---
title: "Digester diameter from 500 m³ sed. tank"
subject: "Sanitary Engineering"
chapter: 8
key: "semester_6/sanitary_engineering:sludge-digester-diameter-500m3"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sludge-digester-diameter-500m3
site: BCT Engineering
---

# Digester diameter from 500 m³ sed. tank

Sanitary Engineering · Chapter 8: Sludge Treatment and Disposal

- Asked in: 2069 Bhadra, 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-digester-diameter-500m3
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-8

## Question

Sewage with a suspended solid content of 200 mg/l flows continuously in a sedimentation tank of 500 m³ capacity. Sewage is detained in the sedimentation tank for 4 hours. Sixty percentages of solids are removed in the sedimentation tank during its detention. The sludge produced in the sedimentation tank has moisture content of 98% and specific gravity of 1.02. The sludge from sedimentation tank is fed to digester for its digestion. The volume of sludge is reduced to 40% of its original volume during digestion. Calculate the diameter of digester if its effective depth is 6 m. Assume detention period in the digester is 30 days.

## Short answer

Flow 3000 m³/d, dry solids removed 360 kg/d. Fresh sludge 17.65 m³/d, digested sludge 7.06 m³/d (40% of original). Digester volume 318 m³, diameter 8.2 m for 6 m depth.

## Answer

#### Flow

$$
Q = \frac{\text{Volume}}{\text{detention time}} = \frac{500}{4} = 125\ \text{m}^3\text{/h} = 3000\ \text{m}^3\text{/d}
$$

#### Sludge from the sedimentation tank

- Solids in sewage: $3000\times1000 \times 200\times10^{-6} = 600$ kg/d
- Removed (60%): $W_s = 360$ kg/d

$$
V_1 = \frac{360}{1000 \times 1.02 \times (1-0.98)} = 17.65\ \text{m}^3\text{/d}
$$

#### Digested sludge

$$
V_2 = 0.40 \times 17.65 = 7.06\ \text{m}^3\text{/d}
$$

#### Digester volume (T = 30 days)

$$
\begin{aligned} V &= \left[V_1 - \tfrac{2}{3}(V_1 - V_2)\right] T \\ &= \left[17.65 - \tfrac{2}{3}(17.65 - 7.06)\right] \times 30 \\ &= 10.59 \times 30 = 317.6\ \text{m}^3 \end{aligned}
$$

#### Diameter (effective depth 6 m)

$$
\frac{\pi}{4}D^2 \times 6 = 317.6 \Rightarrow D = \sqrt{\frac{4 \times 317.6}{\pi \times 6}} = 8.21\ \text{m}
$$

**Answer: diameter of digester $\approx 8.2$ m (say 8.3 m), digester volume 318 m³.**

## Check yourself

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

1. A 500 m3 tank with 4 h detention gives flow
   A. 125 m3/d
   B. 3000 m3/d
   C. 2000 m3/d
   D. 500 m3/d
   Answer: B. Q = 500/4 = 125 m3/h = 3000 m3/d. (Key point: Flow)

2. Fresh sludge volume V1 (98% moisture, 360 kg/d dry) is
   A. 7.06 m3/d
   B. 353 m3/d
   C. 17.65 m3/d
   D. 7.2 m3/d
   Answer: C. V1 = 360/(1000 x 1.02 x 0.02) = 17.65 m3/d. (Key point: Fresh sludge)

3. The volume is reduced to 40% of its original value, so V2 =
   A. 0.60 x 17.65 = 10.59 m3/d
   B. 17.65/0.4 = 44.1 m3/d
   C. 360/(1000 x 1.02 x 0.2) = 1.76 m3/d
   D. 0.40 x 17.65 = 7.06 m3/d
   Answer: D. The problem gives the reduction directly; V2 = 7.06 m3/d. (Key point: Digested volume)

4. Digester volume is 317.6 m3 at 6 m depth. Diameter is
   A. About 8.2 m
   B. About 52.9 m
   C. About 16.4 m
   D. About 4.1 m
   Answer: A. [17.65 - 2/3(10.59)] x 30 = 317.6 m3; D = sqrt(4 x 317.6/(pi x 6)) = 8.21 m. (Key point: Digester diameter)

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