Refrigeration Unit

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withdrawn and heat in the hot body is expelled from it.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


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


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



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



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




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



Refrigerant Flow Rate, FT2 % 60.5

Refrigerant Pressure (Low), P1 Bar (abs) 1.4

Refrigerant Pressure (High), P2 Bar (abs) 6.8


SAMPLE CALCULATIONS



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



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



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.

24

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.


id=JyGeRoZIy80C&dq=vapour+compression+cycle&source=gbs_navlinks_s ,



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


8. Lambers, K., et al (2008). Isentropic and Volumetris Efficiencies for Compressors.

School of Mechanical Engineering. 19(23-30).

27


http://www.grc.nasa.gov/WWW/k-12/airplane/compth.html

https://books.google.com.my/books?id=sfE6f21oCkAC&pg=PA735&dq=vapour+compression+cycle&hl=en&sa=X&ei=yZ4mVc6YHIvjuQTM3oGoDA&ved=0CCcQ6AEwAg#v=onepage&q=vapour%20compression%20cycle&f=false



https://books.google.com.my/books?id=JyGeRoZIy80C&dq=vapour+compression+cycle&source=gbs_navlinks_s

https://books.google.com.my/books?id=JyGeRoZIy80C&dq=vapour+compression+cycle&source=gbs_navlinks_s

https://books.google.com.my/books?id=ovKrOVRQpkkC&dq=refrigeration+unit+experiment&source=gbs_navlinks_s

https://books.google.com.my/books?id=ovKrOVRQpkkC&dq=refrigeration+unit+experiment&source=gbs_navlinks_s

http://energy.gov/energysaver/articles/heat-pump-systems

APPENDICES

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