---
title: "Conventional ASP design, 96 lpcd"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:asp-100000-96lpcd"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/asp-100000-96lpcd
site: BCT Engineering
---

# Conventional ASP design, 96 lpcd

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2073 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/asp-100000-96lpcd
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Design a conventional activated sludge treatment plant to treat the domestic sewage with diffused air aeration with the following data. (Design up to dimensions of aeration tank only) Population = 1,00,000; Per capita sewage flow = 96 liters/day; Settled sewage BOD5 = 200 mg/L; Food/micro-organisms = 0.3; Concentration of microorganism (MLSS) = 2000 mg/L.

## Short answer

Aeration tank volume 3200 m³ from F/M ratio; detention time 8 h; two tanks each 40 m × 10 m × 4.5 m (liquid depth 4 m, freeboard 0.5 m).

## 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: settled-sewage $S_0=200$ mg/l, $F/M=0.3$/day, MLSS $=2000$ mg/l. Only the aeration tank is designed.

#### Step 1: Design flow

$$
Q = 100000\times96/1000 = 9600.0\ \text{m}^3/\text{day}
$$

#### Step 2: Volume of aeration tank

$$
V = \frac{9600.0\times 200}{0.3\times 2000} = 3200.0\ \text{m}^3
$$

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

#### Step 3: Dimensions

$$
\begin{aligned}
\text{Adopt liquid depth } &= 4.0\ \text{m (usual range 3-4.5 m)}\\
\text{Plan area } A &= \frac{V}{4.0} = 800.0\ \text{m}^2
\end{aligned}
$$

Provide 2 tanks (so one can be taken out of service), each with area $400.0$ m² and L : B about 4 : 1:

$$
B=\sqrt{\frac{400.0}{4}} = 10.00\ \text{m (adopt 10.0 m)},\qquad L=\frac{400.0}{10.0} = 40.0\ \text{m (adopt 40 m)}
$$

Adding 0.5 m free board, the tank depth is 4.5 m. The size of each tank is $40\ \text{m}\times 10.0\ \text{m}\times 4.5\ \text{m}$.

#### Step 4: Checks

$$
\begin{aligned}
\text{Hydraulic retention time} &= \frac{V}{Q} = \frac{3200.0}{9600.0}\times 24 = 8.0\ \text{h}\ (\text{usual 4-8 h; accepted})\\
\text{Volumetric loading} &= \frac{Q\,S_0}{V\times 1000} = 0.60\ \text{kg BOD/m}^3\text{/day}\ (\text{usual 0.3-0.6})
\end{aligned}
$$

**Answer: aeration tank volume $\approx 3200$ m³ (detention 8.0 h, depth 4.0 m); provide 2 tanks, each $40\ \text{m}\times 10.0\ \text{m}$ 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 96 lpcd, what is Q?
   A. 9600 m3/day
   B. 960 m3/day
   C. 96 000 m3/day
   D. 19 200 m3/day
   Answer: A. 100000 x 96/1000 = 9600 m3/day. (Key point: Design flow)

2. What volume is required with F/M = 0.3, MLSS = 2000 mg/l and S0 = 200 mg/l?
   A. 1920 m3
   B. 3200 m3
   C. 6400 m3
   D. 9600 m3
   Answer: B. V = 9600 x 200 / (0.3 x 2000) = 3200 m3. (Key point: Aeration volume)

3. Why is the volume right when Q is in m3/day and S0 and X in mg/l?
   A. mg/l must first be converted to kg/l
   B. Q must be in m3/s
   C. mg/l equals g/m3, so S0 and X cancel in g/m3 and V comes out in m3
   D. The 10^6 factor is automatically included in F/M
   Answer: C. Both S0 and X are in g/m3, so Q S0 / ((F/M) X) has units of m3. (Key point: Units in the formula)

4. For area 800 m2 split into 2 tanks, what size does the answer adopt?
   A. Each 40 m x 10 m x 4.5 m (including 0.5 m free board)
   B. Each 80 m x 10 m x 4 m
   C. Each 20 m x 20 m x 4.5 m
   D. Each 40 m x 20 m x 4 m
   Answer: A. Each tank is 400 m2 with L:B about 4:1: B = 10 m, L = 40 m. (Key point: Tank dimensions)

## Close variants set in other exams

- [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×

## Related in Sanitary Engineering

- [Conventional ASP design, population 1,00,000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aasp-100000-150lpcd) — Sanitary Engineering, chapter 7
- [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, 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