Assume we have potential flow normal to a zero thickness square-shaped flat panel.

The question is "is the surface potential jump at the perimeter of the plate zero or non-zero - and why?"

A little background:

We know the velocities at the edges are infinite due to the geometric singularity. However, it is not immediately obvious whether the potential jump there must have a value or not.

In standard 2D potential flow books the potenial at the edges is found to be zero (for flow normal to a flat plate). However, I have a problem with this - the solution is obtained as a degenerate case of flow normal to an ellipse which implicitly assumes single-valued and continuous potential on the boundary.

Thanks in advance for any input

Adrin Gharakhani

(1). I am somewhat confused about these mathematical terms of "infinite", "singularity", and "flat plate with zero thickness". I am sure that you are going to receive messages like " overflow", "divide by zero" or something like that. (2). I think, these terms represent logical processes instead of actual values. (3). So, if that is the case, then one can start with a finite size, finite thickness flat plate with rounded ( circular arc) edges and rounded corners. ( It can be a deformed sphere.) Then, by reducing the thickness parametrically, you will get a almost zero thickness, flat rectangular plate, with almost sharp corner. (4). I don't think one should be able to get pointed edge with velocity reaching infinity. Faster than the speed of light at a point? , with acceleration also reaches infinity? (5). I guess, one can stop at a very small number and still have good results. It is hard for me to understand the world where the velocity is infinite at a point.

The question is one of mathematical modeling - not implementation at this stage, so there will be no division by zero, etc. to be worried about.

A zero thickness plate is just that - zero thickness. This is a model for very thin plates. For example assume flow over a "thin" paper within a room. For all practical purposes the paper has zero thickness compared to the room. You cannot possibly assume some (sub)-micron level thickness for the plate and be able to simulate the flow (without being trapped in under/overflow messages during the run). And for all practical purpuses the numbers you get out of such a simulation would be "identical" to those of a correctly modelled zero thickness plate.

So, while physically the velocity is not infinite at the edges (because of viscosity and finite size curvature at the edge), mathematically potential flow (no viscosity) especially over a zero thickness plate will be singular at the edges - i.e. the (potential) velocity will be infinite there.

The question is though, while the velocity is "mathematically" infinite at the edges, what is the potential jump value there (the jump being the potential above minus the potential below the plate). Is the jump zero or does it have a finite (not infinite) value?

Note that the potential jump is equivalent to the pressure drop at the edges and it is quite relevant to "traditional" CFD simulations of flow around bodies with "zero thickness"

Adrin Gharakhani

(1). So, you are not talking about the actual implementation of the potential flow problem. That is fine with me. But, for the example of flow over a thin "paper" within a room, I can easily calculate the size of a room, say, it is 25 feet wide ( 25x12x25.4=7620 mm.). And the thinkness of a thin paper can be 0.1 mm. The ratio is 1 : 76,200. It is large, but not very large. (2). I can understand the velocity of the potential flow. As long as I stay away from the wall, I can always make the assumption that the flow is initially inviscid, uniform or irrotational. (3). Let's jump into the conclusion that there is a solution of potential flow over (normal to) this paper thin plate, then you will have flow moving toward the paper plate. Also, by continuity, you also will have flow moving away from the paper plate. (4). So, as long as we stay away from this singularity, we should have finite values of velocity field. We will also assume that the normal velocity of the fluid particle on the paper plate is "zero". (this must be a physical constraint, because zero thickness plate can not have this property.)(5). At the center of the plate,one can integrate the wall velocity outward, and represent the result by { f(y)- f(y=0) }. Where f is the potential function. On the opposite side of the paper plate, we can do the same thing, and the result will be { F(y)- F(y=0) }. Again, here F is the potential function on the opposite side of the paper plate. And as long as the velocity on the wall can be derived from these two expressions, one can always pick any reference values of f(y=0), and F(y=0). (6). Now, we have solved the potential function distribution on both sides of the paper plate, from the center (y=0) outward. This is fine as long as we stop just ahead of the singularity point.(7). Now it depends on how we model the singularity region. We probably have to zoom into this region and then define the detail geometry and flow behavior there, so that both side of the wall can meet ( a zero thickness paper plate still has two independent walls. This is necessary, because, on one side the flow can move upward, and on the opposite side, the flow can move downward.) (8). We don't know exactly how the flow will behave in the singularity region. If there is a singularity in the flow field, then it is possible that another singularity can also exist in the flow filed. So, a singularity can grow into a line of sheet from the edge outward. In this case, you have f- distribution on one side and F-distribution on the other side of the singularity sheet. In this case the flow on both sides will likely flow outward. (9). If you don't like this model, you can assume that the flow on the front side moves outward and the flow on the opposite side moves toward the center. So, the flow has to make a very sharp turn around the singularity point ( edge) to get the both side connected. In this model, the singularity will be confined to a line representing the edge.(10). Once the model for the singularity region is selected, one can then work out the constraints on the outer boundary. Basically, one has to build up the solution first, the define the boundary conditions. (11). After all these are finished, then one can try to solve the mathematical problem with consistent boundary conditions. (12). Without the help of the viscous effect, one has to define both the solution model and the corresponding boundary conditions. Nothing is unique in the potential flow virtual world. (13). I think this is my approach. ? .

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(9). If you don't like this model, you can assume that the flow on the front side moves outward and the flow on the opposite side moves toward the center. So, the flow has to make a very sharp turn around the singularity point ( edge) to get the both side connected. In this model, the singularity will be confined to a line representing the edge.(10). Once the model for the singularity region is selected, one can then work out the constraints on the outer boundary. Basically, one has to build up the solution first, the define the boundary condition

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OK. We are getting close to the original question, which is "what is the correct model at the edges in a potential flow scenario?" (The question is one of finding the proper BC's)

The flow does take a sharp turn around the edges. And in the "zoomed" version of the flow single-valuedness of the potential does tell us that the upper and lower potentials are equal, which means that the potential jump is zero at the edges.

