---
title: "Is treatment necessary? Industry near river"
subject: "Sanitary Engineering"
chapter: 6
key: "semester_6/sanitary_engineering:treatment-industry-river-1p72"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/treatment-industry-river-1p72
site: BCT Engineering
---

# Is treatment necessary? Industry near river

Sanitary Engineering · Chapter 6: Wastewater Disposal

- Asked in: 2076 Bhadra, 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/treatment-industry-river-1p72
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6

## Question

An industry is proposed to be established near the river. The characteristics of raw sewage and the characteristics of river in which the sewage is discharged are as follows: a) Sewage discharge of a town = 1.72 m³/s; b) BOD5 at 20°C of sewage = 225 mg/l; c) Minimum river discharge = 7.24 m³/s; d) BOD5 at 20° of river water = 1.2 mg/l; e) Temperature of the river after mixing = 20°C; f) De-oxygenation constant = 0.1/day at 20°C; g) Re-oxygenation constant = 0.5/day at 20°C; h) Initial DO deficit just at mixing = 2.16 mg/l; i) Critical DO deficit = 4.77 mg/l.
Is treatment necessary? If so, what will be the percentage reduction of BOD of sewage?

## Short answer

Initial deficit 2.16 mg/L, ultimate BOD 64.59 mg/L gives critical deficit 8.95 mg/L, exceeding permissible 4.77 mg/L, so treatment is needed. Maximum permissible mixture ultimate BOD is 33.06 mg/L; required BOD reduction of sewage is 49.9% (to 112.72 mg/L BOD5).

## Answer

Treatment is necessary if the critical deficit of the mixture exceeds the permissible value, 4.77 mg/l (the limit given). Rates are base 10: $K=0.1$, $R=0.5$, so $f=R/K=5$.

#### Step 1: Deficit with no treatment

$$
\begin{aligned}
BOD_5^{mix} &= \frac{1.72\times 225+7.24\times 1.2}{8.96} = 44.16\ \text{mg/l}\\
L_0 &= \frac{44.16}{1-10^{-0.5}} = 64.59\ \text{mg/l}\\
t_c &= \frac{1}{0.1(5-1)}\log_{10}\left[5\left\{1-4\times\frac{2.16}{64.59}\right\}\right] = 1.591\ \text{days}\\
D_c &= \frac{64.59}{5}\,10^{-0.1\times 1.591} = 8.95\ \text{mg/l}
\end{aligned}
$$

Since $D_c = 8.95 > 4.77$ mg/l, **treatment is necessary**.

#### Step 2: Maximum permissible BOD of the mixture

The ultimate BOD $L_m$ of the mixture that just gives $D_c = 4.77$ mg/l (with $D_0=2.16$) is found from $D_c=\dfrac{L_m}{f}10^{-Kt_c}$ and the $t_c$ equation by trial. Solving gives $L_m = 33.06$ mg/l (check: $t_c=1.419$ days, $D_c=4.77$ mg/l).

#### Step 3: Permissible BOD of the sewage

River $L_r = 1.2/0.6838 = 1.75$ mg/l. From $Q_sL_s + Q_rL_r = L_m(Q_s+Q_r)$:

$$
L_s = \frac{33.06\times 8.96 - 7.24\times 1.75}{1.72} = 164.85\ \text{mg/l}\quad (BOD_5 = 112.72\ \text{mg/l})
$$

Raw sewage $L_s = 225/0.6838 = 329.06$ mg/l.

$$
\text{BOD reduction} = \frac{329.06-164.85}{329.06}\times 100 = 49.9\%
$$

**Answer: treatment is necessary; BOD of the sewage must be reduced by about 49.9% (to about 112.72 mg/l BOD5).**

## Check yourself

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

1. Why is treatment judged necessary here?
   A. The sewage BOD is above 200 mg/L
   B. The untreated critical deficit 8.95 mg/L exceeds the permissible 4.77 mg/L
   C. The river flow is less than the sewage flow
   D. The initial deficit is 2.16 mg/L
   Answer: B. Treatment is needed when Dc without treatment exceeds the permissible value. (Key point: Need for treatment)

2. How is the permissible mixture L_m found?
   A. By dividing 4.77 by f
   B. By direct formula L = Dc
   C. By trial using the tc and Dc equations so that Dc = 4.77 mg/L with D0 fixed
   D. By assuming tc = 5 days
   Answer: C. tc and Dc both depend on L, so L_m = 33.06 mg/L is solved by trial. (Key point: Permissible mixture BOD)

3. How is the permissible sewage BOD obtained from L_m?
   A. Ls = L_m
   B. Ls = L_m x Qr/Qs
   C. Ls = L_m - Lr
   D. From Qs Ls + Qr Lr = L_m (Qs + Qr)
   Answer: D. The BOD loads of sewage and river add to the mixture load, so Ls = (L_m x 8.96 - 7.24 x 1.75)/1.72 = 164.85 mg/L. (Key point: Mass balance)

4. How is the percentage reduction computed?
   A. (raw Ls - permissible Ls)/raw Ls x 100 = 49.9%
   B. (permissible - raw)/permissible
   C. Dc/D0
   D. (225 - 4.77)/225
   Answer: A. Comparing ultimate BODs: (329.06 - 164.85)/329.06 = 49.9%. (Key point: Percent reduction)

## Related in Sanitary Engineering

- [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
- [Degree of treatment, town with dairy and industries](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Adegree-treatment-city-industries-71) — 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
- [Critical DO location, industry 2.0 m³/s](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Acritical-do-location-industry-2-20) — Sanitary Engineering, chapter 6
- [Critical DO location, industry 1.8 m³/s](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Acritical-do-industry-22-river-70bh) — Sanitary Engineering, chapter 6
- [Critical DO deficit, 300 l/s into river](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Acritical-do-city-300ls-48kmday) — Sanitary Engineering, chapter 6

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