Exact Solutions of Navier-Stokes Equations

Plane Couette and Poiseuille flow

Plane Couette flow and plane Poiseuille flow are the two elementary exact solutions of the Navier–Stokes equations for steady, incompressible, fully developed flow between two infinite parallel flat plates. The flow is unidirectional, so the transverse velocity is zero and the streamwise velocity u depends only on the normal coordinate y. Continuity then gives \partial u/\partial x=0, and the pressure does not vary across the gap. The x-momentum equation reduces to 0=-\frac{dp}{dx}+\mu\frac{d^2u}{dy^2}, so the curvature of the velocity profile is set entirely by the streamwise pressure gradient. The no-slip condition at both walls closes the problem, and the velocity is a polynomial in y of degree at most two.\n\nCouette flow is driven by the relative motion of the walls, with no imposed pressure gradient. If the lower plate is fixed and the upper plate moves at speed U, the profile is linear, u=Uy/h, and the shear stress \tau=\mu\,du/dy=\mu U/h is the same at every height. The same physics governs a belt skimming over a thin oil film: the wall shear times the belt area gives the drag force, and the drag force times the belt speed gives the power, P=\mu V^2 L b/h for a film of depth h. Because the shear stress is uniform across the gap, the ratio of the shear stress at the two walls is unity. The wall shear on the moving plate is simply \mu U/h, so a thin gap with a slow wall gives a small stress. The vorticity of plane Couette flow is constant, with magnitude |\omega_z|=du/dy=U/h. By Stokes' theorem, the circulation around a closed contour equals the vorticity multiplied by the enclosed area, so a square element of side a has circulation (U/h)a^2. The stream function satisfies u=\partial\psi/\partial y, so the difference of \psi between the two plates equals the volume flow per unit depth. For a linear profile across a gap L with top speed U, this difference is the area of a triangle, UL/2.\n\nPlane Poiseuille flow is driven by a streamwise pressure gradient G=-dp/dx between stationary plates. The profile is parabolic, u=\frac{G}{2\mu}y(h-y) measured from the lower wall, with maximum u_{max}=Gh^2/(8\mu) at mid-gap. The mean velocity is \bar u=\frac{2}{3}u_{max}=\frac{Gh^2}{12\mu}, so the gap can be recovered from a measured mean velocity and pressure gradient through h^2=12\mu\bar u/G. If the profile is written as u=u_0(1-ay^2) with y measured from the centreline, the second derivative is the constant -2au_0, and the momentum equation gives a pressure gradient magnitude of 2a\mu u_0. For a channel of half-width h with u=u_{max}(1-y^2/h^2), the local velocity equals the mean at y=\pm h/\sqrt{3}, since 1-y^2/h^2=2/3 there.\n\nWhen a pressure gradient and wall motion act together, the solutions superpose. For a lower wall at rest and an upper wall moving at U, the profile is u(y)=\frac{1}{2\mu}\frac{dp}{dx}(y^2-hy)+\frac{Uy}{h}. The net flow per unit width is Q=-\frac{h^3}{12\mu}\frac{dp}{dx}+\frac{Uh}{2}, so zero net flow requires |dp/dx|=6\mu U/h^2. The pressure gradient must cancel the Couette flux.\n\nFor two immiscible liquids stacked in the same channel, the shear stress is continuous across the interface, but the velocity gradients differ because the viscosities differ. The velocity maximum occurs where the local shear stress vanishes. Because the stress varies linearly across the passage, this zero lies in the less viscous layer, and when the lower fluid is the more viscous one, the maximum lies above the interface.