wnetin is required.
Figure 1: Schematic diagram on refrigeration unit
Performance of refrigerator is defined as coefficient of performance (COP) (Ameen, 2006).
Coefficient of performance(COP) of refrigerator can be expressed as:
4
QL
QH
WNet,in
Cold
Hot
Desired outputRequired Input
= QL
W net ,∈¿¿
The most common type of refrigerator used work input and operates in vapour compression
cycle. There are 4 main components in refrigeration cycle which are compressor, condenser,
expansion valve and evaporator. The temperature at which liquid will evaporate is depends on
pressure, thus if a fluid that has low boiling point will evaporate at a low temperature in the
low pressure evaporator and will condense at a higher temperature in high pressure condenser.
Figure 2: Vapour compression cycle
Due to the flow of refrigerant through the condenser, evaporator, compressor and expansion
valve thus there will be a pressure drop. (Aurora, 2001). Besides, the temperature difference
between refrigerant and surrounding will cause some heat losses or heat gains. Furthermore,
compression will be polytropic with friction and heat transfer instead of isentropic.
5
Figure 3: P-h diagram of refrigeration cycle
The effect of evaporator temperature on performance of a system is obtained by
keeping the condenser temperature (pressure) and compressor displacement rate and clearance
ratio is being fixed. (Sawhney, 2009). Thus, by reducing the pressure ratio the volumetric
efficiency increases. The formula of the compressor ratio and volumetric is:
Figure 4: Compressor pressure ratio equation
Where :
Pt = total pressure
Tt = total temperature
Ht = specific stagnation enthalpy
Cp = specific heat
6
ᵧ = specific heat ration
nc = adiabatic efficiency
Volumetric efficiency equation is as follows:
nvol=induced volumeswept volume
7
APPARATUS
Figure 6: SOLTEQ Refrigeration UnitSOLTEQ Refrigeration Unit
8
PROCEDURE
EXPERIMENTAL PROCEDURE
1.1 GENERAL START-UP PROCEDURES
1. The unit and all instruments that were checked in proper condition.
2. The both water source and drain were checked and connected then opened the water
supply and the cooling water flowrate was set at 1.0 LPM.
3. The drain hose at the condensate collector was checked if connected.
4. The power supply was connected and the main power was switched on follows by main
switch at the control panel.
5. The refrigerant compressor was switched on. The unit was now ready for the experiment
as soon as temperature and pressure are constant.
EXPERIMENT 1 : DETERMINATION OF POWER INPUT, HEAT OUTPUT
AND COEFFICIENT OF PERFORMANCE
1. The general start-up procedures were performed.
2. The cooling water flowrate was adjusted to 40 %.
3. The system were allowed to run for 15 minutes.
4. All necessary readings were recorded into experimental data sheet.
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EXPERIMENT 3 : PRODUCTION OF VAPOUR COMPRESSION CYCLE ON p-
h DIAGRAM AND ENERGY BALANCE
1. The general start-up procedures were performed.
2. The cooling water flow rate were adjusted to 40% and the system were allowed to run for
15 minutes.
