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Potential theory for water waves

  1. Nov 30, 2013 #1
    hi :)

    http://ocw.mit.edu/courses/mechanic...s-fall-2009/course-text/MIT2_017JF09_ch06.pdf

    In page 37 they use newton's 2nd law for a fluid element (while ignoring viscous forces) to derive the bernoulli equation for unsteady flow.

    Well, what I am confused about is the last step: They integrate through the x direction, y direction and then z direction to get 3 separate bernoulli equations,,, but then they just add it all up to one??? I don't understand the mathematics behind this as I'm completely new to potential theory.
     
    Last edited: Nov 30, 2013
  2. jcsd
  3. Nov 30, 2013 #2
    Furthermore, on page 40 they equate the dynamical pressure {for waterwaves} to be equal to the change in the potential function?? To see what I mean:

    From this (Bernoulli) ##p + 0.5 \rho V^2 (\approx 0) + \rho g z +\rho \frac{\partial(\phi)}{\partial(z)} = C##, they go to this: ##p_d = p+\rho g z = -\rho \frac{\partial(\phi)}{\partial(z)}##

    How did they do that?
     
  4. Dec 1, 2013 #3
    This is nothing mysterious, it is just that z=0 for the equations in [itex]\hat{x}[/itex] and [itex]\hat{y}[/itex], so that equation works for all three cases. If that doesn't make sense, do the three integrals and compare the results.
     
  5. Dec 1, 2013 #4

    AlephZero

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    It works through "The first force balance (x direction)" and gets to the result

    "Bermouilli forrmula" = ##C_1## when y and z are constant

    The same process in the y direction gives
    "Bermouilli forrmula" = ##C_2## when z and x are constant

    and in the z direction
    "Bermouilli forrmula" + ##\rho g z## = ##C_3## when x and y are constant

    But ##\rho g z## is a constant when z is constant, so you can merge it into ##C_1## and ##C_2##.

    In other words the final equation is actually valid in the x, y, and z directions.

    This seems like a tortuous way to get to the result, but I guess that depends what you already know and what the teaching objectives for this part of the course are.
     
  6. Dec 1, 2013 #5
    Oh how stupid of me. This is just the same as integrating partial derivatives of a function and then comparing the results to find the actual function. OK, thanks!!
     
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