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Say you make a U-shaped glass tube.

If you fill it with water you are able to measure gas pressure by looking at the height difference of the meniscuses.

In an inverted one, by turning it upside down and putting both ends in water (one end in reservoir, and one end in system under test), you should be able to measure the water pressure of the system under test.

How do you make the water reservoir end to exert a known pressure at its meniscus (i.e. calibrate the reservoir end of the U-tube to a known reference pressure, such as 1 atm)?

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  • $\begingroup$ There are too many questions here. An inverted u-tube connected to two bodies of water would measure the pressure differential, and only if the two bodies were at the same height. The delta pressure would be limited by the height of the tube (otherwise you just made a bypass). $\endgroup$
    – Tiger Guy
    Jun 4 at 7:08
  • $\begingroup$ Removed my thoughts of a solution to only have one question. I don't really understand your comment, do you mean if one end of the U was longer than the other? Like a J? $\endgroup$
    – Emil
    Jun 4 at 11:06
  • $\begingroup$ Your title says "calibrate" but the text doesn't seem to address that. Please rewrite, preferablly with a drawing to explain what your setup is and what it is you intend to measure. Further- all you can measure is pressure differences . $\endgroup$ Jun 4 at 12:02
  • $\begingroup$ No, I mean calibrate. The water reservoir end should be connected to the atmosphere somehow so I can compute the difference relative to the atmosphere pressure. And then when the water in the reservoir has been calibrated I can start measuring other test liquids pressure. Or is that not how they work? I have not seen any examples how they work it is purely guesswork drawn from one image, almost impossible to find info (hence my question). $\endgroup$
    – Emil
    Jun 4 at 15:28
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    $\begingroup$ 1barg water = 9.8m height difference, that part is easy. But the difference vs the mirror image is that the air in the "inverse U" is compressible... $\endgroup$
    – Pete W
    Jun 4 at 17:11

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