Irodov 1.338: The Largest Sphere That Stays Laminar – Amit & Deepali Sinha Foundation

Problem Irodov 1.338

A lead sphere sinks steadily through glycerin of viscosity $\eta = 13.9$ P. Given that the flow around the sphere turns turbulent at a Reynolds number $Re = 0.5$ (using the sphere’s diameter as the characteristic length), find the largest sphere diameter for which the sinking flow remains laminar.

A lead sphere sinking steadily through the glycerin at its constant velocity v. — A lead sphere sinking steadily through the glycerin at its constant velocity $\vect v$ .

Given

$\eta = 13.9$ P
$Re = 0.5$ (transition value, based on diameter)

Find

Find

d_{max}

, the largest sphere diameter for which the sinking flow stays laminar

Solution

Step 1Terminal velocity from Stokes’ drag

The free body this step balances: the net weight (the sphere's weight less the buoyancy of the glycerin it displaces) pulling down, against Stokes' viscous drag pushing up. The velocity v is drawn alongside the body, not on it. — The free body this step balances: the net weight (the sphere’s weight less the buoyancy of the glycerin it displaces) pulling down, against Stokes’ viscous drag pushing up. The velocity $\vect v$ is drawn alongside the body, not on it.

At terminal velocity the sphere (radius $r$ , density $\rho$ , in glycerin of density $\rho_0$ ) no longer accelerates, so the net weight $(m-m_0)g$ , with $m$ the sphere’s mass and $m_0$ that of the glycerin it displaces, is balanced by Stokes’ viscous drag $6\pi\eta r v$ :

\frac{4}{3}\pi r^3(\rho-\rho_0)g = 6\pi\eta r v.

Solving for $v$ by dividing both sides by $6\pi\eta r$ ,

v = \frac{4\pi r^3(\rho-\rho_0)g}{18\pi\eta r} = \frac{2(\rho-\rho_0)gr^2}{9\eta}. \tag{1}

Step 2Reynolds number in terms of the radius

With diameter $d=2r$ as the characteristic length, substituting the terminal velocity (1) into the definition of $Re$ gives it purely in terms of $r$ :

\begin{align*}Re &= \frac{\rho_0 v d}{\eta} \\ &= \frac{\rho_0}{\eta}\cdot\frac{2(\rho-\rho_0)gr^2}{9\eta}\cdot 2r\end{align*}

Re = \frac{4\rho_0(\rho-\rho_0)g}{9\eta^2}\,r^3

(2)

Step 3Set

Re=Re_c

and solve for the radius, then the diameter

Setting (2) equal to the critical value and solving for $r^3$ isolates the transition radius:

Re_c = \frac{4\rho_0(\rho-\rho_0)g}{9\eta^2}r^3 \quad\Longrightarrow\quad r^3 = \frac{9\,Re_c\,\eta^2}{4\rho_0(\rho-\rho_0)g}. \tag{3}

The diameter is $d=2r$ , so $d^3=8r^3$ ; substituting the radius from (3) and simplifying the numerical factor gives the diameter directly:

d^3 = 8r^3 = 8\left(\frac{9\,Re_c\,\eta^2}{4\rho_0(\rho-\rho_0)g}\right) = \frac{18\,Re_c\,\eta^2}{\rho_0(\rho-\rho_0)g}.

d_{max} = \sqrt[3]{\dfrac{18\,Re\,\eta^2}{(\rho-\rho_0)\rho_0 g}}

(4)

Step 4Evaluate

With $\rho_0=1.26\times10^3$ kg/m $^3$ (glycerin), $\rho=11.3\times10^3$ kg/m $^3$ (lead), $\eta=1.39$ Pa $\cdot$ s and $Re=0.5$ ,

d_{max} = \sqrt[3]{\frac{18(0.5)(1.39)^2} {(11.3\times10^3-1.26\times10^3)(1.26\times10^3)(9.8)}}\ \text{m} \approx 5\ \text{mm}.

Check

A larger $\eta$ raises $d_{max}$ (a thicker fluid keeps the sphere slow and the flow laminar up to a bigger size), while a larger density contrast $\rho-\rho_0$ lowers it (a heavier sphere in the same fluid falls faster and turns turbulent sooner); both trends are visible directly in the formula.

Remark

Because $v\propto r^2$ by (1) while $Re\propto vr\propto r^3$ by (2), the transition is controlled by the sphere’s size, not directly by its speed: the speed is not a free parameter here, it is fixed by the size through the terminal-velocity condition itself. Any lead sphere larger than about $5$ mm across cannot settle into steady Stokes sinking; its terminal velocity would already carry it past $Re_c$ . Stokes’ law is therefore self-limiting: the faster free fall it predicts for a larger sphere is exactly what disqualifies the law from applying to that sphere.