# Navier-stokes flow around a sphere

1. Apr 28, 2014

### joshmccraney

hi pf!

basically, i am wondering how to find the velocity profile of slow flow around a sphere in terms of a stream function $\psi = f(r,\theta)$ where we are in spherical coordinates and $\theta$ is the angle with the $z$-axis. (i think this is a classical problem).

i understand the situation like this: if we are trying to satisfy continuity then we have $\nabla \cdot \vec{V} = 0$. thus, since we can describe the flow in two parameters $r, \theta$, we say $\vec{V} = \nabla \times \psi \hat{\phi}$ where $\hat{\phi}$ is the phi unit vector. doing this in spherical coordinates gives me a different solution than the proposed $\vec{V} = -\frac{1}{r^2 sin \theta} \frac{\partial \psi}{\partial \theta}\hat{r} + \frac{1}{r sin \theta} \frac{\partial \psi}{\partial r} \hat{\theta}$

can someone shed some light on this? thanks so much!

2. Apr 28, 2014

### Staff: Mentor

This problem is solved in detail in Transport Phenomena, by Bird, Stewart, and Lightfoot. If you don't already have this book, Josh, you will find it highly worthwhile to get a copy.

Chet

3. Apr 28, 2014

### joshmccraney

thanks for the fast response! i actually do have the book, but it is not showing a derivation of where they came up with the velocity profiles. they just present them in a table. i cross referenced this with other materials and the above approach seemed to work with rectangular coordinates, although i am having troubles with it in spherical coordinates.

do you have any suggestions here?

thanks!

4. Apr 28, 2014

### joshmccraney

nevermind, chet. after checking more sources it seems they are letting $\vec{V} = \nabla \times \frac{\psi}{r \sin \theta}$. i'm assuming to make the algebra easier.

thanks again!

5. Apr 28, 2014

### Staff: Mentor

Chapter 4. Worked Example 4.2.1

Chet

6. Apr 28, 2014

### joshmccraney

Yea they definitely use the velocity here as boundary condition but they don't explain rigorously where the b.c. Comes from. But it's ok another source had it in full.