0
$\begingroup$

Integration of static pressure over wing we get lift.

and

Formula for lift is L= 1/2 x Cl x A x V2 x ro; that implies that static pressure (p - p∞) change exponentially(as square of velocity), becuase A, Cl and ro dont change with velocity in attached flow regime.

One end of differential manometer I connect to small hole that I drill at upper wing surface(hole is drilled perpedicular to wing surface), other end connect to static port. Manometer show -400Pa at 100km/h, does it mean at 150km/h will be -900Pa, at 200km/h = -1600Pa etc ?

Does static pressure (p - p∞) change as square of velocity, in attached flow regime?

$\endgroup$
1
  • 1
    $\begingroup$ Static pressure is constant w.r.t. velocity (by definition). Do you mean dynamic pressure? Or total pressure? $\endgroup$
    – Bianfable
    Mar 22 at 19:31

1 Answer 1

2
$\begingroup$

enter image description herePic source

Indicated static pressure changes proportionally to the square of velocity. Static pressure around an airfoil depends on which point the pressure is measured, and is described as pressure coefficient at a location on the chord. The pressure coefficient is:

$$c_p = \frac{p - p_∞}{q_∞} => p = c_p * q_∞ + p_∞$$

p is the local pressure a particular point on the wing surface. The dynamic pressure $q_∞$ changes with the square of the velocity, but the static pressure has quite a large contribution to the equation. At sea level, with $p_∞$ = 101325 N/m^2 and $\rho$ = 1.225 kg/m$^3$:

enter image description here

$\endgroup$
4
  • $\begingroup$ Is pressure coeficient constant with velocity? If yes, differential manometer show "delta pressure"(static pressure wing - static pressure freestream) = Cp x dynamic pressure, so If Cp is constant then " delta pressure" rise with square of velocity. Am I correct? $\endgroup$
    – Jurgen M
    Mar 25 at 18:13
  • $\begingroup$ Indeed, delta pressure rises with the square of velocity, and then the surrounding pressure is added. $\endgroup$
    – Koyovis
    Mar 26 at 5:00
  • $\begingroup$ Yes only if we assume that Cp remain constant with velocity. But Cp change with Re number so this is not 100% accurate.. $\endgroup$
    – Jurgen M
    Mar 26 at 7:02
  • $\begingroup$ You can edit your answer because differential manometer show delta pressure (p - p∞). $\endgroup$
    – Jurgen M
    Mar 27 at 7:04

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.