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Fluids Problem

  1. Mar 27, 2007 #1
    1. The problem statement, all variables and given/known data
    The inside diameters of the larger portions of the horizontal pipe in Figure P9.40 are 2.80 cm. Water flows to the right at a rate of 2.00 10-4 m3/s. Determine the inside diameter of the constriction. Answer in (cm)

    Here is the picture http://www.webassign.net/sf5/p9_40.gif" [Broken]

    Could you guys please show me how to do this? Not just the answer I want to be able to grasp the concept. Thanks
    2. Relevant equations
    I'd Assume
    1/2pv^2 + pgy = 1/2 pv^2 + pgy
    Has something to do with it, but I don't really know where to start.
    3. The attempt at a solution
    Honestly I have no idea where to start. How does the info given relate to anything?
    From the above equation I can get
    1/2(1000)(2e-4)^2 + 1000 *9.8*.1?= 1/2 1000 (2e-4)^2? + 1000 *9.8* .25?

    But this doesnt use a lot of the info. Any help would be appreciated.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 27, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That equation is Bernoulli's equation, but you are missing a term for pressure. (Look it up!) Hint: The heights of the fluid in those vertical segments will tell you the pressure in the fluid at those points.

    Read this: http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html" [Broken]
    Last edited by a moderator: May 2, 2017
  4. Mar 27, 2007 #3
    How exactly does the heights give me the pressure? I though pressure=F/Area?
    And also is the velocity constant for the entire pipe?
  5. Mar 29, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Realize that the fluid in the vertical sections is not moving--so pressure just depends on depth.
    That's still true.
    If the velocity were constant throughout the pipe, you'd have some explaining to do. That would mean that more fluid flowed through the wide section of pipe per second than flowed through the narrow section--but that can't happen, right? (The amount of fluid flowing per second must be the same everywhere--otherwise where is the water going?) Hint: The volume flow rate is given by area*speed.
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