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G. H. Raisoni College of Engineering , Nagpur.
Department of Mechanical Engineering
Refrigeration and Air conditioning Lab Manual
Semester- VIII
PRACTICAL: REFIRIGERATION AND AIR CONDITIONING
Aim : Study of various tools and equipments used in Air Conditioning
Introduction:
1. Controllers
i) Temperature Controller
ii) Humidity Controller
2. Cooling Coils and Heating Coils
3. Air Filter
4. Electric Filter
5. Air Duct
6. Grills and Diffusers as air outlet
7. Air Washers
8. Adiabatic Dehumidifier
Conclusion
PRACTICAL: REFIRIGERATION AND AIR CONDITIONING
Aim: Study of various types of compressor
Introduction:
Classification of Compressor
1. Hermetic Compressor
• Introduction
• Working Principle with neat sketch
• Application
2. Rotary Compressor
• Introduction
• Working Principle with neat sketch
• Application
3. Reciprocating Compressor
• Introduction
• Working Principle with neat sketch
• Application
Conclusion
PRACTICAL: REFIRIGERATION AND AIR CONDITIONING
Aim: Study of various condensers, evaporator and expansion devices used in
Vapour Compression Refrigeration System.
Introduction:
CONDENSER
Classification of condensers
1. Air cooler condenser
Explain with working principle and neat sketch
2. Water cooled condenser
Explain with working principle and neat sketch
3. Evaporative Condenser
Explain with working principle and neat sketch
EVAPORATOR
Classification of Evaporator
1. According to types of construction
• Base tube coil evaporator
• Plate evaporator
• Fixed tube
• Shell and tube
• Shell and coil
• Tube in tube
2. According to manner in which liquid refrigerant is fed
• flooded evaporator
• Dry expansion evaporator
3. According to mode of heat transfer
• Actual natural convection
• Forced convection
4. According to operating condition
• Frosting evaporator
• Non – frosting evaporator
• Departing evaporator
EXPANSION DEVICES
Types of Expansion device
1. Capillary tube
2. Hand operated expansion valve
3. Thermocoustic expansion valve
4. Low side float valve
5. High side float valve
6. Automatic or constant pressure expansion valve
Conclusion :
Aim : Study and demonstration of various controls used in Refrigeration and air
Conditioning
Introduction :
1. Water level Indicator
2. High pressure control
3. Bulb and bellow element
4. Solenoid or electromagnetic element
5. Humidity sensitive element
6. Single phase induction motor
7. High and low pressure cutouts
8. Control room conditions at Partial load
9. On-Off Control
10. Volume Control
Conclusion :
Aim : Study of window air conditioning, packaged air conditioning and automotive
air conditioning
Introduction :
1. Window Air conditioning
• Classification
• Working Principle
• Application
2. Packaged Air conditioning
• Working Principle
• Application
3. Automotive Air conditioning
• Working Principle
• Application
Conclusion :
Aim : Study of domestic Refrigerator
Introduction :
Working Principle with neat sketch
Various Components used
• Compressor
• Condenser
• Expansion device
• Evaporator
Conclusion :
Aim : Study of Testing and Charging of VCR system.
Introduction :
1. Methods for Testing and Charging
• Evaculation
• Dehydration
• Charging of refrigerant
• Testing for leaks
- Soap bubble method
- Sulphur candle method
- Wet litmus paper
- Halide torch method
- Electric defector
- Use of cold trap
Conclusion :
DESERT COOLER ( EXPERIMENTAL UNIT )
INTRODUCTION :
Evaporative coolers (also called air coolers or desert coolers) are cooling devices
which uses simple evaporation of water in air. They differ from refrigeration or absorption air
conditioning, which use the vapor-compression or absorption refrigeration cycles. small-scale
evaporative coolers are called swamp coolers by some users due to the humid air conditions
produced. Air washers and wet cooling towers utilize the same principles as evaporative
coolers, but are optimized for purposes other than air cooling.
Evaporative cooling is especially well suited for climates where the air is hot and
humidity is low. , if sufficient potable water is available. In dry climates, the installation and
operating cost of an evaporative cooler can be much lower than air conditioning, often by
80% or so. But evaporative cooling and vapor-compression air conditioning are sometimes
used is combination to yield the optimal performance. Some evaporative coolers may also
serve as humidifiers in the heating season.
In moderate humidity locations there are many cost-effective uses for evaporative
cooling, in addition to their widespread use in dry climates. For example, industrial plants,
commercial kitchens, laundries, dry cleaners, greenhouses, spot cooling (loading docks,
warehouses, factories, construction sites, athletic events, workshops, garages, and kennels)
and confinement farming (poultry ranches, hog, and dairy) all often employ evaporative
cooling. In highly humid climates, evaporative cooling may have little thermal comfort
benefit beyond the increased ventilation and air movement it provides.
DESCRIPTION :
Experinmental Desert cooler unit consists of water spray arrangement over the
cooling pads at the centre and duct made of M.S. sheet on both sides . Inlet duct consistes of
inlet orifice with inclind manometer , for air flow ( Q ) measurement .
Inlet section having orifice at air entry and Air heater ( fin type, for simulation of
summer condition ) after that and , DBT / WBT sensor at the centre . Water sprey with
cooling pades after inlet section . Water sump is provided at the bottom . Pump sucks the
water and discharges through nozzle on top of the cooling pades through water flow meter (
rotameter ) . After water sprey section outlet section follows .
Outlet section contains DBT / WBT sensor for outlet air condition measurement .
wheel type anemometer is placed at the centre of the duct for air velocity measurement . At
the end of the outlet section axial fan is fitted to suck the air .
All ducts ( inlet / outlet ) are having superlon insulation to reduce the heat losses .
Front side of the duct is covered by acrylic sheet .
Necessary controls and computer interface converters and switches are fitted over the
main stand in panels .
SPECIFICATION :
1. Axial fan : 250 watts .
2. Duct inlet : 250 mm x 250 mm
3. Outlet duct : 250 mm x 250 mm
4. Inlet orifice : 125 mm
5. Inlet / oulet air condition : DBT / WBT sensor
6. Air Heater : 250 watts x 2 nos with fins .
7. Water sprey chamber : 300 mm width .
8. Cooling pades : 3 Nos.
9. Sprey pipes : 3 nos with 10 no. each holes of 1.2 mm
10. Water sump tank : 8 ltrs
11. Water pump : ½ HP
12. Rotameter : 60-600 LPH
13. Ball valve : 1 No.
14. Anemometer : wheel type ( 30 m/sec max )
15. Temperature indicator : 6 channel , serial interface.
16. Heater voltage / current converter .
Direct Evaporative Cooling ( open circuit ) :
DEC is used to lower the temperature of air by using latent heat of evaporation,
changing water to vapor. In this process, the energy in the air does not change. Warm dry air
is changed to cool moist air. Heat in the air, and the water too, is used to evaporate water.
