Solution

The steady Euler equations for an inviscid fluid are

ρ (u \cdot \nabla) u = - \nabla p - \nabla χ .

(31)

Taking the scalar product with

u

and using

u \cdot (u \cdot \nabla u) = u \cdot \nabla (\frac{1}{2} ∣ u ∣^{2})

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gives

u \cdot \nabla (\frac{1}{2} ρ ∣ u ∣^{2} + p + χ) = 0.

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Thus

u \cdot \nabla H = 0 .

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The Bernoulli function

H

is constant along each streamline: fluid particles in steady flow exchange pressure, kinetic, and potential energy without changing their total mechanical energy density.

Let

y (t)

be the downward displacement of the free surface from its initial level, and let

U

be the speed in the tube. Conservation of volume gives

a U = A \overset{y}{˙} .

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The free surface and the outlet are both at atmospheric pressure, and their vertical separation is

H + h_{0} - y

. Applying the Bernoulli equation between them, while retaining the small free-surface speed, gives

\frac{1}{2} (U^{2} - \overset{y}{˙}^{2}) = g (H + h_{0} - y) .

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Therefore

\overset{y}{˙} = \frac{2 g ( H + h _{0} - y )}{A ^{2} / a ^{2} - 1} .

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The surface reaches the upper tube end when

y = h_{0}

. Hence the draining time of a uniform tank through a siphon is

t = \frac{A ^{2} / a ^{2} - 1}{2 g} \int_{0}^{h_{0}} \frac{d y}{H + h _{0} - y} = 2 (\frac{A ^{2}}{a ^{2}} - 1)^{1/2} \frac{H + h _{0} - H}{g} .

(38)

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 3 / 7D / Solution

 Ancestors (10)