# Constant velocity in siphon?

I was recently reading the explanation for the behaviour of a siphon on Wikipedia, which uses Bernoulli's equation in its proof. The argument is generally pretty easy to follow, except for one key point: how can we assume that the velocity of the water is constant at all points throughout its flow through the tube? This tacit assumption is made at various point, not least in the terming of "the siphon velocity" (implying there's only one). I would be grateful if someone could clarify this.

Note: I've done a hefty amount of classical mechanics in my time, but barely any fluid mechanics I'm afraid. First principles explanations would be appreciated.

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Because in that wiki incompressibility of the fluid is not mentioned, this conclusion is not really correct. Such a siphon could be used for a gas as well. The writer had better used the term liquid in place of fluid. – Georg Nov 13 '11 at 8:38
@Georg: Yep, you are right. With this extra assumption, it now all makes sense, as shown in the derivation in the accepted answer. – Noldorin Nov 13 '11 at 21:40

The assumption of constant velocity throughout the tube arises from the following:

• Assuming constant density
• Conservation of mass

In addition to some other stipulations that should be found in the problem description (like the fact that the tube has constant area), and a few other hydrodynamics simplifying assumptions (like assuming the velocity is constant over the cross sectional area, or an equivalent assumption). One way to argue for the constant velocity assumption is to employ the common $\dot{m}=\rho v A$ equation, which gives flow rate given density, velocity, and area, assuming constant velocity over the cross sectional area. Out of these, area is constant from the basic specifications of the tube, and density can be written to have two dependencies, pressure and temperature, $\rho=\rho(P,T)$. Both the pressure and temperature change over the length of the tube, but it's small enough that if you're working under ordinary "bathtub" like conditions they don't matter.

The fact that $\dot{m}$ is constant over the length of the tube can be established by simply noting that the system is at steady state, and conservation of mass dictates that the mass flow rate going in is the same mass flow rate going out, since the tube has a constant amount of fluid in it.

To do a little bit more overkill, $d\rho/dP$ is a constant that plays a role in determining the speed of sound in the fluid, but is very small for "incompressible" fluids, although incompressibility would mean that value is zero. Temperature can have a significant effect, as $d\rho/dT$ is very much nonzero, but the frictional heating over the pipe just isn't much.

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Thanks for the reply. But how can you be sure that the system is in a steady state? – Noldorin Nov 13 '11 at 17:13
@Noldorin You can relax that, but it probably will not matter. If we add in the possibility that the fluid in the pipe is accelerating, then the velocity is still constant along its length, although it changes over time. If {constant area, constant density} is held, then the answer should still be "yes" because constant volume $\times$ constant density is a constant mass inventory so $\dot{m}_{in}=\dot{m}_{out}$. – Alan Rominger Nov 13 '11 at 18:38
Thanks, all is clear now. It is basically enough to say conservation of mass together with constant cross-section and the incompressibility of water lead to constant velocity. Someone should definitely add this on the Wikipedia page. :-) Answer accepted in any case. – Noldorin Nov 13 '11 at 21:39