3.2 The drag force for 10^3 < Re

For 10^3 < Re < 10^5, let CD have the constant value 0.5. Then the drag coefficient is

In[23]:=

  fdmid = fdeqn[[1,1,2]] /. f[d rho v/m] -> 0.5

Out[23]=

         2   d rho v
  a rho v  f[-------]
               mu
  -------------------
           2

Then FD equals k2 v^2, where

In[24]:=

  k2 = 0.25 a rho

Out[24]=

  0.25 a rho

or, expressing the area of a circle, a, in terms of the diameter, d

In[25]:=

  k2 = k2 /. a -> 1/4 Pi d^2

Out[25]=

          2
  0.0625 d  Pi rho