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Sizing a family biogas plant

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.

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.

Data

Dung per day =6×10=60= 6 \times 10 = 60 kg.

(a) Gas production

Vg=60×0.036=2.16 m3/dayV_g = 60 \times 0.036 = 2.16\ \text{m}^3/\text{day}

(b) Digester volume

Slurry = dung + equal water =60+60=120= 60 + 60 = 120 kg/day. Taking slurry density about 1000 kg/m³, the feed is 0.12 m³/day.

Vd=feed rate×retention time=0.12×40=4.8 m3V_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

Vh=0.60×2.16=1.30 m3V_h = 0.60 \times 2.16 = 1.30\ \text{m}^3

(d) Persons served

2.160.34=6.35≈6 persons\frac{2.16}{0.34} = 6.35 \approx 6\ \text{persons}

Energy content (at about 22 MJ/m³): 2.16×22=47.52.16 \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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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