Question 6 of 7: Question 6 (Part B, Question B2): Fully Developed Annular Flow Driven by a Rod Pulled Through a Vertical Pipe

Question 6 (Part B, Question B2): Fully Developed Annular Flow Driven by a Rod Pulled Through a Vertical Pipe (29 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 with inner radius \(a=d/2\) (the rod) and outer radius \(b=D/2\) (the pipe). The fluid is incompressible and Newtonian, density \(\rho\), dynamic viscosity \(\mu\). The rod translates at \(U_p\) in the \(+z\) (upward) direction; the pipe is stationary. The flow is steady, laminar, axisymmetric, fully developed (\(\partial/\partial z = 0\) for velocity) and has no swirl. The axial pressure gradient is zero, \(\partial P/\partial z = 0\), and gravity acts downward, \(g_z=-g\). All walls are non-porous and wetted.

Find. (a) the boundary conditions, (b) \(u_r(r)\), (c) \(u_z(r)\), and (d) the axial force per unit length required to pull the rod and the axial force per unit length transmitted to the pipe.

Approach. Reduce the cylindrical-polar continuity equation to obtain \(u_r\), then reduce the \(z\)-momentum equation to a linear ODE in \(r\) whose two integration constants are fixed by the no-slip conditions; differentiate the resulting profile to get the wall shear stresses and multiply by the wetted perimeters.

1. Part (a) — state the boundary conditions. The walls are solid, non-porous and wetted, so both no-penetration and no-slip apply at each surface, and axisymmetry with no imposed swirl kills the azimuthal component throughout: \[\text{at }r=a=\tfrac{d}{2}:\quad u_r=0,\quad u_\theta=0,\quad u_z=U_p,\] \[\text{at }r=b=\tfrac{D}{2}:\quad u_r=0,\quad u_\theta=0,\quad u_z=0.\] In addition, "very long" and "fully developed" mean \(\partial u_z/\partial z=0\), and steadiness means \(\partial/\partial t=0\). These six wall conditions plus the two symmetry statements are all the information the problem supplies, and they are exactly enough.
2. Part (b) — reduce continuity to find the radial velocity. For an incompressible, steady, axisymmetric flow with no swirl the aid-sheet continuity equation collapses to \[\frac{1}{r}\frac{\partial}{\partial r}\left(r\,u_r\right)+\frac{\partial u_z}{\partial z}=0 .\] Fully developed flow makes the second term vanish, so \(r\,u_r=\text{constant}\). Applying the no-penetration condition at either wall makes that constant zero, and therefore \[\boxed{u_r(r)=0\quad\text{everywhere in the annulus}}\] The result is worth stating plainly: because the geometry never changes along \(z\), there is nowhere for fluid to go radially, and the flow is purely axial and rectilinear.
3. Part (c) — reduce the axial momentum equation. With \(u_r=u_\theta=0\), \(\partial u_z/\partial z=0\) and \(\partial u_z/\partial t=0\), every convective and unsteady term in the \(z\)-momentum equation vanishes and the whole equation reduces to a balance between the body force and the viscous term: \[0=-\frac{\partial P}{\partial z}+\rho g_z+\frac{\mu}{r}\frac{d}{dr}\left(r\frac{du_z}{dr}\right).\] With \(\partial P/\partial z=0\) as stated and \(g_z=-g\), \[\frac{1}{r}\frac{d}{dr}\left(r\frac{du_z}{dr}\right)=\frac{\rho g}{\mu}.\] The flow is thus a superposition of a gravity-driven falling film and a rod-dragged Couette flow.
4. Integrate twice. Multiplying by \(r\) and integrating, \[r\frac{du_z}{dr}=\frac{\rho g r^{2}}{2\mu}+C_1 \qquad\Longrightarrow\qquad \frac{du_z}{dr}=\frac{\rho g\,r}{2\mu}+\frac{C_1}{r},\] and once more, \[u_z(r)=\frac{\rho g\,r^{2}}{4\mu}+C_1\ln r+C_2 .\] The logarithm is the signature of cylindrical geometry; a plane-channel version of this problem would give a straight line plus a parabola.
5. Apply the no-slip conditions. Imposing \(u_z(b)=0\) and \(u_z(a)=U_p\) and eliminating \(C_2\), \[C_1=\frac{U_p+\dfrac{\rho g}{4\mu}\left(b^{2}-a^{2}\right)}{\ln (a/b)},\qquad C_2=-\frac{\rho g b^{2}}{4\mu}-C_1\ln b,\] so that the profile can be written compactly about the outer wall: \[\begin{aligned}&\boxed{u_z(r)=\frac{\rho g}{4\mu}\left(r^{2}-b^{2}\right)+\left[U_p-\frac{\rho g}{4\mu}\left(a^{2}-b^{2}\right)\right]\frac{\ln (r/b)}{\ln (a/b)}}\\&\text{with }a=\tfrac{d}{2}\text{ and }b=\tfrac{D}{2}.\end{aligned}\] Substituting \(r=a\) returns \(U_p\) and \(r=b\) returns zero, as required. If gravity is neglected the profile reduces to the classical annular Couette result \(u_z=U_p\ln(r/b)/\ln(a/b)\), a pure logarithm.
6. Read the profile. The gravity term is a downward-pulling parabola pinned to zero at the pipe wall, and the Couette term is a logarithm running from \(U_p\) at the rod to zero at the pipe. Their sum can be entirely upward (small \(\rho g b^2/\mu U_p\)), or can contain a downward return flow in the outer part of the annulus when gravity dominates — the physically important case for a rod being withdrawn slowly from a heavy oil. The plot below shows both regimes for \(d/D=0.4\) using the dimensionless group \(G=\rho g b^{2}/(\mu U_p)\).

