Fluid velocity in a vertical pipe Consider a pipe with length $L$ and uniform radius $A$ is held vertically. According to the continuity equation, the velocity of water going into the pipe seems to be the same as the velocity of water coming out. But according to Bernoulli's equation:
$$P_{atm}+\frac{1}{2}\rho v_1^2+\rho gL=P_{atm}+\frac{1}{2}\rho v_2^2$$
$$v_2=\sqrt{v_1^2+2gL}$$
Which means that the e water would come out faster, which makes much sence. What is wrong with my equations?
 A: You cannot have the same pressure at both ends of the pipe. The two velocities must be the same (assuming constant pipe diameter) due to mass conservation, so you'll have just the hydrostatic equation, $$p_2-p_1=\rho\,g\,L.$$ Note that your use of the Bernoulli equation for this problem implies that the flow is assumed to be ideal, with no friction. Also notice that the velocity in this problem remains indeterminate.
A: The diameter of the pipe is likely included to mislead you.
Obviously the fluid should gain speed, but what's also obvious is that it can't increase in mass, and it shouldn't expand since water is basically incompressible. 
I had to picture a stream of water allowed to pour freely.  What will happen is that the stream diameter will become smaller to account for the faster velocity.  If the water were a bar of elastic material this would be the same as stretching it.  It's mass flow is the same, it just elongates as it becomes faster.  By the time it reaches the bottom of the vertical pipe the stream will not have a diameter of A.
A: Imagine two very large (the liquid levels don't vary), very shallow (hydrostatic pressure is negligible at entrance and outlet) reservoirs, connected by a vertical pipe:

It's very tempting to apply Bernoulli here but as the OP observed that would violate the continuity equation.
In reality:
$$P_{atm}+\frac{1}{2}\rho v_1^2+\rho gL=P_{atm}+\frac{1}{2}\rho v_2^2,$$
is not applicable here because it neglects the resistance to flow (viscous drag) exerted by the pipe. In reality $v_1=v_2$ and we need to introduce a head loss term $\Delta P_{\mathrm{friction}}$:
$$P_{atm}+\frac{1}{2}\rho v_1^2+\rho gL=P_{atm}+\frac{1}{2}\rho v_1^2 +\Delta P_{\mathrm{friction}}$$
Or:
$$\rho gL=\Delta P_{\mathrm{friction}}$$
In the case of laminar flow through the pipe the Darcy-Weisbach equation can be used to calculate $\Delta P_{\mathrm{friction}}$:
$$\Delta P_{\mathrm{friction}}=f_{D}\frac{\rho}{2}\frac{v_1^2}{D}L$$
So that $v_1$ is not indeterminate.
For turbulent flow the calculation of $\Delta P_{\mathrm{friction}}$ is more complex.
A: Following the schematics offered by Gert, you have to consider two stages of water flow:
1 - First stage while water flow velocity goes from zero to a certain max that with acceptable negligencies is given by gravity laws. Max velocity magnitude is function of the pipe length...
2 Second stage stable velocity stage if the water  level of the feeding and hosting reservoirs remains constant or such fluctuations should be considered. Bernoulli equation applies considering also head losses from first entrance and length losses. 
A: Your version of Bernoulli's equation only applies to steady flow -ie. to motion in which the velocity at any point is indepenedent of time. In the case of your pipe open at both ends the fluid will accelerate. In that case you must use the time-dependent version of Bernoulli. For the irrotational flow of in incompressible fluid (so that we can write ${\bf v}=\nabla \phi$) Bernoulli says that
$$
\frac 12 |{\bf v}|^2 + \frac{P}{\rho} +\frac{\partial \phi}{\partial t}= const.
$$ 
You are omitting the  $\partial \phi/\partial t$.
A: Consider a vertical pipe connected to a shallow reservoir:
There are mainly 3 points of interest:

*

*The outlet of the vertical pipe:
The heads are:
$P_{atm}+\frac{1}{2}ρv^2$


*The inlet of the vertical pipe:
By continuity, the stream has same velocity at inlet and outlet
$P_{in}+\frac{1}{2}ρv^2 + ρgL = P_{atm}+\frac{1}{2}ρv^2$
$P_{in} = P_{atm} - ρgL$


*At the surface of the reservoir:
$P_{atm} + ρgL = P_{in} + \frac{1}{2}ρv^2 + ρgL $
$v = \sqrt{2gL}$
which is the text book answer of Torricelli's law
https://en.wikipedia.org/wiki/Torricelli%27s_law
The principle behind Bernoulli's equation is conservation of energy. The problem can be understood this way:
On water surface of the reservoir, the velocity is zero, but when water is entering the pipe, it gains velocity. Since there is no change in elevation in a shallow reservoir, energy must be converted from pressure head to kinetic head.
As water travels along the pipe, the elevation head decreases but the kinetic head has to remain unchanged due to continuity, so energy is converted from elevation head to pressure head.
