Is equation of continuity valid, while dealing with vertical pipes? Let try an experiment.
If water enter through an end $A$ with some velocity say $v_1$,and leaving end $B$ with speed $v_2$ in a UNIFORM cylindrical tube $AB$ (which is completely filled with water).
If we consider 3 cases.

*

*tube is horizontal

*tube is vertical with $A$ upward

*tube is vertical with $B$ upward

And we go through the experiment and we got our results, which is $v_1~=~v_2$
So for which case this is valid to?
Edit
Due to lack of logical answers
If we consider Bernoulli's equation
$$P+\rho gh+\frac{1}{2}\rho v^2= \text{Constant}$$
So in the case of vertical pipe, we consider two points $A$ and $B$ and now applying Bernoulli's equation here as follows.
Let's assume $A$ to be upward, and take $B$ as reference level and applying Bernoulli's equation.
$$P_a+\rho gh+\frac{1}{2}\rho v_1^2 = P_b+\frac{1}{2}\rho v_2^2$$
If both are exposed to the atmosphere, then $P_a = P_b = $P_{\text{atm}}$
Then we get
$$\rho gh = \frac{1}{2}\rho(v_2^2 - v_1^2)$$  which implies that $2gh +v_1^2 = v_2^2$
So, finally this will prove that $v_1$ never equal to $v_2$ in vertical pipe, but if we consider equation of continuity then mass going in is same as mass coming out so according to equation of continuity
$$\Delta m = \rho A_1 v_1\Delta t = \rho A_2 v_2 \Delta t$$ which implies $A_1 v_1 = A_2 v_2$, and in our case $A_1 = A_2$ then according to equation of continuity $v_1 = v_2$.
Thus, according to continuity $v_1 = v_2$ in vertical pipe case and according to Bernoulli's equation $v_1$ never equal to $v_2$.
How can this be possible? Please guys help me out. Please go through the question and then answer it?
 A: If the tube is vertical, the fluid can't be filling the tube, and the area of the stream must be changing, thus allowing a velocity change to be consistent with mass conservation. In the case where A is at the bottom, if the fluid fills the tube, the pressure can't be atmospheric at the bottom. Therefore, you can't assume both the same pressure at the ends of the tube as well as a constant area of the fluid stream.
A: I think BioPhysicist's answer is correct, and addresses the problem.  I just have a few points I want to add to possibly help clear it up (and I got pretty into the paint diagrams I was making).
I outlined the two possible situations I can think of where the tube is vertical, with A over B.
In the first situation, the velocity is non-zero, and the tanks at A and B are both exposed to atmosphere.  If the velocity is non-zero though, this means that fluid is going from tank A to tank B.  For this to happen the water in tank B must be pushing against the atmosphere.  If it were just at atmospheric pressure, the water wouldn't be able to move, because the surface of point B would also be suspended by atmospheric pressure.  So basically, even if both sides are exposed to atmosphere, if the water is flowing I don't think it's safe to say both sides of the pipe are actually at atmospheric pressure, the one on the receiving side of the flow should be above atmosphere so that the water can actually get into that vessel and flow out of B (or raise it's surface).
Here is a bad diagram I drew:

Or, if the fluid isn't flowing, the situation becomes hydrostatic, and you can clearly see the pressure difference, because if one end is above the other and there is no flow, hydrostatics dictates that $\Delta P = \rho g h$, so $P_A \neq P_B$, I drew another bad diagram:

So basically as Biophysicist said, the assumptions just cannot all be held at once.  Hopefully this shows why it wouldn't make a lot of sense for them all to hold.
Here is how it must look if pressure is atmospheric on both sides and the flow is vertical:

A: If air cannot enter the tube all 3 are valid, with the possible exception of case 2. In case 2 if the height of the cylindrical tube is greater than 10 meters vacuum or vapor cavities could form allowing some water to accelerate downward.
