Question 3 of 5: Gravity-driven laminar flow in a vertical annulus with one shear-free wall

Question 3: Gravity-driven laminar flow in a vertical annulus with one shear-free wall (25 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.

Find. u_r(r), u_z(r), and the magnitude and direction of the force per unit length that the water exerts on the inner rod.

Approach. Continuity plus the non-porous walls kills the radial component outright; the z-momentum equation from the cylindrical-polar aid sheet then collapses to an ordinary differential equation in r driven by gravity alone, whose two constants are set by the zero-shear condition at the Teflon wall and the no-slip condition at the rod. The rod force is the wall shear times its circumference, and a global force balance on the annulus checks it.

1. Part (a) — eliminate the radial velocity from continuity. For a steady, axisymmetric, incompressible flow the aid-sheet continuity equation reduces to $$\frac{1}{r}\frac{\partial}{\partial r}\left(r\,u_r\right)+\frac{\partial u_z}{\partial z}=0$$ Fully developed flow in a long pipe means $ \partial u_z/\partial z = 0 $, so $ \partial(r u_r)/\partial r = 0 $ and therefore $ r\,u_r = C $, a constant across the gap. Applying the non-porous condition at either wall — $ u_r(b)=0 $ — forces C = 0, and since $ r \geq b > 0 $ everywhere in the annulus, $$\boxed{\;u_r(r)=0\quad\text{throughout the gap}\;}$$ The streamlines are exactly parallel to the axis; nothing crosses between radii.
2. Part (b) — reduce the axial momentum equation. With $ u_r = u_\theta = 0 $, $ \partial/\partial\theta = 0 $, $ \partial u_z/\partial z = 0 $ and steady flow, every convective term in the aid-sheet z-momentum equation vanishes and it leaves a balance between pressure, gravity and viscous diffusion: $$0=-\frac{dP}{dz}+\rho g+\frac{\mu}{r}\frac{d}{dr}\left(r\frac{du_z}{dr}\right)$$ The inlet and outlet are both open to the same atmosphere, so the static pressure does not change along the pipe: $ dP/dz = 0 $. Gravity is therefore unbalanced and is the only thing driving the flow — the pipe is running as a gravity drain, not as a pressure line: $$\frac{1}{r}\frac{d}{dr}\left(r\frac{du_z}{dr}\right)=-\frac{\rho g}{\mu}$$
3. Integrate twice. Multiplying by r and integrating, $ r\,du_z/dr = -\rho g r^{2}/(2\mu)+C_1 $, and integrating again $$u_z(r)=-\frac{\rho g\,r^{2}}{4\mu}+C_1\ln r+C_2$$ The logarithm is the signature of the annular geometry; a plain circular pipe would reject it because the velocity must stay finite on the axis, but here the axis is occupied by the rod.
4. Apply the Teflon condition at r = R. A non-wetting wall carries no shear, so $ du_z/dr = 0 $ there: $$-\frac{\rho g R}{2\mu}+\frac{C_1}{R}=0\quad\Longrightarrow\quad C_1=\frac{\rho g R^{2}}{2\mu}$$ This is the mathematical statement that the outer wall behaves like a free surface: the profile arrives at it flat.
5. Apply no slip at the rod, r = b = R/4. Setting $ u_z(b)=0 $, $ C_2=\rho g b^{2}/(4\mu)-C_1\ln b $, and substituting both constants gives the profile $$\boxed{\;u_z(r)=\frac{\rho g}{4\mu}\left(b^{2}-r^{2}\right) +\frac{\rho g R^{2}}{2\mu}\,\ln\!\frac{r}{b},\qquad b=\frac{R}{4}\;}$$ which is a Poiseuille parabola bent upward by a logarithm. It is positive everywhere in $ b < r \leq R $ (the parabolic term is negative there, but the logarithm dominates), and it rises monotonically from zero at the rod to its maximum at the shear-free outer wall: $$u_{z,\max}=u_z(R)=\frac{\rho g R^{2}}{\mu}\left[\frac{1}{4}\left(\frac{1}{16}-1\right) +\frac{1}{2}\ln 4\right]=0.4588\,\frac{\rho g R^{2}}{\mu}$$
6. Part (c) — evaluate the shear stress the water applies to the rod. Differentiating the profile, $$\begin{aligned} \frac{du_z}{dr}&=-\frac{\rho g r}{2\mu}+\frac{\rho g R^{2}}{2\mu\,r}\ \frac{du_z}{dr}\bigg|_{r=b}&=-\frac{\rho g R}{8\mu}+\frac{2\rho g R}{\mu} =\frac{15}{8}\frac{\rho g R}{\mu} \end{aligned}$$ The traction the fluid exerts on the rod surface is $ \tau_{rz}(b)=\mu\,du_z/dr|_b=\tfrac{15}{8}\rho g R $, acting in the +z direction, i.e. with the flow.
7. Multiply by the wetted perimeter to get force per unit length. The rod presents a circumference $ 2\pi b=2\pi R/4 $, so $$F' = \tau_{rz}(b)\,(2\pi b)=\frac{15}{8}\rho g R\cdot\frac{\pi R}{2} =\frac{15}{16}\pi\rho g R^{2}$$ $$\boxed{\;F'=\frac{15}{16}\pi\rho g R^{2}\approx 2.945\,\rho g R^{2}\ \text{N per metre, directed downward (with the flow)}\;}$$
8. Audit the answer with a global force balance. Take the whole annular column of fluid, per unit length, as a free body. Its weight is $ \rho g\,\pi(R^{2}-b^{2})=\rho g\pi R^{2}(1-\tfrac{1}{16})=\tfrac{15}{16}\pi\rho g R^{2} $; the pressure forces on the two ends cancel because the pressures are equal; the Teflon wall transmits no shear at all. The rod is therefore the only thing holding the water up, and it must carry the entire weight — which is exactly the number obtained in step 7. Any slip in the algebra would break this identity immediately, so it is worth writing down.
9. Put representative numbers on the result. Water at laboratory scale would be turbulent here, so take a light lubricating oil, $ \rho = 900 $ kg/m³, $ \mu = 0.50 $ Pa·s, in a pipe of $ R = 10 $ mm with a 2.5 mm rod. Then $ u_{z,\max}=0.4588(900)(9.81)(0.010)^{2}/0.50=0.810 $ m/s, the mean velocity is 0.657 m/s, the hydraulic diameter is $ 2(R-b)=15 $ mm and $ Re = \rho \bar{u} D_h/\mu = 17.7 $ — comfortably laminar, as assumed. The rod then carries $ F' = \tfrac{15}{16}\pi(900)(9.81)(0.010)^{2}=2.60 $ N per metre, identical to the weight of the 0.265 kg of oil each metre of annulus contains.