# Analysis/design of a siphon

#### stinlin

Hi all -

I'm working on the analysis of a siphon assemble for a water tank. I've attached a crude picture of what I'm working with.

As you can see, the water starts in the tank as given by the water table at about 8’-8” from the outlet pipe at the bottom. The entrance into the suction end is about 4’-10” from the outlet, while the highest point in the pipe rests at 6’-1.5” from the outlet. The larger pipe is 3” in diameter and the smaller is 1.5”. There is a stop check valve between the two.

The idea is that when the tank drains via the typical outlet pipe, there will still be about 3’ of standing water. It is proposed to siphon this water out into an overflow valve by opening the stop check valve after the water is done draining by other means.

At this point, I’ve ignored losses to figure what a cavitation height for the pipe bend would be (about 40’, so it’s no worry). While attempting to analyze the system in terms of “Will it work,” I get a bit confused. I can run iterations and such to determine pressures and velocities throughout the pipes, but I’m not sure how to figure out if the tank will drain via the siphon setup.

Maybe I don’t remember my fluid courses from school well enough, but I seem to not recall discussing siphons in this regard. I’m not asking for a solution to this problem as I enjoy figuring these things out, but I am asking that someone give me a bit of knowledge on how to classify this siphon as will or will not work. What kinds of things make a siphon work and not work? I’ve drained my pool by a hose and starting the flow by sucking the water out, but I’m not sure what’s actually going on in terms of the pressure differential and why the water gets up and over the hose in entirety at (just about) any water level.

Again, I’d really appreciate some sort of theoretical information that would be helpful in determining if this siphon is going to work or not, and what I can do to MAKE it work if I find it to be not capable of draining the tank up to the bottom of the suction end of the pipe. Thanks a bunch folks.

Edit: Talked to a guy here at the office and we approximate the valve as a double gate, not a stop check. :P

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#### Nick Bruno

im not sure if i understand your quesiton fully, but I think your answer lies with the bernoulli eqn.
0.5*rho*v^2 + rho*g*z + p = const, where the first term is the dynamic pressure (velocity based) and th second term is the head.

In your case, i think the dynamic pressure is the same on both ends of your tube, but the elevational pressure is different, therefore, u get a pressure gradient and some flow.

Not sure if this is what you are asking... but heres a helpful site.

http://en.wikipedia.org/wiki/Bernoulli's_principle

from this eqn u can get pressure for a specified velocity or vice verca and use it to find flow rate.

#### stinlin

Yeah, that much I understand. I'm trying to determine at what point the water level in the tank will not make it over the hump in the pipe.

After some running some numbers with assumed values and approximate friction factors, I'm getting an exit pressure of about -824 psf full, and at 1' of water in the tank, I want to say about -350 psf. This would suggest pressure, would it not?

So does that mean after the water starts flowing through, a negative pressure would develop at the outlet and begin pulling the water out of the reservoir?

#### Nick Bruno

hmm, well first I'm a bit confused about the negative pressure. Normally it is positive and a scalar unless you are dealing with gauge or relative pressures.

Just glancing at this, i think as long as the pressure head from rho*g*(z1-z0) is larger than the pressure it takes to raise the fluid the initial "hump" (P = F/A) you can drain the whole tank.

F = the force it takes to raise that amount of liquid in the tube
A = area of tube

Though keep in mind the pressure head is defined by atmosphereic pressure (if opened) and the pressure of the existing fluid in the tank, and how deep the syphon goes.

With this information you should be able to peice a relation together (depending on the height of your tank and volume of fluid in the tube; which is a function of tube depth) to say when the flow will stop and what height the tank will be at in the end.

#### stinlin

I think what you're describing would be the idea of pressure equilibrium...Which is great and all, but I know that suction develops, hence why you can use a hose to drain a pool or a gas tank. :P

E: I'm working in gage pressure, so the negative pressure is actually just some pressure below atmospheric, which is the assumed pressure value in the tank.

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#### Bob S

Yes, the siphon will work, but only if the valve is at the bottom, under the tank. The siphon will not work unless the pipe is filled with water when you open the valve, this canot be guaranteed unless the valve is at the bottom. There are two ways to prime it: 1) suck on the outpuit, or 2) reverse flush it slowly with a hose, and close the valve with water in the siphon.

#### stinlin

Interesting - could you explain WHY it won't work Bob S?

#### Bob S

Interesting - could you explain WHY it won't work Bob S?
If the valve is at the top, like in your drawing, the water in the drain pipe could drain out after the valve is closed, and then when you want to start the siphon, there is no water in the drain pipe to start draining and create a suction. You must keep water in the drain pipe at all times to create suction, or else you will have to prime it every time.

#### stinlin

That makes some sense - my only question is that if there is water in the pipe up to the gate valve (There should be, right? Because that portion is below the water line?), why won't the water force itself down the second section of the drain pipe (after the gate valve) and start the suction automatically?

I'm starting to see why this COULD be an issue, but I just want to have a solid grasp on the issue before talking to my boss and just saying "it won't work." Besides, this is a very intriguing issue to me, too! :)

E: And even the the entire siphon is empty before the valve is open, would the pressure difference not force water through the siphon once the gate valve is opened, too? I'm just having a tough time grasping this - hence why I'm a structural engineer and don't have to deal with this type of stuff too often, haha.

EE: Another edit after doing a bit of research...Will the siphon not be considered self-priming since the crest is below the initial water level? I would only be worried about two different diameter pipes really.

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#### Bob S

The siphon in your drawing will always work if the water level in your tank is higher than the highest point in your siphon (6' 1.5") when the valve is opened. If the water level in the tank is about 3', i.e., about 3' BELOW the highest part of your siphon when the valve is opened, there is nothing to create a vacuum when you open the valve, and the water that is behind (not below) the valve will drain back into the tank. So there has to be a valve (below the outlet level of the tank) that keeps water in the pipe, so that when the valve is opened, the water in the pipe above the valve will flow downwards and create suction in the siphon. The easiest way to prime the siphon is to let water into it and close the valve when the water level in the tank is above the highest level of the siphon.
There is a potential problem even in this case. If the siphon valve is below the tank outlet, and air bubbles travel upward (backward) through the valve when it is opened, they will break the vacuum needed for the siphon to work. This is more likely (I think) if the diameter of the siphon pipe is large.

#### stinlin

Well the water would always be above the crest, but what ended up being the deciding factor was that the weight of water pulling was never going to be greater than the weight of water needing to BE pulled. So after the water line dropped to below the crest, the siphon would break.

Found a solution that eliminates the siphon (since there was no easy way to increase the diameter or length of pulling pipe).

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