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In a discussion with my father, he argued that a siphon can only pull water up to a limit of ~33 ft (10.1 m). I understand that this limit would apply to water with a density of 1.0 g/cm at sea level (101,325 Pa) being pulled with a perfect vacuum (0 Pa) and ignoring the water's vapor pressure. I feel this limit only applies to cylindrical tube siphons, and that the limit could be increased by using a conical siphon with the wider portion towards the top and narrow portion towards the bottom. Is this true and if so, how can I explain it to my father?

My reasoning is that in theory, the water's surface area at the interface with the vacuum does not determine how far water will travel up the siphon, but instead how much water mass the siphon will support. Being that a cone's volume is 1/3 that of a cylinder with the same diameter and height, I would hypothesize that a conical siphon would be able to lift water nearly 3x as high as a standard siphon.

Being that a conical siphon isn't exactly useful and easy to use in everyday applications, I believe this could actually be accomplished by carefully calculating gradually increasing stepped cylindrical tubes as you increase height to perform the same task. That is start with a tube diameter of x, length 5 m, connect to another tube diameter 2x length 5 m, which is again connected to another tube diameter 4x, length 5 m. This 15 m setup, filled with water, would contain the same mass of water as 1 * 5 m + 1/4 * 5 m + 1x/16 * 5 m = 6.5625 m of the top section of tube, and thus could work as a siphon, as it's well under 10.1 m.

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    $\begingroup$ Possible duplicate of Does the weight of fluid in a conical container act entirely on the base? $\endgroup$ – John Rennie Aug 25 '18 at 8:24
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    $\begingroup$ You can prove for yourself whether this is right or wrong by doing the experiment "in reverse". Make a series of tubes that reduce in diameter, calculate how high you think the water could go, and try it. You only need two tubes, one large and one small - e.g. a large jar with a drinking straw sealed into its lid, or something similar. If the sizes decrease fast enough, you should be able to do the experiment within a manageable height - e.g. 1 meter. $\endgroup$ – alephzero Aug 25 '18 at 9:33
  • $\begingroup$ Related to (or possible duplicate of) $P=F/A$ and $P=\rho*g*h$ give different answers $\endgroup$ – sammy gerbil Aug 25 '18 at 11:24
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Your father is correct in this case. The pressure of a fluid in hydrostatic equilibrium does not depend on the shape of the container, only the depth, the density, and the pressure at the surface. If a fluid is exposed to atmospheric pressure on one side and vacuum on the other side then the vacuum side will only rise high enough for the fluid pressure to equal the atmospheric pressure. Again, this height is independent of the shape.

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