Summary
This experiment is carried out to study the refrigeration cycle. The real objective of conducting this experiment are to determine the coefficient of performance of a refrigerator unit, study the effect of evaporating and condensing temperature on the refrigeration rate and condenser heat output, investigate the effect of compressor pressure ratio on system performance and determine the overall heat transfer coefficient between R141b and water in the evaporator and condenser.
Based on methodology section in the report, it tells about the steps involved while conducting the experiment while the results section shows the recorded value of COP for difference condenser pressure, heat transfer in evaporator and condenser for difference condenser pressure , compressor pressure ratio and overall heat transfer. In addition, it also shows the enthalpy based on P-h diagram for five different pressure, H. Instead of that, the discussion section shows the graph plotted using the result obtained and discussed it more detail based on chemical reaction engineering theory. The error and recommendation to avoid mistakes while doing the experiment is are also shared in the discussion. The conclusion section concludes all the objectives and calculations of this experiment.
Aim
To determine the coefficient of performance of a refrigeration unit.
To study the effect of evaporating and condensing temperature on the refrigeration rate and condenser heat output.
To investigate the effect of compressor pressure ratio on system performance.
To determine the overall heat transfer coefficient between R141b and water in the evaporator and condenser.
Introduction
The term ’refrigeration’ may be defined as the process of removing heat from a substance under controlled conditions. It also includes the process of reducing and maintaining the temperature of a body below the general temperature of its surroundings. In other words, the refrigeration means a continued extraction of heat from a body whose temperature is already below temperature of its surroundings. In a refrigerator, heat is virtually pumped from a lower temperature to a higher temperature. According to Second Law of Thermodynamics, this process can only be performed with the aid of some external work. It is thus obvious that supply of power is regularly required to drive a refrigerator. Theoretically, a refrigerator is a reversed heat engine or a heat pump which pumps heat from a cold body and delivers it to a hot body. The substance which works in a pump to extract heat from a cold body and to deliver it to a hot body is known as refrigerant.
Theory
In 1850, the German scientist had proposed the Clausius statement that heat can never pass from a colder region to hot region. The process of heat transfer from cold to hot region cannot occur spontaneously without aid from the external work. In refrigerator, heat flows from cold to hot when forced by an external agent, the refrigeration system. The vapour compression cycle is used in refrigerator.
VAPOUR COMPRESSION CYCLE
The fluid that is used in refrigerator is called refrigerant. The refrigerants generally used for this purpose are ammonia (NH3), carbon dioxide (CO2) and sulphur-dioxide (SO2). The
refrigerant will undergo the process of condensation and evaporation alternately. During evaporation process, the refrigerant absorbs latent heat from the water which is used for circulating it around the cold chamber and in condensing; it gives out its latent heat to the circulating water of the cooler.
Figure 1.1 Vapour compression cycle
It consists of the following essential parts:
Compressor
The low pressure and temperature vapour refrigerant from evaporator is drawn into the compressor through the inlet or suction valve 1, where it is compressed to a high pressure and temperature. This high pressure and temperature vapour refrigerant is discharged into the condenser through the delivery or discharge valve 2.
Condenser
The condenser or cooler consists of coils of pipe in which the high pressure and temperature vapour refrigerant is cooled and condensed. The refrigerant, while passing through the condenser, gives up its latent heat to the surrounding condensing medium which is normally air or water.
Expansion Valve
It is also called throttle valve or refrigerant control valve. The function of the expansion valve is to allow the liquid refrigerant under high pressure and temperature to pass at a controlled rate after reducing its pressure and temperature. Some of the liquid refrigerant evaporates as it passes through the expansion valve, but the greater portion is vaporized in the evaporator at the low pressure and temperature
Evaporator
An evaporator consists of coils of pipe in which the liquid-vapour. refrigerant at low pressure and temperature is evaporated and changed into vapour refrigerant at low pressure and temperature. In evaporating, the liquid vapour refrigerant absorbs its latent heat of vaporization from the medium (air, water or brine) which is to be cooled.
