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The bore on the river

  1. Jul 11, 2009 #1

    physicsworks

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    Gold Member

    Hi.
    Have you ever heard about tidal bore? Such pretty thing occures on some rivers after the tide, away from river's mouth (on the Amazone river it's called "pororoca"). A few days ago I asked myself: can I define the speed of the bore using high school physics. This is what I've got. Please show me the mistakes in my solution if any. :shy:

    Suppose we have a stable bore which means that its form doesn't change when the bore trevells along the river. Also we ignore any nonlinear things. The height of the bore is [tex]h[/tex] and the river depth is [tex]H[/tex] (see the picture below, the bore is moving to the right!). Also the speed of the river (with respect to the shore) is [tex]v[/tex]. Lets find the speed of the bore [tex]u[/tex] (also with respect to the shore).
    _______________
    -----------------\
    ------------------\ h
    -------------------\____________
    ---------------------------------
    ---------------------------[tex]v[/tex]-----
    ---------------------<<-------- H
    ---------------------------------
    ---------------------------------
    ==========================

    Lets see the problem in the reference frame which is moving with the bore. In this frame, water approaches on it with the spped [tex]v+u[/tex].
    Using Bernoulli theorem for the upper linestream (continuous line in the picture) one gets:
    [tex]\frac{(v+u)^2}{2}=gh+\frac{v_0^2}{2}[/tex]
    continuity equation gives:
    [tex](v+u)H=v_0(H+h)[/tex]
    From these two equations for [tex]v+u[/tex] one has
    [tex]v+u=\sqrt{\frac{g(H+h)^2}{H+\frac{h}{2}}}[/tex]
    and finally
    [tex]u=\sqrt{\frac{g(H+h)^2}{H+\frac{h}{2}}}-v[/tex]
    For [tex]H=3~m, h=1.5~m, v=1~m/sec[/tex] we have [tex]6.3~m/sec[/tex] which is not far from the truth. :smile:
     
  2. jcsd
  3. Jul 15, 2009 #2

    physicsworks

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    Gold Member

    First of all, sorry for my terrible English. :frown: May be this is a problem why no one replied for the message :confused:, may be not. But I'm going to rewrite slightly my question and hope some of you would help me regardless of my poor language. Actually it's not even a question but a kind of thing about Bernoulli's principle that bothers me. May be it's better to make a new thread for this.
    So, let us say we have a steady flow of non-compressible ideal (without any dissipation) liquid. One can derive Bernoulli's equation from the Euler equation (equation of motion)
    [tex]\frac{\partial \boldsymbol{v}}{\partial t}+ (\boldsymbol{v}{\nabla} )\boldsymbol{v} = - grad~ w + \boldsymbol{g}[/tex]
    in its form:
    [tex]\frac{\partial \boldsymbol{v}}{\partial t}-[\boldsymbol{v},rot\boldsymbol{v}]=-grad\left(w+\frac{v^2}{2}\right)+\boldsymbol{g}[/tex]
    this one can be easily derived from
    [tex]\frac{1}{2}grad v^2=[\boldsymbol{v},rot \boldsymbol{v}]+(\boldsymbol{v} \nabla \boldsymbol{v})[/tex]
    being multiplyed by the unit vector [tex]\boldsymbol{e}[/tex] along the tangent to the streamline:
    [tex]\frac{\partial}{\partial e}\left( \frac{v^2}{2}+w+gz\right)=0[/tex]
    where z axis is parallel to [tex]\boldsymbol{g}[/tex] but has an opposite direction.
    This one tells us that
    [tex]\frac{v^2}{2}+w+gz[/tex]
    is a constant along the streamlines:
    [tex]\frac{v^2}{2}+w+gz=const[/tex] (*)
    where [tex]w[/tex] is a thermal function:
    [tex]dw=Tds+\frac{1}{\rho}dp[/tex]
    since [tex]s=const[/tex] (s is specifiс entropy) along streamline (and even it is constant everywhere throughout the liquid if it was a constant everywhere initially) one has
    [tex]dw=\frac{1}{\rho}dp[/tex]
    and
    [tex]grad~w = \frac{1}{\rho} grad~p[/tex]

    In many textbooks, especially in those bieng written for high school classes, authors derive Bernoulli's pricniple using simple energy conservation. They consider a tube, constrained by streamlines, which has different cross-section along its length. And for the
    height [tex]z[/tex] they take a height of the point which is the center of the given cross-setion, i.e. the point _inside_ this tube. But as for me it's wrong because the constant [tex]const[/tex] in Bernoulli's equation (*) is not the same for the different streamlines inside the tube. It would be the same only (?) if we have parallel streamlines and forget about gravity. But as I know for the potential flow, when [tex]rot~\boldsymbol{v}[/tex] is zero everywhere, this constant in Bernoulli's equation has the same value everywhere throughout the liquid. So these things bother me... Questions:

    1. Why is that so? (why is the constant in Bernoulli's equation for potential flow has the same value everywhere throughout the liquid?)
    2. Is it OK to derive Bernoulli's principle in such way described above?


    Thank you for any suggestions.
     
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