# Laminarflow in a two-dimensional passage

1. Oct 31, 2009

### lizzyb

Exact Question from Book
For laminar flow in a two-dimensional passage, find the relation between the average and maximum velocities.

Relevant Equations, etc
I think for circular pipes we have:

u = u_max - kr^2 (1)

u: velocity
u_max: maximum velocity at center of the pipe

Work Done So Far
I played around with multiplying (1) by dA = 2 pi r dr and integrating but didn't really achieve much.

I'm not sure what they mean by "two-dimensional passage" - do they mean a square or rectangle?

2. Nov 1, 2009

### viscousflow

If its two dimensional its cross sectional shape doesn't matter right?

Set up a two dimensional flow with uniform flow at the entrance and and a fully developed profile at the exit of your control volume.

Use conservation of momentum to find a relation between the velocities.

3. Nov 1, 2009

### lizzyb

like this?
http://img691.imageshack.us/img691/3605/21166978.jpg [Broken]

how is this so different than the parabola flow we're used to seeing? because its not a 3-d tube?

so u_max is at the center and the median of a triangle is 2/3 the height;

u = 2/3 u_max.

Last edited by a moderator: May 4, 2017
4. Nov 1, 2009

### viscousflow

No a fully developed profile in a 2d passage will be parabolic.

5. Nov 1, 2009

### viscousflow

Hint: the answer is $$u_{max} = 2u_{avg}$$