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.)
www.rgpv.ac.in
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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.
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Refrigeration
Definition by ASHRAE –
It is defined as Science of providing and
maintaining temperature below that of
surrounding .
How do things get colder?
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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.
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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
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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
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A Glass of Iced Water
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ICE REFRIGERATION
Ice
Storage
Storage
Insulation
water
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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
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Pallets of Dry Ice sublimating into CO Gas
2
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Crystalline structure of solid carbon dioxide
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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
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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.
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Throttle
Valve
Expansion
High Pressure Gas
Throttling process
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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.
(n-1/n)
0.4/1.4
T2= T1X (P2/P1)
= T1X(P
2/P1)
0
Let us take initial temp. T1 = 313 K ( i. e. 40°C)
&
Pressure Ratio (P2/P1) =6.5
The temp. T2 is found to be, - 89.5°C
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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 .
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Compressed Gas
Hot Gas
Nozzle
Cold Gas
Orifice
Throttle
Valve
Hot Gas
Ranque hilsch Vortex tube
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Methods of Refrigeration
6. Thermocouple Effect :
Peltier Effect
Hot End
Cold End
+
D C Source
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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 .
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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” .
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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
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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.
aseemctiwari@yahoo.com
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Thermodynamics of a
Refrigerator
T< T
1
2
COP (coefficient of
Performance)= Q1/ W
Source T
Q
2
2
Work W
Refrigerator
Q
Sink T
1
1
• 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.
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Simple Vapour Compression
System
Evaporator
Throttle
valve
Low pr gauge
condenser
High pr gauge
compressor
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Practical Vapour Compression Cycle.
Components of the Practical Cycle
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A home refrigerator with its door open
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Vapour compression Refrigeration test rig at
thermal Engg lab of UIT, RGTU Bhopal
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Multi Evaporator Vapour
Compression test rig at thermal
Engg lab of UIT ,RGTU Bhopal
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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.
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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)
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Total heat removed =6.94x418.74
=2906 KJ /min
COP of the plant =
heat removed/work of compressor
2.5
W
= 2906/W
=1162.4 KJ/min
= 1162.4/60 KW
=21.5 KW
Ans
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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 .
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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
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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
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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.
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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
aseemctiwatri@yahoo.com
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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.
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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
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burner
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
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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 .
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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
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Environmental Aspects with
Refrigerants
Halo Carbons depletes Ozone layer.
Green House effect caused by freons.
1987 –Montreal ProtocolA time schedule was chalked out to Control
release of chloroflouro hydo carbons to
atmosphere.
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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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