A study of two high temperature pumps and

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Published: 17.12.2019 | Words: 2050 | Views: 515
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Heat pumping systems are a gadget that are used in lots of different applications in the automobile and home industries. These are the simple reason as to why a home could be heated or maybe a refrigerator cooled. A temperature pump is made up of 5 key components: the evaporator, fondre, expansion device, compressor as well as the refrigerant inside the system. The capacity of the heat pump is dependent greatly within the volume and pressure from the refrigerant. The warmth pump uses an external electrical power source to be able to transfer the power from the temperature source for the heat sink. Heat pumping systems that provide high temperature for a goal, and high temperature pumps that remove temperature for a purpose are fundamentally the same gadget. The difference is available in the configuration of the program. To cool a certain place, the heat is definitely removed from that area. To heat the, heat is taken from an outdoor source and pumped into that area.

The objective of this test was to identify the POLICEMAN of the high temperature pump through the use of two distinct procedures. These types of measurements had been then when compared with each other, and any mistakes were noted. The initially procedure is by using the assessed amount of heat rejection on the condenser and heat transfer at the evaporator to estimate the POLICE OFFICER. The second treatment will be to make use of the heat of rejection at the condenser as well as the compressor strength input to calculate the COP.


The goal of the test was to make use of two several procedures to calculate the COP of the heat pump in the laboratory. This was performed on the warmth pump that was offered in the research laboratory, which contained an evaporator, expansion valve, refrigerant filtration, condenser, compressor, water heat transfer system and a choose number of receptors. The condenser and evaporator in this program were immersed in water so that the volume of heat change could be tested for each. These kinds of measurements had been then accustomed to calculate the COP in the heat pump. This was completed using two different methods. The initial procedure used the scored amount of warmth rejection in the condenser and heat copy at the evaporator to compute the POLICEMAN. The second process used heat of being rejected at the fondre and the converter energy type. The results from the two types of procedures were located and in contrast. Any discrepancies between the info was noted and analyzed.


A heat pump system was presented to this research laboratory. The objective, since described previously mentioned, is to determine the POLICE OFFICER of this system using two different tactics. Knowledge of thermodynamic properties and processes will be used to find the information needed to estimate the COP in these other ways.

The COP of the heat pump can be computed by computing the rate of warmth rejection on the condenser and related the change to how much heat copy at the evaporator. For this method, the temp of the water tank can be graphed, and the incline used to calculate the POLICE OFFICER.

The COP of any heat pump can also be computed by calculating the rate of warmth rejection with the condenser and relating it to the power of the air compressor. The rate with the heat denial at the fondre can be used by taking the slope of the chart. The power required by the converter is simply scored by a Wattmeter.

The results from the procedures will probably be compared to the other person and any discrepancies will probably be noted. There is also a potential for mistake when it comes to the resolution from the temperature psychic readings. The accuracy of the temperatures readings could also have mistake because the drinking water reservoirs not necessarily uniform temperatures.


The warmth collected by the heat pump can be displayed by:

Heat Accumulated = Q_Evaporator

Heat expelled by heat pump can be showed by:

Heat Removed = Q_Condensor

The job input for the compressor may be represented simply by:

Operate Input sama dengan W_Compressor

Some of the “Q” conditions that seem as inches Q’ inches are a charge at which high temperature is transmitted, since the plain “Q” term represents temperature in itself.

Gear and Process

Heat pump program in the Thermal-Fluids lab will be used to perform the experiment. This product consists of a great evaporator, condenser, compressor, expansion valve, refrigerant filter, drinking water heat copy system and a select quantity of sensors.

Figure 1 illustrates the setup with the heat pump for this research. The evaporator and condenser, as shown, are held in containers of water. This kind of water can then be pumped throughout the system and transfers heat in the program. The system includes a multitude of different sensors. You will find pressure gauges next to both the evaporator and the fondre to regulate the refrigerant challenges in these respective areas. There is a variety of thermocouples inside the system in order to take temperatures readings for all of you important places in the program. The conditions of the normal water reservoirs for the fondre and evaporator were taken using straightforward liquid thermometers.


Figure a couple of illustrates the compressor which is used in the program. This is a basic comrpressor, as well as the main variables are found in Table 3, shown under.

Evaporator Condensor

The evaporator and condenser are made up of identical pieces. Each element is made up of 12 turns of 3/8″ To. D. water piping pipe that is certainly coiled in an internal diameter of 13 cm. Both equally components are placed in their very own 1 gallon buckets which can be filled with normal water. Each container has control valves to manage the circulation in and out of the bucket.

