Operation of a Refrigeration System - Term Paper Example

ME209 Experimental Methods Laboratory Report THF 21 - Refrigeration Name: your name Group: your group Summary The primary objective was to understand the fundamental operation of a refrigeration system and the different energy transfers in the system and to calculate from empirical test results the COP and the compressor efficiencies. The experiment also aims to study the effects of varying the refrigeration load on the cycle. To achieve the set objectives, a literature review, focusing mostly on the theoretical part of the experiment is conducted to aid in the understanding of the refrigerator systems followed by the laboratory test; a refrigerant model R134a gets operated at an initial voltage of 160V and the various measurement quantities recorded. Analysis based on the results then gets conducted and a P-h chart drawn with the state points. Outcomes of the experiment indicate that the coefficient of performance of the refrigerator based upon indicated power, motor power, and input electric power is 11.77, 4.33 and 2.14 respectively with the COP generated by the computer program Coolpack being 213.51. The compressor volumetric efficiency is computed as 140.9% while the isentropic efficiency is 5.18% however these vary considerably with computer generated values of 0.00 and 1.00 % respectively. The discrepancies in the results can get attributed to the possible errors in measuring equipment and fundamental human error. Key words: Coefficient of performance (COP), P-h chart, compressor efficiency, state points. Introduction “In any iterative process the entropy will either remain the same or increase over time until equilibrium gets attained” - the second law of thermodynamics. This postulates that naturally heat flows from high temperature regions to low temperature regions. An energy input is required, such as a refrigerator or heat pump to transfer heat from a low-temperature region to a high-temperature region. Refrigeration involves the extraction and transfer of heat from a low-temperature region to a high-temperature heat sink, thereby maintaining the temperature of the heat source below the surrounding temperature. Refrigeration systems can either be vapour compression systems, absorption systems or gas expansion systems. Vapour compression systems are the most commonly used. Figure 1 below illustrates the fundamental processes in a vapour compression cycle: Figure 1: Components of the vapour compression system The refrigerant gas gets compressed from low to high pressure, and then condensed from gas to liquid while discharging heat into surroundings. In the evaporator, the liquid refrigerant changes from liquid to gas, while removing heat from the surroundings to produce refrigerating effect during evaporation. The expansion valve functions to separate the high-pressure side from the low-pressure side. Objectives The objectives of this laboratory are to: Gain an understanding of the operation of a refrigeration system. Calculate the Coefficient of Performance, COP, based on electrical input, motor input and indicted power. Determine the various energy transfers in the system. Study the effects of varying the refrigeration load on the cycle parameters. Determine the compressor efficiencies. Compare the results manually with those obtained by the computer. Theory A refrigeration cycle is composed of evaporation, compression, condensation, throttling, and expansion processes, and can either be open or closed. If the cycle extracts heat from a lower temperature region and transfers it to a higher temperature reservoir in such a way that the final state is equal to the original form, it then has undergone a closed refrigeration cycle. In the open refrigeration cycle, the final state is not equal to the initial state. The reverse Carnot Cycle The reverse Carnot refrigeration engine cycle operates between