Question Bank • QB_APE_009 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 horizontal screw conveyor of outer flight diameter $D = 300\text{ mm}$ and central shaft diameter $d = 60\text{ mm}$ has a screw pitch of $250\text{ mm}$ and rotates at $60\text{ rpm}$. If the volumetric trough loading efficiency is $40\%$, the conveying capacity in $\text{m}^3\text{/h}$ is ________ (round off to 2 decimal places). Take $\pi = 3.1416$.

Answer Key:

24.43

Step-by-Step Solution & Explanation

1. Effective cross-sectional area of flight annular space: $A = \frac{\pi}{4}(D^2 - d^2) = \frac{3.1416}{4}\left[(0.30)^2 - (0.06)^2\right] = 0.7854 \times (0.09 - 0.0036) = 0.067858\text{ m}^2$ 2. Volumetric conveying capacity ($Q$ in $\text{m}^3\text{/h}$): $Q = A \times p \times N \times 60 \times \phi$ Where: • Pitch $p = 0.25\text{ m}$ • Speed $N = 60\text{ rpm}$ • Filling factor $\phi = 0.40$ $Q = 0.067858 \times 0.25 \times 60 \times 60 \times 0.40 = 24.429\text{ m}^3\text{/h} \approx 24.43\text{ m}^3\text{/h}$
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