---
title: "Critical DO deficit, city 100 l/s"
subject: "Sanitary Engineering"
chapter: 6
key: "semester_6/sanitary_engineering:critical-do-city-100ls"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/critical-do-city-100ls
site: BCT Engineering
---

# Critical DO deficit, city 100 l/s

Sanitary Engineering · Chapter 6: Wastewater Disposal

- Asked in: 2073 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-city-100ls
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6

## Question

A city is discharging sewage of 100 l/s in the river having discharge of 1000 l/s and a velocity of 60 km/day. The BOD5 of sewage and river water are 450 mg/l and 4 mg/l respectively. The DO of sewage is zero. The DO in river is 70% of saturation DO value. And, the saturation DO at 20°C is 9.17 mg/l. Take deoxygenation constant (K1) = 0.1/day (base 10) and reaeration constant (k2) = 0.5/day (base 10). Calculate the value of critical DO deficit.

## Short answer

Critical DO deficit is 9.23 mg/l at 1.50 days (89.9 km downstream). Mixed BOD5=44.55 mg/l, ultimate BOD=65.15 mg/l, initial deficit=3.33 mg/l. Minimum DO would be negative, indicating anaerobic conditions without treatment.

## Answer

River DO $=0.7\times 9.17=6.419$ mg/l. Rates are base 10: $K_1=0.1$/day, $K_2=0.5$/day, so $f=5$. Flows are in l/s.
#### Step 1: Conditions just after mixing

$$
\begin{aligned}
Q_{mix} &= Q_s+Q_r = 100 + 1000 = 1100\ l/s\\
BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{100\times 450 + 1000\times 4}{1100} = 44.55\ \text{mg/l}\\
DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{100\times 0 + 1000\times 6.419}{1100} = 5.84\ \text{mg/l}\\
D_0 &= DO_{sat}-DO_{mix} = 9.17 - 5.84 = 3.33\ \text{mg/l}
\end{aligned}
$$

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

$$
\begin{aligned}
L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{44.55}{1-10^{-5\times 0.1}} = 65.15\ \text{mg/l}\\
f &= \frac{R}{K} = \frac{0.5}{0.1} = 5.00
\end{aligned}
$$

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

$$
\begin{aligned}
t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\
&= \frac{1}{0.1(5.00-1)}\log_{10}\left[5.00\left\{1-(5.00-1)\frac{3.33}{65.15}\right\}\right] = 1.499\ \text{days}\\
D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{65.15}{5.00}\times 10^{-0.1\times 1.499} = 9.23\ \text{mg/l}
\end{aligned}
$$
**Answer: critical DO deficit $D_c=9.23$ mg/l** (at $t_c=1.499$ days, about 89.9 km downstream at 60 km/day). The lowest DO is $9.17-9.23=-0.06$ mg/l.

*Note:* the computed deficit is larger than the saturation DO (9.17 mg/l). In reality DO cannot fall below zero, so the stream would go anaerobic near the critical point and the effluent needs treatment before discharge. The negative value is simply what the formula gives for these data.

## Check yourself

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

1. What is the DO of the mixture?
   A. 6.419 mg/L
   B. (1000 x 6.419)/1100 = 5.84 mg/L
   C. 3.21 mg/L
   D. 9.17 mg/L
   Answer: B. Sewage DO is zero, so only the river contributes, weighted by flow. (Key point: Mixture DO)

2. What is the tc for this problem?
   A. 0.5 days
   B. 3.33 days
   C. 1.499 days
   D. 5 days
   Answer: C. With f = 5, D0 = 3.33 and L0 = 65.15, tc = 1.499 days. (Key point: Critical time)

3. Dc is obtained from which relation?
   A. Dc = L0 f
   B. Dc = D0 10^(-K1 tc)
   C. Dc = L0/tc
   D. Dc = (L0/f) 10^(-K1 tc)
   Answer: D. Dc = (65.15/5) x 10^(-0.1 x 1.499) = 9.23 mg/L. (Key point: Critical deficit)

4. Dc = 9.23 mg/L exceeds the saturation DO 9.17 mg/L. What does this mean?
   A. The stream goes anaerobic near the critical point and the effluent needs treatment
   B. The stream is healthy
   C. The formula is invalid and should be ignored
   D. Reaeration is stronger than expected
   Answer: A. DO cannot be below zero, so oxygen is exhausted near the critical point. (Key point: Interpretation)

## Close variants set in other exams

- [Time of critical DO deficit, 50 l/s city](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Acritical-do-time-city-50ls) — Sanitary Engineering, chapter 6, asked 2×

## Related in Sanitary Engineering

- [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
- [DO deficit 60 km downstream, 2 m³/s stream](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Astream-do-deficit-60km) — Sanitary Engineering, chapter 6, asked 2×
- [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
- [DO deficit 50 km downstream, 1.9 m³/s stream](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Astream-do-deficit-50km) — Sanitary Engineering, chapter 6, asked 2×
- [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

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