Question

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

(a) the shape of the free surface of the water;

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

along its radius provided the pressure at the central point is equal to Po. ​

Answer 1

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

`h(r) = h_0 + ( \Omega^2)/(2g) r^2`

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

`\rho * g * (h_0 + ( \Omega^2)/(2g) r^2)` where r is the distance from the axis.

Step-by-step explanation:

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

`\rho ((\partial u)/(\partial t) -\textbf{u} * \textbf{\omega} ) = -\rho \textbf{u} * \omega = -\bigtriangledown (p + (1)/(2) \rho \parallel \textbf{u} \parallel^2) + \rho \textbf{g}`ω) = -ρu×ω =
`- \\abla`(P +
`(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 =
`(\partial P)/(\partial r) - (\rho)/(2)(d u_(\theta)^2)/(dr) = -(\partial P)/(\partial r) - \rho r\Omega^2`

Therefore,

`(\partial P)/(\partial r) = \rho r\Omega^2` =
`\rho (u_(\theta)^2)/(r)`

Integrating gives

P(r, z) =
`( \rho \Omega^2)/(2) r^2 +f_1(z)`

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

`(dp)/(dz) = -\rho g \Rightarrow (df_1)/(dz) = -\rho g \Rightarrow f_1(z) = -\rho g z+C_3`

Therefore

P(r, z) =
`( \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)` =
`( \rho \Omega^2)/(2) R^2 + -\rho g z_R+C_3`

From which we have

P(r, z) -
`p_(atm)` =
`( \rho \Omega^2)/(2) r^2 + -\rho g z+C_3` -
`(( \rho \Omega^2)/(2) R^2 + -\rho g z_R+C_3)`

P(r, z) -
`p_(atm)` =
`( \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 =( \Omega^2)/(2g) (R^2 -r^2)`,

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

Therefore,
`z_R-z =h_0 + ( \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 * g * ( \Omega^2)/(2g) (R^2 -r^2)` =
`\rho * g * (h_0 + ( \Omega^2)/(2g) r^2)`

Answer 2

Step-by-step 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.