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Homework Help: 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,

    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
    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.

  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
    $$\frac{\partial [\int_0^{h(x)}udz]}{\partial x}=\int_0^{h(x)}{\frac{\partial u}{\partial x}dz}+u(x,h)\frac{dh}{dx}$$
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