---
title: "Critical oxygen deficit and location (table)"
subject: "Sanitary Engineering"
chapter: 6
key: "semester_6/sanitary_engineering:critical-do-effluent-stream-table-78"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/critical-do-effluent-stream-table-78
site: BCT Engineering
---

# Critical oxygen deficit and location (table)

Sanitary Engineering · Chapter 6: Wastewater Disposal

- Asked in: 2078 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-effluent-stream-table-78
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6

## Question

A wastewater treatment plant dispose of its effluents into a stream at a point A. The characteristics of effluents and stream water are given below.

| Item | Effluent | Stream |
|---|---|---|
| Flow (m³/s) | 0.18 | 0.46 |
| DO mg/lit | 1.5 | 8.3 |
| Temp (°C) | 25 | 22 |
| BOD at 20°C mg/lit | 32 | 2 |

Assume that deoxygenating constant K' at 20°C (base e) = 0.2 /day and oxygenation constant R' at 20°C (base e) = 0.4 /day for the mixture. Equilibrium of dissolved oxygen for fresh water is as follows:

| Temp (°C) | 18 | 20 | 22 | 23 | 24 | 25 | 26 |
|---|---|---|---|---|---|---|---|
| DO mg/lit | 9.54 | 9.17 | 8.99 | 8.83 | 8.53 | 8.38 | 8.22 |

The velocity of the stream D/S of the point A is 0.16 m/sec. Determine the critical oxygen deficit and its location.

## Short answer

After mixing, initial deficit 2.47 mg/L, ultimate BOD 16.51 mg/L; temperature-corrected k'=0.2279/day, r'=0.4185/day. Critical oxygen deficit Dc = 5.10 mg/L, occurring 34.40 km downstream (2.49 days).

## Answer

The BOD is taken as the 5-day BOD at 20°C. Rates are base $e$ and are corrected to the mixture temperature with $K_T=K_{20}(1.047)^{T-20}$ and $R_T=R_{20}(1.016)^{T-20}$.

#### Step 1: Mixture properties

$$
\begin{aligned}
Q &= 0.18+0.46 = 0.64\ \text{m}^3/\text{s}\\
T_{mix} &= \frac{0.18\times 25+0.46\times 22}{0.64} = 22.84^\circ\text{C}\\
DO_{mix} &= \frac{0.18\times 1.5+0.46\times 8.3}{0.64} = 6.39\ \text{mg/l}\\
BOD_5^{mix} &= \frac{0.18\times 32+0.46\times 2}{0.64} = 10.44\ \text{mg/l}\\
DO_{sat}(22.84^\circ\text{C}) &= 8.855\ \text{mg/l (interpolated between 22 and 23}^\circ\text{C in the table)}\\
D_0 &= 8.855-6.39 = 2.47\ \text{mg/l}
\end{aligned}
$$

#### Step 2: Ultimate BOD and corrected rate constants

$$
\begin{aligned}
L_0 &= \frac{BOD_5}{1-e^{-5k'}} = \frac{10.44}{1-e^{-5\times 0.2}} = 16.51\ \text{mg/l}\\
k'_T &= 0.2(1.047)^{2.84} = 0.2279\ \text{/day}\\
r'_T &= 0.4(1.016)^{2.84} = 0.4185\ \text{/day}
\end{aligned}
$$

#### Step 3: Critical time, deficit and location

$$
\begin{aligned}
t_c &= \frac{1}{r'-k'}\ln\left[\frac{r'}{k'}\left\{1-\frac{D_0(r'-k')}{k'L_0}\right\}\right] = 2.488\ \text{days}\\
D_c &= \frac{k'L_0}{r'}e^{-k't_c} = 5.10\ \text{mg/l}\\
x_c &= v\,t_c = 0.16\times 86400\times 2.488/1000 = 34.40\ \text{km}
\end{aligned}
$$

**Answer: critical oxygen deficit $D_c = 5.10$ mg/l, occurring about $34.40$ km downstream of point A (after $2.488$ days).** The minimum DO is $8.855-5.10 = 3.75$ mg/l.

## Check yourself

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

1. How is the mixture temperature found?
   A. Arithmetic mean of 25 and 22
   B. Flow-weighted average of the two temperatures
   C. Temperature of the stream
   D. Temperature of the effluent
   Answer: B. T = (0.18 x 25 + 0.46 x 22)/0.64 = 22.84 C. (Key point: Mixture temperature)

2. How are the rate constants corrected from 20 C to the mixture temperature?
   A. Both multiplied by T/20
   B. Both unchanged
   C. K_T = K20 x 1.047^(T-20) and R_T = R20 x 1.016^(T-20)
   D. K_T = K20 x 1.016^(T-20) only
   Answer: C. Deoxygenation uses theta = 1.047 and reaeration 1.016, giving k' = 0.2279 and r' = 0.4185 per day. (Key point: Temperature correction)

3. How is DO_sat for the mixture found from the table?
   A. Use the value at 20 C
   B. Use the value at 25 C
   C. Average the whole table
   D. Interpolate at 22.84 C between 22 and 23 C
   Answer: D. DO_sat is read at the mixture temperature by interpolation, 8.855 mg/L, so D0 = 8.855 - 6.39 = 2.47 mg/L. (Key point: Saturation DO)

4. How is the location of the critical deficit found?
   A. x = v x tc, with v converted to km/day
   B. x = tc / v
   C. x = Dc x v
   D. x = L0 / K
   Answer: A. Distance = 0.16 m/s x 86400 s/day x 2.488 days = 34.4 km. (Key point: Location)

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## Related in other subjects

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