0
$\begingroup$

I have an object that is launched from the side of a plane (it is a flare).

The question is: what does the flare do while it is proximate to the plane? Will it hit the tail?

After accounting for the motion of the aircraft, I have the motion of the flare decently understood and plotted, according to wind resistance without accounting for the propeller wash.


Then it occurs that the flare will be in the prop wash of the aircraft while it is in motion near the plane.


Understanding that the prop wash will be turbulent, and assuming that I find a way to estimate the wind speed of the prop wash, what will be the effect on the force of wind resistance due to the prop wash?

I.e., how will the turbulence of the prop wash mitigate the increased wind resistance due to the airspeed of the prop wash?


To expand on this, and the flexibility of the answer, the parameters are:

  • The velocity of the prop wash is an estimate of the upper bound of the average velocity of the prop wash
  • The purpose of this is to verify that an object coming out of the side of the plane will not hit the tail
  • It is necessary simply that I establish an upper bound on what the wind resistance conditions are going to be, as I already have a code written for the position estimate, relative to the aircraft geometry, speed, altitude, etc.

My current solution is as follows (not ready to post as an answer):

  1. Get an upper bound on power output of the engine (~1900 kilowatts for the turboprop in question here)
  2. Multiply the watts by the reciprocal of the forward velocity of the aircraft and reciprocal of area affected by the propeller to get a force per sq meter
  3. Solve for $v_w$ in the equation $F = \rho A_{prop} v_o (v_w - v_o)$, which is a correction to the estimate in response 3 of the linked forum thread.
$\endgroup$
5
  • $\begingroup$ "Wind speed of the propwash" - what speed are you looking for - speed relative to ground speed, relative to aircraft speed, relative to actual wind speed or the rotational speed ? $\endgroup$
    – Solar Mike
    Jun 27 '17 at 20:15
  • $\begingroup$ You need to edit your question to reflect that then - the hitting the tail part. $\endgroup$
    – Solar Mike
    Jun 27 '17 at 20:20
  • $\begingroup$ @SolarMike updated, cleaning up comments $\endgroup$
    – Chris
    Jun 27 '17 at 20:34
  • $\begingroup$ Sounds like a theoretical question. But for practical purposes, flares ejected from aircraft have an eject vector below and behind the aircraft so that the object cannot impact the aircraft structure. Furthermore, it really doesn't matter what the flare velocity is in absolute terms, it is for all practical purposes falling (gravity) as soon as it is released and for the very brief instant the flare is ejected it could be considered to have a relative (vs. aircraft) horizontal velocity of zero. Prop wash is not a factor is a real world system. $\endgroup$ Sep 26 '17 at 20:54
  • $\begingroup$ @DonaldGibson Except this is incorrect for the CASA in the Jordanian airforce, appreciate it though: the engineers working on it felt the need for an analysis... and, tbh, after completing it and accounting for everything, the worst case scenario did bring the flares reasonably close. $\endgroup$
    – Chris
    Sep 26 '17 at 23:18
0
$\begingroup$

Starting with thrust,

http://nautilus.fis.uc.pt/personal/pvalberto/aulas/cef_mestrado/Air.Resistance.pdf

$$ T_{upper} = P_{shaft} * \frac{1}{v_o} $$

Assuming propeller efficiency is 100% (worst case, in the context of this question).

Next, the relation of thrust to propeller wash velocity with respect to the turbo prop in question is described by the following equation (ignoring exhaust):

$$ T_{upper} = Force = \dot{m}(v_w - v_0) $$

Source: https://www.grc.nasa.gov/WWW/K-12/airplane/turbprp.html

$\dot{m}$ is only as large as the air which the airplane can reach. Assuming that the velocity remains constant (for a worst case analysis) at whichever thrust level is chosen (presumably, max thrust):

$$ \dot{m} = v_o \rho_{air} A_{prop} $$

And, solving for $v_w$,

$$ v_w = \frac{F}{\dot{m}} + v_o $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.