---
title: "Critical DO location, industry 1.8 m³/s"
subject: "Sanitary Engineering"
chapter: 6
key: "semester_6/sanitary_engineering:critical-do-industry-22-river-70bh"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/critical-do-industry-22-river-70bh
site: BCT Engineering
---

# Critical DO location, industry 1.8 m³/s

Sanitary Engineering · Chapter 6: Wastewater Disposal

- Asked in: 2070 Magh, 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-industry-22-river-70bh
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6

## Question

An industry is going to be established in an urban area near the river side. The river water and industrial effluent characteristics are as follows:

| | Industrial effluent | River water |
|---|---|---|
| Flow (m³/s) | 1.8 | 22 |
| DO (mg/l) | 0 | 8.7 |
| BOD5, 20° | 350 | 6.0 |

kd, 20° = 0.25 d⁻¹; kr, 20° = 0.11 d⁻¹; DO saturation = 9.1 mg/l.
At what location in the river critical DO deficit would occur if the flow velocity in the river is 0.20 m/s? Also find out DO at the end of 1 and 3 days.

## Short answer

Critical deficit occurs at 99.7 km (5.77 days) using base-e constants (kd=0.25, kr=0.11). Mixed BOD5=32.0 mg/l, ultimate BOD=44.9 mg/l, initial deficit=1.06 mg/l. DO after 1 and 3 days is effectively zero (formula gives negative), river turns anaerobic.

## Answer

Rates are taken as base $e$: $k_d=0.25$/day, $k_r=0.11$/day (20°C, so no temperature correction). Since $k_r<k_d$, the ratio $f=k_r/k_d=0.44<1$. $DO_{sat}=9.1$ mg/l.

#### Step 1: Mixing

$$
\begin{aligned}
Q &= 1.8+22 = 23.8\ \text{m}^3/\text{s}\\
BOD_5^{mix} &= \frac{1.8\times 350+22\times 6}{23.8} = 32.02\ \text{mg/l}\\
DO_{mix} &= \frac{1.8\times 0+22\times 8.7}{23.8} = 8.04\ \text{mg/l}\\
D_0 &= 9.1-8.04 = 1.06\ \text{mg/l}\\
L_0 &= \frac{BOD_5}{1-e^{-5k_d}} = \frac{32.02}{1-e^{-1.25}} = 44.87\ \text{mg/l}
\end{aligned}
$$

#### Step 2: Critical time and location

$$
t_c = \frac{1}{k_r-k_d}\ln\left[\frac{k_r}{k_d}\left\{1-\frac{D_0(k_r-k_d)}{k_dL_0}\right\}\right] = 5.770\ \text{days}
$$

$$
\begin{aligned}
D_c &= \frac{k_dL_0}{k_r}e^{-k_dt_c} = 24.10\ \text{mg/l}\\
x_c &= v\,t_c = 0.20\times 86400\times 5.770/1000 = 99.7\ \text{km}
\end{aligned}
$$

#### Step 3: DO after 1 and 3 days

$$
D_t = \frac{k_dL_0}{k_r-k_d}\left(e^{-k_dt}-e^{-k_rt}\right)+D_0e^{-k_rt}
$$

| t (days) | $D_t$ (mg/l) | $DO = 9.1 - D_t$ (mg/l) |
|---|---|---|
| 1 | 10.33 | -1.23 |
| 3 | 20.52 | -11.42 |

**Answer: the critical deficit occurs about 99.7 km downstream ($t_c=5.770$ days).** The computed deficits (10.33 and 20.52 mg/l) exceed the saturation value 9.1 mg/l, so the DO after 1 day and 3 days is effectively **zero** (the formula gives -1.23 and -11.42 mg/l, which is physically not possible): the river would turn anaerobic, and the effluent needs treatment.

## Check yourself

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

1. How is the DO of the river-effluent mixture computed?
   A. Flow-weighted average: (1.8x0 + 22x8.7)/23.8 = 8.04 mg/l
   B. Sum of the two DO values
   C. Equal to the river DO, 8.7 mg/l
   D. Simple average of 0 and 8.7 mg/l
   Answer: A. Mixture values are flow-weighted averages, giving DO = 8.04 mg/l. (Key point: Mixing)

2. What is the initial deficit D0 after mixing?
   A. 9.1 - 8.7 = 0.4 mg/l
   B. 9.1 - 8.04 = 1.06 mg/l
   C. 8.04 - 0 = 8.04 mg/l
   D. 9.1 + 8.04 = 17.14 mg/l
   Answer: B. Deficit is saturation DO minus actual DO: 9.1 - 8.04 = 1.06 mg/l. (Key point: Initial deficit)

3. How is the distance to the critical deficit found?
   A. x_c = L0 x v
   B. x_c = v x t_c = 0.20 m/s x 5.77 d, about 99.7 km
   C. x_c = D_c x v
   D. x_c = t_c / v
   Answer: B. Distance is velocity times travel time to the critical point: 0.20 x 86400 x 5.77 / 1000 = 99.7 km. (Key point: Critical location)

4. Why is the DO after 1 and 3 days reported as zero rather than -1.23 and -11.42 mg/l?
   A. The negative values are correct and DO can be negative
   B. The formula gives deficits larger than DO_sat (9.1), but DO cannot be negative so the river goes anaerobic
   C. Temperature correction removes the negatives
   D. kr is larger than kd, so DO is always zero
   Answer: B. The deficit cannot exceed saturation DO; a computed deficit above 9.1 mg/l means oxygen is exhausted and the river turns anaerobic. (Key point: Result for 1 and 3 days)

## Related in Sanitary Engineering

- [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
- [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
- [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
- [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
- [BOD1 at 22°C to BOD5 at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-1day-22c-310-to-5day-30c) — Sanitary Engineering, chapter 5

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