---
title: "Degree of treatment, domestic + industrial"
subject: "Sanitary Engineering"
chapter: 6
key: "semester_6/sanitary_engineering:degree-treatment-town-40000"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/degree-treatment-town-40000
site: BCT Engineering
---

# Degree of treatment, domestic + industrial

Sanitary Engineering · Chapter 6: Wastewater Disposal

- Asked in: 2068 Bhadra (old course), 10 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-town-40000
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6

## Question

The sewage of a town is a mixture of domestic sewage and industrial sewage. The sewage is to be treated at the sewage treatment plant before discharging into river. Determine the degree of treatment required for the following data: Population = 40000; Domestic sewage = 175 lpcd; BOD of domestic sewage = 50 gm/capita/day; Flow of industrial waste = 4 × 10⁶ liters/day; BOD of industrial waste = 4000 mg/l; DO of both domestic and industrial sewage = 0; River discharge = 8500 liters/sec; BOD of river water = 0; DO of river water = 8 mg/l; k1 = 0.1/day and k2 = 0.1/day. Assume other data as required.

## Short answer

The combined sewage BOD5 is 1636.4 mg/l. To keep river DO above 4 mg/l, the treated effluent BOD5 must be reduced to 588.35 mg/l, requiring about 64% BOD removal.

## Answer

#### Step 1: Combined sewage

| Source | Flow (m³/d) | BOD load (kg/d) |
|---|---|---|
| Domestic ($40000\times 175$ l) | 7000 | 2000 ($40000\times 0.05$) |
| Industrial | 4000 | 16000 |
| **Total** | **11000** | **18000** |

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

#### 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).

## Check yourself

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

1. What combined sewage BOD5 results from domestic (2000 kg/d, 7000 m3/d) and industrial (16000 kg/d, 4000 m3/d) waste?
   A. Average of 285.7 and 4000 mg/l = 2143 mg/l
   B. 4000 mg/l, the highest value
   C. 2000 + 16000 = 18000 mg/l
   D. 18000 kg/d / 11000 m3/d = 1636.4 mg/l
   Answer: D. Total load divided by total flow gives 1636.4 mg/l. (Key point: Combined BOD)

2. What does K = R (f = 1) change in the oxygen sag calculation?
   A. BOD5 equals the ultimate BOD
   B. The river never reaches a critical deficit
   C. The critical time uses the special form t_c = (Lm - D0)/(2.303 K Lm)
   D. Deoxygenation stops
   Answer: C. When f = 1 the general log formula is indeterminate, so the special-case expression for t_c and Dc is used. (Key point: Effect of f = 1)

3. With assumed DO_sat = 9.17 mg/l and minimum DO 4 mg/l, what is the allowed Dc?
   A. 5.17 mg/l
   B. 4.0 mg/l
   C. 13.17 mg/l
   D. 1.29 mg/l
   Answer: A. Dc allowed = 9.17 - 4.0 = 5.17 mg/l. (Key point: Permissible deficit)

4. How does the calculation reach the 64% degree of treatment?
   A. Lm is found by trial so Dc = 5.17, then Ls is back-calculated from the mixture balance and compared with raw ultimate BOD
   B. By dividing industrial BOD by domestic BOD
   C. Directly from the river DO
   D. By assuming 85% removal in primary treatment
   Answer: A. Permissible Lm gives permissible effluent Ls (860 mg/l) which is compared with raw L (2393 mg/l): 64.0%. (Key point: Final efficiency)

## Related in Sanitary Engineering

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- [River flow required for 1500 population](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ariver-flow-for-1500-population) — 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
- [Is treatment necessary? Town wastewater 300 l/s](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atreatment-needed-town-300-ls) — Sanitary Engineering, chapter 6
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