Liquid flow in a vertical tube A cylindrical vertical tube has uniform cross section $A_1$, and length $l$. It is open at both ends. Water enters from the top with a constant velocity $v_1$, and allowed to flow out from the bottom. Now, from the continuity equation, its velocity remains constant throughout its flow (the product of cross-section area and velocity remains constant, and the area of cross section is uniform throughout the tube). This means that it flows out from the bottom with same velocity $v_1$.
However, from the Bernoulli equation, $P + {1}/{2} \rho v^2 + h \rho g$  is constant. The pressure at the top and at the bottom of the tube is the same - both are equal to the atmospheric pressure. The two ends differ by a height $l$. This seems to imply that the velocity at the bottom should be equal to $(v_1^2 + 2gl)^{1/2}$. But the equation of continuity gives a different, and the correct, result. How are these two results reconciled?
One explanation is that the flow is not steady, so Bernoulli's equation is not valid, while the equation of continuity still is. 
Another is, assuming steady flow, that perhaps the pressure at the bottom end is not the atmospheric pressure, but equal to $P_{atm} + l \rho g$. As water leaves the tube, the pressure quickly falls back to $P_{atm}$, increasing the water's velocity and decreasing its cross sectional area of flow as soon it is out of the tube. What is the real explanation?
 A: Bernoulli's equation, in the form you are applying it, ignores all friction in the pipe. That friction manifests itself as a pressure loss as given by the Darcy-Weisbach equation.
If that pressure loss is called $\Delta P$ then the Bernoulli equation in your situation becomes simply:
$$\Delta P=\rho gl$$
From it, using the pipe's characteristics and whether flow is laminar or turbulent, $v_1$ can then be calculated. 
For example for laminar flow:
$$\Delta P=\frac{32 \mu l v_1}{D^2}$$
With $\mu$ absolute viscosity and $D$ the pipe's diameter.
You can't however just impose some value of $v_1$ on the system because there no difference in external pressure between inlet and outlet.
The problem is almost identical to the following one:

A very shallow but very large basin feeds the vertical outlet pipe of length $l$.
Pressure at both ends is atmospheric pressure $p_0$ (hydrostatic pressure in the basin is negligible). Throughput is then defined by pipe length and Darcy-Weisbach pressure loss:
$$\frac12 v_1^2=gl-\frac{\Delta P}{\rho}$$
