Presentation on theme: "Mechanical Engineering Air Port Road Gandhi Nagar Bhopal, (M.P.)"— Presentation transcript:
1Mechanical Engineering Air Port Road Gandhi Nagar Bhopal, (M.P.) R.G.P.V.- BHOPALizKkuaczãMadhya PradeshBasic Mechanical EngineeringDr A.C. TiwariProfessor & HeadMechanical EngineeringDepartmentUIT, RGPV BhopalRajiv Gandhi Proudhyogiki VishwavidhyalayAir Port Road Gandhi Nagar Bhopal, (M.P.)
2If those who think to achieve, Have a firm and focused mind, They will realize what they thought ofAnd even as they have thought of .--Thirukkural“Indeed one’s faith in one’s plan and methods are truly tested when the horizon before one is the blackest “.-- Mahatma Gandhi.
3Refrigeration Definition by ASHRAE – It is defined as Science of providing and maintaining temperature below that of surrounding .How do things get colder?
4Necessity 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.
5Methods of Refrigeration 1. 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
6Methods of Refrigeration 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.hsfhsf = Enthalpy of fusion of ice = 335 KJ /kg
9Methods of Refrigeration 2. 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 /kgThis can maintain a temp. of -78.5o C
12Methods of Refrigeration 2. 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
13Methods of Refrigeration 3. 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 < TiWith careful design of throttling valve condition (iii) can be obtained for Refrigeration Effect.
15Methods of Refrigeration 4. 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<P1) the temp ,T2 after expansion is lowered.T2= T1X (P2/P1)(n-1/n) = T1X(P2/P1)0.4/1.4Let us take initial temp. T1 = K ( i. e. 40°C)&Pressure Ratio (P2/P1) =6.5The temp. T2 is found to be, °C
16Methods of Refrigeration 5 . 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 .
19Methods of Refrigeration 6. Thermocouple Effect : Peltier EffectHot EndCold EndD C Source+-
20Methods of Refrigeration 7 . 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 .
21Unit 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” .
22DefinitionA 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/minThis is approximated to 50 Kcal/min and it is called One Ton of Refrigeration .In SI Unit 210 KJ/min or 3.5 KW
23Assignment 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.
24Thermodynamics of a Refrigerator T1< T2Source T2COP (coefficient of Performance)= Q1/ WQ 2Work WRefrigeratorQ 1Sink T1For 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.
25Simple Vapour Compression System EvaporatorThrottle valveLow pr gaugecondenserHigh pr gaugecompressor
29Vapour compression Refrigeration test rig at thermal Engg lab of UIT, RGTU Bhopal
30Multi Evaporator Vapour Compression test rig at thermal Engg lab of UIT ,RGTU Bhopal
31Numerical ProblemsExample 1 : An ice plant produces 10 tonnes of ice per day at 0°C using water at room temp of 20°C .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.
32T1= 0°C=273 K ,T2 = 20°C=293KCOP= 2.5 .Let W= work required to drive the compressor /minAmount of the heat removed from the 1 kg water of 20°C to convert it into 1 kg ice of 0°C.= 1x4.18x(20-0)+335 = KJ/kg( latent heat of ice=335kj/kg)
33Total heat removed =6.94x418.74 =2906 KJ /min W =1162.4 KJ/min COP of the plant = heat removed/work of compressor2.5 = 2906/WW = KJ/min= /60 KW=21.5 KW Ans
34Example 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 Find the refrigeration capacity of the plant .
35Latent heat of freezing Soln : Given data :mass of the fruits ‘m’= 500 kgsT2 = 20°C=293 K ,T1= -5°C = 268K.latent heat of freezing , hfg = 105 KJ/KgSp heat of fruit cf = 1.26Heat removed from the fruit in 10 hrs.Q1 = mcf (T2-T1) =500X1.26( )=15750 kJLatent heat of freezingQ2= m hfg =500X105=52500 kJ
36Total heat removed in 10 hrs. Q= Q1+Q2 = = kJTotal heat removed per minute ,= 68250/ 10X60 = kJ/minRefrigeration capacity of the plant in tons= / 210 = tons Ans
37Home Assignment No 2Qu 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 -6°C , calculate power required to drive the compressor . Latent heat of ice = 335 kJ/Kg and specific heat of ice = 2.1 kJ/Kg K.
38Suggested Project Work : See the domestic refrigerator in your house and chalk out locations of compressor ,condenser, evaporatorand capillary throttling device .Try also to chalk out path of flow of refrigerant .Feed back can be given to my ID
39Vapour Absorption System No moving partsLow grade thermal energy like solar energy can be the input energy.Load variation does not affect system performance.Environmental friendly.
40Vapour Absorption Aqua Ammonia System condenserOne way valveevaporatorNH 3 VapourHydrogen+ NH3 downCooling effectseparatorWorm H2 upWorm H2 upOne way valveWeak sol+ NH3vapourWater+NH3NH3 dissolves into water H2 is leftgeneratorabsorberburner
41RefrigerantsDefined: 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
42Primary Refrigerants are further classified as below: Halocarbon Compounds-Trade Names---- freon,mefron,isotron,genetron ,halidesF-11 Trichloro monofluro –methene –CCl3FF-12 Dichloro difluro –methene –CCl2F2F-22 Monochloro diflouro methene – CHClF2Azeotropes : 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 .
43Properties 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
44Environmental 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.
45Future Refrigerants to Replace CFCs R-502 , replacing R11 and R12R-123A , Promising future refrigerant replacing R11 .R143a another promising refrigerant replacing R12.R69S ,replacing R22 and R502.Hydro fluoro carbons.Hydro fluoro ethers .