BOD and activated sludge (F/M, SRT) on the PE Civil WRE exam

BOD is the load secondary treatment is built to remove, and activated sludge is the process most problems use to remove it. Wastewater Collection and Treatment carries 7–11 of the 80 questions on the WRE exam. A handful of ratios covers most of it, if you know what each one controls.

What you actually need to own

• BOD exertion: BOD_t = L₀(1 − e^(−kt)), where L₀ is the ultimate carbonaceous BOD and k is the rate constant.
  • Check the base: a base-10 constant K converts as k = 2.303K.
  • BOD₅ is only a fraction of L₀. With k = 0.23 day⁻¹, BOD₅ ≈ 0.68·L₀.
• Temperature correction: k_T = k₂₀·θ^(T − 20). Temperature changes the rate constant. L₀ is normally treated as independent of temperature.
• The BOD test: BOD = (D₁ − D₂)/P for an unseeded dilution. When the dilution water is seeded, subtract the seed correction first.
• HRT vs. SRT:
  • HRT is θ = V/Q, usually in hours. It sizes the tank.
  • SRT (mean cell residence time, sludge age) is θc = VX/(Q_w·X_w + Q_e·X_e), usually in days. It sets the biology: how much sludge the plant makes, whether it nitrifies, how the floc settles.
  • The two differ because the clarifier and the return line hold solids back while the water passes through.
• F/M: Q·S₀/(V·X), in lb BOD per day per lb of biomass. Long-SRT processes such as extended aeration run at low F/M, and high-rate processes run high. When the plant makes more sludge, you raise the wasting rate to hold the SRT.
• Biomass: X = θc·Y(S₀ − S_e)/[θ(1 + k_d·θc)]. It links the yield and decay coefficients to the MLSS a design can carry.
• Return sludge: a solids balance around the clarifier, with the effluent and waste solids taken as small, gives R = Q_R/Q = X/(X_R − X).
• Wasting is the operator's control on SRT. More wasting gives a shorter SRT and less biomass under aeration.

Where people lose points

• MLSS vs. MLVSS. Active biomass is the volatile fraction. The Handbook's F/M relation and its design-parameter table are written on MLSS, but many problems and texts use MLVSS. Use the basis the problem gives, and don't mix them inside one ratio.
• SRT terms.
  • Include effluent solids when the stem gives them.
  • Sludge wasted from the return line is at the underflow concentration X_R, not X.
  • SRT ≈ V/Q_w holds only when you waste straight from the basin and neglect effluent solids.
• Hours vs. days. HRT comes out in hours, and SRT and F/M work in days. V in million gallons over Q in MGD gives days. Multiply by 24 before comparing with an HRT range.
• The temperature coefficient. Use the θ the problem gives. Otherwise take the Handbook's BOD value for the range that holds the temperature you are correcting to: 1.135 for 4–20 °C and 1.056 for 21–30 °C. Both differ from the θ ≈ 1.047 many courses quote, and the choice moves the answer. Leave L₀ as is, and keep this θ distinct from the HRT, which uses the same letter.
• BOD₅ vs. ultimate BOD. A stream or oxygen-demand balance may need L₀, while loading and F/M normally run on BOD₅. Convert when the problem asks for the other one.
• The pounds conversion. Loading in lb/day is 8.34 × Q(MGD) × C(mg/L). See the 8.34 mass-loading conversion.

How to study it

Start with the vocabulary: for each ratio (HRT, SRT, F/M, R, SVI), say what it sizes or controls before you compute it. Then work one plant end to end. Take influent BOD, set a target SRT, find the biomass and basin volume, then the wasting rate from the basin and from the return line, and check F/M. Sketch the flow schematic every time, because the exam draws plants that don't always match the textbook layout. The same BOD kinetics reappear in stream oxygen-sag problems. Effluent chlorination rests on the same concentration-times-time idea as drinking-water CT.

Part of the Wastewater Collection and Treatment area of the PE Civil WRE exam. → Start practicing free — the whole practice bank, free with an account.