Question Bank • QB_APE_EXP_039 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Size Reduction and Material Handling

Subtopic: Fluidization of granular solids – pneumatic, bucket, screw and belt conveying

Question Statement:

A spherical agricultural grain of diameter $4.0\text{ mm}$ and density $1200\text{ kg/m}^3$ falls in air (air density $\rho_a = 1.20\text{ kg/m}^3$, $g = 9.81\text{ m/s}^2$) in the Newton's turbulent drag regime where drag coefficient $C_d = 0.44$. The terminal velocity $V_t = \sqrt{\frac{4 g d (\rho_p - \rho_a)}{3 C_d \rho_a}}$ in $\text{m/s}$ is ________ (round off to 2 decimal places).

Answer Key:

10.89

Step-by-Step Solution & Explanation

Formula for terminal settling velocity in Newton's regime: $V_t = \sqrt{\frac{4 g d (\rho_p - \rho_a)}{3 C_d \rho_a}}$ Given: • $d = 0.004\text{ m}$ • $\rho_p = 1200\text{ kg/m}^3$ • $\rho_a = 1.20\text{ kg/m}^3$ • $\rho_p - \rho_a \approx 1200 - 1.2 = 1198.8\text{ kg/m}^3$ • $C_d = 0.44$ • $g = 9.81\text{ m/s}^2$ 1. Numerator: $4 \times 9.81 \times 0.004 \times 1198.8 = 0.15696 \times 1198.8 = 188.1636$ 2. Denominator: $3 \times 0.44 \times 1.20 = 1.584$ 3. Fraction: $\frac{188.1636}{1.584} \approx 118.79$ $V_t = \sqrt{118.79} \approx 10.899\text{ m/s} \approx 10.90\text{ m/s}$ (Range min 10.60, max 11.20).
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