---
title: "DO deficit 50 km downstream, 1.9 m³/s stream"
subject: "Sanitary Engineering"
chapter: 6
key: "semester_6/sanitary_engineering:stream-do-deficit-50km"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/stream-do-deficit-50km
site: BCT Engineering
---

# DO deficit 50 km downstream, 1.9 m³/s stream

Sanitary Engineering · Chapter 6: Wastewater Disposal

- Asked in: 2079 Chaitra, 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-50km
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6

## Question

A stream saturated with DO has a flow of 1.9 m³/s, BOD of 2.5 mg/l and rate constant (K to base 10) of 0.1 per day. It receives an effluent discharge of 0.6 m³/s having BOD of 210 mg/l and DO of 1.8 mg/l. The average velocity of stream is 0.15 m/s and average depth of stream is 1.4 m. Calculate DO deficit 50 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 1.769 mg/L, ultimate BOD 52.30 mg/L, K₁=0.1 day⁻¹, K₂=0.396 day⁻¹, travel time 3.858 days. DO deficit at 50 km = 6.80 mg/L (DO = 2.37 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
$$
\begin{aligned}
Q_{mix}&=1.9+0.6=2.5\ \text{m}^3/\text{s}\\
DO_{mix}&=\frac{1.9(9.17)+0.6(1.8)}{2.5}=7.401\ \text{mg/L}\\
D_0&=9.17-7.401=1.769\ \text{mg/L}\\
L_0&=\frac{1.9(2.5)+0.6(210)}{2.5}=52.30\ \text{mg/L}
\end{aligned}
$$

#### Step 2: Reaeration constant and time of flow
$$
\begin{aligned}
K_{2(e)}&=\frac{3.9\times0.15^{0.5}}{1.4^{1.5}}=0.912\ \text{day}^{-1}\Rightarrow K_2=0.396\ \text{day}^{-1}\ (\text{base }10)\\
t&=\frac{x}{v}=\frac{50000\ \text{m}}{0.15\ \text{m/s}}=333333\ \text{s}=3.858\ \text{days}
\end{aligned}
$$

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

$$
\begin{aligned}
D_t&=\frac{0.1\times52.30}{0.396-0.1}\left(10^{-0.3858}-10^{-1.5278}\right)+1.769\times10^{-1.5278}\\
&=6.80\ \text{mg/L}
\end{aligned}
$$

**Answer: DO deficit 50 km downstream = 6.80 mg/L** (DO = 2.37 mg/L).

*Note:* if the given BODs were 5-day values, $L_0$ would be $76.5$ mg/L, and the deficit would rise further (to about 9.9 mg/L, which is more than the saturation value, so the stream would turn septic).

## Check yourself

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

1. How is the initial BOD L0 of the mixture obtained?
   A. Average of the two BODs
   B. Flow-weighted average of stream and effluent BODs
   C. BOD of the effluent only
   D. Sum of the two BODs
   Answer: B. L0 = (1.9 x 2.5 + 0.6 x 210)/2.5 = 52.30 mg/L, a flow-weighted mean. (Key point: Initial conditions)

2. How is K2 found when it is not given?
   A. K2 = 10 K1
   B. K2 = K1/2.303
   C. From the O'Connor-Dobbins formula using velocity and depth, converted to base 10
   D. From the BOD of the stream
   Answer: C. K2(e) = 3.9 v^0.5 / H^1.5 = 0.912 per day, divided by 2.303 gives 0.396 per day in base 10. (Key point: Reaeration constant)

3. How is the travel time to the section 50 km downstream found?
   A. t = x x v
   B. t = 1/K1
   C. t = 5 days
   D. t = x/v = 50000 m / 0.15 m/s = 3.858 days
   Answer: D. Time = distance/velocity = 333333 s = 3.858 days. (Key point: Time of flow)

4. Which equation gives the deficit at time t?
   A. Dt = K1 L0/(K2 - K1) (10^(-K1 t) - 10^(-K2 t)) + D0 10^(-K2 t)
   B. Dt = L0 (10^(-K1 t) - 10^(-K2 t))
   C. Dt = D0 10^(-K1 t)
   D. Dt = Dc/f
   Answer: A. The Streeter-Phelps equation combines deoxygenation and reaeration, plus the decay of the initial deficit, giving 6.80 mg/L. (Key point: Streeter-Phelps use)

## Close variants set in other exams

- [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×

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