Typically, residential and industrial evaporative coolers use direct evaporation and
can be described as an enclosed metal or plastic box with vented sides containing a
centrifugal fan or 'blower', electric motor with pulleys (known as 'sheaves' in HVAC), and a
water pump to wet the evaporative cooling pads. The units can be mounted on the roof (down
draft, or downflow), or exterior walls or windows (side draft, or horizontal flow) of buildings.
To cool, the fan draws ambient air through vents on the unit's sides and through the damp
pads. Heat in the air evaporates water from the pads which are constantly re-dampened to
continue the cooling process. Thus cooled, moist air is then delivered to the building via a
vent in the roof or wall.
Because the cooling air originates outside the building, one or more large vents must
exist to allow air to move from inside to outside. Air should only be allowed to pass once
through the system, or the cooling effect will decrease. This is due to the air reaching the
saturation point. Often 15 or so air changes per hour (ACHs) occur in spaces served by
evaporative coolers.
Cooler pads
Traditionally, evaporative cooler pads consist of excelsior (wood wool) (aspen wood
fiber) inside a containment net, but more modern materials, such as some plastics and
melamine paper, are entering use as cooler-pad media. Wood absorbs some of the water,
which allows the wood fibers to cool passing air to a lower temperature than some synthetic
materials. The thickness of the padding media plays a large part in cooling efficiency,
allowing longer air contact. For example, an eight-inch-thick pad with its increased interface
will be more efficient than a one-inch pad.
OBSERVATION TABLE :
SR.No. Manometer
reading (
h) mm
Heater
Volts
Heater
Amp.
Water
flow
rate
Inlet
DBT
(T1)
Inlet
WBT
(T2)
Outlet
DBT
(T3)
Outlet
WBT
(T4)
CALCULATION :
1) Amount of water added :-
mw = mair x (∆ω)
mw = mair x (ω1 – ω2)
where,
ω2 = water content in air at inlet
ω1 = water content in air at outlet
mair = Cd x (π/ 4) d2 x √2 x g x (ρw / ρg )x h
2) Desert Cooler Efficiency (η)
η = T1 - T3
T1 - T2
Where, T1 = Dry bulb temperature at inlet
T2 = Wet bulb temperature at inlet
T3 = Dry bulb temperature at outlet
Performance
Understanding evaporative cooling performance requires an understanding of
psychrometrics. Evaporative cooling performance is dynamic due to changes in external
temperature and humidity level. Under typical operating conditions, an evaporative cooler
will nearly always deliver air cooler than 27°Celsius (80°Fahrenheit). A typical residential
'swamp cooler' in good working order should cool air to within 3°C - 4°C (6°F - 8°F) of the
wet-bulb temperature.
Advantages
Less expensive to install
1 Estimated cost for installation is 1/8 to 1/2 that of refrigerated air conditioning
Less expensive to operate
1 Estimated cost of operation is 1/4 that of refrigerated air.
2 Power consumption is limited to the fan and water pump vs. compressors, pumps, and
blowers.
Ventilation air
1 The constant and high volumetric flow rate of air through the building reduces the
age-of-air in the building drammatically.
Disadvantages
Performance
1 High temperature, high humidity outside conditions decrease the cooling capability of
the evaporative cooler.
2 No dehumidification. Traditional air conditioners remove moisture from the air,
which is usually a design requirement except in very dry locations. Evaporative
cooling adds moisture, which, in dry climates, may improve thermal comfort.
Comfort
1 The air supplied by the evaporative cooler is typically 80-90% relative humidity.
2 Very humid air reduces the evaporation rate of moisture from the skin, nose, lungs,
and eyes.
3 High humidity in air accelerates corrosion. This can considerably shorten the life of
electronic and other equipment.
4 High humidity in air may cause condensation. This can be a problem for some
situations (e.g., electrical equipment, paper/books, old wood).
Water
1 Evaporative coolers require a constant supply of water to wet the pads.
2 Water high in mineral content will leave mineral deposits on the pads and interior of
the cooler. Water softeners, bleed-off, and refill systems may reduce this problem,
however.
3 The water supply line needs protection against freeze bursting during off-season,
winter temperatures. The cooler itself needs to be drained too, as well as cleaned
periodically and the pads replaced.
THERMOELECTRICTRIC COOLING
In 1822 seebeck observed that, “ if a closed circuit were made of two dissimilar
metals, an electrical current flows in the circuit when the two junctions were maintained at
different temperatures “ . His investigations covered a wide range of elements and
compounds. This resulted in publication of a series in which the investigated materials were
arranged in the order of magnitude of their effect. However, he failed to realize the
significance of his discovery. In 1834.
Peltier observed the inverse effect, if an electrical current flows across the junction
between two dissimilar material, heat was either absorbed or evolved. Peltier did not realize
the significance of his discovery either, and moreover, failed to recognize the connection
between his discovery and that of Seebeck.
It is alleged that Lenz ended all conjecture surrounding these discoveries by freezing a
small quantity of water placed in the vicinity of the junction between a bismuth and antimony
rod through which a direct current was passed. These were two of the materials in which
Seebeck had observed the most pronounced effect.
A third effect was pointed out by Thomson (Lord Kelvin) in 1851. It related heat
absorbed or evolved in a single conductor to the temperature gradient along it and the current
flowing through it. This effect takes place in addition to the Joule (FR) heating. However, in
thermoelectric cooling materials, Kelvin’s is a second order effect when compared with
Peltier’s and Seebeck’s, and will not be further considered.
For many years. Practical application of these thermoelectric effects was almost
exclusively restricted to thermocouples for temperature measurement, because metals exhibit
a comparatively small Seebeck effect. However, the Seebeck effect in semiconductors can be
considerably greater. The advent of the transistor and other semiconductor devices have
stimulated research pertaining to properties of semiconductors in general, from this, materials
have been developed in which thermoelectric effects are of sufficient magnitude so that
fabrication of useful devices has become a reality.
