0
$\begingroup$

I'd like to find the pressure difference between two points $X$ and $Y$ on the Earth's surface given that wind blows steadily with speed $U$ from (say) West to East. Here's a picture: ($z_i$ is the height above sea-level in metres)

enter image description here

I have seen

  • The Bernoulli equation for potential flow:

$\rho \frac{ \partial \phi}{\partial t} + \frac{1}{2} \rho \vec{u}^2 + p + \rho g z = f(t)$

and

  • the Euler equation for a rotating fluid:

$\frac{D\vec{u}}{Dt} + 2 \rho \vec{\Omega} \times \vec{u} = -\nabla p + \rho \vec{g}$

Attempt 1:

If I neglect the rotation of the Earth and use Bernoulli, then I get

$p_X + \rho g z_1 = p_Y + \rho g z_2$,

since

  • if the flow is steady the potential function $\phi$ is independent of $t$, so $\partial \phi/\partial t=0$
  • the $\frac{1}{2} \rho \vec{u}^2$ terms cancel.

This gives a pressure difference of $\rho g(z_1-z_2)$ (where $\rho$ is the air density).

This solution doesn't seem right as I haven't used the extra information about the wind speed.

Attempt 2:

If we further assume that the air is incompressible ($\nabla U = 0$) then $Du/Dt = 0$.

If I can write $\vec{\Omega} \times \vec{u}$ in terms of known quantities, then I can compute $\nabla p$ from the Euler equation and I'm essentially done.

However, I'm not sure how to handle the $\vec{\Omega} \times \vec{u}$ term.

Any ideas? I think I'm overcomplicating this.

$\endgroup$
0
$\begingroup$

Attempt 1 is right. If you want to see a contribution from the Coriolis term, you should make the wind blow into or out of the page rather than along the direction from X to Y.

$\endgroup$
2
  • $\begingroup$ Thanks for the answer; what about the wind speed though? Could you elaborate a bit please? $\endgroup$ – Vadim Mar 20 at 18:19
  • $\begingroup$ The wind speed is irrelevant if its direction is along XY. If you want an answer that involves wind speed, change the wind direction in the question. $\endgroup$ – Ben51 Mar 20 at 18:33

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.