two specified temperature levels and consists of processes that are all reversible. It is the most efficient refrigeration cycle and thus has the highest theoretical COP (Bahrami, 2015). A schematic diagram and cycle of a Carnot cycle refrigerating system is shown in figure 2 below: Figure 2: Carnot refrigeration cycle (a) schematic diagram; (b) Working fluid cycle (courtesy Wroclaw University of Technology) The cycle is composed of four reversible processes; an isothermal expansion process (4-1) in which heat, q1 is extracted at constant temperature T1 per kg of working substance, an isentropic compression process of the working fluid (1-2) with the aid of external work where the temperature of the fluid rises from T1 to T2, an isothermal compression process (2-3) in which q2 gets rejected at constant high-temperature T2 per kg of working substance, and an isentropic expansion process where the temperature of the working fluid falls from T2 to T1 (3-4) (Shet et al., 2015). The reversed Carnot cycle is theoretical and cannot be applied practically as the isentropic process requires very high speed, while the isothermal process requires a low-speed operation (Shet et al., 2015). Application of the 1st law of thermodynamics To analyse the refrigeration cycle, consider the pressure-enthalpy (P-h) diagram below: Figure 3: P-h diagram (courtesy Boles, 2004) Where process 1 – 2 is the isentropic compression process in the compressor, 2 – 3 is the constant pressure heat rejection process in the condenser, 3 – 4 is the throttling of the expansion valve, and 4 – 1 is the constant pressure heat addition in the evaporator. Applying the 1st Law of Thermodynamics in each process, we have: i) Isentropic compression (in compressor) h1 + 0 = h2 +W - W = h2 – h1 Input of work to the compressor Win, Btu/lb, is therefore given as: Win = h2 – h1 ii) Heat rejection (in condenser) Similarly, h2 + Q = h3 + 0 The heat that gets dissipated by the refrigerant in the condenser is therefore given as: QH = h2 – h3 iii) Throttling (in expansion valve) Assumption: negligible heat loss Therefore, h3 + 0 = h4 + 0 h3 = h4 iv) Constant pressure heat addition (in evaporator) The mass flow rate of refrigerant mr = Qrc qrf Where Qrc is the refrigerating capacity, Btu/h and qrf is the refrigerating effect given as (h1 – h4) Performance of the cycle The performance of any thermodynamic cycle is measured using an index referred to as the coefficient of performance, COP. The coefficient of performance is preferred to “thermal efficiency” because the COP can be greater than 1. The COP is used in the analysis of the following (Wroclaw, 2015): A refrigerator that is used to produce a refrigeration effect only, (COPref) A pump in which the heating effect is obtained by rejected heat, (COPhp) A heat recovery system in which both the refrigeration effect and the heating effect are used at the same time, (COPhr) COP of Refrigerator The coefficient of performance of a refrigerator is defined as the ratio of the refrigeration effect (heat absorbed) to the work input; COP = Heat absorbed = Heat absorbed Work input Heat rejected – Heat absorbed = T1 (s1 – s4) = T1 . T2 (s1 – s4) – T1 (s1 – s4) T2 – T1 Where T1 and T2 are as described under fig. 2, S1 and S4 are the entropy at state points 1 and 4, respectively as in figure 2 above. From the expression, higher COP is achieved with higher evaporator temperatures and lower condenser temperatures but doesn’t put into consideration the type of compressor. The practical COP used is calculated as (UNEP, 2006): COP = Cooling effect (kW) . Power input to compressor (kW) Experimental Procedure Safety Read the COSSH safety information concerning chemicals used in this lab, wear a white coat and take all the standard precautions in the lab. Apparatus The rig used is the Hilton Computer-Linked Refrigeration Laboratory Unit RC712. A diagram of the unit, showing the various