At the bottom end of the tube, if it were closed off, the pressure (P1) would be equal to the weight of the column of water above it, whereas the pressure at that same level outside the tube (P2) is the weight of the column of water above it, in the reservoir. P1 > P2, so flow is out of the tube into the reservoir.
(In both cases, we also have the weight of the air column above, which is nearly equal, and ignored here.)
No doubt there’s a more sophisticated way of looking at it that’s more puzzling, but this type of analysis works for dealing with normal situations.
But first, Learjeff, air pressure is not required, although you do need gravity. This was addressed in previous posts. Atmospheric pressure is not part of the calculation for differential head. At any value for atmospheric pressure the answer is the same. And as previously pointed out, there is atmospheric pressure on both ends of the siphon so it cancels out.
Liquid head = unit weight of liquid X liquid column height
ABSOLUTE PRESSURE inside the siphon tubes = [atmospheric pressure - (unit weight of liquid X height in the liquid column).
Now onto a question as yet unaddressed:
Why does the water flow down the discharge tube from lower to higher pressure?
Let’s look at the water siphon mentioned above with one foot of liquid head. Look at the differential pressure between the supply tube and the discharge tube. At any points of equal elevation between the supply and discharge tubes the pressure inside the discharge tube is lower than the pressure inside the supply tube. So, for example, at the reservoir water level in the supply tube the pressure is higher (by one foot of liquid head in this example) than at that same elevation in the discharge tube. Same for points 1, 2 or any inches higher, all the way to the apex of the siphon. The differential between the supply and discharge tubes is always one foot of liquid head, higher in the supply tube.
I need to put all this in a concise summary and post it.
So it’s not so much that the water flows from lower to higher pressure in the discharge tube but that it’s “driven” by the differential pressure between the high pressure side (point of equal elevation in the supply tube) to the lower pressure side (point of equal elevation in the discharge tube).
Atmospheric pressure is required in order to use a siphon over a tiny height limited by the cohesiveness of water. In practice, it’s what pushes the water up the tube, just as Cecil says.
You’re right that it’s not part of the equation for the differential head. It is part of the equation for the limit of the height of the tube over the higher water level.
No air pressure differential is required, so no point mentioning it further.
Let’s stick to water as our example. As long as the atmospheric pressure is above 0.26 PSI or so (whatever the vapor pressure of water is at whatever temperature the water is) a water siphon will work. The only reason it won’t work much below that pressure is because the water boils. Note that in the explanation below the atmospheric pressure is not required. You could set atmospheric pressure at any value and you will get the same results. And the siphon will work at exactly the same rate. The rate of flow is proportional to the differential head.
Our example is a water siphon. The reservoir is a large bucket, the suction tube has one end submerged in the water. The tube drapes over the edge of the bucket one foot above the water level. The discharge tube ends one foot below the water level in the bucket.
Let’s look at the absolute pressure at any equal elevation in tubes. Let’s start at the water level. The pressure inside the supply tube is atmospheric pressure. The pressure inside the discharge tube at that same elevation = [atmospheric pressure – (unit weight of water * elevation from the end of the tube)]. In this case that’s about [14.7 PSI – (0.43 PSI per foot head * one foot)], or about 14.27 PSI. So the internal pressures imply that the water will flow from the supply (14.7 PSI) to the discharge (14.27 PSI) tube. The differential pressure is about 0.43 PSI.
What if we pick a spot six inches above the water level in the bucket?
Absolute pressures:
Suction side = [14.7 PSI – (0.43 PSI per foot head * ½ foot)] = 14.485 PSI
Discharge = [14.7 PSI – (0.43 PSI per foot head * 1.5 feet)] = 14.055 PSI
Differential pressure = 0.43 PSI
What if atmospheric pressure is 5 PSI?
Absolute pressures:
Suction side = [5 PSI – (0.43 PSI per foot head * ½ foot)] = 4.785 PSI
Discharge = [5 PSI – (0.43 PSI per foot head * 1.5 feet)] = 4.335 PSI
Differential pressure = 0.43 PSI
You can substitute different atmospheric pressures and different equal elevations get exactly the same driving force, the differential pressure between the tubes.
For any equal elevation the absolute pressure inside the discharge tube will be less than the pressure in the supply tube by whatever the liquid head is. This is true even if the suction tube is larger in diameter than the discharge tube. That’s why siphons work. It’s all gravity and hydraulics, atmospheric pressure not required (except to keep the liquid from boiling).
