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Homework Help: Measuring velocity in a pitot-static tube and manometer

  1. Apr 4, 2012 #1


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    1. The problem statement, all variables and given/known data

    A Pitot-static probe connected to a water manometer is used to measure the velocity of air. If the deflection (the vertical distance between the fluid levels in the two arms) is 7.3 cm, determine the air velocity. Take the density of air to be 1.25 kg/m3.

    2. Relevant equations

    http://composmentisconsulting.com/12-36.jpg [Broken]

    Pressure in a manometer = ρfluidgh

    Dynamic pressure from the Bernoulli equation = [itex]\rho \frac{V^2}{2}[/itex]

    3. The attempt at a solution

    I reason that the height change in the water of the manometer is due only to the dynamic pressure entering the device. Therefore,

    [tex]\rho_{air} \frac{V^2}{2}=\rho_{H_{2}o}gh[/tex]


    [tex]V=\sqrt{2\left (\frac{1000}{1.25} \right )\left (9.81m/s^2 \right )(.073m)}=33.8m/s[/tex]

    I think I'm doing the math correctly. I just want to be sure my original assumption (that the pressure in the manometer = dynamic fluid pressure) is correct.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 4, 2015 #2
    But, how can I relate this to the Bernoulli's equation?
  4. Nov 4, 2015 #3

    rude man

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    Well, if you don't get any other replies, my take is that you can't. Bernoulli applies to a streamline only, and the pitot and static pressure sensors are not in the same streamline.
    Other takes solicited!
  5. Mar 27, 2018 #4
    Assume your manometer is water. If you look at the manometer on both ends, then you'll see that
    P01 = P02
    Ps1 + 1/2*rhowater*V12 + rhowater*g*h1 = Ps2 + 1/2*rhowater*V12 + rhowater*g*h2
    assuming that the change in height is only dependent on the static pressure,
    dPs,water = Ps2 - Ps1 = rhowater*g*(h1 - h2)

    Now if you assume that the wind streamline starts at zero (1), and compare to the wind at the inlet to the manometer (2)
    P01 = P02
    Ps1 + 1/2*rhoair*V12 + rho*g*h1 = Ps2 + 1/2*rhoair*V12 + rho*g*h2
    assume that the height is negligible
    dPs,air = Ps2 - Ps1 = 1/2*rhoair*(V12 - V22)

    Since the change is static pressure of the wind is directly affecting the change in static pressure at the manometer,
    dPs,air = dPs,water
    rhowater*g*(h1 - h2) = 1/2*rhoair*V22
    If you measure the manometer's change in height, then you can find the wind velocity. So yes, your assumption is correct.
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