refrigerant unit

UNIVERSITI TEKNOLOGI MARA FAKULTI KEJURUTERAAN KIMIA

PROCESS ENGINEERING LABORATORY 1(CPE435)

No. Title Allocated Marks (%) Marks

1 Abstract/Summary 5  2 Introduction 5  3 Aims 5  4 Theory 5  5 Apparatus 5  6 Methodology/Procedure 10  7 Results 10  8 Calculations 10  9 Discussion 20  

10 Conclusion 5  11 Recommendations 5  12 Reference / Appendix 5  13 Supervisor’s grading 10  

TOTAL MARKS 100  

Remarks:

Checked by: Rechecked by:

--------------------------- ---------------------------

Date: Dat

1

NAME : MUHAMMAD MUKMIN BIN ZAHARISTUDENT ID. : 2011461308GROUP : EH220 2CEXPERIMENT : REFRIGERANT UNITDATE PERFORMED : 02TH APRIL 2012SEMESTER : 2PROGRAMME / CODE : EH220SUBMIT TO : CIK SITI NOR ADIEB BT IDRIS

Table of content

Table of content...............................................................................................................................2Abstract............................................................................................................................................3Introduction......................................................................................................................................4Aims.................................................................................................................................................5Theory..............................................................................................................................................6Apparatus.........................................................................................................................................9Experimental procedure.................................................................................................................10RESULTS......................................................................................................................................12

Experiment 1..............................................................................................................................12Experiment 2..............................................................................................................................14Experiment 3..............................................................................................................................20Experiment 4..............................................................................................................................24Discussion..................................................................................................................................27

Conclusion.....................................................................................................................................30Recommendations..........................................................................................................................31References......................................................................................................................................32Appendices....................................................................................................................................33

2

Abstract

This experiment is carried out to study the mechanical heat pump through the operating

of the equipment SOLTEQ Mechanical Heat Pump (Model: HE165). This experiment is

conducted by doing a series of tests by manipulating the delivery temperatures and flow rates of

the cooling water. Plus, to study the mechanism of a heat pump, coefficient of performance and

heat pump performance curves. Besides, this experiment is investigating the effects of

compressor pressure ratio on the volumetric efficiency. The purpose of conducting this

experiment is to study the thermodynamics concept of energy balance and coefficient of

performance. The heat transfer is been able to determine by applying the energy balance concept.

Next, able to determine the coefficient of performance of the heat pump and how the delivery

temperatures affect the coefficient of performance. Lastly, to produce the performance curves of

the heat pump and vapour compression cycle on a p-h diagram.

3

Introduction

In the experiment of refrigerant unit, the Mechanical Heat Pump is used to study the

vapour compression cycle. This Mechanical Heat Pump (Model: HE 165) has been designed to

provide students with a practical and quantitative demonstration of the vapour compression cycle.

Refrigerators and heat pumps are both use the vapour compression cycle, and although the

applications of these machines differ, the components are essentially the same. The HE 165 is

capable of demonstrating the heat pump application where a large freely available energy source,

such as the atmosphere, is to be upgraded for water heating. The unit will be of particular interest

to those studying Mechanical Engineering, Energy Conservation, Thermodynamics, Building

Services, Chemical Engineering, Plant and Process Engineering, and Refrigeration and Air

Conditioning.

The heat pump consists of a hermetic compressor, a water cooled condenser, a

thermostatically controlled expansion valve and an air heated evaporator. The arrangements of the

components are in a manner similar to that used for many domestic air-water heat pumps where

they are visible from the front of the unit. During the operation, slightly superheated refrigerant

(R134a) vapour enters the compressor from the evaporator and its pressure is increased. Thus, the

temperature rises and the hot vapour then enters the water cooled condenser. Heat is given up to

the cooling water and the refrigerant condenses to a liquid before passing to the expansion valve.

Upon passing through the expansion valve the pressure of the liquid refrigerant is reduced.

This causes the saturation temperature to fall to below that of the atmosphere. Thus, as it flows

through the evaporator, there is a temperature difference between the refrigerant and the air being

drawn across the coils. The resulting heat transfer causes the refrigerant to boil, and upon leaving

the evaporator it has become slightly superheated vapour, ready to return to the compressor.

