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I am studying fluid dynamics at university and have been working on the following problem:

A flat-bottomed barge moves very slowly through a closely fitting canal but generates a significant velocity $U$ in the small gap beneath its bottom. Estimate how much lower the barge sits in the water compared to when it is stationary if $U = 5 \, \text {ms}^{−1}.$

Considering the problem in the rest frame of the barge, I've deduced that, by conservation of mass, if the draught of the barge is $d$, its clearance above the canal bed $h$ and speed through the water $V$, then $Vd = Uh$. This doesn't seem helpful though, as we don't know what $V$ is.

I've thought about using the Bernoulli Streamline Theorem on a streamline along the riverbed and I get $$\frac{V^2}{2}+gh=\frac{U^2}{2}+gd$$ but it would seem that, when the barge is at rest, we have $h = d$, which doesn't seem to make sense (for every conceivable barge).

I can't seem to use any information on buoyancy as I know nothing about the weight of the barge.

Please help me understand how to solve this question, but also why the approach works with such little information.

  • 1
    This is more of a [physics](http://physics.stackexchange.com/) or [engineering](http://engineering.stackexchange.com/) question, than a math question. Have you tried your question at those sites? – 2017-02-25

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Your idea of using the Bernoulli's streamline theorem is correct. The theorem states that, in any arbitrary point along a streamline of an incompressible fluid with steady flow, if we neglect the friction by viscous forces, the following equation holds:

$${\displaystyle {\frac {V^{2}}{2}}+gh+{\frac {P}{\rho }}={\text{constant}}} $$

where $V $ is the fluid flow velocity, $g$ is the gravity acceleration, $h $ is the level of the point above a reference plane, $P $ is the pressure, and $ρ$ is the density of the fluid.

In this problem, we are asked to determine how $h $ changes between a situation of steady flow with $V= \text {5 ms}^{-1} \,\,$ and a situation of rest with $V=0 \,$. So, calling $h'$ the changed level of the point at rest and setting $g= \text {9.8 ms}^{-2} \,\,$, we have to solve

$$\displaystyle \frac {5^2}{2}+9.8\,h+\frac {P}{\rho } \\ = \frac {0^2}{2}+9.8\,h'+\frac {P}{\rho }$$

from which we get

$$\displaystyle h'-h= \frac {25}{2 \cdot 9.8} \approx 1.275 \, \text {m}$$

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    Why does the pressure remain the same in the two cases? – 2017-06-01