1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Askew Pitot tube, airplane velocity measurement

  1. Dec 14, 2015 #1
    1. The problem statement, all variables and given/known data
    A horizontal Pitot tube (with one end open for airflow and another closed) is accidentally installed into an airplane askew, so that the tube is not horizontal. Instead the closed tube end is higher (in the y-direction of a xy-plane) than the inlet end. How does this affect the measurement of the velocity of the airplane?

    2. Relevant equations
    Bernoulli's equation. Solving for velocity it becomes v= √((2*(p_t - p_s)) / ρ_air)
    , where p_t is the stagnation pressure (total pressure), p_s is the static pressure (pressure of the environment, atmospheric pressure at the height), and ρ_air is the density of air.

    3. The attempt at a solution
    I assume that in order to work properly, the Pitot tube needs to be horizontal. In this case, I suppose that the streamlines of the flow hits the wrong surface area (at the opening of a tube), and the flow isn't able to move to the closed end of the tube (which it should do in order to determine the stagnation pressure), because of the inclination.
    As a consequence, the stagnation pressure at the end of the closed tube cannot be measured, since the stagnation pressure occurs at the point when the streamlines hit the inner surface of the tube (here v=0, p=max, which should ideally be measured at the closed end of the tube).

    I assume that when there's no stagnation pressure to be measured (stagnation pressure at the measurement site in the closed end of the tube is 0, p_t = 0), the reduced Bernoulli equation for determining the velocity only uses the static pressure p_s, not the difference between the stagnation pressure and the static pressure (p_t - p_s).
    So, the equation for the velocity reduces to v = √((2*(0-p_s)) / ρ_air) or v = √((2*(p_s)) / ρ_air) ?
    This leads to a wrong velocity measurement, since the stagnation pressure can't be measured.
    Am I on the right track with this issue, or am I in deep trouble?

    Another concern I worked out related to this regards the laminar flow that comes into the tube. Does the laminar flow become turbulent after the point where the streamlines of the flow hits the inner surface of the pitot tube, and does this in turn affect the velocity measurement, since even here stagnation pressure can't be measured?

    Thanks in advance for all your advice!
  2. jcsd
  3. Dec 20, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. Dec 21, 2015 #3
    Well looking at the the full equation and considering two points, p1+row(gh1)+(row*v1^2)/2= p2+row(gh2)+(row*v2^2)/2. You where using the equation assuming no height difference, but in the equation it says the closed end is higher than the open end. So I think you need to take into account the height difference. Also the stagnation pressure wont be zero, the stagnation velocity will be zero.
    Note: I am not completely sure about this, but I think it makes more sense to consider height difference, and also the stag pressure wont be zero.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted