Metrological aspects of air pollution dispersion

Environmental (ambient) lapse rate (ELR, $\Gamma$) is the actual rate at which air temperature falls with height at a place and time, $\Gamma = -dT/dz$.

Dry adiabatic lapse rate (DALR, $\Gamma_d$) is the rate at which a rising dry parcel of air cools because it expands in the lower-pressure air, without exchanging heat with its surroundings:

$$\Gamma_d = \frac{g}{c_p} = \frac{9.81}{1005} \approx 9.8\ ^\circ\text{C/km} \approx 1\ ^\circ\text{C}/100\ \text{m}$$

(The saturated adiabatic rate is lower, about 4-6 $^\circ$C/km, because latent heat is released.)

Stability principle

A parcel displaced upwards cools at $\Gamma_d$. If it becomes warmer than its surroundings it is lighter and keeps rising (unstable); if it becomes colder it sinks back (stable).

 height                                   
  |  \  superadiabatic   \  neutral   \   subadiabatic
  |   \   (ELR>DALR)      \ (ELR=DALR)  \  (ELR<DALR)
  |    \  ..ELR           \             \   ..ELR
  |     \   /DALR          \  ELR=DALR    \  /DALR
  +------------------------------------------> temp

Unstable conditions occur on sunny days with light winds; stable ones on clear, calm nights. Pasquill classes A (very unstable) to F (stable) express the same idea.

A temperature inversion is a layer of the atmosphere in which temperature increases with height (negative lapse rate), so cold dense air lies below warmer lighter air. The layer is extremely stable and prevents vertical mixing.

Radiation (nocturnal) inversion

• On clear, calm nights the ground loses heat by long-wave radiation and cools quickly, cooling the air next to it.
• The surface layer becomes colder than the air above, forming a ground-based inversion.
• It is deepest before sunrise and is destroyed after sunrise when the ground heats up and the "inversion lid" lifts.
• Common in valleys such as the Kathmandu valley in winter, where cold air also drains in from the hills.

Subsidence inversion

• Occurs in high-pressure (anticyclone) regions where a large air mass slowly sinks.
• The sinking air is compressed and warms adiabatically, so it becomes warmer than the air below.
• The inversion forms at some height above the ground (several hundred metres to 1-2 km) and can persist for days, e.g. Los Angeles.

 height
  |        warm air
  |  ======================  <- inversion base (lid)
  |        cooler air
  |  (pollutants trapped below)
  +------------------------> T

Effects on air pollution

• Vertical mixing is suppressed, so pollutants accumulate in the layer below the inversion and concentrations rise (smog episodes).
• Plumes released inside the inversion layer "fan"; plumes released above are kept away from the ground, while plumes released below are "trapped".
• Photochemical smog and fog-smog (London type) episodes are linked to inversions.
• Low-level sources such as vehicles and domestic fires cause the highest exposure.

The shape of a plume is determined by the lapse rate (stability) of the atmosphere around the stack, as in the sketch of temperature profile versus height.

 Looping        Coning        Fanning
  ~\_/~\_/~     ------->      ------>  (flat, narrow)
 chimney        chimney        chimney

 Lofting        Fumigation          Trapping
 inversion      ====inversion====   ====inversion====
 ----> up       ~~~~~> mixes down   -----> (confined)
 ----           chimney             ====inversion====

Fumigation and looping are of greatest concern for ground-level pollution; chimney height is normally selected for the worst of these cases.

Take the dry adiabatic lapse rate $\Gamma_d = 9.8\ ^\circ$C/km $= 0.98\ ^\circ$C/100 m.

(a) Environmental lapse rate

$$\Gamma = \frac{T_1 - T_2}{z_2 - z_1} = \frac{30.0 - 27.9}{270 - 20} = \frac{2.1}{250}\ ^\circ\text{C/m} = 0.84\ ^\circ\text{C}/100\ \text{m} = 8.4\ ^\circ\text{C/km}$$

Since $\Gamma = 8.4 < \Gamma_d = 9.8\ ^\circ$C/km, the atmosphere is subadiabatic (stable).

(b) Parcel temperature

$$T_p = 30.0 - 9.8 \times 0.25 = 30.0 - 2.45 = 27.55\ ^\circ\text{C}$$

The surroundings at 270 m are 27.9 $^\circ$C. The parcel (27.55 $^\circ$C) is colder and denser than the surrounding air, so it sinks back to its original level. The atmosphere resists vertical motion, so it is stable.

(c) New reading

$$\Gamma = \frac{30.0 - 26.7}{250} \times 100 = 1.32\ ^\circ\text{C}/100\ \text{m} = 13.2\ ^\circ\text{C/km} > 9.8$$

The parcel would arrive at 27.55 $^\circ$C, warmer than the surroundings (26.7 $^\circ$C), so it keeps rising: unstable (superadiabatic).

