Question

A cylindrical vessel with water is rotated about its vertical axis with a constant angular velocity co. Find:

Answer 1

Answer:

(a) The shape of the free surface of the water is a parabola of revolution as follows;

$h(r) = h_0 + \frac{ \Omega^2}{2g} r^2$

(b) The water pressure distribution over the bottom of the vessel is

 $\rho \times g \times (h_0 + \frac{ \Omega^2}{2g} r^2)$  where r is the distance from the axis.

Explanation:

To solve the question, we solve the Euler's equation of the form

$\rho (\frac{\partial u}{\partial t} -\textbf{u} \times \textbf{\omega} ) = -\rho \textbf{u} \times \omega = -\bigtriangledown (p + \frac{1}{2} \rho \parallel \textbf{u} \parallel^2) + \rho \textbf{g}$ω) = -ρu×ω = $- \nabla$(P + $\frac{1}{2}$ρ ║u║²) + ρg

When in uniform rotation, we have

u = $u_{\theta}\hat{e}_{\theta}$ , ω = $\omega_z \hat{e}_{z}$ where $u_{\theta}$ = rΩ and   $\omega_z$ = 2Ω

Therefore, u × ω = 2·r·Ω²· $\hat{e}_{r}$

From which the radial component of the vector equation is given as

-2·p·r·Ω² = $\frac{\partial P}{\partial r} - \frac{\rho}{2}\frac{d u_{\theta}^2}{dr} = -\frac{\partial P}{\partial r} - \rho r\Omega^2$

Therefore,

$\frac{\partial P}{\partial r} = \rho r\Omega^2$ = $\rho \frac{u_{\theta}^2}{r}$

Integrating gives

P(r, z) = $\frac{ \rho \Omega^2}{2} r^2 +f_1(z)$

By substituting the above into the z component of the equation of motion, we obtain;

$\frac{dp}{dz} = -\rho g \Rightarrow \frac{df_1}{dz} = -\rho g \Rightarrow f_1(z) = -\rho g z+C_3$

Therefore

P(r, z) = $\frac{ \rho \Omega^2}{2} r^2 + -\rho g z+C_3$

From the boundary conditions r = R and z = $z_R$, we find C₃ as follows

P(r = R, z = $z_R$)  = $p_{atm}$

Therefore  $p_{atm}$ =  $\frac{ \rho \Omega^2}{2} R^2 + -\rho g z_R+C_3$

From which we have

P(r, z) -  $p_{atm}$ =  $\frac{ \rho \Omega^2}{2} r^2 + -\rho g z+C_3$ -  $(\frac{ \rho \Omega^2}{2} R^2 + -\rho g z_R+C_3)$

P(r, z) -  $p_{atm}$  = $\frac{ \rho \Omega^2}{2} (r^2 -R^2) -\rho g (z-z_R)$

We note that at the surface, the interface between the air and the liquid

P = $p_{atm}$, the shape of the of the free surface of the water is therefore;

$z_R-z =\frac{ \Omega^2}{2g} (R^2 -r^2)$,

Given that at r = 0 we have the height = h₀

Therefore, $z_R-z =h_0 + \frac{ \Omega^2}{2g} r^2 = h(r)$

The shape of the of the free surface of the water is a parabola of revolution.

(b) The water pressure distribution over the bottom of the vessel is given by

ρ × g × z  

= $\rho \times g \times \frac{ \Omega^2}{2g} (R^2 -r^2)$ =  $\rho \times g \times (h_0 + \frac{ \Omega^2}{2g} r^2)$

Answer 2

Answer:

Explanation:

The question is one that examine the physical fundamental of mechanics of a cylindrical vessel .

We would use the Euler' equation and some coriolis and centripetal force formula.

The fig below explains it.