An Efficient Vortex Tube with Max C.O.P

An Efficient Vortex Tube with Max C.O.P

S. Kasim Vali1*, K. Prabhakar2, C. Mohan Naidu3, B. Pavan Bharadwaja4

1. M.Tech Student Dept. Of Mechanical Engineering Gates Institute of technology, Gooty, A.p, India.

2. Assistant professor, Dept. Of Mechanical Engineering Gates Institute of technology, Gooty, A.p, India

3. Assistant professor, Dept. Of Mechanical Engineering Gates Institute of technology, Gooty, A.p, India

4. M.Tech student,Dept. Of Mechanical Engineering Amrita school of engineering ,Coimbatore,T.N,India

1. INTRODUCTION:

   The vortex tube, also known as the Ranque-Hilsch vortex tube (RHVT) is a device which generates separated flows of cold and hot gases from a single compressed gas source. The vortex tube was invented quite by accident in 1931 by George Ranque, a French physics student, while experimenting with a vortex-type pump that he had developed, and then he noticed warm air exhausting from one end, and cold air from the other. Ranque soon forgot about his pump and started a small firm to exploit the commercial potential for this strange device that produced hot and cold air with no moving parts. However, it soon failed and the vortex tube slipped into obscurity until 1945 when Rudolph Hilsch, a German physicist, published a widely read scientific paper on the device.[3]

   Much earlier, the great nineteenth century physicist, James Clerk Maxwell postulated that since heat involves the movement of molecules, we might someday be able to get hot and cold air from the same device with the help of a "friendly little demon" who would sort out and separate the hot and cold molecules of air.[4]

   Thus, the vortex tube has been variously known as the "Ranque Vortex Tube", the "Hilsch Tube", the "Ranque-

   Hilsch Tube", and "Maxwells Demon". By any name, it has in recent years gained acceptance as a simple, reliable and low cost answer to a wide variety of industrial spot cooling problems.

   When high-pressure gas 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.[9]

2. COMPONENTS OF VORTEX TUBE:

   The systematic diameter gram of vortex tube is shown in the figure.

   Fig:1.2 Vortex Tube.

   It consists of the following parts.

   1. Nozzle.

   2. Diaphragms.

   3. Control valve.

   4. Hot air side.

   5. Cold air side

   The actual diameter gram of vortex tube is shown below.

   Fig: 1.2.1 Sectional view of Vortex Tube.

3. WORKING OF VORTEX TUBE:

A compressed air is passed through the nozzle as shown in figure. Here air expands and acquires high velocity due to particular shape of the nozzle. A vortex flow is created in the chamber and air travels in spiral motion along the periphery of the hot side. Then, the rotating air is forced down the inner walls of the hot tube at speeds reaching 1,000,000 rpm.

The control valve restricts this flow. When the pressure of the air near the valve is made more than the outside by partly closing the valve, a reversed axial flow through the core of the hot side starts

from high-pressure region. During this process, energy transfer takes place between reversed stream and forward stream and therefore air stream through the core gets cooled below the inlet temperature of the air in the vortex tube while the air stream in forward direction gets heated. The cold stream is escaped through the diaphragms hole into the cold side, while hot stream is passed through the opening of the control valve. By controlling the opening of the valve, the quantity of the cold air and its temperature can be varied.[6]

Fig:1.3 3D Sectional view of Vortex tube.

1. 2.1 ASSUMPTIONS:

   1. Mass flow rate of air entering into compressor is equal to mass flow rate of air entering in to the vortex tube maintaining at constant pressure in receiver tank.[2]

   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. [5]

1. ADIABATIC EFFICIENCY OF AIR COMPRESSOR:

   The coefficient of performance of the vortex tube is the product of adiabatic efficiency of the air compressor and adiabatic efficiency of the vortex tube and [(Pa/Pi)(-1)/], the adiabatic efficiency of the Air compressor is to be calculated first.[3]

