= Solution
Let $w(x)$ be the downward velocity between the vertical plates. A steady vertical momentum balance gives
$$
\frac{d\tau_{xz}}{dx}=-\rho g.
$$
Symmetry requires $\tau_{xz}(0)=0$, so the stress magnitude is
$$
|\tau_{xz}(x)|=\rho g|x|.
$$
The wall stress first reaches the yield stress when $\rho gl_c=\tau_y$. The onset measurement therefore gives
$$
\boxed{\tau_y=\rho gl_c}.
$$
For $l>l_c$, define $\Delta=l-l_c$. The central region $|x|\leq l_c$ is a <plug flow of a yield-stress fluid>, while the layers $l_c<|x|\leq l$ are yielded. The <Herschel–Bulkley fluid> law gives, for $x\geq l_c$,
$$
\left|w'(x)\right|
=\left[\frac{\rho g(x-l_c)}K\right]^{1/n}.
$$
Integrating from the no-slip wall $w(l)=0$ gives
$$
w(x)=\left(\frac{\rho g}{K}\right)^{1/n}
\frac{n}{n+1}
\left[\Delta^{1+1/n}-(x-l_c)^{1+1/n}\right]
$$
in the yielded layer, and the plug moves at
$$
w_p=\left(\frac{\rho g}{K}\right)^{1/n}
\frac{n}{n+1}\Delta^{1+1/n}.
$$
The <volumetric flow rate> per unit span is consequently
$$
\boxed{
Q=2\left(\frac{\rho g}{K}\right)^{1/n}
\left[
\frac{n}{2n+1}\Delta^{2+1/n}
+\frac{nl_c}{n+1}\Delta^{1+1/n}
\right]}.
$$
Near onset, the plug term dominates and
$$
Q\sim(l-l_c)^{1+1/n}.
$$
The 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 as
$$
Q\sim l^{2+1/n}.
$$
The 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:
$$
\tau-\tau_y\sim K_1\dot\gamma
\qquad(\dot\gamma>0\text{ small}).
$$
It then crosses smoothly to the concave high-rate behavior
$$
\tau-\tau_y\sim K_2\dot\gamma^{1/2}
\qquad(\dot\gamma>0\text{ large}),
$$
with the negative-rate branch fixed by odd symmetry. This combines the two <power laws> inferred independently from the measured flux limits.
Back to article page
