# Understanding what a stagnation point is in compressible flow

1. Jan 8, 2012

### Urmi Roy

Suppose we have a flow of a fluid and let it flow (in x direction) into a stationary wall..the x velocity becomes zero....is this an example of stagnation point?

If not, please cite an example of when that is true.

2. Jan 8, 2012

### Studiot

Hello, Urmi, your example is correct.

You get a stagnation point when the local velocity in one direction is zero.
Of course, it cannot be zero in all directions or there would be a build up of fluid at that point!
So the fluid approaching the wall at right angles 'stagnates' at the face of the wall and turns (exits) parallel to the wall.

Streamlines can terminate at a stagnation point.

Other examples would be the leading (front) edge of an aircraft wing and
the central point of a pitot tube inlet.

go well

3. Jan 8, 2012

### Urmi Roy

But in that case, the velocity of the fluid there isn't exactly zero (it might have a y direction velocity)....moreover, why don't we usually consider the fact that there might be back-flow in the x direction...like in a pitot tube, the fluid might smash into the vertical face and bounce back? We're indirectly assuming that the interaction is inelastic....

4. Jan 8, 2012

### AlephZero

For inviscid flow I would define a stagnation point as where the velocity is zero. Forget about the idea of "in one direction".

The velocity at every point in any flow field is zero in one direction (actually, in an infinite number of directions, in 3D flow) - namely, the direcition(s) perpendicular to the velocity vector!

For inviscid flow, the velocity normal to a boundary must be zero everywhere, otherwise the fluiid would be flowing through the boundary. So on a boundary you could define a stagnation point as zero velociity along the boundary. But you can have stagnation points that are NOT on boundaries, if you ignore turbulence.

The notion of a stagnation point gets much harder to pin down for viscous flow, because the velocity is zero everywhere at the flow boundary. But informally, you can say the stagnation point is "where it would be if there were no viscosity".

5. Jan 9, 2012

### Urmi Roy

Right...so its kinda directional..I'm concentrating on the inviscid flow, since that's what we have in our curriculum....it doesn't have to have zero velocity along all directions...along any one direction of interest will do....

6. Jan 9, 2012

No. A stagnation point is zero in all directions. Think about a sphere moving through a fluid. There will be a point on the sphere where the local surface normal is parallel to the movement direction where the flow will come to a complete stop locally.

If you did it with a flat plate, you could have a jet impinging on the flat plate as you describe. Now look at the flat plat directly and imagine plotting all the velocity vectors along the surface. At some point near the center, there will be one point that is a singularity. In fact, it will be a 2-D source. That point is the stagnation point and has zero velocity in all directions.

This is absolutely false. The concept or stagnation conditions is built on the flow at a given point being brought to rest isentropically. That means brought to rest in all directions. If it is still moving in any direction, you end up with all sorts of calculation errors.

This is not true in general, but in cases where this does happen, we are now talking about a 3-D flow and you will get a stagnation line where the fluid has zero velocity in all directions. This is the same concept as the stagnation line in an unswept wing where the flow along the leading edge line has zero velocity.

Can and do terminate at stagnation points. In fact, this is the only place a streamline can terminate.

Again, at the leading edge of an unswept aircraft wing, you have a stagnation line where the flow does, in fact, move with zero velocity in all directions. For a swept wing, there is no stagnation point. Instead we have what is called an attachment line, which is the analogue of the stagnation line, only the fluid is not actually stagnant so it is not a stagnation point or line.

For a Pitot tube, the flow velocity is zero in all directions at the tip.

Listen to AlephZero, guys. He knows his stuff.

Not directional at all. It has to have zero velocity in all directions.

7. Jan 9, 2012

### Studiot

Hello, Urmi, fluid mechanics can get very mathematical very easily.
There are two approaches to stagnation points. One is engineering one via Bernoulli and Momentum balance. The other is purely mathematical via potential theory or similar.

I am guessing that you are taking the engineering approach?

The difficulty in saying that the fluid velocity is precisely zero is as I have already said.

In a fluid in motion fluid is always moving along the streamlines. In particular it is always moving along any streamline that leads to a 'stagnation point'.
So what happens to this fluid when it reaches a stagnation point?
And how does the next element of fluid arrive if the train (last element) is still in the station?
Clearly the first train (element) must leave the station (stagnation point) before the next can arrive.
But to leave the station it must have a velocity in some direction.

As an engineer we can blithely ignore this dilemma by saying that the fluid element spend only an infinitesimal time at the stagnation point at zero velocity, and then moves off with some velocity.

