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Particulate Control Devices Questions and Answers

31.

Dust of resistivity \( 5\times10^{11} \) ohm cm is fed to a precipitator whose workable range is \( 10^7 \) to \( 10^{10} \). The dust is

Answer: C

Correct answer: Too resistive, so back corona is likely

At \( 5 \times 10^{11} \) ohm cm the dust is far above the workable window of \( 10^7 \) to \( 10^{10} \). Charge cannot leak away through the collected layer, so it accumulates until the trapped field breaks the layer down and a reverse discharge - back corona - begins, injecting positive ions that neutralise the charge on incoming particles. Efficiency collapses. Conditioning the gas with sulphur trioxide is the standard remedy.

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32.

A settling chamber's minimum collectable particle size varies with the square root of the required settling velocity. Halving the gas velocity therefore reduces that size by a factor of

Answer: D

Halving the gas velocity doubles the residence time and halves the settling velocity needed, and the size falls as the square root of that, giving \( \sqrt{2} = 1.41 \). Doubling the chamber length would achieve exactly the same thing.

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33.

A baghouse of \( 900 \) m\(^2\) at an air to cloth ratio of \( 0.033 \) m/s must handle a flow increase to \( 42 \) m\(^3\)/s. The fabric area now required at the same ratio is

Answer: A

Correct answer: 1273 m\( ^2 \)

Holding the ratio constant, the area required rises with the flow:

\[ A = \frac{Q}{v} = \frac{42}{0.033} \approx 1273 \ \mathrm{m}^2 \]

The alternative - keeping the existing 900 square metres and accepting a ratio of 0.047 - would raise the pressure drop, push dust into the weave and shorten bag life, which is why fabric area is added rather than the ratio being allowed to drift up.

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34.

Overall penetration through three devices of \( 80 \), \( 90 \) and \( 99 \) per cent efficiency is

Answer: B

The penetrations multiply, giving \( 0.20 \times 0.10 \times 0.01 = 0.0002 \). Arranging a coarse pre-cleaner ahead of a fine collector in this way protects the expensive device and is standard practice on a dirty gas stream.

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35.

Statement I: A gravity settling chamber is unsuitable for fine particulate.

Statement II: Stokes settling velocity varies with the square of the particle diameter.

In the light of the above statements, choose the correct answer.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both are true and the second explains the first. Stokes settling velocity is

\[ v = \frac{(\rho_p - \rho)gd^2}{18\mu} \]

so halving the diameter quarters the velocity. A 100 micrometre particle settles in seconds; a 10 micrometre particle takes a hundred times as long, and a chamber long enough to catch it would be absurdly large. Settling chambers therefore serve only as pre-cleaners for coarse material.

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36.

Statement I: A fabric filter's collection is done chiefly by the fabric itself.

Statement II: The dust cake that accumulates on the fabric is a far finer filter than the weave.

In the light of the above statements, choose the correct answer.

Answer: D

Correct answer: Statement I is incorrect but Statement II is correct

Statement II is correct and it is why statement I is wrong. The fabric’s weave has openings far larger than the particles it must catch; what does the collecting is the dust cake built up on its surface, whose interstices are of particle scale. The fabric is the support on which that cake forms. This is why efficiency is lowest immediately after cleaning and why a filter is never cleaned back to bare fabric.

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37.

Statement I: Efficiency in an electrostatic precipitator rises exponentially with plate area.

Statement II: The Deutsch-Anderson equation expresses efficiency as one minus an exponential in area over flow.

In the light of the above statements, choose the correct answer.

Answer: A

Both statements hold and the second is the relationship itself. Its practical consequence is that each further increment of efficiency costs disproportionately more steel, which is why precipitators are sectionalised into fields.

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38.

Statement I: Very low dust resistivity is as troublesome as very high resistivity.

Statement II: Dust that loses its charge immediately is re-entrained into the gas stream.

In the light of the above statements, choose the correct answer.

Answer: B

Both are true and the second describes the low resistivity failure. There is a workable window of roughly ten to the seventh to ten to the tenth ohm centimetres, and dust outside it in either direction collects poorly.

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39.

Statement I: A venturi scrubber can collect fine particulate at low pressure drop.

Statement II: The contacting power principle relates wet scrubber efficiency to the power expended in contact.

In the light of the above statements, choose the correct answer.

Answer: C

Correct answer: Statement I is incorrect but Statement II is correct

Statement II is correct and it contradicts statement I. The contacting power principle holds that the efficiency of a wet scrubber depends chiefly on the power expended in bringing gas and liquid into contact, almost regardless of how that power is applied. Fine particle collection therefore requires a high pressure drop - a venturi runs at 5 to 20 kPa - and cannot be had cheaply.

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40.

Statement I: A cyclone with a smaller body achieves a smaller cut diameter.

Statement II: A smaller radius gives a greater centrifugal acceleration at the same tangential velocity.

In the light of the above statements, choose the correct answer.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are true and the second explains the first. Centrifugal acceleration is \( v^2/r \), so reducing the radius at the same tangential velocity increases the force driving particles to the wall, and smaller particles can be captured. Hence the multiclone: many small cyclones in parallel achieve a far smaller cut diameter than one large one. The penalty is a higher pressure drop.

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