Boundary condition for a flow past a spherical obstacle

[happyparticle]

Homework Statement

Consider the steady flow pattern produced when an impenetrable rigid spherical obstacle is placed in a uniformly flowing, incompressible, inviscid fluid.

Relevant Equations

$r(r,\theta, \phi) = V \hat{z} , r \to\infty$
$\vec{v} = - \nabla \Phi$

I'm trying to find how the author finds the boundary condition at $r\to\infty$ is $\Phi(r,\theta, \phi) = - V r cos \theta$.

Using the spherical coordinates.

$- V \hat{z} = \nabla \Phi$

$- V ( cos \theta \hat{r} - sin \theta \hat{\theta}) = \frac{d \Phi}{dr}\hat{r} + 1/r \frac{d \Phi}{d \theta} \hat{\theta} + \frac{1}{r sin \theta} \frac{d \Phi}{ d \phi} \hat{\phi}$

I'm not sure to understand why most of the terms vanishes.

 

[renormalize]

I'm trying to find how the author finds the boundary condition at $r\to\infty$ is $\Phi(r,\theta, \phi) = - V r cos \theta$.

What is the pattern of a uniformly flowing, incompressible, inviscid fluid before the obstacle is inserted into the flow? That's the same pattern the flow must approach at long distances ($r\rightarrow\infty$) after insertion of the object.

 

[happyparticle]

What is the pattern of a uniformly flowing, incompressible, inviscid fluid before the obstacle is inserted into the flow? That's the same pattern the flow must approach at long distances ($r\rightarrow\infty$) after insertion of the object.

I understand that part. However is more the mathematical part that I don't really understand. Where $\Phi(r,\theta, \phi) = - V r cos \theta$ come from.

 

[Chestermiller]

What are the velocity components far from the sphere?

 

[pasmith]

I understand that part. However is more the mathematical part that I don't really understand. Where $\Phi(r,\theta, \phi) = - V r cos \theta$ come from.

The background velocity is V\mathbf{e}_z.

The calculation is <br /> \Phi = -\int \mathbf{v} \cdot d\mathbf{x}. You can either do in cartesians, which is straightforward, or you can do in spherical polars; this however is much trickier, since the polar basis vectors are not constant, but vary with position, so that you cannot integrate component by component as you can with cartesians.

 

[happyparticle]

I think I'm even more confuse.

Using cartesians coordinates I have:

$-\vec{v} = \frac{d \Phi}{dx}\hat{x} + \frac{d \Phi}{dy}\hat{y} + \frac{d \Phi}{dz}\hat{z}$

$-V \hat{z} = \frac{d \Phi}{dx}\hat{x} + \frac{d \Phi}{dy}\hat{y} + \frac{d \Phi}{dz}\hat{z}$

It seems like you kept only the $\hat{x}$ component. Why?
Also, to have $z = r cos \theta$, I must use polar coordinates.

 

[Chestermiller]

What are the components of the far-field velocity in spherical coordinates?

 

[happyparticle]

What are the components of the far-field velocity in spherical coordinates?

$V (cos \theta \hat{r} - sin \theta \hat{\theta})$

 

[Chestermiller]

$V (cos \theta \hat{r} - sin \theta \hat{\theta})$

So, at large r, $$\frac{\partial\Phi}{\partial r}=V\cos{\theta}$$ and $$\frac{\partial \Phi}{\partial \theta}=-Vr\sin{\theta}$$

 

[happyparticle]

It seems like there is a issue with the sign. Is there an error on the page linked above?

 

[Chestermiller]

It seems like there is a issue with the sign. Is there an error on the page linked above?

Oops. I assumed that v was equal to the gradient of phi rather than minus the gradient of phi. So just flip the signs in my previous post.

 

[pasmith]

It seems like you kept only the $\hat{x}$ component. Why?

The spherical object has no preferred axes, but the background flow does have one: the direction of the flow. In spherical polar coodinates, that direction is usually aligned with the \theta = 0 ray, which is the positive z axis.

We are only interested in the gradient of the potential, so we can take the constant of integration to be zero.

 

[happyparticle]

Thank you!