We can say that the approaching streamline splits at the stagnation point and apply Bernoulli along this compound streamline so that we can calculate the stagnation pressure as I'm sure your course has shown. So perhaps I should have said the velocity is zero along the streamline at a stagnation point - it's just that your streamline was in the x direction.

No the fluid does not 'bounce back' from a pitot tube. If you consider the fluid element impacting on the stationary wall, no the collision is not elastic. Momentum is converted to a measurable force applied to the wall. So when you consider the momentum balance (in the x direction) there is momentum available in the direction of fluid motion, the wall is stationary so contributes zero momentum, and there is no momentum source in the reverse direction, so no 'bounce back' can occur.
If you wish we can explore this idea further.

There is one case where the point actually has zero velocity in all directions. This is the centre point of a body of fluid rotating in a cylinder. The streamlines are concentric circles and the SP the centre. This also illustrates an alternative definition of the SP – A point where the direction of the streamline is indeterminate.

8. Jan 9, 2012

Studiot, that simply isn't true. If you want to illustrate it, take a piece of sheet metal and spread oil (or some other viscous fluid) on it quite thin and then have a blower blow down on it. There will be a spot in the middle where you can clearly see the stagnation point. That is about as non-mathematical as you can get.

The problem is that you are explaining a fluid as if it is a constant stream of particles along a streamline. That isn't how it works. In real life, the molecules are not only undergoing the bulk motion of the fluid, but also their own Brownian motion as well. Fluid mechanics, as a branch of continuum mechanics, only deals with the bulk motion of the fluid. For that reason, you most certainly can have a stagnation point, even in an "engineering" sense. Of course individual molecules will never quite stop there because even if a molecule did follow that infinitely thin stagnation streamline, its own Brownian motion would be enough for it to whiz right off of that streamline and therefore onto a streamline that diverges from that point. Still, if you look at all the molecules, they will fly off of this point in random directions, meaning the vector sum of their velocities is zero. Zero velocity in the bulk fluid equals stagnation point.

9. Jan 9, 2012

### Studiot

What isn't true?

I think we are talking about the same mathematical models but with a slightly different viewpoint.

10. Jan 9, 2012

What isn't true is the idea that a stagnation point need not have zero velocity in every direction. All stagnation points have zero velocity in every direction. All of them. That is the definition of a stagnation point. That is the only way the mathematics work out. That is the only way the physics work out. There are several analogous concepts such as the stagnation line (an infinite array of stagnation points) or attachment lines (which don't actually exhibit stagnation conditions because they aren't stagnant), but to be a stagnation point, the velocity must be zero in all directions.

11. Jan 9, 2012

### Ken G

Post #8 is an excellent "physical" description of why "trains don't pile up" at the stagnation point, and if one wanted to add a more mathematical description in the fluid picture (where you have a density and velocity at every point), one could make the point that "trains piling up at the station" require the density to increase with time. The fluid continuity equation tells us when density will pile up (in steady state)-- it is when the velocity field has a negative divergence. But the divergence of a velocity field talks about how the velocity is changing with position, not what the velocity is at one position-- so we cannot state that the velocity being zero somewhere implies that trains will pile up. We only need a velocity field that has zero divergence-- such a field can still have zeroes in its velocity, consider for example a 2D flow that satisfies dvx/dx = - dvy/dy via vx=-x and vy=y. This has v=0 at the origin, and streamlines along the x axis that lead right to the origin, yet no trains pile up.

We should probably also mention that we are talking about steady-state flows so there is some kind of fixed boundary that v is being measured relative to, giving meaning to a v=0 point. Otherwise we can always enter an arbitrary frame and make any point a v=0 point, but if the boundaries are moving in that frame, the flow won't appear to be steady in that frame.

Also, the OP mentions compressible flow, so if that is really intended, then it is not even necessary for the divergence of the velocity to be zero. However, you have to decide if you are going to allow infinite densities at the stagnation points in your model. If not, then a steady flow must be effectively incompressible at any stagnation point, because advection could not be present to compensate for the piling up due to a nonzero divergence there. In that case, it seems to me that questions about stagnation points that refer to compressible flows are not any different from those restricted to incompressible flows.

Last edited: Jan 9, 2012
12. Jan 9, 2012

### Studiot

Since you don't like trains let me rephrase in more erudite terms.

Let there be a stagnation point S with a streamline terminating or splitting at S, depending upon the text you read.

Now consider an element of fluid on the streamline leading to S, a distance δs away from S.

The velocity of this element is v, directly towards S along the streamline, by definition.

