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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.)

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Madhya Pradesh

Basic Mechanical Engineering

Dr A.C. Tiwari

Professor & Head

Mechanical Engineering


UIT, RGPV Bhopal

Rajiv Gandhi Proudhyogiki Vishwavidhyalay

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


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 .


“Indeed one’s faith in one’s plan and methods are truly tested when the horizon before one is the blackest “.

-- Mahatma Gandhi.


Definition by ASHRAE –

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

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.
methods of refrigeration
Methods of Refrigeration

1. Dissolution of Certain Salts in Water :

-Salts likeCaCl2 , 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


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








methods of refrigeration9
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

methods of refrigeration12
Methods of Refrigeration

2. Change of Phase :

(c) LiquidtoVapour :

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 refrigeration13
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.

Throttle Valve


High Pressure Gas

Throttling process

methods of refrigeration15
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 = 3130 K ( i. e. 40°C)


Pressure Ratio (P2/P1) =6.5

The temp. T2 is found to be, - 89.5°C

methods of refrigeration16
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 .


Compressed Gas

Hot Gas


Throttle Valve


Cold Gas

Hot Gas

Ranque hilsch Vortex tube

methods of refrigeration19

Hot End

Cold End



D C Source

Methods of Refrigeration

6. Thermocouple Effect : Peltier Effect

methods of refrigeration20
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 .
unit of refrigeration
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” .
  • 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

assignment no 1
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.

thermodynamics of a refrigerator
Thermodynamics of a Refrigerator

T1< T2

Source T2

COP (coefficient of Performance)= Q1/ W

Q 2

Work W


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.

Simple Vapour Compression System


Throttle valve

Low pr gauge


High pr gauge



Practical Vapour Compression Cycle.

Components of the Practical Cycle

numerical problems
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.

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 =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 KWAns

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 = 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(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 kJ

Total heat removed per minute ,

= 68250/ 10X60 = 113.75 kJ/min

Refrigeration capacity of the plant in tons

= 113.75/ 210 = 0.542 tons Ans

home assignment no 2
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.

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

vapour absorption system
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.

Vapour Absorption Aqua Ammonia System


One way valve


NH 3 Vapour

Hydrogen+ NH3 down

Cooling effect


Worm H2 up

Worm H2 up

One way valve

Weak sol+ NH3



NH3 dissolves into water H2 is left





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


Secondary Ref

primary refrigerants are further classified as below
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 .
properties of refrigerants
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
environmental aspects with refrigerants
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 cfcs
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 .