---
title: "Sludge volume before and after digestion"
subject: "Sanitary Engineering"
chapter: 8
key: "semester_6/sanitary_engineering:sludge-volume-before-after-600kg"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sludge-volume-before-after-600kg
site: BCT Engineering
---

# Sludge volume before and after digestion

Sanitary Engineering · Chapter 8: Sludge Treatment and Disposal

- Asked in: 2078 Chaitra, 2+6 marks
- Video: [Sludge Thickening and Digestion | Waste Water Engineering](https://www.youtube.com/watch?v=hlsdeYCCFrw) — APSEd
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sludge-volume-before-after-600kg
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-8

## Question

Determine the sludge volume before and after digestion and percentage reduction for 600 kg (dry basis) of primary sludge having following characteristics:

| | Primary Sludge | Digested Sludge |
|---|---|---|
| Solids (%) | 6 | 12 |
| Volatile Matter (%) | 65 | Digested removes 65% of volatile matter. |

Take specific gravity of volatile and fixed solids 1.0 and 2.5 respectively for both primary and digested sludge.

## Short answer

For 600 kg dry solids, primary sludge: volatile 390 kg, fixed 210 kg, water 9400 kg, volume 9.874 m³. After digestion removing 65% volatile, dry solids 346.5 kg, water 2541 kg, volume 2.761 m³; reduction 72.0%.

## Answer

#### Method
Sludge volume $=$ volume of water $+$ volume of volatile solids $+$ volume of fixed solids, with solids $=600$ kg dry. Water is 1000 kg/m³; volatile solids $G=1.0$ (1000 kg/m³); fixed solids $G=2.5$ (2500 kg/m³).

#### Primary sludge (6% solids, 65% volatile)

$$
\begin{aligned}
VM &= 0.65\times600=390\ \text{kg},\quad FM=210\ \text{kg}\\
\text{Mass of wet sludge} &= \frac{600}{0.06}=10000\ \text{kg}\ \Rightarrow\ \text{water}=9400\ \text{kg}=9.400\ \text{m}^3\\
\text{Volume of solids} &= \frac{390}{1000}+\frac{210}{2500}=0.4740\ \text{m}^3\\
V_{primary} &= 9.400+0.4740=9.874\ \text{m}^3
\end{aligned}
$$

#### Digested sludge (12% solids, 65% of volatile matter removed)

$$
\begin{aligned}
VM_d &= 390\times(1-0.65)=136.5\ \text{kg},\quad FM=210\ \text{kg}\ (\text{unchanged})\\
\text{Dry solids} &= 136.5+210=346.5\ \text{kg}\\
\text{Mass of wet sludge} &= \frac{346.5}{0.12}=2887.5\ \text{kg}\ \Rightarrow\ \text{water}=2541.0\ \text{kg}=2.541\ \text{m}^3\\
\text{Volume of solids} &= \frac{136.5}{1000}+\frac{210}{2500}=0.2205\ \text{m}^3\\
V_{digested} &= 2.541+0.2205=2.761\ \text{m}^3
\end{aligned}
$$

#### Percentage reduction

$$
\frac{9.874-2.761}{9.874}\times100=72.0\%
$$

**Answer: volume before digestion $=9.874$ m³, after digestion $=2.761$ m³, reduction $=72.0\%$.**

## Check yourself

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

1. How is the total volume of sludge obtained in this method?
   A. Dry mass divided by solids fraction only
   B. Mass of wet sludge divided by 2500
   C. Volume of water only
   D. Water volume + volatile solids volume + fixed solids volume
   Answer: D. Each part's volume is its mass divided by its density (1000 kg/m3 water and volatiles, 2500 kg/m3 fixed solids). (Key point: Volume components)

2. 600 kg dry solids at 6% solids has how much water?
   A. 9400 kg
   B. 600 kg
   C. 10000 kg
   D. 564 kg
   Answer: A. Wet mass = 600/0.06 = 10000 kg, so water = 10000 - 600 = 9400 kg (9.4 m3). (Key point: Primary sludge water)

3. What happens to the solids during digestion in this problem?
   A. Both volatile and fixed solids drop by 65%
   B. 65% of volatile matter is removed; fixed matter (210 kg) is unchanged
   C. Fixed solids are destroyed, volatiles remain
   D. All solids are retained
   Answer: B. Digested VM = 390 x 0.35 = 136.5 kg; fixed 210 kg stays, giving 346.5 kg dry solids. (Key point: Digestion effect on solids)

4. The volume reduction after digestion (9.874 m3 to 2.761 m3) is about
   A. 28.0%
   B. 80%
   C. 72.0%
   D. 65%
   Answer: C. (9.874-2.761)/9.874 x 100 = 72.0%. (Key point: Result)

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