2. SPECIFICATIONS OF THE AIR COMPRESSOR USED :

   Compressor H.P = 3

   No of cylinders = 2

   Diameter of two L.P cylinders = 70 mm

   H.P cylinders = 50 mm

   Stroke length = 85 mm

   Number of stages = 2 Coefficient of discharge = 0.62 Working pressure = 120 lbs

   Orifice diameter = 20 mm

   S.No	Pressure Pi, bar	Cold Temp (Tc),0C	Hot Temp (Th), 0C	Difference T= Th- Tc 0C	Cold Temp Drop Tc, 0C	Hot Temp Drop Th, 0C	Cold mass Fraction	Adiabatic Efficiency	COP
   1	2	23	36	13	7	6	0.4615	0.6099	0.0798
   2	3	21	38	17	9	8	0.4705	0.5294	0.0967
   3	4	21	39	18	9	9	0.5	0.4622	0.1022
   4	5	20	40	20	10	10	0.5	0.4550	0.1111
   5	6	18	42	24	12	12	0.5	0.5020	0.1311

   FOR CYLINDRICAL TUBE:

   Table:2.1

   2.2 AIR COMPRESSOR ADIABATIC EFFICIENCY CALCULATION:

   CALCULATIONS:

   P1atm pressure = 1 bar P2 delivery pressure = 6 bar

   Energy input =

   3600 x no .of revolutions of energy meter time taken in sec x 100 x0.8

   =

   3600 x10

   87.28x100x0.8

   =

   5.156 kW

   Theoretical volume V1 =

   x d x d x L x 2 x N

   4 x 60 =

   x 0.07 x 0.07 x 0.085 x 2 x 886

   4 x 60

   =0.0096m3/sec

   1

   Adiabatic Work Done = P

   1

   V1[(p2/p1)(-1/)-1]

   = 1.4 * 1*105 *0.0096 [(6/1)(0.4/1.4) – 1]/0.4

   = 2239.02 J

   Adiabatic Efficiency of a Compressor

   – = (Adiabatic

   Work Done)/(Energy Input )

   =2239.02/(5.156*1000)

   Adiabatic Efficiency of a Compressor

   =0.4342 43%

   2.2SPECIMEN CALCULATIONS:

   SPECIMEN CALCULATIONS FOR THE INLET PRESSURE OF AIR

   Pi = 6 bar

   1. Static Temperature Drop Due To Expansion

      Tc = Ti Tc = Ti[-(Pa/Pi)(-1)/]

      OBSERVATIONS:

      1. Atmospheric pressure Pa

         = 1 bar.

         ( 1/6 ) {1.4-1}/1.4]

         12.010C

         Tc = 30[1-

         =

      2. Inlet pressure of air Pi

         = 6 bar

      3. Inlet temperature of air Ti

         = 300C

      4. Cold air exit temperature Tc

         = 180C

      5. Hot air exit temperature Th = 420C

      6. Relative Temperature Drop (

   Trel) = Tc/( Tc)

1. GRAPHS

After conducting the experiment we noted the tabulated results and the following graphs are plotted.

1. COLD END TEMPERATURE VARIATION AT DIFFERENT PRESSURES

   CALCULATIONS:

   1. Cold drop temperature Tc = Ti

      – Tc

      Tc

      = 30 18 = 120C

   2. Hot raise temperature Th = Th

      Ti

      Th = 42 30 = 120C

   3. Temperature Drop at the two

      T 25

      E

      M 20

      P

      E 15

      R

      A 10

      T

      U 5

      R

      E 0

      2 3 4 5 6

      pressure in bar

      cyl in d

      ends T = Th – Tc

      42 18 = 240C

   4. Cold mass fraction =

      +

      T =

      Fig..3.1 COLD END TEMPERATURE VARIATION AT DIFFERENT PRESSURES

      From the above Fig. 6.3.1 it is clear that at any given pressure the temperature of the conical hot tube is better when compared to cylindrical hot tube and the temperature difference between them is proportional to pressure i.e., the temperature difference is increasing progressively with pressure.

