How is the BOD of the combined waste (domestic + dairy + industry) found?

Total BOD load (kg/d) divided by total flow (m3/d), then converted to mg/l

Domestic BOD plus industrial BOD concentrations added directly

Average of the three BOD concentrations

Highest of the three BOD concentrations

What is the maximum allowed critical deficit Dc in the river?

DO_sat - DO_min = 8.0 - 4.5 = 3.5 mg/l

Why is the initial deficit D0 after mixing only about 0.11 mg/l?

Because the waste has zero DO but its flow (0.107 m3/s) is tiny against the river's 8 m3/s, so mixed DO is about 7.89 mg/l

Because saturation DO is assumed zero

2071 Magh · 8 marks

Degree of treatment, town with dairy and industries

Short answer

Combined waste BOD5=1779.5 mg/l, flow=0.107 m³/s. With river (8 m³/s, DO=8 mg/l), permissible mixture ultimate BOD=22.1 mg/l. Effluent BOD5 must be reduced to 1145 mg/l, requiring 35.6% BOD removal.

The population of a town is 30,000 and domestic sewage is 175 lpcd. The per capita BOD is 50 gm/day. The dairy waste of the town is 2.2 × 10⁶ liters/day with BOD of 5000 mg/l and the waste from other industries is 1.80 × 10⁶ liters/day with BOD of 2200 mg/l. DO of both domestic and industrial wastes are zero. The effluent from the sewage treatment plant is to be discharged in the natural river having minimum discharge of 8000 liters/sec, a dissolved oxygen content of 8.0 mg/l and BOD of zero. The minimum DO content in the river to be maintained is 4.5 mg/l. Determine the degree of treatment required to the sewage. Assume saturation DO in the river after mixing with waste is equal to DO content of river before mixing. Assume any other data not given.

Step 1: Combined waste

Source	Flow (m³/d)	BOD load (kg/d)
Domestic ( $30000\times 175$ l)	5250	1500 ( $30000\times 0.05$ )
Dairy	2200	11000
Other industries	1800	3960
Total	9250	16460

Q_s=\frac{9250}{86400}=0.1071\ \text{m}^3/\text{s},\qquad BOD_5 = \frac{16460\times 1000}{9250} = 1779.5\ \text{mg/l}

Step 2: Assumptions

River flow $Q_r=8000$ l/s $=8$ m³/s, $DO_r=8.0$ mg/l, BOD $=0$ . As stated, $DO_{sat}$ after mixing $=8.0$ mg/l. I assume the usual constants at 20°C: $K=0.1$ /day and $R=0.4$ /day (base 10), $f=4$ .

Step 3: Mixing and permissible deficit

\begin{aligned} DO_{mix} &= \frac{0.1071\times 0 + 8.0\times 8.0}{8.107} = 7.89\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 8.0-7.89 = 0.11\ \text{mg/l}\\ D_{c(allowed)} &= DO_{sat}-DO_{min} = 8.0-4.5 = 3.50\ \text{mg/l} \end{aligned}

Step 4: 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 3.50 mg/l. Solving these two equations for $L_m$ by trial (with $D_0=0.11$ fixed) gives $L_m=22.12$ mg/l, with $t_c=1.986$ days (check: $D_c=3.50$ mg/l).

Step 5: Permissible BOD of the effluent

\begin{aligned} L_{raw} &= \frac{BOD_5}{1-10^{-5K}} = \frac{1779.5}{0.6838} = 2602.42\ \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{22.12\times 8.107 - 8.0\times 0.00}{0.1071} = 1674.79\ \text{mg/l} \end{aligned}

So the treated effluent can have $BOD_5 \approx 1145.18$ mg/l.

\text{Degree of treatment} = \frac{2602.42-1674.79}{2602.42}\times 100 = 35.6\%

Answer: the combined waste BOD5 must be reduced from 1779.5 mg/l to about 1145.18 mg/l, i.e. a treatment efficiency of about 35.6%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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