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Steady state extensional rheology of a dilute suspension of spheres in a dilute polymer solution (2208.13110v2)

Published 28 Aug 2022 in cond-mat.soft and physics.flu-dyn

Abstract: We investigate the steady-state extensional rheology of a dilute suspension of spherical particles in a dilute polymer solution. For a particle-free polymeric fluid, in addition to the solvent viscosity, the extensional viscosity due to the polymers, $\mu\text{poly}$, contributes to the total non-dimensionalized extensional viscosity $1+\mu\text{poly}$. When a small volume fraction, $\phi$, of spheres is added to a polymeric fluid, the stress is altered by the Einstein viscosity of 2.5$\phi$ and two additional stress contributions: the interaction stresslet and the particle-induced polymer stress (PIPS). The net interaction stress is positive at lower Deborah numbers (product of extension rate and polymer relaxation time), $De\lesssim0.5$, and negative at large $De$. Relative to undisturbed flow, the presence of spheres in uniaxial extensional flow creates larger and smaller local stretching regions. Below the coil-stretch transition ($De<0.5$), the polymers far from the particles are in a coiled state, and a wake of stretched polymers forms downstream of the particle as they are stretched by the large stretching regions around the particle. This leads to a positive interaction stresslet (surface) and the PIPS (stretched wake). Beyond the coil-stretch transition, polymers far from the particle are highly stretched, but they collapse closer to the coiled state as they arrive at the low-stretching regions near the particle surface. Therefore, a negative PIPS results from the regions of collapsed polymers. When $De\gtrsim0.6$, the changes in extensional viscosity from the interaction stresslet and the PIPS are $\phi\mu\text{poly}$ and approximately -1.85$\phi\mu\text{poly}$, respectively. At large $De$, the polymer extensional viscosity, $\mu\text{poly}$, is large. Therefore, adding particles reduces the extensional viscosity of the suspension ($(2.5-0.85\mu\text{poly})\phi<0$).

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