The temperature at which heat is delivered in the condenser is controlled by the water flow rate

and its inlet temperature. The evaporating temperature is l a r g e l y d e t e r m i n e d b y t h e

ambient conditions. However, this can be limited, either by restricting the air intake to the

evaporator, or by directing warmed air towards the intake. Instrumentations are all provided

for the measurement of flowrates of both the refrigerant and cooling water, power input to the

4

compressor, and all relevant temperatures.

Aims

To determine the power input, heat output and coefficient of performance of a

vapour compression heat pump system

To produce the performance of heat pump over a range of source and delivery

temperatures

To plot the vapour compression cycle on the p-h diagram and compare with the

ideal cycle

To perform energy balance for the condenser and compressor

To determine the compression ratio and volumetric efficiency

5

Theory

Refrigerators and heat pumps are devices that absorb heat at a low temperature and reject heat at a

higher temperature. Both refrigerators and heat pumps operate in the same way using a reversed

heat engine cycle, but are labeled separately according to their purpose. Devices that are used to

maintain a temperature below ambient are known as refrigerators (or air conditioners) while

devices that are used to supply heat at a higher temperature than ambient are known as heat

pumps. Some devices, in particular those used for space cooling, may be used as both a heat pump

and a refrigerator (as is the case with the Christchurch Town Hall heating/cooling system).

The vapour compression cycle is the most commonly used refrigeration/heat pump cycle and

involves the same four processes as a heat engine cycle but in the reverse order (i.e. evaporation –

compression – condensation – expansion/throttling). Figure 1 gives a schematic representation of

the four essential mechanical components in this cycle; figure 2 shows the ideal cycle on a

pressure-enthalpy diagram.

Heat out

3 2

Condenser

Throttle Work in

Valve

Evaporator Compressor

4 1

6

3 2

4 1

Superheating

Subcooling Critical Point

LiquidPressure

Enthalpy

Heat in

Figure 1: Vapour Compression Cycle components

Figure 2: Ideal cycle on pressure – enthalpy diagram.

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). In an actual cycle states 1 and 3 should not lie on the saturation line

as there is subcooling (helpful to obtain state 3 in the diagram) and superheating (necessary to

avoid droplets (two phase region!) in the compressor).

The usual measure of performance of a refrigerator or heat pump is the Coefficient of Performance

COP which for a refrigerator COPR is defined as:

7

(1a)

For a heat pump COPH:

(1b)

where E, C, R, H stand for Evaporator, Compressor, Refrigeration, and Heat pump respectively.

Note: The cooling water and refrigerant flow rate displays is in percentage (%). Below are the

formula to convert cooling water and refrigerant flow rate to LPM

1. Coolingwaterflowrate ( LPM )=Coolingwaterflowrate (% )100 %

x5 LPM

2. Refrigerantflowrate ( LPM )= Refrigerantflowrate ( %)100 %

x 1.26 LPM

8

Apparatus

SOLTEQ Mechanical Heat Pump Model: HE165

9

Experimental procedure

General Start-up Procedure

1. The unit and all instruments are checked in proper condition.

2. Both water source and drain are checked connected then the water supply is open and

the cooling water flow rate is set at 1.0LPM.

3. The drain hose at the condensate collector is checked connected.

4. Power supply is connected and the main power is switched followed by main switch

at the control panel.

5. The refrigerant compressor is switched on. As soon as temperature and pressure were

constant, the unit was ready for experiment.

General Shut-down Procedure

1. The compressor is switched off, followed by main switch and power supply.

2. The water supply is closed and water is ensured not left running.

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

Objective:

To determine the power input, heat output and coefficient of performance of a vapour

compression heat pump system

Procedures:

1. The general start-up procedures are performed.

2. The cooling water flow rate is adjusted to 40%.

3. The system is allowed to rub for 15 minutes.

4. All necessary reading is recorded into experimental data sheet.

Experiment 2: Production of heat pump performance curves over a range of source and

delivery temperatures

Objective:

To produce the performance of heat pump over a range of source and delivery

temperatures

Procedures:

10

1. The general start-up procedures are performed

2. The cooling water flow rate is adjusted to 80%.

3. The system is allowed to run for 15 minutes.

4. All necessary readings are recorded into the experimental data sheet.

5. The experiment with reducing flow rate is repeated so that the cooling water

outlet temperature increases by about 3℃ (40% and 60%).

6. The experiment might be repeated at different ambient temperature.

Experiment 3: Production of water vapour compression cycle on p-h diagram and energy

balance study

Objective:

1. To plot the vapour compression cycle on the p-h diagram and compare with the ideal

cycle

2. To perform energy balances for the condenser and compressor

Procedures:

1. The general start-up procedures are performed

2. The cooling water flow rate is adjusted to 40% and the system is allowed to run for

15 minutes.

