One question about stream function in fluid mechanics

  • #1
Hi....

Suppose we consider a circular cylinder moving with constant velocity U in x-direction in a two-dimensional unbounded, irrotional, incompressible, inviscid fluid. If the motion of the fluid is completely resulted from the motion of the body, we know the velocity field of fluid can be described by ( \psi_y, -\psi_x), where \psi is the stream function which satisfys the boundary condition

\psi = y + constant, on the cirlce

and also

( \psi_y, -\psi_x) goes to zero as (x,y) goes to infinity

and it satisfies the laplace equation as well.

My question is , is this stream function unique up to a constant?

Actually if the radius of the circle is R, one solution of \psi is

\psi = y U R^2/(x^2+y^2).

But I think

\psi = y U R^2/(x^2+y^2) + log ((x^2+y^2)/R^2)

is also a solution since it satisfies all the conditions.

Thanks a lot.
 

Answers and Replies

  • #2
Clausius2
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aqualonebear said:
\psi = y + constant, on the cirlce

That does not represent the surface of the circle. Use polar coordinates instead.

My question is , is this stream function unique up to a constant?

That is true. Usually the constant is chosen to be zero. It has no influence on the flow field.

\psi = y U R^2/(x^2+y^2) + log ((x^2+y^2)/R^2)

I really don't see how is that a solution of irrotational flow. Maybe I'm wrong, can you show that your stuff is indeed the solution to your problem?
 
  • #3
Thank you for the reply.

I think I didn't make the problem clear. The fluid is assumed to be irrotional, inviscid, and incompressible. So when there is a body moving in the fluid, there is no friction on the surface. In other words, when the circular cylinder (x^2+y^2 =R^2) has a constant velocity U along x-axis, the boundary condition requires that the fluid velocity is tangent to the surface of the body, or the fluid velocity and the body should have the same normal component along the boundary. If we denote the stream function by \psi, then the velocity field of the fluid is (\psi_y, -\psi_x) for (x,y) in the fluid domain, which is outside the circle. Here, \psi_y is the partial derivative of \psi with respect to y. From the boundary condition on the circle, we should have along the boundary

\psi_y n_x - \psi_x n_y = U n_x

From this equation, it can be solved that in general (see "theoretical hydrodynamics" by milne-thomson)

\psi(x,y) = U y + constant, on the boundary


For this simple case here, the stream function can be found as

\psi = y U R^2/(x^2+y^2), for (x,y) in the fluid domain

We can directly verify that (\psi_y, -\psi_x) is tangent to circle. However if we add f(x,y) = log((x^2+y^2)/R^2) to \psi, the flow is still irrotional since f satisfies laplace equation in the fluid domain and the singular point is inside the body. Basically f(x,y) adds a circulation around the body so that all the fluid particles are rotating about the origin but at each point the vorticity is still zero.

So is

\psi = y U R^2/(x^2+y^2) + f

a stream function corresponding to the translation of the circular cylinder?

Thanks.
 
  • #4
Clausius2
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That's a good question. [tex]log(r)[/tex] is still a solution of the laplace equation, but in this context it is an Eigensolution. You can add poles inside the cylinder, and still the solution outside is still well posed. On the other hand, the uniqueness of the solution comes from the matching with the INNER solution (the solution of the flow inside the boundary layer). A multipole in the origin is surely unmatchable with the solution of the boundary layer. Do I make myself clear?.
 
  • #5
Clausius2 said:
That's a good question. [tex]log(r)[/tex] is still a solution of the laplace equation, but in this context it is an Eigensolution. You can add poles inside the cylinder, and still the solution outside is still well posed. On the other hand, the uniqueness of the solution comes from the matching with the INNER solution (the solution of the flow inside the boundary layer). A multipole in the origin is surely unmatchable with the solution of the boundary layer. Do I make myself clear?.


Thank you. But could you explain it in detail? I don't quit understand. Should the stream function satisfy the lapace equation both inside and outside the boundary? I think when we assume the flow is inviscid, we don't consider the boundry layer thus there is no "friction" between the fluid and the boundary. In other words, in the ideal fluid, we allow the fluid to have nonzero velocity relative to the boundary but it should be tangent to the boundary.

