Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 352 (Non-Newtonian Fluid Mechanics)

Letbe the rate-of-strain tensor and let $\dot{\mathbf{\gamma }}=2\mathbf{D}$. The invariant magnitude convention that reproduces the ordinary scalar rate in simple shear flow isThe tensor contraction makes this definition independent of a rigid rotation of coordinates.

A generalized Newtonian fluid responds instantaneously with stress determined only by the current scalar shear rate. It therefore cannot represent viscoelasticity, including stress relaxation, elastic recoil, memory, and rate-dependent normal stresses. It also cannot represent thixotropy, in which microstructure and viscosity evolve with the duration and history of shearing. Other omitted effects include hysteresis and genuine finite-time yielding.

Put $G=\Delta p/L>0$, with flow in the positive $z$ direction, so $dp/dz=-G$. Balancing pressure forces and wall shear on a cylindrical fluid volume of radius $R$ and length $L$ givesThus the wall stress is fixed before any constitutive equation is specified.

For fully developed pipe flow, the axial Cauchy equation reduces toRegularity at the axis removes the $1/r$ integration constant, soIts magnitude rises linearly from zero at the axis to ${\tau }_R$ at the wall.

The axial velocity $u(r)$ decreases toward the wall, so $\dot{\gamma }=-du/dr\ge 0$. For a power-law fluid, the stress magnitude isHenceIntegrating inward from the no-slip boundary condition $u(R)=0$ givesFor $n=1$ this reduces to the parabolic Hagen--Poiseuille profile.

The volumetric flow rate is $Q=2\pi {\int }_0^{R}u(r)rdr$. Integration by parts, using finite $u(0)$ and $u(R)=0$, givesBecause the stress magnitude is $\tau ={\tau }_{R}r/R$, change variables from $r$ to $\tau$:Therefore the required power is $n=3$. The fundamental theorem of calculus gives the pipe form of the Weissenberg–Rabinowitsch equation:Finally $\eta ({\dot{\gamma }}_R)={\tau }_R/{\dot{\gamma }}_R$, soA measured pressure-drop--flow-rate curve therefore recovers the wall viscosity without assuming a constitutive form.

In the Giesekus model, $\lambda$ is the viscoelastic relaxation time: after deformation stops, polymeric stress relaxes on that timescale. The parameter $\eta$ has dimensions of dynamic viscosity and is the model's zero-rate viscosity scale. Since $\lambda {\mathbf{\tau }}^2/\eta$ already has dimensions of stress, $\alpha$ is dimensionless.

The symbol $\stackrel{▽}{\tau }$ is the upper-convected derivativeThe convective and velocity-gradient terms account for translation, rotation, and affine stretching of material elements. They make the constitutive law objective under time-dependent rigid changes of observer.

Setting $\alpha =0$ gives the Upper-convected Maxwell modelIn steady incompressible uniaxial extension of rate $\dot{ϵ}$, its tensile and transverse stresses giveThe extensional viscosity rises above the Trouton ratio value $3\eta$ and diverges at $\lambda \dot{ϵ}=1/2$, the ideal model's extensional catastrophe. An extensional rheometer can locate that rapid growth and estimate $\lambda \simeq 1/(2{\dot{ϵ}}_c)$. More robustly, impose a small deformation, stop the flow, and fit the exponential stress decay $\tau \propto e^{-t/\lambda }$.

Write $g=\dot{\gamma }$ andThe flow and stresses are steady and homogeneous, so the material derivative vanishes. Direct multiplication givesandSince ${\dot{\gamma }}_{xy}={\dot{\gamma }}_{yx}=g$, the four independent component equations areThe branch continuous from equilibrium has ${\tau }_{zz}=0$.

At order ${\alpha }^0$, the preceding equations giveAt order ${\alpha }^1$, the quadratic term is evaluated on ${\mathbf{\tau }}^0$. Solving first the $yy$ equation, then $xy$, then $xx$, givesThus the apparent shear viscosity isso positive $\alpha$ produces shear thinning. The expansion requires $\alpha \ll 1$ and $\alpha (\lambda g{)}^2\ll 1$; it cannot describe arbitrarily high shear rates even when $\alpha$ is numerically small.

Using the normal-stress difference definitionswe findIn 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.