5A (Fluid Dynamics)

z (t)

is the water depth and

A (z) = π r (z)^{2}

is the horizontal cross-sectional area, Torricelli's law gives the volume flux through the hole as

q_{0} = π r_{0}^{2} 2 g z .

Conservation of volume gives

A (z) \overset{z}{˙} = - π r_{0}^{2} 2 g z .

For a prescribed constant fall rate

\overset{z}{˙} = - α

α > 0

, this requires

A (z) = \frac{π r _{0}^{2} 2 g}{α} z .

Consequently

r (z) = r_{0} (\frac{2 g}{α})^{1/2} z^{1/4},

so the container radius must be proportional to the fourth root of height.

b

Since

z (t) = h_{I} - α t

A (t) = \frac{π r _{0}^{2} 2 g}{α} h_{I} - α t

until the emptying time

t_{e} = h_{I} / α

. Equivalently,

\frac{A ( t )}{A ( 0 )} = 1 - \frac{t}{t _{e}} .

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