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Bernoullis Principle Question

  1. Apr 3, 2008 #1
    Hey...

    I have a question about what is causing the Bernoulli effect...a moving fluid (say water) will exert lower pressure on the walls of a container (say a pipe) than a stationary fluid will....and since the water molecules move about randomly...

    There seems to be only two possibilities:

    1. The increase in velocity causes a decrease on the number of molecules in the pipe

    2. The increase in velocity causes a decrease in the energy of the collisions between the water molecules and the walls of the pipe....

    Number one doesn't make as much sense as #2, so I am assuming that #2 is correct...

    So how would a horizontal velocity increase affect the vertical collisions between the water and the pipe?

    What is happening to the water molecules to cause the decreased energy per collision with the pipe?


    Thanks.
     
  2. jcsd
  3. Apr 3, 2008 #2
    #2 is correct, kinda.

    Think of it this way, there is only a fixed amount of energy in the flow (if there is no heat transfer or work done) therefore if some of the energy is acounted for in a kinetic term (the dynamic pressure) then this must effect the other term (static pressure in level flow). So dynamic pressure goes up, statice pressure goes down and visa versa.

    You can actually write out the energy equation and solve it for isentropic flow along a streamline and you will end up with Bernoulli.
     
  4. Apr 3, 2008 #3

    stewartcs

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    Ther Bernoulli effect is a consequence of the law of conservation of energy. So #2 is obviously wrong. The pressure must decrease if the velocity increases so as to conserve energy.

    Here is some more information.

    http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html#beq

    Hope this helps.

    CS
     
  5. Apr 3, 2008 #4
    I'm familiar with the various mathematical methods we can use to derive Bernoulli's law...you can start by defining work or pressure or you can begin with Euler's formula etc....

    I'm looking for a more physical explanation to be honest...

    Since pressure is really just the water molecules hitting the inside of the pipe then the energy or force (or whatever term you want to use) of those collisions MUST be less (than when the fluid was stationary) otherwise the pressure would be the same.

    I'm trying to get a mental picture here of the inter-molecular forces that cause the average collisions to have less energy/force when the fluid is in motion...

    Are you thinking of a situation where we have increased velocity because of a narrowing of the pipe?
     
  6. Apr 3, 2008 #5
    Obviously since you can derive the Bernoulli effect by starting with the conservation of energy.....

    Actually no, you are wrong. The very link you give me (which is hyperphysics, a very watered down website as far as physics explanations go) has this comment in it...

    In the high velocity flow through the constriction, kinetic energy must increase at the expense of pressure energy.

    Please don't just post a hyperphysics link as an answer....especially if you haven't read the article and then post comments that disagree with the very link you provided as an answer!

    The phrase "so as to conserve energy" is not an explanation of what is happening....it is merely a description.

    It doesn't.
     
  7. Apr 3, 2008 #6
    I guess if you wanted to hash together some kind of miscropic description of what is going on, it's possible to think about it in these terms (although I admit this is actually probably wrong, it just sounds nice and intuitive... :D)

    So assume that the fluid particles are specularly reflected from the walls of the pipe, and that they all move approximately the same speed (all have the same mass, non-interacting).

    If you take any point on the inside of the pipe, and over some long time interval [itex]\tau[/itex], measure the angle of incidence of each incident particle at this point on the pipe, you can construct a probability distribution function as a function of the angle of incidence (from 0, to [itex]\pi[/itex]) - so this tells you the probability that a particle hits that one point on the pipe with a specific angle of incidence. If there is no net flow in the pipe, then this probability distribution would be symmetric about the angle [itex]\pi/2[/itex] (and probably should something like Gaussian, or maybe flat - I'm not quite sure).

    The normal force exerted on the pipe at that point (which would translate into pressure) is given by the change in the normal component of the momentum of each particle incident on the pipe. Now given that the particles are specularly reflected, then the particles which exert the maximum force on the pipe are those which are incident with the angle [itex]\pi/2[/itex] (so the particles that approach the point on the pipe "normally").

