Extra dissipation due to a rigid inclusion ($D-D_0\ge 0$)

At fixed outer boundary velocity, extending the inclusion velocity rigidly through its interior makes the particle flow an admissible competitor for the particle-free Minimum-dissipation theorem for Stokes flow. Its interior rate-of-strain tensor vanishes. For a linear background, the increase in viscous dissipation is force times relative translation, torque times relative rotation, plus the contraction of the particle stresslet tensor with background strain.

At fixed sphere rotation in an annular Stokes flow, add rigid particles with zero net force and torque. Extend their actual rigid velocities through their interiors. The extension is an admissible incompressible competitor in the particle-free annulus, with zero interior rate-of-strain tensor. The Minimum-dissipation theorem for Stokes flow gives $D\ge D_0$. The boundary-work identity gives $D=\mathbf{\Omega }\cdot \mathbf{G}$, because the stationary outer wall and the freely moving particles contribute no power. Equality would require the extension to equal the unique particle-free flow, whose strain cannot vanish on a finite particle volume when $\Omega \ne 0$. Thus the resistance component along the imposed rotation strictly increases.

A dilute suspension of identical rigid spheres raises the dynamic viscosity by a fraction $5\varphi /2$, where $\varphi$ is the particle volume fraction. At imposed trace-free rate-of-strain tensor $E$, a sphere of radius $a$ adds viscous dissipation $20\pi \mu a^3(E:E)/3$. Multiplication by the number density $3\varphi /(4\pi a^3)$ gives $5\mu \varphi E:E$, which added to $2\mu E:E$ proves the formula. This is the leading noninteracting-sphere term; hydrodynamic interactions enter at higher concentration.