3. All necessary readings are recorded into experimental data sheet.

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

Objective:

To determine the compression ratio and volumetric efficiency

Procedures:

1. The general start-up procedures are performed.

2. The cooling water flow rate is adjusted to 40%.

3. The system is allowed to run for 15 minutes.

4. All necessary readings are recorded into experimental data sheet.

5. The experiment might be repeated at different compressor delivery pressure.

11

RESULTS

Experiment 1

Cooling water flow rate, FT 1 % 40.0

Cooling water inlet temperature, TT 5 °C 26.9

Cooling water outlet temperature, TT 6 °C 31.7

Compressor power input W 194

For reading 1,

Cooling water flow rate, LPM = coolingwaterflowrate (%)100 %

x 5 LPM

= 40.0 x5

100

= 2.0 LPM

2.0 L1min x 1m3

1000 L x 1min60 s = 3.333 x 10-5 m3/s

Find h5 and h6 from table saturated water-temperature using interpolation

T (°C) h (kJ/kg)

25 104.83

26.9 112.78

30 125.74

T (°C) h (kJ/kg)

30 125.74

31.7 132.85

35 146.64

h5 = 112.78 kJ/kg h6 = 132.85 kJ/kg ρwater = 1000kg/m3

12

Mass flow rate = 3.333 x 10−5 m 3

s x 1000 kg

m3 = 0.0333 kg/s

Ein = Eout

Mh5 = QH + mh6

QH = m(h5 – h6)

= 0.0333 kg/s (112.78 – 132.85) kJ/kg

= -0.6683 kJ/s

COPH = Q HW

= 0.6683 kJ /s

194 kJ /s

= 0.00339

13

Experiment 2

Reading 1 Reading 2 Reading 3

Cooling water flow rate, FT 1 % 80.0 60.0 40.0

Cooling water inlet temperature, TT

5

°C 27.1 27.1 27.1

Cooling water outlet temperature,

TT 6

°C 30.3 31.1 32.5

Compressor power input W 193 196 198

Reading 1,

Cooling water flow rate, LPM = coolingwaterflowrate (%)100 %

x 5 LPM

= 80.0 x5

100 = 4.0 LPM

4.0 L1min x 1m3

1000 L x 1min60 s = 6.6700 x 10-5 m3/s

14

Find h5 and h6from table saturated water-temperature using interpolation.

T (°C) h (kJ/kg)

25 104.83

27.1 113.61

30 125.74

T (°C) h (kJ/kg)

30 125.74

30.3 127.00

35 146.64

h5 =113.61kJ/kg

h6 = 127.00 kJ/kg

ρwater = 1000kg/m3

Mass flow rate = 6.67 x10−5 m3s

x 1000 kg

m3 = 0.0667 kg/s

Ein = Eout

mh5 = QH + mh6

QH = m(h5 – h6)

= 0.0667 kg/s (113.61–127.00) kJ/kg

= -0.8931 kJ/s

COPH = Q HW

= 0.8931 kJ / s

193 kJ / s

15

= 0.004627

Reading 2,

Cooling water flow rate, LPM = coolingwaterflowrate (%)

100 %x 5 LPM

= 60.0 x 5

100 = 3.0 LPM

3.0 L1min x 1m3

1000 L x 1min60 s = 5.000 x 10-5 m3/s

Find hB2,1 and hB2,2from table saturated water-temperature using interpolation.

T (°C) h (kJ/kg)

25.0 104.83

27.1 113.61

30.0 125.74

T (°C) h (kJ/kg)

30.0 125.74

31.1 130.34

35.0 146.64

h5 = 113.61 kJ/kg h6 = 130.34 kJ/kg ρwater = 1000kg/m3

Mass flow rate = 5.000 x 10−5 m3s

x 1000 kg

m3 = 0.0500 kg/s

Ein = Eout

mh5 = QH + mh6

QH = m(h5 – h6)

= 0.0500 kg/s (113.61 – 130.34) kJ/kg

= -0.8365 kJ/s

COPH = Q HW

16

= 0.8365 kJ /s

196kJ /s

= 0.004268

Reading 3,

Cooling water flow rate, LPM = coolingwaterflowrate (%)

100 %x 5 LPM

= 40. 0 x5

100

= 2.0 LPM

2.0 L1min x

1m31000 L x

1min60 s = 3.333 x 10-5 m3/s

Find hB3,1 and hB3,2from table saturated water-temperature.