3. All necessary readings were recorded into the experimental data sheet.
EXPERIMENT 5 : EXTIMATION THE EFFECT OF COMPRESSOR PRESSURE
RATION ON VOLUMETRIC EFFICIENCY
1. The general start-up procedures were performed.
2. The cooling water flow rate were adjusted to 40%.
3. The system were allowed to run for 15 minutes.
4. All necessary readings were recorded into experimental data sheet.
5. The experiment were repeated at different compressor delivery pressure.
1.2 GENERAL SHUT-DOWN PROCEDURES
1. The compressor ware switched off, follows by main switch and power supply.
2. The water supply were closed and the water was ensured that was not left running.
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RESULTS
EXPERIMENT 1 : DETERMINATION OF POWER INPUT, HEAT OUTPUT
AND COEFFICIENT OF PERFORMANCE
Cooling Water Flow Rate, FT1 % 40.0
Cooling Water Inlet Temperature, TT5 °C 29
Cooling Water Outlet Temperature, TT6 °C 30.6
Compressor Power Input W 159
11
EXPERIMENT 3 : PRODUCTION OF VAPOUR COMPRESSION CYCLE ON p-
h DIAGRAM AND ENERGY BALANCE
Refrigerant Flow Rate, FT2 % 60.5
Refrigerant Pressure (Low), P1 Bar (abs) 1.4
Refrigerant Pressure (High), P2 Bar (abs) 6.8
Refrigerant Temperature, TT1 °C 26.7
Refrigerant Temperature, TT2 °C 78.8
Refrigerant Temperature, TT3 °C 29.4
Refrigerant Temperature, TT4 °C 22.2
Cooling Water Flow Rate, FT1 % 40.0
Cooling Water Inlet Temperature, TT5 °C 29
Cooling Water Outlet Temperature, TT6 °C 30.5
12
Compressor Power Input W 159
13
EXPERIMENT 5 : EXTIMATION THE EFFECT OF COMPRESSOR PRESSURE
RATION ON VOLUMETRIC EFFICIENCY
Refrigerant Flow Rate, FT2 % 60.5
Refrigerant Pressure (Low), P1 Bar (abs) 1.4
Refrigerant Pressure (High), P2 Bar (abs) 6.8
Refrigerant Temperature, TT1 °C 26.7
SAMPLE CALCULATIONS
EXPERIMENT 1 : DETERMINATION OF POWER INPUT, HEAT OUTPUT
AND COEFFICIENT OF PERFORMANCE
A. Power Input
1. Cooling water flow rate (LPM) = Cooling water flow rate (%) x 5 LPM
100 %
= 40.0 % x 5 LPM
100 %
14
= 2 LPM
Mass flow rate; ρwater = 1000 kg/m3
Convert =
= 0.333 kg/s
2. Power Input = 159 W
3. Heat Output
Ein = Eout
ṁhin = QH + ṁhout
QH = ṁ(hout - hin)
Refer to table A-4 (Saturated water - Temperature table) ; interpolate
TEMPERATURE, °C ENTHALPHY, KJ / KG
25 104.83
28.8 hin
30 125.74
30.1 hout
35 146.64
15
2 L 1 m3 1000 kg 1 min
min 1000 L m3 60 s
30 - 28.8 = 125.74 - hin
28.8 - 25 hin - 104.83
hin = 120.72 KJ/KG
35 - 30.1 = 146.64 - hout
30.1 - 25 hout - 125.74
hout = 126.158 KJ/KG
QH = ṁ(hout - hin)
= 0.0333 kg/s x (126.158 - 120.72) KJ/KG
= 0.181 KJ/s
= 0.181 KW
4. Coefficient of performance
COP = QH = 0.181 W = 1.138 of desired output/ W of required input
W 0.159 W
EXPERIMENT 3 : PRODUCTION OF VAPOUR COMPRESSION CYCLE ON p-
h DIAGRAM AND ENERGY BALANCE
1. Determination enthalpy of refrigerant
hTT1 at 26.7 °C and 1.4 bar (Refer property table A-13; superheated refrigerant-134a)
1.4 bar × 100000 pa = 0.14 MPa
1 bar
16
T, °C h, enthalpy KJ/kg
20 271.38
26.7 hTT1
30 279.97
hTT1 = 277.13 KJ/kg
hTT2 at 78.8 °C and 6.8 bar; 0.68 MPa
P, MPa
T, °C
0.6 0.68 0.7
h, enthalpy KJ/kg
70 309.73 308.61 308.33
78.8 - hTT2 -
80 319.55 318.534 318.28
hTT2 = 317.44 KJ/kg
17
hTT3 and hTT4 at 29.4 °C and 22.2 °C (Refer property table A-11; saturated refrigerant-134a)
Pressure, kPa h, enthalpy KJ/kg
hg
650 263.77
680 hTT3
700 265.03
hf
650 85.26
680 hTT4
700 88.82
hTT3 = 264.47 KJ/kg
hTT4 = 87.38 KJ/kg
18
Figure 1 : Graph of pressure vs. Enthalpy
19
Figure 2 : p-H Diagram of Refrigeration Cycle
Source : Retrieved on April 8, 2015 from
http://www.arca53.dsl.pipex.com/index_files/phrefrig_files/image002.gif
2. Energy Balance on the condenser
Refrigerant flow rate, LPM = cooling water flow rate (%) x 1.26 LPM
100%
= 60.5 % x 1.26 LPM = 0.7623 LPM
100%
Convert =
= 0.0127 kg/s
20
0.7623 L 1 m3 1000 kg 1 min
min 1000 L m3 60 s
Ein = Eout
(qin - qout) + (win - wout) = he - hi
ṁhTT3 = QH + ṁhTT4
QH = ṁ(hTT3 - hTT4)
= 0.0127 kg/s (264.47 - 87.38)KJ/kg
= 2.25 KW
3. Energy Balance on the Compressor
Ein = Eout
(qin - qout) + (win - wout) = he - hi
ṁhTT3 = QH + ṁhTT4
QH = ṁ(hTT2 - hTT1)
= 0.0127 kg/s (317.44 - 277.13)KJ/kg
= 0.512 KW
EXPERIMENT 5 : EXTIMATION THE EFFECT OF COMPRESSOR PRESSURE
RATION ON VOLUMETRIC EFFICIENCY
1. Compressor Ratio
CPR = Refrigerant pressure 2 / Refrigerant pressure1
= 6.8 bar / 1.4
= 4.857
21
2. Volumetric Efficiency in term of enthalphy
ηvol = Induced Volume / Swept Volume
= (317.44 - 87.88) / (317.44 - 277.13)
= 5.695
22
DISCUSSION
The refrigeration unit experiment were conducted in three different experiment in order to
study how the mechanical heat pump and thermodynamic refrigeration unit works. The
objective of first experiment was to determine compressor power input, heat output and
coefficient of performance of a vapour compression heat pump system. It was recorded at 159
W. At the same time, the enthalpy, h, (from thermodynamic property table) at recorded
temperature; 28.8 °C was used to calculate the heat output, QH, of 0.181 KW was obtained.
Besides, the coefficient of performance was determined which is the ratio of heat output to the
amount of energy input of a heat pump, and is valued about 1.138.
