Table of content Table of content ... ... Abstract ... ... Introduction ... ... Aims ... ... Theory ... ... Apparatus ... ... Experimental procedure ... Results... ...……. Experiment 1...…. Experiment 2...…. Experiment 3...…. Experiment 4...…. Discussion ... ...…. Conclusion ... ... Recommendations ... ... References ... ... Appendices ... ...
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.
Introduction
In the experiment of refrigerant, the Mechanical Heat Pump is used to study the vapour compression cycle. This Mechanical Heat Pump has been designed to provide students with a practical and quantitative demonstration of the 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 liqud 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 refrigerator 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 largely determined by the 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 flow rates of both the refrigerant and cooling water, power input to the compressor, and all relevant temperatures.
Aim
1) To determine the power input, heat output and coefficient of performance of a vapour compression heat pump system.
2) To produce the performance of heat pump over a range of source and delivery temperatures
3) To plot the vapour compression cycle on the p-h diagram and compare with the ideal cycle
4) To perform energy balance for the condenser and compressor 5) To determine the compression ratio and volumetric efficiency
Theory
Devices that absorbs heat at a low temperature and reject heat at a higher temperature are known as refrigerators and heat pumps. Both operates the same way using a reversed heat engine cycle. Refrigerators that are used to maintain a temperature below ambient and heat pumps are used to supply heat higher than ambient. Some device, in particular those used for space cooling, may be used as both a heat pump and a refrigerator.
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. Figure below gives a schematic representation of the four essential mechanical components in this cycle, figure 2 shows the ideal cycle on a pressure-enthalpy diagram
Figure 1 figure2
In ideal vapour compression cycle the refrigerant vapour is compressed isentropically to a higher temperature and pressure(1->2). The compressed vapour is then condensed isobarically which results in heat rejection to the surroundings (2->3). The next step is adiabatic throttling of the refrigerant to the low temperature and pressure(3->4). The final step is the adiabatic Throttling of the refrigerant of the low temperature and pressure, which results in the absorption of heat from its surroundings (4->1). 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 coefficient of performance COP which for a refrigerator COPR is defined as :
COPR = heat absorbed at the lower temperature/compressor net work (1a) For a heat pump COPH :
COPH = heat rejected at the higher temperature/compressor net work (1b)
Below are the formula to convert cooling water and refrigerant flow rate to LPM 1. Coolingwater flowrate (LPM)=coolingwater flowrate(%)/100% X 5 LPM 2. Refrigerant flowrate (LPM) = Refrigerant flowrate(%)/100% X 1.26 LPM
Experimental procedure General Start-up Procedure
1. The unit and all instruments were checked in proper condition.
2. Both water source and drain were checked connected then the water supply was open and cooling water flow rate was set at 1.0LPM.
3. The drain hose at the condensate collector was checked connected.
4. Power supply was connected and the main power was switched followed by main switch at the control panel.
5. The refrigerant compressor was switched on. As soon as temperature and pressure were constant, the unit was ready for experiment.
General Shut-down Procedure
1. The compressor was switched off, followed by main switch and power supply.
2. 2. The water supply was closed and water was 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 were performed. 2. The cooling water flow rate was adjusted to 40%. 3. The system was allowed to rub for 15 minutes.
4. All necessary reading was 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:
1. The general start up procedures were performed.
2. The cooling water flow rate was adjusted to 80%. 3. The system was allowed to run for 15 minutes.
4. All necessary readings were recorded into the experimental data sheet. 5. The experiment with reducing flow rate was repeated so that the cooling
water outlet temperature increases by about 3oc (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 were performed.
2. The cooling water flow rate was adjusted to 40% and the system was allowed to run for 15 minutes.
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 were performed. 2. The cooling water flow rate was adjusted to 40%. 3. The system was allowed to run for 15 minutes.
4. All necessary reading were recorded into experimental data sheet. 5. The experiment might be repeated at different compressor delivery
pressure.
