Microbiology, BOD Kinetics & Monod Growth

Key Takeaways

  • BOD measures oxygen demand of biodegradable organics.
  • First-order BOD kinetics: dL/dt = -kL.
  • Monod growth: μ = μmax S/(Ks+S).
  • Activated sludge design uses SRT, HRT, F/M ratio.
  • Nitrifiers grow slower than heterotrophs — need longer SRT.
Last updated: July 2026

Quick Answer: BOD is oxygen demand. First-order decay for carbonaceous BOD. Monod describes growth vs substrate. SRT controls biomass and nitrification.

BOD Kinetics

[ L_t = L_0 e^{-k_d t} ]

Ultimate BOD L₀; deoxygenation rate k_d (day⁻¹). NBOD from ammonia oxidized separately.

Monod Equation

[ \mu = \mu_{max} \frac{S}{K_s + S} ]

At low S, growth is first-order in substrate; at high S, μ → μmax.

Activated Sludge Parameters

ParameterRole
SRT (θc)Solids retention — nitrification needs ~8–15 d at 20°C
HRTHydraulic detention in aeration basin
F/MFood to microorganism ratio — loading rate

Oxygen Requirements

1 mg/L NH₃-N oxidized ≈ 4.6 mg/L O₂ stoichiometrically.

Exam Traps

  • BOD₅ is not ultimate BOD — only fraction in 5 days.
  • Lower temperature → slower kinetics → longer SRT needed.

Multi-Step Workflow

List givens with units, select the governing relationship, convert to a consistent unit set, solve, and compare to a rough estimate.

BOD Kinetics and Monod Growth

ModelEquation ideaUse
First-order BOD$L_t=L_0 e^{-kt}$Stream/BOD remaining
Ultimate BOD$L_0$Long-term oxygen demand
Monod$\mu=\mu_{max}S/(K_s+S)$Activated sludge growth
F/Mfood/microorganismsAeration tank loading

Worked BOD

If $k=0.23,d^{-1}$ (base e) and $L_0=200$ mg/L, BOD remaining at t=5 d: $L_5=200 e^{-0.23\times5}\approx 63$ mg/L. BOD exerted = $200-63=137$ mg/L.

On the Exam: Confirm whether $k$ is base-10 or base-e; NCEES problems state the convention. Mixing them flips the answer.

BOD Kinetics and Monod Growth

ModelEquation ideaUse
First-order BOD$L_t=L_0 e^{-kt}$Stream/BOD remaining
Ultimate BOD$L_0$Long-term oxygen demand
Monod$\mu=\mu_{max}S/(K_s+S)$Activated sludge growth
F/Mfood/microorganismsAeration tank loading

Worked BOD

If $k=0.23,d^{-1}$ (base e) and $L_0=200$ mg/L, BOD remaining at t=5 d: $L_5=200 e^{-0.23\times5}\approx 63$ mg/L. BOD exerted = $200-63=137$ mg/L.

On the Exam: Confirm whether $k$ is base-10 or base-e; NCEES problems state the convention. Mixing them flips the answer.

BOD Kinetics and Monod Growth

ModelEquation ideaUse
First-order BOD$L_t=L_0 e^{-kt}$Stream/BOD remaining
Ultimate BOD$L_0$Long-term oxygen demand
Monod$\mu=\mu_{max}S/(K_s+S)$Activated sludge growth
F/Mfood/microorganismsAeration tank loading

Worked BOD

If $k=0.23,d^{-1}$ (base e) and $L_0=200$ mg/L, BOD remaining at t=5 d: $L_5=200 e^{-0.23\times5}\approx 63$ mg/L. BOD exerted = $200-63=137$ mg/L.

On the Exam: Confirm whether $k$ is base-10 or base-e; NCEES problems state the convention. Mixing them flips the answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Test Your Knowledge

Longer SRT in activated sludge generally:

A
B
C
D