Question 4 of 6: Gravity-Driven Laminar Flow in an Annulus with One Free-Slip Wall

Question 4: Gravity-Driven Laminar Flow in an Annulus with One Free-Slip Wall (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Fully developed ($L/R\gg10$), laminar, axisymmetric water flow in the annulus $\delta\le r\le R$, $\delta=R/4$. Outer wall $r=R$: Teflon, zero shear (free-slip). Inner rod $r=\delta$: wetted, no-slip. Equal end pressures $\Rightarrow \partial p/\partial z=0$; motion driven by gravity alone.

Find. $v_r(r)$, $v_z(r)$, and the axial force per unit length on the rod.

Approach. Continuity plus the non-porous walls force $v_r=0$; the axial Navier–Stokes equation with $\partial p/\partial z=0$ and body force $-\rho g$ integrates to $v_z(r)$ under the two wall conditions; the rod force follows from the wall shear (and equals the water weight by a global balance).

1. Radial velocity, part (a). Fully developed means $\partial v_z/\partial z=0$, so continuity $\tfrac1r\partial(r v_r)/\partial r+\partial v_z/\partial z=0$ gives $\partial(r v_r)/\partial r=0$, i.e. $r v_r=$const. The non-porous walls require $v_r=0$ at $r=\delta$ and $r=R$, so the constant is zero: $$\boxed{v_r=0.}$$
2. Axial momentum, part (b). With $v_r=0$, $\partial p/\partial z=0$ and gravity $-\rho g$ (taking $z$ upward), the $z$-momentum equation reduces to $$\mu\,\frac1r\frac{d}{dr}\!\left(r\frac{dv_z}{dr}\right)=\rho g .$$ Integrating twice, $v_z=\dfrac{\rho g}{\mu}\dfrac{r^2}{4}+C_1\ln r+C_2$. The two conditions are zero shear at $r=R$, $\left.\dfrac{dv_z}{dr}\right|_{R}=0\Rightarrow C_1=-\dfrac{\rho g}{\mu}\dfrac{R^2}{2}$, and no-slip at $r=\delta$, $v_z(\delta)=0$. Hence $$\boxed{\,v_z(r)=\frac{\rho g}{\mu}\left[\frac{r^2-\delta^2}{4}-\frac{R^2}{2}\ln\frac{r}{\delta}\right]\, }$$ which is negative (downward) throughout $\delta\lt r\lt R$: water falls under gravity, fastest at the free-slip outer wall and zero on the rod.
3. Wall shear on the rod, part (c). The shear stress the fluid exerts on the rod surface follows from $$\left.\frac{dv_z}{dr}\right|_{\delta}=\frac{\rho g}{2\mu}\Big(\delta-\frac{R^2}{\delta}\Big)=-\frac{15\rho g R}{8\mu}\quad(\delta=R/4),$$ so the wall shear magnitude is $\tau_w=\mu\,|dv_z/dr|_\delta = \tfrac{15}{8}\rho g R$.
4. Force per unit length on the rod. Multiplying by the rod circumference $2\pi\delta=\pi R/2$, $$f_{\text{rod}}=\tau_w\,(2\pi\delta)=\frac{15\rho g R}{8}\cdot\frac{\pi R}{2}=\boxed{\frac{15}{16}\pi\rho g R^2\ \text{(downward)}.}$$ This equals the weight per unit length of the annular water column, $\rho g\,\pi(R^2-\delta^2)=\tfrac{15}{16}\pi\rho g R^2$ — the whole weight is carried by the rod because the Teflon wall transmits no shear.