Results
Experiment 1
Cooling water flow rate , FT1 % 40.0
Cooling water inlet temperature, TT5 oC 28.3
Cooling water outlet temperature, TT6 oC 29.4
Compressor power input W 161.0
Experiment 2
Cooling water flow rate , FT1 % 60.0 40.0 20.0
Cooling water inlet temperature, TT5
oC 30.0 28.3 29.7
Cooling water outlet temperature, TT6
oC 29.4 29.4 32.3
Compressor power input W 161.0 161.0 163.0
Experiment 3
Refrigerant Flow Rate, FT2 % 61.0
bs) Refrigerant Pressure (High), P2 Bar(a
bs) 7.2 Refrigerant Temperature, TT1 oC 28.8 Refrigerant Temperature, TT2 oC 79.0 Refrigerant Temperature, TT3 oC 30.8 Refrigerant Temperature, TT4 oC 25.0
Cooling Water Flow Rate, FT1 % 40.0
Cooling Water Inlet Temperature, TT5 oC 29.9
Cooling Water Inlet Temperature, TT6 oC 31.5
Compressor Power Input W 161.0
Experiment 4
Refrigerant Flow Rate, FT2 % 61.0
Refrigerant Pressure (Low), P1 Bar(ab s)
2.0 Refrigerant Pressure (High), P2 Bar(ab
s) 7.2 Refrigerant Temperature, TT1 oC 28.8 Calculations Experiment 1 For reading 1,
Cooling water flow rate LPM = coolingwater flowrate%/100% X 5LPM = (40.0 X 5)/100 = 2.0 LPM
2L/1 min X 1m3/1000L X 1min/60s = 3.33 X 10exp-5 m3/s
For h5 and h6 from table saturated water-temperature using interpolation h5 = 118.63 kJ/kg h6 = 123.23 kJ/kg density water = 1000kg/m3 mass flow rate = 3.33exp-5 x 1000 = 0.0333kg/s
Ein = Eout Mh5 = QH + mh6 QH = m(h5-h6) = 0.0333 kg/s(-4.6) kJ/kg = -153.18exp-3kJ/s COPH = QH/W =153.18exp-3 kJ/s / 161 kJ/s =951.43exp-3 Experiment 2 For reading 1,
Cooling water flow rate LPM
= coolingwater flowrate%/100% X 5LPM = (60.0 X 5)/100
= 3.0 LPM
3L/min X 1m3/1000L X 1 min/60s = 50exp-6 m3/s
Find h5 and h6 from table saturated water-temperature using interpolation h5=125.74
h6=123.23
density water= 1000kg/m3
Ein = Eout Mh5 = QH + mh6 QH = m(h5-h6) = 50exp-3 kg/s(2.51) kJ/kg = 125.5exp-3 kJ/s COPH = QH/W =125.5exp-3 kJ/s / 161 kJ/s =779.50exp-6 For reading 2,
Cooling water flow rate LPM = coolingwater flowrate%/100% X 5LPM = (40.0 X 5)/100 = 2.0 LPM
2L/1 min X 1m3/1000L X 1min/60s = 3.33 X 10exp-5 m3/s
For h5 and h6 from table saturated water-temperature using interpolation h5 = 118.63 kJ/kg h6 = 123.23 kJ/kg density water = 1000kg/m3 mass flow rate = 3.33exp-5 x 1000 = 0.0333kg/s
Ein = Eout
Mh5 = QH + mh6 QH = m(h5-h6)
= 0.0333 kg/s(-4.6) kJ/kg = -153.18exp-3kJ/s
COPH = QH/W
=153.18exp-3 kJ/s / 161 kJ/s =951.43exp-3
Experiment 3
Find hc1 and hc2 using interpolation from superheated R – 134a tbale At T = 28.8 degree celcius P(MPa) h(kJ/kg) 0.2 277.84 At T = 79 degree celcius P(MPa) h(kJ/kg) 0.7 317.29
Find hc3 and hc4 from saturated R- 134a table at give T and P = 0.7MPa
T(degree celcius) h(kJ/kg) 30.8 269.45 25.0 95.47 h1= 277.84 h2 = 317.29
h3 = 269.45 h4 = 95.47
Consider energy balance Refrigerant flow rate, LPM
= coolingwater flowrate%/100% X 1.26 LPM = 61/100 X 1.26 LPM
= 786.6exp-3 LPM
786.6exp-3/1000 X 1/60s =1.311exp-5 m3/s
Mass flow rate = 1.311exp-5 x 1000 = 0.01311 kg/s Ein=Eout
QH = (0.01311)(173.98) =2.28 kg/s
Compressor energy balance W= 0.01311(221.89)
Experiment 4
Compressor pressure ratio
= suction pressure of refrigerant/discharge pressure of refrigerant = (2/7.2) bar/bar
=0.278
Volumetric efficiency = actual volumetric flowrate/theoretical volumetric flowrate Refrigerant flowrate, LPM = refrigerant flowrate %/100% X 1.26 LPM =61/100 X 1.26 =0.769 LPM Change LPM to kg/s Mass flowrate = 0.769 x 1000 x (1/100exp3) x (1/60) = 5.45exp-5 kg/s
