Consider a horizontal rod of radius $aR$ ($a < 1$) in a stationary pipe of radius $R$. The rod is pulled coaxially at a constant velocity $V$ as shown in the figure. The annular region is filled with a Newtonian incompressible fluid of viscosity $\mu$. The steady–state fully–developed axial velocity profile in the fluid is \[ u(r) = V\frac{\ln(r/R)}{\ln(a)}, \] where $r$ is the radial coordinate. Ignoring end effects, the magnitude of the pulling force per unit rod length is
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In annular flows, drag on the inner cylinder is obtained by evaluating viscous shear at its surface and multiplying by the circumference. Always check the sign of $\ln(a)$ when $a<1$.
To find the pulling force per unit length on the rod, we compute the viscous shear stress acting on its surface. Since the rod is moving through a fluid-filled annulus, the resistance to motion arises only from viscous shear. The given velocity profile is fully developed and depends only on $r$.
Step 1: Shear stress expression
For a Newtonian fluid, the shear stress is:
\[
\tau_{rz} = \mu \frac{du}{dr}.
\]
We will evaluate this at the rod surface $r = aR$.
Step 3: Evaluate shear at rod surface
At $r = aR$:
\[
\tau_{rz}(aR) = \mu \frac{V}{\ln(a)}\frac{1}{aR}.
\]
Step 4: Force per unit length on rod
The viscous drag per unit length is:
\[
F = \tau \times (\text{surface area per unit length})
= \tau_{rz}(aR) \cdot (2\pi aR).
\]
Substituting:
\[
F = \left(\mu \frac{V}{\ln(a)} \cdot \frac{1}{aR}\right)(2\pi aR)
= \frac{2\pi \mu V}{\ln(a)}.
\]
Step 5: Sign convention
Because $\ln(a)$ is negative for $0 < a < 1$, the expression becomes:
\[
F = \frac{-2\pi \mu V}{|\ln(a)|}.
\]
Thus the force needed to pull the rod is:
\[
F = \frac{-2\pi \mu V}{\ln(a)}.
\]
This exactly matches option (B).
