Question 4 of 6: Vertical annulus with a shear-free outer wall

Question 4: Vertical annulus with a shear-free outer 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. A vertical annulus of outer radius R and inner rod radius δ = R/4, length L with L/R ≫ 10; the outer pipe wall Teflon-coated so it carries negligible shear, the inner rod uncoated so the no-slip condition applies there; both walls non-porous; inlet and outlet both open to atmosphere so the static pressures are equal; steady, laminar, axisymmetric flow of water of density ρ and dynamic viscosity μ. From Figure 4, z is measured downward, in the direction of gravity and of the flow.

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

[Figure not reproduced: Figure 4 (redrawn) with the resulting velocity profile. The no-slip rod pins the velocity to zero at r = δ; the shear-free coated wall forces d v z /d r = 0 at r = R , so the maximum migrates all the way out to the coated surface. See the official exam paper.]

Approach. Continuity plus the non-porous walls kills the radial component outright. The sentence about both ends being open to atmosphere is load-bearing rather than decorative — it makes dP/dz = 0, so gravity is the only driver, and the z-momentum equation reduces to a balance between the viscous stress gradient and the weight of the fluid.

1. Part (a) — the radial velocity. For steady axisymmetric flow with no swirl, continuity reads $$\frac{1}{r}\frac{\partial\left(r v_{r}\right)}{\partial r}+\frac{\partial v_{z}}{\partial z}=0 .$$ With L/R ≫ 10 the flow is fully developed, so ∂v_z/∂z = 0 and therefore rv_r = constant. The walls are non-porous, so v_r(δ) = 0, which forces that constant to be zero: $$\boxed{\;v_{r}(r)=0\ \ \text{everywhere in }\ \delta\le r\le R\;}$$ The result is worth stating explicitly rather than assumed, because it is what licenses every simplification that follows.
2. Establish the driving term. Both ends of the pipe are exposed to the atmosphere at the same static pressure, so along the axis $$\frac{dP}{dz}=\frac{P_{\text{out}}-P_{\text{in}}}{L}=0 .$$ The flow is therefore driven purely by gravity. With z pointing downward as in Figure 4, the body force is +ρg in the z direction.
3. Part (b) — reduce the z-momentum equation. Everything convective vanishes (v_r = 0 and ∂v_z/∂z = 0), leaving $$\frac{\mu}{r}\frac{d}{dr}\left(r\frac{dv_{z}}{dr}\right)+\rho g=0 .$$ Integrating once, $$r\frac{dv_{z}}{dr}=-\frac{\rho g r^{2}}{2\mu}+C_{1},$$ and again, $$v_{z}(r)=-\frac{\rho g r^{2}}{4\mu}+C_{1}\ln r+C_{2}.$$
4. Apply the two boundary conditions. The Teflon-coated outer wall carries negligible shear, so dv_z/dr = 0 at r = R: $$-\frac{\rho g R}{2\mu}+\frac{C_{1}}{R}=0\quad\Longrightarrow\quad C_{1}=\frac{\rho g R^{2}}{2\mu}.$$ No slip on the uncoated rod, v_z(δ) = 0, then fixes C₂. Collecting terms, $$\boxed{\;v_{z}(r)=\frac{\rho g}{4\mu}\left(\delta^{2}-r^{2}\right) +\frac{\rho g R^{2}}{2\mu}\ln\!\frac{r}{\delta}\;}$$ which is a parabola plus a logarithm — the logarithm being the fingerprint of the annular geometry.
5. Locate and size the maximum. Since the shear vanishes only at r = R, the velocity maximum sits at the coated wall rather than somewhere inside the gap. Putting δ = R/4, $$v_{z,\max}=v_{z}(R)=\frac{\rho g R^{2}}{\mu}\left[\frac{1}{4}\left(\frac{1}{16}-1\right) +\frac{1}{2}\ln 4\right]=0.4588\,\frac{\rho g R^{2}}{\mu}.$$ The profile therefore climbs monotonically from zero at the rod to its peak at the pipe wall, the opposite of the familiar pipe profile and a direct consequence of which surface was coated.
6. Part (c) — wall shear on the rod. Differentiating and evaluating at r = δ, $$\tau_{\delta}=\mu\left.\frac{dv_{z}}{dr}\right|_{r=\delta} =-\frac{\rho g\delta}{2}+\frac{\rho g R^{2}}{2\delta}=\frac{\rho g}{2\delta}\left(R^{2}-\delta^{2}\right).$$ The force per unit length follows by multiplying by the rod perimeter 2πδ: $$F'=2\pi\delta\,\tau_{\delta}=\pi\rho g\left(R^{2}-\delta^{2}\right),$$ and with δ = R/4, $$\boxed{\;F'=\frac{15}{16}\pi\rho g R^{2}\ \ \text{per unit length, directed DOWNWARD (with the flow)}\;}$$
7. The free global check. A shear-free wall transmits no axial force, so the rod must be carrying the entire weight of the water column. The weight per unit length of the annulus is ρgπ(R² − δ²), which is exactly the expression obtained above. That agreement is not a coincidence to be admired but a check to be run: any algebraic slip in C₁ or C₂ breaks it immediately.
8. Put numbers on it, and check the laminar premise. Water at 20 °C (ρ = 998 kg/m³, μ = 1.002 × 10⁻³ Pa·s) in an annulus with R = 0.5 mm gives $$v_{z,\max}=1.12\ \text{m/s},\qquad \bar V=0.909\ \text{m/s},\qquad F'=7.21\times 10^{-3}\ \text{N/m},$$ with a hydraulic-diameter Reynolds number ρV̄D_h/μ = 679 — comfortably laminar, as the question requires.