---
title: "Conventional ASP design, population 50,000"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:asp-50000-80lpcd"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/asp-50000-80lpcd
site: BCT Engineering
---

# Conventional ASP design, population 50,000

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2075 Bhadra, 8 marks
- Video: [activated sludge process solved problem | ASP Solved Examples | Waste Water Engineering | ASP](https://www.youtube.com/watch?v=tTgmlF76EH8) — 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/asp-50000-80lpcd
- 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 = 50,000; Per capita sewage flow = 80 liters/day; Settled sewage BOD5 = 200 mg/L; Food/micro-organisms = 0.3; Concentration of microorganisms (MLSS) = 2000 mg/L.

## Short answer

Aeration tank volume 1333 m³ calculated using F/M ratio; hydraulic retention time 8 h; provide two tanks each 26 m × 6.5 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

$$
Q = 50000\times80/1000 = 4000.0\ \text{m}^3/\text{day}
$$

#### Step 2: Volume of aeration tank

$$
V = \frac{4000.0\times 200}{0.3\times 2000} = 1333.3\ \text{m}^3
$$

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

#### Step 3: Dimensions

$$
\begin{aligned}
\text{Adopt liquid depth } &= 4.0\ \text{m (usual range 3-4.5 m)}\\
\text{Plan area } A &= \frac{V}{4.0} = 333.3\ \text{m}^2
\end{aligned}
$$

Provide 2 tanks (so one can be taken out of service), each with area $166.7$ m² and L : B about 4 : 1:

$$
B=\sqrt{\frac{166.7}{4}} = 6.45\ \text{m (adopt 6.5 m)},\qquad L=\frac{166.7}{6.5} = 25.6\ \text{m (adopt 26 m)}
$$

Adding 0.5 m free board, the tank depth is 4.5 m. The size of each tank is $26\ \text{m}\times 6.5\ \text{m}\times 4.5\ \text{m}$.

#### Step 4: Checks

$$
\begin{aligned}
\text{Hydraulic retention time} &= \frac{V}{Q} = \frac{1333.3}{4000.0}\times 24 = 8.0\ \text{h}\ (\text{usual 4-8 h; accepted})\\
\text{Volumetric loading} &= \frac{Q\,S_0}{V\times 1000} = 0.60\ \text{kg BOD/m}^3\text{/day}\ (\text{usual 0.3-0.6})
\end{aligned}
$$

**Answer: aeration tank volume $\approx 1333$ m³ (detention 8.0 h, depth 4.0 m); provide 2 tanks, each $26\ \text{m}\times 6.5\ \text{m}$ 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 50 000 people at 80 lpcd, what is Q?
   A. 400 m3/day
   B. 4000 m3/day
   C. 40 000 m3/day
   D. 4 000 000 m3/day
   Answer: B. 50000 x 80/1000 = 4000 m3/day. (Key point: Design flow)

2. With F/M = 0.3 and MLSS = 2000 mg/l, what volume results?
   A. 800 m3
   B. 2000 m3
   C. 1333 m3
   D. 2667 m3
   Answer: C. V = 4000 x 200 / (0.3 x 2000) = 1333.3 m3. (Key point: Aeration volume)

3. If F/M were lowered with all else the same, what happens to V?
   A. V increases, since V is inversely proportional to F/M
   B. V decreases
   C. V is unchanged
   D. V becomes independent of MLSS
   Answer: A. V = Q S0/((F/M) X); a lower F/M means more biomass is needed per unit load. (Key point: Effect of F/M)

4. What are the HRT and volumetric loading for this design?
   A. HRT = 4.0 h and loading = 0.3
   B. HRT = 8.0 h and loading = 1.2
   C. HRT = 12 h and loading = 0.6
   D. HRT = 8.0 h and loading = 0.60 kg BOD/m3/day
   Answer: D. 1333.3/4000 x 24 = 8 h; 4000 x 200/(1333.3 x 1000) = 0.60. (Key point: Check results)

## Close variants set in other exams

- [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×

## 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
- [Degree of treatment, domestic + industrial](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Adegree-treatment-town-40000) — Sanitary Engineering, chapter 6

## Related in other subjects

- [Rapid sand filter design, 50,000 population](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Arsf-c72) — Water Supply Engineering, chapter 6