T (°C) h (kJ/kg)25.0 104.8327.1 113.6130.0 125.74

T (°C) h (kJ/kg)30.0 125.74

32.5 136.19

35.0 146.64

h5 = 113.61 kJ/kg h6 = 136.19 kJ/kg ρwater = 1000kg/m3

Mass flow rate = 3.333 x 10−5m 3

s x 1000 kg

m3 = 0.0333 kg/s

Ein = Eout

Mh5 = QH + mh6

QH = m(h5 – h6)

= 0.0333 kg/s (113.61 – 136.19) kJ/kg

= -0.7519 kJ/s

17

COPH = Q HW

= 0.7519 kJ /s

198kJ /s

= 0.003797

From the value that calculated, graph can be constructed

T6(°C) Power input(kJ/s)

Qh(kJ/s) COP

Reading 1 30.3 193 0.8931 0.004627

Reading 2 31.1 196 0.8365 0.004268

Reading 3 32.5 198 0.7519 0.003797

Graph 1

30 30.5 31 31.5 32 32.5 330

0.0005

0.001

0.0015

0.002

0.0025

0.003

0.0035

0.004

0.0045

0.005 0.0046270.004268

32.5,0.003797

graph of COP vs temperature

Temperature (°C)

CO

P

18

Graph 2

30 30.5 31 31.5 32 32.5 330.65

0.7

0.75

0.8

0.85

0.9

0.95

0.8931

0.8365

32.5, 0.7519

graph of QH vs temperature

Temperature(°C)

Hea

t tra

nsfe

r,Q

H(k

J/s)

Graph 3

19

30 30.5 31 31.5 32 32.5 33190

191

192

193

194

195

196

197

198

199

193

196

32.5, 198graph of power input vs temperature

Temperature (°C)

pow

er (k

J/s)

Experiment 3

Refrigerant flow rate, FT2 % 35.2

Refrigerant pressure (low), P1 Bar (abs) 3.1

Refrigerant pressure (high), P2 Bar (abs) 8.0

Refrigerant temperature, TT1 ⁰C 5.9

Refrigerant temperature, TT2 ⁰C 34.2

Refrigerant temperature, TT3 ⁰C 31.3

Refrigerant temperature, TT4 ⁰C 20.4

Cooling water flow rate, FT1 % 40.0

Cooling water inlet temperature, TT5 ⁰C 27.0

Cooling water outlet temperature ,

TT6

⁰C 32.4

Compressor power input W 182

Find hc1 and hc2 using interpolation from superheated R-134a table

At T = 5.9 ⁰C

p (MPa) h (kJ/kg)

0.28 256.05

20

0.31 255.26

0.32 254.99

At T = 34.2 ⁰C

p (MPa) h (kJ/kg)

0.80 270.34

Find hc3 and hc4 from saturated R-134a table at given T and P = 0.8MPa

T (°C) h (kJ/kg)

31.3 267.29

20.4 95.47

h1 = 255.26 kJ/kg

h2 = 270.99kJ/kg

h3 = 267.29/kg

h4 = 95.47kJ/kg

Condenser energy balance

Refrigerant flow rate, LPM = coolingwaterflowrate (%)100 %

x 1.26 LPM

= 35.2% x1.26

100

= 0.44 LPM

0.44 L1 min x

1 m31000 L x

1min60 s = 7.333 x 10-6 m3/s

Mass flow rate = 7.333 x 10−6 m3s

x 1000 kg

m3 = 0.00733kg/s

Ein = Eout

QH = m h3−m h4

QH = m(h3−h4)

=0.00733 kg/s (267.29 –95.47) kJ/kg

21

= 1.259kJ/s

Compressor energy balance

W ¿=m (h2−h1 )

¿ 0.00733 kgs

(270.99−95.47 )kJ /kg

¿128.87 kJ /s

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

h

(kJ/kg)

Pressure

(Mpa)

255.26 0.31

270.99 0.8

267.29 0.8

95.47 0.8

Graph 1Experimental graph

22

80 100 120 140 160 180 200 220 240 260 2800

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.31

0.8 0.80.8

0.31

graph of pressure vs enthalphy

Enthalphy,h (KJ/Kg)

