Streeter–Phelps Oxygen Sag

D = \frac{k _{1} L _{a}}{k _{2} - k _{1}} (e^{- k_{1} t} - e^{- k_{2} t}) + D_{0} e^{- k_{2} t}

DO deficit D=DOsat−DO. k1 deoxygenation (BOD), k2 reaeration. Critical time when dD/dt=0. Base-e rates; if given as base-10, k=2.303 k10.

Step-by-step solved example

k1=0.3 /d, k2=0.6 /d, La=20, D0=2 mg/L. D at t=1 d?

DO sag: deficit peaks at the critical time then recovers.

Answer: D ≈ 4.9 mg/L

0/10 correct

1.Larger k2 (faster reaeration) makes the sag

2.DOsat at 20°C fresh water is about

3.If k2=k1 the formula needs

4.Critical deficit occurs when

5.BOD remaining L=La e^{−k1 t}

6.Temperature raises k roughly

7.A tributary mix: La after mix is

8.DO = DOsat − D so a D of 4.9 with sat 9.1 gives DO

9.k given as log10 days^{−1} must be

10.Initial deficit D0=0 means