Degree of treatment, industrial wastewater

Convert flows to m³/s: $Q_s = 2\times 10^6/(86400\times 1000) = 0.0231$ m³/s and $Q_r = 5.0$ m³/s. The river is at about 20°C, so I assume the standard constants $K=0.1$/day and $R=0.4$/day (base 10), $f=4$. Saturation DO after mixing is 9.2 mg/l, and the minimum DO to be kept is 4 mg/l.

Step 1: Mixing and permissible deficit

$$\begin{aligned} DO_{mix} &= \frac{0.0231\times 0 + 5.0\times 7.4}{5.023} = 7.37\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.2-7.37 = 1.83\ \text{mg/l}\\ D_{c(allowed)} &= DO_{sat}-DO_{min} = 9.2-4.0 = 5.20\ \text{mg/l} \end{aligned}$$

Step 2: Permissible BOD of the mixture

With $K=0.1$ and $R=0.4$ (base 10), $f=R/K=4.00$. The critical deficit is

$$t_c=\frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_m}\right\}\right],\qquad D_c = \frac{L_m}{f}10^{-Kt_c}$$

and $D_c$ must not exceed 5.20 mg/l. Solving these two equations for $L_m$ by trial (with $D_0=1.83$ fixed) gives $L_m=30.93$ mg/l, with $t_c=1.723$ days (check: $D_c=5.20$ mg/l).

Step 3: Permissible BOD of the effluent

$$\begin{aligned} L_{raw} &= \frac{BOD_5}{1-10^{-5K}} = \frac{5000}{0.6838} = 7312.38\ \text{mg/l}\\ L_r &= 0.00\ \text{mg/l (river)}\\ L_s &= \frac{L_m(Q_s+Q_r)-Q_rL_r}{Q_s} = \frac{30.93\times 5.023 - 5.0\times 0.00}{0.0231} = 6712.02\ \text{mg/l} \end{aligned}$$

So the treated effluent can have $BOD_5 \approx 4589.49$ mg/l.

$$\text{Degree of treatment} = \frac{7312.38-6712.02}{7312.38}\times 100 = 8.2\%$$

Answer: the effluent BOD5 must be brought from 5000 mg/l down to about 4589.49 mg/l, i.e. about 8.2% treatment (BOD removal) is required. The required degree is small only because the river dilution is very large (5 m³/s against 0.023 m³/s of waste); if the river flow were lower, much higher treatment would be needed.