---
title: "Sizing a family biogas plant"
subject: "Energy Resources"
chapter: 3
key: "semester_6/energy_resources:biogas-plant-sizing-cattle-dung"
source: https://www.bctengineer.com/q/semester_6/energy_resources/biogas-plant-sizing-cattle-dung
site: BCT Engineering
---

# Sizing a family biogas plant

Energy Resources · Chapter 3: Bio energy

- Asked in: year not given, 6 marks
- Video: [Design  of  Biogas  Plant | Renewable Energy | Complete Calculations](https://www.youtube.com/watch?v=YhqOOBRxljk) — Therm Science
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/energy_resources/biogas-plant-sizing-cattle-dung
- Whole chapter: https://www.bctengineer.com/md/semester_6/energy_resources/chapter-3

## Question

A farmer keeps 6 cattle producing 10 kg of fresh dung per animal per day. The gas yield is 0.036 m³ per kg of dung. The dung is mixed with an equal mass of water, the retention time is 40 days and the gas holder is to store 60 % of the daily gas production. Estimate (a) the daily gas production, (b) the digester volume, (c) the gas holder volume, and (d) the number of persons whose cooking needs (0.34 m³ per person per day) can be met.

## Short answer

Daily gas production = 60 kg × 0.036 m³/kg = 2.16 m³/day. Digester volume = (120 kg slurry/day ≈ 0.12 m³/day) × 40 days = 4.8 m³. Gas holder volume = 0.6 × 2.16 = 1.30 m³. At 0.34 m³/person-day, the biogas satisfies cooking needs of about 6 persons.

## Answer

#### Data
Dung per day $= 6 \times 10 = 60$ kg.

#### (a) Gas production
$$
V_g = 60 \times 0.036 = 2.16\ \text{m}^3/\text{day}
$$

#### (b) Digester volume
Slurry = dung + equal water $= 60 + 60 = 120$ kg/day. Taking slurry density about 1000 kg/m³, the feed is 0.12 m³/day.

$$
V_d = \text{feed rate} \times \text{retention time} = 0.12 \times 40 = 4.8\ \text{m}^3
$$

About 10 % extra may be added for gas space and scum, giving about 5.3 m³ in practice.

#### (c) Gas holder volume
$$
V_h = 0.60 \times 2.16 = 1.30\ \text{m}^3
$$

#### (d) Persons served
$$
\frac{2.16}{0.34} = 6.35 \approx 6\ \text{persons}
$$

Energy content (at about 22 MJ/m³): $2.16 \times 22 = 47.5$ MJ per day.

**Answer:** gas = 2.16 m³/day; digester = 4.8 m³ (about 5.3 m³ with allowance); gas holder = 1.30 m³; cooking for about 6 persons.

## Check yourself

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

1. 6 cattle x 10 kg x 0.036 m³/kg gives...
   A. 21.6 m³/day
   B. 2.16 m³/day
   C. 0.36 m³/day
   D. 12.96 m³/day
   Answer: B. Dung is 60 kg/day; 60 x 0.036 = 2.16. (Key point: Daily gas)

2. How is the digester volume found?
   A. Daily slurry volume x retention time = 0.12 x 40 = 4.8 m³
   B. Gas production x 40
   C. Dung mass x 0.036
   D. Slurry mass / 40
   Answer: A. Slurry is 60 kg dung + 60 kg water = 120 kg, i.e. 0.12 m³/day. (Key point: Digester volume)

3. What is the gas holder volume?
   A. 0.40 x 2.16 = 0.86 m³
   B. 2.16 m³
   C. 4.8 m³
   D. 0.60 x 2.16 = 1.30 m³
   Answer: D. It stores 60% of daily gas production. (Key point: Gas holder)

4. How many persons can be served at 0.34 m³ per person per day?
   A. About 12
   B. About 1
   C. About 6 (2.16/0.34 = 6.35)
   D. About 63
   Answer: C. Divide daily gas by per-person need and round down. (Key point: Persons served)

## Related in Energy Resources

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