Siphon Effect in diesel fuel pipeline. 1

[inginear]

I am installing an explosion vent safety device (per NFPA) on a 12000 gallon diesel fuel tank. The tank is fed by a 30000 gallon storage tank up on a hill with a submersed centrifugal pump. The difference in elevation is 33 feet.

I do not think that a siphon effect can happen with this change in elevation. However, I would like to know for sure it will not happen.

Can anyone give me a little help on how to calculate whether or not a siphon effect will occur? Is it just as simple as looking at the potential energy?

Thanks for any help!

 

[katmar]

It would depend to some extent on the length of the line as well as the height difference, but for all practical purposes you can assume that a siphon will be established.

The actual analysis is fairly complicated and involves keeping the Froude number below about 0.31. The theory is well described by PD Hills, Chem Eng, Sept 1983, pgs 111-114

For rough estimates, if the flowrate in US GPM is less than 1.2 x (Diam^2.5) no siphon will form. The diameter is in inches in this relationship.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[BigInch]

Where's your siphon?

If your being feed by a tank on a hill, presumedly the piping is sloping down at all times.

The only place you might have a siphon that I can imagine is if you had an inverted "U" pipe running over the top edge of the tank on the hill and connecting to your feed line to your 12000 gal tank down at the bottom. Is this what you have?

You can get a siphon effect in the inverted "U", if the distance between the top of the inverted "U" and the liquid level of the tank is around 41 feet or less, because diesel has about the same vapor pressure as water, but is only about 85% as heavy.

With a diesel vapor space at the top of the inverted "U", at normal type ambient temps, there would be about 0.5 psi of saturated diesel vapor there, so let's call that zero.

You will have atmospheric pressure at the surface of the diesel in the tank, 14.7 psia

The pressure difference of atmospheric at one end and 0 or 0.5 psi at the top can lift a column of diesel by,

H = (14.7-0) psi * 144 lbs/in2 /0.85 /62.4 pcf = 40 ft

If you pump into that inverted "U", as soon as the diesel gets to the top and flows over the neck, you've got a siphon started, and if the outlet of the inverted "U" outside that tank, is below the liquid level in the tank, you're virtually assured to quickly get some siphon action.

 

[katmar]

One extra point for clarification of my previous post. The analysis by BigInch is correct and you will virtually certainly get a siphon formed for any flowrate if the discharge point is submerged in the lower tank. If the discharge is not submerged, and the flowrate is below the threshhold calculated from the equation I gave before, then air will flow up the pipe and break the siphon.

Like BigInch, I assumed the pipe from the submerged pump ran over the top edge of the tank wall.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[JStephen]

The fluid will flow downhill without being pumped if that's what you mean.

You could theoretically develop a vacuum in the uphill tank and suck it back up into that tank. In real life, with that much elevation difference, you'd probably collapse the upper tank before that happened.

If you're siphoning over a wall or to the top of the tank or whatever, that's a different issue, but you give no information concerning that.

 

[BigInch]

Believe me, somewhere along the joint between the tank wall and the roof will collapse first, if there's no vacuum breaker on that tank roof, or if the vacuum breaker is installed upside down, as I've seen it happen.

 

[blackwed]

I'm confused. One tank is 33 feet above another tank. You move fuel down to the lower tank and the concern is whether or not it will flow back up to the higher tank? I don't see how any piping configuration would cause this to happen as long as the liquid level in the lower tank is lower than the liquid level in the higher tank. Doesn't a siphon work on gravity instead of pressure?

DB

 

[BigInch]

Don't worry, you're not confused. There is no way any fluid particle of diesel will flow back up hill, provided it gets as low as 40' below the top of the inverted "U". Once the liquid goes over the top of the "1/U" and flows down to an elevation 40 ft down below it, that fluid is subject to the same reasoning as before, where I originally said it would not develop a siphon if the top of the 1/U was greater than 40 ft. above the liquid surface in the higher tank. The maximum elevation that atmospheric pressure differential can lift any fluid particle of diesel is 40 ft +/- full stop. If the lower tank has a surface level 40 ft below the 1/U, it requires a pump to get it back into the higher tank, even if there is a complete vacuum in the higher tank. No way it will siphon. (although it is theoretically possible to backflow the first 40 feet of diesel with a complete vacuum in the tank, but as JS says, it would collapse first). JS overly complicated the issue by mentioning that remote possibility. I think maybe he just has a strange sense of humor and didn't think he would confuse you by saying that.

If anyone wants to prove its impossible, put a pump at the high point of the 1/U at the lip of the higher tank and flowing into the high tank, feed the pump from the lower tank with a diesel surface level 44 feet below that pump's top of volute elevation and do a NPSH calculation using NPSHR of 0.5 feet. You will find that NPSHA is less than NPSHR and the pump will sit there and cavitate rather than pump diesel.

 

[cheongcm]

To eliminate siphon effect some people drill a very small hole on top of inverted U pipe (within the tank) to equalise the pressure.

