Gaseous Pollutant Control Questions and Answers

Answer: B

Correct answer: Upward gas prevents the liquid draining

As the gas velocity rises, the drag it exerts on the descending liquid increases until the liquid can no longer drain freely. It accumulates in the packing, the void space fills, the pressure drop rises very steeply, and liquid is carried out of the top of the tower. Towers are therefore designed at 60 to 70 per cent of the flooding velocity, leaving margin for the load variations of normal operation.

Answer: C

Steam raises the bed temperature and lowers the partial pressure of the adsorbate, driving it off so it can be condensed and recovered. Recovering a valuable solvent in this way often pays for the installation on its own.

Answer: D

Reacting the dissolved gas keeps its concentration in the liquid near zero and so maintains the driving force through the whole tower. The penalty is that the reagent is consumed and the pollutant is no longer recoverable in its original form.

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.

Answer: B

Correct answer: 14.4 t/day

The sulphur burnt is

and every atom of sulphur, of mass 32, becomes a molecule of sulphur dioxide, of mass 64, so the mass doubles:

The 7.2 t/day answer stops at the sulphur and forgets the oxygen picked up in combustion - the commonest error in this calculation.

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.

Answer: D

Correct answer: 24 hours

The bed can hold

and at 9 kg/h that is reached after

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.

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.

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.

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.