A scrubber reduces sulphur dioxide from \( 2400 \) to \( 96 \) mg/m\(^3\). Its removal efficiency is
Answer: A
The efficiency is \( (2400-96)/2400 = 0.96 \). Wet limestone systems routinely exceed this figure, while a dry sorbent injection retrofit on the same duty would typically reach only sixty to seventy per cent for far less capital outlay.
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A boiler burns coal of \( 1.8 \) per cent sulphur at \( 400 \) t/day. Complete conversion gives a sulphur dioxide emission of
Answer: B
Correct answer: 14.4 t/day
The sulphur burnt is
\[ 400 \times 0.018 = 7.2 \text{ t/day} \]and every atom of sulphur, of mass 32, becomes a molecule of sulphur dioxide, of mass 64, so the mass doubles:
\[ 7.2 \times \frac{64}{32} = 14.4 \text{ t/day} \]The 7.2 t/day answer stops at the sulphur and forgets the oxygen picked up in combustion - the commonest error in this calculation.
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Stoichiometric limestone for \( 14.4 \) t/day of sulphur dioxide, with calcium carbonate at \( 100 \) and sulphur dioxide at \( 64 \) g/mol, is
Answer: C
One mole of carbonate captures one mole of the dioxide, so the mass ratio is \( 100/64 = 1.5625 \) and \( 14.4 \times 1.5625 = 22.5 \) tonnes. Real plants feed a stoichiometric excess of five to ten per cent to reach their design removal.
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An adsorption bed holds \( 1800 \) kg of carbon with a working capacity of \( 0.12 \) kg of solvent per kg. Treating a stream carrying \( 9 \) kg/h, breakthrough occurs after about
Answer: D
Correct answer: 24 hours
The bed can hold
\[ 1800 \times 0.12 = 216 \text{ kg of solvent} \]and at 9 kg/h that is reached after
\[ \frac{216}{9} = 24 \text{ hours} \]The 216 hours answer mistakes the capacity in kilograms for a time. Note that the working capacity is already less than the equilibrium capacity, since the bed is switched at breakthrough rather than at saturation.
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Selective catalytic reduction at \( 85 \) per cent removal treats \( 21 \) t/day of nitrogen oxides. The residual emission is
Answer: A
The removal is \( 21 \times 0.85 = 17.85 \) tonnes, leaving \( 3.15 \) tonnes a day. Pushing the removal to ninety per cent would require more catalyst volume and would raise the ammonia slip unless the injection control were improved with it.
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Ammonia is injected at a molar ratio of \( 1.02 \) to the nitrogen oxides present. The excess available to become slip is
Answer: B
Correct answer: 2 per cent
A molar ratio of 1.02 means two per cent more ammonia is injected than the stoichiometry requires, and that excess is what may pass through unreacted as slip. Ratios slightly above unity are used because mixing is never perfect and some reagent finds no nitrogen oxide to react with; pushing the ratio higher improves removal but raises the slip, which forms ammonium bisulphate and fouls the air heater downstream.
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A thermal oxidiser must hold gas for \( 0.75 \) s at \( 800 \) degrees. Treating \( 12 \) m\(^3\)/s at that temperature, the chamber volume required is
Answer: C
Volume is flow times residence time, \( 12 \times 0.75 = 9.0 \) cubic metres. The flow must be evaluated at the operating temperature, since a gas expands by roughly a factor of three between ambient and eight hundred degrees.
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Catalytic oxidation at \( 350 \) degrees replaces thermal oxidation at \( 820 \) degrees. For a gas entering at \( 40 \) degrees, the temperature rise required falls by
Answer: D
The rises are \( 780 \) and \( 310 \) degrees, so the reduction is \( (780-310)/780 = 0.603 \). The fuel saving is close to that proportion, which is what justifies the catalyst's cost and its vulnerability to poisoning.
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A packed tower needs \( 6.4 \) transfer units with a height of a transfer unit of \( 0.55 \) m. The packed height required is
Answer: A
The height is the product, \( 6.4 \times 0.55 = 3.52 \) metres. A more efficient packing reduces the height of a transfer unit without changing the number of units, which is fixed by the difficulty of the separation itself.
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A tower is designed at \( 65 \) per cent of its flooding velocity of \( 2.8 \) m/s. Its design gas velocity is
Answer: B
Correct answer: 1.82 m/s
The design velocity is the specified fraction of the flooding velocity:
\[ 0.65 \times 2.8 = 1.82 \text{ m/s} \]Operating at 60 to 70 per cent of flooding gives a reasonable pressure drop and leaves headroom for load swings. Working closer to flooding buys a smaller tower diameter but risks a steeply rising pressure drop and liquid carryover the moment the gas rate increases.
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