more on Siphon Design

Hello all -
I read with interest the recent conversations initiated by Bob Jordan about
his (potential) siphon in New Zealand. I read this, of course, because I
have a siphon problem of my own. Unfortunately, the previous thread's
discourse didn't solve my problem, so I'm hoping "you" can give me a little
direction. My problem is considerably simpler than Bob's, but I'm still
stumped.
My project is discharge from a reservoir. Way back when the dam was
constructed and a creek cut off, the owners installed a siphon pipe to
provide continuing flow in the downstream creek. It consists solely of
6-inch galvanized irrigation pipe laid over the top of the dam and three
gate valves for 'charging' the system. Total head differential is less than
30 feet at an elevation of around 500 feet. It's worked fine for the last
50 years, but that didn't stop a regulatory agency from requiring its
upgrade. That's where I came in. I've never designed a siphon before, but
I thought, Heck, it's _just_ Bernoulli's equation, right? (Well, to be
honest, I thought, Hmmm, I remember something about Bernoulli; I can dig out
my undergrad hydraulics book.but I digress.)
In all the years of operation, the siphon has never been lost and no one has
complained about the quantity of water, so they want essentially the same
system re-installed with the engineering to back it up. My particular
wrinkle is that the pipe outlet is not submerged.
I need to calculate the flow rate they are *actually* providing (to prove
they are meeting the minimum under all operating reservoir levels) and I
need to calculate the maximum negative pressure inside the pipe to verify
the pipe class selection.
All my textbooks, handbooks, and online literature research assume either
the pipe end is submerged and/or that the pipe is flowing full. Since my
outlet is on a slope (as opposed to pointing straight down) with a free
outfall (the pipe invert is at least 6 inches above the downstream water
surface), my flow area is not the full pipe diameter.
So.I can calculate outlet velocity from Bernoulli's, but I don't know the
area to find Q. Should I use a culvert or orifice equation? Is that
appropriate since I have a negative pressure upstream inside the pipe? How
do I apply it?
And speaking of that negative pressure, it seems like I have too many
unknowns. Velocity and area are not constant through the pipe, but I can't
see how to calculate them while also not knowing the pressure.
Am I missing something just totally obvious here? Would a few more details
help (like elevations, lengths, equations, etc)?
Let me know, and my thanks for any and all assistance!
Selene
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Reply to
Selene
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In order to start the siphon, you have to close the discharge, then evacuate all the air, then open the discharge. So, you're starting with the pipe flowing full.
The depression in the pipe may be less, but can't exceed 1 bar.
-Mike-
Reply to
Mike Halloran
Is it impossible to actually measure the flow rate? At some point, even if you did a calculation, it would be "nice" to verify the equation with a measurement. If you do the measurement, you may be able to avoid the calculation altogether.
Selene wrote:
Reply to
charliew2
"charliew2" wrote
The current pipe is 40-year old galvanized, so it's probably tuberculated, and the regulatory agency has required it's replacement. The owners like the flow rate, so we're putting back in 6-inch pipe. But they want a "calc package" to show that it's been done "right." And I need to check the negative pressure, also a calculation. Hence my earlier questions.
All the books say, "Using Bernoulli, calculate between the known (upstream reservoir) location, and the point of interest." But I have one Bernoulli equation and two unknowns: pressure and velocity. That's where I'm stuck.
Selene
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Reply to
Selene
"Mike Halloran" wrote
Right, I'm starting the pipe that way, but then I want to make sure it stays that way. So I need to check the pressure at several points throughout the profile, but as I said to Charlie, I have one equation (Bernoulli) and two unknowns (velocity and pressure) and I can't see how to solve the problem.
Right, but I need to check that.
Brater & King says, "Upon inserting [the] loss terms in [the Bernoulli equation (from upstream to downstream end)], the velocity, and hence the dicharge, may be computed."
So, assuming full flow at entrance and exit, I can get Q. B&K continues, "Knowing the discharge, the pressure can be determined by writing the Bernoulli equation from point 1 to the point where it is desired to know the pressure, including only the loss between point 1 and the point in question."
Using Bernoulli, my pressure equation becomes Px = w ( Z1 - Zx - V^2/2g (1 + hL + hf) ), where w is specific weight, Z1 and Zx are elevations of upstream reservoir and point of interest, V is velocity, g is gravity, hL is head losses (due to entrance, etc), and hf is friction loss.
So, I know Q, but at the point of negative pressure, I can't assume the area is full flow, so I don't know V. That's why I'm stuck.
I'm wondering, though, if my idea about the area is wrong. In a reference I have from Canada, it says, "It can be seen that the energy equation can be solved for the pressure at some location, as long as the discharge, pipe diameter, pipe profile, and pressure at one point are known."
I know all of these things - the only thing I don't have is the relationship of area and velocity. But I'm not a pressurized flow person (I'm mostly a stormwater, open channel type), so maybe in positively or negatively pressurized flow, the area is _always_ full pipe flow????
Selene
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Reply to
Selene
well, you could always put a pressure tap into the existing pipe someplace convenient. A reasonably competent plumber should be able to do that for you on the charged pipe.
Reply to
Michael
I'm thinking a bit of dye and a stopwatch could get you a decent measure of flow rate...
And if you wanted to get fancy, a small neutral buoyancy datalogger could get you a nice pressure profile if you decide there are no obstructions to slow it down (or hang it up! =).
Al...
Reply to
Alan Adrian
Selene,
there may be a way out of your "fix". If you have a pressure tap at the point of interest (hot tap?) you could install a gauge and actually measure the pressure. Also, if there is a "basin" of known volume on the discharge end of the pipe, note the rate of level rise to allow back-calculation of a flow rate.
