Water Resources HubPE Civil: Water Resources & Environmental — free theory
Surface Water & Groundwater Quality
This page closes the PE WRE “Surface Water and Groundwater Quality” gap: the dissolved-oxygen sag material, water quality parameters and standards, and groundwater contaminant transport, with a worked 10-question quiz. Related background: Water Treatment (disinfection demand and dose, effluent limits), Environmental (related PE WRE coverage), and Groundwater (flow and Darcy's law behind the transport section).
1. Water quality parameters — what gets measured and why
Water quality questions on the PE WRE exam start from a small set of parameters. Know what each one actually measures, because the exam loves to test whether you can tell them apart.
- Dissolved oxygen (DO) — the oxygen gas dissolved in the water, in mg/L. Aquatic life depends on it directly. Saturation DO falls as temperature rises (about 9.1 mg/L at 20 °C in fresh water at sea level), so warm water holds less oxygen before anything pollutes it.
- BOD (biochemical oxygen demand) — the oxygen microorganisms consume while decomposing the biodegradable organic matter in a sample. The standard test is BOD5: oxygen used in 5 days at 20 °C. BOD is a load measure, not a direct measure of organic concentration.
- COD (chemical oxygen demand) — oxygen equivalent of everything a strong chemical oxidant can oxidize, biodegradable or not. COD is therefore almost always ≥ BOD for the same sample, and the test takes hours instead of 5 days. A high COD with a low BOD signals poorly biodegradable (industrial) waste.
- Ultimate BOD (L0) — the total oxygen demand once decomposition is complete. BOD exerted by time t follows first-order decay: BODt = L0(1 − e−kt), and the BOD remaining is Lt = L0·e−kt.
- Nutrients — nitrogen (ammonia, nitrite, nitrate, organic N) and phosphorus. They are not toxic at typical levels; their damage is indirect, through eutrophication (Section 3). Ammonia also exerts its own oxygen demand as it nitrifies, and unionized ammonia is toxic to fish.
- Pathogen indicators — pathogens themselves are hard to assay, so exams and regulations use indicator organisms: total coliform, fecal coliform, and E. coli. Their presence signals fecal contamination and the possible presence of pathogens; they are not themselves necessarily harmful.
- Turbidity and TSS — turbidity is an optical measure (NTU) of how cloudy the water is; TSS (total suspended solids) is a gravimetric measure (mg/L) of the solids caught on a filter. They correlate but are not interchangeable: turbidity responds to fine colloidal particles that contribute little mass.
2. The dissolved oxygen sag (Streeter–Phelps)
When a biodegradable discharge enters a stream, two processes compete:
- Deoxygenation — bacteria consuming the waste use up DO at a rate proportional to the BOD remaining: rate = k1·L.
- Reaeration — the atmosphere replenishes DO at a rate proportional to the oxygen deficit D (saturation DO minus actual DO): rate = k2·D.
Because the waste load is largest at the outfall, deoxygenation wins at first: DO falls and the deficit grows. As the BOD is used up, deoxygenation slows while reaeration keeps pushing back, so the deficit peaks and then recovers. Plotted against travel time downstream, the deficit traces the classic sag curve. (Try the interactive calculator: Dissolved-Oxygen Sag.)
The Streeter–Phelps equation gives the deficit at time t:
D(t) = [k1L0 / (k2 − k1)] · (e−k1t − e−k2t) + D0·e−k2t
where L0 is the ultimate BOD of the mixed water just below the outfall, D0 is the initial deficit there, k1 is the deoxygenation rate constant, and k2 is the reaeration rate constant (k2 > k1 in any stream that recovers).
Key exam facts:
- The critical point is where the deficit is maximum (DO is minimum). It occurs where deoxygenation exactly balances reaeration: k1L = k2D.
- At the critical point, the critical deficit can be computed directly: Dc = (k1/k2)·L0·e−k1·tc.
- If the minimum DO falls below the water quality standard (often 4–5 mg/L for warm-water fisheries), the discharge needs more treatment — this is the whole point of a wasteload allocation.
- Rate constants are temperature-corrected with kT = k20·θ(T−20), with θ ≈ 1.047 for BOD decay (k1) and ≈ 1.024 for reaeration (k2).
3. Eutrophication
Eutrophication is the enrichment of a water body with nutrients, driving excess algal growth. When the algae die and decompose, the resulting BOD can crash DO — the same oxygen story as Section 2, but internally generated.
- The limiting nutrient is the one in shortest supply relative to algal needs; controlling it controls growth. In most fresh waters that is phosphorus; in many coastal/marine waters it is nitrogen.
- Point sources (wastewater effluent) are controlled with effluent limits; nonpoint sources (fertilizer runoff, septic systems) are usually the harder term and are managed with best management practices.
- Symptoms the exam may describe: algal blooms, large day–night DO swings (supersaturated by day, depleted at night), taste-and-odor problems in drinking water supplies, and fish kills.
