Reynolds number in 2D simulation

Wotild · November 5, 2010, 05:53

First of all I am new here, so I'd like to say "hello" to all users. So: HELLO!!!

I am trying to get familiar with simulationg flows in ANSYS CFX, but my question concern some general physics problem:
How do I fdefine the Reynolds number in an 2D case, so I can compare it to the one dafined for a circular pipe?
As in CFX it is not possible to make a real 2D run, I at leat need to nodes depth, my idea was to make $Re = \frac{4 \cdot A}{\text{perimeter}}$ since my 2D case has a depth. But the front and rear walls are symmetric, so the area is infinite?
Or do I just use the hight/length of my grid as the characteristic lenght in the Reynolds number?

I used the search but none seems to have a similar problem. So, thank you all for your help!

Geetz

mettler · November 5, 2010, 10:37

I don't think you can use that to define the Reynold's number. Reynolds is the ratio of two forces. You might want to try

Re = (mdot*4)/(mu*pi*D)

Wotild · November 5, 2010, 12:35

Hey mettler,
thanks for the equation. But am I right that D ist meant to be a diameter? Because that is the problem I am dealing with, I don't habe a circular pipe.
The Re-equation I gave I found in literature (D. Surek, Angewandte Strömungsmechanik) and in some other books also, so it is not made up by myself.
But because the Re is a relation between two forces I must be able to choose a length on which the turbulence is growing. In case of a 2D layer it is the length of the layer an in case of a surrounded cylinder it should be the diameter f the cilinder, or not?

mettler · November 5, 2010, 13:19

will the hydraulic diameter work?

Wotild · November 5, 2010, 13:55

It won't, cause it is 2 dimensional and i don't think I can use the depth of the grid.
But I have the clue I made some mistake in mind. I was trying to make Re equal in every case by choosing the right length. But of course I have to take the length of the layer in case 1 and the diameter of the cylinder in case 2. Than I have to choose the velocity, so both Re are equal. That I am able to compare both flows.
I think I try this one. Thanks for your help!

mettler · November 5, 2010, 14:53

maybe you should state exactly what you are trying to do. Are you trying to do a 2-d simulation of flow in a pipe? If you are doing that, and your 2-d plane is the centerline, you would use the pipe diameter in the Re number.

Wotild · November 8, 2010, 04:19

I simulate a layer flow of infinite depth. No circular pipe. So there is no diameter. The pic attatched shows my boundries. I will use the x-length as my Re-length, because this is the way the turbulence flow accures.

Wotild · November 10, 2010, 09:15

Does anyone know, why ANSYS CFX defines the Reynolds number with [LaTeX Error: Syntax error]?
This leads to very strange values for y+.

Greetings.

November 5, 2010, 05:53	Reynolds number in 2D simulation	#1
Wotild New Member Witold Join Date: Nov 2010 Posts: 13 Rep Power: 16	First of all I am new here, so I'd like to say "hello" to all users. So: HELLO!!! I am trying to get familiar with simulationg flows in ANSYS CFX, but my question concern some general physics problem: How do I fdefine the Reynolds number in an 2D case, so I can compare it to the one dafined for a circular pipe? As in CFX it is not possible to make a real 2D run, I at leat need to nodes depth, my idea was to make $Re = \frac{4 \cdot A}{\text{perimeter}}$ since my 2D case has a depth. But the front and rear walls are symmetric, so the area is infinite? Or do I just use the hight/length of my grid as the characteristic lenght in the Reynolds number? I used the search but none seems to have a similar problem. So, thank you all for your help! Geetz

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November 5, 2010, 10:37		#2
mettler Senior Member Andrew Join Date: Mar 2009 Location: Washington, DC Posts: 211 Rep Power: 18	I don't think you can use that to define the Reynold's number. Reynolds is the ratio of two forces. You might want to try Re = (mdot4)/(mupi*D)

November 5, 2010, 12:35		#3
Wotild New Member Witold Join Date: Nov 2010 Posts: 13 Rep Power: 16	Hey mettler, thanks for the equation. But am I right that D ist meant to be a diameter? Because that is the problem I am dealing with, I don't habe a circular pipe. The Re-equation I gave I found in literature (D. Surek, Angewandte Strömungsmechanik) and in some other books also, so it is not made up by myself. But because the Re is a relation between two forces I must be able to choose a length on which the turbulence is growing. In case of a 2D layer it is the length of the layer an in case of a surrounded cylinder it should be the diameter f the cilinder, or not?

November 5, 2010, 13:19		#4
mettler Senior Member Andrew Join Date: Mar 2009 Location: Washington, DC Posts: 211 Rep Power: 18	will the hydraulic diameter work?

November 5, 2010, 13:55		#5
Wotild New Member Witold Join Date: Nov 2010 Posts: 13 Rep Power: 16	It won't, cause it is 2 dimensional and i don't think I can use the depth of the grid. But I have the clue I made some mistake in mind. I was trying to make Re equal in every case by choosing the right length. But of course I have to take the length of the layer in case 1 and the diameter of the cylinder in case 2. Than I have to choose the velocity, so both Re are equal. That I am able to compare both flows. I think I try this one. Thanks for your help!

November 5, 2010, 14:53		#6
mettler Senior Member Andrew Join Date: Mar 2009 Location: Washington, DC Posts: 211 Rep Power: 18	maybe you should state exactly what you are trying to do. Are you trying to do a 2-d simulation of flow in a pipe? If you are doing that, and your 2-d plane is the centerline, you would use the pipe diameter in the Re number.

November 10, 2010, 09:15		#8
Wotild New Member Witold Join Date: Nov 2010 Posts: 13 Rep Power: 16	Does anyone know, why ANSYS CFX defines the Reynolds number with [LaTeX Error: Syntax error]? This leads to very strange values for y+. Greetings.