Water Resources HubFE section 16 of 16 · free theory
Environmental Engineering
The FE Civil environmental breadth: mass balance in reactors, the oxygen sag curve, water quality parameters, air quality, solid waste, and noise — with the treatment-plant detail living on its own pages.
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Mass balance: the account book of environmental engineering
Nearly every environmental question is a mass balance in disguise: what enters a system either leaves, accumulates, or is created/destroyed by reaction. At steady state the accumulation term is zero, but the reaction term is not — that is the whole game.
Accumulation = In − Out + Generation − Consumption
| Steady state | accumulation = 0, so In + Generation = Out + Consumption |
Q·C0 = Q·C + k·V·C ⇒ C = C01 + k·td
| C0, C | influent and effluent concentration of a completely mixed reactor |
| k | first-order decay rate constant (time−1, base e) |
| td = V/Q | detention time — volume divided by flow |
| The k·V·C term | first-order loss spread over the whole mixed volume — it survives at steady state |
kT = k20 θT−20 k = 2.303 K
| θ | temperature coefficient: ≈ 1.047 for deoxygenation (k1), ≈ 1.024 for reaeration (k2) |
| k = 2.303K | converts a base-10 rate constant K into the base-e k the equations expect |
The mixed-reactor result C = C0/(1 + k·td) is not the plug-flow result C = C0e−kt. A completely mixed tank dilutes the influent instantly into the whole volume, so it removes less than a plug-flow reactor with the same detention time. The exam loves swapping them.
BOD, COD, TOC — what “oxygen demand” really means
Three different lab tests answer three different questions about the organics in water. Keep them straight and half the conceptual questions answer themselves:
COD ≥ BOD BOD5 ≈ 0.68 × L0 (k = 0.23 day−1, base e, 20°C)
| BOD | biochemical oxygen demand — oxygen microbes consume breaking down the waste; BOD5 is the 5-day test, L0 the ultimate (total biodegradable) demand |
| COD | chemical oxygen demand — oxygen needed to chemically oxidise everything oxidisable, biodegradable or not; always ≥ BOD |
| TOC | total organic carbon — a direct carbon count, no oxygen chemistry involved |
The BOD exertion curve itself, y = L0(1 − e−kt), lives on the Wastewater Treatment page — this page takes that result and drops it into a stream, which is where the oxygen sag begins.
The Streeter–Phelps oxygen sag
When organic waste enters a stream, two processes fight over the dissolved oxygen. Deoxygenation (rate k1) pulls oxygen out as microbes consume the BOD; reaeration (rate k2) pulls oxygen back in from the atmosphere. The oxygen deficit D = (saturation DO) − (actual DO) first grows, reaches a worst point, then recovers — the sag curve. The exam's favourite question is the worst point: the critical time tc and the critical deficit Dc.
D(t) = k1L0k2 − k1(e−k1t − e−k2t) + D0e−k2t
| D(t) | oxygen deficit at time t downstream of the discharge |
| L0 | ultimate BOD of the mixture just below the outfall |
| D0 | initial deficit at the mixing point (often small, rarely zero in real streams) |
| k1, k2 | deoxygenation and reaeration constants, same time base (day−1), base e |
tc = 1k2 − k1 · ln [k2k1(1 − D0(k2 − k1)k1L0)]
| tc | time of maximum deficit — set dD/dt = 0 and solve; the logarithm is the natural log |
| Requires k2 > k1 | reaeration must outpace deoxygenation or the sag never recovers |
DOmin = DOsat − Dc where Dc = D(tc)
| DOsat | saturation DO at stream temperature (≈ 9.1 mg/L at 20°C) |
| Dc | the critical deficit — not the answer; subtract it from saturation to get the minimum DO |
If the initial deficit is zero, the critical-time formula collapses to tc = ln(k2/k1)/(k2 − k1). And a useful check: with D0 = 0, Dc = (k1/k2)·L0·e−k1tc — the same number the full formula gives, from one line of arithmetic.
