3.1 The drag force for Re < 0.5

In[20]:=

  fdlowre = fdeqn[[1,1,2]] /. f[d rho v/mu] -> 24/re

Out[20]=

            2
  12 a rho v
  -----------
      re

Substitute the definition for the Reynolds number, Re = r v D/m, and express the cross-sectional area, a, in terms of the diameter, D, using the formula for the area of a circle,

In[21]:=

  fdlow = fdlowre /. {re -> rho v d/mu, a -> 1/4 Pi d^2}

Out[21]=

  3 d mu Pi v

Then the drag force, FD, equals k1 v, where

In[22]:=

  k1 = 3 d mu Pi

Out[22]=

  3 d mu Pi

Up to 3. Two Models for the Drag Force