Understanding Reynolds Number: Flow around a Cylinder

C.E
Messages
100
Reaction score
0
Could somebody please clarify a few things about Reynolds number for me? In class the lecturer approximated |u. \nabla |u by (u^2)/L where L is the characteristic length. In a specific example considering flow around a cylinder of diameter D, it was said that L=D.

What does the characteristic length actually represent and how do you know what it is in different situations? Also, in the specific example of flow around a cylinder why is |u. \nabla |u approximately (u^2)/L?
 
Physics news on Phys.org
In the case of cylinders, the characteristic length is diameter. In the case of an airfoil, it is cord length. And for a pipe it is the diameter of the pipe. So you have to talk about the Reynolds number for each. The Reynolds number for each will be different.
 
How do you know what the characteristic length is in a particular case? Is it selected so that |u. \nabla |u is approximately (u^2)/L? In the case of a cylinder of diameter D I cannot see why |u. \nabla |u is approximately (u^2)/D.
 
C.E said:
How do you know what the characteristic length is in a particular case? Is it selected so that |u. \nabla |u is approximately (u^2)/L? In the case of a cylinder of diameter D I cannot see why |u. \nabla |u is approximately (u^2)/D.

I had the same question during a fluids class. It's a matter of convention. The characteristic length of a cylinder could have been chosen as the circumference, or the the radius and the Reynolds number curves would be different by a factor of 2 and just as correct. You just have to know the convention chosen.
 
I think I understand it a bit more now, could the characteristic length be chosen to be any length which is uniquely determined by the cylinder?
If in the case of a cylinder you select radius instead of the diameter would it then be that u. \nabla |u is approximately equal to (u^2)/2L or would we still approximate |u. \nabla |u by (u^2)/2?

One thing I still do not understand is why for a cylinder |u. \nabla |u is approximately equal to (u^2)/D. How could I verify this mathematically?
 
Last edited:
I don't follow.

To use latex on this forum enclose your equation in [ tex ] ... [ / tex ] without the spaces.
You can go here, https://www.physicsforums.com/showthread.php?t=8997" and click on the examples. A window pop up widow will display the generating text.
 
Last edited by a moderator:
Thanks for that, I have now corrected my previous post.
 
I am not familiar with the approximation you are talking about or else I am just having a brain fart about it. I know that you can show that:

$\nabla \frac{u^2}{2}=\frac{\nabla u \bullet u}{2}+\frac{u \bullet \nabla u}{2}= \left(u \bullet \nabla \right)u$

As for the length scale of the Reynolds number, in some ways it is picked by convention, but it is slightly more involved than that. What it really comes down to is dimensional analysis and the Buckingham Pi Theorem. When the equations are nondimensionalized, each dimensional variable is nondimensionalized by some characteristic value. For example, on an airfoil the x, y, and z variables are often nondimensionalized with the chord length, so when the nondimensionalization is carried out, the characteristic length that shows up in the Reynolds number is c, the chord length. The same thing goes with the aforementioned examples (cylinders and D, pipes and D, etc.). Occasionally the quantities are chosen relatively arbitrarily, but that doesn't happen very often. Usually, you just need to remember what the characteristic length is for a given situation as it is much easier than nondimensionalizing every time you need to know.
 
C.E said:
I think I understand it a bit more now, could the characteristic length be chosen to be any length which is uniquely determined by the cylinder?
If in the case of a cylinder you select radius instead of the diameter would it then be that u. \nabla |u is approximately equal to (u^2)/2L or would we still approximate |u. \nabla |u by (u^2)/2?

One thing I still do not understand is why for a cylinder |u. \nabla |u is approximately equal to (u^2)/D. How could I verify this mathematically?

OK. I still don't know what the vertical bars mean, but I don't think your units match on both sides of your equation. If this is the case, then there no relationship, but an artifact of the units you're using. For instance, if you were to express you velocity in kilometers per second, and your diameter in kilometers any approximate equality would disappear.

By the way, the Latex interpreter has a bug. If you modify your equation, it may not change. After hitting the Preview Changes button, allow the operation to complete then hit the Refresh or Resend tool on your browser. The Preview text should then be corrected.
 
Last edited:
  • #10
I meant to say that |(u. \nabla )u| is approximately (u^2)/D. Here the vertical lines just mean absolute value of.
 
  • #11
C.E said:
I meant to say that |(u. \nabla )u| is approximately (u^2)/D. Here the vertical lines just mean absolute value of.

OK. Convert to kilometers as your unit of distance instead of meters. If you are using English units, convert feet to miles. What values do you have on the right hand side and the left hand side of your equation? It's not that hard. Just plug in some factors of 1000, or 5280.
 
  • #12
I can see that if I change the distance by a scale factor s then both the right and left hand sides change by a factor (1/s)^2 (assuming the units of time remain the same).
 
  • #13
Is what I written in my previous post correct? If so, how does it help?

I have had to submit this as a new reply as for some reason it won't let me edit my previous post.
 
  • #14
The right and left side of what? I am not 100% sure what you are referring to.
 
  • #15
Left hand side: |(u. \nabla )u|
Right hand side: (u^2)/D (where D is the diameter of the cylinder).

I can now see that if I change the units of distance by a scale factor s then the left hand side changes by a factor of (1/s)^2 and the right actually changes by a factor of (1/s). How does this help?
 
Last edited:
  • #16
You must be doing something wrong. If you change the scale factor D by a factor s, then the left hand side would change by 1/s. Unless maybe I am missing something.
 
  • #17
C.E said:
Left hand side: |(u. \nabla )u|

Do you have a definition for your U\cdot\nabla operator?
 
Back
Top