Where does this element go in time δt = δs/v?

Secondly consider another element, distance δs' along the exit streamline from S.

Where has this element come from?

13. Jan 9, 2012

What text says a streamline splits at a stagnation point? That is a mathematical impossibility. Streamlines don't cross, touch or split.

The issue here is that you don't have streams of fluid packets just following along streamlines. Instead, the whole fluid is a continuum and the streamlines are a way of representing the velocity field. Like I mentioned previously, if you had some hypothetical particle of fluid sitting on a streamline, it's own Brownian motion would take it off that streamline. Of course, looking at a fluid in this sense breaks the continuum assumption so the concept of a streamline doesn't even make sense. That is why you won't have fluid "piling up" but can still have a stagnation point in the sense of a continuum.

In a compressible flow, you actually do get a bit of the piling up effect you speak of because the fluid does indeed compress as it approaches that stagnation point, eventually reaching stagnation conditions.

14. Jan 9, 2012

### Ken G

The problem is not with the metaphor, it is with the physics. Streamlines can point to a stagnation point without any piling up, i.e., any need for density to increase with time. I just gave a simple example of that.
Well, let's consider the simple example I just gave, where v along the x axis is equal to -x. So in time dt=ds/v=-ds/x, the parcel finds itself ds/x closed to the stagnation point. Let us now calculate how long it will take to reach the stagnation point: t = integral over dx of 1/v, which is integral over dx of -1/x from s to 0. I'm sure you see that integral is infinite, so the train never reaches the station.
Certainly not from S, because again in the simple example I gave, it would take the element an infinite time to get from S to the point you have in mind.

15. Jan 9, 2012

### Studiot

So now I have two responders avoiding the issue in different ways. No offence meant to either.

Would the professors of Mechanical and Aeronautical Engineering at Glasgow University (Duncan) and Queen Mary College, University of London (Young) do you for a quote from p 60 of their book?

Of course, Ken G (post#14) has shown the mathematical difficulty with this method as has boneh3ad with his (her?) comment about a singularity. In fact the mathematical model just can't hack it. Returning to the real 3D world, in what dimension would the singularity appear?

I am sure we are describing the same model but it is not perfect and capable of interpretation. Except that brownian or other random motion is not admitted in the strict continuum model.

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16. Jan 9, 2012

Like I said, the reason the math works is because the streamlines represent the velocity field. In that velocity field, the velocity at a stagnation point is exactly zero. Individual molecules are not covered under traditional fluid mechanics because they are not a continuum. You don't see fluid piling up infinitely because the random motion of particles prevents any single particle or stream of particles from just following a streamline. That is borne out by reality without violating any of the math that describes stagnation points and streamlines.

Furthermore, even in a real flow where the fluid has to go somewhere, the vector sum of all of those particles leaving from the stagnation point is zero. Plain and simple, a stagnation point has zero velocity in all directions, and this does, in reality, happen.

17. Jan 9, 2012

### Studiot

So do you still disagree with the good Professors?

18. Jan 9, 2012

### Studiot

Ken,

Yes I understand your model which basically says that the closer you approach S the slower the fluid flows (v = -x with S at x=0).

However this is not an artifical flow not a real world one.

What happens when we use Urmis real world example and v = a constant, say χ?

Are you suggesting that the jet does not strike (ie never reaches ) the wall?

Last edited: Jan 9, 2012
19. Jan 9, 2012

Without seeing the figure they reference, I don't want to draw conclusions. It may just be semantics and I don't own a copy of that book and I don't want to overstep without actually reading through what they claim. I can't say that I have ever come across an author who has claimed anything other than streamlines splitting is impossible. It may be that Duncan and Young are treating the stagnation streamline as a special case, but I really just can't infer their meaning without seeing what they did.

While I agree that the flow posited by Ken G is not necessarily a real flow, I also should point out that the flow that you suggest here does not allow for a wall to exist without specifying that $v=x$ online in the free stream. If that is the case, then his flow changes fundamentally in its description as well if you want to only include the free stream.

From a potential flow standpoint, flow impinging on a flat plate can be described by
$$F = Az^{2}$$
Where $F$ is the complex potential, $A$ is a constant scaling parameter (generally assumed to be real, i.e. flow is irrotational), and $z = x + iy$.

Last edited: Jan 9, 2012
20. Jan 9, 2012

### Studiot

Does this not form a flow pattern with streamlines of rectangular hyperbolae about the stagnation point Z0?

And do the axes not form meeting or crossing streamlines, outward on x and inward on iy?