Question 15.15: A spherical metallic ball of diameter 3 mm and density 8500 ......

Given data:

Diameter of ball                    D = 3 mm = 0.003 m

Density of sphere                  \rho_s=8500 \mathrm{~kg} / \mathrm{m}^3

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

Viscosity of liquid                \mu=1.6 \mathrm{~N}-\mathrm{s} / \mathrm{m}^2

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=8500 \times 9.81 \times \frac{\pi}{6}(0.003)^3=0.001179 \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}(0.003)^3=0.000125 \mathrm{~N}

(a) The total drag force is obtained using Eq. (15.20) as

W=F_D+F_B

or                          F_D=W-F_B=0.001179-0.000125=0.001054 \mathrm{~N}

The skin friction drag is \frac{2}{3} F_D=\frac{2}{3} \times 0.001054=0.000703 \mathrm{~N}

The pressure drag is \frac{1}{3} F_D=\frac{1}{3} \times 0.001054=0.000351 \mathrm{~N}

(b) Let U_{\infty} be the terminal velocity 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

or                                  0.001054 = 3π x 1.6 x 0.003 x U                            \left[\because F_D=0.001054 \mathrm{~N}\right]

or                          U=\frac{0.001054}{3 \pi \times 1.6 \times 0.003}=0.023 \mathrm{~m} / \mathrm{s}