4. Surface-water quality standards — the framework
- States set designated uses for each water body (aquatic life, recreation, drinking water supply, etc.) and adopt water quality criteria — numeric limits (e.g., minimum DO, maximum toxicant concentrations) or narrative statements — protective of those uses.
- A water body that fails its standards is listed as impaired (the Section 303(d) list under the Clean Water Act), which triggers a TMDL (total maximum daily load): the maximum pollutant load the water can receive and still meet standards, divided among point sources (wasteload allocations), nonpoint sources (load allocations), and a margin of safety.
- NPDES permits translate those limits into enforceable discharge conditions for individual point sources.
- Antidegradation policy protects water that is already cleaner than standards require from being degraded without justification.
5. Groundwater quality
Groundwater moves slowly, is hard to observe, and is expensive to clean — so the exam emphasizes sources, transport, and natural attenuation.
- Contaminant sources: leaking underground storage tanks, landfills, septic systems, surface spills, agricultural chemicals, and leaking sewers. A dissolved contaminant forms a plume that spreads from the source by advection (moving with the groundwater), dispersion (spreading along and across the flow path), and diffusion (minor at field scale).
- Advection speed is the average linear (seepage) velocity, v = K·i / n — the Darcy flux divided by porosity. The Darcy flux itself is not the speed of the water; dividing by porosity is the step everyone forgets.
- Attenuation slows or removes contaminants: sorption onto soil (captured by the retardation factor R = 1 + ρb·Kd / n, so the contaminant moves at v/R), biodegradation, dispersion/dilution, and volatilization. Chlorinated solvents and metals mostly sorb and persist; petroleum hydrocarbons also biodegrade.
- Dense non-aqueous phase liquids (DNAPLs, e.g., chlorinated solvents) sink below the water table; light ones (LNAPLs, e.g., gasoline) float on it — a favorite conceptual question.
- Protection tools: wellhead protection areas, aquifer classification, and monitoring wells placed downgradient of potential sources.
6. Regulatory drivers (definitional level)
- Clean Water Act (CWA) — governs surface-water discharges: NPDES permits, water quality standards, 303(d) impaired-waters lists, and TMDLs.
- Safe Drinking Water Act (SDWA) — governs public drinking water (see Water Treatment for the treatment side): national primary drinking water regulations set enforceable MCLs (maximum contaminant levels) measured at the tap/entry point, plus treatment-technique rules where a contaminant can't be measured directly. It also runs the underground injection control and wellhead protection programs.
- Sampling follows from these drivers: permit compliance sampling at outfalls, ambient monitoring for standards attainment, and drinking-water sampling at prescribed frequencies — with chain-of-custody and approved analytical methods so results are legally defensible.
Free 10-question mini-quiz
Surface Water & Groundwater Quality mini-quiz
Choose your answer, then check it to see the result and explanation. SI units are used unless stated otherwise.
1. A wastewater sample has a BOD5 of 150 mg/L at 20 °C. The BOD rate constant is k = 0.23/day (base e). The ultimate BOD is most nearly:
Answer: D. BOD5 = L0(1 − e−k·5) = L0(1 − e−1.15) = L0(1 − 0.317) = 0.683·L0. So L0 = 150 / 0.683 = 219.5 ≈ 220 mg/L. Trap: BOD5 is only the oxygen used in 5 days — roughly 68% of the ultimate demand here. Answer A treats BOD5 as the ultimate BOD. (Answer C comes from using the base-10 rate constant K = 0.10 in the base-e formula.)
2. A waste has an ultimate BOD of 300 mg/L and k = 0.20/day (base e). The BOD exerted after 3 days is most nearly:
Answer: A. Exerted BOD = L0(1 − e−kt) = 300(1 − e−0.6) = 300(1 − 0.549) = 300 × 0.451 = 135.4 ≈ 135 mg/L. Trap: 165 mg/L (answer B) is the BOD remaining at day 3 (300·e−0.6 = 164.6). “Exerted” and “remaining” are complements — read which one the question asks for.
3. Below an outfall, the mixed water has ultimate BOD L0 = 25 mg/L and initial deficit D0 = 2.0 mg/L. With k1 = 0.30/day and k2 = 0.50/day, the oxygen deficit 2.0 days downstream (travel time) is most nearly:
Answer: C. D(t) = [k1L0/(k2−k1)]·(e−k1t − e−k2t) + D0·e−k2t. k1L0/(k2−k1) = (0.30 × 25)/0.20 = 37.5. e−0.6 = 0.549, e−1.0 = 0.368. D(2) = 37.5 × (0.549 − 0.368) + 2.0 × 0.368 = 6.78 + 0.74 = 7.52 ≈ 7.5 mg/L. Trap: subtracting the exponentials in the wrong order gives a negative first term and a nonsense answer; also note the answer is the deficit, not the DO itself (the DO would be 9.1 − 7.5 = 1.6 mg/L).
