Syphoning Down a Lake 2

[ChristopherFehr]

As part of our process I will have to reduce the level in a small lake that forms on top of our waste rock containment by about 1 meter. The average inflow is 900 US GPM and the water will only have to be piped a short distance overt he final spill way. How would I get the flow started? I assume I will have to run the discharge to a lower elevastion than the suction, just enough to cover the head loss in the pipe?

 

[CESSNA1]

CHRISTOPHERFEHR: To siphon the water the discharge of the hose must be lower than the inlet side of the hose. The amount of flow is dependent, among other things, on the difference in elevation between the inlet and discharge of the hose. To sart the flow put the inlet side in the water and the outlet side below the level if the inlet and them pull a suction on the discharge side of the hose. This can be done by moth (on small hoses) of with a suction pump.

Can you dig a channel? That would be better or you could put in a permanent drain pipe. You could alos pump it out. Flow in a siphon is usually fairly slow, in the order of 2-5 gpm, depending on the installation. The larger ID of the pipe and the greater the difference in elevation the larger the flow.

Regards
Dave

 

[TheTick]

This question might be better posed in a civil engineering forum.

Definitely sounds like a job too big for rubber hose and mouth suction. Still, the principles would be the same. Discharge lower than outlet, get air out of the pipe.

Perhaps a pipe over the spillway, with a small booster pump to fill the pipe, and vents to eliminate air in the pipe. Once air is out of the pipe, flow should go naturally (you could leave the pump on, if desired).

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[ivymike]

You could undulate the hose (manually, in the vertical direction) to get the flow started.

The max "hump" that you can siphon over might be worth checking as well - you don't want to vaporize the water.

 

[JStephen]

You can fill the hose with water, close off one or both ends, put it in place, and reopen the end.

If a hose is too large in diameter for the length, or flow is too slow, the hose won't run "full", and air can go back up the hose and release the vacuum. If this happens, you can submerge the outlet end of the hose in some small container that is filled with water.

The maximum height you can siphon over is determined by the lowest pressure you can get in the water without boiling it. For 70 degrees F (arbitrary number), this 0.37 PSI, and you should be able to siphon up to 33' or so, maximum. Somewhat less if you're at higher elevations.

This all assumes the hose is good for full vacuum (as opposed to a fire hose, which collapses flat without pressure in it).

 

[MintJulep]

Don't forget to give some thought to how you are going to stop the flow once you get it started.

 

[Philrock]

If possible, put the high point (hump) of the siphon pipe at the same elevation as the highest water level you want in the lake. Obviously, the siphon pipe inlet will have to be somewhat lower than this, (1 foot?, 1 meter?), and the siphon pipe outlet will have to be lower still. Once the level in the lake rises to the height of the hump, the siphoning will start. When it falls to the level of the inlet, it will stop.

 

[tlee123]

Phil,

Your idea is interesting. But will the siphoning really start as soon as the water lever reaches the hump? What happens to the air that has entered the pipe? Have you tried this or is this just your theory?

 

[zeusfaber]

Phil's idea is well proven - this is the principle behind most automatic urinal flush systems.

One quibble with an earlier post - the requirement isn't "discharge lower than inlet" - it's "discharge lower than surface of the reservoir"

A.

 

[TheTick]

zeusfaber:

So, what you're saying is maybe it's okay for the inlet to be above the surface of the reservoir? If you can accomplish that, then you are truly an amazing engineer.

 

[ivymike]

no, I think it's pretty clear he's saying that the inlet can be lower than the discharge as long as the discharge is lower than the surface of the reservoir. Obviously the inlet would have to be lower than the surface of the reservoir. (If it weren't, then in this case it wouldn't be lower than the discharge, would it?)

 

[quark]

Phil's idea is right. The other idea given in another forum is also good and it is the easiest way to do it.

For people who didn't visit other forum, it was suggested by a member to put valves at both ends and a vent valve(also a provision for priming) at the highest point of the pipeline.

Regards,

 

[TheTick]

Consider me humbly corrected. It occured to me last night as I was pondering cleaning my fish tank. Outlet lower than surface!

 

[lilliput1]

To drain, your outflow must be more than the inflow of 900 gpm. At this flowrate you will need to pump! Check how much time you need to get the job done & calculate the required outflow flowrate taking into account the volume to be drained plus the inflow during the time period. Calculate your pump horsepower:

pump bhp = (GPM x Ft wg. pump head)/(3960 x .65)

The above formula assumes 65% pump efficiency.

You would be in the 10 hp range. You should pump & get the job over with as quick as possible since time cost money!