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Water flowing out of a conical tank

  1. Sep 13, 2013 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    For this problem, it reminds me of related rates problems, except I'm not given a constraint such as dr/dt = something at r = something.

    I tried using partial derivatives to solve it, but I'm not seeing a way to get rid of the r terms, or its derivative
     

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  3. Sep 13, 2013 #2

    SteamKing

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    Since you know the dimensions of the conical tank, you can eliminate 'h' as a separate variable by making it a function of 'r'. This will simplify your calculations.
     
  4. Sep 13, 2013 #3

    haruspex

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    Use geometry to find the relationship between r and h.
     
  5. Sep 13, 2013 #4

    SteamKing

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    I'm going to amend my earlier post and say you should eliminate 'r' instead of 'h'.
     
  6. Sep 13, 2013 #5
    Do you mean to say that

    r=√3V/(hπ)?? Because then V is reintroduced into the equation. Or is there another geometrical relationship that I am missing that will remove both V and r?

    or rather plugging in the initial values given

    V = (1/3)πr^2h

    r = 3 and h =5

    (1/3)π(3)^2(5) = 47.12

    47.12 = (1/3)πr^2h
    45 = (r^2)(h)
    r = √(45/h)

    If this is the case, what is dr/dt? Is it just √(45/h) dh/dt?
     
    Last edited: Sep 13, 2013
  7. Sep 13, 2013 #6

    SteamKing

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    Nope. Look at the tank itself. Imagine that the cone is closed at the bottom, for argument's sake. At the bottom, h = 0, and r = 0 (essentially). At the top of the tank, h = 5 m, and r = 3 m. Don't you think you can find a relationship for 'r' in terms of 'h' using these two pieces of information?
     
  8. Sep 13, 2013 #7

    verty

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    Woopydalan, it looks like you are heading towards having a differential equation to solve. There is a better way, although it takes some ingenuity to see. If you know your Fundamental Theorem, you should be able to find a more direct method.
     
  9. Sep 13, 2013 #8
    r = 3/5h?

    would that mean dr/dt = 3/5h dh/dt or is it 3/5 dh/dt? I think its the latter, but not sure. I'm rusty with the math, summer came and went =(
     
    Last edited: Sep 13, 2013
  10. Sep 13, 2013 #9
    Which Fundamental Theorem?
     
  11. Sep 13, 2013 #10
    Here is my 2nd attempt, I'm interested in the method that doesn't rely on a differential equation, but is this correct?
     

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  12. Sep 13, 2013 #11

    SteamKing

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    It seems OK. The original rate of flow out of the tank was given in units of m^3/min. Where did you convert this to m^3/s? Otherwise, the time for the outflow will be in minutes rather than seconds.
     
  13. Sep 13, 2013 #12
    Upon further inspection, in my calculator I had it in degree mode, and in actuality the integral evaluates as -1.64 if in radian mode, which means time is negative, which shouldn't be the case.

    Did something go wrong in my solving for the differential equation?
     
    Last edited: Sep 13, 2013
  14. Sep 14, 2013 #13
    bumping, this problem is still incomplete as far as I know. Anyone have some clues for other avenues I should look or where in my analysis I have gone wrong?
     
  15. Sep 15, 2013 #14

    verty

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    I think, because ##dh## was negative, the left hand side should be negative.
     
  16. Sep 15, 2013 #15
    I mean shouldn't doing the integral from 5 -> 3.15 m take care of that though?
     
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