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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.

        ________
--->    ________|

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.

        ________
--->    _______ |
              | |
              | |
              |_|

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 intoin 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 atnear 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.)

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.

        ________
--->    ________|

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.

        ________
--->    _______ |
              | |
              | |
              |_|

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. If the closed side of the L is long relative to its diameter and there is no external flow, the apparent flow at 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.)

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.

        ________
--->    ________|

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.

        ________
--->    _______ |
              | |
              | |
              |_|

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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    Bounty Started worth 100 reputation by addaon
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Stagnation pressure in rotating reference frame

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.

        ________
--->    ________|

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.

        ________
--->    _______ |
              | |
              | |
              |_|

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. If the closed side of the L is long relative to its diameter and there is no external flow, the apparent flow at 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.)