Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 39C (Fluid Dynamics II)

The Navier-Stokes equation for constant mass density $\rho$, dynamic viscosity $\mu$, and body force per unit mass $\mathbf{f}$ iswhere the Newtonian fluid stress tensor isTaking the dot product with $\mathbf{u}$ givesIncompressibility impliesand converts the left side into a local time derivative plus the divergence of kinetic-energy flux. The divergence theorem therefore gives the kinetic-energy balance for an incompressible Newtonian fluidThe four terms are respectively outward advective transport of kinetic energy, mechanical power supplied by surface traction, power supplied by the body force, and irreversible viscous dissipation into heat.

Let $a$ denote the bubble size, $U$ its speed, and $\mathrm{Re}=Ua/\nu \gg 1$ the Reynolds number. Outside a thin boundary layer, inertia dominates and the flow is an irrotational flow with velocity scale $U$ and strain scale $U/a$. Its viscous dissipation has orderin three dimensions.

The boundary-layer thickness isFor a clean bubble the stress-free boundary condition prescribes tangential stress rather than tangential velocity. The outer flow already has tangential strain of order $U/a$, so cancelling that stress requires a velocity correction of onlynot a correction of order $U$. Its gradient remains $O(U/a)$, and its boundary-layer dissipation iswhich is asymptotically smaller.

In steady translation, the drag power $DU$ balances viscous dissipation. The leading estimate therefore follows from the known outer flow alone:in three dimensions. This is the dissipation estimate for high-Reynolds-number bubble drag.

Work in the frame translating with the bubble. The bubble is the fixed circle $r=a$, while the velocity tends to $-U{\mathbf{e}}_x$ at infinity. The potential flow around a circular cylinder has velocity potentialIt is harmonic for $r>a$ and givesThus $u_r=0$ on $r=a$, satisfying the kinematic boundary condition, and $\mathbf{u}\to -U{\mathbf{e}}_x$ as $r\to \infty$. The laboratory-frame flow is obtained by adding $U{\mathbf{e}}_x$.

For the outer flow from part (i), the rate-of-strain tensor in polar coordinates has componentsThereforeThe leading dissipation per unit axial length isBalancing this with drag power $DU$ gives the drag on a two-dimensional circular bubble from outer-flow dissipationper unit axial length, directed opposite to the bubble velocity.

A rigid body obeys the no-slip boundary condition. Its outer potential flow generally has an $O(U)$ tangential slip velocity at the surface, so the boundary layer must make an $O(U)$ velocity correction across thickness $\delta$. Its strain is therefore $O(U/\delta )$, and in two dimensions its dissipation per unit axial length iswhich exceeds the $O(\mu U^2)$ outer dissipation by $a/\delta \sim {\mathrm{Re}}^{1/2}$.

The dominant drag consequently depends on the detailed boundary-layer velocity field and possibly on boundary-layer separation. It cannot be obtained by integrating the outer irrotational strain alone.