Water quality parameters at a glance
The exam tests whether you know what each parameter means and roughly what healthy looks like — not lab procedure. The short list:
- Dissolved oxygen (DO): the master variable for stream health. Saturation falls as temperature rises (≈ 9.1 mg/L at 20°C, ≈ 7.5 mg/L at 30°C). Many states require ≥ 4–5 mg/L to protect fish.
- pH: drinking water typically 6.5–8.5. Low pH water is corrosive (it leaches lead and copper from plumbing); high pH encourages scale.
- Turbidity (NTU): cloudiness from suspended particles — it shields pathogens from disinfection, which is why filtration precedes chlorination.
- Solids: TSS (suspended, caught on a filter) vs TDS (dissolved, passes through). Volatile solids burn off at 550°C and approximate the organic fraction.
- Nutrients (N and P): nitrogen and phosphorus drive eutrophication — algal blooms, then oxygen depletion when the algae die. Phosphorus usually limits freshwater systems.
- Coliforms / E. coli: indicator organisms. Their presence signals faecal contamination and the possible presence of pathogens — they are the alarm, not the fire.
Treatment trains: the 30-second version
The process-level calculations — detention times, overflow rates, F/M, MCRT, CT — live on their own pages. What the environmental section wants is the order of operations:
- Drinking water: coagulation → flocculation → sedimentation → filtration → disinfection. Each step protects the next: settling removes what would clog the filter, filtration removes what would shield pathogens from the disinfectant. Details: Water Treatment.
- Wastewater: preliminary (screening, grit) → primary settling → secondary biological treatment (e.g. activated sludge) → disinfection, with solids thickened, digested, and dewatered on the side. Details: Wastewater Treatment.
Air quality: criteria pollutants and dispersion
The Clean Air Act directs the EPA to set National Ambient Air Quality Standards (NAAQS) for six criteria pollutants: carbon monoxide (CO), lead (Pb), nitrogen dioxide (NO2), ozone (O3), particulate matter (PM10 and PM2.5), and sulfur dioxide (SO2). Primary standards protect public health. Note what is not on the list: carbon dioxide and methane are greenhouse gases, and benzene is a hazardous air pollutant — none has a NAAQS.
For dispersion, the exam works conceptually with the Gaussian plume: a stack emits at rate Q into wind of speed u, and the plume spreads downwind. Ground-level concentration rises with emission rate, falls as wind speed increases (more dilution), and falls with distance as the plume spreads and mixes. The controlling height is the effective stack height H = physical stack height + plume rise — a taller effective stack pushes the maximum ground-level concentration farther downwind and lowers it.
C(x, 0, 0) = Qπ u σyσz · e−H²/2σz²
| C | ground-level concentration on the plume centreline at downwind distance x |
| Q | emission rate (mass/time) |
| u | mean wind speed at stack height — in the denominator: more wind, more dilution |
| σy, σz | plume spread coefficients — grow with downwind distance and with atmospheric instability (class A, very unstable, spreads fastest; class F, very stable, spreads slowest) |
| H | effective stack height = physical height + plume rise |
Know the shape of this equation, not its arithmetic: the exam gives you σ values from stability-class curves or tables and asks what happens when a variable changes, far more often than it asks you to evaluate the exponential.
Solid and hazardous waste
Municipal solid waste (MSW) generation in the US runs roughly 4–5 lb per person per day (≈ 2 kg/person/day) — the exam's favourite per-capita rate for “size the landfill cell” arithmetic. A modern sanitary landfill is an engineered containment system, not a hole in the ground:
- Liner + leachate collection: an impermeable liner (clay and/or geomembrane) with drains above it captures leachate — the contaminated liquid that percolates through the waste — before it reaches groundwater.
- Daily cover: soil spread over each day's waste controls odour, vectors, and windblown litter.
- Gas collection: decomposing organics generate methane (and CO2); modern landfills collect it for flaring or energy recovery rather than letting it migrate.
