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I Conservation of Momentum Degrees of Freedom

  1. Sep 11, 2016 #1
    Hi I have been dealing with a fluid mechanics pressure gradient problem and from a statistical view point I can see how it resolves itself but am puzzled as to how it can occur at the molecular scale from a conservation of linear momentum perspective if Momentum is a conserved quantity

    While the pressure gradient must resolve itself in the direction of the flow this requires a change of momentum vector within the flow without a resultant force. (without striking the container surface).

    I don't wish to fully explain the gradient I am looking at as it would take too long and is largely irrelevant to my question about Linear and Rotational momentum.

    The only way I can see this occurs is the use of Degrees of freedom to momentarily change linear to rotational momentum. then back again in the direction of the flow.
    Molecule Degree of freedom.jpg

    Tennis Ball.jpg
    The best analogy I can think of is a tennis ball on string (representing a molecular degree of freedom) is hit against in one direction it rotates around center mass and it hits another ball on the opposite side and imparts momentum in the reverse direction
    If we were to equate that to molecules (P representing momentum)

    rotational momentum.png

    Of course I acknowledge that the exact opposite is equally likely to occur so overall Momentum is Conserved but if you could examine on an individual event basis can rotational momentum be used to reverse a linear momentum vector without a resultant force.

    What am I not considering. Does the center mass go backwards ?
     
    Last edited: Sep 11, 2016
  2. jcsd
  3. Sep 12, 2016 #2

    Dale

    Staff: Mentor

    If you have a pressure gradient then you automatically have a resultant force.
     
  4. Sep 12, 2016 #3
    Thanks Dale could you expand a bit on how a fluid can provide a resultant force can be with respect to Momentum conservation.

    If we treated the fluid as a system then the Momentum (and its vectors) is conserved within that system , there is no way to change the vectors from their original values - without an external force (provided by the walls of the container/pipe)

    Yet we see exactly this in the reordering of momentum in a Bernoulli's pressure difference
    If we have some random motion- Static Pressure, including some going momentum backwards against the flow (-px) that becomes ordered along the flow in a constriction (+px) , then this backwards momentum -px must experience a resultant force. If this this resultant force is within the fluid, then it must equate to the original -px. The negative momentum has just travelled further upstream, it doesn't change the overall -px momentum within the system, just the position of the it .
    ∑Δpx=0

    upload_2016-9-13_10-22-16.png

    In a straight pipe I could envisage this negative momentum will continue to travel upstream in the liquid against the flow until it reaches a solid surface or whatever is providing the pressure difference (represented by my black piston) at this point it can be reflected by the external force and resolve itself/ and the pressure difference in the direction of the flow.

    In this situation I can see that ∑Δpx≠0

    upload_2016-9-13_10-21-59.png

    What am I getting wrong ?
     
  5. Sep 12, 2016 #4

    Dale

    Staff: Mentor

    Hi @Tom79Tom that is a good question, but answering all of that would "hijack" @Quentin_C thread. It would be better to post your detailed question in a new thread.
     
  6. Sep 12, 2016 #5
    Hi Dale,
    No I'm happy for it to be 'hijacked', if you can answer how the Momentum change is resolved 'within' the liquid then I don't need to find the resultant force . My degrees of freedom proposition is only my attempt to understand what's happening to the momentum balance.

    Cheers Quentin
     
  7. Sep 13, 2016 #6

    Dale

    Staff: Mentor

    If you have a pressure gradient, say high pressure on the left and low pressure on the right, then if you consider a small section of fluid there will a large force to the right on the left face and a small force to the left on the right face. So there will be a net force to the right and, in the absence of any other forces in that direction, that small section of fluid will accelerate to the right, gaining momentum to the right.

    You seem to be concerned about the conservation of momentum, but momentum is only conserved in an isolated system, and the small section of fluid is not isolated. Neither is the fluid as a whole.
     
    Last edited: Sep 13, 2016
  8. Oct 7, 2016 #7
    Thanks @Dale

    Yes I very much am confused by thinking of it as an isolated system

    I was envisaging that if the force causes an acceleration in the system I would need to see an equal and opposite force on the system (say the whole system moving backwards) Your saying this does not need to occur ?
    So I can understand a pressure gradient to be an internal net force and I am confusing myself by then still trying to see the Newtons third law - force pair ?
     
  9. Oct 8, 2016 #8

    Dale

    Staff: Mentor

    If the system is not isolated from the environment then, by Newton's third law if there is a force on the system from the environment then there is an equal and opposite force on the environment from the system.

    It is very important to be clear and consistent about your system definition. If you define your system as a small piece of fluid then the 3rd law reaction forces are forces acting on other neighboring pieces of fluid. If you define your system as all the fluid together then the 3rd law pairs will be forces acting on the pipes. Etc.

    So, first clearly define your system. Then it should be clear which forces are internal and which are external.
     
  10. Oct 8, 2016 #9
    Thanks @Dale

    If the system is defined as the whole inviscid fluid

    I don't want to pre-empt your answer but this is what I don't get if it is the internal pressure gradient causing the acceleration.
     
  11. Oct 8, 2016 #10

    Dale

    Staff: Mentor

    If the system is the whole fluid then the forces causing the acceleration are from outside the system, I.e. the walls of the pipe.

    The internal forces due to the pressure gradient are internal forces and cannot cause acceleration of the center of mass.
     
    Last edited: Oct 8, 2016
  12. Oct 8, 2016 #11
    Thanks @Dale Thornton
    That really brings us back to something like the initial question. If for the fluid to be accelerated (at a convergence ) the resultant force must eventually be from outside the fluid where does it come from ?

    I could think of a few alternatives

    1. The walls of the pipe thru friction (but we know that Bernoulli's applies to Inviscid flow)
    2. The walls of the pipe thru some other action. I don't know how considering inviscid stress is always normal where such a force would need to be applied steamwise (parallel) ?
    3. If the pipe had bends the force that was originally parallel would act normally on the wall sections past the bend ?

    1,2,3 all must have the same result . If Center mass must stay unchanged. Then fluid is accelerated then the wall must be "decelerated" -

    4. If not from the walls somehow , then it must be all the way up the pipe from the origin of the pressure difference causing the flow itself - Lets call it a pump

    ΔSP=SP Initial -SP Final

    4 makes the most sense too me but as result we now have added an additional force RF opposing the pump (along with the final pressure SP Final) acting all the way forward to the point of acceleration.

    ΔSP=SP Initial -SP Final- RF

    This would suggest that the pump has to work harder (increase SP Initial) or the flow to the point of acceleration would be slower.

    I haven't seen this suggested by anything I have read on Bernoulli's Theorem
     
    Last edited: Oct 8, 2016
  13. Oct 9, 2016 #12

    Dale

    Staff: Mentor

    It comes from all of the stresses on all of the surfaces interacting with the system. You cannot neglect any of them a priori.
     
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