Question 15.13: A spherical ball of diameter 6 mm and density 8000 kg/m³ is ......

Given data:
Diameter of ball                                            D = 6 mm = 0.006 m

Density of ball                                              \rho_s=8000 \mathrm{~kg} / \mathrm{m}^3

Density of oil                                                \rho_o=900 \mathrm{~kg} / \mathrm{m}^3

Terminal velocity                                        U = 50 mm/s = 0.05 m/s

Weight of the ball is W = Density of sphere x g x Volume of sphere

=\rho_s \times g \times \frac{\pi}{6} D^3=8000 \times 9.81 \times \frac{\pi}{6}(0.006)^3=0.00887 \mathrm{~N}

Buoyant force is given by

F_B=\text { Density of oil } \times g \times \text { Volume of sphere }

=\rho_o \times g \times \frac{\pi}{6} D^3=900 \times 9.81 \times \frac{\pi}{6}(0.006)^3=0.000998 \mathrm{~N}

Let μ be the viscosity of the ball.
The drag force is found from Stokes’ formula (Eq. (15.15)) as

F_{D}=3\pi\mu D U_{\infty}    (15.15)

F_D=3 \pi \mu D U

=3\pi\times\mu\times0.006\times0.05=0.002827\;\mu\mathrm{~N}

Using Eq. (15.20), we have

W=F_D+F_B

or                                      0.00887 = 0.002827μ + 0.000998

or                                \mu=\frac{0.00887-0.000998}{0.002827}=\frac{0.007872}{0.002827}=2.78 \mathrm{~N}-\mathrm{s} / \mathrm{m}^2

Reynolds number is found to be

R e=\frac{\rho U D}{\mu}=\frac{900 \times 0.05 \times 0.006}{2.78}=0.09<1  

Since Re < 1, Stokes’ formula for drag force is valid.