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Liquids: Pascal's Principle and fluid collisions

  1. Jan 4, 2017 #1
    Take a U-shaped piece of pipe, standing upwards. Fill the bottom part with liquid and leave the 'prongs' open. Now slam a parcel of liquid into one end at high speed. The other end will experience some momentum exchange and go pushed upwards.

    Now take the same test case and add some piping that comes from the bottom of the U. That piping is full of fluid and stretches out into a complex branch of piping. That piping has various openings where water could get pushed upwards if it received a force. So the piping looks like this from the side:

    | |
    | |
    |---|---------|----------------| ... complicated pipe branch with some openings.

    Only the bottom level is full in this scenario.

    My question is, if you send in a high velocity parcel of liquid into one of the prongs, it will have a very small effect on the other prong. That is because the force of the collision gets spread out to the entire piping system. So if there are a total of 100 openings along the pipe then the open prong of the U shape will only get 1/100 of the force. It will barely move.

    But in real life you would imagine that dynamic pressure force would somehow move the fluid that is closest to it. So it would have a large impact on the close prong and not much impact further down the complicated piping.

    I know I'm doing a lot wrong here and this is for a simplified simulation, but would appreciate any guidance.
     
  2. jcsd
  3. Jan 5, 2017 #2

    boneh3ad

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    Is there a question here somewhere?
     
  4. Jan 5, 2017 #3
    Sorry I should've made it more clear. The summary of the question is what happens during a liquid 'collision'? Does the momentum from the collision go to the nearest openings or does it get spread out evenly for all openings, due to Pascal's Principle?

    Because if the momentum / energy does get spread evenly then I can think of test cases where that wouldn't seem natural.
     
  5. Jan 5, 2017 #4

    boneh3ad

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    Imagine just slowly pouring liquid into on of the tubes. The velocities involved would be tiny and the propagation speed would be much faster than the rate you are pouring the liquid in and all of the branches would react at the same time and rise to the same height until you stop pouring. If instead you do it very quickly, it means two things could become important:
    1. The speed of adding additional fluid may become comparable to the speed at which its effects propagate through the system (i.e. the speed of sound)
    2. The speed of liquid moving through the system in reaction to the additional liquid becomes large enough that viscous effects become non-negligible.
    Both of these would tend to mean that branches of the system would react sooner and would rise higher than branches farther away from where you added the fluid. However, both of them area also transient, and the whole system will eventually reach equilibrium with all fluid columns coming to rest at the same height.
     
  6. Jan 5, 2017 #5
    Maybe I'm getting mixed up with viscosity and also turbulence. If I have a U shaped pipe and run one end under the kitchen faucet then it's easy to spray water out the other side.

    If you have that same U shaped pipe but it now has a horizontal branch coming off it at the bottom, it will still be easy to spray water through the U section. Yet the tap is only traveling at a small fraction of the speed of sound. So I can only assume this effect would be from viscosity? I wonder if it would be difficult to simulate in a very simplified simulation that I'm attempting. I wouldn't want to spray fluid in the simulation and get virtually no rise in the U shape, it just wouldn't feel right to me.
     
  7. Jan 5, 2017 #6

    boneh3ad

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    Turbulence is an effect of viscosity. Without viscosity, turbulence is meaningless. At any rate, I am not even sure how that worked its way into your discussion.

    If you had a U-shaped pipe and put it under the kitchen sink, then of course you can fill it up and eventually it will overflow on the other end. If you had the same pipe with a horizontal branch at the bottom, then the reaction of any additional vertical tubes coming off of that horizontal branch depends on whether they are far enough away that the propagation speed (sound) becomes important (unlikely), the rate of the fill is large enough that the propagation speed becomes important (even more unlikely) or the rate of the fill is large enough that the water moving through the pipes experiences viscous losses, which will increase with distance from the source of the water (much more likely to become important).

    If instead you had something like a municipal water system and the two portions of your U were located, say, 3 km apart, then it would take the far end about 2 seconds to react to the addition of water on the other end, though it would eventually all reach equilibrium with both sides at the same final height.
     
  8. Jan 5, 2017 #7
    Ok that makes a lot of sense thanks.

    Is there a situation where the kitchen water is traveling so fast that is smashes through the water at the bottom of the U pipe? So effectively the water at the bottom is not able to slow the incoming jet down enough to 'capture' it?

    If so I assume it's a complicated momentum equation. But I would like to model it in a very simplified way if possible. For example all pipes are on a fixed grid and only ever hold 1 liter of liquid, and the physics is run once a second.
     
  9. Jan 5, 2017 #8

    boneh3ad

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    You'd be moving one group of molecules through another in a (mostly) confined space and eventually all of the water molecules from one group would be evenly intermixed with those from the other group. If you had a system that wasn't confined, such as trying to shoot a garden hose through a waterfall sheet, then you could get the stream to come out the other end without fully mixing with the waterfall, but some of the waterfall's molecules will come along for the ride with the stream from the hose, and some of the hose's molecules will be left behind in return to follow along with the waterfall instead.

    What you are asking is essentially a mixing problem. Mixing occurs over a finite length of time and depends on a number of factors, not the least of which is motion (and turbulence!), so the real question you are asking is how fast do these two groups of molecules mix, and that's not a simple question, nor does it have a simple answer. Perhaps a better way to illustrate this is not with a pipe, but water pouring into a large sink. If you had a large, full sink filled with water, and you then started to add water with the faucet, that's a similar situation. Imagine that you had dyed the faucet water a different color so that you can tell it apart from the water that was already in the sink.

    You could certainly add faucet water fast enough that the column of colored water impacts the bottom of the sink if the sink was shallow, but at some point, that water mixes in with the sink water and you end up with a sink full of water that is all a lighter version of the color of the faucet water because all of the molecules from both populations are intermixed. It all occurs due to a mix of diffusion and mixing, and there really is no simple way to model it.
     
  10. Jan 5, 2017 #9
    So if a 1 kg parcel of liquid collides with some other body of static liquid , you could simulate it with:

    1. Convert the velocity of the colliding parcel into kinetic energy using kineticEnergy = mass * squared(velocity) / 2 .

    2. Divide that kinetic energy evenly across each parcel / cell in the entire body, including the colliding parcel.

    3. Convert that kinetic energy to pressure for each cell, which is the same thing in such a simple model.

    Then you can work out static pressure and work out velocity from total pressure and run the physics cycle. Sound reasonable?
     
  11. Jan 5, 2017 #10

    boneh3ad

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    Honestly, what you've just described is essentially Bernoulli's equation (though you didn't account for the pressure energy already present in your 1 kg slug of water). It wouldn't capture nearly all the physics involved, though, as it would neglect viscosity and all of its consequences (in particular, a lot of energy would be dissipated due to turbulent mixing). It also wouldn't handle a transient process like that properly.
     
  12. Jan 5, 2017 #11
    I wanted to use Bernoulli's equation initially but unfortunately it seems to 'fail' with one of the basic test cases:

    If you have an L shaped pipe filled with liquid, then each parcel of liquid along the bottom will have pressure = height . So if the vertical part of the L was 10 high the pressure along the bottom would be 10 for each cell.

    So then you try to calculate velocity for each block which is proportional to pressure. But that now means every block along the bottom has got this new momentum / energy. If you had 100 blocks along the bottom you'd have lots of energy, yet only 10 blocks in the column are causing this.

    The way I see it, I need some way to measure force times *distance* when the column moves down and actually does work on the liquid below. But Bernoulli seems to need constant flow which I guess is what you meant by transient.
     
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