7. Part (d) — wall shear stress from the profile. For this rectilinear flow the only non-zero viscous stress on a surface of constant \(r\) is \(\tau_{rz}=\mu\,du_z/dr\). From step 4, \[\tau_{rz}(r)=\mu\left(\frac{\rho g\,r}{2\mu}+\frac{C_1}{r}\right)=\frac{\rho g\,r}{2}+\frac{\mu C_1}{r}.\] Evaluating at the two walls gives everything part (d) needs.
8. Force per unit length on the rod, and the force needed to pull it. The fluid acts on the rod over a perimeter \(2\pi a\); the outward normal of the rod points in \(+r\), so the axial traction on the rod is \(+\tau_{rz}(a)\) and \[F'_{\text{fluid}\to\text{rod}}=2\pi a\,\tau_{rz}(a)=\pi\rho g\,a^{2}+2\pi\mu C_1 .\] The applied pull must balance this (the rod moves at constant speed), so, taking upward as positive and excluding the rod's own weight, \[\begin{aligned}F'_{\text{pull}}&=-\pi\rho g\,a^{2}-2\pi\mu C_1\\&=\boxed{\frac{2\pi\mu\left[U_p+\dfrac{\rho g}{4\mu}\left(b^{2}-a^{2}\right)\right]}{\ln (b/a)}-\pi\rho g\,a^{2}}\end{aligned}\] where \(\ln(b/a)>0\). With gravity neglected this collapses to the familiar \(F'_{\text{pull}}=2\pi\mu U_p/\ln(b/a)\), which is positive as it must be: the fluid resists the upward motion.
9. Force per unit length on the pipe. At \(r=b\) the pipe material lies outside the fluid, so its outward normal points in \(-r\) and the axial traction on the pipe is \(-\tau_{rz}(b)\): \[\boxed{F'_{\text{pipe}}=-2\pi b\,\tau_{rz}(b)=-\pi\rho g\,b^{2}-2\pi\mu C_1}\] Substituting \(C_1\) gives, explicitly, \[F'_{\text{pipe}}=\frac{2\pi\mu\left[U_p+\dfrac{\rho g}{4\mu}\left(b^{2}-a^{2}\right)\right]}{\ln (b/a)}-\pi\rho g\,b^{2}.\]
10. Check the global force balance. Adding the two forces the fluid exerts on its boundaries, \[F'_{\text{fluid}\to\text{rod}}+F'_{\text{fluid}\to\text{pipe}}=\pi\rho g\,a^{2}-\pi\rho g\,b^{2}=-\rho g\,\pi\left(b^{2}-a^{2}\right),\] which is exactly minus the weight of fluid per unit length. By Newton's third law the walls push back on the fluid with \(+\rho g\pi(b^2-a^2)\), so the fluid column is in equilibrium — as it must be with no axial pressure gradient and no acceleration. This one line catches almost every sign or algebra slip in parts (c) and (d).