Critical DO location, industry 2.0 m³/s

Rates are base 10: $K=0.1$/day, $R=0.3$/day. The velocity is 2 km/h $=48$ km/day.

Step 1: Conditions just after mixing

$$\begin{aligned} Q_{mix} &= Q_s+Q_r = 2.0 + 20.0 = 22\ m^3/s\\ BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{2.0\times 350 + 20.0\times 3.0}{22} = 34.55\ \text{mg/l}\\ DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{2.0\times 0.8 + 20.0\times 7.140}{22} = 6.56\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17 - 6.56 = 2.61\ \text{mg/l} \end{aligned}$$

Step 2: Ultimate BOD and self-purification constant

$$\begin{aligned} L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{34.55}{1-10^{-5\times 0.1}} = 50.52\ \text{mg/l}\\ f &= \frac{R}{K} = \frac{0.3}{0.1} = 3.00 \end{aligned}$$

Step 3: Critical time and critical deficit

$$\begin{aligned} t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\ &= \frac{1}{0.1(3.00-1)}\log_{10}\left[3.00\left\{1-(3.00-1)\frac{2.61}{50.52}\right\}\right] = 2.149\ \text{days}\\ D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{50.52}{3.00}\times 10^{-0.1\times 2.149} = 10.27\ \text{mg/l} \end{aligned}$$

Location of the critical deficit

$$x_c = v\,t_c = 48\ \text{km/day}\times 2.149\ \text{days} = 103.2\ \text{km}$$

Answer: the critical DO deficit ($D_c=10.27$ mg/l) occurs about $103.2$ km downstream of the industry, after $2.149$ days. The computed deficit is larger than the saturation DO (9.17 mg/l), so in reality the DO would drop to zero near this point and the effluent needs treatment.