---
title: "Sludge digestion tank, 6.5 MLD"
subject: "Sanitary Engineering"
chapter: 8
key: "semester_6/sanitary_engineering:sludge-digestion-tank-6p5mld"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sludge-digestion-tank-6p5mld
site: BCT Engineering
---

# Sludge digestion tank, 6.5 MLD

Sanitary Engineering · Chapter 8: Sludge Treatment and Disposal

- Asked in: 2075 Baisakh, 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-digestion-tank-6p5mld
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-8

## Question

Design a sludge digestion tank to treat sludge of primary sedimentation tank from the following data: a) Average flow of the sewage = 6.5 MLD; b) Total suspended solids in raw sewage = 250 mg/l; c) Water content of fresh sludge = 95%; d) Water content of digested sludge = 85%; e) Specific gravity of sludge = 1.02; f) Digestion period = 2 months; g) Primary settling tank removes 55% of suspended solids.

## Short answer

Dry solids removed are 893.8 kg/day; fresh sludge volume 17.52 m³/day and digested sludge 5.84 m³/day. Using parabolic digestion over 60 days gives capacity 584 m³; adopt a circular tank of diameter 11.5 m and side water depth 6 m.

## Answer

#### Data and assumptions
Flow $Q=6500$ m³/day; suspended solids (SS) $=250$ mg/l; removal in the primary tank $=55\%$; fresh sludge moisture $=95\%$; digested sludge moisture $=85\%$; specific gravity of sludge $=1.02$; digestion period $=60$ days (2 months). The digested sludge is taken to hold the same dry solids as the fresh sludge (solids loss in digestion is ignored, which is conservative); no extra storage period is added.

#### Step 1: Dry solids removed per day

$$
W_s=Q\times SS\times\text{removal}=6500\times250\times0.55\times10^{-3}=893.8\ \text{kg/day}
$$

#### Step 2: Volume of fresh sludge

$$
V_1=\frac{W_s}{(1-0.95)\times S\times1000}=\frac{893.8}{0.05\times1.02\times1000}=17.52\ \text{m}^3/\text{day}
$$

#### Step 3: Volume of digested sludge (volume-moisture relation)

$$
V_2=V_1\times\frac{100-95}{100-85}=17.52\times\frac{5}{15}=5.84\ \text{m}^3/\text{day}
$$

#### Step 4: Capacity of the digester (parabolic digestion)

$$
V=\left[V_1-\frac{2}{3}(V_1-V_2)\right]t=\left[17.52-\frac{2}{3}(17.52-5.84)\right]\times60=584\ \text{m}^3
$$

#### Step 5: Dimensions
Adopt a circular digester with a side water depth of 6 m (usual 6-12 m), plus 0.5 m free board and a hopper bottom with a 1 : 1 slope:

$$
A=\frac{584}{6}=97.3\ \text{m}^2,\qquad D=\sqrt{\frac{4A}{\pi}}=11.13\ \text{m}\ (\text{adopt 11.5 m})
$$

**Answer: digester capacity $\approx 584$ m³, circular, diameter 11.5 m, side water depth 6 m (plus free board, hopper and gas dome).**

## Check yourself

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

1. Dry solids produced per day are found from
   A. Q x SS only (1625 kg/d)
   B. Q x SS x 0.45
   C. Q x 0.05 x 1.02
   D. Q x SS x removal fraction (6500 x 250 x 0.55 = 893.8 kg/d)
   Answer: D. Only the 55% of suspended solids removed in the primary tank end up in the sludge. (Key point: Dry solids per day)

2. Fresh sludge volume V1 = Ws / ((1 - 0.95) x S x 1000) is about
   A. 17.52 m3/day
   B. 893.8 m3/day
   C. 0.88 m3/day
   D. 87.6 m3/day
   Answer: A. 893.8 / (0.05 x 1.02 x 1000) = 17.52 m3/day. (Key point: Fresh sludge volume)

3. Going from 95% to 85% moisture, the digested volume V2 is
   A. V1 x 85/95 = 15.7 m3/day
   B. V1 x 5/15 = 5.84 m3/day
   C. V1 x 15/5 = 52.6 m3/day
   D. V1 x 10/95 = 1.84 m3/day
   Answer: B. V2/V1 = (100-95)/(100-85) = 5/15. (Key point: Digested sludge volume)

4. The digester capacity formula used is
   A. (V1 + V2)/2 x t = 700 m3
   B. V1 x t = 1051 m3
   C. [V1 - 2/3 (V1 - V2)] x t = 584 m3
   D. V2 x t = 350 m3
   Answer: C. Sludge volume falls along a parabolic curve, so the average volume is V1 - 2/3(V1-V2); times 60 days gives 584 m3. (Key point: Digester capacity)

## Related in Sanitary Engineering

- [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
- [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
- [Yearly sludge volume, 6 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-volume-yearly-6mld) — Sanitary Engineering, chapter 8
- [Sedimentation tank and oxidation pond design](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ased-tank-and-oxidation-pond-10000) — Sanitary Engineering, chapter 7
- [Circular sedimentation tank for 58 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ased-tank-58mld-circular) — Sanitary Engineering, chapter 7

## 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