AIR COOLING IN AUTOMOBILES
USING VORTEX TUBE
REFRIGERATION SYSTEM
B.SREENIVASA KUMAR REDDY
B.Tech., M.Tech(R&A/C) JNTUA College of Engineering,
Anantapur – 515002, Andhra Pradesh, India. sreenivasreddy8@gmail.com
Prof. K.GOVINDARAJULU
M.Tech., Ph.D., F.I.E., M.I.S.T.E., C.E. Professor of Mechanical Engineering Department,
Director of Evaluation, JNTUniversity, Anantapur.
Andhra Pradesh, India. govindjntua@gmail.com
Abstract:
Refrigeration plays an important role in developing countries, primarily for the preservation of food, medicine, and for air conditioning. Conventional refrigeration systems are using Freon as refrigerant. As they are the main cause for depletion of ozone layer, extensive research work is going on alternate refrigeration systems. Vortex tube is a non conventional cooling device, having no moving parts which will produce cold air and hot air from the source of compressed air without affecting the environment. When a high pressure air is tangentially injected into vortex chamber a strong vortex flow will be created which will be split into two air streams, one hot stream and the other is cold stream at its ends.
Keywords: Refrigeration, Vortex Tube Refrigeration, C.O.P, VCRS, R134a 1. Introduction
The system (VCRS) using in automobiles for getting air conditioning, the refrigerant R134a is having ozone depleting potential (ODP) and global warming potential (GWP) and toxic. Leakage of the refrigerant during accidents is also a problem causing deaths due to its inhalation. The alternative cooling system in automobiles is getting by Vortex Tube Refrigeration System. In Vortex Tube Refrigeration System the medium used is compressed air and having zero ODP and GWP. When high-pressure air is tangentially injected into the vortex chamber via the inlet nozzle, a swirling flow is created inside the vortex chamber. In the vortex chamber, part of the gas exists via the cold exhaust directly, and another part called as free vortex swirls to the hot end, where it reverses by the control valve creating a forced vortex moving from the hot end to the cold end. Heat transfer takes place between the free end and the forced vortices there by producing two streams, one hot stream and the other is cold stream at its ends.
To run the system some power from the vehicle engine shaft is diverted by a pulley arrangement to run an air compressor. The high pressure created in the air compressor is used as inlet to a vortex tube which is placed inside the vehicle cabin. The cold air from the vortex tube is used to cool the vehicle cabin and the hot air is exhausted to atmosphere.
2. Analysis of Vortex Tube
Vortex tube gets high pressure air from an air compressor through a tangential nozzle. Assume suffixes i, h, c stands for inlet to the nozzle, hot end and cold end, respectively then the mass and energy conservation of control volume given by
Mass balance mi = mc + mh ....2.1
Steady flow energy balance
mi.hi = mc.hc + mh.hh ....2.2
Assuming the kinetic energies are negligible.
The cold gas temperature difference or the temperature drop of the cold air tube is defined as
ΔTc = Ti−Tc. ....2.3
ΔTh = Th−Ti. ….2.4 If the system is isentropic then the heat lost by the cold stream is equal to heat gained by hot stream. mc (Ti – Tc) = mh (Th – Ti) = (mi - mc).(Th – Ti)
Ti – Tc = [(mi/mc)-1].(Th - Ti) = (
μ-1).(Th – Ti) ....2.5 where μ is the ratio of cold air to the air supplied, called as cold mass fraction.
From equation 2.3 we get.
