Topic 10 of 10 — free theory
Wastewater Treatment
BOD, activated sludge (F/M, MCRT), and clarifier loading rates.
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BOD — the currency of wastewater
Biochemical oxygen demand measures the oxygen that microorganisms will consume breaking down the waste. Two numbers matter: BOD5 (the 5-day lab test, the regulatory workhorse) and ultimate BOD L0 (everything biodegradable, eventually):
BOD exerted at time t: y = L0(1 − e−kt) (base e) or y = L0(1 − 10−Kt) (base 10)
| y | oxygen consumed by time t |
| L0 | ultimate BOD |
| k, K | deoxygenation rate constants — k = 2.303K, so check which base the question uses |
BOD5 ≈ 0.68 × L0 (for the standard k = 0.23 day−1, base e, at 20°C)
| Handy when a question gives one and asks for the other. |
Activated sludge — F/M and MCRT
The aeration tank is a managed ecosystem: wastewater (food) meets microorganisms. Two ratios run the process. The food-to-microorganism ratio (F/M) sets how hard the biomass works; the mean cell residence time (MCRT), or sludge age, sets how long biomass stays in the system:
F/M = Q · S0V · X
| Q | influent flow |
| S0 | influent BOD5 |
| V | aeration tank volume |
| X | MLSS — mixed liquor suspended solids |
| Units: day−1; conventional plants run ≈ 0.2–0.5 day−1. |
MCRT = mass of solids in the systemmass of solids wasted per day = V·XQwXw + QeXe
| Qw, Xw | waste sludge flow and concentration |
| Qe, Xe | effluent flow and suspended solids (often negligible) |
| MCRT in days; conventional ≈ 5–15 days, extended aeration 20–30+. |
In English units the 8.34 lb/gal conversion appears top and bottom of F/M and cancels — work in mg/L and MG throughout and it vanishes.
Clarifier loading
Surface overflow rate = QA Solids loading rate = (Q + Qr) · XA
| Qr | return sludge flow |
| Design SOR ≈ 400–700 gpd/ft²; solids loading ≈ 20–30 lb/day/ft² for conventional plants. |
SVI = (settled volume in 30 min, mL/L) × 1000 / MLSS (mg/L) (mL/g)
| SVI < 100 good settling; > 150 suggests bulking. |
Solids loading includes the return sludge — the clarifier sees Q + Qr, not just Q.
PE depth: collection systems, I/I, and nutrients
A gravity sewer is checked with Manning's equation using the hydraulic radius for the actual depth. A full force main is a pressure conduit and belongs under Darcy-Weisbach or Hazen-Williams. Peak design flow must include sanitary flow plus infiltration/inflow (I/I) using the basis given in the problem.
Mass load (lb/day) = 8.34 × Q(MGD) × C(mg/L)
Removal efficiency = Cin − CoutCin × 100%
For biological nutrient removal, distinguish carbon removal, nitrification (ammonia to nitrate), and denitrification (nitrate to nitrogen gas). Nitrification consumes oxygen and alkalinity; solids age and temperature determine whether nitrifiers can remain in the system.
PE trap: concentration removal and mass removal are equal only when influent and effluent flows are effectively the same. Return and recycle flows affect internal process loading even though they do not create new external mass.
Worked example F/M ratio and sludge age
Given:
- Q = 4 MGD, influent BOD5 S0 = 220 mg/L.
- Aeration volume V = 1.2 MG, MLSS X = 2,800 mg/L.
- Wasting: Qw = 0.06 MGD at Xw = 8,000 mg/L; effluent solids negligible.
Solution:
- F/M = (Q·S0)/(V·X) = (4 × 220)/(1.2 × 2,800) = 880/3,360 = 0.262 day−1. (The 8.34 conversions cancel — a useful check on your algebra.)
- Solids inventory = V·X = 1.2 × 2,800 = 3,360 (in MG·mg/L units).
- Solids wasted/day = Qw·Xw = 0.06 × 8,000 = 480.
- MCRT = 3,360/480 = 7.0 days — a healthy conventional sludge age.
Answer: F/M ≈ 0.262 day−1; MCRT ≈ 7.0 days.