Time of critical DO deficit, 50 l/s city

The critical point is where the oxygen deficit is maximum (Streeter-Phelps oxygen sag). The river DO is 80% of saturation, so $DO_r = 0.8\times 9.17 = 7.336$ mg/l. Rates are base 10: $K=0.1$/day, $R=0.5$/day.

Step 1: Conditions just after mixing

$$\begin{aligned} Q_{mix} &= Q_s+Q_r = 50 + 500 = 550\ l/s\\ BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{50\times 400 + 500\times 4}{550} = 40.00\ \text{mg/l}\\ DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{50\times 0 + 500\times 7.336}{550} = 6.67\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17 - 6.67 = 2.50\ \text{mg/l} \end{aligned}$$

Step 2: Ultimate BOD and self-purification constant

$$\begin{aligned} L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{40.00}{1-10^{-5\times 0.1}} = 58.50\ \text{mg/l}\\ f &= \frac{R}{K} = \frac{0.5}{0.1} = 5.00 \end{aligned}$$

Step 3: Critical time and critical deficit

$$\begin{aligned} t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\ &= \frac{1}{0.1(5.00-1)}\log_{10}\left[5.00\left\{1-(5.00-1)\frac{2.50}{58.50}\right\}\right] = 1.544\ \text{days}\\ D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{58.50}{5.00}\times 10^{-0.1\times 1.544} = 8.20\ \text{mg/l} \end{aligned}$$

Answer: critical time $t_c \approx 1.54$ days (about 74.1 km downstream at 48 km/day), with critical deficit $D_c = 8.20$ mg/l, so the lowest DO is $9.17-8.20=0.97$ mg/l.