Experimental Method
1. The unit is started and ensured that the unit is air free by venting air from the condensor.
2. The evaporator water flow is set to a mid range value and the unit is allowed to run approximately 15 - 20 minutes.
3. All system parameters is recorded.
4. The experiment is repeated for 5 different condenser pressures by adjusting the condenser flow rate from 50 g/s to 10 g/s.
Test No. 1 2 3 4 5
Gauge Evaporator Pressure, Pe (KN/m2) -19 -19 -19 -20 -20
Absolute Evaporator Pressure, Pe (KN/m2) 82.325 82.325 82.325 81.325 81.325
Evaporator Temperature, T5 (oC) 27 26 25 25 25
Evaporator Water Flow Rate, ms (g/s) 25 25 25 25 25
Evaporator Water Inlet Temp, T1 (oC) 35 33 32 31 31
Evaporator Water Outlet Temp, T2 (oC) 31 30 29 28 28
Condensed Liquid Temp, T8 (oC) 44 43 43 44 47
Gauge Condensor Pressure, Pc (KN/m2) +82 +81 +80 +82 +102
Absolute Condensor Pressure, Pc (KN/m2) 183.325 182.325 181.325 183.325 203.325
Condensor Temp, T6 (oC) 45 44 44 45 48
Condensor Water Flow Rate, mc (g/s) 50 40 30 20 10
Condensor Water Inlet Temp, T4 (oC) 35 33 32 31 31
Condensor Water Outlet Temp, T3 (oC) 36 36 35 36 42
Compressor Discharge Temp, T7 (oC) 68 73 76 79 82
Discussion
a) i) Calculate COP by using this formula : COP = = Test No 1 2 3 4 5 h1 455.56 455.56 452.78 455.56 455.56 h2 483.36 488.92 491.70 494.48 500.00 h3 258.34 261.12 255.56 258.34 261.12 h4 258.34 261.12 255.56 258.34 261.12 COP 7.0942 5.7386 5.0673 5.0673 4.3753
b) i) Heat transfer in evaporator for different condensor pressures, Q. Using this formula to calculate Q :
Q = msCpdT , where Cp = 4.18
b ii) Heat transfer in condensor for different condensor pressures,Q. Using this formula to calculate Q :
Q = mcCpdT , where Cp = 4.18
Test No 1 2 3 4 5
Condensor Water Flow Rate, mc (g/s) 0.05 0.04 0.03 0.02 0.01
W work, Compressor Qc , effect or Refrigerat 1 2 1 2 h h h h K kg kJ . Test No 1 2 3 4 5
Evaporator Water Flow Rate, ms (g/s) 0.025 0.025 0.025 0.025 0.025
Evaporator Water Outlet Temp, T2 (K) 304 303 302 301 301
Evaporator Water Inlet Temp, T1 (K) 308 306 305 304 304
dT (T2-T1) -4 -3 -3 -3 -3 Q ( J/s) -418 -313.5 -313.5 -313.5 -313.5 K kg kJ .
Condensor Water Outlet Temp, T3 (K) 309 309 308 309 315
Condensor Water Inlet Temp, T4 (K) 308 306 305 304 304
dT (T3-T4) 1 3 3 5 11
Q ( J/s) 209 501.6 376.2 418 459.8
iii) Compressor Pressure Ratio Pc/Pe
Test No 1 2 3 4 5
Absolute Condensor Pressure, Pc (KN/m2) 183.325 182.325 181.325 183.325 203.325
Absolute Evaporator Pressure, Pe (KN/m2) 82.325 82.325 82.325 81.325 81.325
Compressor Pressure Ratio Pc/Pe
v) Overall heat transfer, U.