Specifications :
1. Module assembly : KRYOTHERM , Russia make .
Frost – 74 HT
1.05 ohm resistance at 295 K
∆ Tmax > 740 C
Height : 3.85 mm
Parallel difference : 0.025 mm
2. Control panel : 1. Separate controls to module and heater
3. Multi pole switch .
4. Two variable power supplies DC .
5. Two Ammeter
6. Two Voltmeter
7. Temperature indicator ( digital ) with serial interface
.
8. Voltage and current module for computer
interface .
THERMOELECTRIC EFFECTS
Seebeck Effect :
For a small temperature difference between the two junction of materials A and B, the
open circuit voltage developed is proportional to the temperature difference and is given by:
TE AB ∆=∆ α
Where,
∆ E = open circuit voltage developed
αAB = relative Seebeck coefficient (the difference between absolute Seebeck
Coefficients for materials A and B).
∆ T = temperature difference between junctions of materials A and B.
In practice, the absolute Seebeck coefficient, α , of a material is determined with
respect to a material such as lead, in which the Seebeck coefficient is negligible.
In metals, α does not exceed 0.00005 volt/deg C in semiconductors available for
thermoelectric applications in 1966, α is typically 0.0002 to 0.00025 volt/deg C.
Peltier Effect :
The same circuit can be considered to be made up of materials A and B, into which a
battery is introduced to provide a direct current, (I). At the junction between the two
dissimilar materials, the heat evolved or absorbed in unit time is proportional to the current
flowing and is given by :
IQ ABπ=
where
Q = heat evolved or absorbed in unit time, watts.
πAB = relative Peltier coefficient for materials A and B.
I = direct current flowing, amperes.
Lord Kelvin, by performing a thermodynamic analysis of such a
Thermoelectric circuit, showed a relationship exists between α and π
Tαπ =
Where
T = the absolute temperature, Kelvin.
Hence, heat absorbed or evolved per unit at the junction between two dissimilar
materials is given by:
ITQ ABα=
Although thermodynamics do not accept Kelvin’s derivation of his well-known
relations because he considered the effects reversible, a more rigorous treatment, by
application of irreversible thermodynamic, leads to the same results.
Thermodynamics and Refrigeration Cycles :
The relative hot and cold surfaces are the result of the basic heat transport
phenomenon called the Peltier effect, and are not necessary to make it occur.
Investigations in solid-state physics have shown that thermoelectric refrigeration is
means of heat pumping which utilizes the energy level changes in electrons for transporting
thermal energy. Electrons flowing across a junction of two dissimilar thermoelectric
materials, i.e., materials with different available electron energy levels, must undergo an
energy change which results in either the evolutions or absorption of heat. The direction of
current flow determines which will occur . The bibliography lists several sources of detailed
information on solid – state theory.
PERFORMANCE OF THERMOELECTRIC COUPLES
A couple circuit consisting of two dissimilar thermoelectric materials is referred to as
a couple. The two thermoelectric materials are represented by n and p. The n-type has a
negative Seebeck coefficient and an excess of electrons. The p-type has a positive Seebeck
coefficient and a deficiency of electrons actually flow in the opposite direction. While there
are a total of four connections between the thermoelectric junctions; the upper or cold one, at
which heat is a absorbed; and the lower or hot one, at which heat is evolved. If the direction
of current were reversed, the upper junction would evolve heat, the lower would absorb it.
The temperature gradient between hot and cold junctions is not linear because of production
of Joule heat within each leg.
The conducted heat arriving at the cold junction is given by:
)(5.0 2 RITC +∆
Where C = thermal conductance of couple legs, watts per degree Celsius.
APPLICATIONS OF THERMOELECTRIC COOLING :
Thermoelectric cooling is different from conventional compression Refrigeration in
that there are no moving parts in the process of producing cold. However, as energy is being
transferred as heat in the cooling process, heat must be removed from the hot side of the
couples in order that cold can be produced. For improved efficiency on the cold side, heat
transfer surfaces are frequency used. These heat transfer surfaces take the forms of fins and
fans in gas systems, or with heat transfer to fluid systems using pumps for circulation.
Two different systems for cooling :
1) by using vapour compression system , 2) Thermoelectric module
VAPOR COMPRESSION THERMOELECTRIC
COMPRESSOR HOT COLD BATTERY HOT COLD
SYSTEM ANALOGES
System can be interchanged by reversing the polarity of the director current applied to
it. Since there are no moving parts, there is nothing to generate noise. There is no refrigerant
to contain, so the problem of handling a two-phase changeover is simplified. The pressure
tight tubing is replaced by electrical wiring.
Since the cooing capacity of a single thermoelectric couple is small, it is practical to
make systems of low refrigerating capacity. In the cooling of infrared detectors, small
thermoelectric assemblies can cool to temperatures as low as 145 K, using cascading.
Thermoelectric systems have been built in capacities up to ten tons of air conditioning by
using many couples.
Capacity control in a thermoelectric system can be achieved by varying the voltage
applied to the couples either by a variable voltage control or by switching series and parallel
circuits. As the voltage drops, the temperature difference between hot and cold side reduces.
Another advantage is that a thermoelectric system will operate under zero gravity, or
many times the force of earth gravity, and will operate in any orientation. This has been an
important consideration in selecting thermoelectric cooling application in the U.S. space
program. A thermoelectric system is capable of operating satisfactorily at temperatures levels
of C.
For many applications, the advantage of thermoelectric cooling outweigh its chief
disadvantage of low coefficient of performance. The lack of moving parts and eliminations of
cooling liquids are appealing features.
For military applications, thermoelectric prove advantageous because of
noiselessness, ruggedness, and lack of
Thermodynamic Analysis of Thermoelectric Refrigeration System
Fig shows an enlarged view of thermoelectric elements with control volume. In order
to analyze the system to obtain refrigeration effect COP, etc., the following assumptions have
been made:
(i) Heat transfer takes place through semi – conductors at the ends only.
(ii) No energy exchange between the elements through space separating
them, and
(iii) Properties such as conductivity, etc., are invariant with temperature.
Now cooling and heating due to the thermoelectric effect is given by (Peltier effect):
qc = abITl
qh = abITh
ADVANTAGES OF THERMOLECTRIC REFRIGERATION SYSTEM
It possesses several advantages as listed below:
(i) Absence of moving parts eliminates vibration problem as well as regular
attendance. Therefore, it can be best suited for system where vibration is
undesirable. In addition there is no were due to rubbing as such the life is
expected to be almost infinite compared with other systems.
(ii) It is easy to overload if desired, just by increasing the current to a certain
limit.
