Discussion
A pneumatic system conveys \( 4 \) tonnes of grain per hour with an air flow of \( 0.55 \) m\(^3\)/s at \( 1.2 \) kg/m\(^3\). The solids to air loading ratio is
Answer: B
Correct answer: 1.68 to 1
Convert both streams to mass per second and divide:
\[ \dot{m}_{air} = 0.55 \times 1.2 = 0.66\ \mathrm{kg/s} \]
\[ \dot{m}_{solids} = \frac{4000}{3600} = 1.11\ \mathrm{kg/s} \]
\[ \text{ratio} = \frac{1.11}{0.66} = 1.68 \]
A loading ratio below about 5 is characteristic of dilute phase conveying, where the particles are fully suspended. Dense phase systems run at ratios of 20 or more, using far less air but requiring much higher pressure.
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