Complex potential describing an inviscid flow

Question

I'm working through a homework sheet for a Fluid Mechanics module. The question is given:

Consider the flow described by the complex potential $$w=4z+\frac{8}{z}.$$

1. Determine $\psi$, $\phi$, $u$ and $v$ in plane polar coordinates $(r,\theta)$.
2. Determine the location of the stagnation points.
3. Show that this complex potential describes an inviscid flow around a solid object, What is the shape of the object?
4. Sketch the streamlines for the flow outside the object.

My working out so far for the question is:

1. Let $z=re^{i\theta}$, and therefore \begin{align} w&=4re^{i\theta}+\frac{8}{r}e^{-i\theta} \\ &=4r(\cos(\theta)+i\sin(\theta))+\frac{8}{r}(\cos(\theta)-i\sin(\theta)) \\ &=(4r+\frac{8}{r})\cos(\theta)+(4r-\frac{8}{r})i\sin(\theta). \end{align} Using the Cauchy-Riemann equations, write $w=\phi+i\psi$, $\phi=(4r+\frac{8}{r})\cos(\theta)$ and $\psi=(4r-\frac{8}{r})\sin(\theta)$. Also, we have that $$u=\frac{\partial\phi}{\partial r}\implies u=(4-8r^{-2})\cos(\theta)$$ and $$v=\frac{1}{r}\frac{\partial\phi}{\partial\theta} \implies v=-(r+8r^{-2})\sin(\theta).$$
2. Stagnation points are given by $u=0$ and $v=0$. So, from $u=0$, we have that $r^2=2$ or $\cos(\theta)=0$. Similarly from $v=0$, we have that $r^2=-2$ and $\sin(\theta)=0$. Since $r\in\mathbb{R}$ (as it is a distance), we have that $r^2=2$ (from $u=0$) and $\sin(\theta))=0$ (from $v=0$). Therefore, the stagnation points occur at $(r,\theta)=(\sqrt{2},0),(\sqrt{2},\pi)$.

From here (ie 3 onwards), I fall down. I think that I should use that $\textbf{u}\cdot\textbf{n}=0$, but I'm not too sure how to use this information. Should I be using Bernoulli's theorem for pressure? Is there some assumption I am missing?

Any help would be much appreciated!

Answer 1

To determine the shape of the obstacle, you just look at the streamlines (they're everywhere tangent to the flow). Since you have a potential flow, it holds that $\Delta u = 0$, where $u$ is the (real) velocity. So, the term $\mu\Delta u$ in the Navier-Stokes equation will vanish. Obviously, this doesn't itself imply $\mu = 0$ so the fluid itsf may not be inviscid, but the flow should be.