Transcript
  • * Basic Mechanical EngineeringDr A.C. Tiwari Professor & Head Mechanical Engineering DepartmentUIT, RGPV Bhopal

    Rajiv Gandhi Proudhyogiki VishwavidhyalayAir Port Road Gandhi Nagar Bhopal, (M.P.)www.rgpv.ac.in

  • * If those who think to achieve, Have a firm and focused mind, They will realize what they thought of And even as they have thought of . --Thirukkural

    Indeed ones faith in ones plan and methods are truly tested when the horizon before one is the blackest . -- Mahatma Gandhi.

  • * Refrigeration

    Definition by ASHRAE It is defined as Science of providing and maintaining temperature below that of surrounding .

    How do things get colder?

  • * Necessity of Refrigeration Food Preservation .Poultry Forms .Development of certain Scientific Instruments .Weaving in textile Industry .Improvement in production in shop floor.Medical Science , Surgery.Customer delight in theaters & shops.

  • *Methods of Refrigeration1. Dissolution of Certain Salts in Water :

    -Salts like CaCl2 , Nacl , Salt Petre , NH4Cl .etc. are dissolved in water , they absorb heat . This property is used to Produce refrigeration .

    CaCl2 can lower the temp of water up to 50 C. Nacl can lower the temp of water up to 20 C. This Method is not feasible for Commercial purpose

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    2.Change of Phase :(a) Solid into Liquid : If a substance such as ice is available it is possible to get refrigeration effect due to phase change from Solid to Liquid . Qc = m.hsf hsf = Enthalpy of fusion of ice = 335 KJ /kgMethods of Refrigeration

  • * A Glass of Iced Water

  • *ICE REFRIGERATION

  • *Methods of Refrigeration2.Change of Phase : (b) Solid into Vapour : Can also produce refrigeration effect :Example : Dry Ice (Solid Carbon dioxide CO2 ) Qc = m.hsv. hsv = Enthalpy of fusion of ice = 573 KJ /kg This can maintain a temp. of -78.5o C

  • *Pallets of Dry Ice sublimating into CO2 Gas

  • *Crystalline structure of solid carbon dioxide

  • *Methods of Refrigeration2.Change of Phase : (c) Liquid to Vapour : If a substance such as alcohol is available it is possible to get refrigeration effect due to phase change from Liquid to vapour . Liquid N2 is sprayed inside the cargo space of a truck . Liquid N2 changes phase from liquid to gas and produce refrigeration effect . Qc = m.hfg

  • *Methods of Refrigeration3.Throttling Process: It is fluid at high pressure is expended through a valve or constriction , either of three effects are expected depending upon initial and final conditions . (i) Te (exit temp) > Ti (inlet temp) (ii) Te = Ti (iii) Te < Ti With careful design of throttling valve condition (iii) can be obtained for Refrigeration Effect.

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  • *Methods of Refrigeration4. By Expansion of Gases : Refrigeration effect can be obtained by Expansion of a Gas through a turbine or behind a piston . If a gas at pressure P1 and temp. T1 expands behind a piston to pressure P2 (P2
  • *Methods of Refrigeration5 . Ranque Hilsch Effect : When a high pressure gas is allowed to expand through a nozzle fitted tangentially to a pipe ,this causes simultaneous discharge of the cool air core and hot air periphery .

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  • *Hot GasThrottle ValveRanque hilsch Vortex tube Hot GasCompressed Gas NozzleCold Gas Orifice

  • *Methods of Refrigeration 6.Thermocouple Effect : Peltier Effect

  • *Methods of Refrigeration7 . Demagnetization : Magnetic materials show that magnetization increase temp and sudden demagnetization lowers the temp. If this process is repeated . One can achieve as low temp. as 0.001K by this method .

  • *Unit of Refrigeration .In olden days refrigeration effect was first produced by Ice , so the effect of refrigerating machines was compared by the refrigeration effect produced by Ice. The refrigeration effect is measured by Tons of Refrigeration .

  • *DefinitionA ton of refrigeration is defined as the quantity of latent heat required to be removed from one ton of water of 0 C temp. to convert it into ice of 0 C temp within 24 hours . Ton in metric unit- (1000 kgX80Kcal/kg)/(24X60) = (10,000/3)= 55.4 Kcal/min This is approximated to 50 Kcal/min and it is called One Ton of Refrigeration . In SI Unit 210 KJ/min or 3.5 KW

  • *Assignment No 1 Qu 1 : Define the following terms Ton of Refrigeration, Refrigerating effect .Qu 2 : Name five means of producing cooling effect ?Qu 3 :Refrigeration can be produced either by expansion of a gas or throttling of gas ,discuss the above statement ?

    Feed back can be given on following ID. [email protected]

  • *Thermodynamics of a Refrigerator T1< T2

    Sink T1Source T2RefrigeratorWork WQ 1Q 2 For refrigerator maximum Q1 should be taken out with minimum expense of W ,so performance of refrigerator is evaluated by COP (coefficient of Performance)= Q1/ W.COP (coefficient of Performance)= Q1/ W

  • *compressorLow pr gaugeEvaporatorThrottle valvecondenserHigh pr gaugeSimple Vapour Compression System

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  • *Components of the Practical Cycle Practical Vapour Compression Cycle.