instruments used and the measuring stations, is presented in Figure 4. All the instruments are linked to a desktop computer. A set of programs are used in order to display: A schematic diagram and system parameters at 60 s intervals. Transient data. The refrigeration cycle diagram and update it at 60 s intervals. The specification of the refrigerator is given in Table 1. Table 1: Refrigerator specification. Refrigerant R134a Compressor Twin cylinder. Belt-driven from electric motor. Bore = 40 mm; Stroke = 30 mm. Total swept volume, Vswept = 75.5x10-6 m3 per rev. Belt pulley ratio, Pr = 3.17 Torque arm radius, r = 0.165 m Condenser Shell and coil type. Heat transfer area = 0.075 m2. Evaporator Compact once-through concentric tube with refrigeration load supplied by two concentric heating elements. Refrigerant flowmeter calibration factor 13.3 Water flowmeter calibration factor 4.78 Compressor friction force 5 N Figure 4 RC712 Computer Linked Refrigeration Laboratory Unit. Procedure a) Turn on the water supply and adjust the water control valve to allow a moderate flow of water through the condenser coils, say 30 g/s. b) Switch on the unit and the computer. c) Load the computer program and follow the displayed instructions. d) Select Option 1 and after setting the evaporator load to 30% obtain a computer print-out of the data and take manually the following set of results: Refrigerant inlet temperature to compressor T1= ºC Refrigerant inlet temperature to condenser T2= ºC Refrigerant inlet temperature to expansion valve T3= ºC Refrigerant inlet temperature to evaporator T4= ºC Water inlet temperature to condenser T5= ºC Water outlet temperature from condenser T6= ºC Evaporating pressure p1= kN/m2 (gauge) Condensing pressure p2= kN/m2 (gauge) Refrigerant flow rate mr= g/s Water flow rate mw= g/s Brake force Fb= N Compressor speed Nc= rpm Compressor current Ic= A Supply voltage V= V Evaporator voltage VE= V Evaporator current IE= A e) Using Option 3, increase the evaporator load to 80% and observe the changes in the evaporator cycle until steady state is reached. Results Table 2 below shows the results recorded. Table 2: Measurement Results Measurement Quantity Absolute Refrigerant inlet temperature to compressor, T1 5.61 0C 278.76K Refrigerant inlet temperature to condenser, T2 54.99 0C 328.14K Refrigerant inlet temperature to expansion valve, T3 26.16 0C 299.31K Refrigerant inlet temperature to evaporator, T4 -5.06 0C 268.09K Water inlet temperature to condenser, T5 16.84 0C 289.99K Water outlet temperature from condenser, T6 23.81 0C 296.96K Evaporating pressure, P1 2.23KN/m2(gauge) 102230Pa Condensing pressure, P2 7.68KN/m2 (gauge) 107680Pa Refrigerant flow rate, mr 4.24g/s 0.00424kg/s Water flow rate, mw 21.66g/s 0.02166kg/s Brake force, Fb 8.5N 8.5N Compressor speed, Nc 448.78rpm 47rad/s Compressor current, Ic 3.46A 3.46A Supply voltage, V 250V 250V Evaporator voltage, VE 160V 160V Evaporator current, IE 4.4A 4.4A The refrigerant properties (from P-h curve, see figure 5) are shown in table 3 below: Table 3: Refrigerant Properties Quantity Value p1 1.0223 bar (abs) p2 1.0768 bar (abs) h1 381.74 kJ/kg h2 399.87 kJ/kg h2’ 382.68 kJ/kg h3 168.27 kJ/kg h4 168.27 kJ/kg v1 0.18760 Analysis Calculations 1. Shaft Power of motor: Wshaft = FbrNM = FbrNCPr = 8.5N x 0.165m x 47rad/s x 3.17 = 208.96W From the performance curve of refrigerator (see appendix), Wshaft = 208.96W Pf = 0.51 thus electric power of the motor is: Welec = ICVPf = 3.46 x 240V x 0.51 = 423.504W 2. Heat transfer rate to the cooling water in the condenser: QWC = mW cPW (T6 – T5) = 0.02166kg/s x 4180J/kgK x (296.96K – 289.99K) = 631.06W 3. Evaporator heat input: QeE = IEVE = 4.4A x 160V = 704W 4. Energy transferred into the refrigerant by compressor: WIN = mr (h1 – h2) = 0.00424kg/s x (381.74kJ/kg – 399.87kJ/kg) = -0.0769kW = -76.9W Heat transfer rate from condenser to surrounding: QrC = mr (h3 – h2) = 0.00424kg/s