Simple expression for transitional pressure force Consider a simple, circular orifice with an upstream, high gas pressure and downstream, normal atmospheric pressure. Consider also a flat circular plate that can be positioned anywhere along the perpendicular axis of the orifice.
At one extreme the plate occludes the orifice and there is no flow of gas. At this position the force on the plate is just the static pressure times the area of the orifice.
At the other extreme the plate is positioned far from the orifice and gas flows at its maximum rate.
Between each of these extremes the force on the plate decreases as it moves away from the orifice. Computational fluid dynamics can solve for the forces, but is there a simple, even approximate way to express the force as a function of the plate's position, the area of the orifice and plate, and the upstream total pressure?
third party edit:
This is roughly what the situation might look like:

original image from http://www.usbr.gov/pmts/hydraulics_lab/pubs/wmm/fig/F14_04L.GIF , but adapted. Question is: what is the force F on the red disk, as a function of distance d from the orifice? Assume that the pipe diameter $D_1$ is very large compared to size of orifice $D_2$ or downstream length of pipe.
 A: I propose the following order-of-magnitude, very rough line of reasoning. The rate of momentum escaping from the aperture is $$\frac{dp}{dt}=\rho\pi(D_2/2)^2 v^2,$$ all in the horizontal direction. I assume that a fraction $(1-\cos\theta)$ of the horizontal momentum will be lost by the fluid since the change of direction of its motion. This fraction of horizontal momentum per unit time is precisely the force applied over the fluid by the obstacle, which by action and reaction is the same that the fluid applies over the obstacle. (I would not expect that the velocity of the fluid change drastically by the presence of the obstacle).

Since $$\cos(\theta)=\frac{d}{\sqrt{d^2+(D_2/2)^2}}=\frac{1}{\sqrt{(D_2/2d)^2+1}}, $$ we get
$$ F \sim \rho\pi(D_2/2)^2 v^2 \left(1-\frac{1}{\sqrt{(D_2/2d)^2+1}}\right).$$
Note that $F\propto d^{-2}$ a $d$ becomes large, and in the limit of small $d$, $F\sim\rho\pi(D_2/2)^2 v^2.$ By Bernoulli you know that, ignoring constants depending of the equation of state,  $\rho v^2 \propto P$, therefore, for small $d$ we conclude that the equation tends to $$ F\sim \pi(D_2/2)^2 P$$ as it should. 
