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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A biofilter treats \( 4.5 \) m\(^3\)/s with a bed volume of \( 90 \) m\(^3\). The empty bed residence time is
Answer: C
Correct answer: 20 seconds
The empty bed residence time is the bed volume divided by the volumetric gas flow:
\[ EBRT = \frac{90}{4.5} = 20 \text{ seconds} \]It is called empty bed because the volume used is the whole bed, not the void fraction, which makes it a simple and comparable design parameter. Twenty to sixty seconds is typical for odour control; readily degraded compounds need less, and recalcitrant ones more.
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Spray dryer absorption achieves \( 85 \) per cent removal against wet limestone's \( 96 \) per cent. On \( 14.4 \) t/day the difference in residual emission is
Answer: D
Correct answer: 1.58 t/day
What matters is the residual, not the removal:
\[ 14.4 \times 0.15 = 2.16 \text{ t/day} \quad \text{and} \quad 14.4 \times 0.04 = 0.576 \text{ t/day} \]\[ 2.16 - 0.576 \approx 1.58 \text{ t/day} \]Eleven percentage points of removal separate the two routes, but the residual emission differs by nearly a factor of four - which is the general point that as removal approaches 100 per cent, each further point costs disproportionately and matters disproportionately.
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An adsorption bed of \( 216 \) kg capacity is regenerated with steam at \( 3 \) kg of steam per kg of solvent. The steam required per cycle is
Answer: A
Multiplying gives \( 216 \times 3 = 648 \) kilograms of steam. The steam cost and the condensate handling are the principal running costs of a carbon adsorption system, and they set the value below which solvent recovery does not pay.
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Gas at \( 40 \) degrees expands on heating to \( 800 \) degrees. Taking absolute temperatures, the volumetric flow rises by a factor of about
Answer: B
The ratio is \( 1073/313 = 3.43 \). Sizing an oxidiser chamber on the cold flow would make it less than a third of the volume actually needed, and the residence time would fall well short of what the reaction requires.
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Statement I: Chemical absorption maintains a higher driving force than physical absorption.
Statement II: Reacting the dissolved gas keeps its concentration in the liquid near zero.
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. The rate of absorption depends on the difference between the actual and equilibrium concentrations at the interface. In physical absorption the dissolved gas accumulates and that difference shrinks; in chemical absorption a reagent consumes it as fast as it dissolves, holding the free concentration near zero and the driving force near its maximum. This is why sulphur dioxide is scrubbed with an alkaline slurry rather than with water.
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Statement I: An adsorption bed is switched at saturation rather than at breakthrough.
Statement II: The mass transfer zone reaches the outlet before the bed is fully saturated.
In the light of the above statements, choose the correct answer.
Answer: D
The second is correct and it explains why beds are switched early: continuing past breakthrough buys the last of the capacity at the price of an emission. Working capacity is therefore well below the equilibrium figure.
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Statement I: Catalytic incineration saves fuel compared with thermal incineration.
Statement II: A catalyst allows oxidation to proceed at three to four hundred degrees rather than eight hundred.
In the light of the above statements, choose the correct answer.
Answer: A
Correct answer: Both Statement I and Statement II are correct
Both are true and the second gives the reason. A catalyst lowers the activation energy so that complete oxidation proceeds at 300 to 400 degrees instead of the 700 to 900 a thermal unit requires. Far less fuel is needed to bring the stream to temperature, and the smaller temperature rise also means less thermal nitrogen oxide is formed. The offsets are the cost of the catalyst and its vulnerability to poisoning by halogens, sulphur and heavy metals.
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