Maximum liquid velocity in a siphon According to wikipedia: Siphon, the velocity of liquid in a siphon is equal to $\sqrt{2gh}$, where h is the difference in level between the water surface and the opening of the siphon.
But it also mentions that there exists a maximum velocity which can be attained; hence the velocity of the liquid coming out of the opening of the siphon cannot be increased beyond a particular limit by merely increasing $h$. 
But wouldn't this violate Bernoulli's equation if applied for the water surface and the opening if $h$ is taken beyond the value at which maximum velocity is obtained?
 A: That formula is simplified by


*

*ignoring viscosity

*ignoring that $h$ could be greater than what could be supported by atmospheric pressure (about 10 meters in the case of water), at which point the fluid separates.
A: Ok let me explain... First of all I regret for not being able to upload an image of siphon... So please follow my words. I hope I can clear your doubts...
So consider a working siphon. We know that the flow is very much steady in the siphon tube. Ok. Now I am asking you to consider the longer right portion of this u-tube. As we can see the diameter of the tube is constant throughout and so is the velocity according to the equation of continuity. Now tell me what conditions do you need to ensure this. Ok let us consider a ball falling at a constant rate(meaning a constant velocity). What exactly do you need for this is a force exactly equal to mg acting opposite. Now consider the case of the siphon. The pressure at the opening of the longer end is Pa. That means for a steady flow you need the pressure at the top to be equal to Pa - $\rho$gh2. Let's imagine now that you increase this height h2 to a point where Pa is exactly equals $\rho$gh2. Ok so now the  pressure at the top is zero but still the balance is well maintained. Now imagine what if you increase h2 more. The pressure at the top can't be less than zero. So what happens now is the same as when in the case of the ball, it's weight is more than the opposing force. Yes, now the weight of the liquid column is more than the forces trying to balance it. So what will happen is an accelerated motion. The fluid stream will as a result discontinue at the top. Due to this discontinuity, more and more air will enter the tube ultimately halting the process.
So for a given h3, because the maximum value of h2 is $\frac{P_a}{\rho g}$ ,so, the maximum value of h1 should be $\frac{P_a}{\rho g}$ - h3. Input this value into the equation v = $\sqrt{2gh_1}$ , and you will get vmax as $\sqrt{ 2(\frac{P_a}{\rho} - h_3 g)}$.
Hope it's helpful...
Note,h2 is the height of the longer right portion of the u-tube, h3 is the shorter end, and h1 is the difference of the two...
A: Air entering a siphon is an issue.
If the outlet is submerged in a pool and the inlet is sufficiently positioned beneath the surface of the source liquid, air can not enter a properly primed and purged siphon as it flows.
