A belt carries a load of cross-sectional area \( 0.045 \) m\(^2\) at \( 1.6 \) m/s with a bulk density of \( 750 \) kg/m\(^3\). Its capacity is
Answer: A
The volumetric rate is \( 0.045 \times 1.6 = 0.072 \) m\(^3\)/s, and multiplying by density gives \( 54 \) kg/s, or \( 194 \) tonnes an hour. Capacity depends on the load area and the speed, not on the length of the run.
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A screw conveyor of \( 0.25 \) m diameter with a full pitch runs at \( 60 \) rev/min at a loading factor of \( 0.35 \) and a bulk density of \( 700 \) kg/m\(^3\). Its capacity is
Answer: B
The swept volume per revolution is \( (\pi/4)(0.25)^2 \times 0.25 = 0.01227 \) m\(^3\), which at 60 rev/min gives \( 44.2 \) m\(^3\) an hour. At 35 per cent loading that is \( 15.5 \) m\(^3\), and multiplying by 700 kg/m\(^3\) gives about 10.8 tonnes an hour.
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A bucket elevator has buckets of \( 2.0 \) litre capacity spaced at \( 0.30 \) m on a belt running at \( 1.5 \) m/s, filled to \( 75 \) per cent with material of density \( 720 \) kg/m\(^3\). Its capacity is
Answer: C
Correct answer: 19.4 t/h
Work from buckets per second to mass per hour:
\[ \frac{1.5}{0.30} = 5\ \mathrm{buckets\ per\ second} \]
\[ 5 \times 0.0020 \times 0.75 = 0.0075\ \mathrm{m^3/s} \]
\[ 0.0075 \times 720 \times 3600 = 19\,440\ \mathrm{kg/h} = 19.4\ \mathrm{t/h} \]
The fill factor of 75 per cent matters: buckets are never filled completely, since the material must enter as they sweep the boot and must stay in as they climb. Ignoring it gives \( 25.9 \) t/h, one of the distractors.
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Lifting \( 20 \) tonnes per hour through \( 18 \) m requires a theoretical power of
Answer: D
Correct answer: 0.98 kW
The theoretical power is the rate of doing work against gravity:
\[ P = \frac{mgh}{t} = \frac{20\,000 \times 9.81 \times 18}{3600} = \frac{3\,531\,600}{3600} = 981\ \mathrm{W} \]
so about \( 0.98 \) kW. This is the irreducible minimum, ignoring friction, bucket digging resistance and drive losses. A real installation might need three or four times as much, which is why an efficiency figure must be applied before a motor is selected.
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A belt conveyor \( 60 \) m long carries \( 150 \) t/h against an effective friction force of \( 1.8 \) kN at \( 1.5 \) m/s. The power required is
Answer: A
Correct answer: 2.7 kW
Power is force times velocity:
\[ P = F v = 1800 \times 1.5 = 2700\ \mathrm{W} = 2.7\ \mathrm{kW} \]
Neither the length nor the tonnage enters directly - both are already accounted for in the effective friction force, which is what the belt drive must overcome. For a horizontal run this friction is the whole of the duty; an inclined belt would add the lifting power \( mgh/t \) on top.
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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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A bin of \( 3.0 \) m diameter is filled to \( 8 \) m with grain of bulk density \( 760 \) kg/m\(^3\). The mass stored is
Answer: C
Correct answer: 43.0 tonnes
Find the volume, then multiply by bulk density:
\[ V = \pi r^2 h = \pi (1.5)^2 (8) = 56.5\ \mathrm{m^3} \]
\[ m = 56.5 \times 760 = 42\,977\ \mathrm{kg} = 43.0\ \mathrm{tonnes} \]
Bulk density is the right figure to use, since it already accounts for the voids between kernels. The commonest error is to use the diameter where the radius belongs, which would quadruple the answer.
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A grain heap forms at an angle of repose of \( 28 \) degrees on a bin of \( 3.0 \) m radius. The height of the surcharge cone is
Answer: D
Correct answer: 1.60 m
The heap forms a cone whose slope is the angle of repose, so its height follows from the radius:
\[ h = r \tan\theta = 3.0 \times \tan 28^\circ = 3.0 \times 0.532 = 1.60\ \mathrm{m} \]
The surcharge cone is worth calculating because it adds real capacity - here about \( 15 \) m\(^3\), or a fifth of what a cylinder of the same radius and height would hold - and because it determines the clearance needed between the fill point and the roof.
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Each transfer in a handling line breaks \( 0.4 \) per cent of the grain. After \( 6 \) transfers the cumulative breakage is about
Answer: A
Summing gives \( 6 \times 0.4 = 2.4 \) per cent, and compounding the survivals gives almost exactly the same figure at these small values. Reducing the number of transfers is the most direct way to protect grain quality.
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A screw conveyor consuming \( 8.5 \) kW moves \( 25 \) t/h. Its specific energy is
Answer: B
Correct answer: 0.34 kWh per tonne
Divide the power drawn by the throughput:
\[ \frac{8.5\ \mathrm{kW}}{25\ \mathrm{t/h}} = 0.34\ \mathrm{kWh\ per\ tonne} \]
Expressing energy this way makes machines of different size and duty comparable. A screw conveyor's figure is characteristically high for the distance covered, because the material is dragged against the trough throughout - which is the practical reason screws are used for short runs and belts for long ones.
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