The <volumetric flow rate> is $Q=2\pi\int_0^R u(r)r\,dr$. <Integration by parts>, using finite $u(0)$ and $u(R)=0$, gives
$$
Q=-\pi\int_0^Rr^2\frac{du}{dr}\,dr
=\boxed{\pi\int_0^R\dot\gamma(r)r^2\,dr}.
$$
Because the stress magnitude is $\tau=\tau_Rr/R$, change variables from $r$ to $\tau$:
$$
\boxed{Q=\frac{\pi R^3}{\tau_R^3}
\int_0^{\tau_R}\dot\gamma(\tau)\tau^2\,d\tau}.
$$
Therefore the required power is $n=3$. The <fundamental theorem of calculus> gives the pipe form of the <Weissenberg–Rabinowitsch equation>:
$$
\boxed{\dot\gamma_R
=\frac1{\pi R^3\tau_R^2}
\frac d{d\tau_R}(Q\tau_R^3)
=\frac{3Q+\tau_R,dQ/d\tau_R}{\pi R^3}}.
$$
Finally $\eta(\dot\gamma_R)=\tau_R/\dot\gamma_R$, so
$$
\boxed{\eta(\dot\gamma_R)
=\frac{\pi R^3\tau_R}
{3Q+\tau_R,dQ/d\tau_R}}.
$$
A measured pressure-drop--flow-rate curve therefore recovers the wall viscosity without assuming a constitutive form.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 352 / 1 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-352/1/e/solution