Pres

sure

(MPa

)

Theoretical graph

23

24

Experiment 4

Refrigerant flow rate , FT2 % 36.0 35.5

Refrigerant pressure (low) , P1 Bar (abs) 3.2 3.1

Refrigerant pressure (high) , P2 Bar (abs) 8.1 8.1

Refrigerant temperature ,TT1 C⁰ 6.4 6.1

Reading 1

Compressor pressure ratio = suction pressure of refrigerant

discharge pressure of refrigerant

= 3.2bars8.1 bars

= 0.3951

Volumetric efficiency = Actualvolumetricflowrate

T h eoreticalvolumeofcompressor

Refrigerant flow rate, LPM = refrigerantflowrate(%)100 %

x 1.26 LPM

= 0.36 x 1.26 LPM

= 0.4536 LPM

Change LPM to the kg/s

Mass flow rate= 0.4536 LPM x 1000 cm3

1L x1m3

¿¿ x

1min60 s x 4.25

kgm3

= 3.213 x 10-5 kgs

Density of refrigerant 134a = 4.25 kgm3

Actual volume flow rate =massflowrate

densityofrefrigerant 134 a

25

= 3.213 x 10-5 kgs x m3

4.25 kg

= 7.56. x 10-6 m3

s

Volumetric efficiency = Actualvolumetricflowrate

T h eoreticalvolumeofcompressor

= 7.56. x 10-6 m3

s x

s1.61 x10−5m3

= 0.4696

Reading 2

Compressor pressure ratio = suction pressure of refrigerant

discharge pressure of refrigerant

= 3.1bars8.1 bars

= 0.3827

Volumetric efficiency = Actualvolumetricflowrate

The oreticalvolumeofcompressor

Refrigerant flow rate, LPM = refrigerantflowrate(%)100 %

x 1.26 LPM

= 0.355 x 1.26 LPM

= 0.4473 LPM

Change LPM to the kg/s

Mass flow rate= 0.4473 LPM x 1000 cm3

1L x1m3

¿¿ x

1min60 s x 4.25

kgm3

= 3.168 x 10-5 kgs

26

Density of refrigerant 134a = 4.25 kgm3

Actual volume flow rate =massflowrate

densityofrefrigerant 134 a

= 3.213 x 10-5 kgs x m3

4.25 kg

= 7.454x 10-6 m3

s

Volumetric efficiency = Actualvolumetricflowrate

T h eoreticalvolumeofcompressor

= 7.454 x 10-6 m3

s x

s1.61 x10−5m3

= 0.4627

Time Min 15 20Refrigerant flow rate , FT2 % 36.0 35.5Refrigerant pressure (low) , P1

Bar (abs)

3.2 3.1

Refrigerant pressure (high) , P2

Bar (abs)

8.1 8.1

Refrigerant temperature ,TT1 ⁰C 6.4 6.1Compressor pressure ratio - 0.3951 0.3827Volumetric efficiency - 0.4696 0.4627

27

Discussion

This experiment was carried out to calculate the performance each of the equipment in

the refrigerant unit. In the first experiment, the power input of the heat pump was recorded at

194kJ/s in order to absorb 0.6683kJ/s heat from the surroundings. The value of enthalpy at given

temperature was calculated using the interpolation method based on the value at Table A-4. The

value obtain was used to calculate the amount rate heat transfer in the system. The coefficient of

performance of the heat pump used is 0.00339.

For the second experiment, the same step as the first experiment was repeated at different

cooling water flow rate which is at 80%, 60% and 40%. From the experiment, the power input

for the heat pump is different for each water flow rate, which are, 193kJ/s, 196kJ/s and 198kJ/s

respectively. The power input varies as the cooling water flow rate decreases. The same method

was used to calculate the rate of heat transfer and the coefficient of performance (COP) for the

heat pump. The COP calculated for cooling water flow rate at 80%, 60% and 40% is 0.004627,

0.004268 and 0.003797 respectively. The COP of heat pump decreases as the cooling water flow

rate decreases.

