1.2 Drag coefficient and Reynolds number

In Subsection 1.1, we mentioned that the drag force, FD, depends on the velocity of the sphere, v, the diameter of the sphere, D, the density of the fluid, rho, and the viscosity of the fluid, mu.

The powerful technique of dimensional analysis shows that these five variables can be combined into two variables without any loss of ability to describe the drag force. These two variables, are the drag coefficient, CD = FD/(1/2 rho v^2 A), and the Reynolds number, defined by Re = rho v D/mu, where A is the cross-sectional area of the sphere.

Both CD and Re are dimensionless. It can be shown that any model for the drag force can be expressed as a relationship involving only these two dimensionless parameters; that is,

In[2]:=

  Clear[rho,v,d,mu,a,fd]
  eqn := fd/(1/2 rho v^2 a) == f[rho v d/mu]

or, solving for FD,

In[3]:=

  Clear[fdeqn]
  fdeqn = Solve[ eqn, fd ]

Out[3]=

                 2   d rho v
          a rho v  f[-------]
                       mu
  {{fd -> -------------------}}
                   2

The function, f, must be determined experimentally.

Exercise 1

Determine what speed a 5.00 cm diameter sphere would have to have, moving through glycerine , in order to have the same Reynolds number as a 1.00 10^-2 cm diameter sphere moving through water at 20oC at 0.200 cm/sec. For glycerine,rho = 1.26 gm/cm3 and mu = 23.3 gm/cm sec. For water at 20oC,
rho = 0.998 gm/cm3 and mu = 1.00 10^-2 gm/cm sec.
Note: Since these two systems have the same Reynolds number, the flow pattern of one system is a scaled-down replica of the other system. An experimenter can determine the flow pattern for one system by constructing an analogous, more tractable system. High velocities and very large or very small sizes are difficult to handle in the laboratory. This strategy of dimensional similitude is also used in constructing scaled-down models of aircraft to test their aerodynamics in wind tunnels.

Solution

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