Question

A small water droplet in a mist of air is approximated as being a sphere of diameter 1.25 mil. Calculate the terminal velocity as it falls through still air to the ground.

Answer 1

The terminal velocity as it falls through still air is 4.65154 in/s.

The diameter of small water droplet is 1.25 mil= 1.25×0.0254×10^-3 m

                                                                            = 3.175 × 10^-5 m

Now the viscosity of still air is η = 1.83× 10⁻⁵ Pa

 So the formula for drag force is:

                Fd = 6πηrv

  where,  v is the velocity.

Now to attain terminal velocity acceleration must be zero.

            →  W = Fd

                ρVg = 6πrηv

                ρ × 4/3 πr³×g = 6πrηv

                           v = 2/9 × ρgr³/ η

                           v =  2/9 × 10³×9.81×(3.175×10^-3) / 18.6×10^-6

                           v = 0.1181  m/s

                           v = 4.65154 in/s

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