---
title: "Completely mixed ASP reactor, 10,000 m³/day"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:asp-completely-mixed-10000"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/asp-completely-mixed-10000
site: BCT Engineering
---

# Completely mixed ASP reactor, 10,000 m³/day

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2072 Asoj, 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-completely-mixed-10000
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

A activated sludge system is to be used for secondary treatment of 10,000 m³/day of municipal wastewater. After primary clarification, the BOD is 250 mg/l and is desired to have not more than 50 mg/l of soluble BOD in the effluent. A completely mixed reactor is to be used. MLSS concentration is of 3000 mg/l. Determine: i) The volume of reactor; ii) Detention time; iii) The recycle ratio.

## Short answer

Reactor volume 2083 m³ using Lawrence-McCarty equation with θc=10 days, Y=0.5, kd=0.06; detention time 5 h; recycle ratio 0.43 assuming return sludge concentration 10,000 mg/L.

## Answer

#### Assumptions (not given, typical values)

Yield $Y=0.5$ kg cells/kg BOD removed; endogenous decay $k_d=0.06$/day; mean cell residence time (sludge age) $\theta_c=10$ days; effluent suspended solids neglected; recycled sludge concentration $X_R=10000$ mg/l (SVI about 100 ml/g).

#### (i) Volume of the reactor

For a completely mixed reactor (Lawrence-McCarty):

$$
V=\frac{\theta_c\,Y\,Q\,(S_0-S)}{X\,(1+k_d\theta_c)}=\frac{10\times0.5\times10000\times(250-50)}{3000\times(1+0.06\times10)}=2083\ \text{m}^3
$$

(Check: $F/M=\dfrac{QS_0}{VX}=\dfrac{10000\times250}{2083\times3000}=0.40$/day, within the usual 0.2-0.6.)

#### (ii) Detention time

$$
t=\frac{V}{Q}=\frac{2083}{10000}=0.208\ \text{day}=5.0\ \text{h}
$$

#### (iii) Recycle ratio

A balance on the biomass in the reactor (no solids in the effluent) gives, for the aeration tank, $X(Q+Q_R)=X_RQ_R$:

$$
r=\frac{Q_R}{Q}=\frac{X}{X_R-X}=\frac{3000}{10000-3000}=0.43
$$

So the return sludge flow is $Q_R=4286$ m³/day. (Waste sludge: $Q_w=\dfrac{VX}{\theta_cX_R}=62.5$ m³/day.)

**Answer: (i) $V\approx 2083$ m³; (ii) detention time $\approx 5.0$ h; (iii) recycle ratio $\approx 0.43$.**

## Check yourself

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

1. Why does the answer state assumptions like Y = 0.5, kd = 0.06/day and theta_c = 10 days?
   A. They are derived from the BOD data
   B. The question did not give them, so typical design values are assumed
   C. They are fixed by law for all plants
   D. They are required only for the recycle ratio
   Answer: B. The Lawrence-McCarty design needs kinetic constants and sludge age that must be assumed when not given. (Key point: Assumptions)

2. Which expression gives the volume for a completely mixed reactor?
   A. V = Q S0 / X
   B. V = Q x theta_c
   C. V = theta_c Y Q (S0 - S) / (X (1 + kd theta_c)) = 2083 m3
   D. V = Y Q S0 / (kd X)
   Answer: C. Substituting 10 x 0.5 x 10000 x 200 / (3000 x 1.6) gives 2083 m3. (Key point: Reactor volume)

3. What is the detention time?
   A. t = V/Q = 0.208 day = 5.0 h
   B. t = 2.08 days
   C. t = 10 days (the sludge age)
   D. t = 0.5 h
   Answer: A. 2083/10000 = 0.208 day = 5.0 h; do not confuse with sludge age. (Key point: Detention time)

4. How is the recycle ratio obtained?
   A. r = Xr/X = 3.3
   B. r = (Xr - X)/X = 2.33
   C. r = X/Xr = 0.30
   D. From a biomass balance X(Q + Qr) = Xr Qr, so r = X/(Xr - X) = 3000/7000 = 0.43
   Answer: D. Balancing solids around the aeration tank with Xr = 10 000 mg/l gives r = 0.43. (Key point: Recycle ratio)

## Related in Sanitary Engineering

- [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
- [1-day BOD 300 mg/l to 5-day at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-1day-300-to-5day-30c) — Sanitary Engineering, chapter 5
- [Degree of treatment, industrial wastewater](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Adegree-treatment-industrial-waste) — Sanitary Engineering, chapter 6
- [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×
- [Is treatment necessary? Town wastewater 300 l/s](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atreatment-needed-town-300-ls) — Sanitary Engineering, chapter 6
- [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

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