      = = 0.5

      12

      12+12

      conical

      0.3

      0.25

      0.2

      0.15

      0.1

      0.05

      0

      S.N o	Pressu re in Bar	COP of cylind er hot tube	COP of conic al hot tube	% increa se in COP of conical hot tube
      1	2	0.0798	0.108 5	35.96
      2	3	0.0967	0.128 9	33.29
      3	4	0.1022	0.119 6	17.25
      4	5	0.1111	0.133 0	19.71
      5	6	0.1311	0.154 0	17.46

2. HOT END TEMPERATURE VARIATION AT DIFFERENT PRESSURES

3. COP VARIATION AT DIFFERENT PRESSURES

pressure in bar

50

40

30

20

10

0

2 3 4 5 6

cylind er

conica l

5 6

4

3

cylindric

al

2

Fig..3.3 COP VARIATION AT DIFFERENT PRESSURES

The above Fig. 3.3 is plotted for pressure V/s COP. From the graph it is noted that the COP of the vortex tube with conical hot end is higher than the vortex tube with cylindrical hot tube.

From the above three graphs it is noted that the performance of the vortex tube with conical hot tube is better than the vortex tube with cylindrical hot tube.

RESULTS:

After evaluating the performance of vortex tube with cylindrical and conical hot tubes it was found that the vortex tube with conical hot tube gives the better performance than the cylindrical hot tube

i.e. there is an increase in cop of about 25%-30%. The COP values obtained for

Fig.3.1 HOT END TEMPERATURE VARIATION AT DIFFERENTPRESSURES

From the above Fig. 6.3.1 the temperature of hot end for conical hot tube is more compared to cylindrical hot tube .from this we can say that temperature difference between them is proportional to pressure i.e., the temperature difference is increasing progressively with pressure.

cylindrical and conical hot tubes at various pressures are:

DISCUSSIONS:

The performance of the vortex tube was evaluated by conducting the experiment by replacing the cylindrical hot tube with a conical hot tube at various inlet pressures

The other parameters like orifice diameter, nozzle is kept unchanged. The

highest COP is obtained at 6bar for taper tube and the value is 0.1540.

The lowest cold temperature for vortex tube with conical hot tube is 14C at 6 bar and with cylindrical hot tube is 18C at 6 bar.

The highest hot temperature for vortex tube with conical hot tube is 44C at 6 bar and with cylindrical hot tube is 42C at 6 bar.

Cold mass fraction obtained is better for the vortex tube with the conical hot tube than the cylindrical hot tube see table 6.2.3.1 & table 6.2.5.1

The maximum of 30C difference between hot and cold ends temperature for vortex tube with the conical hot tube and maximum of 24C difference between hot and cold ends temperature for vortex tube with the cylindrical tube is obtained.

[1].FikretKocabas , Modeling of heating and cooling performance of counter flow type vortex tube by using artificial neural network, Elsevier Journal, February 2010.

[2].K. Dincer , Experimental investigation and exergy analysis of the performance of a counter flow Ranque-Hilsch vortex tube with regard to nozzle cross-section areas, Elsevier Journal, April 2010.

[3].Prabakaran.J , Effect of Diameter of Orifice and Nozzle on the performance of Counter flow Vortex tube, IJEST Journal, 2010.

[4].J. Prabakaran, Effect of orifice and pressure of counter flow vortex tube, IJST, April 2010.

[5].Nader Pourmahmoud, The effect of L/D ratio on the Temperature separation in the Counter flow Vortex tube, IJRRAS Journal, January 2011.

[6].ChengmingGao, Experimental study on the Ranque-Hilsch Vortex Ttube, PhD Thesis.

[7].S.C Arora and S. Domkundwar, A course in refrigeration and air conditioning, DhanapatRai& Sons Publications.

[8]. www.ExAir.Com.

[10]. www.P.A. Hilton.com

[11]. www. Air Tx international.com

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