LAB REPORT REFRIGERATION UNIT PDF

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REFRIGERATION UNITNurfadzwin binti Rozlan (2019406296)Amyira Natasha binti Mahari (2019291162)Siti Nurnhadira binti Sharip (2019405752)Nur Afiza binti Mahmud (2019488502)Muhammad Azrai bin Mohd Dzahari (2019405314)ABSTRACTThere were four methods to perform the experiment. In order to equate it with ...

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REFRIGERATION UNIT

Nurfadzwin binti Rozlan

(2019406296)

Amyira Natasha binti Mahari

(2019291162)

Siti Nurnhadira binti Sharip

(2019405752)

Nur Afiza binti Mahmud

(2019488502)

Muhammad Azrai bin Mohd Dzahari

(2019405314)

ABSTRACT There were four methods to perform the experiment. In order to equate it with the ideal cycle and conduct energy balances for the condenser and compressor, the third experiment is to trace the vapor compression cycle on the p-h diagram and the last experiment is to determine the ratio and volumetric efficiency. The SOLTEQ Mechanical Heat Pump was used to perform those experiments. The experiment began with the application of a general start-up method and concluded with a general shut-down procedure. Experiment 1.3.4 was performed with a 40 percent flow rate, while the flow rate was 80 percent for experiment 2. The data was tabulated for experiment 1 to measure the power input, heat output, and coefficient efficiency of 157 W, 81.93 W, and 0.52, respectively. For experiment two, the efficiency curve revealed a drop in COP and heat delivered as the temperature increases. While for experiment 3, the p-h diagram plotted was similar with the theoretical p-h diagram. The energy balance on the condenser and compressor was 2.3085 kJ/s and 0.4560 kJ/s respectively. Lastly, for experiment 4, the compressor pressure ratio and volumetric efficiency was 3.58 and 80.75% respectively. In conclusion, the objectives are successfully obtained therefore the experiment is successfully done. I. INTRODUCTION The SOLTEQ Mechanical Heat Pump has been developed to provide a realistic and quantitative

vapor compression cycle presentation for students and is ideal for all stages of the course. The vapor compression loop is applied for both fridges and heat pumps. When the flow rate of cooling water is set to a certain percent, the refrigeration machine will demonstrate the result. The Mechanical Heat Pump will illustrate the operation of the heat pump when it is necessary to upgrade a large readily available energy supply, such as the atmosphere, for water heating.

II. OBJECTIVES The main objectives of the experiment is to obtain theheat input, power input and coefficient of performance of vapor compression heat pump system, to get the performance of heat pump over a range of source and delivery temperatures, to plot the vapor compression cycle on the p-h diagram and compare with the ideal cycle, to perform energy balances for the condenser and compressor and to find the compression ratio and volumetric efficiency.

III. THEORY A system that offers heat energy to a destination called a 'heat sink' from a heat source is called as heat pump. By absorbing heat from a cold space and releasing it to a warmer one, heat pumps are

designed to transfer thermal energy opposite the direction of spontaneous heat flow.

energy to provide the required output heat pump can

The term' heat pump 'is more common and refers systems used for space heating or space cooling. It follows the same simple refrigeration-type cycle used for an air conditioner or a refrigerator where a heat pump is used for ventilation, but in the opposite direction, releasing heat into the conditioned room rather than the atmosphere. In this use, heat pumps typically draw heat from the colder external air or from the earth.

the performance curve. Therefore it makes a heat

be calculated by finding the under the curve by using

pump's efficiency curve valuable. An example of a heat pump efficiency curve is below.

The Coefficient of Performance – COP – is the ratio of heat output to the amount of energy input of a heat pump

1.

Cooling water flow rate ( LPM ) = COP can be expressed as:

COP =

=

heating effect heat of compression h2−h 3 h 2−h 1

Cooling water flow r 100 %

2.

Refrigerant flowrate ( LPM )=

Refrigerant flow rate (% 100 %

IV. PROCEDURE

where, COP = Coefficient of Performance

I. General Start-up Procedures

hh = heat produced (Btu/h) hw = equivalent electric energy input (Btu/h) = 3413

1.

Pw where, Pw = electrical input energy (W)

2.

If a heat pump delivers 3 units of heat of every unit of energy input – the COP is 3 1kW = 1000 W = 3413 Btu/h

3. 4.

In the analysis of a heat pump system, a performance curve is also important. The required

5.

Every unit of instruments are checked in right condition. Thewater source and drain are both connected then the water supply opened and the cooling water flow rate is set to be 1.0 LPM. The condensated collecter of the drain hose is connected together. The power supply areswitched on and the main power is connected on then followed by first switch at the control panel. The refrigerant compressor is connected. The unit is ready for experiment as soon

as temperature and pressures are constant.

