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Potential flow uniqueness

by flasherffff
Tags: flow, potential, uniqueness
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Apr25-12, 10:19 AM
P: 10
there is something i cant seem to get about potential flow,

when we work with potential flows we combine some simple potential flows to satisfy some boundary condition (shape of the body and potential at infinity),
we get the resulting flow and we assume that that is the flow in reality

BUT (this is what bothers me) the potential flow that satisfies those boundary condition (shape of the body) is not unique.

example 1: the panel sheet method and vortex sheet method can produce the same external airfoil shape ,but have different flow patterns around them

example 2:the rotating vs the non rotating cylinder
both flows satisfy the same boundary conditions but are different

so... how can we assume that some solution to a given geometry is like reality ,when the solution isn't even unique mathematically
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Apr25-12, 07:52 PM
P: 363
In potential flow, the solution to Laplace's equation is unique to within a specified constant. Therefore to obtain a unique solution it is necessary to specify this constant and this constant is the circulation. In the case of the rotating cylinder the circulation is completely arbitrary but must be specified. So even though the boundary conditions of the non-rotating cylinder and the rotating cylinder are the same the flows are different because for the rotating cylinder there is circulation.

For the airfoil you don't directly specify the circulation but rather you require that the circulation is chosen to satisfy the Kutta Condition which requires the flow to leave the trailing edge smoothly. So if you place panels on the airfoil surface and then place sources on the panels you can satisfy the boundary conditions and create the airfoil shape but there will be no circulation and therefore no lift. And if the airfoil is at an angle of attack the stagnation point on the trailing edge will be in the wrong place. If you instead use vortices on the panels and indirectly specify the circulation by using the kutta condition you will get the correct trailing edge stagnation point and lift will be generated.

So in summary you must specify the circulation in order to have a unique solution.
Apr26-12, 07:32 AM
P: 10
thank you!
this helped alot.

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