I’m guessing that Learjeff is asking what the maximum height of the apex of a siphon above the supply level can be. That’s actually a tricky question.
As long as the sum of the lengths of the liquid column on the discharge side (perhaps separated by vapor bubbles?) is greater than the sum of the lengths of the liquid column on the supply side, the siphon will continue working.
The more commonly seen answer is that a siphon will work until
[atmospheric pressure - (the unit weight of the liquid * the liquid head)] is lower than the vapor pressure of the liquid being siphoned. At that liquid head the liquid at the apex of the siphon will boil. You can’t suck the liquid over the apex.
So for water, for example, with a vapor pressure of about 0.26 PSI (yes, I know it varies, I just picked a reasonable value) the siphon could be in trouble when
[14.7 PSI -(0.43 PSI per foot head * feet of liquid head)] < 0.26 PSI, or about 33.5 feet. At that amount of head the water will boil at the apex of the siphon.
However, a siphon can work even under those conditions. As you go down from the apex in both legs of the siphon the pressure increases, so the vapor bubbles will collapse. If the entire siphon tube were filled and capped before bringing the discharge end over the apex and down to its final destination before uncapping, the siphon could work at, say, 36 feet of head. You’d have to experiment.
And of course, different liquids at different temperatures have different vapor pressures.
Whew! I can’t believe there are so many posts to this well understood mechanism! Thank you nolaman for sticking with it. I think Cecil should do a re-write on his answer to this question for clarity.
Yes, agreed. Now, what is the ABSOLUTE PRESSURE when the column is tall enough that the height x density is greater than atmospheric pressure? What does negative pressure mean? Can a negative pressure exist in a liquid?
I think you’re getting sidetracked by thinking about vapor pressure. You’re right in that liquid will evaporate if the pressure above it is less than the vapor pressure (for that temperature); but if the space above it is closed, like the tubes we’re talking about, the newly formed vapor will fill up the space, creating pressure (like any gas in a closed space), until enough has evaporated to create a pressure equal to the vapor pressure (at that temperature). At that point, the water isn’t boiling any more.
It’s been shown thousands of times that with a very long tube, closed at the top, full of water (or mercury) with the lower end in a tub of water (or mercury), the liquid runs out until (at sea level) there’s only about 33 feet of water (or 30 inches of mercury). It’s also been amply demonstrated that this height changes with atmospheric pressure, particularly getting lower at lower pressure (for instance, at high elevations).
So we know that a 40-foot siphon won’t work because you can’t fill up the tube to get it started; we also know that at lower atmospheric pressures, the height of a tube that you can fill is shorter, and therefore the maximum height of a siphon is lower. How can you say that air pressure doesn’t matter?
Quercus, since the pressure inside the vapor space is atmospheric pressure minus something it’s always below atmospheric pressure, right? That pressure is not negative, it’s just less than atmospheric pressure.
You can have negative relative pressure, but not negative absolute pressure. Zero is zero and that’s it.
Now for barometers. The pressure in the void space above the liquid in a barometer is whatever the vapor pressure of the liquid is regardless of atmospheric pressure (within the “normal” range of temperatures and pressures most of us live in). So in a water barometer that would be about 0.26 PSI.
The sum of forces pushing the liquid down inside a barometer tube must equal the sum of forces on the liquid reservoir of the barometer. So if atmospheric pressure is higher, the liquid column must be higher. If the atmospheric pressure is lower, the liquid column must be lower. No matter, the pressure in the void space is still the liquid’s vapor pressure. Within normal atmospheric pressure and temperature variations, liquid barometers are reasonably accurate.
It’s easy enough to write a formula that would calculate liquid height vs. atmospheric pressure with reasonable accuracy.
Atmospheric pressure does change the maximum workable height of the apex of the siphon above the reservoir level. If you had a liquid that did not vaporize at zero absolute pressure you could run a siphon just fine in a vacuum with any apex height.
If you are working with a water siphon, the maximum apex height would be lower at, say 1/2 atmospheric pressure than at full atmospheric pressure, but it would still work. You would have no trouble, for example, siphoning out a 55 gallon drum of water if you just draped the tubing over the edge of the drum. Restating what I’ve said before, the only thing atmospheric pressure is doing for our real-world water siphon is preventing the water from boiling.