Figure some illustrates a condenser/evaporator in the system. The flow in the bucket is definitely controlled by the valve that is quietly of the part. For high quality data, a thermocouple is usually mounted on both entrance and exit with the water movement.

Results and Discussion

The heat pump was run for a total of thirty minutes. The data was taken for one-minute times. This allowed enough time in order to record all the different temperature and pressure blood pressure measurements. The room temperature for the experiment was 69. 8F, while the atmospheric pressure was 1016mBar. The bedroom temperature can influence just how much heat can be gained and lost by the condenser and the evaporator. An improvement in atmospheric pressure may influence the pressure from the refrigerant and therefor affect the efficiency of the system.

Figure 5 illustrates the temperature of the evaporator tank over time. Not surprisingly, the heat of this normal water reservoir went down over time. This is because it is the source of heat becoming collected by the heat pump. A thready line of best suit was match to the info as well. The slope amount from this series was used to calculate the Coefficient of Performance. Your data doesn’t match the linear line of best suit perfectly, which can be most likely since there are errors in certain of the measurements. Human mistake and device error are both at enjoy because of the constantly changing conditions inside the experiment.

The temperature of the condenser reservoir with time is displayed in Number 6. This kind of data would not meet the expected result. Considering that the condenser is where the heat is declined from the program, it was expected that this physique of normal water would finish up at a greater temperature than at which it started. The graph, initially, looks like a similar slope to the temperature in the evaporator, yet this is not accurate. This chart is scaled differently than the other chart, so that its’ own craze could be examined. These outcomes show a much “flatter” line, meaning that the temperature didn’t change as much over the whole experiment when compared with the evaporator. The slope of this collection was also used in the formula to calculate the COP from the heat program.

Determine 7 shows the pressure in the evaporator over time. This curve clearly shows that the pressure reduced over time. This corresponds with all the temperature drop that occurred in the evaporator. As the temperature decreases, the pressure also decreases. There are a few data points through this curve that don’t fit the smooth competition, showing that there was a great amount of error possibly in the heat pump itself, or in the pressure readings that had been taken. The needle for the pressure gauge doesn’t have very small resolution to put into practice, so an important part of the data can be related to that.

Figure eight illustrates the results of the pressure in the condenser over time. This pressure started reduced, shot up to 8. 8 bars, then flatlined for a time. At the end of the experiment, the pressure beginning going down. This can be able to be related to errors inside the experiment. There are clear signs of error for the parts of the curve that look like steps. These steps are the effect of the pressure gauge not having a small enough resolution. The measurements end up getting rounded off to the closest number. This curve has the exact expected result, which would have had a increasing temperature correlating to a increasing pressure.

Figure being unfaithful illustrates the temperatures of the two thermocouples over time. The input to the compressor was Thermo1. This can be fully supported by the data, since it is the reduced temperature and it stayed at very steady throughout the test. The output, Thermo2, on the other hand climbed in heat for the whole research. This is because heat pump gets more efficient as time goes on and every thing gets “up-to-temperature” in the system. This would be supposed to level off into a smooth line the longer the program had operate. The temp will strike a point where it are unable to possibly go any bigger.

Figure 10 demonstrates the intake and exhaust temperatures from the evaporator as time passes. The developments of these lines fully support the prior data for the evaporator tank temperature. The reservoir temperatures declined over the experiment. The inlet temp on this chart clearly decreased over time. This kind of corresponds to the lower overall tank temperature. At first, however , it temperature will not make sense. This kind of stayed similar, and even increased a small amount by the end of the experiment came. In case the overall reservoir temperature would decline, this means that the evaporator collected more heat in the room. At these times, it the actual heat pump work more efficiently. This compares to the increasing temperature outlet of the converter.

Figure 11 displays the data pertaining to the intake and exhaust temperatures through the condenser. As expected, the condenser gained performance over time in the same way the compressor did. The inlet temp to the fondre comes from the compressor, so it is correct that this increased with time. The outlet temperature stayed similar, as expected. This kind of results in a more substantial amount of warmth being removed from the condenser over time because the inlet temp increased with time, and the store temperature was the same. These are generally curves that look extremely smooth, so it will be assumed that there isn’t very much error inside the results.

Figure 12 illustrates the curve to get the power suggestions to the air compressor over time. The power draw was significantly larger at the beginning of the experiment than at the end. This sharp curve at the beginning of the curve refers with many of the other temperature graphs, when the temps were changing fast about those, there is a much bigger power bring on the compressor. This is because the task input towards the compressor needed to make up for the little difference inside the temperatures for the condenser and evaporator.