---
title: "Time of critical DO deficit, 50 l/s city"
subject: "Sanitary Engineering"
chapter: 6
key: "semester_6/sanitary_engineering:critical-do-time-city-50ls"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/critical-do-time-city-50ls
site: BCT Engineering
---

# Time of critical DO deficit, 50 l/s city

Sanitary Engineering · Chapter 6: Wastewater Disposal

- Asked in: 2077 Chaitra, 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-time-city-50ls
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6

## Question

A city is discharging sewage of 50 l/s in the river having discharge of 500 l/s and a velocity of 48 km/day. The 5-day BOD of sewage and river water are 400 mg/l and 4 mg/l respectively. The DO of sewage is zero. The DO in the river is 80% of saturation value. Saturation DO at 20°C is 9.17 mg/l. Consider deoxygenation constant (K) as 0.1/day (base 10) and reaeration constant (R) as 0.5/day. Calculate the time of critical DO deficit.

## Short answer

Streeter-Phelps: mixed flow 550 L/s, mixed BOD₅ 40 mg/L, mixed DO 6.67 mg/L, initial deficit 2.50 mg/L, ultimate BOD 58.50 mg/L, f=5. Critical time t_c = 1.54 days (74.1 km at 48 km/day), critical deficit 8.20 mg/L, minimum DO = 0.97 mg/L.

## Answer

The critical point is where the oxygen deficit is maximum (Streeter-Phelps oxygen sag). The river DO is 80% of saturation, so $DO_r = 0.8\times 9.17 = 7.336$ mg/l. Rates are base 10: $K=0.1$/day, $R=0.5$/day.
#### Step 1: Conditions just after mixing

$$
\begin{aligned}
Q_{mix} &= Q_s+Q_r = 50 + 500 = 550\ l/s\\
BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{50\times 400 + 500\times 4}{550} = 40.00\ \text{mg/l}\\
DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{50\times 0 + 500\times 7.336}{550} = 6.67\ \text{mg/l}\\
D_0 &= DO_{sat}-DO_{mix} = 9.17 - 6.67 = 2.50\ \text{mg/l}
\end{aligned}
$$

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

$$
\begin{aligned}
L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{40.00}{1-10^{-5\times 0.1}} = 58.50\ \text{mg/l}\\
f &= \frac{R}{K} = \frac{0.5}{0.1} = 5.00
\end{aligned}
$$

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

$$
\begin{aligned}
t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\
&= \frac{1}{0.1(5.00-1)}\log_{10}\left[5.00\left\{1-(5.00-1)\frac{2.50}{58.50}\right\}\right] = 1.544\ \text{days}\\
D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{58.50}{5.00}\times 10^{-0.1\times 1.544} = 8.20\ \text{mg/l}
\end{aligned}
$$
**Answer: critical time $t_c \approx 1.54$ days** (about 74.1 km downstream at 48 km/day), with critical deficit $D_c = 8.20$ mg/l, so the lowest DO is $9.17-8.20=0.97$ mg/l.

## Check yourself

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

1. How is the river DO found when it is given as 80% of saturation?
   A. 9.17 - 0.8
   B. 0.8 x 9.17 = 7.336 mg/L
   C. 9.17 / 0.8
   D. 0.2 x 9.17
   Answer: B. 80% of the saturation value 9.17 is 7.336 mg/L. (Key point: River DO)

2. What is f for K = 0.1/day and R = 0.5/day?
   A. 0.2
   B. 4
   C. 5
   D. 0.6
   Answer: C. f = R/K = 0.5/0.1 = 5. (Key point: Self-purification constant)

3. What does tc represent?
   A. Time after which BOD is fully satisfied
   B. Time for the sewage to reach the river
   C. Time for the initial deficit to halve
   D. Time of travel when the DO deficit is maximum
   Answer: D. At tc the deficit is maximum, tc = 1.544 days, about 74.1 km downstream at 48 km/day. (Key point: Critical time)

4. Why is the L0 (58.50) larger than the mixed BOD5 (40.00)?
   A. L0 is the ultimate BOD, obtained by dividing BOD5 by 1 - 10^(-5K)
   B. Because river BOD is added again
   C. Because DO is subtracted
   D. Because of the velocity
   Answer: A. BOD5 is only part of the total oxygen demand; L0 = 40/0.6838 = 58.50 mg/L. (Key point: Conditions after mixing)

## Close variants set in other exams

- [Critical DO deficit, city 100 l/s](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Acritical-do-city-100ls) — 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 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×
- [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×
- [1-day BOD 300 mg/l to 5-day at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-1day-300-to-5day-30c) — Sanitary Engineering, chapter 5
- [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
- [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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