---
title: "Sludge digestion tank, 812.5 m³ sed. tank"
subject: "Sanitary Engineering"
chapter: 8
key: "semester_6/sanitary_engineering:sludge-digestion-tank-812"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sludge-digestion-tank-812
site: BCT Engineering
---

# Sludge digestion tank, 812.5 m³ sed. tank

Sanitary Engineering · Chapter 8: Sludge Treatment and Disposal

- Asked in: 2072 Magh, 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-digestion-tank-812
- 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: Capacity of sedimentation tank = 812.5 m³; Detained time in Sedimentation = 3 hrs; Suspended Solids in raw sewage = 250 mg/lit; Water content in fresh sludge = 95%; Water content in digested sludge = 80%; Specific gravity of sludge = 1.02; Digestion period in digester = 2 months; Primary sedimentation tank removes 55% of suspended solids.

## Short answer

Compute dry solids (894 kg/d), fresh sludge volume (17.5 m³/d), digested sludge volume (4.4 m³/d). Digester capacity using parabolic formula is 526 m³, circular tank 11 m diameter, 6 m side water depth.

## Answer

#### Flow from the sedimentation tank
The tank capacity is 812.5 m³ with detention time of 3 h, so

$$
Q=\frac{812.5}{3}\times24=6500.0\ \text{m}^3/\text{day}
$$

#### 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 $=80\%$; 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)

$$
V_2=V_1\times\frac{100-95}{100-80}=17.52\times\frac{5}{20}=4.38\ \text{m}^3/\text{day}
$$

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

$$
V=\left[V_1-\frac{2}{3}(V_1-V_2)\right]t=\left[17.52-\frac{2}{3}(17.52-4.38)\right]\times60=526\ \text{m}^3
$$

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

$$
A=\frac{526}{6}=87.6\ \text{m}^2,\qquad D=\sqrt{\frac{4A}{\pi}}=10.56\ \text{m}\ (\text{adopt 11.0 m})
$$

**Answer: digester capacity $\approx 526$ m³, circular, diameter 11.0 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. Tank capacity 812.5 m3 with 3 h detention gives a daily flow of
   A. 2437.5 m3/day
   B. 812.5 m3/day
   C. 270.8 m3/day
   D. 6500 m3/day
   Answer: D. Q = 812.5/3 x 24 = 6500 m3/day. (Key point: Flow from tank volume)

2. With digested sludge at 80% moisture, V2 is
   A. V1 x 5/20 = 4.38 m3/day
   B. V1 x 5/15 = 5.84 m3/day
   C. V1 x 20/5 = 70 m3/day
   D. V1 x 80/95 = 14.8 m3/day
   Answer: A. V2/V1 = (100-95)/(100-80) = 5/20 with V1 = 17.52 m3/day. (Key point: Digested volume)

3. The digester capacity is about
   A. 584 m3
   B. 526 m3
   C. 1051 m3
   D. 263 m3
   Answer: B. [17.52 - 2/3(17.52-4.38)] x 60 = 8.76 x 60 = 526 m3 (584 is the 85% moisture case). (Key point: Digester capacity)

4. With 526 m3 and a side water depth of 6 m, the diameter is found from
   A. D = 526/6 = 87.6 m
   B. A = 526 x 6
   C. A = 526/6 = 87.6 m2, D = sqrt(4A/pi) = 10.56 m (adopt 11 m)
   D. D = sqrt(526) = 22.9 m
   Answer: C. Plan area is volume over depth, and diameter follows from A = pi D^2/4. (Key point: Dimensions)

## Related in Sanitary Engineering

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- [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
- [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 and digester, 4 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-volume-4mld-digester) — Sanitary Engineering, chapter 8
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- [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

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