Incompressible fluid with constant density ρ fills the three-dimensional domain below the free surface $z = η(r)$ in cylindrical polar coordinates. The flow is axisymmetric and steady, and the only non-zero velocity component is $u_θ$. Gravity acts upon the fluid. The fluid in $r\lt a$ rotates rigidly about the z-axis with angular velocity $\Omega$ and the fluid $r \ge a$ is irrotational.

Use the radial and vertical components of the Euler equations to show that the pressure $p$ in the region $z < η$, $r < a$ satisfies

$$\frac{p}{\rho} = \frac{1}{2}{Ω^2}{r^2} −gz + \text{constant}$$ and find the constant.

$$\frac{∂u_r}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_r-\frac{(u_θ)^2}{r}+\frac{1}{\rho}\frac{∂p}{∂r}=0 $$

$$\frac{∂u_θ}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_θ+\frac{u_θu_r} {r}=0 $$

$$\frac{∂u_z}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_z+\frac{1}{\rho}\frac{∂p}{∂z}=-g $$

I have the Euler equation in cylindrical coordinate. Then how should I figure out the $\Omega$. The next question is to show that the free surface position in $ r<a $ is $$ η=\frac{\Omega^2a^2}{g}(\frac{r^2}{2a^2}-1) $$ if it is helpful. Thank you so much!

  • $\begingroup$ Are you sure the radial equations aren't available to you? That's a fair amount of effort to derive those. $\endgroup$
    – Kaynex
    May 15 '17 at 19:14

Sincerely, if you derived the Euler equations for cylindrical coordinates, this is a very simple exercise for you.

With the hypothesis, many terms disappear, those involving $u_r=0,\;u_z=0$ and the partials wrt time (steady flow).

Further, the motion is rigid, meaning that all fluid parts have zero relative motion. If they are moving in circles, the angular velocity for all point in the fuid $\Omega$ is the same! You don't need to figure anything, it must be known. With all this, $u_\theta=\Omega r$

$\begin{cases} -\dfrac{(u_θ)^2}{r}+\dfrac{1}{\rho}\dfrac{\partial p}{\partial r}=0\\ \dfrac{1}{\rho}\dfrac{\partial p}{\partial z}=-g \end{cases}$

Integrating the second,

$\dfrac{p}{\rho}=-gz+f(r)\tag 1$

So is $\dfrac{1}{\rho}\dfrac{\partial p}{\partial r}=f'(r)$ or $\dfrac{p}{\rho}+C=f(r)$. With the first equation,

$-\dfrac{(u_θ)^2}{r}+f'(r)=0$ or $-\dfrac{(\Omega r)^2}{r}+f'(r)=0$ or

$-\Omega^2r+f'(r)=0$. Integrating, $f(r) = \dfrac{1}{2}{Ω^2}{r^2}+h(z)$

$\dfrac{p}{\rho}= \dfrac{1}{2}{Ω^2}{r^2}+h(z)-C\tag 2$

Comparing $(1)$ and $(2)$ we have that $h(z)=-gz+C$

$\dfrac{p}{\rho} = \dfrac{1}{2}{Ω^2}{r^2} −gz +C$

I could not recover the given solution for the free surface, although I've found some relation that produces the well known paraboloid for the free surfce of the Newton's bucket. To find $C$, we can consider the heigh $z$ for the free surface ($p=0$) at some $r$. We can set $z=0$ for $r =a$.

$0=\dfrac{1}{2}{Ω^2}{a^2}+C$ or $C=-\dfrac{1}{2}{Ω^2}{a^2}$

$\dfrac{p}{\rho} = \dfrac{1}{2}{Ω^2}{r^2} −gz-\dfrac{1}{2}{Ω^2}{a^2}=\dfrac{1}{2}Ω^2(r^2-a^2)-gz$

For the free surface $z=\eta(r)$, $p=0$

$\eta=\dfrac{1}{2g}Ω^2(r^2-a^2),\;r\lt a$


From other considerations, $\eta$ for $r\ge a$ is $\eta=-\dfrac{\Omega^2a^4}{2gr^2}$. As $\eta$ has to be continuous $\eta(a)=-\dfrac{\Omega^2a^2}{2g}$, for the formula we've found for $r\lt a$

$0=\dfrac{Ω^2a^2}{2}+\dfrac{\Omega^2a^2}{2}+C$ leading to $C=\Omega^2a^2$ and to $\eta=\dfrac{\Omega^2a^2}{g}(\dfrac{r^2}{2a^2}-1)$ for $r\lt a$

$$\eta= \begin{cases} \dfrac{\Omega^2a^2}{g}(\dfrac{r^2}{2a^2}-1)& x\lt a\\ -\dfrac{\Omega^2a^4}{2gr^2}&a\le x \end{cases}$$

I've sketched the function for the free surface (setting $\dfrac{\Omega^2a^2}{2g}=1$ and $a=1$)

free surface

  • $\begingroup$ It is actually the next two question of this one. math.stackexchange.com/questions/2280167/… I try to get the constant and the free surface and the number matches when I set r=a, p=0 and use the free surface when r>=a. I get C=-1/2(\Omega) ^2a^2-(\Omega)^2a^2/2 and free surface as above. Can I just plug in the limit $\endgroup$
    – stedmoaoa
    May 16 '17 at 7:40
  • $\begingroup$ And if I want to draw a sketch for the free surface displacement then it is gonna be a paraboloid but when happen at r=a then? $\endgroup$
    – stedmoaoa
    May 16 '17 at 7:49
  • $\begingroup$ @stedmoaoa, the resultig function is smooth. $\endgroup$ May 16 '17 at 13:47
  • $\begingroup$ Thank you so much it really makes sense. So the graph is like x^2-1 and -1/x^2 Then, how should I describe at r=a? $\endgroup$
    – stedmoaoa
    May 16 '17 at 14:08
  • $\begingroup$ Continuous and differentiable for all quantites. $\endgroup$ May 16 '17 at 17:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.