Tank discharge paradox? 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?
 A: It's not the pipe that's affecting the discharge speed. The only thing that matters in these equations* is the height of the water column above the hole.
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).

*Note that these equations are themselves a simplified model of the way actual fluids work. In particular, they're only reliable for completely inviscid liquids (i.e. those that flow without any internal resistance).
A: 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. 
