How do I find the minimum dissolved oxygen downstream of a discharge with the Streeter-Phelps model?

The model tracks the oxygen deficit D, the difference between saturation and the actual dissolved oxygen. Two processes compete: bacteria consuming the BOD remove oxygen at a rate k_d × L, and the atmosphere replaces it at a rate k_r × D. The solution is D(t) = (k_d × L_0 / (k_r - k_d)) × (e^(-k_d × t) - e^(-k_r × t)) + D_0 × e^(-k_r × t), where t is the travel time below the discharge.

Your starting values are L_0 = 20 mg/L and D_0 = 9.1 - 8.1 = 1.0 mg/L. The deficit first grows, because demand exceeds reaeration, reaches a maximum at the critical point, and then shrinks as the BOD is used up. Plotted as dissolved oxygen against distance, that is the sag curve.

The critical time comes from setting the derivative to zero: t_c = (1 / (k_r - k_d)) × ln((k_r / k_d) × (1 - D_0 × (k_r - k_d) / (k_d × L_0))). With your numbers: (1 / 0.30) × ln(2 × (1 - 1.0 × 0.30 / (0.30 × 20))) = 3.33 × ln(2 × 0.95) = 3.33 × 0.642 = 2.14 days.

At that point demand and reaeration balance, so D_c = (k_d / k_r) × L_0 × e^(-k_d × t_c) = 0.5 × 20 × e^(-0.642) = 10 × 0.526 = 5.26 mg/L. The minimum dissolved oxygen is 9.1 - 5.26 = about 3.8 mg/L.

Distance is velocity times travel time: 0.3 m/s × 86,400 s/day × 2.14 days = about 55 km downstream. That is worth noting in itself. The worst oxygen conditions are not at the outfall but tens of kilometres away, possibly past a tributary, a weir or another discharge that the simple model does not know about.

A minimum of 3.8 mg/L is below the 5 mg/L often used as a target for healthy fish populations, although the applicable limit depends on the water body's classification. To raise it you can reduce the BOD load, since the deficit is nearly proportional to L_0, or raise the effluent's own oxygen, which has a much smaller effect.

Where the inputs come from matters more than the algebra. L_0 and D_0 are flow-weighted mixtures of river and effluent: L_0 = (Q_r × L_r + Q_w × L_w) / (Q_r + Q_w), and the same form for dissolved oxygen and temperature. Use ultimate BOD, not BOD5. The design case is the low-flow condition in warm weather, when dilution is least, saturation is lowest and the bacteria are most active.

Both constants need temperature correction: k_T = k_20 × θ^(T - 20), with θ about 1.047 for deoxygenation and about 1.024 for reaeration. Reaeration also depends strongly on the river itself. Shallow, fast streams have high k_r and deep, slow rivers have low values, and empirical formulas estimate it from velocity and depth.

As for reliability, treat the result as a first estimate. The classical model includes only carbonaceous BOD and atmospheric reaeration. It leaves out nitrification of ammonia, which can consume as much oxygen as the carbonaceous demand in some effluents, oxygen demand from bottom sediments, and photosynthesis and respiration of algae and plants, which make the oxygen swing between day and night. It also assumes steady flow and complete mixing across the river at the outfall, which can take a long distance in a wide river.

For a permit or a design decision, the constants should be calibrated against measured oxygen profiles along the river, and a water quality model that includes those extra terms is normally used. The hand calculation remains a good check on its output.