What Tank Shape Ensures Constant Height Change Rate in Fluid Mechanics?

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imaweirdo2
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Homework Statement


A liquid is to drain through a small exit port at the bottom of an axisymmetric tank that is open at the top. Find a tank shape such that the rate of change of the height of the free surface is constant (at least until the tank is almost empty). Assume that Bernoulli’s eqn. applies and that the free surface diameter is always much larger than that of the exit port.


Homework Equations


dh/dt = const.
from bernoulli's eqn: v_out = (2gh)^(1/2)


The Attempt at a Solution


My best guess at how to solve this is to relate the change in height with the change in radius of the surface of the water, but because I don't know what the volume is I don't know how to do this.

The most I can come up with is:
drainage rate = A*v_out = pi*r^2*(2gh)^(1/2)

I'm not even sure I'm going in the right direction, and I haven't been able to puzzle it out. If anyone can help me out it would be much appreciated.
 
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imaweirdo2 said:

Homework Statement


A liquid is to drain through a small exit port at the bottom of an axisymmetric tank that is open at the top. Find a tank shape such that the rate of change of the height of the free surface is constant (at least until the tank is almost empty). Assume that Bernoulli’s eqn. applies and that the free surface diameter is always much larger than that of the exit port.

Homework Equations


dh/dt = const.
from bernoulli's eqn: v_out = (2gh)^(1/2)

The Attempt at a Solution


My best guess at how to solve this is to relate the change in height with the change in radius of the surface of the water, but because I don't know what the volume is I don't know how to do this.

The most I can come up with is:
drainage rate = A*v_out = pi*r^2*(2gh)^(1/2)

I'm not even sure I'm going in the right direction, and I haven't been able to puzzle it out. If anyone can help me out it would be much appreciated.
$$\frac{\pi D^2}{4}\frac{dh}{dt}=-\pi r^2 (2gh)^{1/2}$$Since dh/dt is constant, it follows that D is proportional to ##h^{1/4}##.