REFRIGERAION AND AIR CONDIIONING For B.E/B.ech Engineering Students As Per Revised Syllabus of Leading Universities in India Including Dr. APJ Abdul Kalam echnological University, Kerala Dr. S. Ramachandran, M.E., Ph.D., Dr. A. Anderson, M.E., Ph.D., Professors - Mech Sathyabama Institute of Science and echnology Chennai - 119 (Near All India Radio) 80, Karneeshwarar Koil Street, Mylapore, Chennai 600 004. Ph.: 2466 1909, 94440 81904 Email: aishram2000@gmail.com, airwalk800@gmail.com www.airwalkbooks.com, www.srbooks.org
First Edition : 08-07-2018 ISBN:978-93-88084-07-9 225/- 978-93-88084-07-9 www.airwalkbooks.com www.srbooks.org Cell: 9600003081, 9600003082
ME405 Refrigeration and Air Conditioning Course Plan Module Contents Hours I II III Introduction Brief history and applications of refrigeration. hermodynamics of refrigeration reversed Carnot cycle heat pump and refrigeration machines, Limitations of reversed Carnot cycle. Unit of refrigeration Air refrigeration systems Reversed Joule cycle, Air craft refrigeration systems, simple bootstrap Regenerative and reduced ambient system. Vortex tube refrigeration Very low temperature refrigeration systems (concept only). Adiabatic demagnetization of paramagnetic salts Vapour compression systems-simple cycle representation on -s and P-h Diagrams. COP Effect of operating parameters on COP methods of improving COP of simple cycle super-heating, under cooling, Liquid suction heat exchanger, actual cycle. FIRS INERNAL EXAM Multi pressure system multi compression and multi evaporator, systems. Inter cooling flash inter cooling and flash gas removal Different combinations of evaporator and compressor for different applications, Cascade system Refrigerants and their properties-eco-friendly Refrigerants, mixed refrigerants, selection of refrigerants for different applications Vapour absorption systems Ammonia water system simple system drawbacks-lithium Bromide water system Electrolux comparison with vapour compression system steam jet refrigeration. Sem. Exam Marks 6 15% 8 15% 7 15%
IV V VI Application of refrigeration domestic refrigerators water coolers ice plants. Cold storages food preservation methods plate freezing, quick-freezing. Refrigeration system components Compressors, condensers, expansion devices, evaporators. Cooling towers Different types and their application fields Refrigerant leakage and detection charging of refrigerant system controls. SECOND INERNAL EXAM Air conditioning meaning and utility, comfort and industrial air conditioning. Psychometric properties saturated and unsaturated air, dry, wet and dew point temperature humidity, specific humidity, absolute humidity, relative humidity and degree of saturation thermodynamic equations enthalpy of moisture adiabatic saturation process psychrometers. hermodynamic wet bulb temperature, psychometric chart Psychometric processes adiabatic mixing sensible heating and cooling humidifying and dehumidifying, air washer bypass factor sensible heat factor-rshf and GSHF line Design condition Apparent dew point temperature Choice of supply condition, state and mass rate of dehumidifed air quantity Fresh air supplied air refrigeration. Comfort air conditioning factors affecting human comfort. Effective temperature comfort chart. Summer air conditioning factors affecting cooling load estimation. Air conditioning system room air conditioner split system-packaged system-all air system-chilled water system. Winter air conditioning factors affecting heating system, humidifiers. Year round air conditioning AC system controls-thermostat and humidistat. Air distribution systems duct system and design Air conditioning of restaurants, hospitals, retail outlets, computer center, cinema theatre, and other place of amusement. Industrial applications of air conditioning. END SEMESER EXAM 6 15% 8 20% 7 20%
REFRIGERAION & AIR CONDIIONING JNU K AP SYLLABUS III Year - II Semester UNI I INRODUCION O REFRIGERAION: Necessity and applications unit of refrigeration and C.O.P. Mechanical refrigeration types of ideal cycles of refrigeration. air refrigeration: bell coleman cycle open and dense air systems refrigeration systems used in air crafts and problems. UNI II VAPOUR COMPRESSION REFRIGERAION: Working principle and essential omponents of the plant simple vapour compression refrigeration cycle COP representation of cycle on -S and p-h charts effect of sub cooling and super heating cycle analysis actual cycle influence of various parameters on system performance use of p-h charts numerical problems. UNI III REFRIGERANS Desirable properties classification - refrigerants used nomenclature ozone depletion global warming VCR SYSEM COMPONENS: Compressors general classification comparison advantages and disadvantages. condensers classification working principles evaporators classification working principles expansion devices types working principles UNI IV VAPOR ABSORPION SYSEM: Calculation of maximum COP description and working of NH system and Li Br water (wo shell & Four shell) System, principle of operation three fluid absorption system,salient features. SEAM JE REFRIGERAION SYSEM: Working Principle and basic components. principle and operation of (i) thermoelectric refrigerator (ii) vortex tube. UNI V INRODUCION O AIR CONDIIONING: Psychrometric properties & processes characterization of sensible and latent heat loads need for ventilation, consideration of infiltration load concepts of RSHF,GSHF problems, concept of ESHF and ADP temperature. Requirements of human comfort and concept of effective temperature comfort chart comfort air conditioning requirements of industrial air conditioning, air conditioning load calculations. UNI VI AIR CONDIIONING SYSEMS: Classification of equipment, cooling, heating, humidification and dehumidification, filters, grills and registers, fans and blowers. heat pump heat sources different heat pump circuits.
