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Refrigeration system 2

Jan 19, 2015

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Lecture Notes on Refrigeration System

  • 1. REFRIGERATION Refrigeration: The process of cooling a substance and of maintaining it in a temperature below that of the immediate surroundings. Major Uses of Refrigeration: 1. Air Conditioning 2. Food Preservation 3. Removing of heat from substances in chemical, petroleum and petrochemical plants 4. Special applications in the manufacturing and construction industries. TON OF REFRIGERATION It is the heat equivalent to the melting of 1 ton (2000 lb) of water ice at 0C into liquid at 0C in 24 hours. 2000(144) BTU TR 12,000 24 hr KJ TR 211 min where 144 BTU/lb is the latent heat of fusion of ice CARNOT REFRIGERATION CYCLE: T QR TH3TH 2NetworkTLR41QA TLS HEAT ADDED (T = C) QA = TH(S1-S4) QA = TH(S) HEAT REJECTED (T = C) QR = TL (S2-S3) QR = TL (S) (S) = (S2-S3) =(S1-S4) NET WORK W = QR - QA W = (TH - TL)(S) COEFFICIENT OF PERFORMANCE Q COP A W QA COP QR Q ACOP TL TH TL1

2. where: QA - refrigerating effect or refrigerating capacity QR - heat rejected W - net cycle work VAPOR COMPRESSION CYCLE Components: Compressor Condenser Expansion Valve Evaporator Processes Compression, 1 to 2 (S = C) Heat Rejection, 2 to 3 (T = C) Expansion, 3 to 4 (S = C) Heat Addition, 4 to 1 (T = C)QR 32 Condenser1expansion valve evaporatorcompressor4QAP 3T22 3 S=C4h=C14 h1 SSystem: COMPRESSOR (S = C) m(h2 h1 ) W KW 60 k 1 kP1 V1' P2 k - 1 KW W P1 (k 1)60 P1 V1' mRT1 System: CONDENSER (P = C) QR = m(h2-h3) KJ/min System: EXPANSION VALVE (h = C) h3 = h42 3. h4 hf4 h g4 h f 4 System: EVAPORATOR (P = C) Q A m(h1 h 4 ) KJ/min x4 m(h1 h 4 ) Tons of Refrigerat ion 211 DISPLACEMENT VOLUME (For Reciprocating type of compressor) a. For Single Acting LD 2 Nn' m 3 VD 4(60) sec b. For Double Acting without considering piston rod 2LD 2 Nn' m 3 VD 4(60) sec c. For Double Acting considering piston rod 3 LNn' 2D 2 - D 2 m VD 4(60) sec QA KW PER TON OF REFRIGERATION KW 211(h 2 - h 1 ) KW Ton (60)(h1 - h 4 ) Ton COEFFICIENT OF PERFORMANCE Q COP A W h -h COP 1 4 h 2 - h1 CUBIC METER/min PER TON OF REFRIGERATION A. For Single Acting m3 211LD 2 Nn' m3 min - Ton 4m(h1 - h 4 ) min - Ton B. For double acting without considering piston rod m3 211(2)LD 2 Nn' m3 min - Ton 4m(h1 - h 4 ) min - Ton C. For double acting considering piston rod m3 211LNn' m3 2 2 2D - d min - Ton 4m(h1 - h 4 ) min - Ton where; L - length of stroke, m D - diameter of bore, m d - diameter of piston rod, m N - no. of RPM n' - no. of cylinders VOLUMETRIC EFFICIENCY3 4. v V1' x 100% VD1 P k v 1 C - C 2 x 100 % P 1 where: V1' - volume flow rate at intake, m3/sec VD - displacement volume, m3/sec v - volumetric efficiency 1 - specific volume, m3/kg v = [1 + C - C(P2/P1)1/k] x 100% EFFICIENCY A. Compression Efficiency Ideal Work x 100% cn Indicated Work B. Mechanical Efficiency Indicated Work x 100% m Brake or Shaft Work C. Compressor Efficiency Ideal Work x 100% c Brake or Shaft Work c cn m EFFECTS ON OPERATING CONDITIONS A. Effects on Increasing the Vaporizing Temperature PThS1. The refrigerating effect per unit mass increases. 2.The mass flow rate per ton decreases 3. The volume flow rate per ton decreases. 4. The COP increases. 5. The work per ton decreases. 6. The heat rejected at the condenser per ton decreases.4 5. B. Effects on Increasing the Condensing Temperature PThS1. The refrigerating effect per unit mass decreases. 2. The mass flow rate per ton increases. 3. the volume flow rate per ton increases. 4. The COP decreases. 5. The work per ton increases. 6. The heat rejected at the condenser per ton increases. C. Effects of Superheating the Suction Vapor PThSWhen superheating produces useful cooling: 1. The refrigerating effect per unit mass increases. 2. The mass flow rate per ton decreases 3. The volume flow rate per ton decreases. 4. The COP increases. 5. The work per ton decreases. When superheating occurs without useful cooling: 1. The refrigerating effect per unit mass remains the same. 2. The mass flow rate per ton remains the same. 3. The volume flow rate per ton increases. 4. The COP decreases. 5. The work per ton decreases. 6. The heat rejected at the condenser per ton increases.5 6. D. Effects of Sub-cooling the Liquid PThS1. The refrigerating effect per unit mass increases. 2. The mass flow rate per ton decreases. 3. The volume flow rate per ton decreases. 4. The COP increases. 5. The work per ton decreases. 