  • *A home refrigerator with its door open

  • *Vapour compression Refrigeration test rig at thermal Engg lab of UIT, RGTU Bhopal

  • *Multi Evaporator Vapour Compression test rig at thermal Engg lab of UIT ,RGTU Bhopal

  • *Numerical Problems Example 1 : An ice plant produces 10 tonnes of ice per day at 0C using water at room temp of 20C . Estimate the power rating of the compressor-motor ,if the COP of the plant is 2.5. Soln: Given data : m= 10 t/day=10x1000/24x60 = 6.94 kg/min.

  • * T1= 0C=273 K ,T2 = 20C=293K COP= 2.5 . Let W= work required to drive the compressor /min Amount of the heat removed from the 1 kg water of 20C to convert it into 1 kg ice of 0C. = 1x4.18x(20-0)+335 =418.74 KJ/kg ( latent heat of ice=335kj/kg)

  • * Total heat removed =6.94x418.74 =2906 KJ /min

    COP of the plant = heat removed/work of compressor

    2.5 = 2906/W W =1162.4 KJ/min = 1162.4/60 KW =21.5 KW Ans

  • *Example No 2 : Five hundred kg of fruits are supplied to a cold storage at 20 C . The cold storage is maintained at -5 C and the fruits get cooled to the storage temperature in 10 hours . The latent heat of freezing is 105 kJ/Kg and specific heat of fruit is 1.26 . Find the refrigeration capacity of the plant .

  • *Soln : Given data : mass of the fruits m= 500 kgs T2 = 20C=293 K , T1= -5C = 268K. latent heat of freezing , hfg = 105 KJ/Kg Sp heat of fruit cf = 1.26 Heat removed from the fruit in 10 hrs. Q1 = mcf (T2-T1) =500X1.26(293-268)=15750 kJ

    Latent heat of freezing Q2= m hfg =500X105=52500 kJ

  • *Total heat removed in 10 hrs.Q= Q1+Q2 = 15750+52500 = 68250 kJTotal heat removed per minute ,

    = 68250/ 10X60 = 113.75 kJ/minRefrigeration capacity of the plant in tons= 113.75/ 210 = 0.542 tons Ans

  • *Home Assignment No 2 Qu 1 :Explain Vapour Compression refrigeration system with the help of a neat schematic diagram .Qu 2 :A refrigeration plant is required to produce 2.5 tonnes of ice per day at -4 C from water at 20 C . If the temperature range in the compressor is between 25 C and -6C , calculate power required to drive the compressor . Latent heat of ice = 335 kJ/Kg and specific heat of ice = 2.1 kJ/Kg K.

  • *Suggested Project Work : See the domestic refrigerator in your house and chalk out locations of compressor ,condenser, evaporator and capillary throttling device . Try also to chalk out path of flow of refrigerant . Feed back can be given to my [email protected]

  • *Vapour Absorption SystemNo moving partsLow grade thermal energy like solar energy can be the input energy.Load variation does not affect system performance.Environmental friendly.

  • * Vapour Absorption Aqua Ammonia System burnergeneratorseparatorcondenserevaporatorabsorberWorm H2 upWorm H2 upHydrogen+ NH3 downWeak sol+ NH3vapourNH 3 VapourOne way valveOne way valveCooling effectWater+NH3 NH3 dissolves into water H2 is left

  • *Refrigerants Defined: Any substance capable of absorbing heat from another required substance can be used as refrigerant i.e. ice ,water, brine, air etc. Primary RefRefrigerants:

    Secondary Ref

  • *Primary Refrigerants are further classified as below:Halocarbon Compounds -Trade Names---- freon,mefron,isotron,genetron ,halides F-11 Trichloro monofluro methene CCl3F F-12 Dichloro difluro methene CCl2F2 F-22 Monochloro diflouro methene CHClF2 Azeotropes : mixture of certain refrigerantsHydro carbons : Methane, propene etc.Inorganic Compounds :Ammonia, carbon dioxide, water , air etc.Unsaturated Organic Compounds: ethylene and propylene base hydro carbons .

  • *Properties of Refrigerants Low Boiling PointLow Freezing Point.High Latent Heat.Chemically Inert & stable .Non FlammableNon toxicShould not react with lub oil of comp.Should not be corrosive

  • *Environmental Aspects with Refrigerants Halo Carbons depletes Ozone layer. Green House effect caused by freons. 1987 Montreal Protocol- A time schedule was chalked out to Control release of chloroflouro hydo carbons to atmosphere.

  • *Future Refrigerants to Replace CFCsR-502 , replacing R11 and R12 R-123A , Promising future refrigerant replacing R11 . R143a another promising refrigerant replacing R12. R69S ,replacing R22 and R502. Hydro fluoro carbons.Hydro fluoro ethers .

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