Contents C.1 able of Contents Module I Reversed Carnot Cycle and Air Refrigeration 1 1.1 Introduction....................................... 1.1 1.1.1 Fundamentals of Refrigeration.................. 1.1 1.1.1.1 Briefly history............................... 1.1 1.1.2 Applications of Refrigeration................... 1.2 1.1.3 Important elements of a refrigeration system..... 1.2 1.1.4 ypes of Mechanical Refrigeration system........ 1.3 1.1.5 Unit of Refrigeration: (on of Refrigeration)...... 1.3 1.2 Air Refrigeration System............................ 1.6 1.2.1 Air Refrigeration Cycles....................... 1.7 1.3 Reversed Carnot Cycle............................. 1.7 1.3.1 Heat pump and refrigeration machines.......... 1.9 1.3.1.1 Merits of Reversed Carnot Cycle.............. 1.18 1.3.1.2 Demerits of Reversed Carnot Cycle (Limitations) 1.18 1.4 Bell-Coleman Cycle (Reversed Joule Cycle)............ 1.18 1.5 Aircraft Refrigeration System................... 1.53 1.6 Simple Bootstrap ype............................. 1.71 1.7 Regenerative System............................... 1.78 1.8 Reduced Ambient System........................... 1.84 Module II Vortex tube, Low temperature and Vapour Compression Refrigeration (VCR) Systems 2.1 Vortex ube Refrigeration........................... 2.1 2.2 Very Low emperature Refrigeration Systems.......... 2.4 2.2.1 Cascade refrigeration system................... 2.4 2.2.2 Solid Carbondioxide or dry ice................. 2.7 2.2.3 Liquefaction of Gases......................... 2.12
C.2 Refrigeration and Air Conditioning - www.airwalkbooks.com 2.2.3.1 Linde system and Claude system/air Liquefaction System Hampson-Linde (Gas) Air Liquefaction System. 2.12 2.2.3.2 Claude System for Liquefying air............. 2.13 2.3 Adiabatic Demagnetization of Paramagnetic Salts...... 2.14 2.4 Vapour Compression Refrigeration................... 2.16 2.5 PH Chart........................................ 2.41 Module III 3.1 Multi-pressure Systems.............................. 3.1 3.2 Flash Gas Removal................................. 3.3 3.3 Intercooling........................................ 3.4 3.3.1 Flash Intercooling............................ 3.4 3.3.2 Water intercooling............................ 3.6 3.3.3 Intermediate Pressure......................... 3.6 3.4 Multi Compression System.......................... 3.7 3.5 Different Combinations of Evaporator and Compressor for Different Applications................................. 3.30 3.6 Cascade Refrigeration System....................... 3.38 3.7 Refrigerants And heir Properties................... 3.41 3.7.1 Classification of refrigerants................... 3.41 3.7.2 Various Refrigerants and their properties....... 3.43 3.7.3 Characteristics of good refrigerants............. 3.46 3.7.4 Refrigerants Number......................... 3.47 3.7.5 Desirable Properties for refrigerant selection..... 3.50 3.7.6 Ozone Layer Depletion........................ 3.52 3.7.7 Global Warming............................. 3.53 3.7.8 Ozone Depleting Potential (ODP) and Global Warming Potential (GWP) of Various Refrigerants............. 3.53
Contents C.3 3.8 Eco-Friendly Refrigerants........................... 3.54 3.9 Classification of Refrigerants Based on Number of Chemical Compounds.......................................... 3.56 3.10 Selection of Refrigerants for Different Applications.... 3.60 3.11 Vapour Absorption System......................... 3.66 3.12 Ammonia - Water Absorption System............... 3.72 3.13 Lithium Bromide Water Vapour Absorption System... 3.74 3.14 Comparision Between Vapour Compression and Vapour Absorption System.................................... 3.77 3.15 Application of Cryogenic........................... 3.80 3.16 Electrolux Vapour Absorption System............... 3.81 3.17 Steam Jet Refrigeration System (Ejector Refrigeration System)............................................. 3.84 3.18 hermo-electric Refrigeration (ER) System.......... 3.92 Module IV: Application of Refrigeration and Refrigeration System Components 4.1 Application of Refrigeration.......................... 4.1 4.1.1 Food processing, preservation................... 4.1 4.1.2 Chemical and process industries................ 4.3 4.1.3 Special applications........................... 4.5 4.2 Domestic Refrigerators.............................. 4.7 4.2.1 Parts of the household refrigerator:.............. 4.9 4.3 Water Coolers..................................... 4.12 4.4 Ice Plants........................................ 4.16 4.5 Cold Storage...................................... 4.18 4.6 Food Preservation Methods......................... 4.19 4.7 Contact or Plate Freezing.......................... 4.22 4.8 Quick Freezing.................................... 4.23
C.4 Refrigeration and Air Conditioning - www.airwalkbooks.com 4.9 Refrigeration System............................... 4.25 4.10 Compressor...................................... 4.25 4.11 Condensers...................................... 4.28 4.11.1 Parallel-flow condenser...................... 4.29 4.11.2 Counter flow-shell and tube condensor........ 4.30 4.11.3 Cross flow - Shell and tube condenser......... 4.31 4.11.4 Evaporative surface condenser................ 4.31 4.12 Expansion Devices................................ 4.32 4.13 Evaporator...................................... 4.34 4.13.1 Bare tube coil evaporator.................... 4.34 4.13.2 Finned evaporator.......................... 4.35 4.13.3 Shell and tube evaporator................... 4.36 4.14 Cooling owers................................... 4.37 4.14.1 Capacity of cooling tower.................... 4.37 4.14.2 Natural draft cooling towers................. 4.37 4.14.3 Mechanical draft cooling towers.............. 4.40 4.15 Refrigerant Leakage and Detection................. 4.42 4.15.1 Ammonia refrigerant leakage and detection:.... 4.42 4.15.2 Freon-12 leakage and detection............... 4.43 4.16 Charging of Refrigerant........................... 4.44 4.16.1 Methods of charging refrigerant............... 4.44 4.16.1.1 Charging through suction valve.............. 4.44 4.16.1.2 Charging through discharge valve............ 4.45 4.16.1.3 Charging through charging valve............. 4.47 4.17 System Control................................... 4.48 4.17.1 Starting relay.............................. 4.49 4.17.2 Overload protector.......................... 4.49