x (168.27kJ/kg – 399.87kJ/kg) = -0.982kW = -982W Heat transfer rate into the evaporator: QrE = mr (h1 – h4) = 0.00424kg/s x (381.74kJ/kg – 168.27kJ/kg) = 0.9051kW = 905.1W 5. COP of refrigerator unit: Based upon indicated power: COPIN = QrE = (h1 –h4) WIN (h2 – h1) = (381.74kJ/kg – 168.27kJ/kg) = 11.77 (399.87kJ/kg – 381.74kJ/kg) Based upon motor power: COPshaft = QrE Wshaft = 905.1W = 4.33 208.96W Based upon electrical input power: COPshaft = QrE Welec = 905.1W = 2.14 423.504W 6. Efficiency of the condenser Volumetric efficiency of compressor: ηv = Vr = mrv1 Vswept VNC = 0.00424kg/s x 0.1876m3/kg = 1.409 = 140.9% 75.5 x 10-6m3/rev x 448.78rpm x 1min/60s Isentropic efficiency of compressor: ηv = (h2’–h1) (h2 – h1) = (382.68kJ/kg – 381.74kJ/kg) = 0.0518 = 5.18% (399.87kJ/kg – 381.74kJ/kg) The Pressure-enthalpy (P-h) chart. The figure 5 below shows the pressure enthalpy chart. Figure 5: P-h curve for the refrigerator. Discussion Effect of increasing the refrigeration load Comparison of manually and computer obtained results Computer obtained results are as shown below: Refrigerant: R134a Data: Te [°C] = -25.95 Tc [°C] = -24.80 Isentropic efficiency = 1.00 ------------------------------------------------- Calculated: Qe [kJ/kg] = 213.348 Qc [kJ/kg] = 214.348 W [kJ/kg] = 0.999 COP [-] = 213.51 Pressure ratio [-] = 1.053 ------------------------------------------------- Dimensioning: Qe [kW] = 0.905 Qc [kW] = 0.909 m [kg/s] = 0.00423999 V [m^3/h] = 2.8658 Volumetric efficiency = 0.00 Displacement [m^3/h] = 0 W [kW] = 0.004 Q loss [kW] = 0.000 The manually calculated heat transfer rate from the condenser to surrounding, QrC is -0.982kW, while the computer-generated value, QC is 0.909kW, with an acceptable margin of error. The heat transfer rate into the evaporator, QrE is 0.9051kW manually and 0.905kW from computer obtained value. The manually calculated COP based upon indicated power is 11.77 with the computer generated value being 213.51. The isentropic and volumetric efficiency are manually calculated as 5.18% and 140.9% respectively with the computer values being 1.00 and 0.00 % respectively. The vast discrepancies in the manually calculated and computer produced results could be attributed to errors such as equipment errors, human errors (e.g. when taking measurements) Conclusion The outcomes indicate that the coefficient of performance of the refrigerator based upon indicated power, motor power, and input electric power is 11.77, 4.33 and 2.14 respectively with the COP generated by the computer program Coolpack being 213.51. The compressor volumetric efficiency is computed as 140.9% while the isentropic efficiency is 5.18%. These vary considerably with computer generated values of 0.00 and 1.00 % respectively. The discrepancies in the results have been attributed to the possible errors in measuring equipment and inherent human error. References Bahrami, M. (2015). Refrigeration Cycle. Simon Fraser University. Boles, P. (2004). Refrigeration Cycles. In P. Boles, Thermodynamics (p. Chapter 10). North Carolina State University. Prof, U. S. (2015). Refrigeration Cycles. Madras: Indian Institute of Technology. UNEP. (2006). Energy Efficiency Guide for Industry in Asia. UNEP. Wroclaw University of Technology. (n.d.). Refrigerants, Refrigeration cycles and Refrigeration Systems. Wroclaw. Appendices Refrigerator Performance Curve Figure 6: Typical Performance Curve Read More

A set of programs are used in order to display: A schematic diagram and system parameters at 60 s intervals. Transient data. The refrigeration cycle diagram and update it at 60 s intervals. The specification of the refrigerator is given in Table 1. Table 1: Refrigerator specification. Refrigerant R134a Compressor Twin cylinder. Belt-driven from electric motor. Bore = 40 mm; Stroke = 30 mm. Total swept volume, Vswept = 75.5x10-6 m3 per rev. Belt pulley ratio, Pr = 3.17 Torque arm radius, r = 0.

