8.7 Exercises [Stokes flow top]

1. Stokes flow is a low-Reynolds number limit, and the resulting solution (Equations 8.12 and 8.13) is independent of the Reynolds number. Why, then, in Equation 8.26, is the drag coefficient dependent on the Reynolds number? How could the drag coefficient be defined differently such that the drag coefficient is not a function of the Reynolds number?

2. Given a particle of radius a in an infinite domain and a freestream velocity in the x-direction with magnitude U, solve for the velocity around the particle. Assume that the solution is axisymmetric and write the solution in terms of axisymmetric velocity components u_r and u_ϑ, as is done in Equations 8.12 and 8.13.

   1. Write the continuity equation in spherical coordinates, delete terms involving the azimuthal velocity u_φ or derivatives with respect to φ, and show that the continuity equation can be written as

      (8.52)

   2. Write the radial momentum equation in spherical coordinates (Equation D.16), eliminate the right-hand side owing to the Stokes approximation, delete terms involving the azimuthal velocity u_φ or derivatives with respect to φ, and show that the radial momentum equation for Stokes flow can be written as

      (8.53)

   3. Write the colatitudinal momentum equation in spherical coordinates (Equation D.16), eliminate the right-hand side owing to the Stokes approximation, delete terms involving the azimuthal velocity u_φ or derivatives with respect to φ, and show that the colatitudinal momentum equation for Stokes flow can be written as

      (8.54)

   4. Assume that the solutions for the radial and colatitudinal velocities have the following forms:

      (8.55)

      (8.56)

      These six constants can be fixed by examining (a) the boundary conditions at r = a, which gives two constraints (one for each component of velocity at the wall); (b) the continuity equation, which gives three constraints (one for each term in the assumed form for velocity); and (c) the Navier-Stokes equation, which gives one new constraint (two total, but one is redundant with one from the continuity). With these constants fixed, the solution is complete.

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