 Planar Poiseuille flow with equal wall slip

Now both plates at $z=0$ and $z=H$ are stationary and both obey the Navier slip boundary condition with the same slip length $b$. Drive steady flow of a Newtonian fluid in incompressible flow by the constant pressure gradient $G=-dp/dx>0$, with constant dynamic viscosity $\mu >0$ and no other streamwise body force. The planar Poiseuille flow profile and its volumetric flow rate per unit width areThe full volumetric flow rate for channel width $W$ is $Q=Wq$ when side-wall effects are negligible. Thus $q$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, while $Q$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-1}$. Denote the value at $b=0$ by $q_0=G{H}^3/(12\mu )$.

The two wall checks expose the normal signs:Both Navier slip boundary conditions therefore hold exactly. The two walls exert backward tangential tractions of magnitude $GH/2$ on the fluid. Increasing $b$ at fixed $G$ adds a uniform velocity offset, leaving the velocity gradient and wall shear stress unchanged. This differs from Couette flow with a slipping lower wall, where upper-wall velocity is fixed and shear stress decreases.

For $H=10\mu \mathrm{m}$, $b=2\mu \mathrm{m}$, $G=10^3\mathrm{Pa}{\mathrm{m}}^{-1}$ and $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$, the wall velocity is $10\mu \mathrm{m}{\mathrm{s}}^{-1}$, the centre velocity is $22.5\mu \mathrm{m}{\mathrm{s}}^{-1}$, and $q=1.83\times 10^{-10}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. The enhancement is $q/q_0=2.2$. Equivalently, a prescribed volumetric flow rate needs only $1/2.2$ of the original pressure gradient. This makes slip length useful for estimating pumping requirements in small channels.

At fixed $H$, $G>0$ and $\mu$, $b\to 0$ recovers ordinary planar Poiseuille flow. As $b\to \infty$, the velocity and $q$ grow without bound in this model, although the varying part of the profile stays the same. At exactly zero tangential wall traction, no steady solution can balance this nonzero pressure gradient: $u^{\prime }(0)=u^{\prime }(H)=0$ conflicts with $\mu [u^{\prime }(H)-u^{\prime }(0)]=-GH$. Unlimited enhancement is consequently not a prediction for a real channel; finite wall friction, finite driving capacity and the validity of the assumed steady flow must be assessed. At fixed finite $q$ instead, $G\to 0$ and the profile approaches uniform velocity $q/H$.

A short derivation integrates the reduced Navier-Stokes equation $\mu u^{\prime \prime }=-G$ twice. Equal slip lengths and the opposite inward unit normals give $u=G[z(H-z)+bH]/(2\mu )$. Its integral over the gap supplies the enhancement factor $1+6b/H$.