Contents C.5 4.17.3 hermostat................................ 4.49 Module V Psychrometric Processes and Airconditioning 5.1 Air Conditioning................................... 5.1 5.2 Psychrometry..................................... 5.3 5.3 Psychrometric Parameters.......................... 5.4 5.4 Psychrometric Properties............................ 5.5 5.5 Property Calculations of Air Vapour Mixtures.......... 5.9 5.6 Psychrometric Charts.............................. 5.23 5.7 Psychrometric Processes............................ 5.37 5.8 Sensible Heating Process........................... 5.39 5.8.1 By-pass Factor for Heating........................ 5.40 5.9. Sensible Cooling Process........................... 5.53 5.9.1 By-Pass Factor for Cooling Process............. 5.53 5.10. Cooling and Dehumidification...................... 5.58 5.11 Cooling and Humidification........................ 5.74 5.12 Heating and Dehumidification...................... 5.77 5.13 Heating and Humidification........................ 5.77 5.14. Adiabatic [No Heat ransfer] Saturation Process..... 5.86 5.14.1 Evaporative cooling Process (or) Adiabatic Humidification (or) Desert cooling process............ 5.88 5.14.2 Adiabatic Humidification [Cooling and Humidification] Problem........................... 5.89 5.15 Adiabatic Mixing Process.......................... 5.99 5.16 Air Washer..................................... 5.115 5.17 Sensible Heat Factor (S.H.F)...................... 5.119 5.18 Room Sensible Heat Factor (RSHF)................ 5.120
C.6 Refrigeration and Air Conditioning - www.airwalkbooks.com 5.19 Grand Sensible Heat Factor (GSHF)............... 5.122 5.20 Summer Air Conditioning........................ 5.125 5.20.1 Summer Air Conditioning System - Layout.... 5.125 5.20.2 Explanation............................... 5.126 5.20.3 Importance of Ventilation Air (Outside Air) (Fresh air)............................................ 5.128 5.20.4 Problems in Summer Airconditioning......... 5.131 5.21 Dehumidified Air Quantity cmm d................ 5.146 5.22 Effect of Bypass Factor.......................... 5.147 5.23 Comfort Condition............................... 5.148 5.24 Factors Affecting Human Comfort................. 5.149 5.24.1 Effective emperature...................... 5.149 5.24.1.1 Comfort chart............................. 5.151 5.24.1.2 Factors affecting Effective emperature...... 5.152 5.24.2 Heat produced in human body and dissipation 5.153 5.24.3 Heat and Moisture losses from the Human body5.154 5.24.4 Moisture content of air..................... 5.154 5.24.5 Quality and Quantity of air................. 5.154 5.25 Design Condition................................ 5.156 5.25.1 Inside Design condition..................... 5.156 5.25.2 Outside Design conditions................... 5.158 5.25.3 Choice of supply air design condition......... 5.159 5.26 Cooling Load Estimate........................... 5.160 5.26.1 Room Sensible Heat (Sensible Heat Gain)..... 5.161 5.26.2 Room Latent Heat (Latent Heat gain )..... 5.161
Contents C.7 Module VI Air Conditioning System and Winter Air Conditioning 6.1 Air Conditioning System............................ 6.1 6.1.1 Equipments Used in an Air Conditioning System. 6.1 6.1.2 Classification of Air Conditioning Systems....... 6.2 6.2 Window Room Air Conditioner....................... 6.4 6.3 Split ype Room Air Conditioner..................... 6.6 6.4 Packaged Air Conditioners........................... 6.9 6.5 Central Air Conditioning System.................... 6.12 6.6 Components of AC System......................... 6.14 6.6.1 Filter...................................... 6.14 6.6.2 Fans....................................... 6.14 6.6.3 Air washer.................................. 6.14 6.6.4 Radiators................................... 6.14 6.6.5 Convector................................... 6.16 6.7 All Air Systems................................... 6.16 6.7.1 Single duct system........................... 6.16 6.7.2 Dual duct system............................ 6.18 6.8 All Chilled Water Systems.......................... 6.20 6.9 Winter Airconditioning............................. 6.22 6.10 Ac System Control................................ 6.43 6.10.1 hermostat................................ 6.43 6.10.2 Bimetallic strips............................ 6.44 6.10.3 Gas-filled bellows........................... 6.45 6.10.4 Wax thermostats............................ 6.46 6.10.5 Humidistat................................ 6.47 6.11 Air Distribution Systems.......................... 6.49
C.8 Refrigeration and Air Conditioning - www.airwalkbooks.com 6.11.1 Fans...................................... 6.50 6.11.2 Good and poor air connection................ 6.50 6.11.3 Room Air Distribution....................... 6.51 6.11.4 Principle of air distribution.................. 6.51 6.11.5 Air behaviour.............................. 6.51 6.11.6 erms used in Air distribution............... 6.52 6.11.7 Location for supply air outlet................ 6.52 6.11.8 ypes of supply air outlets (devices)........... 6.53 6.11.9 Selection of Air outlets...................... 6.56 6.12 Duct System..................................... 6.56 6.13 Duct Design (or) Duct Sizing...................... 6.67 6.13.1 Objective of duct design..................... 6.67 6.13.2 Duct design methods........................ 6.68 1. Velocity Reduction Method....................... 6.68 2. Equal pressure (Equal friction loss) method........ 6.68 3. Static regain method........................... 6.69 6.13.3 System Resistance.......................... 6.79 1. Systems in series............................... 6.80 2. Systems in parallel............................. 6.80 6.14 Air Conditioning System in Cinema heatre......... 6.83 6.15 Restaurant Air Conditioning....................... 6.84 6.16 Air Conditioning Requirements in Hospitals......... 6.86 6.17 Air-conditioning of Retail Outlets................... 6.89 6.18 Computer Centre Air Conditioning Unit (CRAC):..... 6.92 6.19 Industrial Applications of Air Conditioning.......... 6.92
Module I Reversed Carnot Cycle and Air Refrigeration Introduction Brief history and applications of refrigeration. hermodynamics of refrigeration- Reversed Carnot cycle-heat pump and refrigeration machines, Limitations of Reversed Carnot cycle. Unit of refrigeration - Air refrigeration systems- Reversed Joule Cycle, Air craft refrigeration system, simple bootstrap Regenerative and reduced ambient system. 1.1 INRODUCION Refrigeration is defined as the science of providing and maintaining temperature below surrounding atmosphere. Refrigeration is a method to achieve and maintain low temperature by supplying work input continuously. 1.1.1 Fundamentals of Refrigeration Refrigeration may also be defined as the process by which the temperature of a given space or a substance is lowered below that of the atmosphere or surroundings. In simple, refrigeration means the cooling of or removal of heat from a system. he equipment employed to maintain the system at a low temperature is termed as refrigerating system and the system which is kept at lower temperature is called refrigerated system. Refrigeration can be generally produced in one of the following ways. (i) By melting of a solid (ii) By sublimation of a solid (iii) By evaporation of a liquid. Generally by evaporation of liquid called refrigerant is used in commercial refrigeration. 1.1.1.1 Briefly history of conditioning and refrigeration system It the first decades of the 20 th century, only fresh foods that could be grown locally were available and that too on daily basis only. his problem was also with milk and meat. If you could be able to get ice blocks, you can keep some perishable food items for 2 to 3 days. As for the non-existence of air-conditioning, it made summers in many southern countries sufferable.