μ (Ti – Tc) = (1 – μ).(Th – Ti)
μ [(Ti – Tc) + (Th – Ti)] = (Th – Ti)
μ = (Th – Ti)/[(Ti – Tc) + (Th – Ti)] = ∆Th/(∆Th + ∆Tc)
μ= mc/mi ....2.6
If the process had undergone an isentropic expansion from inlet pressure Pi to atmospheric pressure Pa at the cold end then the static temperature drop due to expansion is given by
∆T’c = Ti – T’c = Ti [1-(Pa/Pi)(γ-1)/γ] ....2.7
The temperature drop occurred in vortex tube is ∆T’c. The ratio of ∆Tc to ∆T’cis called Relative
Temperature drop ∆Trel = ∆Tc/∆T’c ....2.8
The product of μ and ∆Trel represents the adiabatic efficiency of the vortex tube because it is defined as
ηab = C
ηab = (mc∆TcCp)/(mi∆T’cCp) = μ∆Trel
ηab = [∆Th/(∆Th + ∆Tc)]*(∆Tc/∆T’c) ...2.9
The C.O.P of the vortex tube is defined as the ratio of the cooling effect to the work input to the air compressor.
Cooling effect = mc∆TcCp
Work input to air compressor = (CpmiTi)[(Pi/Pa)(γ-1)/γ-1]/ηac where ηac is the adiabatic efficiency of the compressor.
C.O.P = CW = [(mc∆TcCp).ηac]/(CpmiTi)[(Pi/Pa)(γ-1)/γ-1]
C.O.P = μ(∆Tc.ηac)/{Ti[(Pi/Pa)(γ-1)/γ-1]} ....2.10 Substituting the value of Ti from equation 2.5 in equation 2.8
{ μ(∆Tc.ηac)}
C.O.P = ____________________________
{(∆T’c/[1-(Pa/Pi)(γ-1)/γ])[(Pi/Pa)(γ-1)/γ-1]}
C.O.P = μ.(∆Tc/∆T’c)ηac.[(Pa/Pi)(γ-1)/γ] ....2.11 Substituting the value of μ (∆Tc/∆T’c), from equation 2.7
C.O.P = ηab. ηac. PP γ
γ
....2.12
3. Experimental setup
The experimental set up is having 1. Air inlet pipe with air filter 2. Vehicle Engine
3. Single cylinder, single h.p Reciprocating Air-Compressor. 4. Vortex tube assembly.
5. Pressure reducing valve with two side connectors along with pressure gauge and 25 micron air filter.
6. Ball valve 7. Belt
8. Air receiver tank ( 40 lts) with safety valve and pressure gauge 9. Air delivery pipe
10. Air hoses
11. Compressor seating plate.
Specifications- Vortex Tube
Model SCFM SLPM Btu/hr. Kcal/hr. SIZE
3240 40 1,133 2,800 706 Medium
One SCFM (Standard Cubic Foot per Minute) equals to 0.02831m3/min One SLPM (Standard Liters per Minute) equals to 0.001 m3/min
One ton of capacity equals 12,000 Btu/h (British thermal unit)
4. Observations & calculations Assumptions:
1. Mass flow rate of air entering into compressor is equal to mass flow rate of air entering to the vortex tube maintaining at constant pressure in receiver tank.
2. Assuming no losses i.e inlet mass flow rate of air is equal to mass flow rate of cold air + mass flow rate of hot air.
Specifications of the reciprocating air – compressor. Compressor H.P = 1
No. of cylinders = 1
Free air displacement = 0.0018 m3/sec Working pressure = 5 bar
Compressor R.P.M = 650 4.1 Adiabatic efficiency of the air-compressor
P1 = atmospheric pressure = 1.01325 bar P2 = delivery pressure = 5 bar Energy input = 745.7 Watts Theoretical volume (V1) = 0.0018 m3/sec
Adiabatic Work Done = γ
γ P1 V1
P P
γ
γ 1
…..4.1
= 1.4 * 1*105*0.0018
.
. 1
= 368 J.
∴ Compressor efficiency (ηab) =
= .
= 50%.
4.2 Cooling load calculations (No. of occupants - 9)
The following assumptions were made in quantifying the amount of heating load on a car.
Heat conducted from the outside through the glass is considered.
Heat conducted through the firewall into the passenger compartment has been neglected. It is assumed that enough insulation is provided to minimize this heat transfer.
