Gaseous Pollutant Control Questions and Answers

Q: An operator raises the ammonia injection rate to improve nitrogen oxide removal and finds ammonium salts fouling the air heater downstream. What has happened?

Correct answer: Ammonia slip has risen and formed saltsPushing the molar ratio above what the mixing can use leaves unreacted ammonia in the gas. Downstream, where the gas cools, it combines with sulphur trioxide to form ammonium sulphate and bisulphate, which are sticky solids that deposit on the air heater surfaces, foul them and raise the draught loss. This is the practical limit on how far removal can be pushed, and the reason ammonia slip is itself a monitored and limited quantity.

Q: The point at which the outlet concentration of an adsorption bed begins to rise is called _____.

Correct answer: breakthroughBreakthrough is the point at which the mass transfer zone reaches the outlet of the bed and the effluent concentration starts to rise. It comes before saturation, sometimes well before, so the working capacity used in design is always less than the equilibrium capacity. The bed is switched at breakthrough, since running past it means emitting the pollutant to buy the remaining capacity.

Q: Unreacted reagent leaving with the flue gas after a reduction process is known as ammonia _____.

Correct answer: slipAmmonia slip is the unreacted reagent leaving with the flue gas. It rises as the injection ratio is pushed up for higher removal, and it matters for three reasons: ammonia is itself a pollutant, it forms ammonium salts that foul downstream heat transfer surfaces, and it contaminates the fly ash, which may then be unsaleable to the cement industry. Permits typically limit it to a few parts per million.

Q: The condition in which rising gas prevents liquid from draining through a packed tower is called _____.

Correct answer: floodingFlooding is the condition in which the upward gas flow exerts enough drag on the descending liquid to prevent it draining. Liquid accumulates, the pressure drop rises almost vertically and liquid is carried out of the top. Channelling, weeping and entrainment are distinct troubles: channelling is maldistribution through the packing, weeping is liquid falling through the holes of a tray, and entrainment is droplet carryover.

Q: The law relating a gas's partial pressure to its equilibrium concentration in a liquid is _____ law.

Correct answer: Henry’sHenry’s law states that at equilibrium the partial pressure of a gas above a liquid is proportional to its concentration in the liquid, \( p = Hx \). It fixes the equilibrium line on which every absorption calculation rests, and a small Henry constant means a readily soluble gas needing a small tower. Raoult’s law applies to the solvent rather than a dilute solute, Dalton’s law concerns partial pressures in a gas mixture, and Fick’s law describes diffusion.

Q: Comparison of three flue gas desulphurisation routes. RouteRemoval (per cent)By-productRelative capital cost Dry sorbent injection50 to 70Mixed with fly ashLowSpray dryer absorption80 to 90Dry solidModerateWet limestone95 to 98Saleable gypsumHighWet limestone with polishingover 99Saleable gypsumHighest Which route suits a small plant with no gypsum market and a modest emission limit?

Correct answer: Spray dryer absorptionThe plant is small, so the high capital cost of a wet limestone system is hard to justify; it has no gypsum market, so the wet route’s chief economic advantage disappears; and the limit is modest, so 80 to 90 per cent removal suffices. Spray dryer absorption meets all three: moderate cost, a dry solid product handled with the fly ash, and no liquid effluent or reheat. Dry sorbent injection would probably fall short of the limit.

61.

An operator raises the ammonia injection rate to improve nitrogen oxide removal and finds ammonium salts fouling the air heater downstream.

What has happened?

A.Ammonia slip has risen and formed salts
B.The catalyst has been damaged by the ammonia
C.The gas temperature has fallen
D.The nitrogen oxides have increased

Answer: A

Correct answer: Ammonia slip has risen and formed salts

Pushing the molar ratio above what the mixing can use leaves unreacted ammonia in the gas. Downstream, where the gas cools, it combines with sulphur trioxide to form ammonium sulphate and bisulphate, which are sticky solids that deposit on the air heater surfaces, foul them and raise the draught loss. This is the practical limit on how far removal can be pushed, and the reason ammonia slip is itself a monitored and limited quantity.

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62.

A plant with no nearby gypsum market must choose a desulphurisation route for a modest boiler.

Which consideration favours spray dryer absorption?

A.It has the lowest capital cost of all routes
B.It gives a dry product and no effluent
C.It achieves the highest removal efficiency
D.It produces saleable gypsum

Answer: B

The absence of a by-product market removes wet limestone's main economic advantage, and the spray dryer avoids an effluent stream altogether. Dry sorbent injection would be cheaper still but would use far more reagent for less removal.

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63.

A packed absorption tower's pressure drop rises sharply and its outlet concentration climbs.

What is the most likely cause?

