Determine the terminal velocity of 0.2 mm diameter rain drops in a standard atmosphere where the density and dynamic viscosity are 1.2 \mathrm{~kg} / \mathrm{m}^3 \text { and } 1.1 \times 10^{-5} \mathrm{~N}-\mathrm{s} / \mathrm{m}^2 , respectively.
Given data:
Diameter of rain drops D=0.2 \mathrm{~mm}=0.2 \times 10^{-3} \mathrm{~m}
Density of drop ρ = 1.2 kg/m³
Dynamic viscosity \mu=1.1 \times 10^{-5} \mathrm{~N}-\mathrm{s} / \mathrm{m}^2
Weight of rain drop is W = Density x g x Volume of drop
=\rho \times g \times \frac{\pi}{6} D^3
=1.2 \times 9.81 \times \frac{\pi}{6}\left(0.2 \times 10^{-3}\right)^3=4.93 \times 10^{-11} \mathrm{~N}
Let the terminal velocity of rain drop be U.
The drag force on the rain drop is found from Stokes’ formula (Eq. (15.15)) as
F_{D}=3\pi\mu D U_{\infty} (15.15)
F_D=3 \pi \mu D U
=3 \pi \times 1.1 \times 10^{-5} \times 0.2 \times 10^{-3} \times U=2.073 \times 10^{-8} U \mathrm{~N}
At terminal velocity, the drag force must be same as the weight of the drop. Thus,
F_D=W
or 2.073 \times 10^{-8} U=4.93 \times 10^{-11}
or U=\frac{4.93 \times 10^{-11}}{2.073 \times 10^{-8}}=0.00238 \mathrm{~m} / \mathrm{s}
The Reynolds number is
R e=\frac{\rho U D}{\mu}=\frac{1.22 \times 0.00238 \times 0.2 \times 10^{-3}}{1.1 \times 10^{-5}}=0.05
Since Re < 1, Stokes’ formula for drag force is valid.