    Now if you drive a net flow of these particles through the pipe (say in the +x direction), and assume that you have the same number of collisions in the time interval [itex]\tau[/itex], then you can imagine that the probability distribution function will no longer be symmetric about [itex]\pi/2[/itex], but it will be peaked for some angle [itex]\varphi_p<\pi/2[/itex], and depleted in the range [itex][\sim \varphi_p,\pi][/itex] (since, on average these particles are moving in the +x direction - more particles will be hitting that point on the pipe from the left, than from the right). If you now average your momentum transfer with this shifted distribution function, you'll see that the momentum transferred to the pipe will be less than for the static case because collisions on the walls of the pipe are coming from particles with angles of incidence that transfer less momentum.

    Anyway as I said, this is not a very thorough description and probably leaks more water than the above mentioned pipe does - but at least it gives you some idea what is going on microscopically.
     
  8. Apr 4, 2008 #7

    stewartcs

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    Sorry I misunderstood what you wrote, I thought you were saying that energy was lost in number two. However, my comments about pressure decreasing when velocity increasing is absolutely correct and indeed agrees with the link and the quote you took from it.

    Since you, like most people don't post anything about yourself I assumed you needed a watered downed version based on the question you asked.

    CS
     
  9. Apr 4, 2008 #8

    stewartcs

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    I guess I should have read your question a little closer. The title implies you want the macroscopic view of fluid flow and not the microscopic veiw. However, you apparently wanted the microscopic view.

    On molecular level the pressure is a measure of the linear momentum of the molecules as they collide with the wall of a say a pipe or what ever they are flowing in. The pressure decreases due to some of the velocity components (caused by the motion of the fluid) of the random molecules changing in favor of the direction of the flow which is now away from the walls.

    Hence, the momentum and thus pressure is decreased as the fluid flows.

    Hope that helps.

    CS
     
  10. Apr 4, 2008 #9
    I think there can also be a third possibility. Molecules can be so close to each other that a large number of molecules together pushes against the wall with electrostatic force . Remember that this is incompressible fluid, not ideal gas. So we can't assume that each molecule colides with the wall independently.
     
  11. Apr 4, 2008 #10
    Hi. This is my mental picture of what I think is happening:

    1. Lets look at it as that the walls in the pipe is moving and the water staying still (think this makes it easyer to visualize later when the point is comming)
    2. Lets look at the pressure from the watermolecules as one ball being thrown by a person into a wall, bouncing back to the person over and over again letting this one ball create a certain pressure on that area of the wall.
    3. If the wall is now moving the ball is still hitting it with the same frequency but its energy is now spread over a larger peece of wall. (between every impact on the wall there will bee a certain distance)

    hope this is somewhat right.

    So any watermolecule that is in the pipe would have less collisions with the wall.
     
    Last edited: Apr 4, 2008
  12. Apr 4, 2008 #11
    these molecular explanations can't deal with the fact that it's possible to have a pipe with non-moving air with pressure p, and also one with moving air with exactly the same pressure. There's really nothing wrong with the hyperphysics explanation of the bernouilli principle.
     
  13. Apr 5, 2008 #12
    Yes they can. If you have 2 identical pipes with 2 identical pressures, one with a moving fluid, while the other holds the same (but stationary) fluid, what can you say about the temperature of the fluids in each of the pipes? What does this tell you about the average velocity of a molecule in each of the pipes? Then what does this mean in terms of momentum transfer?
     
  14. Apr 5, 2008 #13
    why should the temperature be any different? I can see no reason for that.
     
  15. Apr 5, 2008 #14
    Look at this way: you have 2 pipes, one open ended, one is closed. You're telling me that in the open ended pipe where there is a flow I can have exactly the same pressure as in the first pipe - despite the fact that the fluid is flowing (and that this in some way show's how a microscopic description won't work).

