---
title: "Sludge volume and digester, 4 MLD"
subject: "Sanitary Engineering"
chapter: 8
key: "semester_6/sanitary_engineering:sludge-volume-4mld-digester"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sludge-volume-4mld-digester
site: BCT Engineering
---

# Sludge volume and digester, 4 MLD

Sanitary Engineering · Chapter 8: Sludge Treatment and Disposal

- Asked in: 2073 Bhadra, 8 marks
- Video: [Design of Sludge Digestion tank | Waster water Engineering | Sludge Digestion tank size & Efficiency](https://www.youtube.com/watch?v=ZgSskDho-MA) — 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/sludge-volume-4mld-digester
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-8

## Question

A raw sewage having suspended solids content of 220 mg/lit is passed through primary sedimentation tank at a flow of 4 MLD. The PST removes 55% suspended solids. Determine the volume of sludge produced per day if moisture content and specific gravity of sludge are 95% and 1.02 respectively. What will be the volume if its moisture content reduces to 80% after digestion? Also design a digester for sludge digestion period of 80 days.

## Short answer

Fresh sludge: 9.49 m³/d (95% moisture). Digested sludge: 2.37 m³/d (80% moisture). Digester volume 380 m³ using 80-day parabolic digestion, circular tank 9.0 m diameter and 6 m effective depth.

## Answer

The mass of dry solids stays the same while the water content falls, so $V = \dfrac{W_s}{\rho_w\,G_{sl}\,(1-p)}$.

#### Fresh sludge (95% moisture)

- Solids in raw sewage: $4\times10^6 \times 220\times10^{-6} = 880$ kg/d
- Removed (55%): $W_s = 484.0$ kg/d

$$
V_1 = \frac{484.0}{1000 \times 1.02 \times 0.05} = 9.49\ \text{m}^3\text{/d}
$$

#### Digested sludge (80% moisture)

$$
V_2 = \frac{484.0}{1000 \times 1.02 \times 0.20} = 2.37\ \text{m}^3\text{/d}
$$

(solids loss during digestion is neglected, and the sludge specific gravity is taken as 1.02 in both cases.)

#### Digester capacity (T = 80 days)

$$
\begin{aligned} V &= \left[V_1 - \tfrac{2}{3}(V_1 - V_2)\right] T \\ &= \left[9.49 - \tfrac{2}{3}(9.49 - 2.37)\right] \times 80 \\ &= 4.75 \times 80 = 380\ \text{m}^3 \end{aligned}
$$

Adopt effective depth 6 m (assumed): 

$$
\frac{\pi}{4} D^2 \times 6 = 380 \Rightarrow D = 8.98\ \text{m} \approx 9.0\ \text{m}
$$

Provide a circular digester of 9.0 m diameter and 6 m effective depth, plus 0.5 m free board and a hopper bottom (slope 1 vertical to 1.5 horizontal) with a sludge withdrawal pipe.

**Answer: fresh sludge 9.49 m³/d; digested sludge 2.37 m³/d; digester capacity 380 m³ (diameter about 9.0 m).**

## Check yourself

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

1. Dry solids mass in the fresh sludge is
   A. 880 kg/d
   B. 396 kg/d
   C. 9.49 kg/d
   D. 484 kg/d (55% of 880 kg/d)
   Answer: D. Solids in sewage = 4 x 10^6 x 220 x 10^-6 = 880 kg/d; 55% is removed. (Key point: Dry solids)

2. The volume at 95% moisture is
   A. 9.49 m3/d
   B. 2.37 m3/d
   C. 47.5 m3/d
   D. 23.7 m3/d
   Answer: A. V1 = 484/(1000 x 1.02 x 0.05) = 9.49 m3/d. (Key point: Fresh volume)

3. The volume at 80% moisture is found using
   A. Solids fraction 0.80, giving 0.59 m3/d
   B. Same dry mass, solids fraction 0.20: V2 = 484/(1000 x 1.02 x 0.20) = 2.37 m3/d
   C. Moisture 0.80, giving 9.49 x 0.8
   D. V1 x 80/95
   Answer: B. The solids fraction is 1 - 0.80 = 0.20, not 0.80. (Key point: Digested volume)

4. Digester capacity for T = 80 days is
   A. 9.49 x 80 = 759 m3
   B. 2.37 x 80 = 190 m3
   C. [9.49 - 2/3(9.49 - 2.37)] x 80 = 380 m3
   D. (9.49+2.37)/2 x 80 = 474 m3
   Answer: C. The parabolic formula gives average 4.75 m3/d; times 80 days is 380 m3 (D about 9 m at 6 m depth). (Key point: Digester capacity)

## 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 from PST and SST](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-volume-pst-sst-20mld) — 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
- [Yearly sludge volume, 6 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-volume-yearly-6mld) — Sanitary Engineering, chapter 8
- [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
- [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

## 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
- [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