# Question 15.16: Determine the terminal velocity of 0.2 mm diameter rain drop......

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.

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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.

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