6. The heat rejected at the condenser per ton decreases. LIQUID-SUCTION HEAT EXCHANGER The function of the heat exchanger is: 1. To ensure that no liquid enters the compressor, and 2. To sub-cool the liquid from the condenser to prevent bubbles of vapor from impeding the flow of refrigerant through the expansion valve. QRWQA6 7. SAMPLE PROBLEMS 1. An air conditioning system of a high rise building has a capacity of 350 KW of refrigeration, uses R-12. The evaporating and condensing temperature are 0C and 35C, respectively. Determine the following: P1 = 308.6 KPa a) 0.22 a) mass of flash gas per kg of refrigerant circulated P2 = 847.7 KPa b) 2.97 kg/sec b) mass of R-12 circulated per second h1 = 351.48 KJ/kg c) 0.1645 m3/sec c) volumetric rate of flow under suction conditions v1 = 0.0554 m3/kg d) 49.06 KW h2 = 368 KJ/kg d) compression work e) 7.14 h3 = h4 = 233.5 KJ/kg e) the COP hf at 0 C = 200 KJ/kg hg at 0 C = 351.48 KJ/kg2. A single cylinder, 6.7 cm x 5.7 cm, R-22 compressor operating at 30 rps indicate a refrigerating capacity of 96.4 KW and a power requirement of 19.4 KW at an evaporating temperature of 5C and a condensing temperature of 35C. Compute: P1 = 584 KPa a) 94.91% a) the clearance volumetric efficiency if c = 5% P2 = 1355 KPa b) 65.6% b) the actual volumetric efficiency h1 = 407.1 KJ/kg c) 63.33% c)the compression efficiency v1 = 40.36 L/kg h2 = 428 KJ/kg h3 = h4 =243.1 KJ/kgACTUAL VAPOR COMPRESSION CYCLE As the refrigerant flows through the system there will be pressure drop in the condenser, evaporator and piping. Heat losses or gains will occur depending on the temperature difference between the refrigerant and the surroundings. Compression will be polytropic with friction and heat transfer instead of isentropic. The actual vapor compression cycle may have some or all of the following items of deviation from the ideal cycle: Superheating of the vapor in the evaporator Heat gain in the suction line from the surroundings Pressure drop in the suction line due to fluid friction Pressure drop due to wiredrawing at the compressor suction valve. Polytropic compression with friction and heat transfer Pressure drop at the compressor discharge valve. Pressure drop in the delivery line. Heat loss in the delivery line. Pressure drop in the condenser. Sub-cooling of the liquid in the condenser or sub-cooler. Heat gain in the liquid line. Pressure drop in the evaporator.7 8. MULTIPRESSURE SYSTEMS A multi-pressure system is a refrigeration system that has two or more Low - Side pressures. The low-side pressure is the pressure of the refrigerant between the expansion valve and the intake of the compressor. A multi-pressure system is distinguished from the single-pressure system, which has but one low-side pressure. Removal of Flash Gas A savings in the power requirement of a refrigeration system results if the flash gas that develops in the throttling process between the condenser and the evaporator is removed and recompressed before complete expansion. The vapor is separated from the liquid by an equipment called the Flash Tank. The separation occurs when the upward speed of the vapor is low enough for the liquid particles to drop back into the tank. Normally, a vapor speed of less than 1 m/sec will provide adequate separation. Inter-cooling Inter-cooling between two stages of compression reduces the work of compression per kg of vapor. Inter-cooling in a refrigeration system can be accomplished with a water-cooled heat exchanger or by using refrigerant. The water-cooled intercooler may be satisfactory for two-stage air compression, but for refrigerant compression the water is not usually cold enough. The alternate method uses the liquid refrigerant from the condenser to do the inter-cooling. Discharge gas from the low stage compressor bubbles through the liquid in the intercooler. Refrigerant leaves the intercooler as saturated vapor. Inter-cooling with liquid refrigerant will usually decrease the total power requirement when ammonia is the refrigerant but not when R-12 or R-22 is used. 2-Evaporators and 1-Compressor condenserEvaporator Pressure reducing valveEvaporator8 9. 2-Compressors and 1-Evaporator CondenserHP Compressor Flash Tank and IntercoolerLP CompressorEvaporator2-Compressors and 2-Evaporatorscondense rHP compressor HP Evaporator Flash Tank and IntercoolerLP EvaporatorLP Compressor9 10. SAMPLE PROBLEMS 1. Calculate the power required by a system of one compressor serving two evaporators. One evaporator carries a load of 35 KW at 10C and the other at a load of 70 KW at -5C. A backpressure valve reduces the pressure in the 10C evaporator to that of the -5C evaporator. The condensing temperature is 37C. The refrigerant is ammonia. What is the COP. h3 = h4 = h7 = hf at 37C = 375.9 KJ/kg h5 = h6 = hg at 10C = 1471.6 KJ/kg h8 = hg at -5C = 1456.2 KJ/kg by energy balance at 10C evaporator m4 = m5 = m6 = 35/(h5 - h4) = 0.0319 kg/sec by energy balance at -5C evaporator m7 = m8 = 70/(h8 - h7) = 0.0648 kg/sec by mass and energy balance at the mixing point as shown on figure above m6h6 + m8h8 = m1h1 h1 = 1461.3 KJ/kg from chart, at S1 = S2 to P2 h2 = 1665 KJ/kg W = m1(h2 - h1) = 19.7 KW COP = 35 + 70 19.7 COP = 5.33 2. Calculate the power required in an ammonia system that serves a 210 KW evaporator at -20C. The system uses two-stage compression with inter-cooling and removal of flash gas. The condensing temperature is 32C. For minimum work and with perfect inter-cooling, the intermediate pressure P2 is equal to P2 = (P1P4)1/2 m7 = m8 = m1 = m

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