1. 2.

II. General Shut-down Procedures 3. 1.

2.

The compressor is switched off then followed by main switch and power supply. The water supply is closed and no water is left running is ensured.

The general start-up procedures are performed. The cooling water flow rate is adjusted 40% and the system is allowed to run for 15 minutes. All necessary readings are recorded into the experimental data sheet.

V. RESULTS AND CALCULATIONS

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

1. 2. 3. 4.

The general start-up procedures are performed. The cooling water flow rate is adjusted to 40%. The system is allowed to run for 15 minutes. All necessary readings are recorded into the experiment data sheet.

Cooling water flow rate, FT 1 (%) 39.2 Cooling water inlet temperat ure, TT5(◦C) 28 Cooling water outlet temper ature, TT6(◦C) 28.6 Compressor power input (W ) 157 Table 1

IV. EXPERIMENT 2: Production of heat pump performance curves over a range of source and delivery temperatures.

Cooling water flow rate (LPM) = Cooling water flow rate (%) × 5 LPM 100%

1. 2. 3. 4. 5.

6.

7.

The general start-up procedures are set up. The cooling water flow rate is adjusted to 80%. The system is allowed to start for 15 minutes. All necessary readings are recorded into the experimental data sheet. The experiment is then repeated with decreasing water flow rate so that the cooling water outlet temperature increases by about 3°C. Similar steps are repeated until the compressor delivery pressure reached around 14.0 bars. The experiment is repeated at different ambient temperature.

= 39.2 x 5 LPM 100 = 1.96 LPM Heat output, Q (W) = mCpdT = [ (1960 mL/min)(1 g/1mL)(1 kg/1000 g) (1 min/60 s)] X [4180 J/kg.K] X (301.75- 301.15)K = 81.93 W COPH = Heat output (W) Power input = 81.93 W 157 W

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

= 0.52

100 = 3.04 LPM

Experiment 2 :

Heat output, Q (W) = mCpdT Trial

1st

2nd

3rd

= [ (3040 mL/min)(1 g/1mL)(1 kg/1000 g) (1 min/60 s)] X

Cooling water flow rate, FT1(%)

72.8

60.7

32.2

[4180 J/kg.K] X (303.15- 301.65)K = 317.68 W

Cooling water inlet temperatur e, TT5(◦C) 28

28.5

28.4

Cooling water outlet temperatu re, TT6(◦C) 29.4

30

32.4

Compressor power input (W)

159

158

154

COPH = Heat output (W) Power input (W) = 317.68 W 159 W

Table 2

= 2.00

1st Trial :

Cooling water flow rate (LPM) = Cooling water flow rate (%) × 5 LPM 100% = 72.8 x 5 LPM

3rd Trial : 100

= 3.64 LPM Heat output, Q (W) = mCpdT = [ (3640 mL/min)(1 g/1mL)(1 kg/1000 g) (1 min/60 s)] X

Cooling water flow rate (LPM) = Cooling water flow rate (%) × 5 LPM 100% = 32.2 x 5 LPM

[4180 J/kg.K] X (302.55- 301.15)K

100

= 355.02 W COPH = Heat output (W) Power input (W) = 355.02 W 154 W

= 1.61 LPM Heat output, Q (W) = mCpdT = [ (1610 mL/min)(1 g/1mL)(1 kg/1000 g) (1 min/60 s)] X [4180 J/kg.K] X (305.55- 301.55)K = 448.65 W

= 2.31

COPH = Heat output (W) Power input (W) 2nd Trial :

= 448.65 W 158 W = 2.84

Cooling water flow rate (LPM) = Cooling water flow rate (%) × 5 LPM 100% = 60.7 x 5 LPM

4th Trial :

Cooling water flow rate (LPM) = Cooling water flow rate (%) × 5 LPM 100% = 10 x 5 LPM 100 = 0.5 LPM Heat output, Q (W) = mCpdT = [ (500 mL/min)(1 g/1mL)(1 kg/1000 g) (1 min/60 s)] X

Figure 2

[4180 J/kg.K] X (306.15- 301.05)K = 177.65W COPH = Heat output (W) Power input (W) = 177.65 W 158 W = 1.12

Figure 3

Experiment 3 :

Refrigerant flow rate, FT2

Figure 1

%

61.9

Bar Refrigerant pressure, low, P1 (abs)

1.9

Bar Refrigerant pressure, high, P2 (abs)

6.7

Refrigerant temperature, TT1 oC

24.8

Refrigerant temperature, TT2 oC

70.9

o

Refrigerant temperature, TT3 C

28.8

Refrigerant temperature, TT4 oC

23.2

Cooling water flow rate, FT1 %

39.7

Cooling water inlet, TT5

o

C

28.4

C

29.7

H1 - 274.75 = 0.19 - 0.18

Compressor power input, W W

160

274.36 - 274.75 0.20 - 0.18

Cooling water outlet, TT6

o

H1 = 274.56 kJ/kg

Table 3 Point

1

2

3

4

Pressure (bar)

1.9

6.7

6.7

1.9

Temperature (oC)

24. 8

70.9

Enthalpy (kJ/kg)

H1

H2

2. H2 at 0.67 MPa and 70.9oC 0.60 P (MPa)

28.8

0.67

0.70

Enthalpy, h (kJ/kg)

23.2 T(oC)

H3

H4

Table 4 (1.0 bar = 0.1 MPa)

70.0

309.73

70.9

X

80.0

319.55

308.33 H2

Y 318.28

Table 6 Determination enthalpy of refrigerant using interpolation from superheated R-134a table

X - 309.73

= 70.9 - 70

319.55 - 309.73 80 - 70 X = 310.61 kJ/kg

o

1. H1 at 0.19 MPa and 24.8 C 0.18 P (MPa)

0.19

Y - 308.33 0.20

= 70.9 - 70

318.28 - 308.33 80 - 70

Enthalpy, h (kJ/kg)

Y = 309.23 kJ/kg H2 - 310.61 = 0.67 - 0.60

T(oC) 20

270.59

24.8

X

30

279.25

270.18 H1

309.23 - 310.61 0.70 - 0.60 H2 = 309.64 kJ/kg

Y 278.89

3. Find H3 and H4 (choose hf ) from saturated R-134a table at 28.8oC and 23.2oC by using interpolation.

Table 5

By using Interpolation : Temperature (oC) X - X1 = Y - Y1 X2 - X1 Y2 - Y1

X - 270.59

H3

= 24.8 - 20

279.25 - 270.59

30-20

X = 274.75 kJ/kg Y - 270.18

= 24.8 - 20

H4

Enthalpy (kJ/kg)

28

90.69

28. 8

H3

30

93.58

22

82.14

23. 2

H4

24

84.98

30-20

Table 7

Y = 274.36 kJ/kg

H3- 90.69 = 28.8 - 28

278.89 - 270.18

93.58 - 90.69 30 - 28

H3 = 91.85 kJ/kg H4 - 82.14

= 23.2 - 22

84.98 - 82.14

24 - 22

H4 = 83.84 kJ/kg

W = m ( h2 - h 1 ) = 0.012999 ( 309.64 - 274.56 ) = 0.4560 kJ/s

Experiment 4 : Refrigerant flow rate, FT2 (%)

61.3

Refrigerant pressure (high), P2 (bar) 6.8 Refrigerant pressure (low), P1 (bar) 1.9 Refrigerant temperature, TT1 (°C) Figure 4

22.2

Table 8 Compressor pressure ratio = Suction pressure of refrigerant = P1

Energy balance

Discharge pressure of refrigerant P2 

= 1.9

On the condenser

6.8 = 0.28

Refrigerant flow rate (LPM) = Cooling water flow rate (%) × 1.26 LPM 100% = 61.9 × 1.26 LPM

Refrigerant flow rate (LPM) = Cooling water flow rate (%) × 1.26 LPM 100%

100

= 61.9 × 1.26 LPM

= 0.77994 LPM Mass flow rate = 0.77994 L x 1 kg x 1000 L x 1 min Min

1000 L 1m3 60 s

= 0.012999 kg/s Ein = Eout QH = mh3 - mh4 Q H = m ( h3 - h 4 ) = 0.012999 ( 264.27 - 86.68 ) = 2.3085 kJ/s

100 = 0.77994 LPM Mass flow rate = 0.77994 L x 1000 cm3 x 1m3 x 1 min x 4.25 kg/m3 min

1L

(100 cm)3 60 s

= 5.525 x 10-5 kg/s Actual mass flow rate = Mass flow rate Density of refrigerant = 5.525 x 10-5 kg/s 4.25 kg/m3 = 1.3 x 10-5



On the compressor

Volumetric efficiency = Actual mass flow rate

Theoretical mass flow rate = 1.3 x 10-5 1.61 x 10-5 = 0.8075 x 100% = 80.75%

DISCUSSION

In this experiment 3 aims were carried out to research the action of the mechanical heat pump and the cooling unit for thermodynamics. The first goal is to evaluate the input of the compressor power and the low pressure coolant vapor was registered at 157 W to compress from the evaporator and compress it into a high pressure vapor. The enthalpy (h) was used to measure the heat output(Qh) from the thermodynamic property table at recorded temperature, and 81.83 W was calculated to provide heat to a warm medium. The ratio of heat output to the quantity of energy input from heat output to the quantity of energy input from a heat pump was obtained as an efficiency coefficient for the first trial and was valued at approximately 0.52.