1.2 Refrigeration and Air Conditioning - www.airwalkbooks.com But by the end of the century, all of this had changed. Fresh foods of all kinds were available anywhere all around the world. On the air-conditioning side, all forms of indoor space - office, hospitals, factories and homes - was climate controlled and made comfortable all over the year. While using ice and thermocol for refrigerating, we couldn t prevent the heat transfer to the surroundings. his was overcome by the modern refrigeration and air conditioning systems. he various types of refrigeration system and their operations can be understood from this lesson. 1.1.2 Applications of Refrigeration Refrigeration has a wide applications in a person s daily life. Some of the important one s are listed below. (i) (ii) (iii) (iv) (v) (vi) For comfort purpose: Air conditioning of residential buildings, offices, cinema houses, restaurants, departmental stores, hospitals, halls etc. For industrial purpose: cotton mills, textile industries, liquefaction of gases, treatment of metals, machine tool industries, marketing industries etc. For medicine purpose: Preservation of drugs, bloods, eyes, preservation of surgical equipments, human tissues etc,. For preservation of food products: Preservation of foods, highly perishable foods, produce ice creams, beverages, cold water, diary products etc. For research work: For research under low temperature application, cryogenics study, rocket, fuels, synthetic rubber and oil factory. For computer functioning: Maintaining low temperature in computer environments. 1.1.3 Important elements of a refrigeration system (i) A low temperature thermal sink (ii) A means of extracting energy from the sink, raising the temperature level of this energy and delivering it to a heat receiver. (iii) A receiver to which heat will be transferred from the high temperature high pressure refrigerant
Reversed Carnot Cycle and Air Refrigeration 1.3 (iv) Means of reducing of pressure and temperature of the refrigerant as it returns from the receiver to the sink. 1.1.4 ypes of Mechanical Refrigeration system he important refrigeration systems are (i) Vapour compression refrigeration system (ii) Vapour absorption refrigeration system (iii) Ice refrigeration system (iv) Air refrigeration system (v) Special refrigeration systems. (a) hermoelectric refrigeration system (b) Adsorption refrigeration system (c) Cascade refrigeration system (d) Vortex tube refrigeration system 1.1.5 Unit of Refrigeration: (on of Refrigeration) he unit used in the field of refrigeration is on of refrigeration. One ton of refrigeration is defined as heat removed from 1000 kg of water at 0C to make (1000 kg) 1 ton of ice at 0C within 24 hours. Heat removal rate is one ton of refrigeration. h fg Latent heat of the fusion = 301.5 kj/kg. (It can be taken from steam table) Q out (1 ton of water) 1000 kg of water at 0 o C water become ice Q out (1 ton) 1000 kg of ice at 0 o C Fig. 1.1 Q out ime duration : 24 hours
1.4 Refrigeration and Air Conditioning - www.airwalkbooks.com 1000 kg 301.5 kj/kg 1 ton of refrigeration 1 R 24 3600 sec kg kj 1 R sec kg kj sec kw 1 R 3.4892 kw So, 1 R Heat removal rate of 3.49 kw or 210 kj/min or 50 kcal/min. For problems, take C p water C pw 4.187 kj/kg K C p ice 2.09 kj/kg K 1 R 3.5 kw Problem 1.1: 2000 kg of water at 30C is cooled into ice at 20C and 6 hours. Determine heat removal rate in kw and in R. ake latent heat of ice as 334 kj/kg. Solution: (m w mass in kg/sec) otal Q removed Q R water Q R water to ice Q R ice 1. Q removed from water at 30C to make it 0C Q R water m w C pw 2000 4.18730 0 6 3600 11.631 kw 2. Q removed to make water at 0C to ice at 0C h fg Latent heat 334 kj/kg (1) (2) (3) ice Q out Q removed= Q R water Q rem oved Water at 30 o C Water at 0 o C Q out =Latent heat ice at 0 o C Q out Q rem oved= Q R ice Q removed m w h fg 2000 6 3600 334 30.9259 kw ice at-20 o C ice at -20 o C
Reversed Carnot Cycle and Air Refrigeration 1.5 3. Q removed to make ice from 0C to 20C: Q R ice otal heat removed Q R 46.4273 kw onne of Refrigeration Q R ice m ice C p ice ice 2000 2.09 0 20 6 3600 3.8704 kw Q R water Q R water to ice Q R ice 11.631 30.9259 3.8704 R Q R 3.5 46.4273 13.265 3.5 R 13.265R [... 1 R 3.5 kw] Problem 1.2: 5000 kg of water at 20C is converted into ice at 30C in 12 hrs. Determine heat removal rate in kw and in R. Latent heat of freezing of ice 335 kj/kg. Solution: Q R Q R water Q R water to ice Q R ice m w [ C pw w h fg C p ice ice 5000 12 3600 [ 4.18720 0 335 2.090 30] 55.722 kw R Q R 3.5 15.921 [... 1 R 3.5 kw] onnes of refrigeration R 15.921 R
1.6 Refrigeration and Air Conditioning - www.airwalkbooks.com Problem 1.3: wo refrigertor manufacturers claim that they developed a new model, working between 10C and 40C. First one claims a COP of 7.0, while the second manufacturer claims a COP of 8.0. Which one you will prefer? Why? (Kerala University, 2013) Solution: 1 40 273 313 K ; 2 10 273 263 K COP for carnot cycle COP for carnot cycle 5.26 2 263 1 2 313 263 I won t prefer both of the manufacturers, because, COP of a refrigerator can t be greater than carnot. 1.2 AIR REFRIGERAION SYSEM An air refrigeration system is a simple refrigeration cycle where in the working fluid is air. It has relatively low coefficient of performance despite its high operating costs. Hence, the usage of air refrigeration system has been predominantly limited to aircraft refrigeration system due to its low weight and the availability of cabin air as per the necessity. he notable feature about this system is that the refrigerant remains in gaseous state throughout the cycle. It can be primarily divided into: Closed system - It is also known as dense air refrigeration system. In this system, the refrigerating air is contained within the components of the system at all times. It usually operates in pressures exceeding the atmospheric pressure. Open system - In this system, the air is not circulated repeatedly within the system. Cooled air from the turbine directly comes in contact with the substances to be cooled, and is released into the atmosphere. Hence, the operating pressure is limited to the pressure inside the refrigerator. Consequently, an open cycle air refrigerating system offers low COP with high operating costs.
Reversed Carnot Cycle and Air Refrigeration 1.7 Closed cycle Open cycle Can operate at high pressure. Limited to atmospheric pressures. Moisture is eliminated. Moisture may choke the valves present in the system. he size of the compressor and expander is reduced due to use of dense air. No fog is formed due to absence of moisture. he size of compressor directly depends upon the pressure inside the refrigerator. Fog formation due to moisture at the turbine. Hence a drier is needed. Based on the principle of operation, an air refrigeration system can be classified into three types. 1.2.1 Air Refrigeration Cycles Refrigeration system is working under the following 1. Reversed carnot cycle 2. Bell-Coleman cycle 3. Aircraft refrigeration cycle 1.3 REVERSED CARNO CYCLE Heat is removed from sink at low temperature to source at high temperature by supplying work input. COP-Coefficient of Performance C.O.P means the ratio of the desired effect (Refrigeration effect) to the work input. C.O.P for reversed carnot cycle Heat removed from cold body Work input per cycle Refrigeration effect W Q 2 W Q 2 Q 1 Q 2 2 1 2
1.8 Refrigeration and Air Conditioning - www.airwalkbooks.com Reversed carnot cycle will give more C.O.P. But this cycle is not practically possible. So this cycle C.O.P is used to rate the other cycles C.O.P. Pv diagram and S diagram of Reversed Carnot Cycle f e (a) Air is the working fluid in the reversed carnot cycle. So it is called air refrigerator. his cycle is used to find the maximum C.O.P. for given temperatures. From the -S diagram Heat abstracted from the cold body Area dcfe 2 cd Workdone per cycle Area bcda 1 2 cd Coefficient of performance, C.O.P Fig. 1.3 Heat extracted from the cold body Workdone per cycle 2 cd 1 2 cd 2 1 2 A Carnot cycle can run either as a refrigerating machine, or a heat pump or as a heat engine. (b)
Reversed Carnot Cycle and Air Refrigeration 1.9 1.3.1 Heat pump and refrigeration machines (i) As a refrigerating machine C.O.P ref Heat extracted from cold body Workdone per cycle 2 1 2 (ii) As a Heat engine C.O.P heat engine Work obtained/cycle Heat supplied/cycle 1 2 cd 1 ba 1 2 1 Hot body Room 1 Source Fig. 1.4 Heat pump Q 1 = Heat supplied Q 2 Cold body Sink 2 atmosphere W... cd ba (iii) As a Heat Pump If the desired effect is heat supplied to hot body, then the device is called heat pump. he C.O.P of heat pump Desired effect Work input Q 1 Q 1 Q 2 Q 1 Heat supplied W Q 1 1 2 C.O.P 1 Q 1 Q 2 1 2 1 2 So COP of heat pump is always greater than COP of refrigerator working on reversed carnot cycle between same temperature limits 1 and 2 by unity (1). Problem 1.4: A/C room is to be maintained at 20C. he atmospheric temperature is at 45C. he power given to the compressor is 3 kw. Determine the ton of refrigeration.