However, the above is correct in the "zoomed" view of things where the geometry is smooth, the normals to the wall are all well-defined, etc. From a non-zoomed perspective it is not obvious to me whether the potenial jump should still be equal to zero or whether it should have some finite value.

This is quite a similar issue to what people have to deal with when modeling flow about airfoils (the Kutta condition/model and such)

Adrin Gharakhani

(1). If we assume that the flow has to follow the wall and move from the front wall to the rear wall , then outside the very thin layer next to the wall, the streamline must be smooth and continuous. Wehn one integrates the velocity along this smooth streamline to obtain the potential function from the center of the front wall, the potential function must be continuous. This property will continue to exist for the streamlines closer to the wall until it is just outside the singular point. (2). By using the limiting processes based on the smooth streamlines and continuous distribution of the potential function outside the singularity point, one can conclude that the surface potential function is continuous in this model.

By using the limiting processes based on the smooth streamlines and continuous distribution of the potential function outside the singularity point, one can conclude that the surface potential function is continuous in this model.

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This is a circular logic. If you assume the potential (and the streamline) is continuous, of course you'll end up with the conclusion that the potential is continuous! If the first part of the latter statement stresses "just outside", then it is not at all self evident that such a continuous potential just outside guarantees a continuous potential _on_ the surface. Case in point: The velocity is very smooth just outside the edge but _at_ the edge it is infinite (the mathematics of potential flow says so). Keep in mind that there are no boundary layers yet at t=0- !

Adrin Gharakhani

(1). That is the best I can do right now. You may have to find another real expert on potential flow for a better answer.

The potential should be continuous everywhere on the plate. However, as the plate has zero thickness, the velocity at the edge is (d(phi)/dx where dx=0.) infinity and the pressure becomes zero. That makes the edge 'singularity'. Mathematically, there is no problem in the potional function. But in the real world, the flow can not turn 180 degrees even within a small finit distance.

HJ

The potential should be continuous everywhere on the plate. However, as the plate has zero thickness, the velocity at the edge is (d(phi)/dx where dx=0.) infinity and the pressure becomes zero. That makes the edge 'singularity'.

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Let me start by admitting that I believe that the potential should be continuous at the edge (so that the potential jump is zero there). However, I have some doubts (recently).

So, essentially what you are suggesting is that the velocity singularity comes from the geometric singularity - I believe this to be the case as well.

However, let me disagree with your "justification" for the sake of argument (justifiably so). First, if the potential jump is zero then according to your equation above the velocity will be zero/zero = undefined (not necessarily infinite)- by the way for zero plates it's the potential jump that defines the velocity. Second, even if we discount the above, it is mathematically feasible to have a (non-zero) jump at the edge and still have infinite velocity.

That is, continuous or non-continuous potentials will both lead to infinite velocity - which is not my concern. Can we _prove_ mathematically that in the limit of zero thickness the potential jump at the edge will go to zero - i.e., the potential will remain continuous there?

Thanks for the input

Adrin Gharakhani

Question: Can the velocity-field modelled by a vortex around the circumference of the plate also be modelled by a vorticity distribution on the surface? If the answer is no and we accept the proposition that the flow around the flat plate can be modelled by a doublet distribution without potential jump at the edges, this automatically excludes a solution WITH a potential jump at the edges.

(1). As I said before, " It is hard to change the world, and it is impossible to change a person". The world can be changed because the rules can be changed. The person can not change because the person does not know how to erase the program which is stored in his brain. The program in one's brain will stay there forever. (2). As I said, "infinite" is the consequence of the logical limiting process. One can approach the " singularity point", but one can never reach the singularity. (3). By this definition, one has to stay away from the singularity point or points, or if it is necessary to go over to the other side, a model must be created to circle around it. (4). Even if the physical distance to the singularity point is finite, the process to get there has no end. As one gets closer to the singularity point, the velocity gets bigger in a linear fashion ( because the field equation is linear, there is no surprise in the process). (5). And since a point in space has no dimension to it, it will take forever to reach the singularity point.(actually this will never happen) (6). So, as long as one stay away from the singularity point at an infinitesimal distance, everything will be all right and linear. ( It is not a black hole, it is a black point, which will take forever to reach. And I don't think there is anything to be gain there. )

(1) First, (zero)/(zero) is not always undefined in mathematical sense. For example lim(x~0)[SQRT(X)/X] will give you infinite. It depends on how the functions approach to zero. (2) For 2-D problem, you can easily find an exact solution by using conformal transformation. You may find it from you fluids text book.

(1).if we want to get a potential function, we have to integrate(vxdr) along a curve connecting two points. ok, let's integrate(vxdr) from a front side point to a back side point, then you will find at the singularity edge the dr=0,so integrate(vxdr)=0 at this point no matter what the velocity is. i.e. the edge has no contribution to potential function building-up. And, just before the edge (front side)and just after the edge (back side) (vxdr) is exactly the same because of the symmetry. Therefore, the potential function from front side to back side is continuous, no potential jump ocurrs at the edge. let me give a concrete example: assuming the v=VxRpower(n-1), then integrate(VxRpower(n-1)xdR)=VxRpower(n)/n, when n<1 the velocity is infinite at R=0, but the potential function is continueous at R=0. (2) when we say velocity is infinite at the edge, we embed that the coming flow velocity infinitely far is infinite, in actual implementation, it is impossible. so in an actual case, the velocity at the edge is not infinite even if the zero-thickness established. concerning of this point, the potential funtion is even further (sorry, this is not a mathematical term) continuous.