Siphon flow 2

[1503-44]

In regards to the disappeared siphon thread, I thought that there was something of real interest concerning the height of the gooseneck over fluid level vs siphon flow rate.

I did the experiment running 2 liters 5 times each through a siphon with gooseneck H over liquid level of
..H.. Qavg L/s
1.25m 0.0131
2.63m 0.0128
4.00m 0.0118
My ladder is only 4m high, so had to stop there.

It is just a few runs at relatively low heads with what seems to be not a lot of variation in Q but, if you plot those Qs vs H in Excel and trend it with a 2nd degree poly I can easily imagine that it hits Q=0 at H = 9.1m (29.85ft)

For now I will believe that flow in a siphon decreases with gooseneck height over fluid level, until someone does this experiment with a taller ladder than mine.



Reality used to affect the way we thought. Now we somehow believe that what we think affects reality.

 

[Compositepro]

If your overall tubing length, and elevation difference between inlet and outlet stays the same, there is no reason to expect any flow change until bubbles form. You may be observing a tubing diameter change due to vacuum in the tube.

 

[Latexman]

Anyone know why the first siphon thread disappearred?

Maybe the OP did it before his delete option went away.

Good Luck,
Latexman

 

[1503-44]

To be clear at what I am attempting to do here is check to see if the maximum velocity equation, as shown at bottom of page

https://en.m.wikipedia.org/wiki/Siphon

has some relation to reality.

Vmax = (2 * (P_atm / rho - g * Hb))^0.5

It seems to calculate the max v & Q at any Hb, when flow is not otherwise restricted by a valve.

Yeah. Could be other effects. The clear PE tubing didn't noticeably change diameter. I doubt that it could appreciably compress the circumference of the 1/2" tube enough to appreciably change the ID & area. The wall is relatively thick. There was no apparent local buckling of the wall. Atmospheric pressure is uniform around the circumference, so it wouldnt get pinched from that without a local buckle. There could have been some ovalling at the gooseneck at times, but that doesn't change x-sectional area very much.

And the result of any of that, if it happened at all, still gives a 0 target at 9.something meters. Which is what it should be if this phenomenon is true. I'd think it would hit 0 far shorter than 9m, if there was some deleterous effect taking place. If the experiment did happen to go bad due to the reasons you mentioned, then if I used stronger copper tubing we might get a much greater head than 9m, no? That would be surprising indeed.

I still have an open mind. Its not my formula. I have no skin in the game. Its just an interesting phenomenon.
So can you prove the equation is not true?
Do you have copper tubing and a ladder taller than mine?






Reality used to affect the way we thought. Now we somehow believe that what we think affects reality.

 

[Compositepro]

There are several inaccuracies in that article (like the pressure in the closed end of a mercury barometer is a partial vacuum. It is actually a full vacuum). But from the article:

"This equation for the velocity is the same as that of any object falling height hC".

This neglects all sources of friction, which is why it is called "maximum" velocity. Your small tube will have significant friction.

The ~9 meters simply is the height of a column of water equivalent to barometric pressure. It seems strange that you are so interested in this.

 

[1503-44]

Since mercury has a vapor pressure, the pressure at the closed end of a mercurial barometer is not a full vacuum, it is equal to the vapor pressure of mercury, so that is a partial vacuum.

Yes the PE tube has friction. 10m long tube. ID 7.5mm Velocities were laminar, f = 64/ Re. Water temp 30C. Velocity estimated at 0.25m/s watching a few tiny bubbles. Of course f changes with velocity:flow rate. Is friction the result of flow rate, or is flow rate the result of friction. There was apparently more friction at higher flows when gooseneck was lowered. Increased friction apparently didn't slow the flow. It increased when flow increased.

9m is also the height where the max flow through the siphon equals zero, which we know to be true. My question was to find out if the equation predicts siphon flow rates at all values of Hb, which it appears to do.



Reality used to affect the way we thought. Now we somehow believe that what we think affects reality.

 

[1503-44]

I've concluded with Vmax being a limit to siphon flow. It is simply the Bernoulli equation solved for V at the siphon's gooseneck. There is only so much total head available there, some of which is the result of static pressure, the rest velocity head. Since atmospheric pressure is essentially the only driver of the fluid entering the siphon, it must represent the maximum total head available at any point within the entrance. That energy must also accelerate the fluid from rest at the siphon's entrance and overcome fluid friction to maintain the velocity up to the top of the gooseneck. So from Patm, subtract Hb and V^2/2/g. What's left must be enough to keep the fluid from vaporising. Kind of like NPSHr for a "siphon pump".

So it is apparent that Hc drives the gravity powered flow rate, but only up to Vmax, at which point the fluid starts to cavitate.

Reality used to affect the way we thought. Now we somehow believe that what we think affects reality.

 

[Latexman]

Thanks for following up.

