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(Inspired by this answer)

The back pressure is the force on the rear side of an airfoil (in the pressure recovery zone). When the air streams reach the pressure recovery area on an airfoil, they are met with a force, the back pressure. (this is due to the pressure recovery)

Basically it's the zone on an airfoil where the lower pressure, fast moving air meets with the surrounding atmospheric air conditions.

My question is will this force stay the same through different speeds? If not, why is that?

I see no reason for it to change, as the atmosphere will still provide the same amount of static back pressure at any speed, I think. I've seen here that the pressure recovery is also caused by the higher pressure under the wing coming into contact with the upper surface, lower pressure air. This might change things to where at higher AoA the pressure recovery gradient is steeper.

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First, Cp changes along the airfoil. So the pressure is different at different positions.

However, lets assume briefly that the Cp distribution stays the same with Mach number (it does not).

When we dimensionalize Cp to Pressure

$C_p=\frac{P-P_\infty}{0.5\,\rho\,{V_\infty}^2}$

We see that for the same $C_P$ and freestream pressure $P_\infty$, then we will get a different pressure $P$ because $V_\infty$ is different.

So, when Mach number changes, the pressure changes.

In addition, the $C_p$ changes.

In aerodynamics, we can seldom assume things stay the same as other things vary. When we can, it is usually a very clearly laid out principle that traces back to conservation of mass, energy, and momentum.

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  • $\begingroup$ Ah okay. So do you mean the center of pressure when you say Cp? $\endgroup$
    – Wyatt
    Commented Apr 22 at 13:25
  • $\begingroup$ Cp is the Pressure Coefficient @Wyatt en.wikipedia.org/wiki/Pressure_coefficient $\endgroup$ Commented Apr 22 at 16:05
  • $\begingroup$ @OrganicMarble Ah, I thought it might be something like that, thanks. $\endgroup$
    – Wyatt
    Commented Apr 22 at 16:07
  • $\begingroup$ I think I get what you're saying. So because of the higher speed, pressure distribution changes, also changing the pressure distribution where the pressure recovery area on the wing is? $\endgroup$
    – Wyatt
    Commented Apr 23 at 2:19

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