Refrigerant Unit Lab Report

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.

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

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Chemical

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.

THEORY

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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.

APPARATUS

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1. 2. 3. 4. 5. 6. 7.

R-134-A Compressor Evaporator Water inlet and water outlet Filter dryer Power supply Water Valve

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.

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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.

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 Cooling water flow rate, FT1 Cooling water inlet temperature, TT5 Cooling water outlet temperature, TT6 Compressor power input Heat output COPH

% LPM °C °C W W No unit

40.3 2.015 27.7 28.9 160 167.95 1.04968

Experiment 2: Production of heat pump performance curves over a range of source and delivery temperatures. Test Cooling water flow rate, FT1 Cooling water flow rate, FT1 Cooling water inlet temperature, TT5 Cooling water outlet temperature, TT6 Compressor power input Heat output

% LPM °C °C W W 7

1

2

3

30.0 1.50 27.9 29.8 158 197.95

50.0 2.50 27.9 29.2 160 225.74

70.0 3.50 28.0 28.7 165 170.17

Coefficient of performance, COPH

NO UINT

1.2528

1.4109

1.0313

Experiment 3: Production of vapour compression cycle on p-h diagram and energy balance study. Refrigerant flow rate, FT2 Refrigerant flow rate, FT2 Refrigerant pressure (low), P1 Refrigerant pressure (high), P2 Refrigerant temperature, TT1 Refrigerant temperature, TT2 Refrigerant temperature, TT3 Refrigerant temperature,TT4 Cooling water flow rate,FT1 Cooling water flow rate, FT1 Cooling water inlet temperature,TT5 Cooling water outlet temperature,TT6 Compressor power input

% LPM Bar(abs) Bar(abs) °C °C °C °C % LPM °C °C W

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60.6 0.76 1.9 6.7 25.6 74.0 28.2 21.1 40.0 2.0 27.9 29.5 157

CALCULATION. EXPERIMENT 1 L VOLUMETRIC FLOWRATE = FLOW RATE (%) × 5 MIN = 0.403 × 5 L = 2.015 MIN OUTPUT HEAT = Q × DENSITY 2.015L 1 m3 1 min 997 kg 4180 J × × × × ×(28.9 – 27.7) 3 = min 1000 L 60 s kg. ˚C 1m = 167.95 W COPH =

=

OUTPUT HEAT POWER INPUT 167.95 160

= 1.04968 EXPERIMENT 2

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L VOLUMETRIC FLOWRATE, Q = FLOW RATE (%) × 5 MIN = 0.3 × 5 L = 1.5 MIN OUTPUT HEAT = MASS FLOW RATE × CpH2O × TEMPEARATURE CHANGE = Q × DENSITY × CpH2O × TEMPEARATURE CHANGE 1.5L 1 m3 1 min 997 kg 4180 J × × × × ×(29.8 -27.9) = min 1000 L 60 s kg. ˚C 1 m3 = 197. 95 W COPH =

=

OUTPUT HEAT POWER INPUT 166.7 158

= 1.0523 *CALCULATION IS REPEATED FOR FLOW RATE 50 % AND 70 % .

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PERFORMANCE CURVES FOR HEAT PUMP (COEFFICIENT OF PERFORMANCE, OUTPUT HEAT AND POWER INPUT) VERSUS WATER OUTPUT TEMPERATUR 250 225.74 PERFORMANCE CURVE FOR HEAT PUMP ( POWER INPUT) VERSUS WATER OUTPUT TEMPERATURER 197.95 200 170.17 165 160 158 150 PERFORMANCE CURVES FOR HEAT PUMP (COEFFICIENT OF PERFORMANCE, OUTPUT HEAT

COEFFICI NT OF PERFORMANCE, OUTPUTOUTPUT HEAT TEMPERATURE AND POWER INPUT AND EPOWER INPUT) VERSUS WATER

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

1.03 1.41 1.25 0 28.5 29 29.5 30

WATER OUTPUT TEMPERATURE, ˚C

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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) TEMPERATURE(˚C

0.18 h (kJ/kg)

) 20 25.6 30

270.60 X 279.27

0.19 h (kJ/kg)

0.2 h (kJ/kg)

h1

270.20 y 278.91

Using interpolation 279.27−x 30−25.6 = x−270.6 25.6−20 1.7858 x = 491.9075 x = 275.45kJ/kg 278.91− y 30−25.6 = y −270.2 25.6−20 1.7858 y = 491.23 y = 274.08kJ/kg 274.08−h 1 0.2−0.19 = h1−275.45 0.19−0.18 2h1 = 549.53 h1 = 274.77kJ/kg * REPEAT CALCULATION TO FIND h2 = 312.72 kJ/kg at T = 74˚C and P = 0.67MPa

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Find hc3 and hc4 (CHOOSE hf) from saturated R-134a table at given T and by using interpolation T (°C) 28.2 21.1

h (kJ/kg) 90.89 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( ) x 1.26 LPM 100

=

60.6 x 1.26 100

= 0.76536 LPM 0.76536 L 1 min

Mass flow rate =

x

1m 3 1000 L

1.2726 x 10−5 m3 s

Ein

=

Eout

QH

=

mh 3−mh 4

QH

= m(h3−h4 )

x

x

1 min 60 s 1000 kg m3

=0.012726kg/s (90.89-80.73) kJ/kg 13

= 1.2726 x 10-5 m3/s

= 0.012726kg/s

= 0.2566 kJ/s

Compressor energy balance W ¿ =m ( h2−h 1)

¿

0.012726 kg ( 312.72−274.77 ) kJ /kg s

¿ 0.483 kJ /s

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

Pressure

(kJ/kg)

(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

GRAPH PRESSURE AGAINST ENTHALPY 0.9 0.8 0.7 0.6

PRESSURE (MPa)

0.5 0.4 0.3 0.2 70

120

170

220

ENTHALPY, h (kJ/kg)

15

270

320

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 R134A 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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