Water flowing out of a conical tank

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Homework Statement





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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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  • #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.
 
  • #3
haruspex
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Use geometry to find the relationship between r and h.
 
  • #4
SteamKing
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I'm going to amend my earlier post and say you should eliminate 'r' instead of 'h'.
 
  • #5
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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?
 
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  • #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?
 
  • #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.
 
  • #8
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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?
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 =(
 
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  • #9
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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.
Which Fundamental Theorem?
 
  • #10
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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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  • #11
SteamKing
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Here is my 2nd attempt, I'm interested in the method that doesn't rely on a differential equation, but is this correct?
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.
 
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  • #12
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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?
 
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  • #13
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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?
 
  • #14
verty
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I think, because ##dh## was negative, the left hand side should be negative.
 
  • #15
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I mean shouldn't doing the integral from 5 -> 3.15 m take care of that though?
 

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