High-pressure compression of liquids

  • #1
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Hello,

My memories of fluid mechanics are very rusty and therefore I was wondering if you could help me with this hypothetical problem. Let's say I want to generate very high hydrostatic pressures in a liquid (of order of 400 kbar). Would it be enough to do the following: take a rod weighting 1 ton and drop it through a narrow tube with cross sectional area of 2.5x10^-4 m2 from a height of 5m under free-fall, such that upon impact with the liquid at the bottom of the tube the velocity of the rod is 10 m/s. According to the impulse theorem, the force exerted by the falling rod on the liquid when its fall is suddenly stopped (e.g. by a crane) is F=m dV/dt, where m=1000 kg, dv is 10 m/s and dt is for example 1x10-3 s. After dividing the force by the cross-sectional area, the resulting pressure should be 4x10^10 Pa, or 400 kbar. Can you see anything wrong with this reasoning until here?

Also let's assume that I want to redistribute this 400 kbar pressure across a much larger area. Can I take advantage of Pascal's theorem and assume that the 400 kbar pressure generated inside the narrow tube will be redistributed equally across a larger area below the tube inlet?

Thanks you in advance.....


Gabriele
 

Answers and Replies

  • #2
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I guess what makes my problem slightly more complicated is that the pressure wave that I generate in the liquid is dynamic and not static. In any case the pressure wave should still propagate everywhere across the liquid, right? You can assume that the liquid remains contained and gets compressed upon impact of the 1 ton load (i.e. the diameter of the load rod is exactly that of the tube)....

Gabriele
 
  • #3
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I have big problems with this. Pascal's law applies only to a fluid in static equilibrium. The compression wave you create in the tube will rapidly dissipate once it leaves the tube. I also have a problem with your 1 millisecond. What makes you think that is accurate? Also, once the piston bounces of the top of the fluid, all you will have left is a small confined pulse in the tube. What about elastic deformation of the tube itself? The expansion could be significant, and reduce the pressure pulse. If you really want to figure out what happens, you should consider modelling this system.

Chet
 
  • #4
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I agree that Pascal law requires static equilibrium. I need to think more carefully about a way to generate very high pressures in a liquid. Will keep you posted about any new ideas I will have!
 
  • #5
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Actually upon second thoughts I don't understand why Pascal's law shouldn't apply to a pressure wave just as well as to static pressure. Imagine launching a square wave pressure pulse into a liquid, which remains at rest: the wave front will first impact the liquid and by Pascal's law will propagate equally across the liquid bringing all the liquid to high pressure temporarily, and the same will happen to the final release wave at the end of the pressure pulse which brings the liquid back to ambient pressure. Hence the overall result should be the application of two consecutive pressures throughout the liquid! I think when they say that Pascal's law applies only to liquids in static equilibrium they mean the liquid has to remain at rest and not flow with constant velocity in a certain direction, which is the case in my example......


G
 
  • #6
21,428
4,831
Actually upon second thoughts I don't understand why Pascal's law shouldn't apply to a pressure wave just as well as to static pressure. Imagine launching a square wave pressure pulse into a liquid, which remains at rest: the wave front will first impact the liquid and by Pascal's law will propagate equally across the liquid bringing all the liquid to high pressure temporarily, and the same will happen to the final release wave at the end of the pressure pulse which brings the liquid back to ambient pressure. Hence the overall result should be the application of two consecutive pressures throughout the liquid! I think when they say that Pascal's law applies only to liquids in static equilibrium they mean the liquid has to remain at rest and not flow with constant velocity in a certain direction, which is the case in my example......


G
If you can maintain the compression pulse long enough for the pressure wave to propagate throughout the region of interest, then that might be possible. But, outside your tube, there is a much larger mass of liquid to compress. I don't think the velocity of propagation in the larger mass will be the same as for a rectilinear wave, because the propagation will travel spherically (radially outward). The wave equation will be different for the spherical propagation.

Chet
 
  • #7
cjl
Science Advisor
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As a quick approximation, you could assume the pressure wave propagates at the speed of sound in the liquid, so you could get a rough guess for how quickly the pressure wave would reach the other end of the container. If your impact time is substantially longer than the pressure wave propagation time, you might roughly be correct. Of course, calculating an impact time isn't an easy proposition either.
 
  • #8
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Thanks, do you know any software for modelling such simple fluid mechanics problems?


G
 
  • #10
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Well I don't necessarily have to use water... I can use a liquid like silicone oil which is commonly used as pressure-transmitting liquid in Diamond Anvil Cell experiments that routinely reach pressures of order 1 Mbar....


G
 
  • #12
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Yes non-hydrostatic pressure components are something I need to think about, but at least silicone oil remains liquid at high pressures.....
 
  • #13
Nidum
Science Advisor
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