(iii) The load can be easily controlled by means of adjusting the current to
meet the situation.
(iv) Very compact in size since even the system boundary may be used as the
cooling surface as exhibited in fig. Here one of the walls of the room forms
the evaporator surface. Therefore, the cost of manufacture of that wall is
taken care by the evaporator surface.
(V) It can be operated in any position in contrast with vapour – absorption,
vapour-compression or steam - jet refrigeration systems. As for example, if
a vertical compressor is used in a refrigeration system, it cannot be
operated in any other position.
(vi) It is lighter in weight for the same capacity of refrigeration.
(vii) Since electric current passes through conductors, there is no problem of
leakage which is most undesirable in other refrigeration systems. Further,
the leakage of refrigeration from the system causes the drastic decrease
in the capacity in addition to extra cost for the refrigeration and charging
operations. Thus a thermoelectric refrigeration system operates at the
same capacity for long and eliminates the cost of charging and extra
materials.
(viii) It is most easy to operate as a heat pump, just by reversing the terminals.
Hence, a thermoelectric refrigeration system can be considered as an
year round air conditioner*.
(ix) Since no refrigerant is used, there is no question of toxication etc. and can
be used directly for air conditioning.
(x) Its design and manufacture are rather simpler than the other refrigeration
systems.
(xi) It is most suitable for the production of cooling suit.
Main disadvantage of thermoelectric refrigeration system is the
unavailability of suitable material of high figure of merit. Presently the total
cost of refrigeration system will be a few times higher that the vapour –
compression or other systems for a few ton capacity. In addition, the
running cost is found t obe much higher compared to vapour –
compression system. That is why the vapour – compression or other
systems of refrigeration are most commonly employed. The overall COP
of a thermoelectric refrigeration system is of the order of 01 to 0.2.
Experimentation :
1. Evaluation of the peltier effect :
DC Power is applied to the module at a variety of settings . When settled both
temperatures are noted . Observe that at higher power inputs I2R and heat conduction factors
tend to overwhelm the cooling effect .
Module power
voltage current
Graph : Module Power v/s cold & hot side Temperature and temp. difference
2. Evaluation of Seebeck effect :
By use of heater power create a temperature differential across the module . When settled
note the open circuit voltage across the module voltmeter . Voltage and temperature can be
seen to have a linear relationship .
Heater power
voltage current
Graph : Seebeck open ckt voltage v/s Temperature gradient ∆T
3. Evaluation of Generating effect :
Create a temperature gradient across the module by applying heater power and switch in
electrical load . When settled , note the generator voltage and amperage .
Heater power Module Power
Voltage Current Voltage Current
Graph : module voltage / current / power v/s Temperature Gradient ∆T
4. Coefficient of performance :
Power to the module is set and heater power adjusted to maintain ambient temperature on the
cold side of the module . Repeat over the whole range . The curve shows that the coefficient
is highest when module has to do the minimum work .
Heater power Module Power
Voltage Current Voltage Current
Graph : Module power v/s COP , COP = Wh/Wm
Wh : Heater power
Wm : Module power
VAPOUR ABSORPTION REFRIGERATION SYSTEM
INTRODUCTION & WORKING PRINCIPLE
The vapor absorption Refrigeration System consists of a vapor absorption unit
mounted on a cabinet fixed on a display frame of the set-up.
The principle works on 3-fluid system. There is no solution circulation pump. Total
pressure is same throughout the system. The third fluid remains mainly in the evaporator thus
reducing partial pressure of refrigerant to enable it to evaporate at low pressure and hence
low temperature.
The schematic diagram of vapor absorption refrigerator working on
NH3-H2O system with H2 as the third fluid is shown in Fig 1. Liquid NH3 evaporates in the
evaporator in the presence of H2. Hydrogen is chosen as it is non-corrosive and insoluble in
water
A thermo siphon bubble pump is used to lift the weak aqua from generator to the separator.
The discharge tube from the generator is extended down below the liquid level in the
generator. The bubbles rise and carry slugs of weak NH3-H2O solution into the separator.
Two U bends are provided as vapour locks to prevent H2 from getting into the high
side or solution circuit.
Partial pressure of H2 provides the pressure difference of NH3 between the condenser
and the evaporator. Accordingly, we have:
.In condenser , Pure NH3 vapour pressure = Total pressure
In evaporator, NH3 vapour pressure = Total pressure – Partial pressure of H2
For example, consider the condenser temperature as 500C, and evaporator temperature as –
150C. the corresponding vapour pressures of NH3 are
Condenser, Pk = 20.33 bar
Evaporator outlet, Po2 = 2.36 bar
The approximate pressures in various parts of the system, then will be as given in the table
below:
(Example for partial pressure effect)
Section NH3 H2O H2 Total
Condenser 20.33 0 0 20.33
Evaporator Inlet 1.516 0 18.814 20.33
Evaporator Exit 2.36 0 17.97 20.33
Generator Top 15.54 4.79 0 20.33
It has been assumed that vapours leaving generator top are in equilibrium with
entering rich solution at 400C, at which temperature saturation pressure of NH3 is 15.54 bar.
It has also been assumed that the temperature at evaporator inlet is
–250C at which temperature saturation pressure of NH3 is 1.516 bar.
DESCRIPTION OF THE SET-UP
The unit is charged with ammonia solution in water and hydrogen gas. The maximum
pressure it can withstand is 30 bar and the ammonia charge is 0.245Kg NH3 + H2O.
The unit works on electric heater 90 watt capacity.
A voltmeter / ammeter is provided to measure the power. The heater works with
intermittent action of 18 sec on and 3 sec off to control the temperature rise. A temperature
indicator is provided for measurement of temperature at 10 different locations as displayed on
the mimic diagram presented on the board.
A thermostatic control is provided for the sake of safety and it controls the heater..
An acrylic window in front of the system and open portion from behind gives the
actual view of the system .
SEQUENCE OF OPERATION
1] Strong ammonia solution flows from the absorber vessel to the boiler thro’ the
inner tube.
2] When the ammonia solution is heated in the boiler bubbles of ammonia gas rise
from the pump.
3] The ammonia vapour passes in to the condenser.
4] Weak ammonia solution flows in to the annular tube.
5] Air circulating over the fins of the condenser cools down the ammonia
vapour, {after 2} condensing it to liquid ammonia.
6] Liquid ammonia flows through the pipe to the evaporator after sub cooling &
throttling.
7] A } The hydrogen in the evaporator lowers the ammonia vapour pressure and
makes it to evaporate.
