Table of content

page

abstract 2

introduction 3

objective 4

theory 4

apparatus 5

procedure 6

result and calculation 7

discussion 17

conclusion 18

recommendation 18

references 18

appendices 19

1

Abstract

A refrigeration apparatus was used to evaluate the performance of a refrigeration cycle.

Four experiments need to be done to get the good result. From the first experiment, heat output and coefficient of performance of a vapor compression heat pump can be calculated. From the second experiment, graph of power input against temperature, coefficient of performance against temperature and heat output against temperature can be plotted. From experiment three, p-h diagram can be constructed.

Example of p-h diagram of ideal cycle.

From the last experiment, the compression ration and volumetric efficiency can be calculated.

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Introduction

A machine or device that moves heat from one location (the 'source') at a lower temperature to another location (the 'sink' or 'heat sink') at a higher temperature using mechanical work or a high-temperature heat source is called heat pump. Food refrigerators and freezers, reversible-cycle heat pumps for providing building space heating, and air conditioners are some of the examples of heat pump. Due to the ability of heat pump that can move heat energy from one environment to another, and in either direction, heat pump can both bring heat into an occupied space, and take it out. A heat pump works the same as an ordinary air conditioner in the cooling mode. A heat pump uses an intermediate fluid called a refrigerant to absorb heat as it vaporizes and releases the heat when it condenses. Evaporator is needed to absorb heat from inside an occupied space and rejects this heat to the outside through the condenser. The reversing valve is the component that differentiates heat pump and air conditioner. The reversing valve allows for the flow direction of the refrigerant to be changed. This allows the heat to be pumped in either direction.

Example of heat pump cycle

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Objective

Experiment 1: Determination of power input, heat output and coefficient of performance

To determine the power input, heat output and coefficient of performance of a vapor compression heat pump system.

Experiment 2: Production of heat pump performance curves over a range of sources and delivery temperatures

To produce the performance of heat pump over a range of sources and delivery temperatures.

Experiment 3: Production of vapor compression cycle on p-h diagram and energy balance study

To plot the vapor compression cycle and the p-h diagram and compare with the ideal cycle.

To perform energy balances for the condenser and compressor

Experiment 4: Estimation of the effect of compressor pressure ratio on volumetric efficiency

To determine the compression ratio and volumetric efficiency.

Theory

Generally there are three types of heat pump system:

a. Mechanical vapor recompression heat pump operating at about 200℃b. Closed cycle vapor compression heat pump (electric and engine driven)c. Heat transformer ( a type of absorption heat pump)

Coefficient of performance (COP) is an expansion 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.

COPH =heat produce inside the room

power consumed by the compressor

Thus, for the theoretical simple cycle, this may be written as:

COPH =heatingeffect

heat of compression

COPH =heatingeffect

heat of compression

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COPH =(h2−h3)(h2−h1)

Cooling water and refrigerant flow rate display is in percentage (%). The formula to convert cooling water and refrigerant flow rate to PM are:

1) Cooling water flow rate (LPM) = coolingwater flow rate(% )

100 % x 5 LPM

2) Refrigerant flow rate (LPM) = refrigerant flow rate(% )

100% x 1.26 LPM

Apparatus

Condenser, compressor, refrigeration unit machine and its instruments, water

Procedure

General start up procedure

1. The unit and all instruments were checked and make sure they were in proper condition.2. Both water sources and drain were checked to be connected then water supply was

opened and the cooling water flow rate was set to be at 40%.3. Drain hose at the condensate was checked to be connected.4. Power supply was connected and main power was switched of follows by main switch at

the control panel.5. Refrigerant compressor was switched on and the unit was ready for experiment as soon as

temperature and pressure are constant.

General shut down procedure

1. Compressor was switched off follow by main switch and power supply.2. Water supply was closed and ensure that water was not lost running.

Experiment 1: Determination of power input, heat output and coefficient of performance

Procedures:

1. The general start up procedure was performed.

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2. The cooling water flow rate was adjusted to 40%.3. The system was allowed to run for 15 minutes.4. The reading was recorded.

