Question

A metallic spherical particle of density \( 7001 \, \text{kg/m}^3 \) and diameter \( 1 \, \text{mm} \) is settling steadily due to gravity in a stagnant gas of density \( 1 \, \text{kg/m}^3 \) and viscosity \( 10^{-5} \, \text{kg m}^{-1} \text{s}^{-1} \). Take \( g = 9.8 \, \text{m/s}^2 \). Assume that the settling occurs in the regime where the drag coefficient \( C_D \) is independent of the Reynolds number, and equals \( 0.44 \). The terminal settling velocity of the particle, in \( \text{m/s} \), rounded off to 2 decimal places, is:

Hint:

In drag force problems, carefully identify the regime (e.g., constant \( C_D \)) and use the appropriate force balance equations.

Answer 1

Step 1: Force balance for terminal velocity. At terminal velocity, the gravitational force is balanced by the drag force: \[ \frac{\pi}{6} d^3 \rho_p g = \frac{1}{2} C_D A \rho_g v_t^2, \] where: - \( d = 1 \, \text{mm} = 0.001 \, \text{m} \) (particle diameter), - \( \rho_p = 7001 \, \text{kg/m}^3 \) (particle density), - \( \rho_g = 1 \, \text{kg/m}^3 \) (gas density), - \( g = 9.8 \, \text{m/s}^2 \), - \( C_D = 0.44 \) (drag coefficient), - \( A = \frac{\pi}{4} d^2 \) (projected area). Step 2: Solve for terminal velocity \( v_t \). Substitute \( A = \frac{\pi}{4} d^2 \): \[ \frac{\pi}{6} d^3 \rho_p g = \frac{1}{2} C_D \frac{\pi}{4} d^2 \rho_g v_t^2. \] Simplify: \[ \frac{d \rho_p g}{6} = \frac{C_D \rho_g v_t^2}{8}. \] Rearrange for \( v_t^2 \): \[ v_t^2 = \frac{4 d \rho_p g}{3 C_D \rho_g}. \] Step 3: Substitute values. \[ v_t^2 = \frac{4 \cdot 0.001 \cdot 7001 \cdot 9.8}{3 \cdot 0.44 \cdot 1}. \] \[ v_t^2 = \frac{274.4392}{1.32} \approx 207.91. \] Step 4: Calculate \( v_t \). \[ v_t = \sqrt{207.91} \approx 14.50 \, \text{m/s}. \] Step 5: Conclusion. The terminal settling velocity is \( 14.50 \, \text{m/s} \).