---
title: "DO deficit 60 km downstream, 2 m³/s stream"
subject: "Sanitary Engineering"
chapter: 6
key: "semester_6/sanitary_engineering:stream-do-deficit-60km"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/stream-do-deficit-60km
site: BCT Engineering
---

# DO deficit 60 km downstream, 2 m³/s stream

Sanitary Engineering · Chapter 6: Wastewater Disposal

- Asked in: 2069 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/stream-do-deficit-60km
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6

## Question

A stream saturated with DO has a flow of 2 m³/s, BOD of 3 mg/l and rate constant (K1 to base 10) of 0.1 per day. It receives an effluent discharge of 0.5 m³/s having BOD of 200 mg/l and DO of 2 mg/l. The average velocity of stream is 0.2 m/s. The average depth of stream is 1.2 m. Calculate DO deficit 60 km downstream of outfall. Assume temperature of 20°C throughout and saturation DO at 20°C is 9.17 mg/l.

## Short answer

Using Streeter-Phelps with O'Connor-Dobbins reaeration: mixed flow 2.5 m³/s, initial deficit 1.434 mg/L, ultimate BOD 42.40 mg/L, K₁=0.1 day⁻¹, K₂=0.576 day⁻¹, travel time 3.472 days. DO deficit at 60 km = 3.93 mg/L (DO = 5.24 mg/L).

## Answer

#### Assumptions
Streeter-Phelps equation with rate constants to base 10; the BOD values given are taken as the ultimate (first-stage) BODs, so $L_0$ is the BOD of the mixture; $K_1=0.1$ day⁻¹; the reaeration constant is found from the O'Connor-Dobbins formula (base $e$, 20°C):

$$
K_{2(e)}=\frac{3.9\,v^{0.5}}{H^{1.5}}\ \text{day}^{-1}\quad(v\text{ in m/s},\ H\text{ in m}),\qquad K_2(\text{base }10)=\frac{K_{2(e)}}{2.303}
$$

#### Step 1: Mixing at the outfall
$$
\begin{aligned}
Q_{mix}&=2+0.5=2.5\ \text{m}^3/\text{s}\\
DO_{mix}&=\frac{2(9.17)+0.5(2)}{2.5}=7.736\ \text{mg/L}\\
D_0&=9.17-7.736=1.434\ \text{mg/L}\\
L_0&=\frac{2(3)+0.5(200)}{2.5}=42.40\ \text{mg/L}
\end{aligned}
$$

#### Step 2: Reaeration constant and time of flow
$$
\begin{aligned}
K_{2(e)}&=\frac{3.9\times0.2^{0.5}}{1.2^{1.5}}=1.327\ \text{day}^{-1}\Rightarrow K_2=0.576\ \text{day}^{-1}\ (\text{base }10)\\
t&=\frac{x}{v}=\frac{60000\ \text{m}}{0.2\ \text{m/s}}=300000\ \text{s}=3.472\ \text{days}
\end{aligned}
$$

#### Step 3: Deficit at the section
$$
D_t=\frac{K_1L_0}{K_2-K_1}\left(10^{-K_1t}-10^{-K_2t}\right)+D_010^{-K_2t}
$$

$$
\begin{aligned}
D_t&=\frac{0.1\times42.40}{0.576-0.1}\left(10^{-0.3472}-10^{-2.0008}\right)+1.434\times10^{-2.0008}\\
&=3.93\ \text{mg/L}
\end{aligned}
$$

**Answer: DO deficit 60 km downstream = 3.93 mg/L** (DO = 5.24 mg/L).

*Note:* if the given BODs were 5-day values, $L_0$ would be $62.0$ mg/L, and the deficit would rise further (to about 5.7 mg/L).

## Check yourself

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

1. How is D0 found?
   A. D0 = effluent DO
   B. D0 = saturation DO minus the flow-weighted DO of the mixture
   C. D0 = stream DO minus effluent DO
   D. D0 = L0/f
   Answer: B. DO_mix = (2 x 9.17 + 0.5 x 2)/2.5 = 7.736, so D0 = 9.17 - 7.736 = 1.434 mg/L. (Key point: Initial deficit)

2. What does 'stream saturated with DO' imply for the calculation?
   A. Stream DO is zero
   B. Stream BOD is zero
   C. Stream DO equals the saturation value 9.17 mg/L
   D. Deficit is zero everywhere
   Answer: C. A saturated stream has DO = 9.17 mg/L, which is used in the weighted DO of the mixture. (Key point: Why the stream DO is 9.17)

3. Which pair is used in the Dt equation?
   A. K2 = 1.327 per day (base e) used directly with base-10 exponents, t = 60 days
   B. K2 = 0.1 and t = 5 days
   C. K2 = 0.576 and t = 0.2 days
   D. K2 = 0.576 per day (base 10) and t = 3.472 days
   Answer: D. O'Connor-Dobbins gives 1.327 (base e), converted to 0.576 (base 10) to match 10^(-K t); t = 60000/0.2 s = 3.472 days. (Key point: Reaeration and time)

4. What is the DO at the section 60 km downstream?
   A. 5.24 mg/L
   B. 3.93 mg/L
   C. 9.17 mg/L
   D. 1.43 mg/L
   Answer: A. DO = saturation - deficit = 9.17 - 3.93 = 5.24 mg/L. (Key point: Result)

## Close variants set in other exams

- [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×

## Related in Sanitary Engineering

- [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 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
- [Rate constant and remaining BOD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-rate-250-325) — Sanitary Engineering, chapter 5
- [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×
- [Ultimate BOD and rate constant](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-ultimate-rate-constant-100-210) — Sanitary Engineering, chapter 5
- [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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