---
title: "Critical DO location, industry 2.0 m³/s"
subject: "Sanitary Engineering"
chapter: 6
key: "semester_6/sanitary_engineering:critical-do-location-industry-2-20"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/critical-do-location-industry-2-20
site: BCT Engineering
---

# Critical DO location, industry 2.0 m³/s

Sanitary Engineering · Chapter 6: Wastewater Disposal

- Asked in: 2075 Bhadra, 8 marks
- Video: [oxygen deficit problem | DO deficit solved problem | oxygen deficit in downstream side of river](https://www.youtube.com/watch?v=4wst5Ot1YIg) — Civil Engineering Exam
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/critical-do-location-industry-2-20
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6

## Question

An industry is going to be established in a rural municipality near the river side. The river water and industrial effluent characteristics are as follows:

| Description | Industrial Effluent | River Water |
|---|---|---|
| Flow (m³/sec) | 2.0 | 20.0 |
| Dissolved Oxygen (mg/l) | 0.8 | 7.14 |
| BOD5 at 20°C (mg/l) | 350 | 3.0 |

De-oxygenation rate constant at 20°C (per day, base 10) = 0.1; Re-oxygenation rate constant at 20°C (per day, base 10) = 0.3; Saturation DO at 20°C (mg/l) = 9.17.
At what location in the river, the critical DO deficit would occur if the flow velocity in the river is 2 km/hr?

## Short answer

After mixing, initial deficit 2.61 mg/L, ultimate BOD 50.52 mg/L; f=3.00, K=0.1/d (base 10). Critical time 2.149 days, velocity 48 km/d. Critical deficit 10.27 mg/L occurs at 103.2 km downstream. Deficit exceeds saturation DO (9.17 mg/L), so DO reaches zero.

## Answer

Rates are base 10: $K=0.1$/day, $R=0.3$/day. The velocity is 2 km/h $=48$ km/day.
#### Step 1: Conditions just after mixing

$$
\begin{aligned}
Q_{mix} &= Q_s+Q_r = 2.0 + 20.0 = 22\ m^3/s\\
BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{2.0\times 350 + 20.0\times 3.0}{22} = 34.55\ \text{mg/l}\\
DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{2.0\times 0.8 + 20.0\times 7.140}{22} = 6.56\ \text{mg/l}\\
D_0 &= DO_{sat}-DO_{mix} = 9.17 - 6.56 = 2.61\ \text{mg/l}
\end{aligned}
$$

#### Step 2: Ultimate BOD and self-purification constant

$$
\begin{aligned}
L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{34.55}{1-10^{-5\times 0.1}} = 50.52\ \text{mg/l}\\
f &= \frac{R}{K} = \frac{0.3}{0.1} = 3.00
\end{aligned}
$$

#### Step 3: Critical time and critical deficit

$$
\begin{aligned}
t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\
&= \frac{1}{0.1(3.00-1)}\log_{10}\left[3.00\left\{1-(3.00-1)\frac{2.61}{50.52}\right\}\right] = 2.149\ \text{days}\\
D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{50.52}{3.00}\times 10^{-0.1\times 2.149} = 10.27\ \text{mg/l}
\end{aligned}
$$
#### Location of the critical deficit

$$
x_c = v\,t_c = 48\ \text{km/day}\times 2.149\ \text{days} = 103.2\ \text{km}
$$

**Answer: the critical DO deficit ($D_c=10.27$ mg/l) occurs about $103.2$ km downstream of the industry, after $2.149$ days.** The computed deficit is larger than the saturation DO (9.17 mg/l), so in reality the DO would drop to zero near this point and the effluent needs treatment.

## Check yourself

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

1. What is 2 km/hr in km/day?
   A. 2 km/day
   B. 48 km/day
   C. 24 km/day
   D. 86.4 km/day
   Answer: B. 2 km/h x 24 h = 48 km/day, which matches the day-based rate constants. (Key point: Velocity conversion)

2. What is the initial deficit D0?
   A. 7.14 - 0.8
   B. 9.17 - 0.8
   C. 9.17 - 6.56 = 2.61 mg/L
   D. 0
   Answer: C. DO_mix = (2 x 0.8 + 20 x 7.14)/22 = 6.56, so D0 = 2.61 mg/L. (Key point: Mixture values)

3. Which formula gives tc?
   A. tc = L0/D0
   B. tc = f/K
   C. tc = 1/(K f)
   D. tc = 1/(K(f-1)) log10[f(1 - (f-1)D0/L0)]
   Answer: D. With f = 3, K = 0.1, D0 = 2.61 and L0 = 50.52 this gives tc = 2.149 days. (Key point: Critical time formula)

4. Where does the critical deficit occur?
   A. 103.2 km downstream (48 x 2.149)
   B. 2.149 km
   C. 48 km
   D. 20 km
   Answer: A. x = v tc = 48 km/day x 2.149 days = 103.2 km. (Key point: Location)

## Related in Sanitary Engineering

- [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 oxygen deficit and location (table 2)](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Acritical-do-effluent-stream-table-71) — Sanitary Engineering, chapter 6
- [Critical oxygen deficit and location (table)](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Acritical-do-effluent-stream-table-78) — Sanitary Engineering, chapter 6
- [Is treatment necessary? Industry near river](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atreatment-industry-river-1p72) — 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 deficit, location, sag curve](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asag-curve-effluent-1p5-stream-7p5) — Sanitary Engineering, chapter 6

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- [Demand of rural municipality 2018, 15,000 pop](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ademand-c74) — Water Supply Engineering, chapter 3
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- [Critical DO deficit and location downstream](https://www.bctengineer.com/question.md?key=semester_7%2Fenvironment_and_pollution_control%3Aoxygen-sag-critical-deficit-numerical) — Environment and Pollution Control, chapter 4