The following calculations estimate the heat gain loss via glass, from occupants, and infiltration.
Q = Q + Q +Q …..4.2
4.2.1 Heat gain via glass
The heat gain through an automobile glass can be expressed as the sum of solar heat gain, due to transmitted and absorbed solar radiation and the heat conducted, due to the difference between outdoor and indoor air temperature.
Q = Q + Q …..4.3
Solar radiation varies temporally and spatially.
The heat conducted through the glass depends on the solar heat gain, as the amount of absorbed radiation affects the temperature distribution of the glass.
Heat loss from the car is taken as negative and heat gain is taken as negative. Solar Heat Gain Q
Solar heat gain is obtained by summing direct normal, diffuse horizontal and solar radiation due to ground reflection.
= 0.8 * 875.4 *3.4 = 2381 Watts.
Qsolar_dffuse = Tglass qdiff Aglass(1 + cosS ) …..4.6
= 0.8 * 59 * 3.4 (1 + cos ) (S = slope of glass =71)
= 53 Watts.
Qsolar_ground = gTglass (qdir + qdiff) Aglass(1 - cosS) …..4.7
(g= ground reflectance = 0.2)
= 0.2 *0.8 (2381 + 53) 3.4 (1-cos )
= 447 Watts.
∴ Qsolar = Qsolar_direct+ Qsolar_dffuse + Qsolar_ground = 2381+53+447
= 2881 Watts. Conduction Load (Qcond)
The following assumptions have been made for calculating conduction load.
Glass is in steady state (thermally).
Clear glass with reflectance rglass= 0.07; transmittance τglass = 0.8 and absorptance β= 0.13 (for normal incidence).
Most of the absorption takes place on the outer surface of the glass.
Solar heat gain is obtained by summing direct normal, diffuse horizontal and solar radiation due to ground reflection.
Qcond = hi (TL-Tin) Aglass …..4.8
= 15 (34-23) 3.4 = 561 Watts.
∴ Q = Q + Q …..4.9 = 2881 + 561
= 3442 Watts. 4.2.2 Cooling load due to Infiltration
The heat gain due to infiltration has both sensible and latent components.
Q = Q _ + Q _ …..4.10
1. Sensible load (Q _
Q _ = N . C . .V ∆T …..4.11
= . . 35-23
= 2608 Watts. 2. Latent load (Q _
Q _ = N . L. .V ∆ …..4.12 = . . 1 (note: Humidity (∆) is const.)
= 491 Watts.
Hence, the total load due to infiltration is
Q = Q _ + Q _ = 2608 +491
∴ Q = 3099 Watts.
4.2.3 Cooling Load due to Occupants
components given by
QOCC = Q _ + Q _ …..4.13
Q _ = Q _ * Nocc
= 65 * 9 persons = 595 Watts
Q _ = Q _ * Nocc = 30 * 9 persons
= 270 Watts.
∴ QOCC = 595 + 270 = 865 Watts. Total Generated Heat Load (Qload):
Qload = Qglass + Qocc + Qinf
Qload = Qsolar + Qcond + Qocc + Qinf
Qload = Qsolar_direct+ Qsolar_diffuse+Qsolar_ground+Qcond + Qocc + Qinf
= 2381 + 53 + 447 + 561 + 865 + 3099
= 7406 Watts.
4.3 Coefficient of performance of vortex tube (c.o.p) Observations:
Atmospheric pressure (Pa) = 1.01325 bar Inlet pressure of air (Pi) = 2 bar Inlet temperature of air(Tin) = 300C Cold air exit temperature(Tc) =230C Hot air exit temperature (Th) =500C Calculations:
1. Cold drop temperature (∆Tc) = Tin-Tc = 30-23 = 70C 2. Hot raise temperature (∆Th) = Th-Tin
= 50-30 = 200C
3. Temperature drop at the two ends (∆T) = Th–Tc = 50-23 = 170C
4. Cold mass fraction (μ) =
=
= 0.74
(74% of cold air is coming from cold side & 26% of hot air is from hot side) 5. Static temperature drop due to expansion (∆T’c) = Ti(1-(Pa/Pi)(γ-1)/γ)
= 30 (1-(1/2))(1.4-1)/1.4) =5.40C.