If you can't do the measurement, you are going to have a difficult time, even if you can find a way to model your system. Assuming that the inside of the pipe is rougher than expected, any calculated flow rate and velocity will be in error to an unknown degree.
(cut)
Reply to
charliew2
You do know one thing for certain for this system. The piping that is experiencing "negative pressure" must necessarily be liquid full, or the siphon will fail. Thus, you will have a full pipe throughout the suction section of the pipe, and for some distance on the downstream side of the pipe. You know the pressure at the pipe entrance, and if you can take an accurate measurement somewhere else on the pipe, near the top of the pipe or somewhere not to far on the downstream side of the pipe, you should have data that will allow you to calculate the pressure drop that the pipe is experiencing. From this, you can get to a flow rate.
Reply to
charliew2
Ah, another siphon problem. This could get fun. We need to get the effective head, somehow. Fortunately, we know that the air is not making it to the top of the siphon, otherwise, it would stop entirely. One important question here would be whether the slope of the pipe is monotonic downwards, or if there might be a dip in it somewhere.
It doesn't matter if your pipe is on a slope, or is vertical. The head is going to be equal to the vertical difference between the level of the reservoir and the water/air interface on the downleg. You can assume that the flow rate will be related to the pressure drop, pipe diameter, and friction factor to that point. The trick is to figure out what is the level of the water/air interface.
Your volumetric velocity is the same throughout your pipe, same with the mass velocity. Water is incompressible.
We also can state that the flow rate in the length of pipe without air is equal to the flow in the pipe with air. This sounds a little silly, but we might be able to solve for the length of fully filled pipe by setting its volumetric flow rate equal to that of the partially filled pipe (with a shorter length). At what point does it become energetically more efficient to flow faster with air in the pipe versus have a solid stream of water which must deal with all the friction?
Just a few thoughts. It's late, so I'm going to think more on it tomorrow.
Michael
Reply to
Herman Family
ok, someone looked it up. Z1 is the reservoir level. Zx is the highest point in the downleg that air reaches. You need the friction loss formula (fanning's equation) for hf, which is going to take into account the length of the pipe.
After the air/water interface, the Bernoulli equation isn't in effect. The flow rate will be for a partially filled pipe. The good news is that the two volumetric flow rates are equal, though the velocities will differ.
Michael
Reply to
Herman Family
Disclaimer: no practical experience with syphons... but.... a diaphram or gate valve on the efflux end could be closed until it reduces the flow at minimal flow (max lift) case. Then opened a fraction. That will keep the efflux pipe air free.
Brian W.
Reply to
Brian Whatcott
Bad Plan! Bad Plan!
You have have to purge the air before you stop or significantly decrease the flow, or the air in the pipe rises to the top, breaking the siphon.
Reply to
Michael
I can visual;ize that problem, as described. It even sounds like a practical problem. Is it, in your experience?
Brian W
Reply to
Brian Whatcott
Not a factor in this discussion but thought that I'd mention that the statement , *water is incompressible* is not a true fact. Having said that, unless you're dealing with control systems you'll probably never have to worry about it. There is such a thing as compressible flow and not taking it into account in hydraulic control systems can result in system instability. It is a primarily a function of the Bulk Modules of the fluid, the trapped volume (such as the volume in a line between a pump and actuator piston) and the change in pressure. Quite important when dealing with the transient response of hydraulic systems. Don't mean to distract from the topic at hand but just wanted to add a FYI. MLD
Reply to
MLD
Good comment. I would suspect that it's also possible that you would need to worry about the elasticity of piping and other equipment under pressure in these systems.
Naturally, the siphon system is operating close to atmospheric pressure, so the small changes that you cite above are not a problem. In addition, I've dealt with control systems in "incompressible" flow at low pressures (a few hundred psi), and the fluid compressibility problem doesn't become important for those systems either. This may of course be more due to the fact that the systems I have seen contained centrifugal pumps rather than positive displacement pumps.
Just for curiosity's sake, what frequency range was important in your application?
Reply to
charliew2
To your comment, in my case, I dealt with jet engine control systems where positive displacement pumps (gear and piston). were the prime drivers. Pressures ranged anywhere from 200 to 900 psi in fuel systems to 5000 psi in fuel/oil systems. Typically, the pressure. e.g. from pump to actuator, could change from 200 to 5000 psi in a few millisecs. Additionally, get a little air in the high pressure line and you can expect almost anything to happen. I can recall measuring a flow of about 2-3 gpm going into an actuator piston without it moving for some small fraction of time due to the compressibility of some trapped air in the line that priming couldn't get out. Effectively. a deadband that delayed system response long enough to be responsible for the failure of compressor blading. A 0.010 in. hole drilled in the top of the piston was the fix. Never did worry about or account for the elasticity of the piping etc. MLD
Reply to
MLD
Your problem is interesting - it consists of high pressures and high frequencies. My experience was with systems that sampled their data at approximately 0.1 second frequencies, meaning that pressure pulses due to hydrodynamics occurred so fast that they weren't seen by the control system.
Reply to
charliew2
1. get a 5 gallon bucket. 2. stick it under the outlet pipe and time how long it takes to fill the bucket in seconds 3. divide 5 gallons by number of seconds, then multiply by 60 to get GPM. 4. take repeated measurements as required to arrive at an average.
Assuming you can take physical measurements; why rely upon theory that requires a lot of empirical constants that must be assumed?
Reply to
John Woollen
"John Woollen" wrote
Can you say "design report"?
I have to have said for replacement design/installation.
Love them regulators, Selene
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Reply to
Selene

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