Question 2 of 6: A2 — Axial drag flow in an annulus (moving rod)

Question 2: A2 — Axial drag flow in an annulus (moving rod) (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. Steady, incompressible, axial flow $u_z(r)$ in the annular gap $\kappa R\le r\le R$; inner rod (radius $\kappa R$) translating at $V$ in the $+z$ direction; outer cylinder fixed. The figure labels the fluid at both ends “at pressure $p_o$”, so in the printed problem there is no axial pressure gradient ($dp/dz=0$) and the fluid is moved solely by the rod (Bird–Stewart–Lightfoot Problem 2B.7). To show where that case sits, the derivation carries a general imposed drop $\Delta P=p_0-p_L$ over a length $L$ ($-\,dp/dz=\Delta P/L$) and then sets $\Delta P=0$ (step 7). Constant $\rho,\mu$; no $\theta$- or $z$-dependence.

Find. The velocity distribution $u_z(r)$ and the volumetric flow rate $Q$.

[Figure not reproduced: Fig. A2: Axial annular flow — the fluid in the gap between a stationary outer cylinder (radius $R$) and an inner rod (radius $\kappa R$) moving at $V$ is dragged along by the rod (Couette contribution). With the same pressure $p_o$ at both ends, as printed, there is no Poiseuille contribu. See the official exam paper.]

Approach. Continuity forces $u_z=u_z(r)$ only; the $z$-component of the equation of motion then reduces to a linear ODE whose general solution is a parabola plus a logarithm. Two no-slip conditions fix the constants, and integrating the profile over the annular cross-section gives $Q$.

1. Continuity. For steady incompressible flow with only an axial component, the cylindrical continuity equation gives $\partial u_z/\partial z=0$; with axisymmetry and no swirl, $u_z=u_z(r)$ alone.
2. $z$-momentum reduces to a linear ODE. The Navier–Stokes $z$-component (Table A2, no time-dependence, no convective terms because $u_z$ depends only on $r$) becomes $$0=-\frac{dp}{dz}+\mu\,\frac1r\frac{d}{dr}\!\Big(r\frac{du_z}{dr}\Big),\qquad -\frac{dp}{dz}=\frac{\Delta P}{L}.$$
3. Integrate twice. Writing $\beta\equiv\dfrac{\Delta P}{\mu L}$, $\dfrac{1}{r}\dfrac{d}{dr}(r\,u_z')=-\beta$ integrates to $$u_z(r)=\frac{\beta}{4}\big(R^2-r^2\big)+C_1\ln\!\frac{r}{R},$$ where the constant of the parabola has been written so that the $C_1\ln(r/R)$ term carries the two boundary conditions.
4. Apply the no-slip conditions. Outer wall $u_z(R)=0$ is satisfied automatically by the form above; inner rod $u_z(\kappa R)=V$ gives $\dfrac{\beta}{4}R^2(1-\kappa^2)+C_1\ln\kappa=V$, hence $$\boxed{\,C_1=\frac{V-\dfrac{\beta R^2}{4}(1-\kappa^2)}{\ln\kappa}\,}\qquad(\ln\kappa<0).$$
5. Velocity distribution. Substituting back, $$\boxed{\,u_z(r)=\frac{\Delta P}{4\mu L}\big(R^2-r^2\big)+\frac{V-\dfrac{\Delta P\,R^2}{4\mu L}(1-\kappa^2)}{\ln\kappa}\,\ln\!\frac{r}{R}\,}$$ — a Poiseuille parabola plus a logarithmic Couette (drag) term.
6. Volumetric flow rate. Integrate over the annulus, $Q=\displaystyle\int_{\kappa R}^{R}u_z\,2\pi r\,dr$. Splitting into the pressure and drag contributions gives the closed form $$\boxed{\,Q=\frac{\pi\,\Delta P\,R^4}{8\mu L}\Big[(1-\kappa^4)-\frac{(1-\kappa^2)^2}{\ln(1/\kappa)}\Big]+\pi V R^2\Big[\frac{1-\kappa^2}{2\ln(1/\kappa)}-\kappa^2\Big].}$$ The first bracket is the classic annular-Poiseuille result; the second is the net drag flow carried by the moving rod.
7. The printed case: equal end pressures ($\Delta P=0$). Both ends of the cylinder are at $p_o$, so $\beta=0$ and $C_1=V/\ln\kappa$. The velocity distribution and flow rate reduce to $$\boxed{\,u_z(r)=V\,\frac{\ln(r/R)}{\ln\kappa}\,},\qquad \boxed{\,Q=\frac{\pi R^2V}{2}\Big[\frac{1-\kappa^2}{\ln(1/\kappa)}-2\kappa^2\Big]\,}.$$ Check: $u_z(R)=0$ and $u_z(\kappa R)=V$; for $\kappa=0.5$, $Q=0.291\,\pi R^2V$; and as $\kappa\to1$, $Q\to\pi R^2V(1-\kappa)$, which is $\tfrac12V$ times the thin-gap area $2\pi R^2(1-\kappa)$ — the plane-Couette result.