Cleaning and Grading Equipment Questions and Answers

Answer: A

Correct answer: 90 per cent

Find the undersize in the feed, then the fraction of it recovered:

\[ 1000 \times 0.40 = 400\ \mathrm{kg/h\ undersize\ in\ feed} \]

\[ \frac{360}{400} \times 100 = 90\ \mathrm{per\ cent} \]

The remaining 10 per cent of the undersize left with the oversize - the usual consequence of a bed too deep or a residence time too short for every fine particle to reach the deck. The distractor \( 36 \) expresses the underflow as a fraction of the total feed rather than of the undersize in it.

Answer: B

Correct answer: 2.5 t/h per m\(^2\)

Specific capacity is throughput divided by screen area:

\[ \frac{6.0}{2.4} = 2.5\ \mathrm{t/h\ per\ m^2} \]

Expressing capacity this way makes machines of different size comparable and lets a required area be calculated from a required throughput. It must be read alongside effectiveness, though: the same machine can be run at a higher specific capacity at the cost of a poorer separation, so the two figures are only meaningful together.

Answer: C

Correct answer: 90 per cent

Find the foreign matter present, then the fraction removed:

\[ 2000 \times 0.03 = 60\ \mathrm{kg/h\ foreign\ matter} \]

\[ \frac{54}{60} \times 100 = 90\ \mathrm{per\ cent} \]

The distractor \( 2.7 \) per cent would express the removal as a fraction of the whole feed, which measures nothing useful. Cleaning efficiency must always be referred to the impurity present, since the same 54 kg removed from a much dirtier feed would represent a far poorer performance.

Answer: D

Correct answer: 5.0 m/s

The air velocity must lie between the two terminal velocities: above \( 2.5 \) m/s so the chaff is lifted, and below \( 9.0 \) m/s so the sound grain is not. A velocity of \( 5.0 \) m/s sits comfortably between them.

The setting is a compromise. Nearer the lower figure, some chaff escapes removal; nearer the upper, sound but light grain is carried away with it. Adjustable air control is therefore standard, and the operator sets it by inspecting the rejected fraction for good grain.

Answer: A

Correct answer: 0.16 m/s

Each complete cycle covers the stroke twice, out and back:

\[ v = \frac{400 \times 2 \times 0.012}{60} = \frac{9.6}{60} = 0.16\ \mathrm{m/s} \]

The distractor \( 0.08 \) counts only one stroke per cycle. The oscillation does two things at once on a gravity table: it fluidises the bed in combination with the upward air, and it conveys the settled heavy fraction uphill along the deck to its outlet.

Answer: B

Correct answer: 34.9 per cent

Take one hole and the square of pitch that belongs to it:

\[ \frac{\pi r^2}{p^2} = \frac{\pi (2)^2}{6^2} = \frac{12.57}{36} = 0.349 = 34.9\ \mathrm{per\ cent} \]

The distractor \( 44.4 \) squares the diameter ratio without the factor for a circle inscribed in a square. Open area governs capacity directly, which is why perforations are set as close as the sheet's strength allows - and why blinding, by reducing the effective open area, hurts capacity so quickly.

Answer: C

Correct answer: 1.41 m/s

Peripheral speed is circumference times rotational speed:

\[ v = \pi D N = \pi \times 0.60 \times \frac{45}{60} = 1.885 \times 0.75 = 1.41\ \mathrm{m/s} \]

Speed matters to the separation. Too slow and the short seeds are not carried high enough to clear the bed and reach the trough; too fast and centrifugal force holds every seed in its indent past the trough, so nothing is separated. Each cylinder therefore has a fairly narrow working range of speed.

Answer: D

Correct answer: 96 kernels per second

Apply the rejection percentage to the inspection rate:

\[ 12\,000 \times 0.008 = 96\ \mathrm{kernels\ per\ second} \]

Each of those must be removed by an air jet firing for a few milliseconds within a window of similar length, which is what sets the demands on the ejector array and its timing. The figure also bears on yield: every good kernel caught alongside a bad one is a loss, so ejector precision is as important as detection sensitivity.

Answer: A

Correct answer: 176 kg/h

Compute the impurity in each stream and take the difference:

\[ \text{in} = 5000 \times 0.04 = 200\ \mathrm{kg/h} \]

\[ \text{out} = 4750 \times 0.005 = 23.75\ \mathrm{kg/h} \]

\[ \text{removed} = 200 - 23.75 = 176\ \mathrm{kg/h} \]

The distractor \( 200 \) assumes complete removal, ignoring the impurity still present in the clean stream. Note also that the reject stream is \( 250 \) kg/h, so about \( 74 \) kg/h of good grain leaves with the impurity - a real loss worth measuring.

Answer: B

Correct answer: 0.827

Overall effectiveness is the product of the two recoveries:

\[ E = 0.94 \times 0.88 = 0.827 \]

Multiplying rather than averaging is the point: a screen must place both fractions correctly, so failing at either penalises the result. The distractor \( 0.910 \) is the arithmetic mean, which flatters a screen that does one job well and the other badly - precisely the case the combined measure exists to expose.