Subtopic: Fluidization of granular solids – pneumatic, bucket, screw and belt conveying
Question Statement:
A screw conveyor has an outer screw diameter $D = 0.30\text{ m}$, shaft diameter $d = 0.06\text{ m}$, screw pitch $p = 0.30\text{ m}$, and rotational speed $N = 60\text{ rpm}$. The loading efficiency of the trough is $\phi = 0.40$ and grain bulk density $\rho_b = 750\text{ kg/m}^3$. Neglecting inclination and slip, the theoretical conveying capacity in tonnes per hour ($\text{t/h}$) is ________ (round off to 2 decimal places).
Answer Key:
22.56
Step-by-Step Solution & Explanation
1. Cross-sectional area available between screw flights:
$A = \frac{\pi}{4} (D^2 - d^2) = \frac{\pi}{4} (0.30^2 - 0.06^2) = \frac{\pi}{4} (0.09 - 0.0036) = \frac{\pi}{4} (0.0864) \approx 0.067858\text{ m}^2$
2. Linear conveying velocity $v$:
$v = p \times \frac{N}{60} = 0.30 \times \frac{60}{60} = 0.30\text{ m/s}$
3. Volumetric conveying capacity $\dot{V}$:
$\dot{V} = A \times v \times \phi = 0.067858 \times 0.30 \times 0.40 \approx 0.008143\text{ m}^3\text{/s}$
4. Mass throughput $\dot{m}$ in $\text{t/h}$:
$\dot{m} = \dot{V} \times \rho_b \times 3600 / 1000 = 0.008143 \times 750 \times 3.6 = 6.10725 \times 3.6 \approx 21.986\text{ t/h}$
Wait, if $A = \frac{\pi}{4} D^2 \phi = \frac{\pi}{4}(0.09)(0.40) = 0.02827$, with shaft subtraction $0.067858 \times 0.30 \times 0.40 \times 750 \times 3.6 = 21.99\text{ t/h}$. Let's set correct_answer: "21.99", range min: 21.50, max: 22.50.