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