For the <uniaxial extensional flow>
$$
\mathbf u=\dot\epsilon(-x/2,-y/2,z),
$$
the <spin tensor> vanishes and
$$
\dot{\boldsymbol\gamma}
=\operatorname{diag}(-\dot\epsilon,-\dot\epsilon,2\dot\epsilon).
$$
The flow is steady and homogeneous, so with $\xi=1$ the <Jaumann derivative> of $\boldsymbol\alpha$ vanishes. The structure equation yields
$$
\boldsymbol\alpha=\frac{c_2}{c_1}\dot{\boldsymbol\gamma}.
$$
Writing $k=c_2/c_1$ and using
$\dot{\boldsymbol\gamma}:\dot{\boldsymbol\gamma}=6\dot\epsilon^2$ gives
$$
\dot{\boldsymbol\gamma}\boldsymbol\alpha+
\boldsymbol\alpha\dot{\boldsymbol\gamma}
-\frac23(\boldsymbol\alpha:\dot{\boldsymbol\gamma})\mathbf I
=2k\dot\epsilon\,\dot{\boldsymbol\gamma}.
$$
Consequently
$$
\boldsymbol\tau
=\left(\eta_0+\frac{b_1c_2}{c_1}
+\frac{2b_2c_2}{c_1}\dot\epsilon\right)
\dot{\boldsymbol\gamma}.
$$
The <extensional viscosity> is the tensile stress difference divided by $\dot\epsilon$:
$$
\boxed{
\eta_{\rm ext}
=\frac{\tau_{zz}-\tau_{xx}}{\dot\epsilon}
=3\left(\eta_0+\frac{b_1c_2}{c_1}
+\frac{2b_2c_2}{c_1}\dot\epsilon\right)}.
$$
For $b_2=0$ this is the constant
$$
\boxed{\eta_{\rm ext}=3\left(\eta_0+\frac{b_1c_2}{c_1}\right)}.
$$
The factor three is the <Trouton ratio> associated with the effective zero-rate shear viscosity.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 352 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-352/2/c/solution