Question Bank • QB_APE_MAT_005 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:

In pneumatic conveying, a spherical grain of diameter $d_p = 5.0\text{ mm}$ and particle density $\rho_p = 1300\text{ kg/m}^3$ is suspended in an upward air stream (air density $\rho = 1.20\text{ kg/m}^3$). Assuming turbulent flow with a constant drag coefficient $C_D = 0.44$, the terminal suspension velocity $v_t = \sqrt{\frac{4 g d_p (\rho_p - \rho)}{3 C_D \rho}}$ in $\text{m/s}$ is ________ (round off to two decimal places).

Answer Key:

12.68

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Terminal Velocity Formula** $v_t = \sqrt{\frac{4 \times 9.81\text{ m/s}^2 \times 0.005\text{ m} \times (1300 - 1.20)\text{ kg/m}^3}{3 \times 0.44 \times 1.20\text{ kg/m}^3}}$ $\text{Numerator} = 4 \times 9.81 \times 0.005 \times 1298.8 = 0.1962 \times 1298.8 = 254.82456$ $\text{Denominator} = 3 \times 0.44 \times 1.20 = 1.584$ $v_t = \sqrt{\frac{254.82456}{1.584}} = \sqrt{160.874} = 12.6836\text{ m/s} \approx 12.68\text{ m/s}$ **Method 2: Force Balance Derivation** At terminal velocity, drag force equals net gravitational force: $\frac{1}{2} C_D \rho v_t^2 \left(\frac{\pi}{4} d_p^2\right) = \frac{\pi}{6} d_p^3 (\rho_p - \rho) g \implies v_t = \sqrt{\frac{4 g d_p (\rho_p - \rho)}{3 C_D \rho}}$
← Previous (QB_APE_MAT_003) Next (QB_APE_EXP_091) →

Practice In Authentic CBT Exam Interface

Test your skills under authentic timed conditions with live scoring and AIR percentile ranking.

Launch CBT Mock Test