The second experiment replicated the process at a different cooling water flow rate of about 72.8% and reported results. The cooling water flow rate is manipulated until the temperature seen rises to about 3 ° C and the heat pump efficiency across the source range is produced. such as COP, heat delivered, power input and delivery temperatures. From data determined, cooling water flow rate percentage was decreased to 60.7 % as the temperature increased and the power input of the compressor is static from 158 W. COP obtained for the first trial is 2.31 . This means that addition of 1 kW of electrical energy is needed to have released of 2.31 kW of heat at the condeser and this also applied to the second trial wherein the temperature was increased. Thus, several graphs against the temperature delivered plotted to observe the behavior of the heat pump as the source was manipulated.

the heat pump depends on a variety of variables and this can be explained. The temperature differential between the source of waste heat and the prospective consumer is a major factor. The temperature differential between the temperature of condensation and evaporation primarily determines the quality, the larger the COP value, which is the smaller difference. In addition, the COP value is often influenced by the system operation and performance of peripheral devices such as fans. The compressor's power input of 158 W was unchanged, though the temperature rose. Therefore the digital instrument for data acquisition was used by the system. This method of capturing data may have some drawbacks as the data should appeared in the digital displays observed minor delay to signal a real data for recording.

Thus, to plot the vapor compression cycle on the p-h diagram, the cooling water flow rate was set up at 39.7 % and all the data recorded. Enthalpy (h) for each of recorded temperature and pressure is referred to the property table of saturated and superheated refrigerant.This diagram is slightly similar to the theoretical diagram. The line 2-2 represent cooling of the superheated gas in the condenser down to the saturated vapor temperature.the remainder of the condensing takes place from 2-3 where latent heat is removed. Thus line 3-4 where the liquid/vapor I passed through an expansion device the pressure decrease without any enthalpy change. Lastly, line 4-1 where the liquid/vapor is evaporated completely to a gas and where enthalpy is extracted from surrounding.

Finally, to observe the heat output from the equipment, the energy balance is determined for the condenser and compressor. Cold air is blown into it at the condenser as the hot vapors join, and because air is colder than refrigerant in around 0.7794 LPM, heat release to the warm atmosphere when 0.4560 kJ/s is required by the compressor to generate the job such that the evaporator's low pressure refrigerant vapor is compressed into a high pressure vapor. CONCLUSION

The temperature of the flow rate of cooling water increases, the heat pump COP reduces. This pattern is identical to the heat that the heat pump provides to the surrounding environment. The efficiency of

This experiment conduct with two different flow speeds with different reported data. The inlet and outlet temperatures are respectively 28 ° C and 28.6 ° C in the first flow rate experiment of 39.2 percent. It also reveals that the cooling water in the coolant unit has undergone a heating phase. This can be believed to have moved the temperature from the

hot surface to the cool surface. The data shows that 39.2% of the resulting power input cooling water flow volume, heat output COP is 81.93 W. For the second trial, the output curve displayed a correct hypothesis as a hypothesis that decreased the COP and heat delivered as the water temperature increased. However the power input did not display the same sequence that was not altered during the experiment. Finally, to observe the equipment output against the theoretical p-h diagram, the p-h diagram is plotted, the outcome showing the same as the hypothesis. The energy balance was also carried out at the condenser and the compressor reveals that, respectively, 2,3085 kJ/s and 0,4560 kJ/s kW. In conclusion, the output curve was very difficult to draw and build during the second trial, as the third test for pressure manipulated to 14 bar was not conducted, as it is difficult for the equipment to reach the necessary pressure. Overall, performance is known to be an experiment.

Retrieved 14 November 2020, from https://berg-group.com/engineeredsolutions/the-science-behind-refrigeration/

APPENDIX

RECOMMENDATIONS When conducting the experiment, there are some precautions that should be taken:

1.

2.

3.

The experiments conducted for a few times and take the average value to obtain more accurate results. Allow the heat pump to warm up for about 15 minutes before beginning the experiment to make sure its condition is in operating state. Perform the experiment at place with standard room temperature as the equipment might affected by the surrounding such as the condenser that operates in the air-conditioned room.

Figure 5....