Question 4 of 5: Mixing Height, Atmospheric Removal Processes, and Groundwater Flow

Problem 4: Mixing Height, Atmospheric Removal Processes, and Groundwater Flow (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Mixing Height

The mixing height (or mixing depth), z_i, is the height above ground up to which relatively vigorous vertical mixing extends, and it bounds the volume of air within which pollutants released at or near the surface are diluted. It is set by buoyancy: a parcel of air rising from the ground cools at the dry adiabatic lapse rate (DALR, ≈9.8 °C/km); as long as that rising parcel stays warmer (less dense) than its surroundings it keeps rising, but once its temperature drops to match the actual environmental (measured) temperature profile, it is no longer buoyant and vertical mixing stops — often right at a temperature inversion.

Estimation: from an early-morning radiosonde/temperature-sounding profile, draw the DALR line starting from the day’s forecast maximum surface temperature T_s, and find the height at which that line intersects the measured environmental lapse-rate curve (as illustrated above) — that intersection height is the estimated (afternoon) mixing height. A shallow mixing height combined with light winds gives poor pollutant dilution and high ground-level concentrations; a deep, well-mixed daytime boundary layer disperses pollutants over a much larger air volume.

(b) Three Physical Removal Processes

1. Dry deposition — the direct transfer of a gaseous or particulate pollutant to the earth’s surface (vegetation, soil, water, structures) by contact, impaction, or adsorption, occurring continuously, without any need for precipitation.

2. Wet deposition (washout/rainout) — scavenging of pollutants by falling precipitation: rainout incorporates the pollutant into cloud droplets as they form, while washout scrubs it out of the air below the cloud as raindrops fall through it; either way the contaminant is delivered to the ground in rain or snow.

3. Gravitational settling (sedimentation) — larger particulates and droplets settle out of the atmosphere under their own weight, following (approximately) Stokes’ law; this is the dominant removal pathway for coarse particulate matter.

(c) Groundwater Flow Between the Two Wells

Given. Well spacing L = 200 m; head at Well #1, h₁ = 20 m; head at Well #2, h₂ = 18 m; hydraulic conductivity K = 10⁻³ cm/s = 1.0×10⁻⁵ m/s; porosity n = 0.30 (unconfined sandy aquifer).

Find. (i) the specific discharge (Darcy flux) q between the wells; (ii) the seepage (linear pore) velocity v of a non-sorbing dissolved chemical.

Approach. Darcy’s law gives the specific discharge directly from the hydraulic gradient; dividing by the effective porosity converts that bulk cross-sectional flux into the true average linear velocity that actually carries a dissolved, non-sorbing solute through the pore space.

1. Hydraulic gradient. $$i = \dfrac{h_1-h_2}{L} = \dfrac{20-18}{200} = 0.0100$$
2. Specific discharge (Darcy flux). By Darcy’s law, q = K·i: $$q = (1.0\times10^{-5}\ \text{m/s})(0.0100) = \boxed{1.00\times10^{-7}\ \text{m/s}}\ \ (=\,0.00864\ \text{m/day})$$
3. Seepage velocity. Dividing by the effective porosity, v = q/n: $$v = \dfrac{1.00\times10^{-7}}{0.30} = \boxed{3.33\times10^{-7}\ \text{m/s}}\ \ (=\,0.0288\ \text{m/day} \approx 10.5\ \text{m/yr})$$ This seepage velocity is the actual travel speed of a non-sorbing dissolved chemical moving from Well #1 toward Well #2 — noticeably faster than the Darcy flux, because the true flow is squeezed through only the connected pore space (30% of the bulk cross-section), not the full aquifer cross-section q itself represents.