---
title: "Yearly sludge volume, 6 MLD"
subject: "Sanitary Engineering"
chapter: 8
key: "semester_6/sanitary_engineering:sludge-volume-yearly-6mld"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sludge-volume-yearly-6mld
site: BCT Engineering
---

# Yearly sludge volume, 6 MLD

Sanitary Engineering · Chapter 8: Sludge Treatment and Disposal

- Asked in: 2071 Magh, 4 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-volume-yearly-6mld
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-8

## Question

A sedimentation tank treats 6 mld of sewage containing 300 mg/l of suspended solids. The tank removes 65% of the suspended solids. Compute the volume of the sludge produced yearly if the moisture content of the sludge is 95%.

## Short answer

Dry solids removed: 1170 kg/d, sludge volume 22.9 m³/d (assuming sp. gr. 1.02, 95% moisture). Yearly volume = 8374 m³.

## Answer

The specific gravity of sludge is not given, so take $G_{sl} = 1.02$ (usual value for sewage sludge).

- Solids in raw sewage: $6\times10^6 \times 300\times10^{-6} = 1800$ kg/d
- Solids removed (65%): $W_s = 1170$ kg/d

$$
V = \frac{W_s}{\rho_w\,G_{sl}\,(1-p)} = \frac{1170}{1000 \times 1.02 \times 0.05} = 22.94\ \text{m}^3\text{/d}
$$

$$
V_{year} = 22.94 \times 365 = 8,374\ \text{m}^3
$$

**Answer: about 22.9 m³/day, i.e. about 8,374 m³ of sludge per year.**

## Check yourself

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

1. The specific gravity of sludge is not given. What is taken?
   A. 1.0 exactly with no solids correction
   B. 2.65
   C. 1.02, the usual value for sewage sludge
   D. 0.95
   Answer: C. Sewage sludge is mostly water, so G of about 1.02 is used. (Key point: Assumed G)

2. Daily dry solids removed are
   A. 1800 kg/d
   B. 630 kg/d
   C. 117 kg/d
   D. 1170 kg/d
   Answer: D. 6 x 10^6 x 300 x 10^-6 = 1800 kg/d; 65% gives 1170 kg/d. (Key point: Solids removed)

3. The yearly volume of sludge is
   A. About 8374 m3 (22.94 m3/d x 365)
   B. About 22.9 m3
   C. About 419 m3
   D. About 83,740 m3
   Answer: A. V = 1170/(1000 x 1.02 x 0.05) = 22.94 m3/d, times 365 days. (Key point: Yearly volume)

## Related in Sanitary Engineering

- [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 volume and digester, 4 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-volume-4mld-digester) — Sanitary Engineering, chapter 8
- [Sludge digestion tank, 6.5 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-digestion-tank-6p5mld) — Sanitary Engineering, chapter 8
- [Sludge volume from PST and SST](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-volume-pst-sst-20mld) — Sanitary Engineering, chapter 8
- [Digester volume for sludge, moisture 92%](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-digester-volume-92) — Sanitary Engineering, chapter 8, asked 2×
- [Sludge digestion tank, 812.5 m³ sed. tank](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-digestion-tank-812) — Sanitary Engineering, chapter 8

## Related in other subjects

- [Rectangular sedimentation tank for 10 MLD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-s71) — Water Supply Engineering, chapter 6
- [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
- [Plain sedimentation tank for 4 MLD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-b78) — Water Supply Engineering, chapter 6