A Sticky Physics Problem

As I often do when I’m too busy to write anything strenuous I thought I’d post something from my back catalogue of physics problems. I don’t remember where this one comes from but I think you’ll find it interesting…

Oil of viscosity η and density ρ flows downhill in a flat shallow channel of width w which is sloped at an angle θ. The oil is everywhere of the same depth, d, where d<<w. The effect of viscosity on the side walls can be assumed to be negligible.

If x is a coordinate that represents the vertical position within the flow (i.e. x=0 at the bottom and x=d at the top), write down a differential equation for the velocity within the flow  v(x) as a function of x. Use physical arguments to derive appropriate boundary conditions at x=0 and x=d and use these to solve the equation, thereby determining an explicit form for v(x). Hence determine the volume flow rate in terms of η, ρ, θ, d and w as well as the acceleration due to gravity, g.

As usual, answers through the comments box please!

 

Advertisements

7 Responses to “A Sticky Physics Problem”

  1. Way over my head. More interesting to me: Say Mars was 36 million miles from earth. How long would it take to travel there on a spaceship accelerating at 1g and then turning around at at the half-way point and decelerating also at 1g?

  2. Is this happening on Jupiter?

    • telescoper Says:

      I doubt if oil would flow on Jupiter…

      • Themos Says:

        I must have completely missed the last bit “as well as the acceleration due to gravity, g”, sorry! Isn’t this a case where a dimensional argument gives you the answer?

      • telescoper Says:

        Only up to dimensionless factors (like, e.g., 1/3)…

  3. Chris Brunt Says:

    I have a million things to do, therefore I am doing this instead, though I can’t do maths symbols or Greek symbols.

    Using big D derivative, with del dot v = 0, assuming x is perpendicular to the inclined plane and nu=constant –

    0 = g sin(theta) + nu d^2v/dx^2

    v = – (g sin(theta) / 2nu) x^2 + c_1 x + c_0

    Assuming the flow is stuck on the bottom (v = 0 at x = 0) and there is no shear at the air boundary (dv/dx = 0 at x = d)

    v = (g sin(theta) / 2nu) ( 2xd – x^2 )

    (fastest at the top)

    The flow rate is integral of this over x, times width w

    = (g sin(theta) / 3 nu) d^3 w

    This looks right dimensionally (m^3 s^-1)

    • Chris Brunt Says:

      What I’ve called nu is your eta over rho, unless you meant kinematic viscosity, in which case my nu is your eta.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: