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Flow speed

  1. Jun 15, 2005 #1
    The horizontal pipe has a cross-sectional area of 40.0 cm^2 at the wider portions and at the constriction. Water is flowing in the pipe, and the discharge from the pipe is 6.0 x 10^-3 m^3/s (6.0 L/s).

    a) Find the flow speed at the wide portion.
    b) Find the flow speed at the narrow portion.

    c) What is the pressure difference between these portions? 1.69×104 Pa
    d) What is the difference in height between the mercury columns in the U-shaped tube? 12.7 cm

    How do I find a and b knowing c and d?

    ** So far I now:
    v1A1 = v2A2
    v2 = A1/A2 x v1
    By applying the principle of conservation of energy:
    p +(0.5*rho*v1^2) + rho*g*y1 = p + (0.5*rho*v2^2) + rho*g*y2
  2. jcsd
  3. Jun 15, 2005 #2


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    How can the pipe have the same cross sectional area at the constriction and the wide portions? Where is the constriction? In the middle of the pipe? At the end of the pipe?
  4. Jun 15, 2005 #3
  5. Jun 15, 2005 #4

    Doc Al

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    Staff: Mentor

    You find a and b before knowing c and d. The volume flow rate is given, as well as the areas of the two pipe sections (at least they are given in your diagram).
  6. Jun 15, 2005 #5
    for part a) v2 = 6.0x10^-3 * (10 cm^3 / 40 cm^3) = 1.50x10^-3 which was wrong
  7. Jun 15, 2005 #6

    Doc Al

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    I'm not sure what you are doing.

    Try this: The flow rate (which is given) = V x Area. So, V = (flow rate)/Area. (Be sure to use proper units.)
  8. Jun 15, 2005 #7
    I converted 10 cm^3 = 1.0 x 10^-5 m^3 and 40 cm^3 = 4.0 x 10^-5 m^3

    v = 6.0 x 10^-3 * (1.0x10^-5/4.0x10^-5) = 1.50x10^-3 m/s ?
  9. Jun 15, 2005 #8

    Doc Al

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    Not good.

    Why are you taking the ratio of the two areas? Realize that the flow rate has units of volume/sec, while speed has units of distance/sec. So, if you checked your units (always a good idea) you would see that this equation cannot hold.

    See my last post for an equation for speed (V) in terms of flow rate and area.
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