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Clepsydra shape using Fourier series

  1. May 28, 2014 #1
    Our Fluid Mechanics professor gave us a challenge: to find the shape of a vessel with a hole at the bottom such that the water level in the vessel will change at a constant rate (i.e. if z is the height of the water in the tank dz/dt=constant).

    I presented a solution assuming that the vessel would be a 3D curve: http://imgur.com/2RhMCgD
    This was correct but apparently not good enough. He responded:

    "You have to show how you come up with the 1/4 power mathematically and rigorously from first principle. For instance, start with a Fourier series with a set of orthogonal functions, and take it from here."

    Does anyone have any idea where to begin?
    Thanks in advance.
     
  2. jcsd
  3. May 29, 2014 #2
    I fail to see what that could have to do with Fourier series. You have shown that ##R^2 = a \sqrt z## if ##z'(t) = \mathrm{const}##. That is all it takes.
     
  4. May 29, 2014 #3

    AlephZero

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    I agree. The way to do this is set up a differential equation and solve it.

    Your solution does the right sort of things, except you seem to have made some arbitrary assumptions like the first line "assume it's a parabola". The solution to the ODE will be whatever it is - you don't need to assume anything.
     
  5. May 29, 2014 #4
    It may actually be the ungrounded assumption in the beginning that made the prof unhappy. It is not necessary anyway, just drop it.
     
  6. May 30, 2014 #5
    I was unsure how the fourier series would come into it as well.
    How would you drop the assumption, allowing for the possiblity that it might be a cone, or cylinder or anything? By assuming it was a curve I was able to solve for the 1/4 power.
     
  7. May 30, 2014 #6
    You used the parabolic function as a placeholder for a unknown function, and then you demonstrated that the unknown function had to satisfy a particular property, which determines it almost completely. You could have started with just an unknown function.
     
  8. May 30, 2014 #7
    OK. So like this then? http://imgur.com/zvrjcVt

    But it sounds like everyone is as puzzled as me as to how one could use a Fourier series..
     
  9. May 31, 2014 #8
    You are still making ungrounded assumptions. Don't. Just use ##R(z)## and transform the equations so that you obtain ##R(z) = ...##. That is all you need.
     
  10. Jun 1, 2014 #9
    OK Thanks guys. I'll rework it and have a little chat with him
     
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