Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 352 (Non-Newtonian Fluid Mechanics)

A Bingham plastic is a yield-stress fluid, and hence a non-Newtonian fluid. It is also an instantaneous generalized-Newtonian model after yielding. It cannot describe memory-dependent viscoelasticity, including stress relaxation, elastic recoil, and time-dependent normal-stress effects.

For scalar shear stress $\tau$ and shear rate $\dot{\gamma }$, its complete ideal constitutive law isEquivalently, in yielded material,

Set $G=\Delta p/L>0$, so $dp/dz=-G$. For the unidirectional velocity $\mathbf{u}=u(r){\mathbf{e}}_z$, the axial inertialess Cauchy momentum equation isThe free surface is shear-free, ${\tau }_{rz}(b)=0$, and integration givesThis result follows from force balance alone, so it is independent of the constitutive equation and of ${\sigma }_y$.

The axial shear force exerted by the liquid on the cylindrical fibre is thereforeIts direction is downstream; the reaction on the fluid is upstream. In particular, the yield stress changes the velocity field but not the total force required by the imposed pressure drop.

The shear stress decreases monotonically from the fibre to zero at the free surface, so its maximum isFlow occurs precisely when this exceeds the yield stress:

When this condition holds, there is one yield surface $r=r_y$ determined byThe region $a\le r<r_y$ is yielded and the outer region $r_y\le r\le b$ is an unyielded plug flow of a yield-stress fluid. In the yielded region, $du/dr=({\tau }_{rz}-{\sigma }_y)/\eta$. The no-slip condition $u(a)=0$ givesThe unyielded layer translates without shearing at the plug speed

The local motion is simple shear with flow direction $z$, gradient direction $r$, and vorticity direction $\theta$. Writing $q=du/dr$, a second-order fluid has first and second normal-stress differencesUp to an isotropic contribution absorbed into pressure, a convenient representation isThus both the flow-direction and gradient-direction normal stresses can be nonzero, while the radial momentum equation becomesThe first normal-stress coefficient affects ${\tau }_{zz}$ but cancels from radial balance.

In the yielded layer,Moreover,Consequently $dp/dr$ has the sign opposite to ${\Psi }_2$ and is independent of ${\Psi }_1$. For the common case ${\Psi }_2<0$, pressure increases radially outward toward the free surface.

An objective time derivative transforms as a tensor under every superposed time-dependent rigid translation and rotation of the observer. A constitutive prediction using it is therefore frame-indifferent. The upper-convected derivative, lower-convected derivative, and Jaumann derivative are objective; the componentwise partial time derivative, and generally the uncorrected material derivative of a tensor, are not objective under rotating observers.

The Johnson--Segalman--Oldroyd model uses the Gordon--Schowalter derivativeHere $\eta$ is the zero-shear viscosity, ${\lambda }_1$ is the stress-relaxation time, and ${\lambda }_2$ is the retardation time associated with the more rapidly responding viscous contribution. Passive Oldroyd-type materials normally have $0\le {\lambda }_2\le {\lambda }_1$.

A low-rate steady-shear experiment measures $\eta$. A step strain followed by stress relaxation determines ${\lambda }_1$ from the exponential decay. Creep and recovery, or a frequency sweep in small-amplitude oscillatory shear, then separates ${\lambda }_2$ from ${\lambda }_1$; fitting both storage modulus and loss modulus is especially direct.

The convected terms are quadratic in the disturbance. Linear response requires small strain and small rate-based Weissenberg numbers, in particularThe constitutive equation then reduces toFor $\dot{\gamma }={\dot{\gamma }}_0\cos \omega t$, its periodic stress response isSince $\gamma =({\dot{\gamma }}_0/\omega )\sin \omega t$, comparison with $\tau ={\gamma }_0[G^{\prime }\sin \omega t+G^{\prime \prime }\cos \omega t]$ givesThe condition ${\lambda }_1\ge {\lambda }_2$ makes the storage modulus nonnegative.

Take steady simple shear $u_x=\dot{\gamma }y$ and define $W_i={\lambda }_i\dot{\gamma }$. Substitution of the symmetric stress components into the full constitutive equation gives, after solving the three coupled algebraic equations,Thus the Steady shear viscosity of a Johnson--Segalman--Oldroyd fluid isFor the stated small positive $a$, so that $0<a<2$, this decreases from $\eta$ at zero rate to $\eta {\lambda }_2/{\lambda }_1$ at high rate exactly whenEquality gives constant viscosity, while ${\lambda }_2>{\lambda }_1$ gives shear thickening.