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Drag on sphere at Re = 0.2

Short answer

For a sphere at Re=0.2 the drag follows Stokes' law: F_D = 3πμVd. The drag coefficient C_D = 3πμVd / (0.5ρV^2 * πd²/4) simplifies to 24/Re, and at Re=0.2 yields C_D = 120.

What is the expression for the drag on a sphere, when Re of the flow is 0.2? Prove that the coefficient of drag for sphere for this range of the Reynolds number is given by CD=24/ReC_D = 24/Re, where Re is the Reynolds number.

At very low Reynolds number (Re<1Re < 1, creeping flow) inertia is negligible compared with viscous forces, and the drag on a sphere is given by Stokes' law:

FD=3πμVdF_D = 3\pi\mu V d

where μ\mu is the dynamic viscosity, VV the velocity of the sphere relative to the fluid, and dd the diameter. This holds for Re=ρVd/μ≲0.2Re = \rho V d/\mu \lesssim 0.2 to 11.

Proof that CD=24/ReC_D = 24/Re

By definition the drag coefficient is the drag divided by the dynamic pressure times the projected area:

CD=FD12ρV2A,A=πd24C_D = \frac{F_D}{\tfrac12\rho V^2 A},\qquad A = \frac{\pi d^2}{4}

Substituting Stokes' drag:

CD=3πμVd12ρV2⋅πd24=3πμVd×8ρV2πd2=24μρVd=24Re\begin{aligned} C_D &= \frac{3\pi\mu V d}{\tfrac12\rho V^2\cdot\dfrac{\pi d^2}{4}} = \frac{3\pi\mu V d\times 8}{\rho V^2\pi d^2} \\ &= \frac{24\mu}{\rho V d} = \frac{24}{Re} \end{aligned}

Hence for Re=0.2Re = 0.2:

CD=240.2=120C_D = \frac{24}{0.2} = 120

Result: FD=3πμVdF_D = 3\pi\mu V d and CD=24/ReC_D = 24/Re for the Stokes range.

✓ Answer checked by Claude Opus 5.5 (high effort)

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