Tài liệu về nguyên lý hoạt động và các cơ chê của tủ lạnh và máy lạnh. Refrigeration and air conditioning is an useful reference which provides detailed information about the principles and mechanisms of refrigeration and airconditioner. All refrigerants and their impacts to the environment are also mentioned in this book
Introduction to Refrigeration and Air - Conditioning Introduction to Refrigeration & Air Conditioning Applied Thermodynamics & Heat Engines S.Y B Tech ME0223 SEM - IV Production Engineering ME0223 SEM-IV Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Outline • Applications of Refrigeration • Bell – Coleman Cycle • COP and Power Calculations • Vapour – Compression Refrigeration System • Presentation on T-S and P-h diagram • Vapour – Absorption System ME0223 SEM-IV Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Refrigeration REFRIGERATION – Science of producing and maintaining temperature below that of surrounding / atmosphere REFRIGERATION – Cooling of or removal of heat from a system Refrigerating System – Equipment employed to maintain the system at a low temperature Refrigerated System – System which is kept at lower temperature Refrigeration – 1) By melting of a solid, 2) By sublimation of a solid, 3) By evaporation of a liquid Most of the commercial refrigeration production : Evaporation of liquid This liquid is known as Refrigerant ME0223 SEM-IV Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Refrigeration Circuit Evaporator Compressor Refrigeration Circuit Expansion Valve ME0223 SEM-IV Condenser Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Refrigeration - Elements High Temp Source Surrounding Air QH Condenser QH Wnet, in Wnet, in Expansion Valve Compressor Evaporator QL QL Refrigerated Space ME0223 SEM-IV Applied Thermodynamics & Heat Engines Low Temp Sink S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Refrigeration - Applications Ice making Transportation of food items above and below freezing Industrial Air – Conditioning Comfort Air – Conditioning Chemical and related industries Medical and Surgical instruments Applications : Processing food products and beverages Oil Refining Synthetic Rubber Manufacturing 10 Manufacture and treatment of metals 11 Freezing food products 12 Manufacturing Solid Carbon Dioxide 13 Production of extremely low temperatures (Cryogenics) 14 Plumbing 15 Building Construction ME0223 SEM-IV Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Refrigeration Systems Ice Refrigeration System Air Refrigeration System Vapour Compression Refrigeration System Refrigeration Systems : Vapour Absorption Refrigeration System Adsorption Refrigeration System Cascade Refrigeration System Mixed Refrigeration System Thermoelectric Refrigeration System Steam Jet Refrigeration System 10 Vortex Tube Refrigeration System ME0223 SEM-IV Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Performance - COP Performance of Refrigeration System : - Measured in terms of COP (Coefficient of Performance) COP – Ratio of Heat absorbed by the Refrigerant while passing through the Evaporator to the Work Input required to compress the Refrigerant in the Compressor If; Then; Rn = Net Refrigerating Effect Rn W COP = Re lative COP = Actual COP W = Work required by the machine Actual COP Theoretical COP = Ratio of Rn and W actually measured Theoretical COP = Ratio of Theoretical values of Rn and W obtained by applying Laws of Thermodynamics to the Refrigerating Cycle ME0223 SEM-IV Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Performance - Rating Rating of Refrigeration System : - Refrigeration Effect / Amount of Heat extracted from a body in a given time Definition : - Refrigeration Effect produced by melting tonne of ice from and at ºC in 24 hours Unit : - Standard commercial Tonne of Refrigeration / TR Capacity Latent Heat of ice = 336 kJ/kg ME0223 SEM-IV Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Air Refrigeration System One of the earliest method Obsolete due to low COP and high operating cost Preferred in Aircraft Refrigeration due to its low weight Characteristic : - Throughout the cycle, Refrigerant remains in gaseous state Air Refrigeration Closed System Open System • Air refrigerant contained within piping or components of system • Pressures above atm Pr ME0223 SEM-IV • Refrigerator space is actual room to be cooled • Air expansion to atm Pr And then compressed to cooler pressure • Pressures limited to near atm Pr levels Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Example 28 tonnes of ice from and at ºC is produced per day in an ammonia refrigerator The temperature range in the compressor is from 25 ºC to -15oC The vapour is dry and saturated at the end of compression and an expansion valve is used Assuming a co-efficient of performance of 62% of the theoretical, calculate the power required to drive the compressor Take latent heat of ice = 335 kJ/kg Given : } ME0223 SEM-IV Temp (ºC) Enthalpy (kJ/kg) Vapour Entropy of Liquid (kJ/kg.K) Entropy of Vapour (kJ/kg.K) Liquid 25 100.04 1319.22 0.3473 4.4852 -15 -54.56 1304.99 -2.1338 5.0585 Tcond = 25 ºC hf1 = -54.56 kJ/kg Tevap = -15 ºC hg1 = 1304.99kJ/kg x2 = 1….dry saturated vapour hf2 = 100.04 kJ/kg COPactual = 0.62 (COPtheor) hg2 = 1319.22 kJ/kg Latent Heat of ice = 335.7 kJ/kg hf3 = h4 = 100.04 kJ/kg Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Example 7….contd h2 = hg = 1319.22 kJ / kg h3 = h4 = 100.04 kJ / kg Isenthalpic process Isentropic Compression : 1-2 Temperature, T s2 = s1 298 K s g = s f + x1 ∗ s fg1 Sat Vapour Line ⇒ 258 K g Sat Liq Line 4.4852 = (−2.1338) + ( x1 ) [ 5.0585 − ( − 2.1338) ] f Entropy, s COP of the Cycle : COPtheoretical = x2 = 0.92 h1 = h f + x1 (h fg1 ) = (−54.56) + (0.92) [1304.99 − (−54.56)] = 1196.23 kJ / kg (1196.23 − 100.04) = 8.91 h1 − h4 = h2 − h1 (1319.22 − 1196.23) S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Example 7….contd Actual Rn = COPactual X Work done Actual COP = ηrel X COPtheor = 5.52 X (h2 – h1) = 0.62 X 8.91 = 5.52 X (1319.22 – 1196.23) Temperature, T = 5.52 298 K = 678.9 kJ/kg Heat extracted from 28 tonnes of water at ºC to form ice at ºC : Sat Vapour Line 258 K g Sat Liq Line 28 (kg ) X 1000 (kg / tonne) X 335 (kJ / kg ) = 24 (hr ) X 3600 (sec/ hr ) = 108.56 kJ / sec (kW ) Mass of refrigerant : f Entropy, s = 108.56 (kJ / sec) = 0.1599 kg 678.9 (kJ / kg ) Total Work done by Compressor : = mrefrig X ( h2 − h1 ) = 0.1599 (kg ) X (1319.22 − 1196.23) kJ / kg = 19.67 kJ / sec (kW ) ….ANS ME0223 SEM-IV Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Example In a standard vapour compression refrigeration cycle, operating between an evaporator temperature of -10 ºC and a condenser temperature of 40 ºC, the enthalpy of the refrigerant, Freon-12, at the end of compression is 220 kJ/kg Show the cycle diagram on T-s plane Calculate: The C.O.P of the cycle The refrigerating capacity and the compressor power assuming a refrigerant flow rate of kg/min You may use the extract of Freon-12 property table given below: Given : } ME0223 SEM-IV