= Solution
Let
$$
G=\frac{\Delta P}{L}>0.
$$
Axial momentum balance gives shear-stress magnitude $|\tau_{rz}|=Gr/2$. The wall stress must exceed the pressure-dependent yield stress, so motion requires
$$
\boxed{\Delta P>\Delta P_{\min}
=\frac{2L\widehat\mu p_s}{R}}.
$$
When this holds, define the plug radius
$$
a=\frac{2\widehat\mu p_s}{G}<R.
$$
The no-slip solid velocity is
$$
\boxed{
w_s(r)=
\begin{cases}
\dfrac{G}{4\eta}(R-a)^2,&0\leq r\leq a,\\[4pt]
\dfrac{G}{4\eta}(R^2-r^2)
-\dfrac{\widehat\mu p_s}{\eta}(R-r),&a<r\leq R.
\end{cases}}
$$
In the sheared annulus,
$$
|w_s'|=\frac{G(r-a)}{2\eta},
\qquad
I_v=\frac{G(r-a)}{2p_s},
$$
so
$$
\boxed{
\phi(r)=
\begin{cases}
\phi_m,&0\leq r\leq a,\\[2pt]
\dfrac{\phi_m}{1+\sqrt{G(r-a)/(2p_s)}},&a<r\leq R.
\end{cases}}
$$
Back to article page
