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Mechanical Engineering Air Port Road Gandhi Nagar Bhopal, (M.P.)

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Presentation on theme: "Mechanical Engineering Air Port Road Gandhi Nagar Bhopal, (M.P.)"— Presentation transcript:

1 Mechanical Engineering Air Port Road Gandhi Nagar Bhopal, (M.P.)
R.G.P.V.- BHOPAL izKkua czã Madhya Pradesh Basic Mechanical Engineering Dr A.C. Tiwari Professor & Head Mechanical Engineering Department UIT, RGPV Bhopal Rajiv Gandhi Proudhyogiki Vishwavidhyalay Air Port Road Gandhi Nagar Bhopal, (M.P.)

2 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 one’s faith in one’s plan and methods are truly tested when the horizon before one is the blackest “. -- Mahatma Gandhi.

3 Refrigeration Definition by ASHRAE –
It is defined as Science of providing and maintaining temperature below that of surrounding . How do things get colder?

4 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.

5 Methods 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

6 Methods 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.hsf hsf = Enthalpy of fusion of ice = 335 KJ /kg

7 A Glass of Iced Water

8 ICE REFRIGERATION Storage Ice water Insulation

9 Methods 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 /kg This can maintain a temp. of -78.5o C

10 Pallets of Dry Ice sublimating into CO2 Gas

11 Crystalline structure of solid carbon dioxide

12 Methods 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

13 Methods 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 < Ti With careful design of throttling valve condition (iii) can be obtained for Refrigeration Effect.

14 Throttling process High Pressure Gas Throttle Valve Expansion

15 Methods 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)0.4/1.4 Let us take initial temp. T1 = K ( i. e. 40°C) & Pressure Ratio (P2/P1) =6.5 The temp. T2 is found to be, °C

16 Methods 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 .


18 Ranque hilsch Vortex tube
Compressed Gas Hot Gas Nozzle Throttle Valve Cold Gas Orifice Hot Gas Ranque hilsch Vortex tube

19 Methods of Refrigeration
6. Thermocouple Effect : Peltier Effect Hot End Cold End D C Source + -

20 Methods 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 .

21 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” .

22 Definition A 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

23 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.

24 Thermodynamics of a Refrigerator
T1< T2 Source T2 COP (coefficient of Performance)= Q1/ W Q 2 Work W Refrigerator Q 1 Sink T1 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.

25 Simple Vapour Compression System
Evaporator Throttle valve Low pr gauge condenser High pr gauge compressor


27 Practical Vapour Compression Cycle.
Components of the Practical Cycle                                                        

28 A home refrigerator with its door open

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

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

31 Numerical Problems Example 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.

32 T1= 0°C=273 K ,T2 = 20°C=293K COP= 2.5 . Let W= work required to drive the compressor /min Amount 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)

33 Total heat removed =6.94x418.74 =2906 KJ /min W =1162.4 KJ/min
COP of the plant = heat removed/work of compressor 2.5 = 2906/W W = KJ/min = /60 KW =21.5 KW Ans

34 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 Find the refrigeration capacity of the plant .

35 Latent heat of freezing
Soln : Given data : mass of the fruits ‘m’= 500 kgs T2 = 20°C=293 K , T1= -5°C = 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( )=15750 kJ Latent heat of freezing Q2= m hfg =500X105=52500 kJ

36 Total heat removed in 10 hrs.
Q= Q1+Q2 = = kJ Total heat removed per minute , = 68250/ 10X60 = kJ/min Refrigeration capacity of the plant in tons = / 210 = tons Ans

37 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 -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.

38 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 ID

39 Vapour Absorption System
No moving parts Low grade thermal energy like solar energy can be the input energy. Load variation does not affect system performance. Environmental friendly.

40 Vapour Absorption Aqua Ammonia System
condenser One way valve evaporator NH 3 Vapour Hydrogen+ NH3 down Cooling effect separator Worm H2 up Worm H2 up One way valve Weak sol+ NH3 vapour Water+NH3 NH3 dissolves into water H2 is left generator absorber burner

41 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 Ref Refrigerants: Secondary Ref

42 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 refrigerants Hydro carbons : Methane, propene etc. Inorganic Compounds :Ammonia, carbon dioxide, water , air etc. Unsaturated Organic Compounds: ethylene and propylene base hydro carbons .

43 Properties of Refrigerants
Low Boiling Point Low Freezing Point. High Latent Heat. Chemically Inert & stable . Non Flammable Non toxic Should not react with lub oil of comp. Should not be corrosive

44 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.

45 Future Refrigerants to Replace CFCs
R-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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