Density of refrigerant 134a = 4.25 kg/m3 Actual volume flow rate
= mass flowrate/density of refrigerant 134a = 5.45exp-5/4.25
=1.28exp-5 m3/s Volumetric efficiency
= actual volumetric flowrate/theoretical volumetric flowrate = 1.28exp-5/1.61exp-5
= 0.795
Discussion
This experiment was carried out to calculate the performance each of the equipment in the refrigerant unit. In the experiment, the power input of the heat pump was recorded at 161kJ/s in order to absorb 0.15318 kJ/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 the performance of the heat pump used is 0.95143.
For the second experiment, the same step as the first experiment was repeated at different cooling water flow rate which is at 60%, 40%, and 20%. From the
experiment, the power input for the input for the heat pump is different for each water flow rate, which are, 161kJ/s, 161kJ/s and 163kJ/s respectively. The power input varies as the cooling water flow rate decreases. The same method was used to calculate the rate of the heat transfer and the coefficient of performance (COP) for the heat pump. The COP calculated for cooling water flow rate at 60%, 40% and 20% is 0.0007795, 0.95143, and 0.000111 respectively. The COP of heat pump decreases as the cooling water flow rate decreases.
In the third experiment, the change in pressure 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 0.2 MPa at 28.8oC
to 0.7 MPa at 79.0oC and the enthalpy calculated is 255.26kJ/s and 270.99kJ/s
respectively. The R-134A enters the compressor superheated then compressed at constant entropy the leaves as superheated. The refrigerant than enter the condenser at temperature of 31.3C at 0.8Mpa and leave the condenser at
temperature at 20.4C at 0.8Mpa. The pressure is constant because the condenser undergoes the rejection process at constant pressure. 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 system are allowed to run at different period of time before the readings were taken, First, the systems are allowed to run for 15 minutes than the steps repeated for 20 minute period. The values of compressor pressure ratio are 0.278. There is slightly difference in the ratio calculated. This is due to the difference in the ratio calculated. This 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 what 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 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.
Conclusion
This experiment is considered success. Firstly, the power input, heat output and coefficient of performance of a vapor compression heat pump system are been determined. In experiment 1, the values Qh and COPh are 0.15318 kJ/s and 0.95143. For second experiment, the performance of heat pump over a range of source and delivery temperature are been able to produce. The values of COPH are 0.0007795, 0.95143, and 0.000111. 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 shown 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 calculate. The ratio is 0.278 and for the volumetric efficiency is 0.795. There is slightly difference due to difference in pressure and refrigerant flow rate.
Recommendations
1. Consultation is compulsory in order to make sure the experiment 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.
References
UiTM faculty of Chemical Engineering, Laboratory Manual CHE 465, Chemical engineering Lab 2
www.tyxerhubpages.com
www.solution.com
www.1.eere.energy.gov
Yunus A. Cengel. Michael A Boles ‘Thermodynamics An Engineering Approach’ 7th edition
