---
title: "Conventional ASP design, population 1,00,000"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:asp-100000-150lpcd"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/asp-100000-150lpcd
site: BCT Engineering
---

# Conventional ASP design, population 1,00,000

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2076 Baisakh, 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/asp-100000-150lpcd
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Design a conventional activated sludge plant to treat settled domestic sewage with diffused air aeration system for the following data; (Design up to dimension of Aeration tank): Population = 1,00,000; Per capita sewage contribution = 150 lit/d; BOD5 of sewage = 200 mg/l; Effluent BOD5 = 30 mg/l; F/M = 0.2 day⁻¹; MLSS = 3000 mg/l.

## Short answer

Flow 15000 m³/d, aeration tank volume 5000 m³. Liquid depth 4.0 m, provide two tanks each 50 m × 12.5 m × 4.5 m. HRT 8.0 h, volumetric loading 0.60 kg BOD/m³/d.

## Answer

#### Data and design basis

Conventional ASP (plug flow, diffused air). The aeration tank volume follows from the food-to-micro-organism ratio:

$$
\frac{F}{M}=\frac{Q\,S_0}{V\,X}\ \Rightarrow\ V=\frac{Q\,S_0}{(F/M)\,X}
$$

Data: $S_0=200$ mg/l, effluent $BOD_5=30$ mg/l (efficiency 85%), $F/M=0.2$/day, MLSS $=3000$ mg/l. Only the aeration tank is designed.

#### Step 1: Design flow

$$
Q = 100000\times150/1000 = 15000.0\ \text{m}^3/\text{day}
$$

#### Step 2: Volume of aeration tank

$$
V = \frac{15000.0\times 200}{0.2\times 3000} = 5000.0\ \text{m}^3
$$

(units: $Q$ in m³/day and $S_0$, $X$ in mg/l = g/m³.)

#### Step 3: Dimensions

$$
\begin{aligned}
\text{Adopt liquid depth } &= 4.0\ \text{m (usual range 3-4.5 m)}\\
\text{Plan area } A &= \frac{V}{4.0} = 1250.0\ \text{m}^2
\end{aligned}
$$

Provide 2 tanks (so one can be taken out of service), each with area $625.0$ m² and L : B about 4 : 1:

$$
B=\sqrt{\frac{625.0}{4}} = 12.50\ \text{m (adopt 12.5 m)},\qquad L=\frac{625.0}{12.5} = 50.0\ \text{m (adopt 50 m)}
$$

Adding 0.5 m free board, the tank depth is 4.5 m. The size of each tank is $50\ \text{m}\times 12.5\ \text{m}\times 4.5\ \text{m}$.

#### Step 4: Checks

$$
\begin{aligned}
\text{Hydraulic retention time} &= \frac{V}{Q} = \frac{5000.0}{15000.0}\times 24 = 8.0\ \text{h}\ (\text{usual 4-8 h; accepted})\\
\text{Volumetric loading} &= \frac{Q\,S_0}{V\times 1000} = 0.60\ \text{kg BOD/m}^3\text{/day}\ (\text{usual 0.3-0.6})
\end{aligned}
$$

**Answer: aeration tank volume $\approx 5000$ m³ (detention 8.0 h, depth 4.0 m); provide 2 tanks, each $50\ \text{m}\times 12.5\ \text{m}$ in plan with liquid depth 4.0 m (tank depth 4.5 m).**

## Check yourself

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

1. For 100 000 people at 150 lpcd, what is Q?
   A. 15 000 m3/day
   B. 1500 m3/day
   C. 150 000 m3/day
   D. 100 000 m3/day
   Answer: A. 100000 x 150 / 1000 = 15 000 m3/day. (Key point: Design flow)

2. What is V for S0 = 200 mg/l, F/M = 0.2, MLSS = 3000 mg/l?
   A. 10 000 m3
   B. 5000 m3
   C. 2500 m3
   D. 15 000 m3
   Answer: B. V = 15000 x 200 / (0.2 x 3000) = 5000 m3. (Key point: Aeration volume)

3. Why does the design provide at least two aeration tanks?
   A. Because one tank cannot exceed 1000 m3
   B. To double the retention time
   C. So one can be taken out of service for maintenance while the plant keeps running
   D. To halve the F/M ratio
   Answer: C. Units in parallel give flexibility; each tank here is 50 m x 12.5 m x 4.5 m. (Key point: Why provide 2 tanks)

4. Area is 1250 m2 for 4 m depth. How does a 4:1 tank layout come out?
   A. Each of two tanks has 625 m2: B = 12.5 m, L = 50 m, depth 4.5 m with free board
   B. One tank of 1250 m2: 25 m x 50 m
   C. Each tank 625 m2 as 25 m x 25 m x 4 m
   D. Each tank 1250 m2 as 50 m x 25 m x 4.5 m
   Answer: A. B = sqrt(625/4) = 12.5 m, L = 625/12.5 = 50 m, plus 0.5 m free board. (Key point: Dimensions)

## Related in Sanitary Engineering

- [Conventional ASP design, population 50,000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aasp-50000-80lpcd) — Sanitary Engineering, chapter 7, asked 2×
- [Conventional ASP design, population 126000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aasp-126000-checks) — Sanitary Engineering, chapter 7
- [Conventional ASP design, 96 lpcd](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aasp-100000-96lpcd) — Sanitary Engineering, chapter 7, asked 2×
- [Conventional ASP design, 0.104 m³/s](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aasp-0p104-conventional) — Sanitary Engineering, chapter 7
- [Activated sludge and ASP treatment process](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aasp-sketch-treatment-process) — Sanitary Engineering, chapter 7

## Related in other subjects

- [Service reservoir of 30 m diameter, 1,00,000 population](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ares-bk79) — Water Supply Engineering, chapter 7