BOD First-Order Kinetics

y_{t} = L_{0} (1 - e^{- k t}), remaining L_{t} = L_{0} e^{- k t}

Carbonaceous BOD remaining decays exponentially. L0 is ultimate CBOD; y_t is the oxygen consumed after time t. Standard 5-day BOD is y_5. Rate k is temperature-sensitive: k_T = k_20 θ^{T−20} with θ ≈ 1.047.

$L_{0}$ (mg/L): Ultimate CBOD
$k$ (1/d): BOD rate constant (base e)
$y_{t}$ (mg/L): BOD exerted at time t

Step-by-step solved example

L0 = 250 mg/L, k = 0.23 d⁻¹ (e-base). BOD5?

First-order decay of BOD in a reactor / stream reach.

1. Exerted
y5 = 250(1 − e^{−0.23×5}) = 250(1 − e^{−1.15}) = 250(1 − 0.317) = 171 mg/L.
$y_{5} = 171 mg/L$

Answer: 171 mg/L

10 practice questions

0/10 correct

1.If k → ∞, BOD5 approaches

2.Remaining ultimate BOD after time t is

3.Typical municipal untreated wastewater BOD5 is on the order of

4.COD is typically

5.DO saturation in fresh water at 20°C is about

6.Streeter–Phelps models

7.k_20 = 0.20 d⁻¹, θ = 1.047, T = 30°C. k_30 ≈

8.NBOD is associated with

9.L0 = 200, k = 0.1 d⁻¹. Remaining at t = 5 d is

10.A BOD bottle with dilution still computes y using