Navier slip boundary condition

{title2=$\mathcal D_w=\mu u_{\mathrm{rel}}^2/b$}

For a <Newtonian fluid> and a flat wall with $0<b<\infty$, the <Navier slip boundary condition> is equivalent to a tangential wall-on-fluid <traction>
$$
t_{w,t}=-\frac{\mu}{b}u_{\mathrm{rel}}.
$$
The sign expresses resisting <friction>: $\partial_nu$ uses the inward <unit normal>, whereas the wall-on-fluid <traction> uses the outward <unit normal> of the fluid domain. The irreversible loss of mechanical <energy> per unit wall <area> and time is
$$
\mathcal D_w=-t_{w,t}u_{\mathrm{rel}}
=\frac{\mu}{b}u_{\mathrm{rel}}^2\ge0,
\qquad [\mathcal D_w]=\mathrm{W\,m^{-2}}.
$$
For two tangential directions replace $u_{\mathrm{rel}}^2$ by the squared <Euclidean norm> of relative tangential <velocity>. At a moving wall this relative-<velocity> loss must be distinguished from the <power> supplied by the wall's motion. Bulk <viscous dissipation> also remains present. A finite <slip length> models reduced interfacial resistance, without declaring every boundary ideal or removing <dynamic viscosity>.

The limiting loss depends on how the flow is driven. In <Couette flow with a slipping lower wall>,
$$
\mathcal D_w=\frac{\mu U^2b}{(H+b)^2}.
$$
It tends to zero for both $b\to0$ and $b\to\infty$ at fixed $H$ and $U$. At $b=0$ impose the <no-slip boundary condition> directly, rather than dividing by zero. In <planar Poiseuille flow with equal wall slip> at fixed $G$, each wall instead dissipates $G^2bH^2/(4\mu)$, which grows with $b$ because relative <velocity> grows. The bulk loss per unit width is $G^2H^3/(12\mu)$; adding both wall losses gives $Gq$, the <pressure> work per unit channel length and width. This <energy> balance shows why small friction coefficients do not guarantee small losses under every driving constraint.


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 Incoming links: Wall friction in the Navier slip boundary condition