Subtopic: Fluidization of granular solids – pneumatic, bucket, screw and belt conveying
Question Statement:
A horizontal screw conveyor of screw flight diameter $D = 0.25\text{ m}$, shaft diameter $d = 0.05\text{ m}$, and pitch $p = 0.25\text{ m}$ rotates at $60\text{ rpm}$. The trough loading efficiency (filling factor) is $\phi = 0.35$. The theoretical volumetric conveying capacity of the conveyor in $\text{m}^3\text{/h}$ is ________ (round off to two decimal places).
Answer Key:
14.84
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Net Cross-Sectional Area and Linear Speed**
1. Net cross-sectional area of screw flight:
$A = \frac{\pi}{4} (D^2 - d^2) = \frac{\pi}{4} (0.25^2 - 0.05^2) = \frac{\pi}{4} (0.0625 - 0.0025) = \frac{\pi}{4} \times 0.060 = 0.047124\text{ m}^2$
2. Linear axial velocity of material:
$v = p \times N = 0.25\text{ m/rev} \times 60\text{ rev/min} = 15.0\text{ m/min} = 900\text{ m/h}$
3. Volumetric conveying capacity with filling factor $\phi = 0.35$:
$Q = A \times v \times \phi = 0.047124\text{ m}^2 \times 900\text{ m/h} \times 0.35 = 14.844\text{ m}^3/\text{h} \approx 14.84\text{ m}^3/\text{h}$
**Method 2: Standard Screw Conveyor Capacity Equation**
$Q = 60 \times \frac{\pi}{4} (D^2 - d^2) p N \phi = 60 \times 0.047124 \times 0.25 \times 60 \times 0.35 = 14.84\text{ m}^3/\text{h}$