TheBrain states that the “pressure difference … is being caused by the difference in fluid velocities in the tube.” Seeing this, and freightliner’s comments about dynamic pressure, forced me to go back and look at the books again, and sure enough, I neglected this effect. However, it only modifies the situation in two ways.
It turns out that if you insert manometer tubes (open to the air) into the water, the height to which the water will rise in them depends on how the opening in the tube is oriented with the flow rate. If the opening is “facing” the direction of flow, the water level in the manometer tube will rise height D, less some amount due to friction losses between D and the measurement point. If the opening is perpendicular to the flow, it will rise to a lowered level determined by the velocity of the flow. This pressure would determine the height at which cavitation occurs.
(Since manometer tubes measure relative to air pressure, it would be more practical to simply orient absolute pressure sensors in the flow rather than manometer tube inlets for when the pressure drops below atmospheric pressure.)
Note that if the diameter of the siphon tube is uniform along its length, the dynamic pressure is a constant offset through the whole tube. Therefore, it is not a driving force for the siphon effect, though it will reduce the height over which the siphon can work. This is evident as well because a siphon will work even when the flow velocity approaches zero.
I don’t think the “marbles” model can be extended to explain dynamic pressure, unfortunately.
The second alteration to my arguments is that now I think that friction losses in the tube are likely much smaller than the turbulent losses in the receiving container, so my arguments for reconciling the hydrostatic equation don’t hold. Bernoulli’s equation probably gets pretty close when applied between D and B, and the large discrepancy in a similar calculation between C’ and B is due to the pressure gradients involved in slowing down the fluid. In the “marbles” model, this would correspond to having them slide through a long, convoluted “skid area” to slow them down at the outlet of the tube.
Chesnakas suggests the use of mercury in these thought experiments, for its low vapor pressure (nonexistent at room temperature). Another key point is that it also doesn’t adhere (meniscus) to most materials we would make containers and tubes with. A fluid at a temperature such that it has no vapor pressure contained in a material to which it will adhere might be able to support siphoning in a vacuum or above the point at which the static pressure is zero. Maybe mercury in uranium?
TheBrain says in a recent message: “If you do a force analysis you should include in it the tension in the water. If there is no movement in the tube and either side of the tube is the same length then a rope analogy is great to discribe the forces. On either side of the tube gravity is pulling down on the water just like a rope on a pully being pulled on both sides by you but it is not moving. You can pull until the rope breaks which in water would be cavitation. So before the water moves in the tubes we have a pre-stressed water under tension caused by gravity.”
This is incorrect. The atmospheric pressure on both ends of the siphon is forcing the water up into the siphon tube, and the entire volume is under pressure, not tension. The pressure varies according to hydrostatic or hydrodynamic principles (depending on the velocities) but nevertheless it is not in tension. Imagine a silly putty rope hung over a stick… pull on it, and its cross-section will shrink. If you attempt to pull on the mercury in a glass tube with a piston you will find that it will simply separate from the glass walls leaving a vacuum… and the mercury will not be under any “tension,” just zero pressure.