# Velocity from velocity potential

I have this homework question and I get a different answer to the solutions.

In Cylindrical polar coordinates $(r,\theta,z)$, the velocity potential of a flow is given by:

$$\phi = -\frac{Ua^2r}{b^2-a^2}(1+\frac{b^2}{r^2})cos\theta$$

Find the velocity.

I get the velocity as:

$$v = (-\frac{Ua^2}{b^2-a^2}(1+\frac{b^2}{r^2})cos\theta + \frac{2Ua^2b^2}{(b^2-a^2)r^2}cos\theta)e_r + (\frac{Ua^2}{b^2-a^2}(1+\frac{b^2}{r^2})sin\theta) e_{\theta}$$

The answer misses out the second term in the $r$ direction, but I can't see where I've gone wrong. Any help appreciated.

For a flow in polar coordinates, the stream function $\phi$ leads to the velocities as $$v_r=\frac{1}{r}\frac{\partial\phi}{\partial\theta}\qquad v_\theta=-\frac{\partial\phi}{\partial r}$$ and not $v_r=\partial_r\phi$ and $v_\theta=\partial_\theta\phi$. Thus, $$v_r=\frac{1}{r}\frac{\partial}{\partial \theta}\left(-\frac{Ua^2r}{b^2-a^2}\left(1+\frac{b^2}{r^2}\right)\cos\theta\right) \\ = \frac{1}{r}\left(\frac{Ua^2r}{b^2-a^2}\left(1+\frac{b^2}{r^2}\right)\sin\theta\right) \\ =\frac{Ua^2}{b^2-a^2}\left(1+\frac{b^2}{r^2}\right)\sin\theta$$ and $$v_\theta=-\frac{\partial}{\partial r}\left(-\frac{Ua^2r}{b^2-a^2}\cos\theta-\frac{Ua^2}{b^2-a^2}\frac{b^2}{r}\cos\theta\right) \\ = +\frac{Ua^2}{b^2-a^2}\cos\theta-\frac{Ua^2}{b^2-a^2}\frac{b^2}{r^2}\cos\theta$$ If $v_0\equiv Ua^2/(b^2-a^2)$, then the vector velocity is $$\vec{v} = v_0\left(1+\frac{b^2}{r^2}\right)\sin\theta\hat{r}+v_0\left(1-\frac{b^2}{r^2}\right)\cos\theta\hat{\theta}$$
• I was told that the velocity is given by $\nabla \phi$? This is how I got to my answer. – Brit Miller Oct 6 '13 at 20:37
• No, the velocities come from the curl of the stream function ($\vec{v}=\nabla\times\phi$). In 2D, $\phi$ is in the $z$ direction, leading to what I wrote. – Kyle Kanos Oct 6 '13 at 23:03
• @KyleKanos I think the OP meant "velocity potential" instead of "velocity profile". For the velocity potential, the velocity is found by $\vec{v} = \nabla\phi$ as the OP stated. – OSE Oct 7 '13 at 20:05
• @OSE: OP states that "the answer misses out the second term in the $r$ direction...". Since he clearly took $\nabla\phi$, then the velocity potential is not what was defined above but the stream function. – Kyle Kanos Oct 7 '13 at 20:08