Sciencemadness Discussion Board

Sipohn flow rate

RogueRose - 27-5-2016 at 17:22

I've always been curious if there is something that governs the speed at which siphoning can occur. If a pump is used to start the process and a rate of 30 GPM is achieved, will that rate be continued when the pump is removed (as long as the outlet is below the source)?

blogfast25 - 27-5-2016 at 17:31

Quote: Originally posted by RogueRose  
I've always been curious if there is something that governs the speed at which siphoning can occur. If a pump is used to start the process and a rate of 30 GPM is achieved, will that rate be continued when the pump is removed (as long as the outlet is below the source)?


You can find a more or less complete mathematical derivation of how siphons work, here:

http://www.sciencemadness.org/talk/viewthread.php?tid=65532&...

In short, once initial flow is established by a pump and you then switch off the pump, the initial rate becomes irrelevant very quickly. The flow rate of a siphon isn't affected by the initial rate but is determined by gravity and limited by friction (viscous) resistance in the siphon's pipe itself.

So if initially a rate of 30 GPM was established by the pump, after switching off the pump, the volumetric flow rate Qv will adjust until it is approx.:

$$Q_v \approx \frac{\pi}{4}D^2\sqrt{2gy}$$

With D the siphon pipe diameter and y the height difference between the surface of the water and the outlet of the siphon. The formula is roughly valid for internally smooth siphons that aren't too long.


[Edited on 28-5-2016 by blogfast25]

Eddygp - 28-5-2016 at 07:46

Quote: Originally posted by blogfast25  

You can find a more or less complete mathematical derivation of how siphons work, here:

http://www.sciencemadness.org/talk/viewthread.php?tid=65532&...


Oh wow! In spite of knowing that there have been many topics discussed here, I would have never thought that this would have been addressed already.

Magpie - 28-5-2016 at 08:40

Quote: Originally posted by blogfast25  


$$Q_v \approx \frac{\pi}{4}D^2\sqrt{2gy}$$

With D the siphon pipe diameter and y the height difference between the surface of the water and the outlet of the siphon. The formula is roughly valid for internally smooth siphons that aren't too long.


I thought this equation looked familiar. It is Toricelli's law, equally valid for a tank draining through an orifice:

blogfast25 - 28-5-2016 at 09:43

Quote: Originally posted by Magpie  
Quote: Originally posted by blogfast25  


$$Q_v \approx \frac{\pi}{4}D^2\sqrt{2gy}$$

With D the siphon pipe diameter and y the height difference between the surface of the water and the outlet of the siphon. The formula is roughly valid for internally smooth siphons that aren't too long.


I thought this equation looked familiar. It is Toricelli's law, equally valid for a tank draining through an orifice:



Yes, in the simple treatment of siphons the head loss in the pipe is neglected, which is OK for short, smooth siphons. Bear in mind though that for siphons y is not the same as the fluid height in a tank.

Taking head loss into account makes it quite complicated, mathematically.

[Edited on 28-5-2016 by blogfast25]