I'm studying fluid dynamics and I came up with an example that I find profoundly counterintuitive. I'd like someone more used to this kind of problems to confirm my guess.

We have two identical tanks with water $$A$$ and $$B$$. The only difference is $$A$$ has a hole, while $$B$$ has a long tube of the same width as the hole in $$A$$. I attach a diagram.

The problem I find is that if one applies Bernoulli, the velocity is different. Therefore, deposit $$B$$ empties much faster than $$A$$. This is very counterintuitive to me. How having a pipe facilitates the discharge? I would expect the opposite (although I acknowledge that this guess is motivated by the presence of friction the tube that I'm explicitly neglecting in this problem).

Is deposit $$B$$ discharging faster in reality? Why is so? How does water "know" that it must flow faster because there is a pipe somewhere below?

• It is not intuitive but it does discharge faster. Try draining a tank with a syphon hose, the lower you put the end the larger the flow – Wolphram jonny Sep 2 at 17:36
• Does it help your intuition if you move tank A down so the holes are at the same altitude? If you wanted to plug those holes. How much force would each require to hold the plug in? – mmesser314 Sep 2 at 19:35

The water at the opening in B has more water pushing down on it from directly above, so it makes sense that it should exit faster. But the fact that there's a pipe surrounding the column doesn't actually matter. To show this, suppose we construct a tank C, which looks the same as tank A (i.e. no pipe) but has a height $$h_1+h_2$$. If you use those same equations, you'll find that the water exiting the hole in the bottom of tank C (with no pipe) has the same velocity as the water exiting the hole in the bottom of tank B (with a pipe).
• This is of course only true for completely inviscid fluids. For viscous flow, viscous losses reduces output of $B$. – Gert Sep 2 at 14:24
I suspect that the velocity of fluid flowing from a container through a hole is $$\sqrt{2 g h}$$ only if the area of the hole is much smaller than the cross-section area of the container, otherwise the pressure cannot be maintained at the hole. Imagine that in case A the entire bottom falls out from the container, then the liquid will not move with the velocity of $$\sqrt{2 g h}$$. Case B is similar to that: the tube has no bottom.
EDIT (9/2/2019): See http://www.physics.princeton.edu/~mcdonald/examples/leaky_tank.pdf : " the velocity V of the eﬄuent stream from the tank is a function of the area ratio $$a/A$$", where $$a$$ is the area of the hole and $$A$$ is the cross-sectional area of the tank.