Past exam of the advanced fluid dynamics course of the University of Oxford / 2016 / 2 / c

Let $g=\dot{\gamma }$ and use $G_{yx}=g$ for the source's velocity gradient convention. Seek the homogeneous stationary stress reached under steady simple shear flow. The material derivative of that stress tensor vanishes. In the $xy$ plane,The $xx$, $yy$ and $zz$ components of the constitutive equation give, respectively,Thus, putting $D=1+\alpha (2\tau -\alpha )g^2$ and using the shear stress calculated below,The out-of-plane shear components vanish on the homogeneous branch. The two normal-stress differences areThe ratio is defined at nonzero shear when $N_1\ne 0$; at $g=0$ both differences vanish, and the displayed constant is their limiting ratio. An arbitrary isotropic pressure in the total Cauchy stress tensor cancels from both differences; it is not an arbitrary addition to the constitutively defined extra stress $\mathsf{T}$ itself.

The $xy$ component of the stationary constitutive equation isInsert $T_{yy}=-\alpha g{T}_{xy}$ and $T_{xx}=(2\tau -\alpha )g{T}_{xy}$. The coefficient of $g^2{T}_{xy}$ becomes $\tau \alpha +\alpha (\tau -\alpha )=\alpha (2\tau -\alpha )$, givingFor $C=\alpha (2\tau -\alpha )>0$, the small-rate shear stress is $T_{xy}=\mu \dot{\gamma }-\mu C{\dot{\gamma }}^3+O({\dot{\gamma }}^5)$. At large positive rate, $T_{xy}\sim \mu /(C\dot{\gamma })$ and the shear viscosity decreases as $\mu /(C{\dot{\gamma }}^2)$. The shear stress is odd in rate, reaches a positive maximum atand decreases thereafter. The original plot displays this homogeneous constitutive curve. Its decreasing branch can permit shear banding; the local model alone supplies neither an interface-selection rule nor a guaranteed stable homogeneous realization.

This is a Johnson-Segalman fluid without an added Newtonian fluid solvent contribution. To identify conventions explicitly, set $a=1-\alpha /\tau$ and $L=G^T=E+W$, with $W$ the spin tensor. The derivative appearing in the constitutive equation becomesthe Gordon--Schowalter derivative. Its usual slip interval $|a|\le 1$ corresponds to $0\le \alpha \le 2\tau$, while a nonsingular genuinely shear thinning curve requiresAt $\alpha =0$ the model is the Upper-convected Maxwell model; at $\alpha =2\tau$ it is the lower-convected derivative counterpart, and both have constant shear viscosity in this test. At $\alpha =\tau$ the derivative is the Jaumann derivative, giving the corotational Maxwell fluid. Merely requiring $\alpha >0$ is insufficient: for $\alpha =3\tau$ the denominator is $1-3{\tau }^2{\dot{\gamma }}^2$, which vanishes at $|\dot{\gamma }|=1/(\sqrt{3}\tau )$ and then gives a negative steady shear viscosity. There is no solvent-induced large-rate plateau in the stated equation.