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Milling of Cereals Questions and Answers

21.

A mill processes \( 1000 \) kg of wheat and produces \( 760 \) kg of flour. The extraction rate is

Answer: A

Correct answer: 76.0 per cent

Extraction rate is flour as a percentage of the wheat milled:

\[ \frac{760}{1000} \times 100 = 76.0\ \mathrm{per\ cent} \]

The distractor \( 24.0 \) is the offal - the bran and germ that did not become flour. A 76 per cent extraction is a typical straight-run white flour: high enough to be commercially efficient, low enough to leave most of the bran and germ behind and give a light, low-ash product.

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22.

Paddy of \( 1000 \) kg yields \( 780 \) kg of brown rice. The husk content was

Answer: B

Correct answer: 22.0 per cent

The husk is what the paddy lost in hulling:

\[ \frac{1000 - 780}{1000} \times 100 = 22.0\ \mathrm{per\ cent} \]

Twenty to twenty-two per cent is the normal range for paddy, and it is a large fraction to transport and handle - which is one reason milling is sited near the growing area rather than near the consumer. The husk itself is siliceous and abrasive but useful as boiler fuel, in board manufacture and as a silica source.

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23.

From \( 780 \) kg of brown rice, whitening gives \( 700 \) kg of white rice. The degree of milling, as bran removed, is

Answer: C

Correct answer: 10.3 per cent

The degree of milling is the bran removed as a percentage of the brown rice entering the whitener:

\[ \frac{780 - 700}{780} \times 100 = 10.3\ \mathrm{per\ cent} \]

The distractor \( 8.0 \) takes the loss as a percentage of the original paddy instead. Degree of milling is a quality decision as much as a technical one: whitening further improves appearance and keeping quality but removes more of the nutritious bran layers and produces more brokens.

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24.

Milled rice of \( 700 \) kg contains \( 574 \) kg of whole kernels. The head rice yield, expressed on the paddy, is

Answer: D

Correct answer: 57.4 per cent

Head rice yield is conventionally expressed on the paddy, so the whole kernels are divided by the paddy that entered the mill:

\[ \frac{574}{1000} \times 100 = 57.4\ \mathrm{per\ cent} \]

The distractor \( 82.0 \) expresses the whole kernels as a percentage of the milled rice rather than of the paddy, which is a different and more flattering figure. Quoting the basis matters commercially, since the two can differ by twenty percentage points on the same batch.

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25.

Wheat at \( 11 \) per cent moisture is conditioned to \( 16 \) per cent. Water to be added to \( 1000 \) kg is

Answer: A

Correct answer: 59.5 kg

Water must be added to raise the moisture while the dry matter stays constant, so the added water enters both the numerator and the total:

\[ W = M \frac{b-a}{100-b} = 1000 \times \frac{16-11}{100-16} = \frac{5000}{84} = 59.5\ \mathrm{kg} \]

The distractor \( 50.0 \) comes from simply taking 5 per cent of the wheat, which ignores that the added water increases the batch mass as well. The difference of nearly 10 kg would leave the wheat short of its conditioning target.

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26.

Break rolls run at \( 500 \) and \( 125 \) rev/min. The differential ratio is

Answer: B

Correct answer: 4.0 to 1

The differential is the ratio of the two roll speeds:

\[ \frac{500}{125} = 4.0 \]

The distractor \( 625 \) multiplies rather than divides. A high differential of this order is characteristic of break rolls, where the difference in surface speed produces the shearing action that opens the kernel and scrapes endosperm from the bran. Reduction rolls run at a much lower differential, around 1.25 to 1, because their action is crushing rather than shearing.

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27.

A roll of \( 250 \) mm diameter turns at \( 400 \) rev/min. Its peripheral speed is

Answer: C

Correct answer: 5.24 m/s

Peripheral speed is the circumference times the rotational speed:

\[ v = \pi D N = \pi \times 0.250 \times \frac{400}{60} = 0.785 \times 6.67 = 5.24\ \mathrm{m/s} \]

The units are the trap: the speed must be converted from revolutions per minute to revolutions per second. Peripheral speed matters because it, rather than the rotational speed, determines the shearing action at the nip and the heat generated there, which is why rolls of different diameter are run at different revolutions.

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28.

Applying Rittinger's law with a constant of \( 12 \) kWh mm per tonne, grinding from \( 4.0 \) mm to \( 0.5 \) mm requires

Answer: D

Correct answer: 21.0 kWh per tonne

Rittinger's law gives the energy as proportional to the new surface created:

\[ E = K_R \left( \frac{1}{x_2} - \frac{1}{x_1} \right) = 12 \left( \frac{1}{0.5} - \frac{1}{4.0} \right) \]

\[ E = 12(2.0 - 0.25) = 12 \times 1.75 = 21.0\ \mathrm{kWh/tonne} \]

Note how the fine end dominates: the \( 1/x_2 \) term contributes \( 24 \) and the coarse term subtracts only \( 3 \). That asymmetry is why grinding finer costs so disproportionately much.

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29.

Applying Kick's law with a constant of \( 8 \) kWh per tonne, reducing from \( 20 \) mm to \( 5 \) mm requires

Answer: A

Correct answer: 11.1 kWh per tonne

Kick's law makes the energy depend on the ratio of sizes rather than on their absolute values:

\[ E = K_K \ln \frac{x_1}{x_2} = 8 \ln \frac{20}{5} = 8 \ln 4 = 8 \times 1.386 = 11.1\ \mathrm{kWh/tonne} \]

Because only the ratio enters, Kick's law predicts the same energy for 20 mm to 5 mm as for 4 mm to 1 mm - which is why it describes coarse crushing well and fine grinding badly, where Rittinger's surface-based law fits better.

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30.

A sieve nest retains \( 12 \), \( 34 \), \( 40 \) and \( 14 \) g on four decks. The mass fraction on the third deck is

Answer: B

The total is \( 100 \) g, so the third deck holds forty per cent of the sample. Cumulative totals from such a nest give the particle size distribution that the miller actually works from day to day in setting the mill.

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