# Stagnation Point in Dynamic Flow

Could somebody please explain why stagnation pressure at stagnation point equals to just the static pressure? That is the Velocity of the flow at stagnation point is 0, and therefore the stagnation pressure equals to static from the Bernoulli's equation:

P(stag) = P(static) + P(dynamic) = P + 0.5(rho)(V)^2;

I don't understand why velocity of the flow at that point is 0. The molecules are hitting the surface as they encounter it with some V(final). If the velocity of molecules is suddenly 0, they wouldn't they just start piling up at the surface?

I don't understand why the values are the way they are. My professor couldn't really explain why. It was more like "believe me" :)

The flow is incompressible by the way!

Here's the picture for better visualization:

I think you are misunderstanding the meaning of the equation. You are picturing the equation as describing the behavior of individual molecules as they flow. It's not; the equation is describing a velocity and a pressure field (in the mathematical sense), not what any given molecule is doing. Molecules ("might") move in and out, but the field stays the same (assuming of course a steady flow, which is one of the conditions of the ideal flow). By picturing molecules, you are assuming that there must be something moving with a non-zero speed because molecules are being slammed into the boundary.

First, remember that the velocity in this situation is a continuous field. That is one of the assumptions of the ideal flow, otherwise you necessarily would have separation of the fluid (a "gap" would form) or the fluid would have to compress (by the way, that's the whole reason why the fluid is assumed incompressible), and it would not be a steady flow.

Second, remember that in the ideal flow the velocity vector on a boundary point is always tangent to the surface of the boundary, otherwise (again) the flow would separate from the boundary (or cross it), and then the "boundary" would not really be a boundary.

Let's look at your example. We know that for at least some points ("higher up") at the front edge boundary, the vertical component of the velocity vector is positive (the flow goes around the top of the wing). For at least some points below those, also at the front-edge boundary, the vertical component of the velocity vector is negative (flow goes around the bottom of the wing). Therefore, since the velocity is a continuous field, there must exist at least one point on the wing boundary between the two regions such that the vertical component of the velocity is zero. And if the vertical component is zero, the horizontal component must be zero as well, because otherwise the vector could not be tangent to the boundary. That is the stagnation point, and yes, the velocity there is zero by definition.

(If you have difficulty with the concept of a zero vector being tangent to anything, I'll replace "tangent" with more formal language: the actual requirement is that the dot product of the velocity vector and the boundary's normal vector is always zero: that is, the fluid cannot flow through - or away from - the surface of the boundary. If it cannot flow "along" the boundary, it cannot flow at all)

If you still want to use a "molecule/particle" view of the equation, picture it this way: at the stagnation point, no fluid molecule enters, and no molecule leaves. There is no flow at all.

If you find this contrary to your intuition about reality, you have to remember that this is an ideal fluid. In a real flow, slamming like that against a wing's front edge, the ideal flow conditions cannot be maintained (except as an approximation as long as the flow speed is low enough). The fluid does separate from the boundary at the front edge, and a wedge region forms in the fluid, were eddies develop and which flows rotationaly and independently of the main flow. The ensuing turbulence complicates the whole thing. You have to understand the ideal flow first, to be able to eventually tackle the more complicated reality.

For incompressible fluids, if the local flow velocity at a point P is zero, then P is a stagnation point in fluid flow. Using bernoullis equation, you can easily see that the term 1/2 *(rho)*v^2 ,is zero at P since v is zero. So obviously, only the static pressure will contribute, not the dynamic pressure.

• Yeah, that wasn't the question :) Sorry, I know I wasn't very clear. Why is the local velocity zero at A? What is happening on molecular level? – Micard Sep 11 '16 at 6:33
• Fluids stick to surfaces. The layer of fluid in contact with a surface is at rest, as you can see from the velocity gradient for viscous flow. This is well demomstrated by the Coanda effect. That is why, in this case, the fluid layer in contact with the surface is at relative rest. – Lelouch Sep 11 '16 at 6:47
• I dont quite know about the molecular level stuff here, but you van search about the relevant effects and terms i mentioned to know more . – Lelouch Sep 11 '16 at 6:48
• I'd say that layer is at rest due to adhesive forces(basically due to its viscosity) – Lelouch Sep 11 '16 at 6:50
• And that's what I figured, okay, one layer sticks to the surface. Now why don't other molecules that keep hitting the surface start piling up in front of it, getting rid of their momentum, and coming to rest? As my professor explained it, "imagine the flow goes around the surface not hitting it". Well I can't do that since that's not what's happening. – Micard Sep 11 '16 at 6:53

The velocity they are talking about here is the average (vector) velocity of the molecules, averaged over each tiny parcel of fluid (containing many many molecules). For example, when a rigid solid is moving, its individual molecules each have a different velocity, yet we have no trouble writing F = mdv/dt. In Newton's 2nd law for a rigid body, the velocity we use is the average velocity of the molecules comprising the body. The same concept applies to a flowing incompressible fluid. What's happening is that the fluid is slowing down toward the stagnation point, so the downstream pressure (at the stagnation point) has to be higher than the upstream pressure in order to slow it down.