Past exam of the fluids course of the University of Oxford / 2022 / 1

Take $x$ along a horizontal gap and $s$ perpendicular to its plates, with $0<s<b$. Assume a Newtonian fluid, stationary plates, a steady flow, a laminar flow and a fully developed flow away from the entrances and side edges. The no-slip boundary condition holds on both plates. Any transverse gravity is absorbed in the hydrostatic pressure; there is no imposed transverse through-flow. Translational symmetry and the no-penetration boundary condition then give $\mathbf{u}=(u(s),0,0)$. The Navier-Stokes equation reduces toTwice integrating gives the plane Poiseuille flow:The volumetric flow rate through one gap isThere are $n-1$ internal gaps. Their total cross-sectional area is $(n-1)wb$, because the plate thickness is negligible, so the Darcy velocity isThus the permeability of a porous medium has the dimensions of squared length. If finite plate thickness were included, the Darcy velocity averaged over the whole manifold would also contain its porosity.

Both the local acceleration and the convective acceleration vanish: ${\partial }_t\mathbf{u}=0$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=u_x{\partial }_x\mathbf{u}=0$. The dominant balance is therefore between the pressure gradient and viscous stress, with no inertia contribution in this fully developed solution. This exact cancellation does not guarantee that arbitrary large-Reynolds number realizations remain laminar flows; entrances, disturbances and turbulence can invalidate the assumptions.

Use a fixed representative volume $V$ containing many pores. Let $\phi$ be the porosity and let $\mathbf{q}$ be the Darcy velocity, defined per unit total area. The conservation of mass isThe solid matrix is rigid and fixed, and the liquid is fully saturating and has constant mass density, so the stored mass on the left is constant. Microscopic internal liquid-solid boundaries contribute no through-flux by the no-penetration boundary condition. Applying the divergence theorem to the external flux givesSince the representative volume is arbitrary, the macroscopic continuity equation isThere is no extra factor of $\phi$ in the flux: that factor is already incorporated in the definition of the Darcy velocity. A compressible liquid or deformable matrix would require retaining the storage term.

Under spherical symmetry, the continuity equation and Darcy's law giveHence $q_r=C/r^2$, and integrating the pressure gradient from infinity givesThe boundary condition at $r=a$ fixes $C={\Pi }_{1}a\Delta p/\eta$. Therefore spherical Darcy injection hasHere $q_r$ is the superficial Darcy velocity, not the pore velocity $q_r/\phi$.

For the comparison, assume geometrically similar pore networks, equal porosity, the same source radius and far boundary geometry, and equal pump efficiency. Dimensional analysis then gives ${\Pi }_1=C_{\mathrm{geom}}{d}^2$ with the same dimensionless factor. ConsequentlyAt a fixed desired volumetric flow rate, the required excess pressure and pumping power areAquifer A requires half the pumping power at equal throughput, or half the energy to inject the same volume over the same time. This compares the excess pressure-volume work dissipated by the Darcy flow; any different baseline compression or other plant costs need separate accounting.

The imposed comparison matters. At fixed $\Delta p$, A takes twice the volumetric flow rate and twice the instantaneous power, but the same excess energy per injected volume. The pore spacing alone does not determine permeability of a porous medium if network geometry or porosity changes; without the geometric assumptions the preference cannot be inferred from $d$ alone.

For the common constant mass density, the background hydrostatic pressure satisfies $d{p}_0/dz=-\rho g$. Define the excess pressure $P=p-p_0(z)$. In Darcy's law, the background pressure gradient cancels gravity:If the source imposes the same excess pressure relative to the local hydrostatic pressure everywhere on its spherical surface, $P(a)=\Delta p$, the spherical Darcy injection result survives exactly:It is important to specify what is maintained on the source. A literally uniform absolute pressure $p_s=p_0(0)+\Delta p$ on a finite sphere centered at $z=0$ instead gives $P(a,\theta )=\Delta p+\rho ga\cos \theta$, where $\theta$ is measured from the upward vertical. The decaying harmonic function outside the sphere is thenand thereforeThis supplies a concrete exception to an unqualified claim that gravity leaves the radial flow unchanged. The additional dipolar contribution has zero net volumetric flow rate, so $Q=4\pi {\Pi }_{1}a\Delta p/\eta$ still holds. Equal injected and resident mass densities are essential to the background cancellation.