ABSTRACT

The Refrigerant Unit experiment is carried out to observe how the mechanical heat

pump and thermodynamic refrigeration unit work. The equipment that is used in the

laboratory to perform the experiment is the SOLTEQ Mechanical Heat Pump (Model:

HE165). The experiment capabilities with different objectives. For experiment 1, the

objective is to determine the power input, heat output and coefficient of performance of a

vapour compression heat pump system while for experiment 2 is to produce the performance

of heat pump over a range of source and delivery temperatures. For experiment 3, there are

two objectives which are to plot the vapour compression cycle on the p-h diagram and

compare with the ideal cycle and to perform energy balances for the condenser and

compressor.

1

INTRODUCTION

The SOLTEQ Mechanical Heat Pump (Model: HE165) has been designed to provide

students with a practical and quantitative demonstration of a vapour compression cycle, and is

suitable for all course levels (intermediate and undergraduate). Refrigerators and heat pumps

both apply the vapour compression cycle, although the applications of these machines differ,

the components are essentially the same.

The Mechanical Heat Pump 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 particularly interest to those studying Mechanical

Engineering, Energy Conservation, Thermodynamics, Building Services, Chemical

Engineering, Plant and Process Engineering, Refrigeration and Air Conditioning.

2

OBJECTIVES

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 vapour

compression heat pump system.

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

delivery temperatures

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

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

study

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

-To perform energy balances for the condenser and compressor.

3

THEORY

When enough heat is released from a glass of water, the water will freeze to ice. When

that heat is absorbed by the ice, the ice will melt. Heat has its own laws, called the laws of

thermodynamics. One of those laws is that heat will move from a place that has a lot of heat

to a place that has less heat, or another way to put it is that heat will move from a place of

higher intensity to a place of lower intensity. From refrigeration theory, air conditioning and

refrigeration equipment is designed to create a cold area that acts as a "heat sponge" that will

soak up heat from air or food. The heat is then moved to a place where it can be released

safely and efficiently. The second point is to understand about refrigeration theory has to do

with why we use evaporators and condensers. When a liquid like water or refrigerant absorbs

enough heat to start boiling, what's happening is that the added heat energy causes the

vibration of the liquid's molecules to speed up to the point where they move far apart from

each other. When the molecules of liquid reach a certain distance from each other, the liquid

changes into a vapor. This is called boiling, evaporating, or vaporizing. A liquid absorbs some

levels of heat as it changes state to a vapor and air conditioning and refrigeration equipment is

designed to use this point of refrigeration theory by keeping a constant flow of refrigerant

vaporizing and absorbing heat in the evaporator. The evaporator is the "heat sponge" area, and

the refrigerant vaporizing inside of it is absorbing the heat. When vapor cools and releases

enough heat energy, it's molecules will slow down and move closer together to the point

where the vapor changes into a liquid. This is called condensation, and it's also a change of

state. To condense, a vapor must release the same level of heat that it absorbed when it

vaporized. Air conditioning and refrigeration uses this point of refrigeration theory by causing

refrigerant to cool and condense in the condensing unit. The refrigerant repeats this cycle

continuously, absorbing heat in the evaporator and releasing it in the condenser.

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APPARATUS

1. R-134-A Compressor

2. Evaporator

3. Water inlet and water outlet

4. Filter dryer

5. Power supply

6. Water

7. Valve

5

PROCEDURES

General Start-up Procedures

1. Both of the water source and the drain were checked before being connected,

then water supply is opened and the flow rate of cooling water was set to be at 1.0

LPM.

2. Checked that the drain hose at the condensate collector is connected.

3. The power supply was connected and switched on the main power follows by main

switch at the control panel.

4. Then the refrigerant compressor was switched on until the pressure and temperature

were in stabilizing condition.

General Shut-down Procedures

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

2. The water supply was closed and make sure that there was no water left running.

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

performance

1. The apparatus was step up.

2. The flow rate of cooling water was adjusted to 40%.

3. The system was run for 15 minutes.

4. Recorded all the data into the experimental data sheet.

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

source and delivery temperatures

1. .By continuing the steps in experiment 1, we adjusted the cooling water flow rate to

60% and the data was recorded.

2. The experiment was repeated with reducing water flow rate so that the cooling.water

outlet temperature increases by about 3°C.

3. The similar steps was repeated until the compressor delivery pressure reaches around

14.0 bars.

4. All the steps were repeated by different ambient temperature.

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

balance study

1. Followed the general start-up steps.

2. The flow rate of cooling water was adjusted to 40% and let the system run for 15

minutes.

