Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 352 / 2

In the Giesekus model, $\lambda$ is the viscoelastic relaxation time: after deformation stops, polymeric stress relaxes on that timescale. The parameter $\eta$ has dimensions of dynamic viscosity and is the model's zero-rate viscosity scale. Since $\lambda {\mathbf{\tau }}^2/\eta$ already has dimensions of stress, $\alpha$ is dimensionless.

The symbol $\stackrel{▽}{\tau }$ is the upper-convected derivativeThe convective and velocity-gradient terms account for translation, rotation, and affine stretching of material elements. They make the constitutive law objective under time-dependent rigid changes of observer.

Setting $\alpha =0$ gives the Upper-convected Maxwell modelIn steady incompressible uniaxial extension of rate $\dot{ϵ}$, its tensile and transverse stresses giveThe extensional viscosity rises above the Trouton ratio value $3\eta$ and diverges at $\lambda \dot{ϵ}=1/2$, the ideal model's extensional catastrophe. An extensional rheometer can locate that rapid growth and estimate $\lambda \simeq 1/(2{\dot{ϵ}}_c)$. More robustly, impose a small deformation, stop the flow, and fit the exponential stress decay $\tau \propto e^{-t/\lambda }$.

Write $g=\dot{\gamma }$ andThe flow and stresses are steady and homogeneous, so the material derivative vanishes. Direct multiplication givesandSince ${\dot{\gamma }}_{xy}={\dot{\gamma }}_{yx}=g$, the four independent component equations areThe branch continuous from equilibrium has ${\tau }_{zz}=0$.

At order ${\alpha }^0$, the preceding equations giveAt order ${\alpha }^1$, the quadratic term is evaluated on ${\mathbf{\tau }}^0$. Solving first the $yy$ equation, then $xy$, then $xx$, givesThus the apparent shear viscosity isso positive $\alpha$ produces shear thinning. The expansion requires $\alpha \ll 1$ and $\alpha (\lambda g{)}^2\ll 1$; it cannot describe arbitrarily high shear rates even when $\alpha$ is numerically small.

Using the normal-stress difference definitionswe findIn the low-rate limit, $-2{\Psi }_2/{\Psi }_1=\alpha +O({\alpha }^2)$. A cone-and-plate or parallel-plate rheometer can measure shear stress and normal thrust over a low-rate range; combining ${\Psi }_1$ and ${\Psi }_2$ then estimates $\alpha$ independently of the viscosity scale.