The Coefficients of Performance of Particular Refrigeration System - Lab Report Example

Summary The following report is based on Refrigeration Unit. The aim of the experiment is to shows students a demonstration of refrigerant to show students an understanding on how refrigeration unit works and the reactions inside the refrigerant. The report explains and shows the readings taken from The Hilton Refrigeration Laboratory Unit and the calculations made from the readings taken. Results is also provided in the report with a discussion explaining about the outcomes and the final conclusion explaining what was achieved and learnt from this lab. Aim The major objectives of undertaking this laboratory experiment is to compute the Coefficients of Performance, COPR of particular refrigeration system ‘Hilton Refrigeration Laboratory Unit’ modelling it as a Refrigerator Unit or as a Heat pump. Introduction Refrigerators are special kind of electrical gadgets used to transfer heat from cold space to warm the surrounding close environment especially in the kitchen. Refrigerators often operate on the natural perceptions and experience that always flows from a region of high temperature towards that of a low temperature. For instance, heat transfer in a refrigerator may also occur from a region of high temperature to that of a low temperature. Refrigerators utilise working fluid referred to as refrigerants which absorb heat and utilize the same absorbed heat in heat exchange cycles. The calculations in this report are majorly concerned with electrical shaft or piston (indicated) power input. Subsequent computations also aim at evaluating energy balances for the various components as well as the complete functional plant. Finally, the report is also concerned with determination of volumetric and mechanical efficiencies of compressor by taking ratios of volume flow rates and power inputs respectively hence subsequently achieving the overall assessment of the motor compressor “losses” . Experimental Data/Results Table 1: Table of results SERIES TEST No. 1 2 3 4 5 6 Specific Enthalpy h1 kJ/kg 413 411 398 435 411 405 Specific Enthalpy h2 kJ/kg 442 450 450 440 451 452 Specific Enthalpy h3 kJ/kg 230 235 239 238 238 238 Specific Enthalpy h4 kJ/kg 212 213 213 220 229 236 Specific Enthalpy h5 kJ/kg 212 213 213 220 229 236 Specific Enthalpy h6 kJ/kg 390 382 380 400 385 383 Specific volume ѵ1 m3/kg 0.27 0.18 0.12 0.5 0.28 0.2 EVAPORATOR Evaporator heat input=VeIe Qe W 82.5 360 504 24 133 216 R12 enthalpy change rate=ṁr(h6-h4) W 0.2225 0.507 0.7014 0.144 0.195 0.294 CONDENSOR Heat transfer to cooling water=ṁrCpw(t8-t7) Qc W 0.21736 0.65208 0.86944 0.43472 0.43472 0.65208 R12 enthalpy change rate=ṁr(h2-h3) W 0.265 0.645 0.8862 0.1616 0.26625 0.428 COMPRESSOR & MOTOR Volume flow rate at compressor suction=ṁrѵ1 V1 m3/sec 3.375*10-4 5.4*10-4 5.4*10-4 4.0*10-4 3.5*10-4 4*10-4 compressor swept volume rate=2*(π/4)d2l*nc/60 Vs m3/sec 9.11*10-3 9.11*10-3 9.11*10-3 9.11*10-3 9.11*10-3 9.11*10-3 electrical input to motor=Vm Im cosѲ Pel W 414 414 414 414 414 414 motor shaft power=0.15F*2π nm/60 Ps W 159.436 227.766 273.319 273.319 273.319 273.319 compressor “indicated power” =0.15( F-Ff)*2π nm/60 Pi W 159.436 227.766 273.319 273.319 273.319 273.319 compressor enthalpy change rate= ṁr(h2-h1) W 0.036 0.117 0.218 0.004 0.050 0.094 EFFICIENCIES Motor efficiency=Ps/Pel η mot. % 38.51 20.71 55.01 66.01 66.01 66.01 Mechanical efficiency of comp. & drive=Pi/Ps η mech. % 100 100 100 100 100 100 Volumetric efficiency of compressor=V1/Vs η vol. % 3.70 5.92 5.53 4.39 3.84 4.39 COEFFICIENTS OF PERFORMANCE (REFRIGERATOR) CoP (based on electrical input)=Qe/Pel 0.199 0.870 1.217 0.058 0.321 0.522 CoP (based on shaft power input)=Qe/Ps 0.517 1.581 1.844 0.088 0.487 0.790 CoP (based on indicated power input)=Qe/Pi 0.517 1.581 1.844 0.088 0.487 0.790 CoP (of ideal cycle) =(h6-h4)/(h2sh1) 6.138 4.333 3.212 36 3.9 3.128 Evaporation temp. = t5 te ºC -35 -18 -11 -36 -25 -23 Condensation temp.=t3 tc ºC 21 23 25 23 24 22 Table 2: Table of results with calculated values 4.0. Discussion of Results To be able to compute coefficient of performance (COP) for refrigeration unit, take in consideration that refrigerator main function is to remove heat from cold space. In this case, the refrigerator will be acting as a heat sink, taking all the heat from the environment to itself and in the process work is done on the environment by the compressor. Therefore, to complete this heat removal task refrigerator needs work input. Let’s start with simple expression of Coefficient of Performance (COP) before attempting any calculations. COP of refrigerator is given by the following formula, COP = COPR = QL/ Wnet, in, to briefly explain this equation: QL represents heat taken off the refrigeration environment (Absorbed by the refrigerants)and Wnet, in is work required to complete removing heat from the cold environment of the refrigerator. Hence the coefficient of refrigerator is basically the ration between heat energy absorbed by the refrigerants and the work done to absorb the same heat. In our case, to effectively apply the formula for coefficient of performance in our subsequent computations, we will be required to replace QL and Wnet, in with evaporation rate of heat input, electrical power input to the motor, motor shaft power and compressor indicated power. Note that, QL is located at the refrigerated space in the evaporator, hence can be taken as the electrical power input to the motor. In this case therefore, the net electrical power input to the shaft transforms into work done to remove heat from the refrigerator environment. I.e. QL = Qe . Wnet, in is the work input at the compressor, equivalent to the heat absorbed by the action of the motor shaft. Let’s now begin off by calculating the COPR, for test no. 1. Taking the heat input from evaporator is given by product of evaporator voltage by evaporator current, VeIe . This is similar to calculating the apparent electrical power of the motor. (The sum of reactive and real power of and electrical motor) The electrical input of the motor is given by VmImcosθ,. VmImcosθ, on the other hand is equivalent to the real or true power developed by the motor. (Power developed at no power factor). The COP of refrigerator is based on the electrical input of the motor. Therefore, COPR = VeIe/ VmImcosθ, (Ratio of apparent power to real power). Substituting collected values gives, COPR = (155V*4.2A)/ (220V*3.8Acos45º) = 1.101. Estimation of COP of refrigeration can also be specify by taking ratio of evaporator heat input on Ps for shaft power input, or on Pi for indicated power input. 5.0. Conclusion Once all the required data had been collected, the refrigerator coefficients were determined. There is a possibility that the generator coefficients determined above had some inherent inconsistencies. Such errors might have easily been passed onto the results from the initial measurements obtained from the ‘Hilton’ refrigeration unit for subsequent use in this laboratory work. Another potential source of errors could be the refrigerant table or chart used to read off specific enthalpy in the process of trying to estimate the COP of ideal cycle. 6.0. References Cengel, Y and Turner, R 2001, Fundamentals of Thermal-Fluid Sciences, McGraw Hill: International Edition, New York. Read More