Question 15.14: A sphere of diameter 30 cm and density 7500 kg/m³ is dropped......

Given data:

Diameter of ball                                      D= 30 cm = 0.3 m

Density of sphere                                    \rho_s=7500 \mathrm{~kg} / \mathrm{m}^3

Density of oil                                            \rho_o=800 \mathrm{~kg} / \mathrm{m}^3

Coefficient of drag                                C_D=0.4

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=7500 \times 9.81 \times \frac{\pi}{6}(0.3)^3=1040.14 \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=800 \times 9.81 \times \frac{\pi}{6}(0.3)^3=110.95 \mathrm{~N}

Let U_{\infty} be the terminal velocity of the sphere.
The total drag force is found to be

F_D=C_D A \frac{1}{2} \rho_o U_{\infty}^2=C_D \frac{\pi}{4} D^2 \frac{1}{2} \rho_o U_{\infty}^2

=0.4 \times \frac{\pi}{4}(0.3)^2 \times \frac{1}{2} \times 800 \times U_{\infty}^2=11.31 U_{\infty}^2

Using Eq. (15.20), we have

W=F_D+F_B

or                        1040.14=11.31 U_{\infty}^2+110.95

or                      U_{\infty}^2=\frac{1040.14-110.95}{11.31}=82.156

or                      U_{\infty}=\sqrt{82.156}=9.06 \mathrm{~m} / \mathrm{s}

Reynolds number is found to be

R e=\frac{\rho U D}{\mu}=\frac{900 \times 0.023 \times 0.003}{1.6}=0.03

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