Temp (ºC) Pr (MPa) hf (kJ/kg) hg (kJ/kg) -10 0.2191 26.85 183.1 40 0.9607 74.53 203.1 Tcond = 40 ºC Tevap = -10 ºC x1 = 1….dry saturated vapour h2 = 220 kJ/kg hf1 = 26.85 kJ/kg hg1 = h1 = 183.1 kJ/kg hf2 = 74.53 kJ/kg hg2 = 203.1 kJ/kg hf3 = h4 = 74.53 kJ/kg Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Example 8….contd COP of Original Cycle : COP = Temperature, T = 40 ºC Rn h1 − h4 = W h2 − h1 (183.1 − 74.53) kJ / kg = 2.94 ( 220.0 − 183.1) kJ / kg ….ANS 2’ Refrigerating Capacity : = m ( h1 − h4 ) = (kg ) X (183.1 − 74.53) kJ / kg -10 ºC Sat Vapour Line g Sat Liq Line = 108.57 kJ / ….ANS f Entropy, s Compressor Power : = m ( h2 − h1 ) = (kg ) X ( 220.0 − 183.1) kJ / kg = 36.9 kJ / = 0.615 kW ….ANS S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Example A Freon-12 refrigerator producing a cooling effect of 20 kJ/sec operates on a simple cycle with pressure limits of 1.509 bar and 9.607 bar The vapour leaves the evaporator dry saturated and there is no undercooling Determine the power required by the machine If the compressor operates at 300 rpm and has a clearance volume of 3% of stroke volume, determine the piston displacement of the compressor For compressor assume that the expansion following the law PV1.3 = Constant Temp (oC) Ps (bar) vg (m3/kg) -20 1.509 } 9.607 40 Given : ME0223 SEM-IV Enthalpy hg (kJ/kg) 178.61 Entropy sf (kJ/kg) 0.073 Entropy sg (kJ/kg) 0.7082 Specific heat (kJ/kg.K) 0.1088 Enthalpy hf (kJ/kg) 17.8 - 74.53 203.05 0.2716 0.682 0.747 Tcond = 40 ºC Tevap = -20 ºC x1 = 1….dry saturated vapour h2 = 220 kJ/kg - hf1 = 17.8 kJ/kg hg1 = h1 = 178.61 kJ/kg hf2 = 74.53 kJ/kg hg2’ = 203.05 kJ/kg hf3 = h4 = 74.53 kJ/kg Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Example 9….contd • Refrigerating Capacity : = m ( h1 − h4 ) • ⇒ 20 kW = m X (178.61 − 74.53) kJ / kg • Temperature, T ⇒ m = 0.192 kg / sec 313 K Isentropic Compression : 1-2 s1 = s2 2’ T2 s1 = s2 ' + C P ln T2' 253 K g Sat Liq Line f Entropy, s T 0.7082 = 0.682 + ( 0.747 ) ln 313 Sat Vapour Line ⇒ T2 = 324.2 K h2 = h2' + C P ( T2 − T2 ' ) = 203.05 (kJ / kg ) + ( 0.747 kJ / kg.K ) ( 324.2 − 313.0 ) K = 211.4 kJ / kg ME0223 SEM-IV Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Example • Power Required : = m ( h2 − h1 ) = 0.192 (kg / sec) X ( 211.4 − 178.61) kJ / kg Temperature, T = 6.29 kW ….ANS Vol Efficiency : 313 K 2’ Sat Vapour Line Sat Liq Line / 1.13 253 K g η vol Pd = + k − k PS 1/ n f 9.607 bar = + 0.03 − 0.03 1.509 bar = 87.6 % • Vol of Refrigerant = m ∗ v g at Intake : = 0.192 (kg / sec) X 0.1088 (m / kg ) = 0.02089 m / sec Entropy, s ( Actual Vol.) 0.02089 (m / sec) ∗ 60 (sec/ min) = = Piston Displ Vol : ME0223 SEM-IV η vol ∗ (rpm) 0.876 ∗ 300 (rpm) Applied Thermodynamics & Heat Engines = 0.00477 m ….ANS S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Example 10 A food storage locker requires a refrigeration capacity of 50 kW It works between a condenser temperature of 35 ºC and an evaporator temperature of -10 ºC The refrigerator is ammonia It is sub-cooled by ºC before entering the expansion valve By the dry saturated vapour leaving the evaporator Assuming a single-cylinder, single-acting compressor operating at 1000 rpm with stroke equal to 1.2 times the bore, determine : The power required The cylinder dimensions Properties of ammonia are : Sat Temp (oC) Pr (bar) Enthalpy (kJ/kg) Entropy (kJ/kg) Sp Vol (m3/kg) Sp Heat (kJ/kg.K) Liquid Vapour Liquid Vapour Liquid Vapour Liquid Vapour -10 2.9157 154.056 1450.22 0.82965 5.7550 - 0.417477 - 2.492 35 13.522 } 366.072 1488.57 1.56605 5.2086 1.7023 0.095629 4.556 2.903 Given : ME0223 SEM-IV Tcond = 35 ºC Tevap = -10 ºC x1 = 1….dry saturated vapour State = Sub-cooled by ºC h1 = 1450.22 kJ/kg h2’ = 1488.57 kJ/kg hf3 = 366.072 kJ/kg Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Example 10….contd h3' = h4 = h f − C P liq ( Tsat − Tsubcool ) Temperature, T = 366.07 (kJ / kg ) − 405.56 ( 308 − 303) (kJ / kg ) 308 K 303 K 2’ = 343.29 kJ / kg Isentropic Compression : 1-2 3’ 263 K g Sat Liq Line f T2 s1 = s2 ⇒ s1 = s2 ' + C P ln T2 ' T 5.755 = 5.2086 + ( 2.903) ln 308 Sat Vapour Line ⇒ T2 = 371.8 K Entropy, s h2 = h2' + C P ( T2 − T2' ) = 1488.57 ( kJ / kg ) + ( 2.903 kJ / kg.K ) ( 371.8 − 308.0 ) K = 1673.8 kJ / kg ME0223 SEM-IV Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Example 10….contd Temperature, T Mass of Refrigerant : 303 K m= 50 (kW ) 50 (kW ) = ( h1 − h4 ) kJ / kg (1450.22 − 343.29) kJ / kg = 0.04517 kg / sec 308 K • 2’ Compressor Power : • = m ( h2 − h1 ) 3’ = 0.04517 (kg ) X (1673.8 − 1450.22 ) kJ / kg 263 K g Sat Liq Line f = 10.1 kW ….ANS Sat Vapour Line Cylinder Dimensions : 1000 (rpm) π N π D L D (1.2 D) Entropy, • s 60 60 m = 0.04517 (kg / sec) = = vg 0.417477 m / kg ( ⇒ ⇒ ME0223 SEM-IV ) D = 0.19 m ….ANS L =1.2 ∗ (0.19 m) = 0.228 m ….ANS Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Vapour Absorption System Solubility of NH3 in water @ ↓ Temp and Pr is MORE than that @ ↑ Temp and Pr NH3 vapour from Evaporator (State 1) is readily absorbed in Absorber ⇒ Heat Rejection This solution is then pumped to ↑ Temp and Pr @ Generator Reduction in stability of solution ⇒ Vapour removed from Solution Vapour passes to Condenser Weak Solution returns to Absorber ME0223 SEM-IV Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Vapour Absorption System COP : Heat extracted in Evaporator Heat sup plied in Generator + Work done on Liquid Pump COP = Merits : Pumping work is much less than work for Compressing vapour Work done on Compression is LESS Demerits : Heat input to the Generator is required Low COP ME0223 SEM-IV Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Vapour Compression Vs Vapour Absorption Sr No Particulars Vapour Compression Systems Vapour Absorption Systems Type of Energy Supplied Mechanical – High Grade Heat – Low Grade Energy Supply Rate Low High Wear & Tear More Less Performance of Part Load Poor Not affected at Part Load Suitability Used where High Grade Mechanical Energy is available Can be used at Remote Places, as can be used with simple Kerosene lamp Charging of Refrigerant Simple Difficult Leakage More chances Damage No chances, as no Compressor or Reciprocating Part No danger ME0223 SEM-IV Liquid traces in Suction Line may damage Compressor Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg Introduction to Refrigeration and Air - Conditioning Thank Y ou ! ME0223 SEM-IV Applied Thermodynamics & Heat Engines S Y B Tech Prod Engg ... Introduction to Refrigeration and Air - Conditioning Refrigeration - Applications Ice making Transportation of food items above and below freezing Industrial Air – Conditioning Comfort Air – Conditioning. .. Engg Introduction to Refrigeration and Air - Conditioning Refrigeration Systems Ice Refrigeration System Air Refrigeration System Vapour Compression Refrigeration System Refrigeration Systems... Introduction to Refrigeration and Air - Conditioning Air Refrigeration System One of the earliest method Obsolete due to low COP and high operating cost Preferred in Aircraft Refrigeration due