1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Volume Flux for a Hydraulic Jump

  1. Sep 12, 2017 #1
    1. The problem statement, all variables and given/known data
    Problem is given in this image,
    https://gyazo.com/454370ff9549dcd7c53604ebfe5df105

    2. Relevant equations

    Continuity or conservation of mass equation:
    [tex] \frac{\partial u}{\partial x} + \frac{\partial w}{\partial z} = 0 [/tex]
    Where u is the horizontal velocity and w is the vertical velocity


    3. The attempt at a solution

    Firstly I integrated the conservation of mass equation with respect to x between the two points:

    [tex] \int_ {x_1} ^ {x_2} \frac{\partial u}{\partial x} \mathrm{d}x + \int_ {x_1} ^ {x_2} \frac{\partial w}{\partial z} \mathrm{d}x = 0
    [/tex]
    Which after evaluating I get,
    [tex] u(x_2) - u(x_1) + x_2 \frac{\partial w}{\partial z} - x_1 \frac{\partial w}{\partial z} = 0 [/tex]
    Firstly here I am not 100% sure I can assume w is just a function of z only, but I have yet to see it as a function of anything else other than t in 2D flow?

    I then integrated the mass equation vertically first from [itex] z = 0 [/itex] to [itex] z = h_1 [/itex] and then from [itex] z=0 [/itex] to [itex] z= h_2 [/itex]

    [tex] \int_{0}^{h_1} \frac{\partial u}{\partial x} dz + \int_{0}^{h_1} \frac{\partial w}{\partial z} dz = 0 [/tex]

    which yields,
    [tex] h_1 \frac{\partial u}{\partial x} - w(0) = 0 [/tex]
    Since [tex] w(h_1) = 0 [/tex]

    Similarly for [itex] z=0 [/itex] to [itex] z= h_2 [/itex] I get,
    [tex] h_2 \frac{\partial u}{\partial x} - w(0) = 0 [/tex]

    This is where I am kind of stuck, from these two equations it appears [itex] h_1 = h_2 [/itex] which doesn't make sense since this is a hydraulic jump and the nature of such is to increase the surface of the liquid.

    Any hints where I may be going wrong here or missing something is appreciated.

    Thanks.
     
  2. jcsd
  3. Sep 12, 2017 #2
    First integrate with respect to z, using the Leibnitz rule for differentiation under the integral sign on the u term.
     
  4. Sep 13, 2017 #3
    I try to integrate with respect to z first, I haven't used the Leibniz rule before but from what I can gather I can treat my bounds as constant thus I get this for the u term.

    [tex] \int_{0}^{h_1} \frac {\partial u}{\partial x} dz = \frac {\partial}{\partial x} \int_{0}^{h_1} u dz [/tex]

    After evaluating this I get,
    [tex] h_1 \frac {\partial u}{\partial x} [/tex]

    I'm not sure if I did this correct but following this doesn't seem to get me much further?
    I suspect my math is off I haven't done any multi variable calc for a while.
     
  5. Sep 13, 2017 #4
    $$\frac{\partial [\int_0^{h(x)}udz]}{\partial x}=\int_0^{h(x)}{\frac{\partial u}{\partial x}dz}+u(x,h)\frac{dh}{dx}$$
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Volume Flux for a Hydraulic Jump
  1. Hydraulic conductivity (Replies: 1)

Loading...