B} This mixture of hydrogen and ammonia passes from the evaporator , to the
Absorber.
C} This process extracts heat from the evaporator, which in turn extracts heat
from the storage space , thereby, the temperature inside the refrigerator is
lowered.
8] As the weak solution is fed from the boiler system as it runs to the absorber
vessel , it absorbs the ammonia from the ammonia – hydrogen mixture and gets
ready for another round of cycle in the boiler.
Principal Operating Line
1 The actual ammonia – water system requires the separation of water vapour as much
as practicable in order to avoid the high operating temperature of evaporator & the
large amount of purging of water into the absorber.
2 To meet the above requirement an analyser & a rectifier are incorporated with the
generator . deflegmator is not used ..
3 Thus the actual absorption system comprises an analyser, rectifier, a pre heater and a
pre cooler in addition to the basic components.
4 The strong solution of ammonia is pumped from the absorber into generator thro’ A
pre heater , a counter flow heat exchange .Thus the hot and weak solution returning
from the generator heats the aqua solution before it entered generator, causing
reduction in heat input to generator.
5 The heat transfer to the generator brings about the separating of ammonia vapour
accompanied by a small fraction of water vapour.
6 This mixture of ammonia and water vapour passes thro’ the analyser where a part of
Water vapour is separated from the vapour mixture.
7 The mixture then enters the rectifier in a simple form and not in the form of
deflegmeter .water particles are drained back to analyzer due to presence of
contraction and bend.
8 The ammonia vapour then condenses in the condenser , which , in turn ,is sub cooled
by the vapour leaving the evaporator in a pre cooler ,a counter flow heat exchanger.
9 Thus the refrigeration effect is increased. The sub cooled condensate is allowed to
expand through the throttle valve down to the evaporator pressure. The energy
transfer to evaporation causes vaporization of ammonia . The ammonia vapor are then
reabsorbed.
10 The analyzer column is an open type counter flow heat – exchanger having baffle
plate effect provided with the dents to the analyzer column , increasing the area of
contact between the vapour leaving the generator and strong aqua ammonia entering
the analyzer at appropriate location.
11 The warm fluid from the pre heater enters the analyzer column where it comes in
direct contact with the hot vapour. The aqua solution gets heated at the same time a
part of water vapour is condensed since the ammonia vapour leaving the generator
Contains small percentage of water vapour . Thus it acts as a column which purifies
the ammonia vapour by reducing the amount of water vapour – a kind of rectifying
action.
12 As soon as strong aqua is heated in the generator , there is separation of vapour of
ammonia leaving behind weak solution of ammonia, corresponding to the generator
temperature and pressure.
13 The vapour leaving the generator with refrigerant concentration ξ va in equilibrium
with the boiling poor solution having concentration ξ L
a enters the analyzer at 2 v
As
it travels upwards , counter flow to the entering rich solution at 1 with concentration
ξ Lr , the vapour encounters heat and mass exchange with the falling rich solution
ultimately leaving the analyzer enriched in the refrigerant with vapour concentration
ξ 5 In equilibrium with the rich solution having concentration ξ Lr.
SPECIFICATIONS
1. 0.245 Kg NH3 + H2O Refrigerant
2. Heater capacity 90 watt,1 ph .Ac Supply
3. Air cooled evaporator
4. Digital temperature indicator (10 channel, Cr-Al)
5. Digital voltmeter
6. Digital ammeter
OPERATING INSTRUCTIONS
1. Start the electric connection-heater.
2. Keep the unit in working for about 4 hours to reach the equilibrium.
3. Take readings when steady state is reached
4. Find out the coefficient of performance COP as per the example illustrated with the
help of Enthalpy-concentration chart and refrigerant properties of ammonia water.
from The standard chart.
5. Keep the unit working for 24 hours.
OBSERVATION TABLE
Temperature Readings
(Thermocouple numbers as shown in diagram)
Heater
input
T10C T2
0C T3
0C T4
0C T5
0C T6
0C T7
0C T8
0C T9
0C T10
0C V
Vol
t
I
amp
Precautions :-
1] Do not start the switch when there is cover on the unit from behind.
2] Remove the cover and then start the unit.
3] Do not disturb the thermocouples placed at various locations.
4]Do not remove the cover wrapped over generator / boiler.
5] Do not operate the thermostatic switch at the right hand bottom side on the main panel.
ACTUAL VAPOUR ABSORPTION CYCLE & ITS
REPRESENTATION ON ENTHALPY-COMPOSITION DIAGRAM.
Figure 2 shows the schematic arrangement of the actual vapor absorption cycle and
figure 3 represents its thermodynamic cycle on the h - diagram. The system consists of
generator G together with analyzer AN, and condenser C on the high-pressure side, and the
evaporator E and Absorber A on the low-pressure side. Pump P, expansion valve VI, and
pressure reducing valve VII separate the two sides. In addition, liquid-vapor heat exchanger
HEI is also provided.
The vapors at 5 distilled from the generator-analyzer, enter the refrigerant circuit. The
vapors are condensed to 8 in the condenser, precooked to 9 in the liquid-vapor regenerative
heat exchanger and throttled to 10 before entering the evaporator. The state 10 is at the same
point as state 9 in the h - diagram, as both enthalpy and composition remain the same
before and after throttling.
The refrigerant entering the evaporator at 10, leaving the evaporator at 11, and the
liquid vapor heat exchanger at 12, comprise a liquid plus vapor mixture. The refrigerant is
finally absorbed by the poor solution at 2 returning from the generator and after being cooled
in the liquid-liquid heat exchanger to 3 and throttled to 3a whereas the rich solution from the
absorber at 4 is pumped to 4 a and heated to 1a before entering the analyzer.
The state points 1,2,3 & 4 can be located on the h - diagram according to their
temperatures and compositions as the enthalpy of liquid is independent of pressure. Also
point 3a lies at 3 only (isenthalpic process) and point 4a lies approximately at 4 itself as the
pump work s very small. State point 5 of the vapor is along the isothermal tie line drawn from
1. Point 8 is the saturation state at Pike. Point 9 after sub cooling of the liquid can be plotted
according to the temperature and composition and point 10 is at 9 itself (isenthalpic process).
Point 11 is on the tie line corresponding to the evaporator leaving temperature t02 = t2 and
pressure P0.the composition is same at 7,8,9,10,11 and 12. Point 12 can be similarly located
by knowing the temperature from the energy balance of the heat exchanger.
The absorber pressure in the absorber system is equal to the evaporator pressure Po &
the generator pressure is similarly equal to condenser pressure Pk . Hence at a given
generator pressure , the poor solution concentration is determined by the heating temperature
t h and at a given absorber pressure , the rich solution concentration is determined by the
cooling temp. t A the absorber temperature.
STEPS FROM START TO FINAL RESULT
OBSERVATIONS
I) 1) Generator Temperature
2) Temperature after Generator outlet.
3) Temperature before condensation
4) Temperature after condensation
5) Temperature after sub cooling
6) Evaporator Inlet Temperature
7) Evaporator outlet Temperature
8) Absorber Inlet Temperature
9) Temperature of Inlet to absorber vessel
10) Temperature of Outlet of Absorber
II) Refer to the Properties of Ammonia - Table (Pk & Po)
1) Note Temperature after condensation th = tc. With this as saturation
temperature refer to the Properties of Ammonia & find out the condenser
pressure = Pk = (Psat)th = tc
2) Note temperature at the inlet of Evaporator = t6 = t0, with this as saturation
temperature refer to the properties of Ammonia & find out the Evaporator
pressure p0 = (Psat) t6 = t0
III) 1) Refer to the Enthalpy – Concentration chart of Ammonia
Note the temperature of generator = t1=th at condenser pressure Pk & call this
point as 2. The poor solution concentration .a is obtained at this point on x axis.
2) Evaporator pressure and absorber temperature, t8 = tA call this point as 4,
The rich solution concentrations �r is obtained at this point .Points 1 & 3 are
obtained as illustrated on the Evlhalpy chart by drawing �r & �a Lines & where they
intersect the pressure pk & P0 are the points resp , 1 &3.
3) Note temperature after condensation & condenser pressure Pk on the chart
and this is the point 8. The concentrations of vapour leaving the analyzer, in
equilibrium with the entering rich solution is �r = vr
This point can also be traced as given below Draw vertical line from
point 4-1 , where it meets the auxiliary line, from that point draw a horizontal
line and where it meets the condenser pressure line that point is 5v & r =
vr= 5v. Is the same as described above.
ar
afξξ
ξξ
−
−= 5
4) Now specific rich solution rate
Specific poor solution rate = (f-1)
5) Temperature after sub cooling = t9
Note this point on the line 5v
6) The vapour state 10v at v
10 = 1 and then join the point 10v to 9 and extend
the line to in tersect the saturated liquid line for evaporator pressure p0 at
10 L which gives the temperature after expansion and the temperature at the
in let to Evaporator as
t01 = t10 & t02 = t11
7) Draw the isothermal tie line for t02 = t11 & p0
The intersection of this line with the 11 is the point 11
8) Refrigerating Effect = q0 = h11- h10
9) Heat transfer in the liquid Heat Exchanger HE II
q = (f-1) (h2 – h3) = f(h1a – h4)
From this h1a is known as h2, h3, h4 are known from the Enthalpy
Concentration chart.
10) Heat added in the generator
qh = h5 +h2 (f-1) – fh1a
qh
qoCOP =
11) Coefficient of Performance
12) h12 = h11 + (h8 – h9)
13) qA = Heat rejected in Absorber
qA = h12 – h3 +f (h3 – h4)
qA + qc = Heat rejected (Total)
14) Heat rejected in the condenser
qo + qh = total heat given
15) Energy balance
qA + qc = qo +qh
Heat rejected in the condenser + Heat rejected in the Absorber = Heat
added in generator +Refrigerating Effect.
Sea that both the LHS & RHS match well showing that balance is achieved well.
Vapour Absorption Refrigeration system
SET No. 1
I = Observations
1) Generator temperature = 154.3 0
c
2) Top of Generator temperature = 145.4 0
c
3) Condenser in let temperature = 46.6 0
c
4) Condenser out let temperature = 35.6 0
c
5) Subcooled temperature = 3.6 0
c
6) .In let of E vapur =..17.6 0
c
7) Out let of E vapur =.. 2.7 0
c
8) In let of absorber = 29.7 0c
9) out let of absorber = 30.7 0c
10) out let of absorber vessel = 30.4 0c
I I = Observations
Referring NH 3 properties from Table
1) PK = Condenser pressure = (p sat) at 35.6 0c
pk = 13.75 kg / cm2
2) PO = Evaporation pressure = (P sat) = 17 0c
po = 2.15 kg /cm2
I I I = Observations
Points on enthalpy conc. diagram
POINT NO 2 – Generator temperature 154.5 0c
Generator pressure 13.75 kg / cm2
14.0=aξ
POINT NO 4 – Evaporator pressure po = 2.15 kg / gm2
4.0=rξ Absorber Temp = 29.7
0c
POINT NO 8 – 8
= 0.96 h8 = 460
POINT NO 5 – 5 = 0.96 h5
= 1850
POINT NO 1- r = 0.4 h1 = 300
POINT NO 3- a = 0.14 h3 = 280
POINT NO 9- 9 = 0.96 h9 = 280
POINT NO 10-10 = 0.96 h10 = 280
POINT NO 11 - 11 = 0.96 h11 = 1400
qo = h11 – h10 = 1400 –280 = 1120
15.326.0
82.0
14.040.0
14.096.0==
−
−=f
f = 3.15 , (f-1) = 2.15
q = (f-1) (h2 – h3) = 2.15 x (600 –280)
= 2.15 x 320 = 688 = f(h1a – h4)
= 688 = 3.15 (h1a –20)
h1a = 238
qh = h5 +h2(f-1)-fh1a = 1850 +600 x 2.15
= 1850 +1290 – 750 = 3140 – 750
qh = 2390
qo + qh = 3110 = Heat given
= 3510
h12 = h11 + h8 – h9
= 1400 +460 – 260
= 1860 – 280
h12 = 1580
qA = h12 – h3 + f(h3 – h4)
= 1580 – 280 +3.15 (280 –20)
= 1580 – 280 +3.15 x 260
= 1300 + 3.15 x 260
= 1300 +819 = 2119
qA = 2119 KJ/Kg
qc = h5 – h8
= 1850 –460
qc = 1390
qA +qc = Heat Rejected
= 2119 +1390
= 3509
qA + qc = 3509 qo +qh = 3510
Heat Balance is checked Well.
Vapour Absorption Refrigeration system
SET No. 2
I = Observations
1)Generator temperature = 151.7 0
c
2)Top of Generator temperature = 138.8 0
c
3)Condenser in let temperature = 42.6 0
c
4)Condenser out let temperature = 33.2 0
c
5) subcooled temperature = -7.5 0
c
6) In let of E vapur =..-19.3 0
c
7) Out let of E vapur =.. -5.2 0
c
8) In let of absorber = 29.8 0c
9) out let of absorber = 30.8 0c
10) out let of absorber vessel = 31.4 0c
I I = Observations
Referring NH 3 properties from Table
1) pk = (Psat) at 32.20cCondenser outlet
pk = 12.83 kg / cm2
2) PO = (P sat)at – 19.30c Evaporation Inlet Temperature
po = 1.19 kg / cm2
I I I = Observations
POINT NO 2 – Generator temperature 151.7 0c
Condenser pressure = pk = 12.83 kg / cm2
h2 = 600 KJ/kg
POINT NO 3 – a = 0.14 & po = 1.96 Kg/cm2
h3 = 280
POINT NO 4 – po = Ev. Pressure & abs. Temperature
Po = 1.96 Kg/cm2
tA = 29.80c
r = 0.4 h4 = 30
15.326.0
82.0
14.04.0
14.096.0==
−
−=f
POINT NO 1 – r = 0.4 pk = 12.83 kg/cm
2
H1 = 300 f-1 = 2.15
POINT NO 5 – h5 = 1850 5
= 0.96
POINT NO 8 –pk = 12.83 kg/cm2 T
Temperature after condenser = 33.2
8 = 0.96 h8 = 470
POINT NO 9 – h9 = 260
9 = 0.96 t subcooled = 7.5
POINT NO 10 – h10 = 260
POINT NO 11 – h11 = 1380
Draw 10v to po= 1.96 kg/cm
2 & temperature at Ev. Outlet = -5.2
0c.
This line intersects 5 at 11
h12 = h11 + h8 – h9
= 1380 +470 – 260
= 1850 – 260
h12 = 1590 Kj/Kg
q = (f-1) (h2 –h3) = 2.15 (600 – 280)
q = 2.15 (320)
q = 688
q = 688 = f (h1a – h4)
= 3.15 (h1a – 30)
h1a = 248
qo = Ref. Effect = h11 – h10
= 1380 – 260
qo = 1120
qh = Heat added in gen.
= h5 + h2 (f-1) – fh1a
= 1850 +600 x 2.15 – 3.15 x 248
= 1850 +1290 – 781
= 3140 – 781
qh = 2359
4744.0=
=
cop
qh
qocop
qa +qh = Heat given = qo + qh
= 3479
qc = heat rejected in condenser
= h5 – h8
= 1850 – 470
qc = 1380
qA = Heat rejected in Absorber
= h12 – h3 + f(h3 – h4)
= 1590 – 280 +3.15 (280 –30)
= 1310 +3.15 x 250
= 1310 + 787.5
qA = 2097.5
qA + qc = Total heat rejected
= 2097.5 + 1380
qA + qc = 3477.5
qo + qh = Total Heat given
= 1120 + 2359
qo +qh = 3479 KJ/Kg
This qA +qc qo +qh
Heat Balance matches well.
ACTUAL VAPOUR ABSORPTION CYCLE & ITS REPRESENTION ON
ENTHALPY – COMPOSITION DIAGRAM.
Figure 2 shows the schematic arrangement of the actual vapour absorption cycle and
figure 3 represents its thermodynamic cycle on the h - diagram. The system consists of
generator G together with analyzer AN, and condenser C on the high pressure side, and the
evaporator E and Absorber A on the low pressure side. Pump P, extension valve VI, and
pressure reducing valve VII separate the two sides. In addition, liquid-vapour heat exchanger
HEI is also provided.
The vapour ‘s at 5 distilled from the generator analyzer, enter the refrigerant circuit.
The vapour ’s are condensed to 8 in the condenser, pre cooled to 9 in the liquid – vapour
regenerative heat exchanger and throttled to 10 before entering the evaporator. The state 10 is
at the same point as state 9 in the h- diagram as both enthalpy and composition remain the
same before and after throttling.
The refrigerant entering the evaporator at 10, leaving the evaporator at 11, and the
liquid vapour heat exchanger at 12, comprise a liquid plus vapour mixture. The refrigerant as
finally absorbed by the poor solution at 2 returning from the generator and after being cooled
in the liquid – liquid heat exchanger to 3 and throttled to 3a whereas the rich solution from
the absorber at 4 is pumped to 4 a and heated to 1a before entering the analyzer.
PLOTTING STATE POINTS
The state points 1,2,3&4 can be located on the h- diagram according to their
temperatures and compositions as the enthalpy of liquid is independent of pressure. Also
point 3a lies at 3 only (isenthalpic process) and point 4a lies approximately at 4 itself as the
pump works very small. State point 5 of the vapour is along the isothermal tie line drawn
from 1. point 8 is the saturation state at Pk. Point 9 after sub cooling of the liquid can be
plotted according to the temperature and composition and point 10 is at 9 itself (isenthalpic
process). Point 11 is on the tie line corresponding to the evaporator leaving temperature t02 =
t2 and pressure P0.the composition is same at 7,8,9,10,11 and 12. Point 12 can be similarly
located by knowing the temperature from the energy balance of heat exchanger.
The absorber pressure in the absorber system is equal to the evaporator pressure Po &
the generator pressure is similarly equal to condenser pressure Pk.