Answer: (a) ELR = 8.4 $^\circ$C/km, stable; (b) parcel 27.55 $^\circ$C < 27.9 $^\circ$C, stable; (c) ELR = 13.2 $^\circ$C/km, unstable.

(a) Assumptions

• Emission rate $Q$ is continuous and constant (steady state).
• Wind speed and direction are constant with height and time; the plume travels in the x-direction.
• Concentration distribution in the y (crosswind) and z (vertical) directions is Gaussian (normal).
• No chemical reaction, deposition or decay of the pollutant (conservative).
• The ground reflects the plume completely (an image source is used).
• Terrain is flat and the diffusion in the x direction is small compared with the wind transport.

General equation (point source, effective height $H$):

$$C(x,y,z) = \frac{Q}{2\pi u \sigma_y \sigma_z}\exp\left(-\frac{y^2}{2\sigma_y^2}\right)\left[\exp\left(-\frac{(z-H)^2}{2\sigma_z^2}\right) + \exp\left(-\frac{(z+H)^2}{2\sigma_z^2}\right)\right]$$

At ground level ($z=0$) on the centreline ($y=0$):

$$C(x,0,0) = \frac{Q}{\pi u \sigma_y \sigma_z}\exp\left(-\frac{H^2}{2\sigma_z^2}\right)$$

Here $u$ is mean wind speed (m/s), $\sigma_y, \sigma_z$ are dispersion coefficients (m) that depend on stability and distance, $H$ is effective stack height $=$ physical height $+$ plume rise.

(b) Numerical

Data: $Q = 80$ g/s $= 80\times10^6\ \mu$g/s, $u = 5$ m/s, $H = 60$ m, $\sigma_y = 36$ m, $\sigma_z = 18.5$ m.

Pre-exponential term:

$$\frac{Q}{\pi u \sigma_y\sigma_z} = \frac{80\times10^6}{\pi \times 5 \times 36 \times 18.5} = \frac{80\times10^6}{10\,461} = 7647\ \mu\text{g/m}^3$$

Exponential term:

$$\exp\left(-\frac{60^2}{2\times18.5^2}\right) = \exp(-5.259) = 0.005199$$ $$C = 7647 \times 0.005199 = 39.8\ \mu\text{g/m}^3$$

Answer: Ground-level centreline concentration of SO$_2$ at 500 m is about 39.8 $\mu$g/m$^3$ ($3.98\times10^{-5}$ g/m$^3$).

Derivation

The ground-level centreline concentration is

$$C = \frac{Q}{\pi u \sigma_y \sigma_z}\exp\left(-\frac{H^2}{2\sigma_z^2}\right)$$

Let $\sigma_y = \sigma_z / k$ where $k = \sigma_z/\sigma_y$ is constant. Then

$$C = \frac{kQ}{\pi u}\,\frac{1}{\sigma_z^2}\exp\left(-\frac{H^2}{2\sigma_z^2}\right)$$

Differentiating with respect to $\sigma_z$ and setting $dC/d\sigma_z = 0$ (let $s=\sigma_z$):

$$\frac{d}{ds}\left[s^{-2}\exp\left(-\frac{H^2}{2s^2}\right)\right] = \exp\left(-\frac{H^2}{2s^2}\right)\left[-2s^{-3} + s^{-2}\cdot\frac{H^2}{s^3}\right] = 0$$ $$\Rightarrow \frac{H^2}{s^2} = 2 \Rightarrow \sigma_z = \frac{H}{\sqrt{2}}$$

Substituting back, $\exp(-1) = 1/e$ and $\sigma_z^2 = H^2/2$:

$$C_{max} = \frac{kQ}{\pi u}\cdot\frac{2}{H^2}\cdot\frac{1}{e} = \frac{2Q}{\pi e\, u H^2}\left(\frac{\sigma_z}{\sigma_y}\right)$$

The maximum occurs at the distance $x$ where $\sigma_z = H/\sqrt{2}$. It falls as $1/H^2$, so raising the stack height is very effective.

Numerical

$Q = 100$ g/s, $H = 80$ m, $u = 4$ m/s, $k = 0.6$.

$$\sigma_z = \frac{80}{\sqrt 2} = 56.6\ \text{m}$$ $$C_{max} = \frac{2\times100}{\pi\times 2.718\times 4\times 80^2}\times0.6 = \frac{200\times0.6}{218\,620} = 5.49\times10^{-4}\ \text{g/m}^3$$

Answer: $C_{max} \approx 5.5\times10^{-4}$ g/m$^3$ = 549 $\mu$g/m$^3$, at the distance where $\sigma_z = 56.6$ m.