Question

How quickly does a 1-micrometer water droplet in the air (20 Celsius, 1atm) relax to terminal velocity? ( Please show steps)

Answer 1

Given

Diameter of the water droplet D = 1 x 10^-6 m

Pressure P = 1 atm = 101325 Pa

Temperature T = 20^oC = 293.15 K

Known

Density of air d_a at 20^oC d_air= 1.2041 kg/m³

Density of water at 20^oC d_water= 1000 kg/m³

The drag coefficient of a a sphere C = 0.5

Solution

Radius of the drop r = D/2 = 1 x10^-6 /2 = 0.5 x 10^-6

Cross sectional area A = πr² = 3.14 x (0.5x10^-6)² = 0.785 x 10^-12

Volume of the drop V = 4πr³/3 = 4 x 3.14 x (0.5x10^-6)³/3 = 0.523 x 10^-18

Mass of the drop m = d_water x V = 1000 x 0.523 x 10^-18 = 523 x 10^-18

Terminal velocity

v_t = {2mg/CAd_air}^1/2

v_t = {2 x 523 x 10^-18 x 9.8 / 0.5 x 0.785 x 10^-12 x 1.2041}^1/2

V_t = 0.147 m/s

Using equations of motion since initial velocity is zero for free falling object

V_t = u + gt

0.147 = 0 + 9.8t

t = 0.015 s

The this droplet will reach terminal velocity in 0.015 s