River flow required for 1500 population

Assumptions

The river upstream is saturated with oxygen (DO = 8.38 ppm) and has no BOD; the sewage has zero DO. The constants are in base 10 and refer to 25°C. Let $Q_r$ be the river flow required (m³/day).

Step 1: Sewage and its ultimate BOD

$$Q_s=1500\times200\ \text{L/d}=300\ \text{m}^3/\text{day}$$ $$L_s=\frac{BOD_5}{1-10^{-K_1\times5}}=\frac{350}{1-10^{-0.15\times5}}=\frac{350}{1-0.1778}=425.7\ \text{ppm}$$

Step 2: Allowable critical deficit

Minimum DO to be kept = 4 ppm, so the maximum (critical) deficit is

$$D_c=8.38-4=4.38\ \text{ppm}$$

Step 3: Mixture (let $x=\dfrac{Q_s}{Q_s+Q_r}$)

$$L_0=x\,L_s=425.7\,x,\qquad D_0=x\,(8.38-0)=8.38\,x$$

Self-purification factor $f=\dfrac{K_2}{K_1}=\dfrac{0.27}{0.15}=1.8$.

Critical time:

$$t_c=\frac{1}{K_2-K_1}\log_{10}\left[\frac{K_2}{K_1}\left(1-\frac{(f-1)D_0}{L_0}\right)\right]$$

Since $D_0/L_0=8.38/425.7=0.01969$ is independent of $x$,

$$t_c=\frac{1}{0.12}\log_{10}\left[1.8\left(1-0.8\times0.01969\right)\right]=2.07\ \text{days}$$

Critical deficit: $D_c=\dfrac{L_0}{f}10^{-K_1t_c}$

$$4.38=\frac{425.7\,x}{1.8}\times10^{-0.15\times2.07}=115.71\,x\ \Rightarrow\ x=0.03785$$

Step 4: River flow

$$Q_r=Q_s\left(\frac{1}{x}-1\right)=300\left(\frac{1}{0.03785}-1\right)=7625\ \text{m}^3/\text{day}$$

Answer: Minimum river flow = about 7625 m³/day (= 0.0883 m³/s or 88.3 l/s) for the 1500 people, i.e. a dilution of about 25.4 times the sewage flow.