 

[BigInch]

don't try it with a floating roof

 

[blackwed]

Beg to differ, but you're very wrong--I am confused!! But I think it's a matter of semantics though. Please bear with me, I am electrical, but the general concepts of this are directly related to my work so I do have more than an academic interest in them.

I always thought a siphon worked off gravity not surface pressure. The principle will work even if both vessels are in a vacuum. So if pressure is developed inside a vessel and that pressure forces its contents out, I wouldn't call that a siphon action whether the fluid is forced up hill or down. It's just fluid being forced from a pressurized vessel.

On the other hand, if inginear's concern is with fluid siphoning from the upper tank to the lower tank, if the pipe connecting them is full and is not large enough to allow air to break the siphon, I don't see why it wouldn't occur.

Why not use the high point vent with a floating roof?

DB

 

[BigInch]

OK, I'll do my best. Siphons work on differential pressure where that differential pressure was once sufficient to overcome gravity and "start" the siphon. Somehow you had to get fluid to the top of the 1/U and get the pressures balanced. You had to fight gravity by "sucking on the end" and getting a mouthfull of gasoline, or you had to put a submersible pump in the tank and pump it up to the lip, over the top and down the downcomer on the outside. Once all pressures are exactly balanced, only a slight perturbation will allow them to continue. That perturbation is momentum, and some surface tension or capilary action if small diameter piping and maybe some kind of unknown (to me) "stickyness" that fluid seems to have, that pulls a little bit of the trailing fluid along with the moving fluid. Not my description, that came from the Physics guys. When you look at it, you see that the fluid, once started flowing, continues to flow, seemingly without a running power source, and its a mystery, but obviously there is still something powering them.

To start a siphon, you must have a differential pressure to raise the fluid to the top of the 1/U. To do that, you must fight gravity for the run up the riser. Once you get to the top and start to go over, gravity helps on that bit as it starts to move down in the downcomer, but there's still a lot of fluid in the riser, where gravity opposes the flow direction. "If you stop sucking, the gas goes back into the tank". At that point you've only cancelled out a little bit of gravity. As the fluid on the downcomer side gets lower, gravity difference between the riser and the downcomer is more balanced and when the fluid in the downcomer finally gets all the way down to the level of the surface inside the tank (ambient pressure on both liquid surfaces (inside the tank and the liquid surface at the interface inside the pipe, being equal), gravity and pressure are completely balanced. "stop sucking now (almost)." If the pipe ended at that point and was completely open to gravity, the fluid in the downcomer would simply tend to fall out. As it fell out, it would evacuate the downcomer which would tend to reduce the pressure at the top of the "1/U", that was just before completely balanced. That slight pressure difference across the top of the U would allow a little bit of fluid in the riser to flow over the top into the downcomer. The faster the liquid fell out of the downcomer, the faster the fluid would also move over the top. The fluid in the riser is no longer balanced because its lost some of the weight of the fluid at its top, so the pressure in the tank is again sufficient to push a little more fluid into the bottom of the riser and force the remaining fluid in the riser to move slightly higher, once again balancing pressure at the top of the "1/U".

Take the example you mention for the evacuated tank. Think in terms of absolute pressure. If internal pressure is 5 psia, and the siphon is open to atmosphere somewhere, you've got 15 psi pushing in and only 5 pushing out. Let's say we ditch the 1/U for a, /\/\ where the left "/\" is full of diesel and the right "/\" is full of air, and we let the fluid oscillate back and forth for a minute or two and see what happens. Eventually, the "\/" in the middle will reach a point where it has apx 21 more feet of diesel in its left leg than it has in the right leg, oscillations will stop and all flows will halt, because the 10 psi differential pressure would be balanced by the 21 feet of differential head of the diesel in the left leg over that of the right leg. You'd have to connect a "vacuum pump" to the right leg's outlet to get any more diesel out of that tank.

Take a look at Bernoulli's equation and see if it makes sense now. Flow Energy at any point = elevation + Pressure head + velocity^2/2g - Head loss due to flow. If not, see if you can find out how a mercury barometer works. A water barometer is the same except instead of 760 mm, its 33 ft, and a diesel barometer needs 40 ft.

And one more thing... remember that in a closed tank, with just one drop of fluid remaining inside, the absolute pressure in that tank is equal to the vapor pressure of that fluid and that's as much vacuum as you can ever get in that tank at that temp.

Don't feel you have to make excuses. I can't understand how to wire a 2-switch-at-each-door-going-to-the-one-light-in-the-middle-of-the-hallway to save my life.

 

[blackwed]

I understand how a siphon works. My confusion was self-inflicted in that I hadn't considered until just now that, as katmar already said, the pipe could be too large or the initial flow rate too small to allow the pipe to be completely filled. My experiences have been with fish tanks, backyard water features, garden hoses and gasoline and in each case my "systems" have been more than adequate to sustain the action. Clearly not all systems will.

DB

 

[BigInch]

Ugh!

 

[JStephen]

The reason for the confusion is that in the initial post, he mentions a siphon effect, but doesn't give any indication that the geometry is right for a siphon. Either he doesn't mean siphon, or there's something about the geometry that isn't stated in problem.

 

[blackwed]

Bingo!

DB