Meteorology and Dispersion Questions and Answers

Answer: C

Correct answer: Concentration goes as height to the minus two

Test the figures against an inverse-square law. Going from \( 50 \) to \( 100 \) m doubles the height and the relative concentration falls from \( 1.00 \) to \( 0.25 \), a factor of four. From \( 50 \) to \( 75 \) m the ratio is \( (75/50)^2 = 2.25 \), and \( 1/2.25 = 0.44 \), matching the table; from \( 50 \) to \( 150 \) m, \( (150/50)^2 = 9 \) and \( 1/9 = 0.11 \), which also matches.

So \( C_{max} \propto H^{-2} \). A simple inverse relation would have given \( 0.50 \) at \( 100 \) m rather than \( 0.25 \), and a direct square dependence would have made concentration rise with height.

Answer: D

Correct answer: Mechanical turbulence drives conditions to neutral

The last row shows that at winds over \( 6 \) m/s the class is C or D whatever the insolation or cloud cover, whereas at low winds the same conditions span A to F. Strong winds generate mechanical turbulence by shear over the surface, and that turbulence swamps the buoyant effects of heating or cooling, pushing the atmosphere towards neutral.

This is why a windy night is not stable and a windy sunny day is not strongly unstable. Wind speed clearly does affect the class - it is one of the two axes of the table - but at high speeds it dominates rather than combining with the radiation term.

Answer: A

Correct answer: Lofting

Lofting keeps pollution furthest from the ground. The profile is stable below the stack and unstable above, so the plume disperses freely upward into the unstable air while the stable layer beneath acts as a floor preventing downward mixing. Ground level concentrations are the lowest of any of the four forms.

Fanning also keeps the plume aloft, but only by suppressing dispersion in every direction, so the material accumulates in a thin ribbon and may come down abruptly when the inversion breaks. Looping brings undiluted plume to the surface intermittently, and coning brings it down gradually with distance.

Answer: B

Correct answer: Both A and R are true, and R is the correct explanation of A

Both statements are true and the reason explains the assertion. Severe pollution episodes do occur under stagnant high pressure systems, and the mechanism is the one given: air descending within the high is compressed and warms adiabatically, forming a subsidence inversion that caps the mixed layer, while the weak pressure gradients of a stagnant high leave the wind light.

Both factors reduce the ventilation coefficient - the product of mixing height and wind speed - so a given emission is diluted into a much smaller volume. And because such systems can persist for days, concentrations accumulate rather than clearing overnight.

Answer: C

Correct answer: Both A and R are true, and R is the correct explanation of A

Both statements are true and the reason is the explanation. Tall stacks do reduce ground level concentrations substantially, and the inverse-square dependence is why: doubling the effective height cuts the maximum to a quarter.

The effect is on ground level concentration near the source, not on the total emitted - the same mass is released, simply spread over a wider area and diluted before it reaches the surface. That distinction is why tall stacks were criticised as exporting rather than solving the problem, and why control at source is preferred where it is available.

Answer: D

Correct answer: Both A and R are true, and R is the correct explanation of A

Both statements are true and the second explains the first. Fumigation does produce some of the highest short-term ground level concentrations recorded, and the reason is the mechanism described: a breaking inversion brings material that accumulated aloft through the night down to the surface within minutes.

The severity comes from the combination of a large accumulated inventory and a very rapid delivery. Once the mixed layer has deepened well beyond the plume the episode ends, so it is short - which is exactly why short-averaging-time standards can be breached at a site whose daily means are unremarkable.

Answer: A

Correct answer: Both A and R are true, and R is the correct explanation of A

Both statements are true and the reason explains the assertion. Concentration in a Gaussian plume is inversely proportional to wind speed, with \( u \) appearing once in the denominator of \( C = \dfrac{Q}{2\pi u \sigma_y \sigma_z}\exp(\cdot) \).

The reason gives the physical basis: a faster wind distributes the same emission over a greater length of plume in a given time, so the mass per unit volume falls in proportion. Doubling the wind speed therefore halves the concentration - a first-power relation, not a square.

Answer: B

Correct answer: Both A and R are true, and R is the correct explanation of A

Both statements are true and the reason is the explanation. The Pasquill class must be established first because the dispersion coefficients are tabulated against stability class and downwind distance; without a class, there is no curve to read.

This fixes the order of a plume calculation: assign the class from wind speed and insolation or cloud cover, read \( \sigma_y \) and \( \sigma_z \) at the distance of interest, compute the effective stack height, and only then evaluate the concentration. It also explains why the class matters so much - it enters the answer through both dispersion coefficients at once.

Answer: C

Correct answer: A is false but R is true

The assertion is false, and the reason is exactly why. Momentum from the exit velocity contributes to plume rise, so reducing the exit velocity reduces the rise, lowers the effective stack height, and therefore raises ground level concentrations.

Since the maximum varies roughly as \( 1/H^2 \), the penalty can be significant. The relation is often misread when fan power is being economised: slowing the exhaust saves energy at the stack but costs dispersion, and a very low exit velocity can also allow the plume to be caught in the wake downwash behind the stack itself.

Answer: D

Correct answer: Both A and R are true, and R is the correct explanation of A

Both statements are true and the reason explains the assertion. Looping conditions can produce high concentrations close to a stack because the convective eddies of a strongly unstable atmosphere are comparable in scale with the plume itself, so whole sections of relatively undiluted plume are carried down to the surface.

The contact is intermittent, giving brief high readings rather than a steady elevated level, which is why looping shows up in short-averaging-time data more than in daily means. Overall dilution in unstable conditions is rapid, so concentrations further downwind are low - the hazard is specifically near the source.