Question 1 of 5: Oxygen-Demanding Wastes, Air Pollutant Classes, and Well Hydraulics

Problem 1: Oxygen-Demanding Wastes, Air Pollutant Classes, and Well Hydraulics (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) Impact of Oxygen-Demanding Wastes on Rivers

Oxygen-demanding wastes are organic (and some inorganic, e.g. reduced iron or ammonia) substances that microorganisms in the receiving stream oxidize, consuming dissolved oxygen (DO) from the water column in the process. When such a load enters a flowing river, the DO concentration does not drop uniformly; it follows the classic oxygen-sag curve first described by Streeter and Phelps: deoxygenation from bacterial respiration on the waste (rate constant k_d) competes with reaeration from the atmosphere across the air–water interface (rate constant k_r), producing a DO minimum some travel time downstream of the outfall before the river recovers. The ecological consequence is severe if the sag drives DO below the roughly 4–5 mg/L threshold most game fish need: cold-water species (trout, salmon) are the first to suffer stress, avoidance, or kill, followed by a general shift in the biological community toward pollution-tolerant, low-oxygen organisms (sludge worms, certain midge larvae) and away from the sensitive mayfly/stonefly/caddisfly taxa used as clean-water indicators. Beyond fish kills, low DO promotes anaerobic conditions in bottom sediments, releasing hydrogen sulphide (odour, further oxygen demand, black “sapropel” deposits) and re-mobilizing phosphorus and ammonia that had been bound in the sediment, which in turn can fuel downstream eutrophication once the river re-aerates. Because the sag response depends on the waste's ultimate BOD, the stream's temperature and velocity (which set k_r), and the travel time to any downstream water intake, oxygen-demanding loads are the single most common driver of minimum-DO effluent limits and river water-quality objectives in Canadian receiving-water assessments.

(b) Primary vs. Secondary Air Pollutants

Primary pollutants are emitted directly from an identifiable source into the atmosphere in the chemical form in which they cause harm — e.g. sulphur dioxide (SO₂) from high-sulphur fuel combustion, or carbon monoxide (CO) from incomplete combustion in vehicle engines. Secondary pollutants are not emitted at all; they are formed in the atmosphere itself by chemical reaction (often photochemical, driven by sunlight) among primary pollutants and normal atmospheric constituents — the textbook example is ground-level ozone (O₃), formed from the photolysis of NO₂ in the presence of volatile organic compounds (VOCs) and sunlight, or peroxyacetyl nitrate (PAN), formed from NO₂ and VOC photo-oxidation products. The distinction matters for control strategy: a primary pollutant is reduced by controlling its own source directly (scrubbers, low-sulphur fuel), whereas a secondary pollutant like ozone can only be reduced indirectly, by controlling its atmospheric precursors (NO_x and VOCs), and its peak concentration is often displaced in both time and distance from the precursor sources — which is why urban smog events peak on hot, sunny afternoons some hours after and downwind of the morning traffic that released the precursor NO_x.

(c)(i) Hydraulic Gradient, Elevation Head, Pressure Head, and Total Head

For groundwater flow, Bernoulli's principle reduces (velocity heads are negligible in porous media) to a statement about three components of head, all expressed as an equivalent height of water above a common horizontal datum. The elevation head, z, is the height of the point of measurement itself above the datum. The pressure head, p/γ, is the height the water would rise to in a piezometer (open standpipe) inserted at that point, driven purely by the fluid pressure there. The total head, h, is their sum, h = z + p/γ — it is the level to which water actually rises in an open well or piezometer screened at that point, and it is total head, not pressure head alone, that drives groundwater flow (from high h to low h), by Darcy's law. The hydraulic gradient is the rate of change of total head with distance along the flow path, dh/dl (dimensionless, m of head drop per m of travel) — it is the driving force per unit weight of water in Darcy's law, q = −K (dh/dl).

For the three wells here, each measured total head (9.6 m, 8.8 m, 8.0 m) already IS the sum z + p/γ at that well — the problem gives no separate elevation/pressure split, so parts (ii) and (iii) below work directly with total head, which is all Darcy's law needs.

(c)(ii)–(iii) Direction of Flow and Hydraulic Gradient

Given. Three wells at the corners of a right-angled isosceles triangle: W₂ at the right-angle corner, W₁ 400 m due east of W₂, W₃ 400 m due south of W₂. Total heads: h₁ = 9.6 m, h₂ = 8.8 m, h₃ = 8.0 m.

Find. The compass direction of groundwater flow and the magnitude of the hydraulic gradient.

Approach. The standard three-point problem: fit the (locally uniform) head field as a plane over the three well positions, then the gradient vector is the plane's slope and flow runs opposite to it, from high head toward low head.

1. Fit the head plane. With W₂ at the origin, assume $h(x,y) = a x + b y + c$ (x = east, y = north, m). At W₂: $c = 8.8$. At W₁ $(400,0)$: $400a + 8.8 = 9.6 \Rightarrow a = 0.00200\ \text{m/m}$. At W₃ $(0,-400)$: $-400b + 8.8 = 8.0 \Rightarrow b = 0.00200\ \text{m/m}$.
2. Gradient vector. $\nabla h = (\partial h/\partial x,\ \partial h/\partial y) = (0.00200,\ 0.00200)$, pointing east-northeast…in fact exactly northeast (45° between the two axes), toward increasing head. Its magnitude is $$i = |\nabla h| = \sqrt{a^{2}+b^{2}} = \sqrt{0.00200^{2}+0.00200^{2}} = \boxed{0.00283}$$ (dimensionless, m of head drop per m of travel).
3. Direction of flow. Groundwater flows down-gradient, i.e. opposite to $\nabla h$: bearing 45° west of due south, a compass bearing of S45°W (225° azimuth) — away from the high-head corner near W₁ and past W₂ toward the low-head side beyond W₃.
4. Cross-check by the classical graphical method. Locate point P on segment W₁–W₃ with the same head as W₂ (8.8 m): linear interpolation gives a fraction $(9.6-8.8)/(9.6-8.0) = 0.50$ of the way from W₁ to W₃, i.e. P is the midpoint, 400/√2 = 283 m from W₂. The segment W₂–P is therefore an equipotential line (both ends at 8.8 m); the perpendicular distance from W₃ to this line is $400/\sqrt{2} = 283\ \text{m}$, giving $$i = \frac{h_2 - h_3}{283\ \text{m}} = \frac{0.8\ \text{m}}{283\ \text{m}} = 0.00283$$ identical to Step 2, confirming the plane-fit result.