Streeter-Phelps DO Sag Calculator

Understanding the Streeter-Phelps DO sag calculator — concept and principle

When organic waste enters a river, bacteria consume dissolved oxygen (DO) as they break it down, while the water surface absorbs oxygen from the air. The result is a dip in DO downstream, the oxygen sag curve. This calculator uses the Streeter-Phelps model to find the deficit and DO after a given travel time, plus the time and level of the lowest DO.

Early on, BOD decay outpaces reaeration; past the critical point the air wins.

How it is calculated

Deficit: D(t) = kd·L₀/(ka − kd) · (e^(−kd·t) − e^(−ka·t)) + D₀·e^(−ka·t).
DO: C(t) = Cs − D(t).
Critical time: tc = 1/(ka − kd) · ln[(ka/kd)(1 − D₀(ka − kd)/(kd·L₀))].
Minimum DO: Cs − Dc, with Dc = (kd/ka)·L₀·e^(−kd·tc).

L₀ is the ultimate BOD just below the outfall, D₀ the initial deficit, kd and ka the base-e deoxygenation and reaeration rates (per day), Cs the saturation DO and t the travel time in days.

The principle

The model treats the river as plug flow with two first-order processes. BOD decays as L = L₀e^(−kd·t) and consumes oxygen at a rate kd·L. Reaeration adds oxygen in proportion to the deficit, ka·D. The deficit therefore obeys dD/dt = kd·L − ka·D, and solving this linear equation gives D(t). Early on, oxygen use outpaces reaeration and the deficit grows; as BOD is exhausted and the deficit rises, reaeration catches up. The lowest DO occurs where dD/dt = 0, which gives tc and the relation Dc = (kd/ka)·L.

Worked example

Inputs: L₀ = 15 mg/L, D₀ = 1 mg/L, kd = 0.35 /d, ka = 0.7 /d, Cs = 9.1 mg/L, t = 1 d.
kd·L₀/(ka − kd) = 5.25 / 0.35 = 15; e^(−0.35) = 0.70469, e^(−0.7) = 0.49659.
D(1) = 15 × (0.70469 − 0.49659) + 1 × 0.49659 = 3.6181 mg/L; DO = 9.1 − 3.6181 = 5.4819 mg/L.
tc = (1/0.35) × ln[2 × (1 − 0.35/5.25)] = ln(1.86667)/0.35 = 1.7833 d.
Dc = 0.5 × 15 × e^(−0.35 × 1.7833) = 7.5 × 0.53571 = 4.0179; minimum DO = 5.0821 mg/L.

Each point is the calculator re-run with only “Initial BOD L₀ (mg/L)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Initial BOD L₀ (mg/L)	Deficit at t (mg/L)
3.75	1.276970751
7.5	2.057356198
11.2	2.837741645
15	3.618127093
18.8	4.39851254
22.5	5.178897987
30	6.739668882

How to read the result

Multiply tc by the river velocity to find where along the river the worst point lies. Many aquatic-life criteria for warm-water fish sit around 5 mg/L, and water below about 2 mg/L is generally called hypoxic. Here the minimum is just above 5 mg/L, so a little more load or warmer water could push it below. Raising ka (shallow, fast, turbulent streams) or lowering L₀ (better treatment) lifts the minimum.

Common mistakes

Mixing base-10 and base-e rate constants.
Using 20 °C rates and saturation values for a summer river without temperature correction.
Entering 5-day BOD as L₀ instead of ultimate BOD.
Trusting a negative critical time; the lowest DO is then at the outfall (ka = kd also needs a separate formula).

Keep in mind: Streeter-Phelps ignores sediment oxygen demand, nitrification, algae, tributaries and dispersion. Regulatory studies use calibrated water-quality models and field data.

Frequently asked questions

What is the difference between deficit and DO?

Deficit is how far DO is below saturation, D = Cs − C. Saturation itself falls as water warms.

How do I convert BOD₅ to ultimate BOD?

For first-order decay, L₀ = BOD₅ / (1 − e^(−5k)), using the laboratory rate k.

Sources: H. W. Streeter and E. B. Phelps, "A Study of the Pollution and Natural Purification of the Ohio River", U.S. Public Health Service Bulletin 146, 1925 · S. C. Chapra, Surface Water-Quality Modeling, McGraw-Hill

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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