3. All data was recorded in the experiment.

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RESULT

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

Cooling water flow rate, FT1 % 40.3

Cooling water flow rate, FT1 LPM 2.015

Cooling water inlet temperature, TT5 °C 27.7

Cooling water outlet temperature, TT6 °C 28.9

Compressor power input W 160

Heat output W 167.95

COPH No unit 1.04968

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

delivery temperatures.

Test 1 2 3

Cooling water flow rate, FT1 % 30.0 50.0 70.0

Cooling water flow rate, FT1 LPM 1.50 2.50 3.50

Cooling water inlet temperature, TT5 °C 27.9 27.9 28.0

Cooling water outlet temperature, TT6 °C 29.8 29.2 28.7

Compressor power input W 158 160 165

Heat output W 197.95 225.74 170.17

Coefficient of performance, COPH NO UINT 1.2528 1.4109 1.0313

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

Refrigerant flow rate, FT2 % 60.6Refrigerant flow rate, FT2 LPM 0.76Refrigerant pressure (low), P1 Bar(abs) 1.9Refrigerant pressure (high), P2 Bar(abs) 6.7Refrigerant temperature, TT1 °C 25.6Refrigerant temperature, TT2 °C 74.0Refrigerant temperature, TT3 °C 28.2Refrigerant temperature,TT4 °C 21.1Cooling water flow rate,FT1 % 40.0Cooling water flow rate, FT1 LPM 2.0Cooling water inlet temperature,TT5 °C 27.9Cooling water outlet temperature,TT6 °C 29.5Compressor power input W 157

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CALCULATION.

EXPERIMENT 1

VOLUMETRIC FLOWRATE = FLOW RATE (%) × 5L

MIN

= 0.403 × 5

= 2.015L

MIN

OUTPUT HEAT = Q × DENSITY

= 2.015Lmin

× 1 m31000 L

× 1 min60 s

× 997 kg1 m3 × 4180 J

kg. ˚C ×(28.9 – 27.7)

= 167.95 W

COPH = OUTPUT HEATPOWER INPUT

= 167.95

160

= 1.04968

EXPERIMENT 2

VOLUMETRIC FLOWRATE, Q = FLOW RATE (%) × 5L

MIN

= 0.3 × 5

= 1.5L

MIN

OUTPUT HEAT = MASS FLOW RATE × CpH2O × TEMPEARATURE CHANGE

= Q × DENSITY × CpH2O × TEMPEARATURE CHANGE

= 1.5Lmin

× 1 m31000 L

× 1 min60 s

× 997 kg1m3 × 4180 J

kg. ˚C ×(29.8 -27.9)

= 197. 95 W

COPH = OUTPUT HEATPOWER INPUT

= 166.7158

10

= 1.0523

*CALCULATION IS REPEATED FOR FLOW RATE 50 % AND 70 % .

28.6 28.8 29 29.2 29.4 29.6 29.8 300

50

100

150

200

250

1.25281.41091.0313

197.95

225.74

170.17

158160165

PERFORMANCE CURVES FOR HEAT PUMP (COEFFICIENT OF PERFORMANCE, OUTPUT HEAT AND POWER INPUT) VERSUS

WATER OUTPUT TEMPERATUR

PERFORMANCE CURVE FOR HEAT PUMP ( POWER INPUT) VERSUS WATER OUTPUT TEMPERATURER

PERFORMANCE CURVES FOR HEAT PUMP (COEFFICIENT OF PERFORMANCE, OUTPUT HEAT AND POWER INPUT) VERSUS WATER OUTPUT TEMPERA-TURE

PERFORMANCE CURVES FOR HEAT PUMP (COEFFICIENT OF PERFORMANCE, OUTPUT HEAT AND POWER INPUT) VERSUS WATER OUTPUT TEMPERA-TURE

WATER OUTPUT TEMPERATURE, ˚C

COEF

FICI

ENT

OF

PERF

ORM

ANCE

, OUT

PUT

HEAT

AND

PO

WER

INPU

T

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EXPERIMENT 3

Find h1 and h2 using interpolation from superheated R-134a table

PRESSURE = 1.9bar = 0.19 MPa

At T = 25.6 ⁰C TO FIND h1

PRESSURE(MPa) 0.18 0.19 0.2

TEMPERATURE(˚C

)

h (kJ/kg) h (kJ/kg) h (kJ/kg)

