Question 4 of 7: Question 4 (Part A, Question A4): Glycerol in a Rotating Beaker — Pressure on the Base and the Shape of the Free Surface

Question 4 (Part A, Question A4): Glycerol in a Rotating Beaker — Pressure on the Base and the Shape of the Free Surface (16 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.

Find. The pressure distribution over the base of the beaker, the equation of the free surface, and the maximum fluid elevation (at the wall).

Approach. Substitute the stated rigid-body velocity field into the cylindrical-polar momentum equations from the aid sheet. Every viscous term vanishes because rigid-body rotation has no rate of strain, leaving a two-dimensional hydrostatic balance whose integration gives the pressure field directly; the free surface is then the isobar $P=P_{\text{atm}}$.

1. Reduce the momentum equations to two ordinary derivatives of pressure. With $u_r=u_z=0$, $u_\theta=\omega r$ and no dependence on $\theta$ or $t$, every convective term except the centripetal one disappears. Crucially the viscous stresses vanish too: the aid sheet gives $\tau_{r\theta}=\mu\left[r\,\partial(u_\theta/r)/\partial r+r^{-1}\partial u_r/\partial\theta\right]$, and since $u_\theta/r=\omega$ is a constant, $\tau_{r\theta}=0$. Rigid-body rotation deforms no fluid element, so the viscosity plays no part in the final answer. The r- and z-momentum equations collapse to $$\begin{aligned} -\rho\frac{u_\theta^{2}}{r}&=-\frac{\partial P}{\partial r} \;\;\Rightarrow\;\; \frac{\partial P}{\partial r}=\rho\omega^{2}r\\ 0&=-\frac{\partial P}{\partial z}-\rho g \;\;\Rightarrow\;\; \frac{\partial P}{\partial z}=-\rho g \end{aligned}$$
2. Integrate to the general pressure field. The two partial derivatives are each functions of one variable only, so the field integrates immediately: $$\boxed{\ P(r,z)=\frac{\rho\omega^{2}r^{2}}{2}-\rho g z+C\ }$$ The constant is fixed by the one point whose pressure and position we know: on the axis at the free surface, $r=0$ and $z=h_0$, where $P=P_{\text{atm}}$. Hence $C=P_{\text{atm}}+\rho g h_0$ and $$P(r,z)=P_{\text{atm}}+\rho g\left(h_0-z\right)+\frac{\rho\omega^{2}r^{2}}{2}.$$
3. Evaluate the pressure along the bottom of the beaker. Setting $z=0$, $$P(r,0)=P_{\text{atm}}+\rho g h_0+\frac{\rho\omega^{2}r^{2}}{2} =P_{\text{atm}}+370.8+9948.6\,r^{2}\ \ \text{Pa}\quad(r\ \text{in m}),$$ using $\rho g h_0=1260\times9.81\times0.030=370.8\ \text{Pa}$ and $\rho\omega^{2}/2=1260\times157.914/2=99\,486\ \text{Pa/m}^{2}$. In gauge terms the base pressure rises parabolically from 370.8 Pa on the axis to $$P_{\text{gauge}}(R)=370.8+\tfrac{1}{2}\times1260\times157.914\times0.025^{2} =370.8+62.2=433.0\ \text{Pa}$$ at the wall — a 17 per cent increase across the base.
4. Obtain the free surface as the atmospheric isobar. The free surface is the locus of $P=P_{\text{atm}}$. Putting that into the pressure field and solving for $z$, $$\boxed{\ z_s(r)=h_0+\frac{\omega^{2}r^{2}}{2g}\ }$$ a paraboloid of revolution. Numerically $\omega^{2}/2g=157.914/19.62=8.049\ \text{m}^{-1}$, so $$z_s(r)=0.030+8.049\,r^{2}\ \ \text{m}.$$ The surface is independent of density — the same paraboloid would form in water or in mercury — because both the centrifugal and gravitational terms carry one factor of ρ.
5. Compute the maximum fluid elevation. The highest point is at the wall, $r=R=0.025\ \text{m}$: $$z_{\max}=0.030+8.049\times(0.025)^{2}=0.030+0.005030,$$ so $$\boxed{\ z_{\max}=0.03503\ \text{m}=35.0\ \text{mm}\ }$$ The surface climbs 5.03 mm above the centre, and the total rise from axis to wall, $\omega^{2}R^{2}/2g$, is exactly the velocity head $u_\theta^{2}(R)/2g$ of the rim — a useful sanity check.
6. Confirm the physical assumptions hold. Two are worth testing. First, the rotational Reynolds number is $\text{Re}=\rho\omega R^{2}/\mu=1260\times12.566\times0.025^{2}/1.49=6.6$, so the flow is deep in the creeping regime and no secondary circulation or turbulence disturbs the rigid-body state. Second, the spin-up time scales as $R^{2}/\nu$ with $\nu=\mu/\rho=1.18\times10^{-3}\ \text{m}^{2}/\text{s}$, giving $0.025^{2}/1.18\times10^{-3}\approx0.53\ \text{s}$: the "initial start-up period" mentioned in the question is well under a second for a liquid this viscous, so steady state is reached almost at once. Finally the surface never spills or exposes the base, since the total rise is only 5 mm on a 30 mm depth.