1.10 Refrigeration and Air Conditioning - www.airwalkbooks.com Solution: 2 C.O.P 1 2 293 318 293 11.72 1 = 45+273 Q 1 = 318 K Also C.O.P Q 2 W 11.72 Q 2 3 Heat removed Work input Q 2 3 11.72 35.16 kw Ref 2 = 20+273 Q 2 W = 293 K = 3 KW Q 2 35.16 10.045 R 3.5 Heat removal rate 10.045 R Problem 1.5: A carnot refrigerator requires 2 kw per ton of refrigeration to maintain a space at 40C. Determine 1. COP of carnot refrigerator. 2. emperature of the hot body. 3. he heat delivered and COP when this device is used as heat pump. Solution: 2 40 273 233 K Work input = Power required = 2 kw Refrigeration effect = 1 on of refrigeration = 3.5 kw C.O.P Refrigeration effect Work input 2 But C.O.P for carnot cycle 1 2 1.75 233 1 233 1 233 233 1.75 133.14 1 366.143 K 3.5 2 1.75
Reversed Carnot Cycle and Air Refrigeration 1.11 Q 1 Q 2 W Q 2 Refrigeration effect 3.5 kw W Power required 2 kw Q 1 3.5 2 5.5 kw C.O.P of heat pump Q 1 W 5.5 2 2.75 Note: C.O.P of heat pump (2.75) = 1 + C.O.P of Refrigerator (1.75). 1 =366.143 Q 1 Heat pump Device Q 2 2 =233 W Problem 1.6: A Carnot refrigerator requires 1.25 KW per tonne of refrigeration to maintain the temperature of 30C. Find (i) COP (ii) emperature at which heat is rejected. [CUSA, April 2017] Solution: 2 30 273 243 K Work input 1.25 kw Refrigeration effect 3.5 kw COP RE W 3.5 1.25 COP 2.8 2 But COP for carnot cycle 1 2 2.8 243 1 243 1 243 243 2.8 1 36.78 243 1 329.78 K Problem 1.7: A reversed carnot cycle is used to deliver 1680 kj/sec to heat the conditioned space. he heat is taken from atmosphere at 10C and supplied to the conditioned space at 25C. Find the followings: power required to run the system. [Kakinada University Nov-2011] Given data: 1 25C 298 K ; 2 10C 283 K Q 1 Heat delivered 1680 kj/sec
1.12 Refrigeration and Air Conditioning - www.airwalkbooks.com Solution: Q 1 COP HP Q 1 Q 2 COP of reverse carnot heat pump 1 1 2 298 298.283 19.87 W 1 =298 Ref Q 1 Q 2 2 =283 Power COP Desired effect Work done/sec Desired effect COP 1680 19.87 work done Power sec Power required 84.55 kj/sec 84.55 kw Problem 1.8: Find the least power to produce 400 kg of ice per hour at 10C from feed water at 20C. Assume specific heat of ice as 2.09 kj/kg-k and latent heat 334 kj/kg. Solution: Least Power means more COP. More COP means it is reversed carnot cycle. Mass of ice produced per hour h fg 334 kj/kg 1 20+273=293 K Q 1 m w m ice 400 kg/hr 400 0.1111 kg/sec 1 3600 Ref Q 2 W C.O.P 2 263 1 2 293 263 8.767 2-10+273 = 263K
Reversed Carnot Cycle and Air Refrigeration 1.13 o Find Least Power C.O.P Net refrigeration effect W Q 2 W Q 2 Q removed from water at 20C to make it ice at 10C Q 2 m w [ Q R water Q R water to ice Q R ice ] 400 1 3600 [ C pw w h fg C p ice ice ] 400 [ 4.187 20 0 334 2.090 10 ] 48.74 kw 3600 Least power W Q 2 C.O.P 48.74 5.56 kw 8.767 Relative COP Relative COP is the ratio of actual COP to the carnot COP. Relative Actual COP COP Carnot COP Actual COP Q 2 W Q R W Q rejected or Q removed Power input 2 Carnot COP 1 2 Problem 1.9: An A/C room is to be maintained at 20C and the atmospheric temperature is 40C. he power required to run the compressor is 5 kw. Determine capacity of the refrigerator when relative COP is 50%. Solution: Capacity of refrigerator is given by ton of refrigeration. 1 = 40+273=313K Q 1 Carnot COP 2 293 1 2 313 293 14.65 Ref Q 2 W =5KW Actual COP Carnot COP Relative COP 2 2 = 20+273 = 293K
1.14 Refrigeration and Air Conditioning - www.airwalkbooks.com Also Actual COP Q 2 W 14.65 0.5 7.325 Q 2 Actual COP W 7.325 5 36.625 kw Capacity of refrigerator Q 2 in R 10.46 R 36.625 kw 3.5 kw Problem 1.10: he capacity of the refrigerator working on reversed Carnot cycle is 280 tonnes when operating between 10C and 25C. Determine (i) quantity of ice produced within 24 hours when water is supplied at 20C (ii) Minimum power required. Latent heat of ice is 335 kj/kg Solution: (i) Heat removed from water at 25C to make it ice at 10C per kg of water Q 2 [Q R Water Q RWater to rice Q R ice ] [C Pw w h fg C Pice ice ] [4.187 20 0 335 2.09 0 10] Q 2 439 kj/kg. Heat extracting capacity of refrigerator 280 tonnes. 1 tonne 3.5 kw Heat extracting capacity 280 3.5 980 kw. Quantity of ice produced in 24 hours M ice 980 24 60 60 439 192874 kgs or 193 tonnes.
