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

Write the velocity gradient aswhere $\mathbf{\Omega }$ is the spin tensor. Expanding the upper-convected derivative in the structure equation givesFor $\xi =0$, the first four terms reproduce the upper-convected derivative. Settingtherefore givesWith polymeric stress ${\mathbf{\tau }}_p=b_1\mathbf{\alpha }$, this is the Oldroyd-B model. For $c_1>0$, its relaxation time and polymer viscosity areIndeed the total stress obeys

For $\xi =2$, the coefficient of
$\dot{\mathbf{\gamma }}\mathbf{\alpha }+\mathbf{\alpha }\dot{\mathbf{\gamma }}$
is $+1/2$, so the objective derivative becomes the lower-convected derivative. Hencerecovers the Oldroyd-A model, again with $c_1>0$ for a finite positive relaxation time. Parameter choices such as $b_1{c}_2=0$ give the degenerate Newtonian limit.

Let $g=\dot{\gamma }$ denote the constant rate in the simple shear flowThenWhen $\xi =1$, the structure equation contains the Jaumann derivative. In a steady homogeneous flow it becomesSolving its component equations gives

The $b_2$ term has no $xy$ component because ${\alpha }_{xx}+{\alpha }_{yy}=0$. Thusand the shear viscosity isIt exhibits shear thinning whenever $b_1{c}_1{c}_2>0$: it decreases from ${\eta }_0+b_1{c}_2/c_1$ at zero shear rate to the solvent plateau ${\eta }_0$ at large shear rate. If the product vanishes, the viscosity is constant.

For the diagonal stresses,The two normal-stress differences are therefore

For the uniaxial extensional flowthe spin tensor vanishes andThe flow is steady and homogeneous, so with $\xi =1$ the Jaumann derivative of $\mathbf{\alpha }$ vanishes. The structure equation yieldsWriting $k=c_2/c_1$ and using
$\dot{\mathbf{\gamma }}:\dot{\mathbf{\gamma }}=6{\dot{ϵ}}^2$ givesConsequentlyThe extensional viscosity is the tensile stress difference divided by $\dot{ϵ}$:For $b_2=0$ this is the constantThe factor three is the Trouton ratio associated with the effective zero-rate shear viscosity.