1
$\begingroup$

Drag increases with the square of airspeed. Is the same true of the pressure that a forward-facing area experiences?

As an aircraft approaches and then exceeds the speed of sound, how does experienced pressure on a forward-facing area increase?

$\endgroup$
1
$\begingroup$

1) Pressure is simply a force divided by an area, so yes: The dynamic pressure experienced by a forward-facing surface will, in an incompressible flow, vary with the square of the speed.

2) The dynamic pressure increases dramatically as one approaches the speed of sound, once going faster than mach one the pressure will begin to decrease. The trans-sonic phase (around mach one) is difficult to describe and can depend on the airflow around the body - but in many cases it can be quite stable.

-edit- For the second question here's a couple of links that can provide more information:

$\endgroup$
0
$\begingroup$

Drag increases with the square of airspeed. Is the same true of the pressure that a forward-facing area experiences?

Drag increases with the square of the airspeed because it is a function of the dynamic pressure $\bar{p}$:

$$D = C_D A \bar{p} = C_D A \frac{1}{2} \rho V^2$$

So the answer is yes, because the drag is a function of the (dynamic) pressure that a forward-facing area experiences.

I do not speak of the static pressure because that is a function of the atmospheric conditions, and is not a function of the movement of the aircraft through the atmosphere.

As an aircraft approaches and then exceeds the speed of sound, how does experienced pressure on a forward-facing area increase?

I think this has been addressed in the answers here.

$\endgroup$
  • $\begingroup$ to whom has downvoted: care to explain? or should we guess? $\endgroup$ – Federico Jun 3 '16 at 7:39
  • $\begingroup$ The answers to aviation.stackexchange.com/questions/25089 don't describe how pressure on a forward-facing area varies with velocity in the transonic/supersonic regimes. $\endgroup$ – Daniele Procida Jun 7 '16 at 11:01

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.