Scaling Interpretation for 2-D Continuity PDE: What Does UH/L Represent?

[member 428835]

Hi PF!

I'm doing some scaling over a PDE and I understand the math side of things but I do not understand the physical side of what we are finding.

For example, suppose we have some PDE, say 2-D continuity for it's simplicity $u_x + v_y = 0$. Let $L$ be the length of a side of a flowing channel and the height of the channel be $H$. Now if $x$ scales as $L$ and $y$ scales as $H$ and if the mainstream velocity coming into the channel is $U$ then we may write $U H/L \sim v$. What is actually being said here? That the vertical velocity is maximized as $U H/L$? Please help!

Thanks a ton!

 

[Chestermiller]

It seems to be saying that the typical y velocity will be smaller than the x velocity by a factor on the order of H/L.

Chet

 

[member 428835]

What do you mean by "typical"?

They also use this technique with a time derivative as well, like in the momentum equation. My professor has said that this technique can save a lot of math, like if you are are doing a force balance for someone jumping out of an airplane, and you are concerned with initial velocity, this technique can eliminate "less important" terms.

 

[Chestermiller]

What do you mean by "typical"?

Maybe typical was a poor choice of term. Maybe it would have been better to say that the y velocities will be on the order of H/L times smaller than the x velocities. It is difficult to be more precise with something like this.

They also use this technique with a time derivative as well, like in the momentum equation. My professor has said that this technique can save a lot of math, like if you are are doing a force balance for someone jumping out of an airplane, and you are concerned with initial velocity, this technique can eliminate "less important" terms.

I don't follow what you are saying here. The way I learned dimensional analysis was taught to me by S. W. Churchill at the University of Michigan in 1963. See the famous paper by Hellums and Churchill in AIChE Journal (1964)

Chet

 

[member 428835]

Thanks, I'll look into it! You're awesome Chet!