Reversed Carnot Cycle and Air Refrigeration 1.15 (ii) Minimum power required: 1 25 273 298 K; 2 10 273 263 K C.O.P C.O.P Power required 2 263 1 2 298 263 7.51 Heat extracting capacity Power required Heat extracting capacity C.O.P 980 130.49 kw 7.51 Problem 1.11: Ice is formed at 0C from water at 20C the temperature of the brine is 8C. Find out the kg of ice formed per kw-hr. Assume that the refrigeration cycle used is perfect reversed carnot cycle. ake latent heat of fusion of ice as 335 kj/kg. (MBCE College Kerala - Dec 2016) Given data: h fg 335 kj /kg ; w 20C 0C 20C 293 K ; ice 0C 8C 8C 281 K Solution: Heat removed from water at 20C to make it ice at 8C per kg of water. Q 2 m [Q R water Q R water to ice Q R ice ] [4.187 20 335 2.09 8] Q 2 435.46 kj/kg 1 kwhr 1 kj s hr kj s 3600s Capacity of refrigerator 3600 kj Quantity of ice produced per kwhr m heat extracting capacity ice 3600 heat removed 435.46 m ice 8.27 kg/hr [m 1 kg]
1.16 Refrigeration and Air Conditioning - www.airwalkbooks.com Problem 1.12: A cold storage plant is required to store 20 tonnes of food. he temperature of the food when supplied is 25C; storage temperature of food is 8C. Specific heat of food above freezing point is 2.93 kj/kgc. Specific heat of food below freezing point 1.25 kj/kgc; Freezing point of food is 3C. Latent heat of food is 232 kj/kg. If the cooling is achieved with in 8 hrs; find (i) Capacity of the refrigeration plant (ii) C.O.P of Carnot cycle (iii) If actual C.O.P is 1 of the Carnot C.O.P, find out the th 5 power required to run the plant. Solution: (i) Heat removed from 1 kg of food Q R m f [Q R above Q R Freeze Q below ] 1 [C P above H fg C P below ] 1 [2.93 25 3 232 1.25 3 8] Q 4 320 kj/kg Heat removed by the plant Q R 20 1000 320 20 1000 222.22 kj/sec 8 60 60 8 3600 Capacity of Refrigeration plant Heat removed by plant 3.5 (ii) COP of Carnot Cycle C.O.P (iii) Power required 222.25 3.5 222.22 kw 63.49 tonnes or R 2 8 273 1 2 25 273 8 273 265 298 265 8.03. Actual COP 1 5 COP 1 8.03 1.606 5
Reversed Carnot Cycle and Air Refrigeration 1.17 Refrigeration or Heat removed Actual C.O.P 222.22 Workdone/min Power W Power required W 222.22 1.606 138.49 kj/sec or kw Problem 1.13: 5600 kg of water available at 22C is to be converted to ice at 4C in one day (24 hrs). If the sp. heat of ice is 2.09 kj/kgk and the latent heat is 335 kj/kg, find the capacity of refrigeration system. (CUSA June - 2011) Given data: Mass of water circulating 5600 kg emperature of water 22C emperature of ice 4C Specific heat of ice 2.09 kj/kgk Latent heat 335 kj/kg Solution: heat removed from 1 kg, Q R m [Q R1 Q R2 Q R3 ] m [C pw w h fg C p ice ice ] 1 [4.187 22 0 335 2.09 0 4 ] 435.5 Heat removed from 1 kg, Q 2 435.5 kj/kg. otal heat removed by plant Q R 5600 435.5 5600 24 60 60 24 60 60 28.23 Heat removed, Q 2 28.23 kw Capacity of the refrigerator, heat removed by plant 3.5 8.06 tonnes 28.23 3.5
1.18 Refrigeration and Air Conditioning - www.airwalkbooks.com 1.3.1.1 Merits of Reversed Carnot Cycle 1. No risk of fire as compared to ammonia refrigerator. 2. Cheaper than other refrigerants. 3. he weight is lesser than other refrigeration system. For this reason, it is used in air-crafts. 1.3.1.2 Demerits of Reversed Carnot Cycle (Limitations) 1. COP of this cycle is very low when compared to other refrigeration system. 2. he weight of circulating air is more in this cycle than other systems. 1.4 BELL-COLEMAN CYCLE (Reversed Joule Cycle) If the isothermal processes in carnot cycle is replaced by constant pressure processes, then the cycle is called Bell-coleman cycle. Process 1-2 Air is compressed isentropically during first part of the stroke. During the remainder stroke, the compressed air (at high temperature) is forced into cooler at constant pressure. water out Closed cycle air-refrigerator working on Bell - Coleman cycle. 3 Cooler 2 3 Heat Exchanger (cooler) 4 Air water in Expander Air Refrigerator 1 2 Compressor Coupling Bearing Fig. 1.5 Bearing Air Electric motor
Reversed Carnot Cycle and Air Refrigeration 1.19 Process 2-3 he cooler cools the air at constant pressure. Process 3-4 he cold air is now drawn into expansion cylinder and expanded isentropically. During isentropic expansion, the air is further cooled below the refrigerator space temperature. Process 4-1 he cold air passes through the refrigerator and absorbs the heat at constant pressure. Conclusion he heat is absorbed from refrigerator and rejected into the circulating water of cooler. So the refrigerator is maintained at low temperature. he bell-coleman cycle consists of 1-2 isentropic compression process [ Pv C ] 2-3 constant pressure cooling process [ P C ] 3 C on st.p r. g p v = c C onst.pr. P 2 2 1 4 Fig. 1.6 g pv =c P 1 S 3-4 isentropic expansion process [ Pv C ] 4-1 constant pressure heat absorbing process [ P C ] 4 C.O.P isentropic process 3 4 But in actual practice, perfect isentropic process is not possible. So the process will be polytropic process. Pv n C. C.O.P polytropic process n n 1 4 1 [ 3 4 ]
1.20 Refrigeration and Air Conditioning - www.airwalkbooks.com P 3 Isentropic =c expansion PV (or) Polytropic n expansion (PV =c) BELL-COLEMAN CYCLE Bell Coleman Cycle Constant pressure cooling 2 Qout Isentropic compression pv =c (or) Polytropic compression (PV n=c) Constant pressure heat absorption 4 1 Q = Cp( - ) absorbed 1 4 Fig. 1.7 V Polytropic Law 2 1 P 2 P 1 n 1 n ; 3 4 P 3 P 4 n 1 n n 1 P 2 n P 1 Heat absorbed from refrigerator (cold chamber) per kg of air C p 1 4 Heat rejected in the cooling tower per kg of air C p 2 3 Work done n n 1 [ P 2 v 2 P 1 v 1 P 3 v 3 P 4 v 4 ] n n 1 R [ 2 1 3 4 ] Heat absorbed C.O.P Work Problem 1.14: A refrigerator working on Bell-coleman cycle operates between pressure limits of 1.05 bar and 8.5 bar. Air is drawn from the cold chamber (refrigerator) at 10C. Air coming out of compressor is cooled to 30C before entering the expansion cylinder. Expansion and compression follow the law Pv 1.35 constant. Determine theoretical C.O.P. of the system. (JNU - May/June 2009) Apr 2016 - Calicut University
Reversed Carnot Cycle and Air Refrigeration 1.21 Solution: P 1 1.05 bar; P 2 8.5 bar; 1 10 273 283 K 3 30 273 303 K; Pv 1.3 C Consider Polytropic Compression 1-2 n 1 1 P 1 n 2 P 2 P P=P 2 3 3 Cooling chamber 2 Pv =C 2 1.05 8.5 0.35 1.35 0.58147 1 0.58147 283 0.58147 P=P 1 4 expansion 1.25 Pv =c 4 com pression Cold chamber 1 v 486.69 K Consider Polytropic Expansion 3-4 n 1 4 P 1 n 3 P 2 0.58147 4 3 0.58147 3030.58147 176.185 K Q a -Heat (absorbed) extracted from cold chamber (refrigerator) per kg of air. Q a C p 1 4 1.005283 176.185 107.348 kj/kg Q r -Heat rejected in the cooling chamber per kg of air. C p 2 3 1.005486.69 303 184.61 kj/kg Since the compression and expansion are not isentropic, the difference between heat rejected and heat absorbed is not equal to work done. So work done is found as follows for polytropic process.