Experiment 2: Production of heat pump performance curves over a range of sources and delivery temperatures

1. The general start up procedure was performed.2. Water flow rate was adjusted to 60%.3. The system was allowed to run for 15 minutes.4. The reading was recorded.5. Experiment was repeated with reduced water flow rate so that the cooling water outlet

increases about 3℃.

Experiment 3: Production of vapor compression cycle on p-h diagram and energy balance study.

1. The general start up procedure was performed.2. The cooling water flow rate was adjusted to 40%.3. The system was allowed to run for 15 minutes.4. The reading was recorded.

Experiment 4: Estimation of the effect of compressor pressure ratio on volumetric efficiency.

1. The general start up procedure was performed2. The cooling water flow rate was adjusted to 40%.3. The system was allowed to run for 15 minutes.4. The reading was recorded.

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Result and calculation

Experiment 1

cooling water flow rate, FT1 % 39.9cooling water inlet temperature, TT5

C 27.8

cooling water outlet temperature,TT6

C 30.3

compressor power input W 159

Cooling water flow rate, LPM = coolingwater flow rate(% )

100 %x 5 LPM

= 39.9x 5

100

= 1.995 LPM ≈ 2 LPM

2L1min

x 1 m 31000L

x 1min60 s

= 3.333 x 10-5 m3/s

Find h1 and h2 from table saturated water-temperature using interpolation

T (°C) h (kJ/kg)25 104.8327.8 h1

30 125.74

T (°C) h (kJ/kg)30 125.7430.3 h2

35 146.64

h1 = 116.65 kJ/kg

h2 = 126.97 kJ/kg

ρwater = 1000kg/m3

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Mass flow rate = 3.333x 10−5 m 3s

x 1000kgm 3

= 0.0333 kg/s

Ein = Eout

mh1 = QH + mh2

QH = m(h1 – h2)

= 0.0333 kg/s (116.65 – 126.97) kJ/kg

= -0.344 kJ/s

COPH = Q HW

= 0.344kJ /s159kJ /s

= 0.00216

Experiment 2

reading 1 reading 2 reading 3cooling water flow rate, FT1

% 61.8 40.2 20.0

cooling water inlet temperature,TT5

C 27.9 27.9 27.9

cooling water outlet temperature,TT6

C 30.0 30.3 31.1

compressor power input

W 158 159 158

Reading 1,

Cooling water flow rate, LPM = coolingwater flow rate(% )

100 %x 5 LPM

8

= 61.8 x5

100 = 3.09 LPM

3.09L1min

x 1 m 31000L

x 1min60 s

= 5.15 x 10-5 m3/s

Find h1 and h2 from table saturated water-temperature using interpolation.

T (°C) h (kJ/kg)25 104.8327.9 h1

30 125.74

T (°C) h (kJ/kg)30 125.74

h1 = 116.95 kJ/kg

h2 = 125.74 kJ/kg

ρwater = 1000kg/m3

Mass flow rate = 5.15x 10−5m3s

x 1000kgm 3

= 0.0515 kg/s

Ein = Eout

mh1 = QH + mh2

QH = m(h1 – h2)

= 0.0515 kg/s (116.95 – 125.74) kJ/kg

= -0.452 kJ/s

COPH = Q HW

= 0.452kJ /s158kJ /s

= 0.00286

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Reading 2,

Cooling water flow rate, LPM = coolingwater flow rate(% )

100 %x 5 LPM

= 40.2 x5

100 = 2.01 LPM

2.01L1min

x 1 m 31000L

x 1min60 s

= 3.35 x 10-5 m3/s

Find h1 and h2 from table saturated water-temperature using interpolation.