165 m Condenser Shell and coil type. Heat transfer area = 0.075 m2. Evaporator Compact once-through concentric tube with refrigeration load supplied by two concentric heating elements. Refrigerant flowmeter calibration factor 13.3 Water flowmeter calibration factor 4.78 Compressor friction force 5 N Figure 4 RC712 Computer Linked Refrigeration Laboratory Unit. Procedure a) Turn on the water supply and adjust the water control valve to allow a moderate flow of water through the condenser coils, say 30 g/s. b) Switch on the unit and the computer. c) Load the computer program and follow the displayed instructions. d) Select Option 1 and after setting the evaporator load to 30% obtain a computer print-out of the data and take manually the following set of results: Refrigerant inlet temperature to compressor T1= ºC Refrigerant inlet temperature to condenser T2= ºC Refrigerant inlet temperature to expansion valve T3= ºC Refrigerant inlet temperature to evaporator T4= ºC Water inlet temperature to condenser T5= ºC Water outlet temperature from condenser T6= ºC Evaporating pressure p1= kN/m2 (gauge) Condensing pressure p2= kN/m2 (gauge) Refrigerant flow rate mr= g/s Water flow rate mw= g/s Brake force Fb= N Compressor speed Nc= rpm Compressor current Ic= A Supply voltage V= V Evaporator voltage VE= V Evaporator current IE= A e) Using Option 3, increase the evaporator load to 80% and observe the changes in the evaporator cycle until steady state is reached.

Results Table 2 below shows the results recorded. Table 2: Measurement Results Measurement Quantity Absolute Refrigerant inlet temperature to compressor, T1 5.61 0C 278.76K Refrigerant inlet temperature to condenser, T2 54.99 0C 328.14K Refrigerant inlet temperature to expansion valve, T3 26.16 0C 299.31K Refrigerant inlet temperature to evaporator, T4 -5.06 0C 268.09K Water inlet temperature to condenser, T5 16.84 0C 289.99K Water outlet temperature from condenser, T6 23.81 0C 296.96K Evaporating pressure, P1 2.

23KN/m2(gauge) 102230Pa Condensing pressure, P2 7.68KN/m2 (gauge) 107680Pa Refrigerant flow rate, mr 4.24g/s 0.00424kg/s Water flow rate, mw 21.66g/s 0.02166kg/s Brake force, Fb 8.5N 8.5N Compressor speed, Nc 448.78rpm 47rad/s Compressor current, Ic 3.46A 3.46A Supply voltage, V 250V 250V Evaporator voltage, VE 160V 160V Evaporator current, IE 4.4A 4.4A The refrigerant properties (from P-h curve, see figure 5) are shown in table 3 below: Table 3: Refrigerant Properties Quantity Value p1 1.0223 bar (abs) p2 1.

0768 bar (abs) h1 381.74 kJ/kg h2 399.87 kJ/kg h2’ 382.68 kJ/kg h3 168.27 kJ/kg h4 168.27 kJ/kg v1 0.18760 Analysis Calculations 1. Shaft Power of motor: Wshaft = FbrNM = FbrNCPr = 8.5N x 0.165m x 47rad/s x 3.17 = 208.96W From the performance curve of refrigerator (see appendix), Wshaft = 208.96W Pf = 0.51 thus electric power of the motor is: Welec = ICVPf = 3.46 x 240V x 0.51 = 423.504W 2. Heat transfer rate to the cooling water in the condenser: QWC = mW cPW (T6 – T5) = 0.

02166kg/s x 4180J/kgK x (296.96K – 289.99K) = 631.06W 3. Evaporator heat input: QeE = IEVE = 4.4A x 160V = 704W 4. Energy transferred into the refrigerant by compressor: WIN = mr (h1 – h2) = 0.00424kg/s x (381.74kJ/kg – 399.87kJ/kg) = -0.0769kW = -76.9W Heat transfer rate from condenser to surrounding: QrC = mr (h3 – h2) = 0.00424kg/s x (168.

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