Calculate Potential [closed]

Question

$\vec{w} = \begin{pmatrix}w_r \\ w_{phi} \end{pmatrix} = \begin{pmatrix}\frac{Q_0}{2 \pi r} \\ 0 \end{pmatrix} $

1) Show that the flow satisfys the continuity equation

2) Show that $\vec{w}$ has a potential $\Phi$

3) Calculate $\Phi$

Task 1

$\nabla \cdot \vec{w} = 0$ I can't derive $\frac{\partial w_r}{\partial r} + \frac{1}{r} \frac{\partial w_{\phi}}{\partial \phi} + \frac{\partial w_z}{\partial z} + \frac{w_r}{r} = 0$ from that. The last part $+ \frac{w_r}{r}$ remains a mystery to me.

Task 2

I would have used $\vec{\nabla} \times \vec{w} = 0$ However if I calculate that vector product I see that the equation is satisfyed, if I assume $w_z$ is zero since it is not given. From the solution however it states that: $w_z = \frac{w_{\phi}}{r} + \frac{\partial w_{\phi}}{\partial r} - \frac{1}{r}\frac{\partial w_r}{\partial \phi} = 0$ This is due to the fact that a plane flow can only have the third compoment $w_z$ different from zero. Neither can I imagine why that component is the only one that can be different from zeor nor where the equation comes from.

Answer 1

Task 1

Find a calculus textbook and look for cylindrical coordinates, you will find that the divergence of a vector in those coordinates is given by: $$\vec{\nabla}\cdot\vec{\omega} = \frac{1}{r}\frac{\partial}{\partial r}(r\omega_r) + \frac{1}{r}\frac{\partial}{\partial \phi}\omega_{\phi}+\frac{\partial\omega_z}{\partial z}$$ Now apply the product rule to the term $\frac{1}{r}\frac{\partial}{\partial r}(r\omega_r)$ and you will arrive at $\frac{\partial\omega_r}{\partial r} + \frac{\omega_r}{r}$. The additional mysterious term is to account for the curvature which is present compared to cartesian coordinates.

I find the unexpanded form generally easier to work with as in this case the $r$ dependence of $\omega_r$ cancels immediately and therefore the derivative is identically zero.

Task 2

The vector $\vec{\nabla}\times\vec{\omega}$ only has a $z$-component because the $\omega_z$-component is assumed zero. $$\vec{\nabla}\times\vec{\omega}=(\frac{1}{r}\frac{\partial}{\partial\phi}\omega_z-\frac{\partial}{\partial z}\omega_{\phi})\hat{e_r}-(\frac{1}{r}\frac{\partial}{\partial r}(r\omega_z)-\frac{\partial}{\partial z}\omega_r)\hat{e_\phi}+(\frac{1}{r}\frac{\partial}{\partial r}(r\omega_{\phi})-\frac{1}{r}\frac{\partial\omega_{r}}{\partial\phi})\hat{e_z}$$ since $\omega_{\phi}=\omega_{z}=0$, and $\omega_r$ is not a function of $z$; we have that only the $z$-component $\frac{1}{r}\frac{\partial}{\partial r}(r\omega_{\phi})-\frac{1}{r}\frac{\partial\omega_{r}}{\partial\phi}$ is non-zero. Expanding this as in Task 1 will give you the same form as in your solution and inserting the given $\omega_r$ and $\omega_{\phi}$ components will result in it being zero.

Answer 2

T1: $w_r/r$ is just $Q_o/2\pi r^2$ which cancels the first term. T2: If $w_z$ is not given, one should assume that the flow is perpendicular to the $z$-axis i.e. that $w_z=0$. In that case, the curl only has a $z$-component $$\frac{1}{\rho}(\frac{\partial(\rho w_\phi)}{\partial\rho}-\frac{\partial w_\rho}{\partial \phi}) $$ which turns out to be $0$.