Cluster article
Flour Mill Capacity Calculation: From Demand to Tonnage in Four Steps
A practical calculation method for specifying flour mill capacity — extraction rate, utilisation headroom, shift pattern and wheat quality all change the number you should buy. Includes worked examples.
Key takeaways
- Capacity in tonnes per day means wheat intake, not flour output. The two differ by the extraction rate.
- Always divide by 0.85 to convert average demand into required capacity — plants need maintenance windows.
- Shift pattern changes the plant you need as much as demand does.
Capacity numbers in the milling industry are quoted in tonnes of wheat per 24 hours. This trips up almost every first-time buyer, because what they know is how much flour they can sell.
The conversion has four steps, and skipping any one of them gives you the wrong plant.
Step 1 — Flour demand into wheat demand
Divide your finished flour requirement by the extraction rate.
| Product mix | Extraction | Wheat per tonne of flour |
|---|---|---|
| Patent only | 68–72% | 1.39–1.47 t |
| Two grades | 75–78% | 1.28–1.33 t |
| Atta / whole wheat | 85–92% | 1.09–1.18 t |
If you need 20 t of flour per day at 76% extraction, you need 26.3 t of wheat.
Step 2 — Add process losses
Cleaning removes 1.5–4% by weight. Tempering adds water but sifting and conveying lose material. Spillage is real.
Multiply by 1.08. That covers 4% cleaning loss, 2% process loss and 2% handling loss.
26.3 × 1.08 = 28.4 t/day
Step 3 — Divide by utilisation
No plant runs at 100% every day. Maintenance, cleaning breaks, power interruptions and demand troughs all take a bite.
Divide by 0.85. That assumes a competent operation with one maintenance day per week and modest seasonal variation. If your market is highly seasonal, use 0.75.
28.4 ÷ 0.85 = 33.4 t/day
You need a 35 t/d plant, or a 40 t/d with comfortable headroom.
Step 4 — Check it against your shift pattern
Published capacity assumes 24-hour operation. If you plan one 10-hour shift, you get 10/24 = 42% of rated capacity.
This is where buyers get badly caught. A 30 t/d plant run for one shift produces 12.5 t of wheat throughput — about 9.5 t of flour.
Two responses:
- Buy bigger. A 60 t/d plant on one shift gives you 25 t/d of wheat throughput.
- Run more shifts. Same plant, three shifts, full output — but you need three crews.
Most new mills start with one or two shifts and buy capacity for three. Labour is easier to add than a building.
A complete example
Target: 15 t of flour per day, two grades, 76% extraction, two shifts.
| Step | Calculation | Result |
|---|---|---|
| Wheat for flour | 15 ÷ 0.76 | 19.7 t |
| Plus losses | 19.7 × 1.08 | 21.3 t |
| Plus utilisation | 21.3 ÷ 0.85 | 25.1 t |
| Shift adjustment | 25.1 ÷ 0.83 (two shifts) | 30.2 t |
| Plant to buy | 30 t/d (6FTDP-30) |
Note the shift adjustment: two shifts out of 24 hours is 83% of rated capacity, so the required rated capacity rises.
Capacity versus building
Once you have the tonnage, the building follows:
| Capacity | Building (L × W × H) | Installed power |
|---|---|---|
| 20 t/d | 12 × 5 × 4 m | 51 kW |
| 25 t/d | 15 × 5 × 4 m | 67 kW |
| 30 t/d | 18 × 6 × 6.5 m | 92 kW |
| 40 t/d | 18 × 8 × 8 m | 100 kW |
| 60 t/d | 24 × 8 × 8 m | 150 kW |
| 100 t/d | 35 × 7.5 × 18.5 m | 308 kW |
| 200 t/d | 50 × 7.5 × 22.5 m | 690 kW |
If your available building cannot accommodate the tonnage you calculated, either the tonnage is wrong or the building is.
The three numbers people get wrong
Using flour tonnage as plant tonnage. Off by 25–30% in the wrong direction.
Forgetting utilisation. Produces a plant that is chronically 15% short and cannot catch up after any disruption.
Forgetting shift pattern. Produces a plant that delivers 40% of what the buyer expected in month one.