---
title: "Degree of treatment, town with dairy and industries"
subject: "Sanitary Engineering"
chapter: 6
key: "semester_6/sanitary_engineering:degree-treatment-city-industries-71"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/degree-treatment-city-industries-71
site: BCT Engineering
---

# Degree of treatment, town with dairy and industries

Sanitary Engineering · Chapter 6: Wastewater Disposal

- Asked in: 2071 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/degree-treatment-city-industries-71
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6

## Question

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.

## 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.

## Answer

#### 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%.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the BOD of the combined waste (domestic + dairy + industry) found?
   A. Domestic BOD plus industrial BOD concentrations added directly
   B. Average of the three BOD concentrations
   C. Highest of the three BOD concentrations
   D. Total BOD load (kg/d) divided by total flow (m3/d), then converted to mg/l
   Answer: D. Concentrations cannot be averaged; loads are added and divided by the total flow, giving 16460 kg/d over 9250 m3/d = 1779.5 mg/l. (Key point: Combined waste BOD)

2. What is the maximum allowed critical deficit Dc in the river?
   A. DO_sat - DO_min = 8.0 - 4.5 = 3.5 mg/l
   B. DO_min = 4.5 mg/l
   C. River DO - waste DO = 8.0 mg/l
   D. DO_sat + DO_min = 12.5 mg/l
   Answer: A. The critical deficit is the gap between saturation DO and the lowest DO allowed: 8.0 - 4.5 = 3.5 mg/l. (Key point: Permissible deficit)

3. Why is the initial deficit D0 after mixing only about 0.11 mg/l?
   A. 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
   B. Because saturation DO is assumed zero
   C. Because the river BOD is large
   D. Because the waste DO is high
   Answer: A. Flow-weighted DO is 8.0x8.0/8.107 = 7.89 mg/l, so D0 = 8.0 - 7.89 = 0.11 mg/l. (Key point: Initial deficit)

4. How is the degree of treatment obtained once the permissible effluent ultimate BOD Ls is known?
   A. (Lm - Ls)/Lm x 100
   B. (L_raw - Ls)/L_raw x 100
   C. Ls/L_raw x 100
   D. (Q_r/Q_s) x 100
   Answer: B. Efficiency is the fraction of the raw ultimate BOD that must be removed: (2602.42 - 1674.79)/2602.42 = 35.6%. (Key point: Degree of treatment)

## Related in Sanitary Engineering

- [Degree of treatment, domestic + industrial](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Adegree-treatment-town-40000) — Sanitary Engineering, chapter 6
- [Conventional ASP design, population 50,000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aasp-50000-80lpcd) — Sanitary Engineering, chapter 7, asked 2×
- [1-day BOD 300 mg/l to 5-day at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-1day-300-to-5day-30c) — Sanitary Engineering, chapter 5
- [Degree of treatment, city 1200 l/s](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Adegree-treatment-city-1200ls) — Sanitary Engineering, chapter 6
- [Degree of treatment, industrial wastewater](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Adegree-treatment-industrial-waste) — Sanitary Engineering, chapter 6
- [Percentage purification, town of 50000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Apercent-purification-town-50000) — Sanitary Engineering, chapter 6

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