1.22 Refrigeration and Air Conditioning - www.airwalkbooks.com Work done 1 283 K; 2 486.69 K; 3 303 K; 4 176.185 K W n n 1 R [ 2 1 3 4 ] 1.35 0.287 [ 486.69 283 303 176.185 ] 0.35 85.1 kj/kg C.O.P Heat absorbed Work done 107.348 85.1 1.261 Problem 1.15: Refrigerator working on reversed Joule s cycle operates between pressure limits of 1.05 bar and 8.5 bar. Air is drawn from the cold chamber at 10C, compressed and, then it is cooled to 30C before entering the expansion cylinder. Determine the theoretical C.O.P of the system. If the expansion and compression follows the law PV 1.3 Constant, what will be the change in COP? (Kerala University, 2013) Given data: P 1 1.05 bar ; P 2 8.5 bar ; 1 10C 283 K ; 3 30C 303 K Solution: For polytropic compression, 1 n 1 P 1 n 2 P 2 P P=P 2 3 3 Cooling chamber 2 Pv =C 283 0.3 1.05 1.3 2 8.5 2 283 0.617 P=P 1 4 expansion 1.25 Pv =c 4 com pression Cold chamber 1 v 2 458.5 K
Reversed Carnot Cycle and Air Refrigeration 1.23 For polytropic expansion, 3 n 1 P 2 n 4 P 4 303 0.30 8.5 1.30 4 1.05 4 303 1.719 Heat absorbed, Q a C p 1 4 1.005 283 176.18 107.35 kj/kg 4 176.18 Work done n n 1 R [ 2 1 3 4 ] 1.3 0.287 [458.5 283 303 176.18] 0.3 60.54 kj/kg heoretical COP Q a W 107.35 60.54 heoretical COP 1.77 (ii) For isentropic process of compression, 1 0.4 0.4 P 1 1.4 283 2 P 1.05 1.4 2 2 8.5 2 514.37 K For isentropic expansion, 3 0.4 0.4 P 3 1.4 303 4 P 8.5 1.4 4 4 1.05 4 166.7 K Q a C p 1 4 1.005 283 166.7 Q a 116.88 kj/kg
1.24 Refrigeration and Air Conditioning - www.airwalkbooks.com Compressor work done 1 mr 2 1 Expander work done Net work done 136.9 50.225 86.675 1.4 1 0.287 514.37 283 0.4 50.225 kj/min 1 mr 3 4 1.4 1 0.287 303 166.7 0.4 136.9 kj/kg COP Q a W 116.88 86.675 Change in COP 1.77 1.35 0.422 COP 1.35 If the process follows the law pv 1.30 C, then the COP is greater than the isentropic process. Problem 1.16: A refrigerator working on Bell-Coleman cycle (Reversed Brayton cycle) operates between 1 bar and 15 bar. Air is drawn from cold chamber at 10C. Air coming out of compressor is cooled to 50C before entering the expansion cylinder. Polytropic law PV 1.3 constant is followed during expansion and compression. Find theoretical C.O.P. of the system. ake 1.4 and C p 1.005 kj/kgc for air. Given data: (CUSA, June 2011) JNU Kakkinada University (May/april 2016) PV 1.3 constant P 1 1 bar ; P 2 15 bar 1 10 273 263 K ; 3 50 273 323 K
Reversed Carnot Cycle and Air Refrigeration 1.25 Solution: From the process 1 and 2 1 n 1 P 1 n 2 P 2 2 0.535 1 15 1 0.535 263 0.535 1.3 1 1.3 P 3 Cooling 2 15 bar Compression Expansion 1 bar 4 Cold 1 V Chamber 2 491.6 K Here polytropic expansion, Process 3 4 4 n 1 P 1 n 3 P 2 4 1 3 15 4 3 0.535 1.3 1 1.3 4 323 0.535 4 172.8 K [ 3 323 K] COP Heat absorbed Work input Heat absorbed, Q a C p 1 4 C p 263 172.8 1.005 263 172.8 Q a 90.651 kj/kg [C p 1.005 kj/kg]
1.26 Refrigeration and Air Conditioning - www.airwalkbooks.com Workdone, Heat Rejected, Q r C p 2 3 1.005 491.6 323 Q r 169.443 kj/kg 1 263 K ; 3 323 K ; 2 491.6 K ; 4 172.8 K R 0.287 W n n 1 R [ 2 1 3 4 ] 1.3 0.287 [491.6 263 323 172.8] 1.3 1 4.3 0.287 78.4 97.5 kj/kg COP Q a W 90.651 97.5 0.93 Problem 1.17: A refrigerator of 6 ton capacity working on Bell-coleman cycle has an upper limit of pressure of 5 bar. he pressure and temperature at the start of the compression are 1 bar and 15C respectively. he compressed air is cooled at a constant pressure to a temperature of 40C enters the expansion cylinder. Assuming both expansion and compression processes to be isentropic with 1.4, calculate (i) C.O.P. (ii) Quantity of air in circulation per minute. (iii) Piston displacement of compressor and expander. (iv) Bore of compressor and expansion cylinders. he refrigerating unit runs at 240 r.p.m. and is double acting. Stroke length = 250 mm. (v) Power required to drive the unit. For air take 1.4 and C p 1.005 kj/kg K. Solution: Capacity 6.R 6 3.5 21 kw Refrigerating effect produced by the refrigeration. P 2 P 3 5 bar; P 1 P 4 1 bar 1 15 273 288 K; 3 40 273 313 K
Reversed Carnot Cycle and Air Refrigeration 1.27 Isentropic Compression Process 1-2 1 2 P 2 1 P 1 1.58382 5 1 0.4 1.4 P(bar) 3 P 2=P 3=5bar 2 Cooling chamber Pv =c Com pressor Pv =c 2 1.58382 1 1.58382 288 456.14 K 2 456.14 K Isentropic Expansion Process 3-4 1 4 P 4 3 P 3 P 1 P 2 1 4 0.6314 3 0.6314 313 197.624 K 0.4 1 1.4 5 0.6314 o Find C.O.P Since both compression and expansion are isentropic, C.O.P isentropic 4 197.624 3 4 313 197.624 1.7128 (ii) o Find Mass of Air in Circulation: m a R.E = Refrigerating Effect R.E per kg of air C p 1 4 1.005288 197.624 90.828 kj/kg Expansion cylinder P 1=P 4=1 bar 4 Cold chamber(refrigerator) 1 R.E. produced by the refrigerator Capacity of refrigerator 6 3.5 21 kw Capacity of the refrigerator m a R.E. per kg of air v(m 3) volume Here C p 1.005 kj/kg