    Take the open ended pipe, and then close the ends (so the flow stops). By Bernouli's equation the pressure should increase. So I'm now left with 2 identical systems, both closed, no fluid flow, yet one has a higher pressure than the other. How can that be? The point is that one fluid is at a higher temperature than the other.
     
  16. Apr 5, 2008 #15

    russ_watters

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    jpr0, in the scenarios you describe, it matters where the pressure and flow are coming from. If you have a large pressure vessel with a pipe attached to it (and no friction), the total pressure is indeed constant and equal regardless of if the end of the pipe is open or closed. That's what Bernoulli's equation says and it is correct.

    In your second scenario, closing the end of a pipe where there is flow (again, assuming the flow comes from a large pressure vessel) will have a transient effect on the flow, but after that settles out the total pressure before and after will be the same.

    The problem here may be just in that you keep saying "the pressure" instead of saying which pressure you are talking about.
     
  17. Apr 5, 2008 #16
    for a gas you have a point. In low pressure areas it will cool. This doesn't work for
    nearly uncompressible fluids such as water. There you could have the following situations.

    1 an open pipe through wich water is pumped.

    2 a closed pipe where we turn the pump down until you get the same pressure.
    The temperatures can be made the same as well.

    according to a theory where the pressure is somehow caused by interactions with the wall this can't happen.

    Bernouilly can only be used along a streamline in the same flow. If you want to compare two different systems, (one open and one closed pipe) you must make sure that there is a point in both systems where the pressure is known to be the same, or you won't be able to draw any conclusions at all.
    For my two pipes I used a pump with a varying output pressure, your conclusion that we'd get two identical pipes with different pressures after closing the open one isn't valid.
     
  18. Apr 5, 2008 #17
    hi firefox, it seems interesting to account for the phenomenon microscopically.

    I have an idea. Take this one as an example :- inside a horizontal pipe with converging cross section, the gas molecules are going downstream. Along the pipe, the gas molecules exchange their momentum with the pipe wall, some in normal-to-wall direction, some in tangential direction. Suppose there is no tangential momentum loss, only normal total-reflection. The molecules, as a whole, tends to reflect back to the upstream due to the geometry, and avoid traveling downstream. This scenario decreases the mean horizontal speed of molecules upstream. If there is nothing driving the molecules, the upstream molecules will eventually stop the mean forward motion. What makes them flows continually? the external pressure gradient is applied to accelerates the molecules. With a proper pressure, the upstream molecules continue going downstream. The acceleration balances the reflection from the converging walls. When going gradually downstream, there should be less and less reflected molecules with opposing velocity. Now, it makes sense that the downstream has a faster mean speed and upstream a slower one.

    For your question, #1 the number of molecules should remain intact by conservation of mass per volume. #2 the downstream molecules have higher mean speed than those of uptream, since the upstream "receives more reflected molecules with opposing velocity". BUT, the upstream molecules have a higher mean molecular speed, since it is at a higher pressure.(Becareful with mean speed(vector mean) and mean molecular speed(scalar mean)) The Bernoulli principle tells you exactly the same thing from a macroscopic view: the pressure energy upstream is converted to downstream kinetic energy of the flow.
     
    Last edited: Apr 5, 2008
  19. Apr 5, 2008 #18
    When I say "pressure" I'm talking about the tangential force extered on the pipe by the fluid divided by the surface area of the pipe.

    If I take an open pipe through which water is pumped, and then a closed pipe where the pressure is turned down until the pressure is equal to that of the first pipe (at the same temperature), then you will have a smaller density of molecules in the second pipe. This decreases the rate at which collisions with the surface occur.

    You can't say that these are incompressible fluids if you can fine tune the pressure from the external pump.
     
  20. Apr 28, 2008 #19
    I think that explanation is a start at the very least....
     
  21. Apr 28, 2008 #20
    Sorry about the tone of my post stewartcs...my apologies.
     
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