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Is it possible to design a simple container with holes of a certain shape such that if it contains any arbitrary volume of a non-viscous Newtonian fluid between 0 and some pre-defined limit it will "leak" at a rate directly proportional to the amount of fluid it contains?

NJE
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  • Wouldn't a straight cylinder do this? If flow rate is proportional to pressure, then the pressure at the bottom of a cylindrical vessel is proportional to the quantity of liquid. How accurately do you want to do this - are you looking for more subtle second order effects? – Floris Nov 12 '17 at 20:49
  • @Floris: in that case the output is proportional to the square root of the volume. – Gert Nov 12 '17 at 21:00
  • @Gert You're right, it depends on the role of viscosity. I had glossed over the "non-viscous" part of the question. See this earlier answer for some relevant analyses (but of a different question). – Floris Nov 12 '17 at 21:03
  • What you do is use a container with a long, low diameter tube attached at the bottom, so that the flow in the tube is laminar and the pressure drop is proportional to the flow rate. The pressure drop has to be high enough for the pressure head in the container to be small compared to the pressure drop in the tube, irrespective of the head depth in the container. This dictates some constraints on the design of the tube. – Chet Miller Nov 12 '17 at 22:30

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Let's say we have a body of revolution defined by $x=f(y)$:

Water tank

It's filled with an inviscid liquid to level $y$. The liquid will flow out of the bottom hole at speed $v$ and acc. Torricelli:

$$v=\sqrt{2gy}$$

The volume output $Q$, if the surface area of the hole is $A$, is:

$$Q=vA=A\sqrt{2gy}$$

The volume of liquid in the tank is given by:

$$V=\pi\int_0^yx(y)^2dy$$

As per the OP, $Q$ and $V$ need to be proportional, so with $k$ a constant:

$$kA\sqrt{2gy}=\pi\int_0^yx(y)^2dy$$ Derive to $y$ to get rid of the integral: $$-kA\frac12 \sqrt{2g}y^{-\frac12}=\pi x(y)^2$$

Note that this means that $k<0$. Grouping all the constants together we get:

$$x(y)=Ky^{-\frac14}=\frac{K}{\sqrt [4]{y}}$$

So for any tank defined by such a function, the OP's requirement is fulfilled.

Gert
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  • Are you sure? I thought this shape of tank would give you a constant rate of change of the height in the container but that is not what was asked. – Floris Nov 12 '17 at 20:57
  • No, I'm pretty sure. I solved the constant rate of height somewhere else. It is different. – Gert Nov 12 '17 at 20:59
  • My bad - you're right. – Floris Nov 12 '17 at 21:04
  • Thanks for the very clear answer. That is what I arrived at as well. Not a very convenient shape for, say, a novelty cup (since the width at the base must be infinite), but interesting all the same. – NJE Nov 19 '17 at 03:11