It's a shame we lost the knowledge in the previous post, which was deleted.

Good Luck,
Latexman

 

[1503-44]

Yes. I'm still PO. Wasted a lot of time looking for that formula somebody posted there. I can't find it anywhere.

There are so many worse questions posted across the forums that somehow don't get deleted, it didn't make sense to single out that one.

Reality used to affect the way we thought. Now we somehow believe that what we think affects reality.

 

[Latexman]

I suspect the OP deleted his first post, which takes down the whole thread.

Good Luck,
Latexman

 

[HTURKAK]

The previous siphon thread disappeared .What was the reason don't know.. However , I got disappointed and doubt to respond again..Maybe the OP did it his delete option went away.


I disagree with this statement.. The velocity is a function of water level height H . H= Water level at inlet - water level at outlet.
However , there is an upper limit for the upstream water level and gooseneck level. The maximum height the water can raise at suction for the siphon theoretically 10.30 m. However, this height reduces for the friction losses, available atm. presssure,density of water, steam pressure of the water.
If the pressure at highest point around - 10.0 meters, the dissolved air will start to escape in the form of bubbles and will be collected at summit , eventually the flow will stop.

In order to keep the flow , the common practice is, to provide safety margin of 2.5-3.0 water pressure to avoid the air build up at summit. That is H max = 10.3- 3.0 = 7.3 m


One of the limit is Vmax.. The other is Patm/(ρg)

If the Atm. pressure 10.3 m and for ideal water, Hmax= 7.3m that is , Hs= P/(ρg) + V**2/2g+ ΔH(local losses) ≤ Hmax= 7.3m.

The altitude at my hometown is around 1300 m and I remember, when I was a child , during summer, the GWL get lower and hand pumps could not obtain water from the well. Bring a bucket of water from somewhere, fill the hand pump with pumping and eventually loose one bucket of water and get nervous..

I think the following doc. is useful to look.
Link

 

[1503-44]

HTURKAK Don't disagree with it. Prove it.

The link you provided states,

The suction head of the siphon is defined from equation (3.36)
H[sub]s[/sub]= P/℘ -v[sup]2[/sup]/(2g) Σ h[sub]l[/sub]

As you can see, the upstream head of the siphon decreases with velocity.
With a given H[sub]s[/sub] and atmos pressure P, v is limited, thereby Q is limited.

v[sup]2[/sup]/(2g) Σ h[sub]l[/sub] = P/℘ - H[sub]s[/sub]

V = √((2g) Σ h[sub]l[/sub] * (P/℘ - H[sub]s[/sub] ))

Case1
For ℘ =1, P = 10, H[sub]s[/sub] =5
V1 = √((2g) Σ h[sub]l[/sub] * 5)

Case2
For ℘ =1, P = 10, H[sub]s[/sub] =1
V2 = √((2g) Σ h[sub]l[/sub] * 9)
V2 = 1.34 * V1

3718m (12198ft)elevation is 15km from here. Do I need to carry my buckets and ladder there?


Reality used to affect the way we thought. Now we somehow believe that what we think affects reality.

 

[HTURKAK]

What is the point that you disagree?

Regarding your calculation ;


Case1
For ℘ =1, P = 10, Hs =5
V1 = √((2g) Σ hl * 5)

Case2
For ℘ =1, P = 10, Hs =1
V2 = √((2g) Σ hl * 9)
V2 = 1.34 * V1

These are the upper limits or ALLOWABLE velocities that CAN be provided by the siphon.

The velocity is the function of H. ( H is the head ; upstream water level- downstream water level).

V= C* SQRT(2gH) . If we neglect the friction and local loses C will be C=1.0 and one can obtain V= SQRT(2gH) which is the free fall formula.

and discharge is, Q=C*A* SQRT(2gH) (look to the formula (3.33)

 

[1503-44]

You said "I disagree with this statement" 22 Oct 20 13:23
Although I do not understand to exactly what you disagree about.

Reality used to affect the way we thought. Now we somehow believe that what we think affects reality.

 

[HTURKAK]

I will believe that flow in a siphon decreases with gooseneck height over fluid level[/b], until someone does this experiment with a taller ladder than mine.

]

I disagree with this statement...

Choose a gooseneck height say 3.0 m, and look to the flow for H=10 m and H= 100 m and see the difference...

 

[1503-44]

I have never been talking about H. The variation of flow with H is not in question.
I have been addressing variation of limit to maximum flow rate with H[sub]s[/sub].

I can rephrase the above a little.

Maximum flow is limited by gooseneck height over upper reservoir surface,H[sub]s[/sub], and the maximum flow limitation decreases with that height.

If you disagree with that above, prove it wrong. Opinions don't count.

Otherwise I dont see the point in doing your multiple calculations in order to prove your opinions about things that we do already apparently agree.

Reality used to affect the way we thought. Now we somehow believe that what we think affects reality.