- Regulations, conceptually: RCRA governs hazardous waste cradle-to-grave (generation → transport → treatment/disposal); CERCLA (“Superfund”) funds cleanup of abandoned or uncontrolled hazardous-waste sites.
Noise: adding what you cannot add arithmetically
Sound level in decibels is logarithmic, so 85 dB + 85 dB is not 170 dB. Combine sources in energy space (10L/10), add, then convert back. Two rules of thumb carry most questions:
Ltotal = 10 log10(Σ10Li/10)
| Two equal sources | add 3 dB — doubling the energy is 10 log10(2) ≈ 3.01 dB |
| 10 dB difference | the quieter source barely matters — combined level rises only ≈ 0.4 dB |
| Distance, point source | −6 dB per doubling of distance (spherical spreading) |
| Distance, line source | −3 dB per doubling of distance (cylindrical spreading, e.g. a highway) |
Levels quoted in dBA are A-weighted to match human hearing — the weighting the exam uses for community and occupational noise questions.
Worked example Oxygen sag: does the stream violate the DO standard?
Given:
- A waste discharge mixes into a stream. Just below the outfall: ultimate BOD L0 = 40 mg/L, initial deficit D0 = 1.5 mg/L.
- Deoxygenation k1 = 0.30 day−1, reaeration k2 = 0.90 day−1 (both base e).
- Saturation DO at stream temperature = 9.1 mg/L; the state standard requires minimum DO ≥ 4.0 mg/L.
Solution:
- Critical time: tc = [1/(0.90 − 0.30)] · ln[(0.90/0.30)(1 − 1.5(0.90 − 0.30)/(0.30 · 40))] = (1/0.60) · ln[3(1 − 0.075)] = 1.667 · ln(2.775) = 1.70 days.
- Critical deficit: Dc = [0.30 · 40/(0.90 − 0.30)](e−0.30·1.70 − e−0.90·1.70) + 1.5e−0.90·1.70 = 20(0.6005 − 0.2164) + 1.5(0.2164) = 7.68 + 0.32 = 8.00 mg/L.
- Minimum DO = 9.1 − 8.00 = 1.10 mg/L.
- 1.1 mg/L is far below the 4.0 mg/L standard — the discharge as modelled would violate it, and the worst point arrives about 1.7 days of travel time downstream.
Answer: tc ≈ 1.70 days, Dc ≈ 8.00 mg/L, minimum DO ≈ 1.1 mg/L — violates the 4.0 mg/L standard.
Worked example Completely mixed lagoon with first-order decay
Given:
- A completely mixed lagoon: volume V = 50,000 m³, flow Q = 2,500 m³/day.
- A pollutant decays by first-order kinetics with k = 0.10 day−1 (base e).
- Influent concentration C0 = 180 mg/L. Steady state.
Solution:
- Detention time: td = V/Q = 50,000/2,500 = 20 days.
- Steady-state balance: accumulation is zero, so what enters (Q·C0) equals what leaves (Q·C) plus what reacts away (k·V·C).
- Solve: C = C0/(1 + k·td) = 180/(1 + 0.10 · 20) = 180/3 = 60 mg/L.
- Removal efficiency = (180 − 60)/180 = 67%.
Answer: Effluent concentration ≈ 60 mg/L (about 67% removal).
Worked example Combining noise sources, then moving away
Given:
- Two machines at a property line produce 80 dB and 74 dB respectively (point sources).
- The receptor then moves to twice the distance from both.
Solution:
- Combine in energy space: L = 10 log10(1080/10 + 1074/10) = 10 log10(108(1 + 10−0.6)) = 80 + 10 log10(1.2512) = 80 + 0.97 = 80.97 ≈ 81 dB.
- Sanity check: the 74 dB source is 6 dB quieter, so it contributes roughly 1 dB — 81 dB is consistent.
- Doubling the distance from a point source subtracts 6 dB: 81 − 6 = 75 dB at the new receptor.
Answer: ≈ 81 dB at the property line, ≈ 75 dB at twice the distance.