Question Bank • QB_APE_EXP_036 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 flat belt conveyor carries grain at a speed of $1.5\text{ m/s}$. The cross-sectional area of the grain stream on the belt is $0.025\text{ m}^2$. If the bulk density of the grain is $720\text{ kg/m}^3$, the mass conveying rate in tonnes per hour ($\text{t/h}$) is ________ (round off to 1 decimal place).

Answer Key:

97.2

Step-by-Step Solution & Explanation

1. Volumetric conveying rate $\dot{V} = A \times v$: $\dot{V} = 0.025\text{ m}^2 \times 1.5\text{ m/s} = 0.0375\text{ m}^3\text{/s}$ 2. Mass conveying rate $\dot{m}$: $\dot{m} = 0.0375\text{ m}^3\text{/s} \times 720\text{ kg/m}^3 = 27.0\text{ kg/s}$ 3. In tonnes per hour: $\dot{m}_{t/h} = \frac{27.0 \times 3600}{1000} = 27.0 \times 3.6 = 97.2\text{ t/h}$
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