---
title: "River flow required for 1500 population"
subject: "Sanitary Engineering"
chapter: 6
key: "semester_6/sanitary_engineering:river-flow-for-1500-population"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/river-flow-for-1500-population
site: BCT Engineering
---

# River flow required for 1500 population

Sanitary Engineering · Chapter 6: Wastewater Disposal

- Asked in: 2081 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/river-flow-for-1500-population
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6

## Question

Determine the flow in river required per 1500 population for disposing off sewage from a residential town with the given data: average temperature of river water = 25°C; 5-day BOD of sewage at 25°C = 350 ppm; Average sewage flow = 200 lpcd; values of de-oxygenation and re-oxygenation constants of the river at 25°C are 0.15 day⁻¹ and 0.27 day⁻¹ respectively. Minimum DO concentration to be maintained in river water = 4 ppm and saturation DO of river water at 25°C = 8.38 ppm.

## Short answer

Using Streeter-Phelps equation with base-10 constants, sewage flow 300 m³/day, ultimate BOD 425.7 ppm, critical deficit 4.38 ppm, self-purification factor 1.8, critical time 2.07 days, mixture fraction x=0.03785. Minimum river flow required = 7625 m³/day (0.0883 m³/s or 88.3 L/s), dilution factor ~25.4.

## Answer

#### Assumptions
The river upstream is saturated with oxygen (DO = 8.38 ppm) and has no BOD; the sewage has zero DO. The constants are in base 10 and refer to 25°C. Let $Q_r$ be the river flow required (m³/day).

#### Step 1: Sewage and its ultimate BOD
$$
Q_s=1500\times200\ \text{L/d}=300\ \text{m}^3/\text{day}
$$

$$
L_s=\frac{BOD_5}{1-10^{-K_1\times5}}=\frac{350}{1-10^{-0.15\times5}}=\frac{350}{1-0.1778}=425.7\ \text{ppm}
$$

#### Step 2: Allowable critical deficit
Minimum DO to be kept = 4 ppm, so the maximum (critical) deficit is

$$
D_c=8.38-4=4.38\ \text{ppm}
$$

#### Step 3: Mixture (let $x=\dfrac{Q_s}{Q_s+Q_r}$)
$$
L_0=x\,L_s=425.7\,x,\qquad D_0=x\,(8.38-0)=8.38\,x
$$

Self-purification factor $f=\dfrac{K_2}{K_1}=\dfrac{0.27}{0.15}=1.8$.

Critical time:

$$
t_c=\frac{1}{K_2-K_1}\log_{10}\left[\frac{K_2}{K_1}\left(1-\frac{(f-1)D_0}{L_0}\right)\right]
$$

Since $D_0/L_0=8.38/425.7=0.01969$ is independent of $x$,

$$
t_c=\frac{1}{0.12}\log_{10}\left[1.8\left(1-0.8\times0.01969\right)\right]=2.07\ \text{days}
$$

Critical deficit: $D_c=\dfrac{L_0}{f}10^{-K_1t_c}$

$$
4.38=\frac{425.7\,x}{1.8}\times10^{-0.15\times2.07}=115.71\,x\ \Rightarrow\ x=0.03785
$$

#### Step 4: River flow
$$
Q_r=Q_s\left(\frac{1}{x}-1\right)=300\left(\frac{1}{0.03785}-1\right)=7625\ \text{m}^3/\text{day}
$$

**Answer: Minimum river flow = about 7625 m³/day (= 0.0883 m³/s or 88.3 l/s) for the 1500 people**, i.e. a dilution of about 25.4 times the sewage flow.

## Check yourself

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

1. How is the ultimate BOD of the sewage obtained from the 5-day BOD?
   A. L = BOD5 x 5
   B. L = BOD5 / (1 - 10^(-5 K1))
   C. L = BOD5 x (1 - 10^(-5 K1))
   D. L = BOD5 / K1
   Answer: B. BOD5 = L(1 - 10^(-5K1)), so L = BOD5/(1 - 10^(-K1 x 5)) = 350/0.8222 = 425.7 ppm. (Key point: Ultimate BOD)

2. If the minimum DO is 4 ppm and saturation is 8.38 ppm, what is the allowable critical deficit?
   A. 4.00 ppm
   B. 8.38 ppm
   C. 4.38 ppm
   D. 12.38 ppm
   Answer: C. Dc = saturation DO - minimum DO = 8.38 - 4 = 4.38 ppm. (Key point: Critical deficit)

3. Why is the critical time tc found without knowing the river flow?
   A. The river flow does not change the BOD
   B. tc is fixed by K1 alone
   C. Because the sewage has zero BOD
   D. D0/L0 is the same for any dilution because both are scaled by the same factor x
   Answer: D. L0 = x Ls and D0 = x(8.38), so D0/L0 = 8.38/425.7 is independent of x, giving tc = 2.07 days directly. (Key point: Independence of x)

4. Once x = Qs/(Qs+Qr) is found from Dc = (L0/f)10^(-K1 tc), how is the river flow Qr obtained?
   A. Qr = Qs (1/x - 1)
   B. Qr = Qs x x
   C. Qr = Qs / (1 - x)
   D. Qr = Qs + x
   Answer: A. From x = Qs/(Qs+Qr), Qr = Qs(1/x - 1) = 300(1/0.03785 - 1), about 7625 m3/day. (Key point: Final flow)

## Related in Sanitary Engineering

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- [Degree of treatment, town with dairy and industries](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Adegree-treatment-city-industries-71) — Sanitary Engineering, chapter 6
- [Percentage purification, town of 50000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Apercent-purification-town-50000) — Sanitary Engineering, chapter 6
- [Oxidation pond for Terai town, 12,000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aoxidation-pond-terai-12000) — Sanitary Engineering, chapter 7
- [Minimum sewer diameter, 80000 population](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asewer-min-diameter-80000) — Sanitary Engineering, chapter 3
- [Sewage sampling and BOD dilution method](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asampling-bod-dilution) — Sanitary Engineering, chapter 5

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