Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Need help with Fluid Dynamics Problem

  1. Apr 27, 2005 #1
    Hi everyone

    I have a relatively simple fluid dynamics problem I need some help with:

    We have two cylinders L and R of crossectional areas A and 2A respectively. Initially, the level of water in X is H and Y is empty. At t = 0, the two cylinders are joined at the bottom by a tube of cross-section a (after a hole of the same cross-section is opened in each cylinder). Find the time at which the water level is equal in both cylinders.

    This is what I've done so far.

    Denote the water levels in the left (L) and right (R) cylinders by [itex]y_{L}[/itex] and [itex]y_{R}[/itex] respectively. So [itex]y_{L}(t=0) = H[/itex] and [itex]y_{R}(t=0) = 0[/itex].

    Mass conservation (or continuity equation) leads to [itex]AH = Ay_{L} + 2Ay_{R}[/itex] or equivalently [itex]H = y_{L} + 2y_{R}[/itex]. This gives [itex]0 = \dot{y_{L}} + 2\dot{y_{R}}[/itex].

    Applying Bernoulli's Theorem to two points at the surface of each meniscus, we get

    [itex]P_{atm} + \frac{1}{2}{\rho v_{1}^2} + \rho g h_{1} = P_{atm} + \frac{1}{2}{\rho v_{2}^2} + \rho g h_{2}[/itex]

    where [itex]P_{atm}[/itex] is the atmospheric pressure, [itex]\rho[/itex] is the density of water, [itex]v_{1} = -\dot{y_{L}}[/itex], [itex]v_{2} = \dot{y_{R}}[/itex], [itex]h_{1} = y_{L}[/itex], [itex]h_{2} = y_{R}[/itex].


    [itex]\frac{1}{2}{\rho \dot{y_{L}}^2} + \rho g y_{L} = \frac{1}{2}{\rho \dot{y_{R}}^2} + \rho g y_{R}[/itex]

    After simplifying a bit, this gives a differential equation in [itex]y_{R}[/itex] with the boundary conditions [itex]y_{R}(t=0) = 0[/itex] and [itex]y_{R}(t = T) = \frac{H}{3}[/itex] where H/3 is the equilibrium height of water level (in each cylinder--this follows from the mass conservation equation above) and T is the time when this happens.

    Now, if you try this out you get an imaginary (and therefore ridiculous) solution for [itex]y_R[/itex]. As far as I think, my algebra is okay so there must be a conceptual fault somewhere. I would be very grateful if someone could offer some advice.

    PLEASE NOTE: This is not a homework problem but I couldn't think of a better place to post it on PF.

    Thanks and cheers,

    EDIT: The answer should involve [itex]a[/itex], the cross-section of the orifice, but it doesn't in my case.
    Last edited: Apr 27, 2005
  2. jcsd
  3. Apr 27, 2005 #2
    Hi again

    I checked my TeX code (it seems okay to me) but most of it isn't working...


    EDIT: Its working now :-D
    Last edited: Apr 27, 2005
  4. Apr 27, 2005 #3
    I figured it out thanks.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook