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1 Basic Mechanical Engineering Basic Mechanical Engineering Dr A.C. Tiwari Professor & Head Mechanical Engineering Mechanical EngineeringDepartment UIT,

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Presentation on theme: "1 Basic Mechanical Engineering Basic Mechanical Engineering Dr A.C. Tiwari Professor & Head Mechanical Engineering Mechanical EngineeringDepartment UIT,"— Presentation transcript:

1 1 Basic Mechanical Engineering Basic Mechanical Engineering Dr A.C. Tiwari Professor & Head Mechanical Engineering Mechanical EngineeringDepartment UIT, RGPV Bhopal Dr A.C. Tiwari Professor & Head Mechanical Engineering Mechanical EngineeringDepartment UIT, RGPV Bhopal Rajiv Gandhi Proudhyogiki Vishwavidhyalay Air Port Road Gandhi Nagar Bhopal, (M.P.) Rajiv Gandhi Proudhyogiki Vishwavidhyalay Air Port Road Gandhi Nagar Bhopal, (M.P.)

2 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 3 Refrigeration Refrigeration Definition by ASHRAE – It is defined as Science of providing and maintaining temperature below that of surrounding. It is defined as Science of providing and maintaining temperature below that of surrounding. How do things get colder?

4 4 Necessity of Refrigeration 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 5 Methods of Refrigeration 1. Dissolution of Certain Salts in Water : - Salts like CaCl 2, Nacl, Salt Petre, NH 4 Cl …….etc. are dissolved in water, they absorb heat. This property is used to Produce refrigeration. CaCl 2 can lower the temp of water up to – 50 °C. CaCl 2 can lower the temp of water up to – 50 °C. Nacl can lower the temp of water up to – 20 °C. Nacl can lower the temp of water up to – 20 °C. This Method is not feasible for Commercial purpose This Method is not feasible for Commercial purpose

6 6 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. If a substance such as ice is available it is possible to get refrigeration effect due to phase change from Solid to Liquid. Q c = m. h sf Q c = m. h sf h sf = Enthalpy of fusion of ice = 335 KJ /kg h sf = Enthalpy of fusion of ice = 335 KJ /kg Methods of Refrigeration

7 7 A Glass of Iced Water

8 8 Storage Ice water Insulation ICE REFRIGERATION

9 9 Methods of Refrigeration 2.Change of Phase : (b) Solid into Vapour : (b) Solid into Vapour : Can also produce refrigeration effect : Can also produce refrigeration effect : Example : Dry Ice (Solid Carbon dioxide CO 2 ) Q c = m. h sv. Q c = m. h sv. h sv = Enthalpy of fusion of ice = 573 KJ /kg h sv = Enthalpy of fusion of ice = 573 KJ /kg This can maintain a temp. of o C

10 10 Pallets of Dry Ice sublimating into CO 2 Gas

11 11 Crystalline structure of solid carbon dioxide

12 12 Methods of Refrigeration 2.Change of Phase : (c) Liquid to Vapour : (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. If a substance such as alcohol is available it is possible to get refrigeration effect due to phase change from Liquid to vapour. Liquid N 2 is sprayed inside the cargo space of a truck. Liquid N 2 changes phase from liquid to gas and produce refrigeration effect. Liquid N 2 is sprayed inside the cargo space of a truck. Liquid N 2 changes phase from liquid to gas and produce refrigeration effect. Q c = m. h fg Q c = m. h fg

13 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. 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) (i) Te (exit temp) > Ti (inlet temp) (ii) Te = Ti (ii) Te = Ti (iii) Te < Ti (iii) Te < Ti W ith careful design of throttling valve condition (iii) can be obtained for Refrigeration Effect. W ith careful design of throttling valve condition (iii) can be obtained for Refrigeration Effect.

14 14 Throttling process High Pressure Gas Throttle Valve Expansion

15 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 P 1 and temp. T 1 expands behind a piston to pressure P 2 (P 2


16 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. 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.

17 17

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

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

20 20 Methods of Refrigeration 7. Demagnetization : Magnetic materials show that magnetization increase temp and sudden demagnetization lowers the temp. 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. If this process is repeated. One can achieve as low temp. as 0.001K by this method.

21 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”. The refrigeration effect is measured by “Tons of Refrigeration”.

22 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- 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. 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 In SI Unit 210 KJ/min or 3.5 KW

23 23 Assignment No 1 Qu 1 : Define the following terms Ton of Refrigeration, Refrigerating effect. 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. Feed back can be given on following

24 24 Thermodynamics of a Refrigerator T 1 < T 2 Sink T 1 Source T 2 Refrigerator Work W Q 1 Q 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

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

26 26

27 27 Components of the Practical Cycle Practical Vapour Compression Cycle.

28 28 A home refrigerator with its door open

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

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

31 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. 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. 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 m= 10 t/day=10x1000/24x60 = 6.94 kg/min. = 6.94 kg/min.

32 32 T1= 0°C=273 K,T2 = 20°C=293K T1= 0°C=273 K,T2 = 20°C=293K COP= 2.5. COP= 2.5. Let W= work required to drive the compressor /min 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. 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 = 1x4.18x(20-0)+335 = KJ/kg ( latent heat of ice=335kj/kg) ( latent heat of ice=335kj/kg)

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

34 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 35 Soln : Given data : mass of the fruits ‘m’= 500 kgs mass of the fruits ‘m’= 500 kgs T2 = 20°C=293 K, T2 = 20°C=293 K, T1= -5°C = 268K. T1= -5°C = 268K. latent heat of freezing, h fg = 105 KJ/Kg latent heat of freezing, h fg = 105 KJ/Kg Sp heat of fruit c f = 1.26 Sp heat of fruit c f = 1.26 Heat removed from the fruit in 10 hrs. Q1 = mc f (T2-T1) =500X1.26( )=15750 kJ Q1 = mc f (T2-T1) =500X1.26( )=15750 kJ Latent heat of freezing Q2= m h fg =500X105=52500 kJ

36 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 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 38 Suggested Project Work : See the domestic refrigerator in your house and chalk out locations of compressor,condenser, evaporator See the domestic refrigerator in your house and chalk out locations of compressor,condenser, evaporator and capillary throttling device. and capillary throttling device. Try also to chalk out path of flow of refrigerant. Try also to chalk out path of flow of refrigerant. Feed back can be given to my ID

39 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 40 Vapour Absorption Aqua Ammonia System burner generator separator condenser evaporator absorber Worm H 2 up Hydrogen+ NH 3 down Weak sol+ NH 3 vapour NH 3 Vapour One way valve Cooling effect Water+NH3 NH 3 dissolves into water H 2 is left

41 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 Primary RefRefrigerants: Secondary Ref Secondary Ref

42 42 Primary Refrigerants are further classified as below: Halocarbon Compounds - Trade Names---- freon,mefron,isotron,genetron,halides - Trade Names---- freon,mefron,isotron,genetron,halides F-11 Trichloro monofluro –methene –CCl 3 F F-11 Trichloro monofluro –methene –CCl 3 F F-12 Dichloro difluro –methene –CCl 2 F 2 F-12 Dichloro difluro –methene –CCl 2 F 2 F-22 Monochloro diflouro methene – CHClF 2 F-22 Monochloro diflouro methene – CHClF 2 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 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 44 Environmental Aspects with Refrigerants Halo Carbons depletes Ozone layer. Green House effect caused by freons –Montreal Protocol –Montreal Protocol- A time schedule was chalked out to Control release of chloroflouro hydo carbons to atmosphere. A time schedule was chalked out to Control release of chloroflouro hydo carbons to atmosphere.

45 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.

46 46


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