From dimensional analysis, we know that the drag coeffiffifficient C_Dof a

smooth sphere moving in a flfluid is a function of Reynolds number:

C_D = f(Re) = f [VD/]

where and are the density and viscosity of the flfluid (e.g., air), D is the diameter of the

sphere, and V is its velocity. The drag force can then be expressed as

F = 1/2C_DAV²

with A as the projected area. Note that you can derive the Stokes drag force formula (valid

when Re < 1) from the above equation by inserting C_D = 24/Re.

Now consider two raindrops, one with diameter 2 mm and another one with diameter

4 mm. Which one falls faster in the air and by how much?

Hints:

(i) Assume that the drops fall at their terminal velocities, which is the velocity at which

the drag force and gravitational forces cancel each other: F = 1/6D³(_S )g, where

1/6D³ is the volume of the droplet, _S is its density ( 1000 kg/m3 ), and g 9.8 m/s2 is

the gravitational acceleration. Also, for the air, use = 1 kg/m3 and = 2105 Pa s.

(ii) Initially assume that the drops fall in the Stokes regime (i.e., Re < 1) and use the

Stokes drag formula. After fifinding the terminal velocity calculate the Re number to

verify your initial assumption.

(iii) In case the Stokes regime is not valid: for high Reynolds number 10³ < Re .< 10⁵ , one

can assume that C_D is constant and not a function of Re. Use CD 0.4 and redo the

above process. Verify if the Reynolds number is within the rang