Digester volume is the daily slurry volume multiplied by the retention time, so a plant fed 40-60 litres of slurry a day for 20-40 days needs 0.8-2.4 m³ of liquid volume, not the 4-6 m³ the parent’s table gives for 2 cows. The table’s sizes imply retention of 53 to 300 days, and its 5-8% daily loading implies 12.5 to 20 days, so the table, the retention rule and the loading rule cannot all be right, and the volume should be computed from your own measured manure.
Part of Biogas Digesters
Sizing a digester is calculating its liquid volume from the daily feed and the retention time, and its gas output from the manure yield.
The Sizing Relation
V = Q × HRT, where Q is the slurry added each day and HRT the retention time in days (Temperature & Retention Time). The parent says to mix manure with an equal weight of water before feeding, so Q = 2 × M, taking 1 kg of slurry as 1 litre (an approximation), where M is the kilograms of manure per day.
Auditing The Parent’s Table
| Row | Manure M (kg/day) | Slurry Q = 2M (L/day) | Parent’s volume (m³) | Implied retention (days) |
|---|---|---|---|---|
| 2 cows | 20-30 | 40-60 | 4-6 | 4,000 ÷ 60 = 67 to 6,000 ÷ 40 = 150 |
| 5 cows | 50-75 | 100-150 | 8-12 | 8,000 ÷ 150 = 53 to 12,000 ÷ 100 = 120 |
| 10 pigs | 15-25 | 30-50 | 4-6 | 4,000 ÷ 50 = 80 to 6,000 ÷ 30 = 200 |
| 50 chickens | 5-8 | 10-16 | 2-3 | 2,000 ÷ 16 = 125 to 3,000 ÷ 10 = 300 |
At the parent’s own 20-40 days the 2-cow row needs only 40 × 20 ÷ 1,000 = 0.8 m³ up to 60 × 40 ÷ 1,000 = 2.4 m³ of liquid. So the table’s volumes correspond to 53-300 days of retention, against the parent’s own 20-40 days for a mesophilic plant (a factor of 1.3 to 15 too large). The parent’s daily loading of 5-8% of the volume is the other extreme: 1 ÷ 0.05 = 20 days and 1 ÷ 0.08 = 12.5 days. Its “0.5-1 m³ per cow” does fit: a cow giving 10-15 kg a day (the table’s rate, 20-30 kg for 2 cows) has Q = 20-30 L/day, and 20 × 20 ÷ 1,000 = 0.4 m³ up to 30 × 40 ÷ 1,000 = 1.2 m³ per cow. So the rule of thumb and the retention figure agree, and the table is the outlier.
Auditing The Gas Column
Gas = manure (tonnes) × yield (m³ per tonne), using the parent’s yields (cow 30-50, pig 40-60, chicken 50-80):
| Row | Computed gas (m³/day) | Parent’s table |
|---|---|---|
| 2 cows | 20 × 30 ÷ 1,000 = 0.6 to 30 × 50 ÷ 1,000 = 1.5 | 1.0-1.5 |
| 5 cows | 1.5 to 3.75 | 2.5-4.0 |
| 10 pigs | 15 × 40 ÷ 1,000 = 0.6 to 25 × 60 ÷ 1,000 = 1.5 | 1.0-2.0 |
| 50 chickens | 5 × 50 ÷ 1,000 = 0.25 to 8 × 80 ÷ 1,000 = 0.64 | 0.5-1.0 |
The table’s upper figures exceed the computed ones by 0.25 (cows), 0.5 (pigs) and 0.36 m³ (chickens); the 2-cow row just fits.
A Second Disagreement: How Much Manure A Cow Makes
The parent’s table uses 10-15 kg per cow per day. The corpus’s manure pages give 25-30 kg (Manure Use), 25-35 kg (Manure Management) and 10-15 tonnes a year (Animal Manure, 27-41 kg a day). Breed, size and feed explain part of the difference, and only manure you can collect counts, since dung dropped in a pasture stays there. At 25-35 kg per cow, two cows give M = 50-70 kg, Q = 100-140 L/day, and V = 100 × 20 ÷ 1,000 = 2.0 m³ up to 140 × 40 ÷ 1,000 = 5.6 m³, which does bracket the parent’s 4-6 m³ for 2 cows; the gas would be 50 × 30 ÷ 1,000 = 1.5 to 70 × 50 ÷ 1,000 = 3.5 m³ a day. Weigh a week of the dung you actually collect before sizing.
What You Need
- Your measured daily manure, in kilograms, from a week of weighing
- A chosen retention time from climate (20-40 days at 30-40°C)
- A calculator or a slate
Method: Size From Measurement
Step 1: Weigh daily collected manure M; take the average.
Step 2: Q = 2 × M litres.
Step 3: V = Q × HRT ÷ 1,000 cubic metres of liquid; add gas space per the design (dome, drum, or the unfilled quarter of a tube).
Step 4: Gas per day = M ÷ 1,000 × yield (30-50 for cow manure).
Step 5: Compare with need: cooking for 4-5 people takes 1.5-2.0 m³ a day (parent). That is 1.5 ÷ 0.050 = 30 kg up to 2.0 ÷ 0.030 = 67 kg of cow manure a day; at 10-15 kg per cow that is 2 to 7 cows, at 25-35 kg per cow, 1 to 3. If your manure is less, the family cooks on gas for part of the day only.
Worked Example
Two cows, 20 kg of manure collected a day (measured). Q = 40 L. HRT 30 days: V = 40 × 30 ÷ 1,000 = 1.2 m³ liquid. Gas: 20 × 30 ÷ 1,000 = 0.6 to 20 × 50 ÷ 1,000 = 1.0 m³ a day, which is 11-25 MJ (at 18-25 MJ/m³), against the 1.5-2.0 m³ a day cooking needs. Warming the 40 L from 10 to 35°C costs 4.2 MJ, 17-39% of that gas.
Common Mistakes
- Copying the parent’s table without checking its retention.
- Counting manure that cannot be collected.
- Sizing for a cooking need the manure cannot meet.
Larger plants hold larger volumes of gas
A bigger digester holds more methane, carbon dioxide and hydrogen sulfide in a bigger pit. The rules do not change: no entry, no flame, a lid or rail on every pit, and a relief path for gas (Pressure Regulation, Anaerobic Digestion Science).
Sources And Assumptions
- The relation V = Q × HRT is the definition of retention time.
- Parent’s figures from the carrier; manure rates from the corpus pages named above; 4.18 kJ/kg·K for water; energy per cubic metre of biogas as on the science page (18-25 MJ).
- 1 kg of slurry taken as 1 litre.
What’s Next
- Learn next: Construction
- Combine with: N Ratio & Feedstock Selection