We can consider the following problem first:

Find a function f(x,y) such that f satisfies the laplace equation in the domain D :={(x,y) | x^2+y^2 >= 1} and on the boundary (x^2+y^2 =1), f is 0.

A trivial solution is that f is constantly zero in D, but f = log(x^2+y^2) is also a solution.

Now consider my original question about the moving cylinder in an unbound ideal fluid. I think the only conditions for \psi is that it should satisfy the laplace equation in D, and on the boundary of D, \psi = y + constant, and also (\psi_x, \psi_y) goes to zero as (x,y) goes to infinity. Those conditions come from the potential flow theory and the assumption that the fluid is ideal, or , it is inviscid, incompressible, and irrotional.

So if \psi_1 is a stream function, why \psi_ + log(x^2+y^2) is not?
 
  • #6
Clausius2
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Sorry for not being too clear. What I am going to do is to translate your formulation to the common one, in polar coordinates. Don't be worried, the results are the same ones that you are obtaining. I only want to clarify the things to the foreign reader.

Your problem is a translating cylinder with velocity [tex]U[/tex] in a fluid at rest. You are considering plane potential flow ([tex]Re\rightarrow \infty[/tex]), with the reference frame attached to the fluid at rest. In stream function formulation, your problem is:

[tex]\nabla^2 \psi=0[/tex]

with [tex]u_r=\frac{1}{r}\frac{\partial \psi}{\partial \theta}=Ucos\theta[/tex] at [tex]r=R[/tex] and [tex]u_r,u_\theta\rightarrow 0[/tex] as [tex]r\rightarrow \infty[/tex].

The solution to this problem, as you point out, is [tex]\psi=-U\frac{R^2}{r}sin\theta[/tex]. This IS the actual solution of YOUR problem.

On the other hand, the radial laplacian operator has a fundamental solution, an Eigensolution: [tex]log(r)[/tex], which turns out to be the Green's Function of the radial laplacian operator. If you add this eigensolution to the former solution the flow field becomes altered, but still the new solution yields the boundary conditions:

[tex]\psi=-U\frac{R^2}{r}sin\theta+A\cdot log\left(\frac{r^2}{R^2}\right)[/tex].

The constant [tex]A[/tex] represents, as you well said, a measure of the circulation in the flow field. The explanation of why [tex]A=0[/tex] in your problem, in which the cylinder only translates, can be stated in two ways, which are equivalent:

1) The circulation in the flow field is 0. There is no reason for thinking that the translation of the cylinder introduces circulation, due to the fact that the problem contains symmetry. By introducing the logarithmic term you are introducing a point vortex at the origin, a singularity. The vortex is inducing azimuthal rotation in the flow field. Due to the fact that the cylinder does not rotate and the fluid is at rest at infinity [tex]A=0[/tex].

2) As it is said, the eigensolution is a measure of the circulation in the flow field. How is the circulation introduced?. In this problem the circulation may be introduced by means of rotating the cylinder. DO NOT forget that potential flow is only an asymptotic expansion of N-S equations for [tex]Re\rightarrow\infty[/tex], and therefore corresponds to the leading order term, not taking into account viscous effects. But these effects do exist on the solid boundaries. The circulation itself is introduced by means of viscosity, inside the boundary layer. The flow field is thus divided in the OUTER (irrotational) solution--- where [tex]\psi[/tex] can have all the eigensolutions that you want and it satisfies the non-penetration boundary condition---, and the INNER (viscous) solution---which corresponds to the boundary layer flow---. Both solutions must MATCH in the overlapping region. If you solve the boundary layer flow around the cylinder, the leading term of the boundary layer solution MUST match with the leading term of the outer potential flow (the stream function calculated above). As the circulation is introduced from the rotational motion of the cylinder and this motion is propagated through the boundary layer via the non slip condition, it can be derived that in the process of matching you would obtain [tex]A=0[/tex].

Hope my answer makes you happier. Actually, this is one of the best questions I have ever found in PF about fluid mechanics. I do not understand how this question is hidden in the HWK forum, it should be posted in one of the main sections.
 
Last edited:
  • #7
Clausius2 said:
Hope my answer makes you happier. Actually, this is one of the best questions I have ever found in PF about fluid mechanics. I do not understand how this question is hidden in the HWK forum, it should be posted in one of the main sections.


Thanks a lot!
 

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