20 270.60 270.20

25.6 X h1 y

30 279.27 278.91

Using interpolation

279.27−xx−270.6

=30−25.625.6−20

1.7858 x = 491.9075

x = 275.45kJ/kg

278.91− yy−270.2

=30−25.625.6−20

1.7858 y = 491.23

y = 274.08kJ/kg

274.08−h 1h 1−275.45

= 0.2−0.190.19−0.18

2h1 = 549.53

h1 = 274.77kJ/kg

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* REPEAT CALCULATION TO FIND h2 = 312.72 kJ/kg at T = 74˚C and P = 0.67MPa

Find hc3 and hc4 (CHOOSE hf) from saturated R-134a table at given T and by using

interpolation

T (°C) h (kJ/kg)

28.2 90.89

21.1 80.73

h1 = 274.77kJ/kg

h2 = 312.72kJ

h3 = 90.89/kg

h4 = 80.73kJ/kg

Condenser energy balance

Refrigerant flow rate, LPM = coolingwaterflowrate (%)

100 %x 1.26 LPM

= 60.6 % x1.26

100

= 0.76536 LPM

0.76536 L1min x

1 m31000 L x

1min60 s = 1.2726 x 10-5 m3/s

Mass flow rate = 1.2726 x 10−5m 3

s x 1000 kg

m3 = 0.012726kg/s

Ein = Eout

QH = mh3−mh4

QH = m(h3−h4)

=0.012726kg/s (90.89-80.73) kJ/kg

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= 0.2566 kJ/s

Compressor energy balance

W ¿=m (h2−h1 )

¿ 0.012726 kgs

(312.72−274.77 ) kJ /kg

¿0.483 kJ /s

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

h

(kJ/kg)

Pressure

(Mpa)

274.77 0.31

312.72 0.8

90.89 0.8

80.73 0.8

80.73 0.31

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Experimental graph

70 120 170 220 270 3200.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

GRAPH PRESSURE AGAINST ENTHALPY

ENTHALPY, h (kJ/kg)

PRES

SURE

(MPa

)

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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

160W in order to absorb 17.95 W heat from the surroundings in order to make sure the

environment temperature is kept at 17.95 W. The coefficient of performance of the heat pump

used is1.049689. From this experiment we know that the function of equipment is to heat up

the temperature of environment.

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

different cooling water flow rate which is at 30%, 50% and 70%. From the experiment, the

power input for the heat pump is different for each water flow rate, which are, 158W, 160W

and 165W respectively. The power input varies as the cooling water flow rate increase. From

the experiment, the flow rate is directly proportional to the input power of compressor. 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 30%, 50% and

70% is 1.2528, 1.4109 and 1.0313 respectively.

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 0.19MPa at 25.6°C to 0.6Mpa

at 74 °C and the enthalpy calculated is 274.77kJ/kg and 312.72kJ/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 21.5°C at 0.8MPa. The pressure is constant because the

condenser undergoes the heat rejection or change in phase from liquid to vapour process at

constant pressure. The enthalpy calculated at 31.3°C and 21.5°C is 131.17/kg and 90.19kJ/kg

respectively. The enthalpy is directly proportional to the temperature change due to heat loss

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to the surrounding when the refrigerant enter the condenser. This will lead to the power output

or heat output will decrease due this process.

CONCLUSION

The power input, heat output and coefficient of performance of a vapour compression

heat pump system, COPH H are 160 W, 167.95 W and 1.04968 respectively for Experiment 1.

In Experiment 2, the purpose was to produce the performance of heat pump over a range of

source and delivery temperatures and it is shown in the calculation section. While Experiment

3 was conducted to plot the vapour compression cycle on the p-h diagram and compare with

the ideal cycle and to perform energy balances for the condenser and compressor. The plotted

graph has been shown in the calculation part and from the energy balance, W ¿ is 0.483 kJ/s.

From all the experiment, it can be said that the higher flow rate of water, the lower the

coefficient of performance. For temperature, the lower the flow rate, the higher the

temperature of refrigerator. The power input is constant for all water flow rates that are

around 160 W to 162 W. The objective of all the experiment has been achieved.

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RECCOMMENDATION

1. The experiment is repeated a few times to get more accurate result.

2. Ensure that the mechanical heat pump had been run and warm up early for 15 minutes

before begin the experiment. It should be notice that the surrounding of the laboratory

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

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REFERENCE

1. Sontag, Borgnakke, Van Wylen. Fundamentals of Thermodynamics. Sixth edition.

John Wiley & Sons, Inc. 2003. 5,7, 434 – 449.

2. http://www.energy.gov/energysaver/heat-pump-systems

3. http://en.wikipedia.org/wiki/Heat_pump?

4. http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/seclaw.html

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