# 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?

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:

$$\frac12 v_1^2=gl-\frac{\Delta P}{\rho}$$

• No, I'm not implying that at all. In the case of the large basin what I'm saying is that the speed of the water in the basin is negligible compared to the speed in the pipe. That can be proved easily, precisely by means of the continuum condition for incompressible fluids. You cannot, however, ignore friction as it would leave $\rho gh$ unaccounted for and $v_1$ incalculable (as per your answer). "This question has been evaluated for many years" is an irrelevant kind of appeal to authority type of fallacy. – Gert Apr 16 '16 at 17:46
• Still, if we do neglect viscous forces as was done by Toricelli, I dont know what do you mean by inablity to calculate v1, because, in that case flow will be"conical", and thus v can be found by eqns of motion or bernoulli theorem – Prayas Agrawal Apr 16 '16 at 18:05
• Torricelli could afford to neglect viscous forces because: a. the flow speed inside the tank is considered zero. b. there's no pipe connected to the outlet. Pipe means viscous losses, by definition. In the shallow basin problem flow speed in the basin is basically zero but the pipe resists flow. Unavoidably. From pipe resistance and $\rho lg$, the speed $v_1$ is then calculated. " thus v can be found by eqns of motion or bernoulli theorem": make it happen! – Gert Apr 16 '16 at 18:51
• So it is only explained if friction exists? What if, in an idealistic situation, there is no friction? – Charles Apr 18 '16 at 4:28
• @Charles: sorry about the delay. Here, if you want to use Bernoulli, you need to take into account friction, otherwise it leads to a mathematical anomaly: $\rho gl=0$!! Considering the problem w/o friction is equivalent to a free falling cylinder of water, w/o pipe. After having fallen a length of $l$, its velocity would then be $v^2=2\rho gl$. Part of the problem is that it isn't very well defined: you say "Water enters from the top with a constant velocity v_1" but where does the water come from? And how can you impose a specific value for v_1? It leads to the confusion you experienced. – Gert Apr 18 '16 at 14:09