Complex Potential describing inviscid fluid flow

I'm a Mathematics student, 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, finite 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 equtaions, $$w=\phi+i\psi$$, we then have that $$\phi=(4r+\frac{8}{r})\cos(\theta)$$ 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 $$v=\frac{1}{r}\frac{\partial\phi}{\partial\theta} \implies v=-(4r+8r^{-2})\sin(\theta)$$.

(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 $$\sin(\theta)=0$$. 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!

An alternative to Trounev's answer. You already found the stagnation points at $$(r,\theta)=(\sqrt{2},0)$$ and $$(\sqrt{2},\pi)$$. Since there are two stagnation points we anticipate a finite solid body (in contrast, Rankine half body has one stagnation point and is semi-infinite). The streamfunction $$\psi$$ evaluates to zero at both stagnation points, which implies that $$\psi=0$$ streamline connects them. A finite solid body is formed by a closed streamline; if $$r=\sqrt{2}$$ then the value of the streamline connecting the two stagnation points remains the same (equal to zero) for any $$\theta$$, which implies that it is closed. Therefore the finite solid body you seek (which is the streamline $$\psi=0$$) is a circle of radius $$\sqrt{2}$$. This approach is more general than Trounev's; he begins by assuming that the body is a circle (by assuming that the normal $$\mathbf{n}$$ to the solid body is the unit radial vector) which assumption is then shown to be consistent from the ensuing result; however his procedure wouldn't work for a body of arbitrary shape, say an ellipse.

P.S. The question should have clarified that the body sought for is finite. Otherwise any pair of distinct streamlines can be taken as the boundary of a solid body because $$\mathbf{u}\cdot\mathbf{n}=0$$ is satisfied on every streamline.

• I discussed the question with my lecturer after posting this, and he advised me to use $\textbf{u}\cdot\textbf{n}=0$, rather than the streamlines and streamfunction. Also, we haven't considered not finite bodies in class, which perhaps I should have made clear in my question, so I did not even think it would be an option (will edit). – Fats Mar 21 '19 at 8:22
• @Fats The problem is that you don't know ahead of time what $\mathbf{n}$ is (the shape of the body isn't given, I presume) and must be guessed. – Deep Mar 21 '19 at 15:09

Let's pretend that this potential describes inviscid flow around cylinder with radius $$r=\sqrt {2}$$. We have $$u = (4 - 8/r^2)\cos {\theta},v = -(4 + 8/r^2)\sin {\theta}$$. Using the no-flow condition on the surface of a solid object $$\vec {u}.\vec {n}=0$$ we find $$(u,v).(1,0)=0$$, and from here we get $$u=0$$, and $$r=\sqrt {2}$$. Therefore, it is proven. To build the streamlines, I used the standard program from Mathematica 11.3. For this, the field $$(u,v)$$ has been converted to Cartesian coordinates. To make sure that this is a 2D flow, and not 3D, we calculate the divergence of the velocity field: $$\nabla .\vec {u}=0$$ in 2D, and $$\nabla .\vec {u}=-16z/(x^2+y^2+z^2)^2$$ in 3D. Therefore it is 2D flow.

• Can the object be a sphere? Why are we sure that it is a cylinder? – Fats Mar 21 '19 at 10:11
• @Fats First, correct the typo in your topic should be $v=-(4+8/r^2)\sin {\theta }$. Second, calculate the divergence in 2D and in 3D, should be $\nabla .\vec {u}=0$. See update to my answer. – Alex Trounev Mar 21 '19 at 12:49
• @ Alex Trounev: Is the complex potential even defined for a 3D flow? – D. Halsey Mar 21 '19 at 13:45
• @D.Halsey Axisymmetric flow around a sphere is also described in coordinates $r, \theta$. It looks like a 2D. Apparently so there was a question for Fats. – Alex Trounev Mar 21 '19 at 14:12
• No one understood the solution, but they vote down:) – Alex Trounev Nov 13 '19 at 13:49