Show that at a fixed rigid solid surface contacting a viscous fluid, the stress vector is given by

Question

[latex]s=-p extbf{n}+\mu extbf{w} imes extbf{n}[/latex] (10.13.7a)

(incompressible case);

[latex]s=\left[-p+(\lambda+2\mu)div\ extbf{v} ight] extbf{n}+\mu extbf{w} imes extbf{n}[/latex] (10.13.7b)

(compressible case). Hence, compute the normal and shear stresses exerted by the fluid on the solid surface.

Accepted Answer

At a fixed rigid solid surface contacting a viscous fluid, we have v = 0, by the no-slip boundary condition (10.13.1)'. Hence if C is an arbitrary simple closed curve chosen on this surface,

v = 0 (10.13.1)'

[latex]\int_{C} extbf{v} imes d extbf{x}=0[/latex] (10.13.8)

By using a consequence of Stokes's theorem, given by (3.6.20), expression (10.13.8) becomes

[latex]\oint_{C} extbf{u} imes extbf{f}\ ds=\int_{s}{[(div\ extbf{u}) extbf{n}-( extbf{n}. riangledown) extbf{u}}- extbf{n} imes curl\ extbf{u}]dS[/latex] (3.6.20)

[latex]\int_{s}\left[(div\ extbf{v}) extbf{n}-( extbf{n}. riangledown) extbf{v}- extbf{n} imes extbf{w} ight]dS=0[/latex] (10.13.9)

where S is the area on the solid surface, bounded by C Since C is arbitrary, it follows that the integrand of the surface integral in (10.13.9) should vanish identically; hence,

[latex]( extbf{n}. riangledown) extbf{v}=(div\ extbf{v}) extbf{n}+ extbf{w} imes extbf{n}[/latex] (10.13.10)

on the solid surface considered.
Substituting for (n · ▽)v from (10.13.10) in (10.13.2a) and noting that divv = 0 for an incompressible fluid, we get the result (10.13.7a).
Similarly, (10.13.10) and (10.13.2b) yield the result (10.13.7b).
The normal stress exerted by the fluid on the solid surface is given by

[latex]s=-p extbf{n}+2\mu( extbf{n}. riangledown) extbf{v}+\mu extbf{n} imes extbf{w}[/latex] (10.13.2a)

[latex]s=-p extbf{n}+\mu extbf{w} imes extbf{n}[/latex] (10.13.7a)

[latex]s=(-p+\lambda\ div\ extbf{v}) extbf{n}+2\mu( extbf{n}. riangledown) extbf{v}+\mu extbf{n} imes extbf{w}[/latex] (10.13.2b)

[latex]s=\left[-p+(\lambda+2\mu)div\ extbf{v} ight] extbf{n}+\mu extbf{w} imes extbf{n}[/latex] (10.13.7b)

[latex]\sigma=-s. extbf{n}[/latex] (10.13.11)

where n is directed into the fluid. For an incompressible fluid, (10.13.7a) and (10.13.11) yield σ = p. Since p is equal to the mean pressure p for an incompressible fluid, it follows that

[latex]\sigma=\bar{p}[/latex] (10.13.12)

Thus, in the case of an incompressible viscous fluid, the fluid elements exert just the mean stress on the surface, along the normal.

For a compressible fluid, we find by using (10.13.7b) in (10.13.11) that

[latex]\sigma=p-(\lambda+2\mu)(div\ extbf{v})[/latex] (10.13.13)

which on using (10.11.6) and (10.11.5) may be rewritten as

[latex]p=\bar{p}+\kappa (div v)[/latex] (10.11.6)

[latex]\kappa =\lambda +\frac{2}{3}\mu [/latex] (10.11.5)

[latex]\sigma=\bar{p}-\frac{4}{3}\mu(div\ extbf{v})[/latex] (10.13.14)

Thus, in the case of a compressible viscous fluid the normal stress exerted
by the fluid on the surface is caused by the mean pressure as well as the rate of change in volume of fluid elements.
If t is a unit vector tangential to the surface, the magnitude of the shear stress exerted by the fluid on the solid surface is

[latex] au=| extbf{s.t}|[/latex] (10.13.15)

By using (10.13.7ab) in (10.13.15), we find that the following expression holds for both incompressible and compressible fluids:

[latex]\begin{matrix} s=-pn+\mu w imes n & (10.13.7a) \ s=[-p+(\lambda +2\mu )div v]n+\mu w imes n & (10.13.7b) \end{matrix} [/latex]

[latex] au=\mu|[ extbf{w,n,t}]|[/latex] (10.13.16)

Thus, τ depends only on the vorticity vector w for both incompressible and compressible viscous fluids.
Form (10.13.16) it also follows that, in irrotational motion, no shear stress is exerted on a fixed solid boundary.

Question 10.13.2: Show that at a fixed rigid solid surface contacting a viscou...

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