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Ideal fluids, bernouilli's law

  1. Oct 16, 2013 #1
    Screenshot_2013-10-16-23-16-47.png

    I do not understand why B is not true. There is much greater fluid molecules above point A, and given P = density x g x depth, pressure should be greater at point A.....making B true. But then again if u use poiseuilles' law for nonideal fluids you see that in order to keep flow rate Q constant, if radius is larger, then pressure must be smaller at point A....making B untrue. Could you pls explain how to recincile these cinflictinf cinclusions? Thank u so much !
     
    Last edited: Oct 16, 2013
  2. jcsd
  3. Oct 17, 2013 #2
    If you wanna get mathematical, pressure can be defined as force per unit area, so

    [tex]P = \frac{F}{A}[/tex]

    In the case of your picture, the cross sectional area of A is larger than B, so that would mean the pressure is less at A than at B
     
  4. Oct 17, 2013 #3
    Thank u for your reply :) but how come we can assume that force is equal at points A @nd B?
     
  5. Oct 17, 2013 #4
    The explanation is misleading. One should not analyze the situation from this perspective because doing so will cause you to hold F constant, which doesn't make sense.
     
  6. Oct 17, 2013 #5
    You are using an equation from fluid statics to analyze a dynamic situation.

    In dynamic fluid, pressure is never simply density x g x depth. You can study bernoulli's equation and realize that there are static pressure, dynamic pressure and pressure caused by gravity.
     
  7. Oct 17, 2013 #6
    Okay I guess it's a little more complicated than that, let's see:

    You have your continuity equation for fluids

    [tex]\rho_{1}A_{1}v_{1}=\rho_{2}A_{2}v_{2}[/tex]

    The problem says that this is an ideal fluid, so the fluid isn't compressible, so

    [tex]\rho_{1}=\rho_{2}[/tex]

    and

    [tex]A_{1}v_{1}=A_{2}v_{2}[/tex]

    And since A = F/P

    [tex]\frac{F_{1}}{P_{1}}v_{1}=\frac{F_{2}}{P_{2}}v_{2}[/tex]

    From a simple thought experiment we can conclude that the velocities are different, think about putting your thumb over a running water hose. And we have already seen that the pressures are the same. This leaves us with

    [tex]F_{1}v_{1}=F_{2}v_{2}[/tex]

    Since the velocities are different, we can conclude that in fact, the forces must be different if both sides of the equation are to be equal.
     
  8. Oct 17, 2013 #7
    ##P_{1} = P_{2}## ? How is it possible?
     
  9. Oct 17, 2013 #8
    Oh wow I just realized I wrote that, allow me to slowly walk away, as I clearly don't know fluid mechanics :sly: I'll go ahead and blame the time of night
     
  10. Oct 17, 2013 #9
    Thank you both the disvussion was very helpful!!!!
     
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