Critical DO deficit, city 100 l/s

River DO $=0.7\times 9.17=6.419$ mg/l. Rates are base 10: $K_1=0.1$/day, $K_2=0.5$/day, so $f=5$. Flows are in l/s.

Step 1: Conditions just after mixing

$$\begin{aligned} Q_{mix} &= Q_s+Q_r = 100 + 1000 = 1100\ l/s\\ BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{100\times 450 + 1000\times 4}{1100} = 44.55\ \text{mg/l}\\ DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{100\times 0 + 1000\times 6.419}{1100} = 5.84\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17 - 5.84 = 3.33\ \text{mg/l} \end{aligned}$$

Step 2: Ultimate BOD and self-purification constant

$$\begin{aligned} L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{44.55}{1-10^{-5\times 0.1}} = 65.15\ \text{mg/l}\\ f &= \frac{R}{K} = \frac{0.5}{0.1} = 5.00 \end{aligned}$$

Step 3: Critical time and critical deficit

$$\begin{aligned} t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\ &= \frac{1}{0.1(5.00-1)}\log_{10}\left[5.00\left\{1-(5.00-1)\frac{3.33}{65.15}\right\}\right] = 1.499\ \text{days}\\ D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{65.15}{5.00}\times 10^{-0.1\times 1.499} = 9.23\ \text{mg/l} \end{aligned}$$

Answer: critical DO deficit $D_c=9.23$ mg/l (at $t_c=1.499$ days, about 89.9 km downstream at 60 km/day). The lowest DO is $9.17-9.23=-0.06$ mg/l.

Note: the computed deficit is larger than the saturation DO (9.17 mg/l). In reality DO cannot fall below zero, so the stream would go anaerobic near the critical point and the effluent needs treatment before discharge. The negative value is simply what the formula gives for these data.