4. For the same stream as Q3, the travel time to the critical point (maximum deficit) is most nearly:
Answer: B. tc = [1/(k2−k1)] · ln{ (k2/k1) · [1 − D0(k2−k1)/(k1L0)] } = (1/0.20) · ln{ (0.50/0.30) · [1 − (2.0 × 0.20)/(0.30 × 25)] } = 5 × ln{ 1.667 × [1 − 0.0533] } = 5 × ln(1.578) = 5 × 0.456 = 2.28 ≈ 2.3 days. Trap: ignoring the initial deficit D0 in the bracket overstates tc; and tc is a travel time, converted to distance downstream only by multiplying by stream velocity.
5. For the stream in Q3–Q4, saturation DO is 9.1 mg/L. The minimum DO in the sag is most nearly:
Answer: A. At the critical point, Dc = (k1/k2)·L0·e−k1·tc = (0.30/0.50) × 25 × e−0.30 × 2.28 = 15 × e−0.684 = 15 × 0.505 = 7.57 mg/L. Minimum DO = 9.1 − 7.57 = 1.53 ≈ 1.5 mg/L. (Cross-check: the full Streeter–Phelps equation at t = 2.28 days gives D = 7.57 mg/L — the two routes agree.) This stream would violate a typical 4–5 mg/L DO standard: the discharge needs a tighter wasteload allocation. Trap: answer D is the critical deficit. Minimum DO is always saturation minus deficit — the last subtraction is the whole question.
6. A river flowing at 8 m³/s with a BOD5 of 2 mg/L receives a treated effluent of 2 m³/s with a BOD5 of 200 mg/L. Assuming complete mixing, the downstream BOD5 is most nearly:
Answer: B. Flow-weighted average: C = (8 × 2 + 2 × 200) / (8 + 2) = (16 + 400)/10 = 41.6 ≈ 42 mg/L. Trap: a simple average of the two concentrations, (2 + 200)/2 = 101 mg/L (answer D), ignores that the river carries four times the effluent flow. Mass balance is always flow × concentration.
7. A treatment plant disinfects a flow of 2.0 MGD with a chlorine dose of 2.5 mg/L. The chlorine feed rate is most nearly:
Answer: A. Feed (lb/day) = 8.34 × flow (MGD) × dose (mg/L) = 8.34 × 2.0 × 2.5 = 41.7 ≈ 42 lb/day. (8.34 is the weight of a gallon of water in pounds, which is what converts MGD × mg/L into lb/day.) Trap: dose is not demand — if the question gives a residual, the dose is demand + residual, and the feed rate is based on the dose actually applied.
8. A BOD test is run at 25 °C on a waste whose rate constant at 20 °C is k20 = 0.23/day. Using θ = 1.047, the rate constant at 25 °C is most nearly:
Answer: B. k25 = k20·θ(T−20) = 0.23 × 1.0475 = 0.23 × 1.258 = 0.289 ≈ 0.29/day. Trap: warmer water means faster decay, so k must increase — answers at or below 0.23 fail the sanity check. Note the consequence: the same waste exerts its BOD faster in summer, exactly when saturation DO is lowest.
9. An aquifer has hydraulic conductivity K = 15 m/day, hydraulic gradient i = 0.004, and porosity n = 0.30. The time for groundwater to travel 100 m is most nearly:
Answer: D. Seepage velocity v = K·i / n = (15 × 0.004)/0.30 = 0.20 m/day. Travel time = 100 / 0.20 = 500 days. Trap: using the Darcy flux q = K·i = 0.06 m/day as the velocity gives ≈1,700 days (answer C). The Darcy flux is discharge per unit total area; the water only moves through the pore fraction, so it travels faster by 1/n.
10. A contaminant in the aquifer of Q9 has distribution coefficient Kd = 0.5 L/kg, and the soil bulk density is ρb = 1.6 g/cm³. The contaminant's travel time over the same 100 m is most nearly:
Answer: C. R = 1 + ρb·Kd / n = 1 + (1.6 × 0.5)/0.30 = 1 + 2.67 = 3.67. Contaminant velocity = v/R = 0.20/3.67 = 0.0545 m/day. Travel time = 100 / 0.0545 = 1,833 ≈ 1,800 days. Trap: retardation can only slow a sorbing contaminant relative to the water — any answer under 500 days (the water's own travel time) is impossible. Chloride, which barely sorbs (Kd ≈ 0, R ≈ 1), is the classic exception that travels at the water's speed, which is why it's used as a tracer.
Preparing for the PE Civil: Water Resources & Environmental exam?
Work the FE→PE Bridge material for this topic, then test yourself under time pressure with the 30-question Water Resources practice set — fully worked solutions and distractor analysis included.