= Solution
For an <incompressible flow>, write the <Newtonian fluid stress tensor> as $\sigma=-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$ is the <rate-of-strain tensor>. The <Stokes equation> gives $\partial_j\sigma_{ij}=0$. Symmetry of the <Cauchy stress tensor> therefore gives
$$
\partial_j(u_i\sigma_{ij})=\sigma_{ij}\partial_j u_i=\sigma:e=2\mu e:e.
$$
Integrating and applying the <divergence theorem> expresses the <viscous dissipation> as boundary power:
$$
\boxed{D=\int_{\partial V}u\cdot\sigma n\,dS.}
$$
At a moving <rigid body>, the surface <velocity> is $U+\Omega\times(x-x_c)$. Its boundary-power contribution is $U\cdot F+\Omega\cdot G$, with the <force> and <torque> evaluated using the fluid's outward <normal vector>. Both resultants vanish for a <force-free>, <torque-free> inclusion. Thus the new <viscous dissipation> comes entirely from the unchanged outer boundary <velocity>. Subtracting the particle-free boundary power gives
$$
D'=\int_{\partial V_0}U_0\cdot\sigma'n\,dS.
$$
Apply the <Lorentz reciprocal theorem> to $(u_0,\sigma_0)$ and $(u',\sigma')$ in the fluid outside the inclusion. On the outer boundary $u'=0$, so
$$
0=D'+\int_A(u_0\cdot\sigma'-u'\cdot\sigma_0)n\,dS.
$$
Writing $n_p=-n$ for the particle's outward <normal vector> gives the required <extra dissipation due to a rigid inclusion>:
$$
\boxed{D'=\int_A(u_0\cdot\sigma'-u'\cdot\sigma_0)n_p\,dS.}
$$
This subtraction already accounts for the fluid volume displaced by the inclusion; it is not just the integral of the disturbance's local <viscous dissipation>.
For the sphere, the ambient <rate-of-strain tensor> $E$ is symmetric and trace free. The <sphere in a uniform straining Stokes flow> has zero translational and rotational <velocity>: inversion symmetry eliminates its <force>, and symmetry of $E$ eliminates its <torque>. To derive the disturbance, put $r=|x|$, $Q=x\cdot Ex$ and seek
$$
u'=f(r)Ex+g(r)Qx,\qquad p'=\mu j(r)Q.
$$
The <incompressible flow> condition and <Stokes equation> reduce to
$$
\frac{f'}r+rg'+5g=0,\qquad
f''+\frac4r f'+4g=2j,\qquad
g''+\frac8r g'=\frac{j'}r.
$$
A decaying family satisfying these equations is $f=Ar^{-5}$, $g=Br^{-5}-5Ar^{-7}/2$, $j=2Br^{-5}$. The <no-slip boundary condition> requires $f(a)=-1$ and $g(a)=0$, so $A=-a^5$ and $B=-5a^3/2$. Consequently
$$
\boxed{u'=-\frac{a^5}{r^5}Ex-\frac{5a^3}{2r^5}\left(1-\frac{a^2}{r^2}\right)(x\cdot Ex)x,\qquad
p'=-\frac{5\mu a^3}{r^5}(x\cdot Ex).}
$$
The total <velocity> is zero at $r=a$, and the disturbance decays as $r^{-2}$. The pressure constant has been set to zero. These boundary and far-field conditions, together with <Uniqueness of Stokes flow>, establish the solution. There is no rotational background in a pure strain flow.
On the sphere $u'=-u_0=-aEn_p$. Contracting the supplied <Newtonian fluid stress tensor> with $n_p$ makes its two terms proportional to $(n_p\cdot En_p)n_p$ cancel, leaving $\sigma n_p=5\mu En_p$. Hence the <extra dissipation due to a rigid inclusion> is
$$
D'=5\mu a\int_A|En_p|^2\,dS
=5\mu a^3 E_{ij}E_{ik}\frac{4\pi}{3}\delta_{jk}
=\boxed{\frac{20\pi}{3}\mu a^3 E:E}.
$$
Here $\int_{S^2}n_jn_k\,d\Omega=4\pi\delta_{jk}/3$. Taking the large outer boundary limit only after using the fixed-boundary power identity avoids replacing that prescribed boundary by an uncontrolled boundary at infinity.
If $N_v$ denotes the number of spheres per unit volume, $N_v(4\pi a^3/3)=\phi$. Thus
$$
\boxed{N_v=\frac{3\phi}{4\pi a^3},\qquad N_vD'=5\mu\phi E:E.}
$$
Add this to the particle-free <viscous dissipation> density $2\mu E:E$. Comparing with $2\mu_{\mathrm{eff}}E:E$ gives the <Einstein viscosity formula for a dilute suspension>:
$$
\boxed{\mu_{\mathrm{eff}}=\mu\left(1+\frac52\phi\right)}
$$
to first order in $\phi$. The <hydrodynamic interactions> neglected here contribute beyond that dilute order.
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