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Suppose I have a pipe, open on one side, with a pressure sensor inside the closed end of the pipe. If the open end of the pipe faces into a uniform airflow blowing towards the pipe, I will measure a positive gauge pressure — this is basically a pitot tube, where I’m converting dynamic pressure into static pressure by stopping the flow.

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--->    ________|

Similarly, if the pipe is turned into an L-shaped tube, nothing changes, as long as one end is sealed so that there is no net flow.

        ________
--->    _______ |
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If I take the L-shaped tube from the previous example and rotate it counter-clockwise around an axis going into the page at the closed end of the pipe, there will be an apparent flow in the same direction as in the previous case, from the reference frame of the opening of the pipe. If the closed side of the L is long relative to its diameter and there is no external flow, the apparent flow near the opening approaches uniform.

Question: In this rotating case, what is the pressure measured at the closed end of the pipe? Is it equivalent to the previous cases (dynamic pressure in the opening’s reference frame converted fully to static pressure), or does the rotating reference frame cause a difference? In particular, is there a centrifugal effect on the air in the pipe that causes a difference? How should I think about this?

(I’m most interested in low speeds, low pressures, incompressible flow in air.)

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  • $\begingroup$ Unclear. Do you mean take that L-shaped pipe and rotate it 90 degrees counter-clockwise, so that the wind is blowing across, not into, the opening? And what do you mean the "flow at the opening approaches uniform"? $\endgroup$
    – Mike Dunlavey
    Commented Dec 5, 2016 at 1:45
  • $\begingroup$ I can try to draw this better. But I'm picturing continuous rotation of the pipe. Here's another way of asking it -- if you had a pitot on the rim of a wheel, with the wheel mounted on a fixed axle, and the pitot opening facing into the apparent flow, the pitot gauge pressure measured at the rim of the wheel would match the dynamic pressure of the apparent flow, yes? Now, if you extended a tube from that pitot to the /hub/ of the wheel, would the measurement be the same or different at the hub? $\endgroup$
    – addaon
    Commented Dec 5, 2016 at 2:03
  • $\begingroup$ My first suspicion as that 1) the pressure would be accurate when pointing directly into the stream, except that 2) centripetal acceleration would decrease it by a predictable amount. If the rotation rate is slow, I suspect you can ignore that. $\endgroup$
    – Mike Dunlavey
    Commented Dec 5, 2016 at 15:23
  • $\begingroup$ Mike, estimating the reduction by centripetal acceleration is exactly what I'd like to understand. For incompressible flow the density is constant, so it's not a buoyancy effect that would cause the acceleration... but it sure feels like there should be something. $\endgroup$
    – addaon
    Commented Dec 5, 2016 at 16:06
  • $\begingroup$ "..If I take the L-shaped tube from the previous example and rotate it counter-clockwise around an axis going into the page at the closed end of the pipe, there will be an apparent flow into the pipe..." Why? Haven't you assumed incompressibility? $\endgroup$
    – Deep
    Commented Dec 7, 2016 at 5:21

1 Answer 1

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Fluid inside the L-tube is stationary. Therefore there is no net force on fluid particles inside the tube. Centrifugal force experienced by fluid particles inside L-tube is being counterbalanced by its inlet walls. Therefore centrifugal effects do not play a role in pressure measurement in the rotating case. But of course there will be pressure variation due to obliqueness of flow which depends on current orientation of rotating pipe w.r.t. flow direction (you must consider relative flow velocities, i.e. switch to the reference frame in which L-tube is stationary).

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