A1 — Combined pressure–drag flow in an annulus: 23-Chem-B1 Transport Phenomena, December 2014, Q1

Question 1: A1 — Combined pressure–drag flow in an annulus

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 Newtonian fluid (constant $\rho$, $\mu$) fills the annular gap between an inner rod of radius $R_1$ moving axially at velocity $V$ and a stationary outer tube of inner radius $R_2$. Flow is horizontal, fully developed, steady and laminar, driven by both the moving rod (drag) and a constant axial pressure gradient $dP/dz$.

Find. (a) the axial velocity profile $u(r)$ for the moving rod; (b) the profile when $V=0$ (rod stationary).

Fig. A1: Longitudinal section of the extruder annulus. The stationary outer tube ($R_2$) and the axially moving rod ($R_1$, velocity $V$) bound a Newtonian film that is simultaneously dragged by the rod and pushed by the pressure gradient $dP/dz$.

Approach. With unidirectional fully developed flow $u=u_z(r)$ only, the cylindrical $z$-momentum equation collapses to an ordinary differential equation; integrate twice and fix the two constants with no-slip on the moving rod and the stationary tube.

Reduce the equation of motion. For steady, fully developed axial flow $u_z=u(r)$, $u_r=u_\theta=0$, the Navier–Stokes $z$-component (Appendix A, Table A.2f) loses every inertial and $\theta,z$ term, leaving a balance of pressure and viscous stress: $$\mu\,\frac{1}{r}\frac{d}{dr}\!\left(r\frac{du}{dr}\right)=\frac{dP}{dz}.$$
Integrate twice. Writing $\beta=\dfrac{1}{\mu}\dfrac{dP}{dz}$ (a constant), $\dfrac{d}{dr}\!\left(r\dfrac{du}{dr}\right)=\beta r$, so $r\dfrac{du}{dr}=\dfrac{\beta r^2}{2}+C_1$ and hence $$u(r)=\frac{\beta r^2}{4}+C_1\ln r+C_2 .$$
Recast about the outer wall. Absorbing $C_2$ so that the constant vanishes at $r=R_2$: $$u(r)=\frac{\beta}{4}\left(r^2-R_2^2\right)+C_1\ln\!\frac{r}{R_2},$$ which already satisfies $u(R_2)=0$.
Apply no-slip on the rod. Setting $u(R_1)=V$ gives $\dfrac{\beta}{4}(R_1^2-R_2^2)+C_1\ln\dfrac{R_1}{R_2}=V$, so $$C_1=\frac{V+\dfrac{\beta}{4}\left(R_2^2-R_1^2\right)}{\ln\!\dfrac{R_1}{R_2}} .$$
Assemble the profile (part a). With $\beta=(1/\mu)\,dP/dz$, $$\boxed{\,u(r)=\frac{1}{4\mu}\frac{dP}{dz}\left(r^2-R_2^2\right)+C_1\ln\!\frac{r}{R_2}\,},\qquad C_1=\frac{V+\dfrac{1}{4\mu}\dfrac{dP}{dz}\left(R_2^2-R_1^2\right)}{\ln\!\dfrac{R_1}{R_2}}.$$ The first term is the pressure-driven (Poiseuille) contribution; the logarithmic term carries the drag of the moving rod plus the annular curvature.
Stationary rod (part b). Put $V=0$: the drag term disappears and only pressure flow remains, $$u(r)=\frac{1}{4\mu}\frac{dP}{dz}\left(r^2-R_2^2\right)+C_1'\ln\!\frac{r}{R_2},\qquad C_1'=\frac{\dfrac{1}{4\mu}\dfrac{dP}{dz}\left(R_2^2-R_1^2\right)}{\ln\!\dfrac{R_1}{R_2}}.$$ The maximum occurs where $du/dr=0$, i.e. at $$\boxed{\,r_{max}=\sqrt{\dfrac{R_2^2-R_1^2}{2\,\ln(R_2/R_1)}}\,},$$ which lies nearer the inner wall than the arithmetic mean radius.

Case	Result
(a) Rod moving at $V$	$u(r)=\dfrac{1}{4\mu}\dfrac{dP}{dz}(r^2-R_2^2)+C_1\ln(r/R_2)$, $C_1=\dfrac{V+\tfrac{1}{4\mu}\tfrac{dP}{dz}(R_2^2-R_1^2)}{\ln(R_1/R_2)}$
(b) Rod stationary	Same form with $V=0$; peak at $r_{max}=\sqrt{(R_2^2-R_1^2)/[2\ln(R_2/R_1)]}$

Concept & Theory

Topic. Superposition of Couette (drag) and Poiseuille (pressure) flow in a concentric annulus.

Key relations. Reduced axial momentum $\mu\frac1r(r u')'=dP/dz$; general solution $u=\beta r^2/4+C_1\ln r+C_2$; no-slip $u(R_1)=V$, $u(R_2)=0$.

Why this works. Because the flow is fully developed and unidirectional, convective acceleration is identically zero and the momentum equation is linear; the pressure and drag responses therefore add. The logarithm is the annulus’s signature — it replaces the parabola’s symmetry axis with a shifted maximum, because equal radial increments enclose unequal areas.

Common pitfall. Forgetting the $\ln r$ term (valid only for a solid pipe, where finiteness at $r=0$ kills it) — in an annulus $r=0$ is excluded, so both constants survive. Also, the peak of pressure-driven annular flow is not at the mid-gap radius $(R_1+R_2)/2$.

Source. Bird, Stewart & Lightfoot, §2.4 (annular flow); Welty et al., Ch. 8 (laminar internal flow).

Question 1 of 6: A1 — Combined pressure–drag flow in an annulus

Question 1: A1 — Combined pressure–drag flow in an annulus