Flow in vertical pipe and Bernoulli's equation If we had a flow in an open vertical pipe kept in atmosphere so that external pressure is equal at both ends and if we try to apply Bernoulli's equation to it would stand to reason that due to pipe being vertical, potential energy is smaller at the lower end so, kinetic energy must be bigger, which is sensible because gravity did positive work and increased fluid's kinetic energy.
But, if pipe's diameter is the same on both ends how can we agree this with continuity principle? From this principle, velocity should have been the same. Is Bernoulli's continuity equation applicable here in its simplest form ? Although I am not quite sure what is missing here.
 A: Consider an open-top   parallel-sided vertical pipe filled with an inviscid fluid and plugged at the bottom. At $t=0$ the plug is  removed.  As  the fluid is being acted on by gravity  and is  now  unconstrained both above and below,  it starts to fall. It starts form rest and accelerates uniformly   downwards so that $v=-gt$. This  velocity is the same for  all vertical heights $z$ in the pipe, so the fluid not have to separate from the walls.
Let us compare this  motion with Bernoulli.  Bernoulli's equation for non-steady flow states that
$$
\frac{\partial \phi}{\partial t} +\frac 12 v^2 + \frac P \rho
+gz
$$
is independent of $z$. Here $\phi$ is the velocity potential defined by $v=\partial_z \phi$, so for $v= -gt$ we have $\phi=- gzt$ and
$$
\frac{\partial \phi}{\partial t}=-gz
$$
Bernoulli now claims that
$$
-gz+\frac 12 g^2t^2 + \frac P \rho +gz
$$
is independent of $z$. This means that he claims that $P$ is independent of $z$.  There is thus no problem with the pressure being atmospheric at both ends of the pipe. Indeed, as the fluid is in free fall, it effectivly sees no gravity, and so $ P= P_0-m g_{\rm effective}z$ with $g_{\rm effective}=0$.
I think that the discussion in the comments shows that people are  not very familar with the non-steady version of of Bernoulli's equation,  so here is a dervation from Euler's equation
$$
\rho\left(\frac{\partial v}{\partial t}+ v\partial_z v\right)=-\partial_z P- \rho g 
$$
for one-dimensional incompressible flow in the $z$ direction.
We set  $v=\partial_x \phi$ and  note that $v\partial_z v= \partial_z(v^2/2)$, so  Euler becomes
$$
\rho \frac{\partial}{\partial z}\left(\frac{\partial \phi}{\partial t}+ \frac 1 2 v^2+ \frac P \rho - gz\right)=0.
$$
A: Bernoulli's equation goes like this:
$$p_1 + h_1\rho_1 g + \frac12\rho_1 v_1^2 = p_2 + h_2\rho_2 g + \frac12\rho_2 v_2^2$$
Continuity principle states: $$A_1V_1 = A_2V_2$$
When the cross-section is same the velocity at both topmost point and bottom most point gets cancelled out from  Bernoulli's equation and the kinetic energy term  vanishes.So we are left with the pressure term  and the $\rho gh$ term. So the difference in potential energy (represented by the $\rho gh$ term) is balanced by the pressure difference between the two points   i.e
$$p_2 - p_1 = h_1\rho _1g - h_2 \rho_2g$$
A: Continuity Theorem is always applicable. When thrown a stream vertically upwards [not in a pipe] it widens its area, and when it flows vertically downwards, it narrows its area.
When fluid gains Kinetic energy in significant amount then, to obey Rate Flow of Mass it narrows the stream-flow area. Just imagine the vertical pipe to be very long, considering that atmospheric pressure is same at both ends, when made to flow a fluid, it will become nearly impossible to get a stream which exits with an area equal to the area of the ends of the pipe.
It is reasonably difficult to clarify this mathematically but needs just some practical examples to justify ourselves. If someone happens to know any mathematical formulation then do share it.
A: The fluid in a vertical pipe which is open at both ends will be in free fall, unless you consider adhesion and viscosity.