For the next experiment, the procedure was repeated until the compressor delivery
pressures reaches around 14.0 bars. and data was recorded. From the data obtained, cooling
water flow rate percentage was maintained at 40.0% as the temperature increased and the
power input of the compressor is static from 159 W. In order words, from the value of COP
obtained in first experiment is 1.138 is means that the addition of 1 kW of electrical energy is
needed to have a released of 1.138 kW of heat at the condenser as by itself heat is always
transferred from an object with high temperature to objects with lower temperatures. Then,
the experimental vapour compression cycle on the p-h diagram of R-134a was plotted. In fact,
in an ideal vapor-compression refrigeration cycle, the refrigerant enters the compressor as a
saturated vapour and is cooled to the saturated liquid state in the condenser (Lambers, K.,
2008).
Furthermore, it has been proven that the point at pressure; 0.68 MPa and enthalpy;
264.47 KJ/kg is represents cooling of the superheated refrigerant vapour in the condenser
down to the saturated vapour temperature as the heat was released to the surrounding during
that time. The remaining process of cooling down is where the latent heat is removed as we
mentioned above, while at point of enthalpy and pressure ; 87.38 KJ/kg and 0.14 MPA is the
point where the liquid/vapor is passed through an expansion device, the pressure is reduced
without any enthalpy change. Finally, the straight line from 87.38 KJ/kg to 277.13 KJ/kg is
23
the point where the liquid/vapor is evaporated completely to a gas and where enthalpy is
extracted from surroundings as the latent heat were entered into the system.
Next, the energy balance for the condenser and compressor is calculated in order to
observe the heat output from the equipment itself. At the condenser, for about 2.25 kW of heat
is released to the environment of surroundings, while 0.512 kW is needed by the compressor
in order to produce the work so that the process of compression of low pressure refrigerant
vapour to high pressure will be accomplished.
Last but not least, compressor pressure ratio (CPR) is the ratio of the air total pressure
(pt) exiting the compressor to the air pressure entering the compressor. To produce the
increase in pressure, the compressor must perform work on the flow. The CPR that have been
calculated is 4.857 while the volume efficiency is around 5.695.
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CONCLUSION
The objective of Experiment 1 is to determine the power input, heat output and
coefficient of performance of a vapour compression heat pump system. The value of power
input, heat output and coefficient of performance is obtained and calculated successfully.
Thus, the objective of the experiment is achieved.
For Experiment 3, the objectives are to plot the vapour compression cycle on the p-h
diagram and compare with ideal cycle and to perform energy balances for the condenser and
compressor. Both experiments’ objective can be achieved depending on the graph. As stated
in the discussion part, the graph shows that the coefficient of performances falls between the
heat output and power input at the beginning and falls below the value of power input at the
end of experiment. Based on the theory, the Coefficient of Performance, (COPH) of a heat
pump cycle is an expression of the cycle efficiency and is stated as the ratio of the heat
removed in the heated space to the heat energy equivalent of the energy supplied to the
compressor. The COPH should maintain in between both heat output and power input in order
for the heat pump to cycle efficiently. Thus, it can be concluded that the experiment only
achieved the objective of showing the performance curves but not theoretically.
For the experiment 5, the main objective is to determine the compression ratio and
volumetric efficiency. This value can be obtained by calculating the compressor pressure ratio
and volumetric efficiency using formula given.
25
RECOMMENDATIONS
1. Make sure the reading is stabilized and waited about 15 minutes before taking the
reading because it will affect the result.
2. The water supply must be in good condition and high in flow rate as it may affect
the result.
3. Any calculation and graph readings must be made repeatedly in order to avoid
error.
4. Ensure that the machine is in good condition and consult with the technician if
there any problem.
26
REFERENCES
1. Chemical Engineering Laboratory Manual. (CGE 536), Faculty of Chemical
Engineering, UiTM Shah Alam .
2. http://energy.gov/energysaver/articles/heat-pump-systems , Retrieved on 9th April
2015.
3. https://books.google.com.my/books?
id=ovKrOVRQpkkC&dq=refrigeration+unit+experiment&source=gbs_navlinks_s ,
Retrieved on 9th April 2015.
4. https://books.google.com.my/books?
id=JyGeRoZIy80C&dq=vapour+compression+cycle&source=gbs_navlinks_s ,
Retrieved on 9th April 2015.
5. https://books.google.com.my/books?
id=sfE6f21oCkAC&pg=PA735&dq=vapour+compression+cycle&hl=en&sa=X&e
i=yZ4mVc6YHIvjuQTM3oGoDA&ved=0CCcQ6AEwAg#v=onepage&q=vapour
%20compression%20cycle&f=false Retrieved on 9th April 2015.
6. http://www.grc.nasa.gov/WWW/k-12/airplane/compth.html , Retrieved on 9th April
2015.
7. http://www.arca53.dsl.pipex.com/index_files/phrefrig_files/image002.gif
Retrieved on 9th April 2015.
8. Lambers, K., et al (2008). Isentropic and Volumetris Efficiencies for Compressors.
School of Mechanical Engineering. 19(23-30).
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APPENDICES
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