At order $\alpha^0$, the preceding equations give
$$
\boxed{\tau_{xy}^0=\eta g,
\qquad \tau_{xx}^0=2\eta\lambda g^2,
\qquad \tau_{yy}^0=\tau_{zz}^0=0}.
$$
At order $\alpha^1$, the quadratic term is evaluated on $\boldsymbol\tau^0$. Solving first the $yy$ equation, then $xy$, then $xx$, gives
$$
\boxed{\tau_{yy}^1=-\eta\lambda g^2,
\qquad
\tau_{xy}^1=-3\eta\lambda^2g^3},
$$
$$
\boxed{\tau_{xx}^1
=-\eta\lambda g^2-10\eta\lambda^3g^4,
\qquad \tau_{zz}^1=0}.
$$
Thus the apparent <shear viscosity> is
$$
\boxed{\eta_{\rm app}(g)=\frac{\tau_{xy}}g
=\eta\left[1-3\alpha(\lambda g)^2\right]
+O(\alpha^2)},
$$
so positive $\alpha$ produces <shear thinning>. The expansion requires $\alpha\ll1$ and $\alpha(\lambda g)^2\ll1$; it cannot describe arbitrarily high shear rates even when $\alpha$ is numerically small.

Using the <normal-stress difference> definitions
$$
N_1=\tau_{xx}-\tau_{yy}=\Psi_1g^2,
\qquad
N_2=\tau_{yy}-\tau_{zz}=\Psi_2g^2,
$$
we find
$$
\boxed{\Psi_1
=2\eta\lambda\left[1-5\alpha(\lambda g)^2\right]
+O(\alpha^2),
\qquad
\Psi_2=-\alpha\eta\lambda+O(\alpha^2)}.
$$
In 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.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-352/2/d/solution