Hence at a given generator pressure, the poor solution concentration is determined by the
heating temperature th and at a given absorber pressure, the rich solution concentration is
determined by the cooling temp. tA the absorber temperate
Sequence of procedure to calculate the performance of
vapour absorption (3 fluid) refrigerator
1) To find the coefficient of performance – c.o. p. = Heat absorbed
Heat supplied
2) To find the enthalpy balance in the process
HEAT ABSORBED + HEAT SUPPLIED ____ HEAT REJECTED IN
CONDENCER + HEAT REJECTED IN ABSORBER
I) To ascertain these two aspects of performance following data is essential
1) Generator temperature
4) Temperature after condensation
5) Temperature after sub cooling
6) Temperature at evaporator inlet
8) absorber inlet temperature
II) properties of NH3 – ( R 717 – refrigerant )
1) pressure – saturation temperature .table
III) ENTHALPY - H / CONCERTRATION CHART
( h = K - cal / Kg ) (ξ = 0.1 to 1.0)
Procedure to plot the process recurrence on the h ξ chart
And obtain the values of h 1 , h 2 , h 3 , h 4 , h 8 , h5 – h 10 v ,h 11
h 9 ,h 10, h 12 ,h ia
I) obtain the partial pressure of NH3 in condenser and boiler
pressure Pk as the pressure at saturation Temperature = condenser
outlet temperature (T4)
ii) obtain the partial pressure of NH 3 in Evaporator & absorber
Pressure Po as the pressure at saturation Temperature = absorber
inlet Temperature
iii) Mark the point 2 at Pk = NH 3 pressure in condenser and boiler
(generator Temperature T1 ) note ξ ah 2
iv) Mark point 4 at Po = NH3 pressure in evaporator Absorber and
Absorber inlet temp. T8 and note ξ of h4
v) Mark point 3 at Po and ξa & note h3
vi) Mark point 1 at Pk and ξa & note h1
vii) Mark point 8 at Pk and temp. after condensation T4 and
note h8, ξ8
viii) mark point 9 at ξ8 and temp. after sub cooling T5 and
Note h9
iv) Mark point 10 at point 9 and note h10
x) Mark 10 v at ξ = 1 and Pk
xi) join 10 v to 9 and draw a line to meet pressure Po and get the
temp.. at inlet to evaporator and mark 10 L
XII) Join 10v to 11
L at Po and temp at inlet to Evaporator
XIII) Where this line 10v – 10
L interserts the line ξ 8 – 8 is the point 11
and make h11
XIV) Find the specific rich solution rate = f
ar
afξξ
ξξ
−
−= 8
XVI) Find the specific poor solution rate = (f – 1)
XVII) Mark point 5 on Pk at E8 and mark h5
XVIII) Refere the Schematic diagram and find h1a from equation –
afhfhhqh 125 )1( −−+=
( )( ) ( )4132 hhfhhifq a −=−−=
XIX) Heat added in the generator = qh
XX) Heat absorbed in Evaporator = qo
qo = h11 – h10
XX) Total heat supplied = qo + qh
XXI) h12 = h11 + (h8 – h9)
)( 43312 hhfhhqA −+−=XXII)
XXIII) Heat rejected in absorber = qA
XXIV) Heat rejected in condenser = qC
qC = h5 – h8
XII) Total heat rejected = qA + qC
XIII) Compare the total heat supplied
= qo +qh with the total heat rejected = qA +qC and These should be
almost the equality of qo +qh = qA + qC showing the energy +balance
VORTEX TUBE APPARATUS
INTRODUCTION :
George J. Ranque a French metallurgist discovered in 1928, that the temperature of
central and peripharal layers of air in a cyclone dust separator are different, the core having
low temperature. This led Ranque to develop the vortex tube which is often called the
Ranque tube, vortex means fluid under rotation. Further development work was done by
Hilsch from 1933 onwards. In 1946 he published a remarkable means of refrigeration.
The Vortex Tube consists of
a) Nozzle – cum – chamber
b) Hot –end tube fitted with
c) Diaphragm and Cold end.
OPERATION OF VORTEX TUBE :-
Compressed air at about 7 atmospheres of pressure and at room temperature is
admitted into vortex tube. The air attains a higher velocity while passing through the nozzle
and its temperature drops. This air then enters the chamber where vortex is formed. The core
has a higher static temperature than the outer layer and hence heat flows core to periphery.
The outer layer gets heated due to heat transfer as well as due to viscous and wall friction
effects. The length of the tube has a optimum value at which maximum cooling of the core
results. The cold air at the core comes out from the cold end through the diaphragm while hot
air escapes through the valve side. The cold end temperature decreases as the inlet pressure
increases. The temperature drop between hot and cold sides is controlled by the valve
position for a given inlet air pressure.
It has been observed that for favorable cooling effect the diaphragm should be as near
to the nozzle as possible.
APPARATUS
The vortex tube test apparatus consists of a vortex tube with air inlet with pressure
gauge and pressure transmitter . Air inlet is provided with pressure regulator . Hot air outlet is
provided with balance valve for cooling effect .
Two separate rotameters are provided to measure air flowrate of cold and hot air
.Differential pressure transmitter are placed in the air steam for air flow measurement .
Temperature scanner with computer interface are provided at diffetent locations of the
system .
SPECIFICATIONS
1) Vortex Tube : Nominal capacity : 0.25 m3/min
2) Inlet Pressure : 0 –10 Kgf/cm2, Gauge and pressure transmitter.
3) Heat exchanger : Concentric tube , counter flow , heat transfer area
0.03 m2
4) Pressure regulator : controls inlet pressure
5) Flowmeters : To measure cold and hot air mass flow rates .
6) Valves : 1) For isolating and pressure control .
2) For varying the mass ratio.
7) Digital temperature indicator with serial interface .
OBSERVATION TABLE
1. inlet pressure - Kg/Cm2
2. Inlet air temperature, t1 – 0C
Sr. No. Cold air temp 0C
t2
Hot air temp. 0C
t1
Hot air
Manometer
cm
Cold air
Manometer
Cm
1
2
3
CALCULATIONS
1. Mass balance
mi = m1 + m2
Where,
mi = total air flowrate , Kg/Sec.
m1 = Hot air flowrate, Kg/Sec.
m2 = Cold air flowrate, Kg/sec.
2. Energy balance
mihi = m1h1 + m2h2
and HT2 = (1- H) T1
where T1 = t1 – ti and T2 = ti – t2
h1 – hi = CD T1
hi – h2 = CD T2
im
m
TT
T 2
21
1 =∆+∆
∆=
The value of H is generally between 0.3 to 0.5. The C.O.P. of vortex tube is low and is of the
order of 0.15 to 0.20
PERFORMANCE :-
The performance of vortex tube can be studied by plotting air temperature and cooling effect
produced against the cold fraction as shown in fig.
From the observations made we can plot these characteristic as under
Total air = Cold air flow rate( m1 ) + hot air flow rate (m2)
2
1
m
mH =
Cold air and Hot air temperature are to be taken from observations table.
Cooling effect can be calculated as N = m1. CP. ∆t
CONCLUSIONS
1) As the mass fraction increases, cold air temperature increases and also the hot end
temperature.
2) As the cold fraction (H) increases, cooling effect increases initially and reaches an
optimum cooling effect value.
APPLICATIONS OF VORTEX TUBE :-
Where compressed air is available vortex tube can be used as a method of producing
refrigerating effect. No moving parts are present which is an additional plus point. The
applications can be in the following areas –
1) Spot Cooling
2) Suits for foundry persons.
3) Biological specimen preservation.
4) Surgical Aids.
