Question 5 of 6: Dimensional Analysis of 2D Creeping-Flow Drag

Question 5: Dimensional Analysis of 2D Creeping-Flow Drag (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. Creeping flow past an infinite cylinder; variables for the drag per unit length $F$ are $U$, $\mu$, $D$ (part a–b), and additionally $\rho$ (part c–d). Dimensions in $M,L,T$: $[F]=MT^{-2}$ (force per length), $[U]=LT^{-1}$, $[\mu]=ML^{-1}T^{-1}$, $[D]=L$, $[\rho]=ML^{-3}$.

Find. The dimensionless relation for $F$ with and without $\rho$, and a physical assessment of each.

Approach. Apply Buckingham Pi: count variables and repeating dimensions to get the number of $\Pi$ groups, form them, and interpret the resulting functional relation physically (this is the classic route to Stokes' paradox).

1. Part (a) — without density. Four variables $(F,U,\mu,D)$ in three dimensions $(M,L,T)$ give $4-3=1$ dimensionless group. The unique group is $$\Pi_1=\frac{F}{\mu U}\quad(\text{dimensionless, since }[\mu U]=MT^{-2}=[F]).$$ A single group must equal a constant, so $$\boxed{F=C\,\mu U,\qquad C=\text{const},}$$ i.e. the 2D creeping drag per unit length is independent of the cylinder diameter $D$.
2. Part (b) — physically plausible? No. A drag that does not depend on the body size is not physically reasonable — a thicker cylinder must disturb more fluid and feel more drag. This is the dimensional face of Stokes' paradox: the two-dimensional creeping-flow (Stokes) equations admit no solution that satisfies both the no-slip condition on the cylinder and a uniform stream at infinity, so the naive analysis cannot produce a size-dependent, self-consistent drag.
3. Part (c) — add density back. With five variables $(F,U,\mu,D,\rho)$ and three dimensions, $5-3=2$ groups. Keeping $\Pi_1=F/(\mu U)$ and forming a second group from $\rho$, $$\Pi_2=\frac{\rho U D}{\mu}=Re_D ,$$ so the relation becomes $$\boxed{\frac{F}{\mu U}=f(Re_D)\quad\Longrightarrow\quad F=\mu U\,f\!\left(\frac{\rho U D}{\mu}\right).}$$ The diameter now re-enters through the Reynolds number.
4. Part (d) — is it better? Yes. Even though $\rho$ was formally dropped from the creeping-flow momentum balance, dimensional analysis shows it is impossible to build a size-dependent 2D drag without it. Restoring $\rho$ supplies the second group $Re_D$, so $F$ can depend on $D$ — physically sensible. This mirrors the real result (Lamb–Oseen): the 2D cylinder drag carries a $1/\ln(Re_D)$ factor, i.e. an unavoidable weak dependence on inertia and size. The three-dimensional sphere has no such trouble ($F=3\pi\mu U D$, one group), which is exactly why the 2D case is singular.