A cone-and-plate rheometer has a cone of radius and cone angle

PHYSICS · Subjective · Fermi University Physics · Fluids & Continuum Physics · Solid continuum mechanics and complex fluids · Solid continuum mechanics and complex fluids

Bending, Ringing, Breaking and Flowing, Ch Non-Newtonian and viscoelastic fluids

Problem

A cone-and-plate rheometer has a cone of radius R=25 mmR = 25\,\mathrm{mm} and cone angle α=0.040 rad\alpha = 0.040\,\mathrm{rad}. With a sample of an unknown fluid in the gap, the torque needed to turn the cone is M1=0.411 mN mM_1 = 0.411\,\mathrm{mN\,m} at Ω1=0.40 rad s−1\Omega_1 = 0.40\,\mathrm{rad\,s^{-1}} and M2=1.03 mN mM_2 = 1.03\,\mathrm{mN\,m} at Ω2=4.0 rad s−1\Omega_2 = 4.0\,\mathrm{rad\,s^{-1}}. Neglect edge effects and inertia. (a) The shear rate in the gap is the same at every radius. Explain why the shear stress τ\tau on the cone is then the same at every radius too, and show that the torque is M=2πR3τ/3M = 2\pi R^3\tau/3. (b) Find the shear rate and the viscosity η=τ/γ˙\eta = \tau/\dot\gamma at each of the two speeds. (c) Fit a power law τ=Kγ˙n\tau = K\dot\gamma^n through the two points to find nn and KK, and say whether the fluid is shear-thinning or shear-thickening.

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