Question 4 of 6: Rigid-body rotation of water in a spinning container

Question 4: Rigid-body rotation of water in a spinning container (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An incompressible Newtonian liquid of constant density ρ and viscosity μ, filling a cylindrical container that has been spinning long enough about the vertical z axis for the liquid to turn with it as a solid body at constant angular velocity ω. The velocity field is therefore $v_r=v_z=0$, $v_\theta=\omega r$, the motion is steady and axisymmetric, gravity acts in the −z direction, and the free surface is exposed to a uniform ambient pressure p_a.

Find. The reduced r- and z-momentum equations with a justification for discarding the viscous terms, the pressure field p(r, z) throughout the liquid, and from it the equation of the free surface.

Approach. Insert the rigid-body velocity field into the cylindrical Navier–Stokes equations, show that the deformation-rate tensor vanishes identically so the viscous term is exactly zero, then integrate the two surviving pressure gradients and apply the constant-pressure condition at the free surface.

1. Part (a) — reduce the r-momentum equation. With $v_r=v_z=0$, $v_\theta=\omega r$, and $\partial/\partial t=\partial/\partial\theta=0$, the only surviving inertia term in the radial direction is the centripetal one: $$-\rho\frac{v_\theta^{2}}{r}=-\frac{\partial p}{\partial r}+\mu\left[\ldots\right] \quad\Longrightarrow\quad \frac{\partial p}{\partial r}=\rho\frac{v_\theta^{2}}{r}=\rho\omega^{2}r .$$ Physically, a radial pressure gradient is what supplies the centripetal acceleration of each fluid ring.
2. Reduce the z-momentum equation. There is no vertical acceleration and no vertical velocity, so the vertical equation is a pure hydrostatic balance, $$0=-\frac{\partial p}{\partial z}-\rho g \quad\Longrightarrow\quad \frac{\partial p}{\partial z}=-\rho g .$$ The θ-momentum equation is satisfied identically because nothing varies with θ and there is no azimuthal forcing.
3. Show that the viscous stresses vanish exactly, not merely approximately. The only candidate viscous term acts on v_θ, and in cylindrical coordinates it is $$\mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r}\frac{\partial (r v_\theta)}{\partial r}\right)\right] =\mu\frac{\partial}{\partial r}\left(\frac{1}{r}\frac{\partial(\omega r^{2})}{\partial r}\right) =\mu\frac{\partial}{\partial r}\left(2\omega\right)=0 .$$ The deeper reason is kinematic: in rigid-body motion every fluid particle keeps its position relative to every other, so the rate-of-strain tensor is identically zero. A Newtonian fluid generates viscous stress only in response to deformation, never to rotation, so a fluid in solid-body rotation carries no viscous stress at all. This is why the answer is exact rather than a high-Reynolds-number approximation, and it is also why μ never appears in the final result.
4. Part (b) — integrate the radial equation. Holding z fixed, $$p(r,z)=\int \rho\omega^{2}r\,\mathrm{d}r = \frac{\rho\omega^{2}r^{2}}{2}+F(z),$$ where F is so far an arbitrary function of z alone.
5. Integrate the vertical equation to pin down F(z). Differentiating the last result with respect to z and matching step 2, $$\frac{\mathrm{d}F}{\mathrm{d}z}=-\rho g \quad\Longrightarrow\quad F(z)=-\rho g z + C .$$ Hence the pressure everywhere in the liquid is $$p(r,z)=\frac{\rho\omega^{2}r^{2}}{2}-\rho g z + C .$$
6. Fix the constant of integration. Place the origin at the vertex of the free surface, where r = 0, z = 0 and the pressure equals the ambient value p_a. Then C = p_a and $$\boxed{p(r,z)=p_a+\frac{\rho\omega^{2}r^{2}}{2}-\rho g z}$$ The result is the familiar hydrostatic law with one extra term, and that term is the centrifugal contribution: at fixed depth the pressure rises quadratically outward.
7. Part (c) — impose the free-surface condition. The free surface is by definition the locus on which the pressure equals the ambient value, so set p = p_a in the boxed result: $$p_a=p_a+\frac{\rho\omega^{2}r^{2}}{2}-\rho g z_{\text{surface}} \quad\Longrightarrow\quad \rho g z_{\text{surface}}=\frac{\rho\omega^{2}r^{2}}{2}.$$
8. Read off the shape. The density cancels, leaving $$\boxed{z_{\text{surface}}=\frac{\omega^{2}r^{2}}{2g}}$$ as required. The surface is a paraboloid of revolution, its shape independent of which liquid is in the container. Note also that the isobars are the family $z=\dfrac{\omega^{2}r^{2}}{2g}-\dfrac{p-p_a}{\rho g}$ — congruent paraboloids stacked vertically, of which the free surface is simply the one at ambient pressure.
9. Put numbers on the result. For a container of radius R = 0.15 m spun at ω = 10 rad/s, the rim of the surface stands $z_s(R)=(10)^{2}(0.15)^{2}/(2\times9.81)=0.1147$ m above the vertex. Because the volume under a paraboloid is half that of the enclosing cylinder, the surface drops $\omega^{2}R^{2}/4g=0.0573$ m at the axis and rises the same 0.0573 m at the wall relative to the liquid level at rest — so at least 57 mm of freeboard is needed if the container is not to overflow. The gauge pressure at the wall in the plane of the vertex is $\rho\omega^{2}R^{2}/2=998(100)(0.0225)/2=1123$ Pa.