T (°C) h (kJ/kg)25 104.8327.9 h1

30 125.74

T (°C) h (kJ/kg)30 125.7430.3 h2

35 146.64

h1 = 116.95 kJ/kg

hB2,2 = 126.97 kJ/kg

ρwater = 1000kg/m3

Mass flow rate = 3.35x 10−5m3s

x 1000 kgm 3

= 0.0335 kg/s

Ein = Eout

mh1 = QH + mh2

QH = m(h1 – h2)

= 0.0335 kg/s (116.95 – 127.97) kJ/kg

= -0.369kJ/s

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COPH = Q HW

= 0.369kJ /s159kJ /s

= 0.00232

Reading 3,

Cooling water flow rate, LPM = coolingwater flow rate(% )

100 %x 5 LPM

= 20x 5100

= 1 LPM

1L1min

x 1 m 31000L

x 1min60 s

= 1.67 x 10-5 m3/s

Find h1 and h2 from table saturated water-temperature using interpolation.

T (°C) h (kJ/kg)25 104.8327.9 h1

30 125.74

T (°C) h (kJ/kg)30 125.7431.1 h2

35 146.64

h1 = 116.95 kJ/kg

h2 = 130.55kJ/kg

ρwater = 1000kg/m3

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Mass flow rate = 1.67 x10−5 m 3s

x 1000kgm 3

= 0.0167 kg/s

Ein = Eout

mh1 = QH + mh2

QH = m(h1 – h2)

= 0.0167 kg/s (116.95 – 130.55) kJ/kg

= -0.227 kJ/s

COPH = Q HW

= 0.227kJ /s158kJ /s

= 0.00144

From the value that calculated, graph can be constructed

reading T COPhp PowerQh (kJ/s)

1 26.3 0.00214 158 -0.4522 26.2 0.002 159 -0.3693 26.1 0.00144 158 -0.227

Graph of COPhp against temperature

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29.8 30 30.2 30.4 30.6 30.8 31 31.20

0.0005

0.001

0.0015

0.002

0.0025

0.003

0.0035

Graph of power against temperature

29.8 30 30.2 30.4 30.6 30.8 31 31.2157.4157.6157.8

158158.2158.4158.6158.8

159159.2

Graph of Qh against temperature

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29.8 30 30.2 30.4 30.6 30.8 31 31.2

-0.5-0.45

-0.4-0.35

-0.3-0.25

-0.2-0.15

-0.1-0.05

0

Experiment 3

refrigerant flow rate,FT2 % 60.8refrigerant pressure (low), P1 bar (abs) 1.9refrigerant pressure (high),P2 bar (abs) 6.8refrigerant temperature,TT1 ℃ 26.9refrigerant temperature,TT2 ℃ 84.4refrigerant temperature,TT3 ℃ 28.5refrigerant temperature,TT4 ℃ 22.5cooling water flow rate,FT1 % 40.4cooling water inlet temperature,TT5

C 27.8

cooling water inlet temperature,TT6

C 30.2

compressor power input W 160

1.9 bar x 101.325 kpa1.01325 ¿̄ ¿ = 190 kPa

6.8 bar x 101.325 kpa1.01325 ¿̄ ¿ = 680 kPa

interpolation from superheated R-134a table

p (kPa) h (kJ/kg)180 207.90190 h1200 206.03

p (kPa) h (kJ/kg)

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650 178.51680 h2700 176.21

Calculation 3 and 4,

Find hc3 and hc4 using interpolation from saturated R-134a table

T (°C) h (kJ/kg)28 90.6928.5 h330 93.58

T (°C) h (kJ/kg)22 82.1422.5 h424 84.98

h1 = 206.97 kJ/kg

h2 = 177.13 kJ/kg

h3 = 91.35 kJ/kg

h4 = 82.79 kJ/kg

From the value that calculated, p-h diagram can be constructed

h (kJ/kg)Pressure

(kpa)206.97 190

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177.13 68091.35 68082.79 190206.97 190

60 80 100 120 140 160 180 200 2200

100200300400500600700800

p-h diagram

entalphy(kj/kg)

pres

sure

(kPa

)

Experiment 4

refrigerant flow rate,FT2 % 60.8refrigerant pressure (low),P1 bar (abs) 19.0refrigerant pressure (high), P2 bar (abs) 6.8refrigerant temperature, TT1 C 27.0

