# 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.

• Could you draw a picture of your contraption? – MariusMatutiae Aug 15 '14 at 15:33
• You mentioned gas - so you are interested in compressible flow, right? – Floris Aug 15 '14 at 15:49
• I created a picture of what I think you are asking - please confirm my understanding is correct (or just roll back the edit if it is not...) – Floris Aug 15 '14 at 15:59
• I would suggest dimensional analysis would be a good place to start to obtain a basic approximation. First write down the variables you think might be involved - keep it short it start with - pressure difference, density, diameter, and spacing $d$. Then write down a general equation in which the force equals a product of unknown powers of these quantities. Apply the requirement that the equation must be dimensionally homogeneous, and solve for the unknown exponents. In this case you have two exact limiting cases - you could try to apply them. – akrasia Aug 15 '14 at 21:35
• Please don't expect too much from this method - but it will give an idea of how far you need to move the plate for the force to diminish - but you will say that you knew it all along! – akrasia Aug 15 '14 at 21:36

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.

• This looks like an incompressible treatment of the problem? Note (also shown in diagram) that actual point of greatest velocity happens after the constriction - the Vena Contracta point is where the flow is most constricted. This is exploited in various flow meters. – Floris Aug 16 '14 at 12:17
• @Floris Well, I would expect this type of reasoning to work in some compressible fluid situations. Since they were suggesting dimensional analysis in the comments, I think this is only but a slight sophistication over that, not much more. – Enredanrestos Aug 16 '14 at 12:46
• @Enredanrestos, one problem with this formulation. As d approaches zero, so does flow velocity. So the expression actually goes to zero. Bernoulli's energy equation also has a term for static pressure which was not included in the original premise. So then I suppose the expression you derived only accounts for the dynamic (pressure) force above the static (pressure) force? – docscience Aug 17 '14 at 1:10
• @Enredanrestos, and this does make it more interesting. By the geometry at some distance the cylindrical flow surface area (radial, not axial) will equal the area of the orifice (axial), and that should be the maximum flow and velocity (ignoring any vena contracta). – docscience Aug 17 '14 at 1:23
• I don't know. Probably you are right: all approximations break down from the first equation if the obstacle gets too close to the orifice, since we cannot longer assume that the velocity is the same across the aperture. Perhaps is just coincidence that at d=0 the equation makes some sense at all. Actually, at very short distance, I would expect the flow to suck the obstacle instead of pushing it. – Enredanrestos Aug 17 '14 at 3:06