6. Relative temperature drop (∆Trel) = (∆Tc/∆T’c)
= .
= 1.3
7. Vortex tube adiabatic efficiency (ηab) =
8. Adiabatic compressor efficiency (ηac) = 0.5
∴Coefficient of Performance (C.O.P) = ηab. ηac.[(Pa/Pi)(γ-1)/γ] = 0.5*0.96 [(1/2)(1.4-1)1.4] = 0.34
4.4 Cooling capacity of vortex tube Observations:
Velocity of air (v) = 350 m/sec.
Area of the cold air outlet (a) = π/4 d2 = π/4 * 0.012 Calculations:
Air mass flow rate (m) = a*a * v (where a= density of air = 1.225 kg/m3.)
= 1.225 * 7.85 * 10-5 * 350
= 0.033 kg/sec.
∴ Cooling Capacity of Vortex Tube = mCp(TH-TC)
= 0.33 * 1*103* (50-23)
= 900 Watts. = 0.9 kW.
= ¼ Ton of refrigeration.
∴ Cooling Capacity of Vortex Tube = ¼ Ton of refrigeration. 4.5 Energy consumption (per Ton of Refrigeration)
Total energy developed i.e Brake Power = 69h.p (TATA –SUMO)
For ¼ Ton of refrigeration the power needed to run the compressor is 1 h.p (Compressor Capacity is 1 h.p )
So per Ton of refrigeration the power needed = 4 h.p
So the consumption energy per Ton of refrigeration is 4 h.p from the developed Brake power (69h.p) by vehicle.
∴ Energy consumption per Ton of refrigeration is 4 h.p 5.Conclusions:
By installing the Vortex Tube Refrigeration System in passenger car to get the required cooling conditions, the following conclusions are observed.
1. The present project work has successfully developed air cooling for a passenger car based on Vortex Tube Refrigeration System, which overcomes the major problems such as global warming and ozone depletion.
2. In Vortex Tube Refrigeration System air is being used in place of refrigerant is of having zero ozone depletion, zero global warming potential, non pollutant, non toxic, inflammable, cheaper and eco friendly to the environment.
3. Cooling conditions met by the Vortex Refrigeration System is healthier than the comfort conditions met by the VCRS system, because air is free from chemicals but refrigerant is a gas which is a chemical mixture.
4. Energy consumption per ton of refrigeration in Vortex Refrigeration System is 4 h.p, whereas energy consumption per ton of refrigeration to run Vapour Compression Refrigeration System in vehicle is 6 h.p. So Vortex Refrigeration System is economical.
5. Vortex Tube Refrigeration System is of having no moving parts, so its durability is more. Installing and maintaining is easier than the current air conditioning system.
6. Performance of the Cooling system is good at high speeds of the vehicle as high pressures are created in the Air compressor. Cooling system works well in the vehicles with less cabin space as the C.O.P of the vortex tube is less.
6.Scope for future work
By installing humidifier along with the Vortex Tube Refrigeration System, the required comfort conditions i.e. temperature & humidity will be maintained in the vehicle cabin.
References
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[4] C.W.Bullard and T. Malik “Literature on heat load calculations” prepared at Air Conditioning and Refrigeration Center, University of Illinois, Mechanical & Industrial Engineering Dept. 1206 West Green Street Urbana, IL.
[5] PRABAKARAN.J, “Effect of Diameter of Orifice and Nozzle on the performance of Counter flow Vortex tube”, IJEST Journal, 2010. [6] J. Prabakaran, “Effect of orifice and pressure of counter flow vortex tube”, IJST, April 2010.
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