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> is
$$
-G=\frac1r\frac{d}{dr}(r\tau_{rz}).
$$
The free surface is shear-free, $\tau_{rz}(b)=0$, and integration gives
$$
\boxed{\tau_{rz}(r)=\frac G2\left(\frac{b^2}{r}-r\right).}
$$
This 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 therefore
$$
F_z=2\pi aL\tau_{rz}(a)
=\pi aLG\left(\frac{b^2}{a}-a\right)
=\boxed{\pi\Delta p\,(b^2-a^2)}.
$$
Its 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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 352 / 1 / b / Solution

{c}

The Cauchy momentum equation balances material acceleration against stress divergence and body force:
$$
\rho\frac{D\mathbf u}{Dt}=\nabla\mathbin\cdot\boldsymbol\sigma+\rho\mathbf b.
$$
For an inertialess flow, the left-hand side is neglected.


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