Meteorology and Dispersion Questions and Answers

Answer: A

Correct answer: 8.0 degrees per km

The environmental lapse rate is the observed fall of temperature with height. Here the temperature falls from \( 22 \) degrees at the surface to \( 14 \) degrees at \( 1000 \) m:

\[ \Gamma = \frac{22 - 14}{1.0\ \text{km}} = 8.0\ \text{degrees per km} \]

The value \( 9.8 \) is the dry adiabatic rate, not the measured one, and \( 14 \) is simply the upper temperature. Since \( 8.0 \) is less than \( 9.8 \), this profile is subadiabatic and therefore stable.

Answer: B

Correct answer: Unstable

Compare the environmental lapse rate with the dry adiabatic rate of \( 9.8 \) degrees per kilometre. Here \( 11.5 > 9.8 \), so the profile is superadiabatic and the atmosphere is unstable.

The reason is the parcel argument: a parcel displaced upward cools at \( 9.8 \) degrees per kilometre, but the air around it is cooling at \( 11.5 \), so the parcel is warmer and less dense than its surroundings and continues to rise. A lapse rate below \( 9.8 \) would be stable and one equal to it neutral; an inversion requires temperature to increase with height, which is not the case here.

Answer: C

Correct answer: 3600 m\(^2\)/s

The ventilation coefficient is the mixing height multiplied by the mean wind speed:

\[ VC = 800\ \text{m} \times 4.5\ \text{m/s} = 3600\ \text{m}^2/\text{s} \]

Dividing instead of multiplying gives \( 178 \), which is the trap in the distractors. A value of \( 3600 \) indicates reasonable dispersion; values below about \( 6000 \) are commonly taken as a warning of poor ventilation, and a shallow mixing height with light winds can drive it to a few hundred, the condition under which episodes develop.

Answer: D

Correct answer: 105 m

The effective stack height is the physical height plus the plume rise:

\[ H = h_s + \Delta h = 60 + 45 = 105\ \text{m} \]

The plume rise is a real gain in dispersion, not a correction to be subtracted: buoyancy and exit momentum carry the plume \( 45 \) m above the stack top before it levels off, and it is from \( 105 \) m that it then travels downwind. Since maximum ground level concentration falls roughly as \( 1/H^2 \), that extra \( 45 \) m matters a great deal.

Answer: A

Correct answer: 3.06

With the maximum ground level concentration varying as \( 1/H^2 \), the ratio of the two maxima is the square of the ratio of the heights:

\[ \frac{C_{60}}{C_{105}} = \left(\frac{105}{60}\right)^2 = 1.75^2 = 3.06 \]

So raising the effective height from \( 60 \) to \( 105 \) m cuts the peak concentration to about a third. The value \( 1.75 \) is the height ratio itself, before squaring - the usual mistake. This inverse-square dependence is why plume rise, and hence gas exit temperature and velocity, is worth as much attention as the physical stack.

Answer: B

Correct answer: 0.5

Ground level concentration is inversely proportional to wind speed, since \( u \) appears once in the denominator of the Gaussian plume equation. Doubling the wind speed therefore halves the concentration:

\[ \frac{C_8}{C_4} = \frac{u_4}{u_8} = \frac{4}{8} = 0.5 \]

The emission rate of \( 80 \) g/s is unchanged and does not enter the ratio. A factor of \( 0.25 \) would follow from an inverse-square dependence, which is not the case here; the wind dilutes in direct proportion to how fast it carries the plume away.

Answer: C

Correct answer: 42400 m\(^2\)

Substituting the two dispersion coefficients:

\[ \pi \sigma_y \sigma_z = \pi \times 150 \times 90 = \pi \times 13500 \approx 42412\ \text{m}^2 \]

which rounds to about \( 42400 \). The distractor \( 13500 \) is the product without \( \pi \), and \( 84800 \) doubles the result - the factor of two belongs in the full expression \( 2\pi u \sigma_y \sigma_z \), not in this product. The quantity has the character of an effective cross-sectional area over which the emission is spread.

Answer: D

Correct answer: 0.607

With \( H = \sigma_z = 100 \) m the exponent simplifies:

\[ \exp\left(-\frac{H^2}{2\sigma_z^2}\right) = \exp\left(-\frac{100^2}{2 \times 100^2}\right) = \exp(-0.5) = 0.607 \]

This is the point at which the vertical spread has grown to equal the release height, so the ground is one standard deviation from the plume centreline and the term has fallen only to about \( 0.61 \) - a substantial fraction of the plume is now reaching the surface. The value \( 0.368 \) is \( \exp(-1) \) and \( 0.135 \) is \( \exp(-2) \), which correspond to larger \( H/\sigma_z \) ratios.

Answer: A

The leading term is \( 80/(4 \times 42412) = 4.72\times10^{-4} \) g/m\(^3\), and multiplying by 0.607 gives \( 2.86\times10^{-4} \) g/m\(^3\), or 286 micrograms. Note that the reflection term would raise this further once the plume is well established at the surface.

Answer: B

Correct answer: 17.0 per cent

The frequency plotted on a wind rose for a sector is simply the proportion of observations from that direction:

\[ f = \frac{62}{365} \times 100 = 17.0\ \text{per cent} \]

So the north-west spoke would be drawn to a length representing 17 per cent, usually subdivided into speed bands. The figure \( 62 \) is the raw count of days rather than a frequency, and the other options do not follow from the data. Sectors of this size are what identify a prevailing wind and hence the most frequently exposed downwind area.