---
title: "Digester volume for sludge, moisture 92%"
subject: "Sanitary Engineering"
chapter: 8
key: "semester_6/sanitary_engineering:sludge-digester-volume-92"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sludge-digester-volume-92
site: BCT Engineering
---

# Digester volume for sludge, moisture 92%

Sanitary Engineering · Chapter 8: Sludge Treatment and Disposal

- Asked in: 2076 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/sludge-digester-volume-92
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-8

## Question

A sewage treatment plant produces sludge having a moisture content of 92% with a dry solids of 500 kg/d in which 70% solids are volatile. The specific gravity of volatile and fixed solids are 1.02 and 2.65 respectively. Assuming parabolic digestion of 30 days and 12 days of monsoon storage, calculate the volume of conventional rate digester required to produce digested sludge with a moisture content of 90%.

## Short answer

For 92% moisture, 500 kg/d dry solids (70% volatile), 50% VS destruction, 30-day digestion, 12-day storage: raw sludge volume V1=6.150 m3/d, digested sludge V2=3.153 m3/d. Using parabolic formula V=[V1-2/3(V1-V2)]td+V2ts, digester volume = 162.4 m3.

## Answer

For a conventional-rate digester the capacity with parabolic variation of sludge volume is

$$
V=\left[V_1-\frac{2}{3}(V_1-V_2)\right]t_d+V_2\,t_s
$$

where $V_1$ = raw sludge volume per day, $V_2$ = digested sludge volume per day, $t_d=30$ days (digestion), $t_s=12$ days (storage during monsoon).

**Assumption:** 50% of the volatile solids are destroyed in digestion (usual 40-60%).

**Raw sludge** (500 kg dry solids/day, 70% volatile, moisture 92% i.e. 8% solids):

$$
\begin{aligned}
VS &= 350\ \text{kg},\quad FS=150\ \text{kg}\\
\text{Mass of wet sludge} &= \frac{500}{0.08}=6250\ \text{kg},\ \text{water}=5.750\ \text{m}^3\\
\text{Volume of solids} &= \frac{350}{1020}+\frac{150}{2650}=0.3997\ \text{m}^3\\
V_1 &= 5.750+0.3997=6.150\ \text{m}^3/\text{day}
\end{aligned}
$$

**Digested sludge** (VS left $=175$ kg, FS $=150$ kg, solids $=325$ kg, 90% moisture so 10% solids):

$$
\begin{aligned}
\text{Mass of wet sludge} &= \frac{325}{0.10}=3250\ \text{kg},\ \text{water}=2.925\ \text{m}^3\\
\text{Volume of solids} &= \frac{175}{1020}+\frac{150}{2650}=0.2282\ \text{m}^3\\
V_2 &= 2.925+0.2282=3.153\ \text{m}^3/\text{day}
\end{aligned}
$$

**Volume of digester:**

$$
V=\left[6.150-\frac{2}{3}(6.150-3.153)\right]\times30+3.153\times12=162.4\ \text{m}^3
$$

**Answer: digester volume $\approx 162.4$ m³ (say 162 m³).**

## Check yourself

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

1. At 92% moisture, 500 kg/day dry solids corresponds to a wet sludge mass of
   A. 625 kg
   B. 6250 kg
   C. 4000 kg
   D. 500 kg
   Answer: B. Solids are 8%, so 500/0.08 = 6250 kg. (Key point: Raw sludge mass)

2. How is the volume of the solids found?
   A. By dividing the total mass by 1000
   B. By dividing dry mass by the mean 1.83
   C. By VS/(1.02 x 1000) + FS/(2.65 x 1000)
   D. By multiplying by specific gravities
   Answer: C. Volatile (350 kg, SG 1.02) and fixed (150 kg, SG 2.65) are treated separately, giving 0.3997 m3. (Key point: Volume of solids)

3. The raw sludge volume V1 is
   A. 8.23 m3/day
   B. 5.75 m3/day
   C. 6.15 m3/day
   D. 3.15 m3/day
   Answer: C. Water 5.750 m3 plus solids 0.3997 m3 = 6.150 m3/day. (Key point: Raw volume V1)

4. The digester volume obtained is about
   A. 162 m3
   B. 183 m3
   C. 98 m3
   D. 6.15 m3
   Answer: A. V = [6.150 - (2/3)(6.150 - 3.153)] x 30 + 3.153 x 12 = 162.4 m3. (Key point: Final digester volume)

## Close variants set in other exams

- [Sludge treatment aims; digester volume (94%)](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-aims-digestion-volume-94) — Sanitary Engineering, chapter 8, asked 2×

## Related in Sanitary Engineering

- [Volume-moisture relation; specific gravity](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-specific-gravity-95) — Sanitary Engineering, chapter 8
- [Sludge volume and digester, 4 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-volume-4mld-digester) — Sanitary Engineering, chapter 8
- [Digester diameter from 500 m³ sed. tank](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-digester-diameter-500m3) — Sanitary Engineering, chapter 8
- [Sludge volume before and after digestion](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-volume-before-after-600kg) — Sanitary Engineering, chapter 8
- [Sludge volume from sedimentation tank, 10 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-volume-10mld-250) — Sanitary Engineering, chapter 8
- [Sludge treatment need; digester sketch](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-necessity-digester-sketch) — Sanitary Engineering, chapter 8

## Related in other subjects

- [Quality, moisture content, specific volume of mixture](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Aquality-moisture-specific-volume) — Fundamental of Thermodynamics & Heat Transfer, chapter 3