Meteorology and Dispersion Questions and Answers

Answer: C

Correct answer: 220 m depth

Once the growing mixed layer reaches the plume, the plume material is distributed through the whole depth of that layer - from the ground up to the mixing height, which has just passed \( 220 \) m. The layer available for dilution is therefore \( 220 \) m deep.

This is the fumigation geometry: material that spent the night confined in a thin stable ribbon aloft is suddenly mixed down through a couple of hundred metres to the surface, giving a brief but severe ground level episode. The \( 150 \) m inversion top is where the mixed layer started, not where it now ends, and \( 70 \) m is merely the difference between the two.

Answer: D

Correct answer: 81.4 m\(^3\)/s

The volumetric flow is the exit velocity times the cross-sectional area of the stack:

\[ A = \frac{\pi d^2}{4} = \frac{\pi (2.4)^2}{4} = 4.524\ \text{m}^2 \]

\[ Q = vA = 18 \times 4.524 = 81.4\ \text{m}^3/\text{s} \]

The distractor \( 43.2 \) is velocity times diameter, omitting the area entirely, and \( 162.8 \) uses the radius as if it were the diameter, doubling the answer. The volumetric flow is what fixes the momentum contribution to plume rise and, with the gas temperature, the buoyancy flux.

Answer: A

The air throughput is \( 4000 \times 20000 = 8\times10^7 \) cubic metres a second, so \( 120/8\times10^7 = 1.5\times10^{-6} \) g/m\(^3\). Box models of this kind give a useful order of magnitude before any detailed calculation.

Answer: B

Correct answer: An inversion

Temperature rises from \( 15.0 \) degrees at \( 100 \) m to \( 16.4 \) degrees at \( 300 \) m - an increase of \( 1.4 \) degrees over \( 200 \) m of ascent. Temperature increasing with height is the definition of an inversion.

The lapse rate is negative, \( -7 \) degrees per kilometre, so the layer is far more stable than the neutral case and vertical motion is strongly suppressed. Superadiabatic and unstable both require the temperature to fall faster than \( 9.8 \) degrees per kilometre, which is the opposite of what is observed here.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both statements are correct, and the second is the mechanism behind the first. Within an inversion the air above is warmer than the air below, so a parcel rising from below - cooling as it goes at the adiabatic rate - arrives colder and denser than its surroundings and is pushed back down.

That restoring force is exactly what suppresses vertical mixing. Pollutants released beneath an inversion are therefore confined to a shallow layer and accumulate, which is why inversions are the meteorological condition most associated with pollution episodes.

Answer: D

Correct answer: Statement I is incorrect but Statement II is correct

Statement I is incorrect: ground level concentration is inversely proportional to wind speed, not proportional to it. The wind speed appears in the denominator of the Gaussian plume equation, so a faster wind gives a lower concentration.

Statement II is correct, and it is the reason the first is wrong. A faster wind carries the same emission over a greater distance in the same time, so the material is spread through a larger volume of air and diluted in proportion. The two statements together are a common trap: a true mechanism paired with a relation stated the wrong way round.

Answer: A

Correct answer: Both Statement I and Statement II are correct

Both statements are correct and the second explains the first. A radiation inversion forms overnight when the ground loses heat by radiation and cools the air immediately above it, leaving cold air beneath warmer air. After sunrise the ground absorbs solar radiation and warms much faster than the air above it, because it is an opaque surface while the air is largely transparent to sunlight.

The warmed ground heats the lowest air, convection begins, and the inversion is eroded from below - usually within a few hours of sunrise. It is during this erosion that fumigation can occur, as any plume trapped aloft overnight is mixed abruptly to the surface.

Answer: B

Correct answer: Statement I is incorrect but Statement II is correct

Statement I is incorrect. Plume rise from a hot combustion source is far from negligible; it commonly adds tens or even hundreds of metres to the physical stack height, and since maximum ground level concentration falls roughly as \( 1/H^2 \), ignoring it would badly overestimate the impact.

Statement II is correct and is the reason: the exhaust is much hotter and hence less dense than the ambient air, so buoyancy accelerates it upward until mixing brings it to the surrounding temperature. Momentum from the exit velocity adds to this, but for a hot source buoyancy dominates.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both statements are correct, and the second describes the mechanism of the first. When a nocturnal inversion breaks up, the growing unstable layer reaches a plume that has been travelling in the stable air above and mixes it rapidly down to the surface.

The episode is brief - typically half an hour to an hour, until the mixed layer has deepened well beyond the plume - but severe, because material that accumulated in a thin ribbon overnight arrives at the ground almost at once. Fumigation is the reason short-averaging-time standards can be breached at sites that comply comfortably on a daily mean.

Answer: D

Correct answer: Statement I is incorrect but Statement II is correct

Statement I is incorrect. The Pasquill class describes the turbulent state of the atmosphere and has nothing to do with what is being emitted; the same class applies whether the source emits a gram per second or a kilogram.

Statement II is correct and gives the actual procedure: the class is read from a table indexed by surface wind speed together with incoming solar radiation by day, or cloud cover by night. The class then selects the dispersion coefficient curves, so it enters the plume calculation through \( \sigma_y \) and \( \sigma_z \), while the emission rate enters separately as \( Q \).