1.28 Refrigeration and Air Conditioning - www.airwalkbooks.com where m a Mass of air in circulation. m Capacity of refrigerator a R.E. per kg of air m a 0.2312 kg/sec 21 90.828 (iii) o Find Piston Displacement of Compressor V V Volume of air in circulation in m 3 /sec Volume corresponding to point 1 i.e. V 1. kj sec kj/kg P 1 V 1 m a R 1 [P 1 in kpa ; 1 bar 1 102 kpa] V 1 m R 1 0.2312 0.287 288 P 1 1 10 2 0.1911 m 3 /sec (iv) o Find Dia. of Compressor Cylinder V 1 V s Swept volume per stroke V s 2 N 60 for double acting compressor. N R.P.M. ; V s Swept volume per stroke 0.1911 60 2 240 m 3 sec sec 0.023887 m 3 Also, V s 0.023887 m 3 V s 4 d c 2 L where L Stroke length d c dia. of compressor cylinder V s 0.023887 4 d c 2 0.25 d c 2 0.12165 d c 0.3488 m
Reversed Carnot Cycle and Air Refrigeration 1.29 o Find Diameter of Expander Cylinder V 4 Volume of air in circulation in m3 /sec. = Volume corresponding to point 4 is V 4. P 4 V 4 m a R 4 V 4 m a R 4 0.2312 0.287 197.624 P 4 1 10 2 0.13113 m 3 /sec V s Swept volume per stroke V 4 for double acting expander. 2 N/60 0.13113 60 2 240 0.01639 m 3 /sec V s 4 d e 2 L 0.01639 4 d e 2 0.25 where d e dia. of expander cylinder. 2 d e 0.08347 d e 0.2889 m (v) o Find the Power Required to Run the Unit C.O.P R.E W 1.7128 6 3.5 W where W Power required W 6 3.5 12.26 kw 1.7128 Power required 12.26 kw
1.30 Refrigeration and Air Conditioning - www.airwalkbooks.com Problem 1.18: In a Bell-Coleman cycle working between pressures of 1 and 6 bar and temperature at the beginning of compression and expansion of 8C and 35C, air flow rate in 30 kg/min. If the compand expansion indices in the polytrophic process are 1.3 and 1.35 respectively. Determine (i) COP (ii) tonnage of the plant (iii) determine the heat transfer rates per kg of air during each process. [May - 2016, JNU Kakinada University] Given data: 1 8 273 281 K P 1 1 bar 3 35 273 308 K P 2 6 bar Air flow rate, m a 30 kg/min. 0.5 kg/s Compression and Expansion indices in the polytropic process are 1.3 and 1.35 (i) C.O.P? (ii) onnage of the plant? (iii) Heat transfer Rates per kg of air Solution: Consider a polytrophic compression 1-2 n 1.3 1 n 1 P 1 n 2 P 2 1.3 1 1 1.3 6 1 0.66 2 1 0.66 2 2 281 0.66 2 425 K Consider a polytropic expansion 3-4 3 Cooling 2 Chamber 1.35 P v = C Expansion P =P 2 3 Compression 1.3 P v = C Cold P 1=P4 4 Chamber 1 4 n 1 P 1 n 3 P 2 Here, n 1.35
Reversed Carnot Cycle and Air Refrigeration 1.31 1.35 1 4 1 1.35 3 6 4 0.628 3 4 0.628 308 4 193.6 K Q a Heat absorbed from cold chamber per kg of air Q a C p 1 4 1.005 281 193.6 Q a 87.4 kj/kg Q a m Q a 0.5 87.4 43.7 kw Q r Heat Rejected in the cooling chamber per kg of air Q r C p 2 3 1.005 425 308 1.005 117.7 Q R 117.6 kj/kg Workdone 1 281 K, 2 425 K, 3 308 K ; 4 193.6 K Workdone n 1 n 1 1 R 2 1 n 2 n 2 1 R 3 4 1.3 0.3 0.287 425 281 1.35 0.35 0.287 308 193.6 179 126.6 52.36 52.36 kj/kg (i) COP COP heat absorbed workdone 87.4 52.36 1.67 (ii) onnage of plant COP 1.67 R Q a 3.5 43.7 3.5 12.5 R 12.5 ton
1.32 Refrigeration and Air Conditioning - www.airwalkbooks.com (iii) Heat transfer rate per kg of air Q a 87.4 kj/kg Q R 117.6 kj/kg Problem 1.19: A Bell-Coleman refrigerator is required to produce 6 tonnes of refrigerating effect with a cooler pressure of 11 bar and a refrigerated space or region at a pressure of 1.05 bar. he temperature of air leaving the cooler is 38C and air leaving the room is 16C. Calculate: (i) Mass of air circulated per minute (ii) Compressor displacement required per minute (iii) Expander displacement required per minute (iv) COP and (v) Power required per tonne of refrigeration. (MBCE Dec-2016) Given data: Capacity 6 ton 6 3.5 kw 21 kw Refrigerating effect So, P 2 P 3 11 bar P 1 P 4 1.05 bar 3 V(m ) Volume 1 16 273 289 K ; 3 38 273 311 K Solution: Isentropic compression process 1 2 1 2 P 2 1 P 11 1 1.05 2 1.96 1 1.4 1 1.4
Reversed Carnot Cycle and Air Refrigeration 1.33 2 1.96 289 2 565.4 K Isentropic Expansion process, 3 4 1 4 P 4 3 P 3 1 P 1 P 2 4 1.05 3 11 4 3 0.511 0.4 1.4 4 0.511 311 4 158.95 K (i) Find mass of air circulated per minute m a : Capacity of the Refrigerator m a R.E per kg of air R.E Refrigerating effect. R.E per kg of air C p 1 4 1.005 289 158.95 130.7 kj/kg R.E produced by the Refrigerator Capacity of Refrigerator 6 3.5 21 kw
1.34 Refrigeration and Air Conditioning - www.airwalkbooks.com hen, m Capacity of Refrigerator 21 a R.E per kg of air 130.7 m a 0.161 kg/sec 9.64 kg/min Q a C p 1 4 1.005 289 158.95 130.7 kj/kg W c 1 R 2 1 1.4 0.287 565.4 289 277.64 0.4 W E 1 R 3 4 1.4 0.287 311 158.95 152.73 0.4 W net W c W E 124.9 kj/kg COP Q a W net 130.7 124.9 1.045 COP isentropic (Or) 4 158.95 3 4 311 158.95 1.045 COP 1.045 Compressor displacement required per minute: P 1 V 1 m a R 1 V 1 volume of air in circulation in m3 /sec V 1 m R 1 0.161 0.287 289 P 1 1 10 2 V 1 0.134 m3 /sec 8.01 m 3 /min Power required per tonne of refrigeration. C.O.P R.E W 1.045 6 3.5 W