Amit & Deepali Sinha Foundation

Matching Two Reynolds Numbers

Irodov 1.337DifficultyUniqueness

Problem Irodov 1.337

A sphere of radius r1=1.2r_1 = 1.2 mm moving through glycerin (viscosity η1=13.9\eta_1 = 13.9 P) stays in laminar flow as long as its speed does not exceed v1=23v_1 = 23 cm/s. Using this and the viscosity of water, η2=0.011\eta_2 = 0.011 P, find the minimum speed v2v_2 at which a larger sphere of radius r2=5.5r_2 = 5.5 cm moving through water is driven into turbulent flow.

The same transition, viewed for two different spheres: a small sphere moving through glycerin at the onset speed v_1, and a larger sphere moving through water at the onset speed v_2 being sought.

The same transition, viewed for two different spheres: a small sphere moving through glycerin at the onset speed v1v_1, and a larger sphere moving through water at the onset speed v2v_2 being sought.

Given
  • r1=1.2r_1 = 1.2 mm
  • v1=23v_1 = 23 cm/s
  • r2=5.5r_2 = 5.5 cm
  • η1=13.9\eta_1 = 13.9 P
  • η2=0.011\eta_2 = 0.011 P
Find
Find v2v_2, the speed at which the flow around the second sphere becomes turbulent

Solution

Step 1The Reynolds number governs the transition

Whether the flow around a sphere is laminar or turbulent is decided by the Reynolds number built from the sphere’s own size,

Re=ρvrη,Re = \frac{\rho v r}{\eta},

and the transition happens at the same critical value RecRe_c for any sphere in any Newtonian liquid, since ReRe is exactly the dimensionless group that makes the transition condition universal.

Step 2Equate the critical Reynolds number in the two cases

The first sphere is on the edge of turbulence at v1v_1, and the second is sought at the speed v2v_2 where it, too, first becomes turbulent, so both correspond to the same RecRe_c:

ρ1v1r1η1=ρ2v2r2η2.(1) \frac{\rho_1 v_1 r_1}{\eta_1} = \frac{\rho_2 v_2 r_2}{\eta_2}. \tag{1}
Step 3Solve for v2v_2

Solving the matching condition (1) for v2v_2 isolates it in terms of the two spheres’ known properties.

v2=ρ1v1r1η1⋅η2ρ2r2.v_2 = \frac{\rho_1 v_1 r_1}{\eta_1}\cdot\frac{\eta_2}{\rho_2 r_2}.
v2=v1 r1ρ1η2r2ρ2η1v_2 = v_1\,\dfrac{r_1\rho_1\eta_2}{r_2\rho_2\eta_1}
(2)
Step 4Evaluate

Converting the two viscosities to SI units and collecting the known values,

ρ1=1.26×103 kg/m3,ρ2=1.00×103 kg/m3,η1=1.39 Pa⋅s,η2=1.1×10−3 Pa⋅s.\rho_1 = 1.26\times10^3\ \text{kg/m}^3,\quad \rho_2 = 1.00\times10^3\ \text{kg/m}^3,\quad \eta_1 = 1.39\ \text{Pa}\cdot\text{s},\quad \eta_2 = 1.1\times10^{-3}\ \text{Pa}\cdot\text{s}.

Substituting these together with r1r_1, r2r_2 and v1v_1 into (2) gives

v2=(0.23)(1.2×10−3)(1.26×103)(1.1×10−3)(5.5×10−2)(1.00×103)(1.39) m/s.v_2 = (0.23)\frac{(1.2\times10^{-3})(1.26\times10^3)(1.1\times10^{-3})} {(5.5\times10^{-2})(1.00\times10^3)(1.39)}\ \text{m/s}.
v2=v1r1ρ1η2r2ρ2η1≈5 μm/sv_2 = v_1\dfrac{r_1\rho_1\eta_2}{r_2\rho_2\eta_1} \approx 5\ \mu\text{m/s}
(3)
Check

Water is nearly 13001300 times less viscous than glycerin while the second sphere is roughly 4646 times larger, and both changes push v2v_2 sharply below v1v_1: a big sphere in a thin liquid turns turbulent at a much lower speed than a tiny sphere in a thick one, which is exactly the enormous drop the numbers show.

Remark

Because only the ratio RecRe_c is shared between the two spheres, this is really a statement that ρvr/η\rho v r/\eta is the single number nature cares about, not vv, rr, ρ\rho or η\eta separately: two wildly different objects, a millimeter bead in glycerin and a fist-sized ball in water, sit at the same point on the same universal curve once their own Reynolds number is used as the coordinate. That is the whole logic behind testing a scale model of a ship or aircraft in a wind tunnel or tow tank at a different size and speed than the real thing.