Article Contents
Article Contents

# Modeling of macroscopic stresses in a dilute suspension of small weakly inertial particles

• * Corresponding author: A. Vibe
• In this paper we derive asymptotically the macroscopic bulk stress of a suspension of small inertial particles in an incompressible Newtonian fluid. We apply the general asymptotic framework to the special case of ellipsoidal particles and show the resulting modification due to inertia on the well-known particle-stresses based on the theory by Batchelor and Jeffery.

Mathematics Subject Classification: Primary: 76Axx, 76Dxx, 76Mxx, 76Txx.

 Citation:

• Figure 1.  Lagrangian description, bijective mapping between reference state and the actual time-dependent state.

Table 1.  Classification of inertial types by means of the density scaling function, cf. (4).

 $\alpha_{\rho}(\epsilon)=\epsilon^r$ Name of type Behavior of density ratio $r\geq1$ light weighted tracer particles $\rho_\epsilon\to0$ $r=0$ normal tracer particles $\rho_\epsilon\equiv const$ $r=-1$ heavy tracer particles $\rho_\epsilon\to\infty$

Table 2.  Balances of the accelerations with the integral terms (given in Remark 2) in (15) for the different inertial regimes. Each row shows the $\mathcal{O}(\epsilon^\ell)$-correction of the momentum equations and each column corresponds to the choice of the density scaling function $\alpha_\rho(\epsilon) = \epsilon^r$.

 $\ell\backslash r$ $-3$ $-2$ $-1$ $0$ 1 $-2$ ${\bf{k}}_0={\bf{0}}$ $-1$ ${\bf{k}}_1={\bf{0}}$ ${\bf{k}}_0={\bf{0}}$ $0$ ${\bf{k}}_2={\bf{m}}_1^v$ ${\bf{k}}_1={\bf{m}}_1^v$ ${\bf{k}}_0={\bf{m}}_1^v$ ${\bf{0}}={\bf{m}}_1^v$ ${\bf{0}}={\bf{m}}_1^v$ $1$ ${\bf{k}}_3={\bf{m}}_2^v$ ${\bf{k}}_2={\bf{m}}_2^v$ ${\bf{k}}_1={\bf{m}}_2^v$ ${\bf{k}}_0={\bf{m}}_2^v$ ${\bf{0}}={\bf{m}}_2^v$ $\ell\backslash r$ $-3$ $-2$ $-1$ $0$ $1$ $-1$ $\mathit{\boldsymbol{\ell}}_0={\bf{0}}$ $0$ $\mathit{\boldsymbol{\ell}}_1={\bf{m}}_1^{\omega}$ $\mathit{\boldsymbol{\ell}}_0={\bf{m}}_1^{\omega}$ ${\bf{0}}={\bf{m}}_1^{\omega}$ ${\bf{0}}={\bf{m}}_1^{\omega}$ ${\bf{0}}={\bf{m}}_1^{\omega}$ $1$ $\mathit{\boldsymbol{\ell}}_2={\bf{m}}_2^{\omega}$ $\mathit{\boldsymbol{\ell}}_1={\bf{m}}_2^{\omega}$ $\mathit{\boldsymbol{\ell}}_0={\bf{m}}_2^{\omega}$ ${\bf{0}}={\bf{m}}_2^{\omega}$ ${\bf{0}}={\bf{m}}_2^{\omega}$
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