Particulate Control Devices Questions and Answers

Q: Pressure drop matters in selecting a control device because

Correct answer: It determines the fan power requiredThe fan must move the whole gas stream against the device’s resistance, and\[ P = \frac{Q\Delta p}{\eta} \]so pressure drop translates directly into shaft power and therefore into the running cost, which over a plant’s life usually exceeds the capital cost. This is why a venturi scrubber at 10 kPa is expensive to run and a precipitator at a few hundred pascals is cheap.

Q: A baghouse treats \( 30 \) m\(^3\)/s with \( 900 \) m\(^2\) of fabric. Its air to cloth ratio is

Correct answer: 0.033 m/sThe air to cloth ratio is gas flow over fabric area:\[ \frac{30}{900} = 0.033 \ \mathrm{m/s} \]It has the units of a velocity and is often called the filtration velocity. Values around 0.01 to 0.02 m/s are usual for reverse air cleaning and 0.03 to 0.05 for pulse jet; too high a ratio drives dust into the fabric, raises the pressure drop and shortens bag life.

Q: A fan moves \( 40 \) m\(^3\)/s against a pressure drop of \( 9 \) kPa at \( 70 \) per cent efficiency. The power required is

Correct answer: 514 kWFan power is flow times pressure drop divided by efficiency:\[ P = \frac{Q\Delta p}{\eta} = \frac{40 \times 9000}{0.7} = \frac{360\,000}{0.7} \approx 514 \ \mathrm{kW} \]Half a megawatt to move the gas is what a venturi scrubber’s high pressure drop costs, and over a year it dominates the whole economics of the installation. Multiplying by the efficiency instead of dividing gives 252 kW and would have the fan produce more than it consumed.

31.

Pressure drop matters in selecting a control device because

A.It fixes the dust disposal route
B.It has no operational significance
C.It determines the fan power required
D.It determines the capital cost only

Answer: C

Correct answer: It determines the fan power required

The fan must move the whole gas stream against the device’s resistance, and

\[ P = \frac{Q\Delta p}{\eta} \]

so pressure drop translates directly into shaft power and therefore into the running cost, which over a plant’s life usually exceeds the capital cost. This is why a venturi scrubber at 10 kPa is expensive to run and a precipitator at a few hundred pascals is cheap.

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32.

A precipitator is often built with several fields in series because

A.Each field collects a different chemical species
B.The gas must be cooled between fields
C.It reduces the total plate area required
D.Efficiency rises exponentially with area

Answer: D

The Deutsch relationship means each field removes a similar fraction of what reaches it, so three or four in series achieve what one very long field could. Sectionalising also lets each field run at the voltage its own dust loading allows.

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33.

A device reduces particulate from \( 8.4 \) g/m\(^3\) to \( 0.21 \) g/m\(^3\). Its collection efficiency is

A.97.5 per cent
B.75.0 per cent
C.2.5 per cent
D.40.0 per cent

Answer: A

The efficiency is \( (8.4-0.21)/8.4 = 0.975 \). Efficiency figures alone can mislead, since a device at 97.5 per cent still passes a fifth of a gram per cubic metre, and it is the outlet concentration that a standard is written against.

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34.

Two devices in series achieve \( 90 \) and \( 95 \) per cent efficiency. The overall efficiency is

A.85.5 per cent
B.99.5 per cent
C.92.5 per cent
D.185 per cent

Answer: B

The penetrations multiply, so \( 0.10 \times 0.05 = 0.005 \) passes and 99.5 per cent is collected. Working in penetration rather than efficiency is much the easier way to handle devices in series, and it makes the diminishing returns obvious.

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35.

A settling chamber \( 6 \) m long and \( 2 \) m high passes gas at \( 0.6 \) m/s. The minimum settling velocity for complete collection is

A.0.60 m/s
B.1.80 m/s
C.0.20 m/s
D.0.05 m/s

Answer: C

Residence time is \( 6/0.6 = 10 \) s, and a particle must fall 2 m in that time, requiring \( 0.20 \) m/s. Settling velocities of that order belong to particles well above a hundred micrometres, which shows how limited the device is.

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36.

Applying the Deutsch-Anderson equation with a migration velocity of \( 0.10 \) m/s, a plate area of \( 4000 \) m\(^2\) and a flow of \( 100 \) m\(^3\)/s gives an efficiency of

A.95.0 per cent
B.63.2 per cent
C.99.9 per cent
D.98.2 per cent

Answer: D

The exponent is \( -Aw/Q = -4000 \times 0.10/100 = -4 \), and \( 1 - e^{-4} = 0.982 \). Because the relationship is exponential, raising the efficiency to 99.5 per cent would require more than a third as much plate area again.

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37.

To raise a precipitator from \( 98.2 \) to \( 99.5 \) per cent at the same flow and migration velocity, the plate area must rise from \( 4000 \) m\(^2\) to about

A.5300 m\(^2\)
B.4400 m\(^2\)
C.8000 m\(^2\)
D.4100 m\(^2\)

Answer: A

The exponent must move from 4 to \( -\ln 0.005 = 5.30 \), so the area rises in the same ratio to 5300 square metres. A third more steel for one and three tenths of a percentage point is the characteristic economics of precipitator design.

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38.

A baghouse treats \( 30 \) m\(^3\)/s with \( 900 \) m\(^2\) of fabric. Its air to cloth ratio is

A.3.3 m/s
B.0.033 m/s
C.30 m/s
D.0.30 m/s

Answer: B

Correct answer: 0.033 m/s

The air to cloth ratio is gas flow over fabric area:

\[ \frac{30}{900} = 0.033 \ \mathrm{m/s} \]

It has the units of a velocity and is often called the filtration velocity. Values around 0.01 to 0.02 m/s are usual for reverse air cleaning and 0.03 to 0.05 for pulse jet; too high a ratio drives dust into the fabric, raises the pressure drop and shortens bag life.

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39.

A fan moves \( 40 \) m\(^3\)/s against a pressure drop of \( 9 \) kPa at \( 70 \) per cent efficiency. The power required is

A.252 kW
B.225 kW
C.514 kW
D.360 kW

Answer: C

Correct answer: 514 kW

Fan power is flow times pressure drop divided by efficiency:

\[ P = \frac{Q\Delta p}{\eta} = \frac{40 \times 9000}{0.7} = \frac{360\,000}{0.7} \approx 514 \ \mathrm{kW} \]

Half a megawatt to move the gas is what a venturi scrubber’s high pressure drop costs, and over a year it dominates the whole economics of the installation. Multiplying by the efficiency instead of dividing gives 252 kW and would have the fan produce more than it consumed.

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40.

A cyclone has a cut diameter of \( 12 \) micrometres. Halving the body diameter, with efficiency scaling as the square root of the body diameter, gives a cut diameter of about

A.6.0 micrometres
B.17.0 micrometres
C.24.0 micrometres
D.8.5 micrometres

Answer: D

The cut diameter falls by \( \sqrt{0.5} = 0.707 \), giving \( 12 \times 0.707 = 8.5 \) micrometres. The pressure drop rises steeply at the same time, which is the standing trade in cyclone design and the reason multiclones exist.

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