Compressor pressure ratio = suction pressureof refrigerantdischarge pressureof refrigerant

= 19bars6.8bars

= 2.79

Volumetric efficiency = Actual volumetric flow rate

Theoreticalvolume of compressor

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Refrigerant flow rate, LPM = refrigerant flow rate(% )

100% x 1.26 LPM

= 0.608 x 1.26 LPM X

= 0.766 LPM

Change LPM to the kg/s

Mass flow rate= 0.766 LPM x 1000cm3

1 L x1m3

¿¿ x

1min60 s

x 4.25 kg

m3

= 5.43 x 10-5 kgs

Density of refrigerant 134a = 4.25 kg

m3

Actual volume flow rate =mass flowrate

density of refrigerant134 a

= 5.43 x 10-5 kgs x m3

4.25kg

= 1.28 x 10-5 m3

s

Volumetric efficiency = Actual volumetric flow rate

Theoreticalvolume of compressor

= 1.28 x 10-5 m3

s x

s

1.61x 10−5m3

= 0.79

Discussion

From experiment one, coefficient of performance of the vapor compression heat pump system can be calculated and the value is 0.00216. The reason for expressing the efficiency of a refrigerator by another term that is coefficient of performance is the desire to avoid the oddity of having efficiencies greater than unity. Most heat pumps in operation today have a seasonally averaged COP of 2 to 3. COP of a refrigerator decreases with decreasing refrigeration temperature. Therefore it is not economical to refrigerate to a lower temperature than needed. The COPs of refrigerators are in the range of 2.6 to 3 for cutting and preparation rooms. It is a

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3 2

4 1

Superheating

Subcooling Critical Point

LiquidPressure

Enthalpy

good energy conservation practice to use separate refrigeration system to meet different refrigeration needs. From experiment two, graph of power input against temperature, coefficient of performance against temperature and heat output against temperature can be plotted. From experiment three, p-h diagram can be constructed.

In ideal vapour compression cycle the refrigerant vapour is compressed isentropically to a higher temperature and pressure (12). The compressed vapour is then condensed isobarically which results in heat rejection to the surroundings (23). The next step is the adiabatic throttling of the refrigerant to the low temperature and pressure (34). The final step is where the refrigerant is evaporated isobarically at low temperature and pressure, which results in the absorption of heat from its surroundings (41). From the last experiment, the compression ratio is 2.79 and the volumetric efficiency is 0.79. There are some errors happen during experiment. The refrigerant compressor had been switched on earlier before checking whether all instruments were in good condition or not. Water supply also was slow and there were not enough water while running the experiment.

Conclusion

From experiment one, coefficient of performance of the vapor compression heat pump system can be calculated and the value is 0.00216. From experiment two, graph of power input against temperature, coefficient of performance against temperature and heat output against temperature can be plotted. From experiment three, p-h diagram can be constructed. From the last

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experiment, the compression ratio is 2.79 and the volumetric efficiency is 0.79. COP of a refrigerator decreases with decreasing refrigeration temperature.

Recommendation

1. We must make sure that all instruments are in good condition before starting the experiment.

2. Make sure that there is enough water while doing the experiment.3. Make sure that the water always continuously running while doing the experiment.4. Make sure that the experiment follows the right time as stated in the procedure.

References

Thermodynamics 6th edition, Yunus A. Cengel. Michael A. Boles Advance engineering thermodynamics, Adrian Bejan Basic engineering thermodynamics, Rayner Joel, B.sc(eng) Lond.C.Eng,

F.I.Mech.E,F.I,Mar.E

Date of search: 26/2/2011

http://en.wikipedia.org/wiki/Refrigeration http://www.master-bilt.com/products/refrigeration.htm

Date of search: 28/2/2011

http://www.heatpumps.org.uk/HowDoesAHeatPumpWork.htm http://www.springerlink.com/content/0q288773655705u5/ http://en.wikipedia.org/wiki/Vapor-compression_refrigeration

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