A.The liquid rate is too low
B.The gas is too soluble
C.The tower has reached flooding
D.The packing has been removed

Answer: C

Liquid can no longer drain against the rising gas and accumulates in the packing, and both symptoms follow from that. Reducing the gas or liquid rate restores operation, and the design margin exists to keep the tower clear of the condition.

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64.

The point at which the outlet concentration of an adsorption bed begins to rise is called _____.

A.saturation
B.flooding
C.slip
D.breakthrough

Answer: D

Correct answer: breakthrough

Breakthrough is the point at which the mass transfer zone reaches the outlet of the bed and the effluent concentration starts to rise. It comes before saturation, sometimes well before, so the working capacity used in design is always less than the equilibrium capacity. The bed is switched at breakthrough, since running past it means emitting the pollutant to buy the remaining capacity.

Discuss in Forum

65.

Unreacted reagent leaving with the flue gas after a reduction process is known as ammonia _____.

A.slip
B.carryover
C.breakthrough
D.excess

Answer: A

Correct answer: slip

Ammonia slip is the unreacted reagent leaving with the flue gas. It rises as the injection ratio is pushed up for higher removal, and it matters for three reasons: ammonia is itself a pollutant, it forms ammonium salts that foul downstream heat transfer surfaces, and it contaminates the fly ash, which may then be unsaleable to the cement industry. Permits typically limit it to a few parts per million.

Discuss in Forum

66.

The condition in which rising gas prevents liquid from draining through a packed tower is called _____.

A.entrainment
B.flooding
C.channelling
D.weeping

Answer: B

Correct answer: flooding

Flooding is the condition in which the upward gas flow exerts enough drag on the descending liquid to prevent it draining. Liquid accumulates, the pressure drop rises almost vertically and liquid is carried out of the top. Channelling, weeping and entrainment are distinct troubles: channelling is maldistribution through the packing, weeping is liquid falling through the holes of a tray, and entrainment is droplet carryover.

Discuss in Forum

67.

The law relating a gas's partial pressure to its equilibrium concentration in a liquid is _____ law.

A.Fick's
B.Dalton's
C.Henry's
D.Raoult's

Answer: C

Correct answer: Henry’s

Henry’s law states that at equilibrium the partial pressure of a gas above a liquid is proportional to its concentration in the liquid, \( p = Hx \). It fixes the equilibrium line on which every absorption calculation rests, and a small Henry constant means a readily soluble gas needing a small tower. Raoult’s law applies to the solvent rather than a dilute solute, Dalton’s law concerns partial pressures in a gas mixture, and Fick’s law describes diffusion.

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68.

The three requirements for complete oxidation in an incinerator are temperature, turbulence and _____.

A.throughput
B.tolerance
C.titration
D.time

Answer: D

Sufficient residence time lets the reaction complete once temperature and mixing have started it. Falling short on any one of the three produces partial oxidation products that may be more harmful than the original stream.

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69.

Comparison of three flue gas desulphurisation routes.

Route	Removal (per cent)	By-product	Relative capital cost
Dry sorbent injection	50 to 70	Mixed with fly ash	Low
Spray dryer absorption	80 to 90	Dry solid	Moderate
Wet limestone	95 to 98	Saleable gypsum	High
Wet limestone with polishing	over 99	Saleable gypsum	Highest

Which route suits a small plant with no gypsum market and a modest emission limit?

A.Spray dryer absorption
B.Wet limestone
C.Dry sorbent injection
D.Wet limestone with polishing

Answer: A

Correct answer: Spray dryer absorption

The plant is small, so the high capital cost of a wet limestone system is hard to justify; it has no gypsum market, so the wet route’s chief economic advantage disappears; and the limit is modest, so 80 to 90 per cent removal suffices. Spray dryer absorption meets all three: moderate cost, a dry solid product handled with the fly ash, and no liquid effluent or reheat. Dry sorbent injection would probably fall short of the limit.

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70.

Methods for volatile organic compounds by concentration.

Concentration	Preferred method	Reason
Very low, large volume	Biofiltration	Combustion would need prohibitive fuel
Low to moderate	Adsorption	Recovery or destruction after concentration
Moderate to high	Catalytic incineration	Self-sustaining with modest preheat
Very high, recoverable	Condensation	Material has value as a liquid

Why is incineration unsuitable for a very dilute stream?

A.The catalyst would be poisoned
B.Heating the whole volume costs more than it yields
C.The pollutant would not oxidise
D.The residence time would be too short

Answer: B

Correct answer: Heating the whole volume costs more than it yields

An incinerator must raise the entire gas stream to the oxidation temperature, and in a dilute stream almost all of that gas is air carrying no calorific value at all. The fuel bill is set by the volume, while the benefit is set by the pollutant mass, so the economics collapse as the concentration falls. The pollutant would oxidise perfectly well; it is simply not worth what it costs to heat the air around it.

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