Compare raindrops

The flow around the raindrop may be laminar or turbulent. For laminar flow the force depends on the viscosity $\eta$ of the air: $F_l=6 \pi r v \eta$ , where $r$ is the radius and $v$ is the velocity of the droplet. For turbulent flow the force is $F_t=\frac{1}{2} c_d \rho v^2 A$ , where $c_d=0.47$ is the drag coefficient, $\rho$ is the density of air and $A=r^2\pi$ . The terminal velocity is obtained by making the drag force equal the the weight, $W=\frac{4}{3}r^3\pi \rho_w g$ , where $\rho_w$ is the density of water and g is the acceleration of gravity. Solving the equations for the two possibilities we get

$v_l=\frac{2}{9} \frac{r^2 \rho_w g}{\eta}$ for the laminar flow and

$v_t=\sqrt{\frac{8}{3} \frac{r\rho_w g}{c_d \rho}}$ for the turbulent flow.

For the droplet of $r=0.1$ mm the two velocities are $v_l=1.2$ m/s and $v_t=2.15$ m/s. For the other droplet the velocities are $v_l=120$ m/s and $v_t=6.81$ m/s.

How do we select the correct combination? As a general rule, if the size of the object is given, at low velocities the flow is laminar and the force is proportional to the velocity. At large velocities the flow is turbulent and the force is proportional to the square of the velocity. The crossover happens when the two forces are approximately equal. It is safe to say that for any given velocity the larger force is the correct choice. Due to the inverse (or inverse quadratic) relationship between the velocity and the drag force, this translates to the smaller terminal velocity. Therefore the terminal velocity for the smaller droplet is $v_l=1.2$ m/s and for the larger droplet $v_t=6.81$ m/s.

The Reynolds number is $Re=\frac{\rho v d}{\eta}=\frac{2r \rho v }{\eta}$ and the crossover from laminar to turbulent flow is happening around $Re=40$ (see note). For the smaller droplet with $v_l=1.2$ m/s we get $Re=16$ , and the flow is, indeed, laminar. For the larger droplet with $v_t=6.81$ m/s we have $Re=900$ , corresponding to the turbulent flow.

Note: For a sphere the turbulent and laminar drag forces are equal at $Re=48$ . This is pretty close the $Re=40$ , where the crossover is observed experimentally.