Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 352 (Non-Newtonian Fluid Mechanics)

In small-amplitude oscillatory shear, expand the measured shear stress asComparison with the definitionof the storage modulus and loss modulus givesand henceThus the measured phase directly gives the loss tangent.

At low angular frequency, a standard viscoelastic fluid has time to relax and responds mainly as a viscous liquid: stress is approximately in phase with strain rate, so $\delta \to \pi /2$ as $\omega \to 0$. At high frequency it cannot relax during one cycle and responds mainly as an elastic solid, so stress is approximately in phase with strain and $\delta \to 0$ as $\omega \to \infty$.

For a material dominated by one viscoelastic relaxation time $\lambda$, the crossover between these regimes occurs when $\omega \lambda$ is of order one. Measure the frequency ${\omega }_c$ at which $G^{\prime }=G^{\prime \prime }$, equivalently $\delta =\pi /4$, and estimateA broad or multi-peaked crossover instead indicates a spectrum of relaxation times.

In a parallel-plate rheometer, a point at radius $r$ on the upper plate moves at speed $\Omega r$. The thin-gap approximation therefore gives the local shear rateDefine the rim valueAn annulus of radius $r$ and width $dr$ has area $2\pi rdr$; its tangential force is $\tau (r)2\pi rdr$, and its moment arm is $r$. The measured torque is consequently

For a generalized Newtonian fluid, $\tau =\eta (\dot{\gamma })\dot{\gamma }$. Changing the integration variable from $r$ to $\dot{\gamma }$ givesThus the requested kernel isMultiplying by ${\dot{\gamma }}_R^3$ and taking a derivative with respect to ${\dot{\gamma }}_R$ turns the upper-limit contribution of the integral into the rim viscosity:ThereforeA measured torque curve and its slope hence determine $\eta$ over the range of imposed rim shear rates.

The experimental law impliesSubstitution givesThe graph is a decreasing rectangular hyperbola with a positive high-rate plateau $2/(\pi \alpha R^3)$ and a $1/\dot{\gamma }$ divergence at the origin. Equivalently,so the inferred material is a Bingham plastic with yield stress $3T_c/(2\pi R^3)$.

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.

At the final state of the slump test for yield stress, an axisymmetric deposit of height $h(r)$ is just able to support itself. Hydrostatic pressure gives radial pressure gradient $\rho g{h}^{\prime }(r)$, while lubrication theory makes the magnitude of the basal shear stressThe marginally yielded final profile therefore satisfiesIntegration givesUsing mass conservation, the known volume isSolving for the yield stress produces the estimateThus one measures the final radius $R$ and inserts it with $V$ and $\rho$. The estimate assumes a thin deposit, negligible surface tension, complete initial yielding, and a spatially uniform yield stress.

Dry sand is a granular material governed primarily by frictional stability. Its final free surface reaches the angle of repose ${\theta }_r$, so the deposit is approximately a cone,This constant-slope profile differs from the square-root edge of the yield-stress-fluid model.

Let $w(x)$ be the downward velocity between the vertical plates. A steady vertical momentum balance givesSymmetry requires ${\tau }_{xz}(0)=0$, so the stress magnitude isThe wall stress first reaches the yield stress when $\rho g{l}_c={\tau }_y$. The onset measurement therefore gives

For $l>l_c$, define $\Delta =l-l_c$. The central region $|x|\le l_c$ is a plug flow of a yield-stress fluid, while the layers $l_c<|x|\le l$ are yielded. The Herschel–Bulkley fluid law gives, for $x\ge l_c$,Integrating from the no-slip wall $w(l)=0$ givesin the yielded layer, and the plug moves atThe volumetric flow rate per unit span is consequently

Near onset, the plug term dominates andThe observed exponent two therefore corresponds to $n=1$, a linear post-yield constitutive law at small shear rate. Far above onset, the fully yielded contribution scales asThe observed exponent four corresponds to $n=1/2$, a shear-thinning square-root law at high shear rate.

A plausible stress curve is therefore odd in $\dot{\gamma }$, starts at the two yield-stress values $\tau =\pm {\tau }_y$, and has finite linear slope just after yield:It then crosses smoothly to the concave high-rate behaviorwith the negative-rate branch fixed by odd symmetry. This combines the two power laws inferred independently from the measured flux limits.