= Solution
{c}
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 to
$$
0=-\frac{dp}{dx}+\eta\frac{d^2u}{ds^2},\qquad
\frac{dp}{dx}=-\frac{\Delta p}{L},\qquad u(0)=u(b)=0.
$$
Twice integrating gives the <plane Poiseuille flow>:
$$
\boxed{u_x(s)=\frac{\Delta p}{2\eta L}s(b-s),\qquad u_s=u_{\mathrm{span}}=0.}
$$
The <volumetric flow rate> through one gap is
$$
Q_{\mathrm{gap}}=w\int_0^b u_x(s)\,ds
=\frac{wb^3\Delta p}{12\eta L}.
$$
There are $n-1$ internal gaps. Their total cross-sectional area is $(n-1)wb$, because the plate thickness is negligible, so the <Darcy velocity> is
$$
\boxed{q=\frac{(n-1)Q_{\mathrm{gap}}}{(n-1)wb}
=\frac{b^2}{12\eta}\frac{\Delta p}{L},\qquad \Pi_0=\frac{b^2}{12}.}
$$
Thus 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.
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