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Height of a fluid in a small tube

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data

    An upper tube of cross sectional area A, atmospheric pressure, and height h1 is connected to a lower tube with cross sectional area A/2 (and height 0). A fluid of uniform density rho flows through the system, with velocity v measured in the upper tube. A small open tube branches up off the lower one, with a column of fluid of height h2 inside it. Find h2.

    (See attached diagram.)

    2. Relevant equations

    Bernoulli's equation,

    P1 + pgh1 +1/2*pv12 = constant

    Continuity equation,

    A1v1 = A2v2

    3. The attempt at a solution

    Applying Bernoulli's equation between the top of fluid of height h2 and the left endpoint of the system, I got:

    (point 1 = left endpoint, point 2 = top of the column with height h2)

    P1 = atmospheric
    P2 = atmospheric
    KE1 = 1/2*rho*v2
    KE2 = 0
    PE1 = rho*g*h1
    PE2 = rho*g*h2

    Therefore h2 = h1 + v2/(2g). Is this right? I would think that a faster moving fluid pushes h2 up even higher (with h2 = h1 when v=0).

    The continuity equation tells us that the speed of the fluid inside the smaller tube is 2v. But does this even matter?

    Attached Files:

    Last edited: Apr 10, 2013
  2. jcsd
  3. Apr 10, 2013 #2


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    Welcome to PF!

    Hi Ampere! Welcome to PF! :smile:
    No, you can't do that … Bernoulli's equation only applies along a streamline

    and the streamline doesn't go up that little tube, does it? :wink:
    Yes, because that is in the streamline!
  4. Apr 10, 2013 #3
    Thanks. So in that case you can find the pressure right under the small tube (with height h2 - did this below). But what equation do I use to travel outside the streamline?

    P1 = atmospheric
    P2 = unknown
    KE1 = 1/2*rho*v2
    KE2 = 1/2*rho*(2v)2
    PE1 = rho*g*h1
    PE2 = 0

    So P2 = Patm + rho*g*h1 - 3/2*rho*v2
  5. Apr 10, 2013 #4


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    Hi Ampere! :smile:
    The fluid in the thin vertical tube is stationary, so just use pressure difference = ρgh :wink:

    (technically, you can use any line in a stationary fluid as a streamline … but that doesn't mean you can then tag it onto a "real" streamline in a moving part of the fluid :wink:)
  6. Apr 10, 2013 #5
    So I get

    dP = rho*g*h2
    P2 - Patm = rho*g*h2


    h2 = h1 - 3v2/(2g)

    So the faster the fluid flows, the lower the height h2 is? Is there actually a flow rate which would make h2 = 0, or even less than 0?
  7. Apr 10, 2013 #6


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    Bernoulli's equation does mean that when the speed is faster the pressure is lower. :smile:
    work it out! :wink:
  8. Apr 10, 2013 #7
    I found that the speed at which h2=0 is √(2gh1/3).

    So if h1 = 1m, then v = 2.556 m/s. What if I had v = 3 m/s, or even 6 m/s? What would be the physical meaning of a negative height?

    For example, if we go with h1 = 1m and v = 6 m/s, we get h2 = -4.51m. Huh?
  9. Apr 10, 2013 #8


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    It would mean you have a vacuum pump! :smile:
  10. Apr 10, 2013 #9
    Ah so air would be sucked into the tube along with the fluid? Interesting!

    Thanks, this makes sense now.
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