Degree of treatment, domestic + industrial

Step 1: Combined sewage

$$Q_s=\frac{11000}{86400}=0.1273\ \text{m}^3/\text{s},\qquad BOD_5=\frac{18000\times 1000}{11000}=1636.4\ \text{mg/l}$$

Step 2: Assumptions

River flow $Q_r=8.5$ m³/s, $DO_r=8$ mg/l, BOD $=0$. Assumed (not given): temperature 20°C so $DO_{sat}=9.17$ mg/l, and the minimum DO to be kept in the river is 4 mg/l (needed for fish life). The rate constants are taken as base 10, $K=R=0.1$/day, so $f=1$ and the critical time is $t_c=\dfrac{L_m-D_0}{2.303\,K L_m}$ with $D_c=(2.303K L_m t_c+D_0)10^{-Kt_c}$.

Step 3: Mixing and permissible deficit

$$\begin{aligned} DO_{mix} &= \frac{0.1273\times 0 + 8.5\times 8.0}{8.627} = 7.88\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17-7.88 = 1.29\ \text{mg/l}\\ D_{c(allowed)} &= DO_{sat}-DO_{min} = 9.17-4.0 = 5.17\ \text{mg/l} \end{aligned}$$

Step 4: Permissible BOD of the mixture

With $K=0.1$ and $R=0.1$ (base 10), $f=R/K=1.00$. The critical deficit is

$$t_c=\frac{L_m-D_0}{2.303\,K L_m}\ (f=1),\qquad D_c = (2.303K L_m t_c + D_0)\,10^{-Kt_c}$$

and $D_c$ must not exceed 5.17 mg/l. Solving these two equations for $L_m$ by trial (with $D_0=1.29$ fixed) gives $L_m=12.70$ mg/l, with $t_c=3.902$ days (check: $D_c=5.17$ mg/l).

Step 5: Permissible BOD of the effluent

$$\begin{aligned} L_{raw} &= \frac{BOD_5}{1-10^{-5K}} = \frac{1636.4}{0.6838} = 2393.14\ \text{mg/l}\\ L_r &= 0.00\ \text{mg/l (river)}\\ L_s &= \frac{L_m(Q_s+Q_r)-Q_rL_r}{Q_s} = \frac{12.70\times 8.627 - 8.5\times 0.00}{0.1273} = 860.45\ \text{mg/l} \end{aligned}$$

So the treated effluent can have $BOD_5 \approx 588.35$ mg/l.

$$\text{Degree of treatment} = \frac{2393.14-860.45}{2393.14}\times 100 = 64.0\%$$

Answer: the sewage BOD5 must be reduced from 1636.4 mg/l to about 588.35 mg/l, i.e. the degree of treatment required is about 64.0%. This needs a secondary biological plant (for example ASP, with the industrial waste pre-treated).