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Is there any way to find the pressure of a liquid inside a closed container without making contact with it?

You can make any assumptions, like transparent container or anything, only rule, the sensing device shouldn't make contact with the liquid.

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  • $\begingroup$ Can you hug your someone without touching her? $\endgroup$
    – user5402
    Apr 17, 2014 at 13:35
  • $\begingroup$ If you know the density of the fluid, then you can calculate pressure based on depth, without having to make physical contact. Does that count? $\endgroup$ Apr 17, 2014 at 13:51
  • $\begingroup$ depth meaning? @DumpsterDoofus $\endgroup$
    – Prashanth
    Apr 17, 2014 at 13:52
  • $\begingroup$ @DumpsterDoofus If the liquid is influenced only by gravity then we may use the depth, but the liquid is already pressurized inside the container and I do not know how much is the pressure - I need to find it from the outside. $\endgroup$
    – Prashanth
    Apr 17, 2014 at 13:56
  • $\begingroup$ @Prashanth You should define making contact. $\endgroup$
    – elv
    Apr 17, 2014 at 13:57

4 Answers 4

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You could measure the deformation of the container. That is, something has to be in contact with the measured medium, except if it behaves different in the electromagnetic spectrum because of a phase change induced by pressure.

There will be miniscule changes in mechanical wave propagation(for small pressure changes), but that is dependent on contact.

Is your container levitatiting in a vacuum, or how do you imagine this?

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  • $\begingroup$ The container is in room conditions, it is pressurized and I can change the pressure inside it. Observing the deformation of the container is a good idea, but to model it, lets say a change in diameter of cylinder under pressure is tough under normal conditions. $\endgroup$
    – Prashanth
    Apr 17, 2014 at 13:51
  • $\begingroup$ If you can access the container from the outside you can add a strain gauge-Wiki. These are usually glued to a surface to measure how much it is strained by a process. Are phyisical modifications on the outside of the container possible? Is the container transparent? $\endgroup$
    – WalyKu
    Apr 17, 2014 at 14:15
  • $\begingroup$ Strain gauge seems to be a good idea! thanks - We can use a transparent container, say some kind of plastic, no problem there. $\endgroup$
    – Prashanth
    Apr 17, 2014 at 14:29
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    $\begingroup$ If you can use different containers and you build the system yourself, I would suggest you install a pressure measurement sensor at the inlet or the outlet and create another sealing point(with container+a little pipe where the sensor is). That way you have an external sensor. Those sensors can for example have capacitive membranes("contatcless"), and would not interfere with your medium. $\endgroup$
    – WalyKu
    Apr 17, 2014 at 14:47
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What about change in refraction index? I imagine that a properly calibrated interferometer could measure absolute refraction index, and taking into account the temperature, you could estimate the pressure. Depending on the liquid, you could also observe dielectric constant (at some frequency, for instance).

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  • $\begingroup$ Dielectric constant seems innovative. Let me experiment with that. $\endgroup$
    – Prashanth
    May 26, 2014 at 9:47
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You can ping the container with, for instance, a shock wave or a high energy laser pulse (for opaque material). Then use an optical or acoustic sensor to measure the vibrational frequencies of the container as it "rings". The frequency will depend on the tension on the container walls due to the internal pressure.

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  • $\begingroup$ That is interesting! do you have any idea about the science behind this technique? $\endgroup$
    – Prashanth
    Apr 18, 2014 at 8:14
  • $\begingroup$ I'm sorry, but I'm not sure I understand. If you set off an explosive, or use a spark gap, you get a shock wave in air. Or, if you use a high power pulsed laser, when it hits a surface it vaporizes a bit of the surface, and blowing off the plasma/debris imparts a mechanical shock to the surface. Is this what you meant? $\endgroup$ Apr 18, 2014 at 18:27
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You can make any assumptions.

My container is leaked near top. A very small hole. This method works best for incompressible and non viscous liquids.

enter image description here

Let the pressure be $P_0$ at top and $P_A$ outside.

$$P_0+\rho g h=P_A+\dfrac1 2 \rho v^2$$

$$H-h=\frac 1 2 gt^2$$

$$R=vt$$

Solve for $v$ from last $2$ and put in first equation to get $P_0$

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  • $\begingroup$ You can make any assumptions, but not change the already assumed scenario, like the closed container and a non-invasive method. Still, this is a nice solution for a slightly different question. good one. $\endgroup$
    – Prashanth
    May 26, 2014 at 9:45

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