I've had problems conceptualising and working through this problem, could someone please just give it a quick check through and let me know if i've done anything majorly wrong.
Consider a vortex with centre along the ${\vec{\hat{\textbf{z}}}}$-axis, Given a flow field of the form $\vec{u}=f(r) \mathbb{\vec{\hat{\theta}}}$ in cylindrical polars.
- Show that $\vec{\nabla} \cdot \vec{u} = 0$
- using euler's equation in cylindrical co-ordinates show that the pressure p(r) satisfies $$\frac{dp}{dr} = \rho \frac{f^2(r)}{r}$$
- Compute the value of $\vec{\omega} = \vec{\nabla} \times\vec{u}$ and show that it vanishes if $f = A/r$. Compute the pressure for this case, take $p_0$ to be the pressure at infinity.
so what i have is this:
Given: $\vec{u} = f(r)\hat{\theta}$, we have $$\vec{\nabla} \cdot \vec{u} = \frac{1}{r}\frac{\partial (ru_{r})}{\partial r}+\frac{1}{r}\frac{\partial u_{\theta}}{\partial \theta}+\frac{\partial u_{z}}{\partial z}$$ substituting in the above values gives $$\nabla \cdot u = \frac{1}{r}\frac{\partial (r\cdot 0)}{\partial r}+\frac{1}{r}\frac{\partial f(r)}{\partial \theta}+\frac{\partial 0}{\partial z} = 0$$
Moving on: from my understanding eulers equation is $$\rho \frac{Du}{Dt} = - \nabla p +f$$ we can sub in and change from cartesian to cylindrical giving
$$ r:~\rho \left[\frac{Du_{r}}{Dt}-\frac{u^{2}_{\theta}}{r}\right]=-\frac{\partial p}{\partial r}+f_{r}$$ $$ \theta:~\rho \left[\frac{Du_{\theta}}{Dt}-\frac{u_{\theta}u_{r}}{r}\right]=-\frac{1}{r}\frac{\partial p}{\partial \theta}+f_{\theta}$$ $$ z:~\rho \left[\frac{Du_{z}}{Dt}\right]=-\frac{\partial p}{\partial z}+f_{z}$$
since $p$ is a function of r alone we consider just the r component, substituting in $$u_{\theta} = f(r),~u_{r},=0~u_{z}=0,~f_{r}=0$$ gives $$ r:~-\rho \frac{f^{2}(r)}{r}=-\frac{\partial p}{\partial r} \implies$$ $$\frac{dp}{dr} = \rho \frac{f^2(r)}{r}$$ as req. Moving on: $$\vec{\nabla} \times \vec{u} = \frac{1}{r}(0)\vec{\hat{r}}+(0)\vec{\hat{\theta}}+\frac{1}{r}\left(\frac{\partial(r f(r))}{\partial r}\right) = \frac{1}{r}\frac{\partial(r f(r))}{\partial r}$$
substituting in $f(r) = A/r$ gives $$\frac{1}{r}\frac{\partial A}{\partial r}=0$$
Finally, Using the above $$\frac{dp}{dr}=\rho \frac{A^{2}}{r^{3}} \implies$$ $$dp = A^2\rho \int \frac{1}{r^3} dr \implies $$ $$p(r)=\left[-\frac{A^2 \rho}{2r^2}\right]$$
now i admit, i'm a little unsure what limits i should be adding to the integral
any help would be greatly appreciated.
thanks!
Here you really want to use $r$ as a limit of integration and use a dummy variable as the variable of integration.
$$\int_r^\infty p'(x)dx = A^2\rho \int_r^\infty \frac{1}{x^3} dx $$
the left side becomes $p_0-p(r)$