Suction caused by an open window of a plane

Question

So I know this question has been asked before, and I have seen other posts. The reason I want to post a new one is because there is something about the Bernoulli's principle I want to understand.

Is it correct to say in this case:

$$P_1+\frac{1}{2}\rho v_1^2+ \rho gh_1=P_2+\frac{1}{2}\rho v_2^2+ \rho gh_2$$

that the term $\rho gh$ would go away since they are at the same height? So

$$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$$

Now, since it's a moving plane, due to the frame of reference, we can say the air inside the plane is moving and outside isn't (or assume there is barely any wind). Is my thinking correct, where the left side of the equation is for the air inside the plane?

$$P_1+\frac{1}{2}\rho v_1^2=P_2$$

But, inside the plane, the air is pressurised, so the pressure is higher than outside. Now, we added another positive term to it? I know something is wrong, but I don't know where.

The change in pressure would be

$$P_2-P_1=\frac{1}{2}\rho v_1^2$$

and the "suction force" would be

$$F=\Delta P*A=\frac{1}{2}\rho v_1^2*A$$

where $A$ is the area of the open window?

PS: I know doors or windows can't be opened mid air. I just want to know if my understanding is correct!

Answer 1

You are misunderstanding Bernoulli. The quantity $P+\frac 12 \rho v^2$ is a constant along streamlines. The Bernoulli value can jump from one streamline to another when they are separated by a surface with with vorticity. Contrary to popular claims, the pressure in a jet of rapidly moving air has the same pressure as the stationary air through which it moves because the jet is bounded by a surface with vorticity.

Similarly when air outside a window is moving faster than inside it does not mean that the pressure inside is lower than that outside. If this was not true the static ports of the Pitot tube system that tells the pilot how fast the plane is moving would not work.