28

In the third experiment, the change in pressure and temperature for refrigerant R-134A

after passing condenser and compressor was recorded. The enthalpy was calculated using

interpolation to calculate the change of enthalpy at compressor and condenser. At the compressor

the superheated refrigerant was compressed from 0.31 MPa at 5.9°C to 0.8Mpa at 34.2 °C and

the enthalpy calculated is 255.26kJ/kg and 270.99kJ/kg respectively. The R-134A enters the

compressor superheated then compressed at constant entropy the leaves as superheated. The

refrigerant then enter the condenser at temperature of 31.3°C at 0.8MPa and leave the condenser

at temperature at 20.4°C at 0.8MPa. The pressure is constant because the condenser undergoes

the heat rejection process at constant pressure. The enthalpy calculated at 31.3°C and 20.4°C is

267.29/kg and 95.47kJ/kg respectively. The enthalpy was decreased because at the condenser,

the heat energy is released to surrounding cause the heat energy in the refrigerant to drop.

Last experiment, the compression ratio and volumetric efficiency are been calculated.

The systems are allowed to run at different period of time before the readings were taken. First,

the systems are allowed to run for 15 minute then the steps repeated for 20 minute period.

The values of compressor pressure ratio are 0.3951 and 0.3827 respectively. There is slightly

difference in the ratio calculated. This is due to the difference in the reading of pressure before

entering the compressor. However, the pressure leaving the compressor for both period of time is

the same which is 0.81MPa. The volumetric efficiencies are then calculated, and the values are

0.4696 and 0.4627 respectively. There is also slightly difference in the value calculated that is

due to the difference in the refrigerant flow rate.

Higher compression ratio allows an engine to extract more mechanical energy from a

given mass of air fuel mixture due to its higher thermal efficiency and is most wanted. Based on

the value of volumetric efficiency, we can compare the actual value of volumetric flow rate in

theory to the value that obtained from this experiment. The differences of both values are high.

A several steps might be conducted inaccurately which results in all experiments in such an

outrageous values. Firstly the water flow rate is not in stable condition while taking reading.

Reading is been taken in the range +6 from the actual reading. Next experiment shows that the

flow rate of water that we should take is 80% but the reading is not constant. Hence, the reading

29

is been taken when it reach 80% and does not increase exceed 80%. This happen because the

water source in laboratory is not enough for this apparatus and the present of pump.

As going through this experiment, time does not give much effect, but a little time

difference could make a little variance from the theory

30

Conclusion

Overall this experiment is considering success. First and foremost experiment, the power

input, heat output and coefficient of performance of a vapour compression heat pump system are

been determined. In experiment 1, the values QH and COPH are- 0.6683kJ/s and 0.00339. For

second experiment, the performances of heat pump over a range of source and delivery

temperature are been able to produce. The values of COPH are 0.004627, 0.004268, and

0.003797. The values are decreasing. From the results, it can be concluded that the heat pump

functioning with a high percent of flow rate have higher efficiency compare to the lower flow

rate. As show in experiment 3, the p-h diagram of vapour compression cycle is been plotted

successfully and able to be compared with the ideal cycle and able to perform energy balances

for the condenser and compressor. It clearly can be seen that the values are differ in small

amount. Last experiment, ratio and volumetric efficiency are been able to be calculated. The

ratio is 0.3951 and 0.3827 and for the volumetric efficiency is 0.4696 and 0.4627. There is

slightly difference due to difference in pressure and refrigerant flow rate.

31

Recommendations

1. Consultation is compulsory in order to make sure the experiments are conducted

properly.

2. To make sure that the machine in good conditions and the experiment are been doing

wrong, we have to understand the general start up and general shut down.

3. Before the experiment begin, ensure that the mechanical heat pump should run and warm

up early for 15 minutes. It should be notice that, surrounding in the laboratory also affect

the result, thus it hard to get an accurate reading.

4. A trial should not be forgotten before running the experiment. This is due to get an

accurate result in experiment 3. Thus, the graph of vapour compression cycle plotted on

p-h diagram of R-134a is in correct order with appropriate ideal cycle.

5. While running the experiment, if some technical problems occur directly ask the

technician to overcome the problems.

32

References

UiTM Faculty of Chemical Engineering, Laboratory Manual CHE 465, Chemical

Engineering Lab 1

www.tyxer.hubpages.com/hub/theory-of-refrigeration

www.solution.com.my/pdf/HE165(A4).pdf

www.1.eere.energy.gov/manufacturing/tech_deployment/pdf/heatpump.pdf

Yunus A. Cengel, Micheal A. Boles “ Thermodynamics An Engineering Approach “,McGraw-Hill Book Company, 7th edition 2011.

33

Appendices

Figure 1: Compressor

Figure 2: Condenser

34

Figure 3: R-134A

Figure 4: Evaporator

35