Question 4 of 6: Rigid-body rotation of a liquid — Navier–Stokes reduction and the free surface

Question 4: Rigid-body rotation of a liquid — Navier–Stokes reduction and the free surface (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 $\rho$ and viscosity $\mu$ filling a cylindrical container that has been spinning long enough about its vertical axis for all relative motion to have died out. The velocity field is therefore that of a rigid body, $$V_r = 0,\qquad V_\theta = \omega r,\qquad V_z = 0$$ with $\omega$ constant, and gravity acts as $\mathbf{g} = -g\,\hat{\mathbf{k}}$. Nothing depends on $\theta$ or on time.

Find. The reduced $r$- and $z$-momentum equations with a justification for discarding the viscous terms; the pressure field $p(r,z)$ obtained by integrating them; and a demonstration that the free surface is the paraboloid $z_s = \omega^{2}r^{2}/2g$.

[Figure not reproduced: Figure 4 (redrawn). Rigid-body rotation. The free surface (blue) and every constant-pressure surface inside the fluid (dashed) are congruent paraboloids of revolution, all with the same curvature $\omega^{2}/2g$. See the official exam paper.]

Approach. Substitute the rigid-body velocity field into the incompressible Navier–Stokes equations in cylindrical coordinates, show that both the convective and the viscous terms collapse, integrate the two surviving partial derivatives, and then impose $p = p_{atm}$ on the free surface.

1. Reduce the $r$-momentum equation (part a). The steady, axisymmetric $r$-component of the Navier–Stokes equations in cylindrical coordinates is $$\rho\left(V_r\frac{\partial V_r}{\partial r} - \frac{V_\theta^{2}}{r} + V_z\frac{\partial V_r}{\partial z}\right) = -\frac{\partial p}{\partial r} + \mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r}\frac{\partial (rV_r)}{\partial r}\right) + \frac{\partial^{2}V_r}{\partial z^{2}}\right]$$ With $V_r = V_z = 0$ the only surviving inertia term is the centripetal one, $-\rho V_\theta^{2}/r$, and the entire viscous bracket vanishes because $V_r \equiv 0$. Hence $$\frac{\partial p}{\partial r} = \rho\,\frac{V_\theta^{2}}{r} = \rho\,\frac{(\omega r)^{2}}{r} = \rho\,\omega^{2} r$$
2. Reduce the $z$-momentum equation. With $V_z \equiv 0$ every inertia term and every viscous term drops, leaving only the pressure gradient and the body force: $$0 = -\frac{\partial p}{\partial z} - \rho g \qquad\Longrightarrow\qquad \frac{\partial p}{\partial z} = -\rho g$$ The vertical direction is therefore purely hydrostatic, exactly as in a stationary tank.
3. Justify discarding the viscous stresses. This is the conceptual heart of part (a), and it is not the usual “high Reynolds number” argument — the viscous terms here are identically zero, not merely small. Viscous stress in a Newtonian fluid is proportional to the rate of deformation, and a rigid-body motion has none: every fluid element translates and rotates but never strains. Formally, the only candidate shear component in this field is $$\tau_{r\theta} = \mu\,r\,\frac{\partial}{\partial r}\!\left(\frac{V_\theta}{r}\right) = \mu\,r\,\frac{\partial}{\partial r}(\omega) = 0$$ because $V_\theta/r = \omega$ is a constant. Likewise the $\theta$-momentum viscous operator gives $$\frac{\partial}{\partial r}\left[\frac{1}{r}\frac{\partial (rV_\theta)}{\partial r}\right] = \frac{\partial}{\partial r}\left[\frac{1}{r}\frac{\partial (\omega r^{2})}{\partial r}\right] = \frac{\partial}{\partial r}\left(2\omega\right) = 0$$ So the fluid behaves inviscidly no matter how viscous it is; viscosity was essential only in the transient that established the rigid-body state, by diffusing the wall’s rotation into the interior. The no-slip condition is satisfied automatically once that state is reached, since the fluid at the wall turns with the wall.
4. Integrate the radial equation (part b). Integrating $\partial p/\partial r = \rho\omega^{2}r$ at fixed $z$, $$p(r,z) = \frac{\rho\,\omega^{2}r^{2}}{2} + f(z)$$ where $f$ is an arbitrary function of $z$ alone — it is a function, not a constant, because the integration was partial.
5. Close it with the vertical equation. Differentiating the result of step 4 with respect to $z$ and matching step 2, $$\frac{\partial p}{\partial z} = f'(z) = -\rho g \qquad\Longrightarrow\qquad f(z) = -\rho g z + C$$ Writing $C = p_0$, the pressure at the origin $(r,z) = (0,0)$, the complete field is $$\boxed{p(r,z) = p_0 + \frac{\rho\,\omega^{2}r^{2}}{2} - \rho\,g\,z}$$ valid everywhere in the fluid. It is the ordinary hydrostatic distribution plus a centrifugal term that raises the pressure quadratically with radius — the mathematical statement of why the liquid piles up at the rim.
6. Impose the free-surface condition (part c). The free surface is the locus on which the pressure equals the atmospheric value. Choosing the origin at the lowest point of the surface, on the axis, makes $p_0 = p_{atm}$; measuring $p$ as a gauge pressure then sets $p_0 = 0$, and the surface is the curve $p(r,z_s) = 0$: $$0 = \frac{\rho\,\omega^{2}r^{2}}{2} - \rho\,g\,z_s$$ Dividing through by $\rho g$ (both non-zero) gives $$\boxed{z_s = \frac{\omega^{2} r^{2}}{2g}}$$ which is the required result — a paraboloid of revolution. Two structural remarks close the question. First, every other isobar has exactly the same shape, merely shifted vertically, because setting $p = $ const in the boxed field of step 5 always gives $z = \omega^{2}r^{2}/2g + $ const. Second, a paraboloid encloses half the volume of its bounding cylinder, so conservation of liquid volume fixes the vertical position of the vertex: the axis level drops by exactly the same amount as the rim level rises, each equal to $\omega^{2}R^{2}/4g$ for a container of radius $R$. For a 150 mm-radius container spinning at 12 rad/s, for example, the rim-to-axis difference is $\omega^{2}R^{2}/2g = 165\;\text{mm}$, of which 82.6 mm is a rise at the rim and 82.6 mm a fall on the axis.