By using this formula to calculate U. Q= UAdT
U=
IWater Coil Surface Area of Evaporator , A = 0.032 m2 i) Overall heat transfer for evaporator
Test No 1 2 3 4 5 Q ( J/s) -418 -313.5 -313.5 -313.5 -313.5 3293 7333 3293 7293 3293 7253 3253 7333 325 . 81 325 . 203 AdT Q
Overall heat transfer , U (W/K.m2) 3265.62 5 3265.62 5 3265.625 3265.625 3265.62 5
a) Overall heat transfer for condensor
Water Coil Surface Area of Evaporator , A = 0.032 m2
Test No 1 2 3 4 5
Q ( J/s) 209 501.6 376.2 418 459.8
Overall heat transfer , U (W/K.m2) 6531.25 5225 3918.75 2612.5 1306.25
Discussion
From table :
1. The COP for each test that we have calculated are 7.0942 , 5.7386 , 5.0673 , 5.0673 , and 4.3753. The variation of the measurement of temperature has an effect on the calculated of COP.
2. When calculating the values for the refrigerant, we read the values on the tables based on pressure and not temperature. This was done to minimize the error in calculations due to the variation involved with the thermometers.
3.The smaller the difference between the condensing and evaporating pressures, or between
condensing and evaporating temperatures, the lower the Win input to the compressor at a specific Q
refrigeration effect , QEF and, therefore, the higher the COP.A higher evaporating pressure and
evaporating temperature or a lower condensing pressure and condensing temperature will always conserve energy.
5. Parallax error .This is because the Pressure gauges must be read from directly in front and not from a slight angle. The gauges should also be calibrated by manufactures. This will cause errors in
the calculations because this is just an estimate of the actual transfer of heat to the
surroundings the values obtained from the actual refrigerator will not be equal to those of the ideal refrigerator
6. The gauge evaporator pressure have negative value because heat loss to surrounding but for the gauge condensor pressure have positive value because heat absorbed to system.
From graph :
1. From the heat transfer against compressor pressure ratio graph, we obtained evaporator curve is increasing.
2. However, the theoritical graph is supposed to parallel to the x-axis while the condenser in our graph is increasing eventhough there is a slightly decrease in heat transfer.
3. The deviation of the experimental curves and some observation of data for which the rate of heat transfer was not identical with other calculate value might cause the presence of air in refrigeration unit causes the compressor pressure to rise, reducing the COP.
4.In P-h diagram it shows different value of COP . For the COP value , test 5 showed the least value of COP which is 4.3753. The highest the value of COP, the more efficient it is.
Conclusion
From the refrigeration laboratory, the pressure, enthalpy, and temperature were calculated at the four state points of the vapour compression refrigeration cycle . With the state point properties and the power drawn by the compressor the COP can be determined. However the COP can be determined from the work input to the compressor, or as the work applied to the working fluid.
In comparison with the ideal vapour compression refrigeration cycle , data obtained and calculated from the laboratory (actual results) differed from those of the ideal cycle. The differences are a result of the assumptions made when working with the ideal cycle to simplify the calculations. The assumption that the compressor is isentropic shows the greatest impact on the variation of the ideal and the actual cycle.
A relevant source of error involves the accuracy of the measuring equipment and of the methods used in obtaining the measured values. The actual cycle would greatly approximate that of the ideal cycle if the assumptions could be reached, such as constant pressure heat transfer across heat exchangers, and the reduction of heat transfer to and from the environment due to insinuated devices.
Reference
-Refrigerant, Refrigerant Cycles and Refrigerant System [online], Retrieved from :
http://www.itcmp.pwr.wroc.pl/~kriogen/Wyklady/PodsChlod/Refrigeration/refrigeration%209.1.pdf on March 3, 2014.
-Norwegian University of Technology and Science. (2002). Manual Operation For Lab Experiment [online], Retrieved from :
http://www.nature.com/nature/journal/v424/n6952/abs/4241013a.html on
