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Thermal-Fluids Lab

REFRIGERATION EXPERIMENT

PROFESSOR: Manuel Magrane

PERFORMED: August 14, 2018

REPORT SUBMITTED: August 24, 2018

BY: Talal Aloumi

Group #4

Abstract

Refrigeration systems are used in several application, such as in air conditioning and fridges. Refrigeration systems has four main components which are compressor, condenser, expansion valve, and evaporator. In this experiment, the objective was to determine the COP and COPideal of a refrigeration system, and verify that the compressor is adiabatic. The pressure, temperature, mass flow rate, and power data of the system were recorded until the steady state has been reached. These data helped to get the specific enthalpies which was used to determine the COP’s. The COPactual, COPideal, and COPCarnot calculated were 2.0, 7.7, and 18.5 respectively. Since the calculated COP was less than COPideal that verifies that the compressor is in fact not adiabatic as assumed.

Introduction

Refrigeration systems are extensively used for the intent of cooling a room and food preservation. In this experiment, R134-A will circulate through an evaporator, or a heat exchanger. The evaporator converts the fluid into vapor and in that process it takes in heats from the medium that to be cooled. After that, the vapor is passes through a compressor which compresses gas which leads to a higher temperature and pressure in the fluid. The extremely hot vapor is then passes through the condenser where it condenses, cool down. Then it passes through the expansion valve where it expands. The fluid keeps circulating in the same process which is then can be called refrigeration. The four main components of the refrigeration are Compressor, Condenser, Expansion Valve, and Evaporator. At these four main points of the refrigeration cycle temperature, pressure, fluid flow rate, and electrical output were recorded. From these values, the specific enthalpy at each of the four points was found and the Coefficient of Performance, COP, was calculated using:

COP = ṁ (h1 – h4) / Ẇelect … (1)

COPideal = (h1 – h4) / (h2 – h1) … (2)

COPCarnot = 1 / [(T2 / T4) – 1] … (3)

Where:

ṁ is the mass flow rate.

h# is the specific enthalpy at point #.

T# is the temperature at point #.

Welect is the electrical input work.

Description of Work

First, the valves that are required to be closed are kept closed, while the other valves are kept open. The system (Figure 1) was turned on and at the identified four main points (Figure 2) of the refrigeration cycle temperature, pressure, fluid flow rate, and electrical output data were recorded before warming up the system. Then it was kept running for 5 minutes to warm up. After that, another recording of the same data has taken place again. The same data were recorded then after every 5 minutes until the cycle reached steady state. Steady state can be identified when the data start showing constant values at all four different points of the cycle. These values helped in getting the thermodynamic state of the fluid at each point, as well as the specific enthalpy associated with the cycle. Finally, a pressure vs. enthalpy graph was constructed to show the cycle, and the coefficient of performance was then calculated.

Figure 1: The entire refrigeration system used in the experiment.

Figure 2: The four main points in the refrigeration process.

Results and Discussion

First, the temperature data that were recorded in Celsius were converted to Fahrenheit, and the pressure data were converted to psia. Then the rotameter readings were converted to lbm/min using the figure included in the lab manual. The electrical work rate or input power data were then converted from Watts to Btu/min by dividing by a factor of 17.58. Table 1 below shows how the pressure, temperature, mass flow rate, and electrical work rate data were changing over intervals of five minutes. At the last interval, or after 35 minutes, all the data were not changing and that means that the steady state has been reached. Therefore, to simplify the data, Table 2 was constructed were it shows the pressure and temperature data of the four main points in the cycle (as show in Figure 2 above). Not that in the cases of points 1 and 2, multiple measurement locations correspond to the same reference point as on figure 2. These locations are separated by a length of tubing and should be approximately the same. The values in Table 2 were used to find the specific enthalpy at these four major points. Then, these specific enthalpy values were put on a table next to their corresponding pressure values, as seen in Table 3. Figure 3 is the graphical representation of Table 3, and it shows the refrigeration cycle. By using equations (1), (2), and (3), The COP, COPideal, and COPCarnot were calculated, as seen in Table 4. The COP (2.0) was far less than COPideal (7.7) meaning that the compressor is not fully adiabatic and it exchanges heat with the surrounding environment.

Table 1: Converted refrigeration data.

Table 2: Pressure and temperature at each of the four main points.

Table 3: Pressure data and the associated specific enthalpy for all four main points

Figure 3: (P-H) diagram showing the loop.

Table 4: Coefficients of performance.

Conclusion

To conclude, the objective of this experiment was to find the specific enthalpies to determine the Coefficient of Performance of the refrigeration system and validate that if the compressor is fully adiabatic or not. The COP was found to be 2.0, while the ideal COP was found to be 7.7 meaning that the compressor is not adiabatic as assumed at the beginning of the experiment. Nevertheless, the COP of the actual cycle is somehow suitable for most of the domestic applications. The only source of error can be the reading of the data as they were not recorded at the very same instant. All in all, the experiment can be found to be successful, the P-H plot has a perfect closed loop or cycle.

Appendix

Table 5: Raw data

Sample Calculation

· To convert pressure from psi to psia:

P = 75 (psi) + 14.7

P = 89.7 psia.

· To convert the temperature from °C to °F:

T = 34 (°C) * 1.8 + 32

T = 93.2 °F.

· To calculate the COPideal = (h1 – h4) / (h2 – h1):

COPideal = (179– 110) / (188 – 179)

COPideal = 7.7

Pressure vs. Enthalpy (R-134A)

179 188 110 110 44.7 153.69999999999999 154.69999999999999 49.7

Enthalpy (Btu/lbm)

Pressure (psi)

8