1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pressure against atmosphere pressure

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data
    See problem 3


    2. Relevant equations



    3. The attempt at a solution
    In the solution, the interval for the integral of pressure is from [itex]P_{0}[/itex] to the required P.

    However, I wonder that why we don't use net pressure as the intervals.
    i.e. From [itex]P_{0}-P_{1}[/itex] to [itex]P-R[/itex]
    Where [itex]P_{1}[/itex] is the pressure against atmosphere pressure and [itex]R[/itex] is the reaction force act on liquid by the bottom of the tube.

    It seems the interval made by me make more sense than the solution since the interval match the meaning of [itex]dp[/itex] which is net pressure.
     

    Attached Files:

  2. jcsd
  3. Sep 24, 2011 #2

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Re: Centrifuge

    For the integral

    [tex]\int dp[/tex]

    Since the integral contains the term dp, then the variable of integration is p, the pressure. Hence the limits of integration should be values of the pressure.

    Also, I'm not sure what you mean by "net" pressure. Do you mean absolute pressure minus atmospheric (what is usually called gauge pressure)?
     
  4. Sep 24, 2011 #3
    Re: Centrifuge

    Since for any mass element of the liquid, there are two pressure acting on them. One it from the left to right, another one is from the right to left.

    Take the mass element of the liquid closest to the center as an example.
    Atmosphere pressure act from the left to it.
    Another pressure act from the rest of the liquid to it in order to provide the force needed for circular motion.(Is it what we call gauge pressure? I can't remember the terms)

    So each mass element has 2 pressure acting on it. Why we only choose the one on the left as the interval? Can I still obtain the answer if the pick another one as the interval? If no, why not?
     
  5. Sep 25, 2011 #4

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Re: Centrifuge

    Okay, I think I better understand what you are asking.

    We do use both pressures, on the left and right sides of the volume element. Letting p be the pressure to the left of the element, then the pressure to the right is [itex]p + dp[/itex]. Both of these pressures are used to come up with the net force acting on the volume element.

    If you prefer to use p for the pressure to the right, and [itex]p - dp[/itex] for the pressure to the left, then you may do so. Either way, the net force on the volume element is [itex]S \ dp[/itex]
     
  6. Sep 25, 2011 #5

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Re: Centrifuge

    Well, I realize I should add something more to this. The pressure p is a function of position x, i.e. it is p(x). So if you use p as the pressure at the right edge of the volume element, then x would be the position at the right edge of the volume element, not the left edge as shown in the solution you posted. The left edge would be located at x-dx in that case.

    Hope that addresses your question -- though I am not quite certain that it does?
     
  7. Sep 25, 2011 #6
    Re: Centrifuge

    I think I understand a bit more now, thank you very much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Pressure against atmosphere pressure
  1. Atmospheric Pressure (Replies: 2)

  2. Atmospheric Pressure (Replies: 3)

  3. Atmospheric pressures (Replies: 10)

  4. Atmospheric pressure (Replies: 3)

Loading...