Forced-Air Cooling & Size the Fan & Airflow
Sizes the airflow (cfm)
Enter the load, airflow per pound, field heat and supply-air temperature and read the total airflow (cfm and m³/s), stack static pressure and fan power (kW and hp) — and whether your design 7/8-cools the produce within the target time.
Size the cooler
Next: lower the supply-air temperature below the 0°C target — forced air cannot pull pulp below the air doing the cooling.
Airflow Q (cfm) = (cfm/lb) × mass(lb); fan power ≈ Q·ΔP/η. 7/8-cooling time falls ~ with airflow^−0.78 toward a product-conduction floor (extra airflow above ~2 cfm/lb buys little). Sources: Thompson et al., UC ANR 21567 'Commercial Cooling of Fruits, Vegetables, and Flowers'; ASHRAE Refrigeration; UC-Davis Postharvest.
Forced-air fan sizing — key facts
- Design airflow
- ≈ 1–2 cfm per lb of product
- Total airflow
- (cfm/lb) × load mass (lb)
- Fan power
- Q (m³/s) × ΔP (Pa) ÷ 1000 ÷ η
- Typical static
- 0.4–0.8 in.w.g (100–200 Pa)
- 7/8 cool point
- 3 half-cooling times (commercially cool)
- Conduction floor
- extra airflow above ~2 cfm/lb buys little
- Speed vs room
- forced-air ≈ 4–10× faster
- Privacy
- Runs in your browser; nothing uploaded
Airflow per pound vs cooling time
More airflow cools faster, but with sharply diminishing returns — the seven-eighths cooling time flattens against the product's internal conduction. This is why over-sizing the fan wastes energy without cooling the centre any faster.
| Airflow (cfm/lb) | 7/8 cooling time (h) | Design note |
|---|---|---|
| 0.25 | 2.95 | under-aired — slow, uneven cooling |
| 0.5 | 1.72 | under-aired — slow, uneven cooling |
| 0.75 | 1.25 | economical but slower |
| 1 | 1 | typical good-practice design range |
| 1.5 | 0.73 | typical good-practice design range |
| 2 | 0.58 | typical good-practice design range |
| 3 | 0.42 | diminishing returns — extra fan energy buys little |
Commodity design airflow & static pressure
| Commodity | Target pulp (°C) | Design airflow (cfm/lb) | Static (in.w.g) | Respiration |
|---|---|---|---|---|
| Strawberry | 0 | 1 | 0.6 | high |
| Blueberry | 0 | 1 | 0.7 | high |
| Raspberry / blackberry | 0 | 1.2 | 0.7 | veryHigh |
| Sweet cherry | 0 | 0.9 | 0.6 | moderate |
| Peach / nectarine | 0 | 0.8 | 0.6 | moderate |
| Plum | 0 | 0.8 | 0.6 | moderate |
| Table grape | 0 | 1 | 0.8 | low |
| Apple | 0 | 0.6 | 0.5 | low |
| Pear | -1 | 0.6 | 0.5 | low |
| Tomato (ripe) | 10 | 0.8 | 0.6 | moderate |
| Bell pepper | 8 | 0.9 | 0.6 | moderate |
| Cucumber | 10 | 0.9 | 0.6 | moderate |
| Broccoli | 0 | 1.2 | 0.7 | veryHigh |
| Sweet corn | 0 | 1.3 | 0.8 | veryHigh |
| Lettuce (crisphead) | 0 | 1 | 0.7 | moderate |
| Green bean | 5 | 1.1 | 0.7 | high |
| Carrot (topped) | 0 | 0.8 | 0.6 | moderate |
| Citrus (orange) | 5 | 0.6 | 0.5 | low |
| Mango | 13 | 0.8 | 0.6 | moderate |
| Cut flowers | 1 | 1 | 0.6 | high |
Sources: Thompson, Mitchell, Rumsey, Kasmire, Crisosto (2008), "Commercial Cooling of Fruits, Vegetables, and Flowers", UC ANR Pub 21567; ASHRAE Handbook—Refrigeration; UC-Davis Postharvest Technology Center; Kader (ed.) UC ANR 3311. Values are representative planning figures.
Forced-air cooling sizes the fan, not just the curve
A half-cooling chart tells you how fast produce cools; it does not tell you how big a fan to buy. Forced-air precooling works by drawing cold air through vented packages — not around them — so the design problem is sizing the airflow (in cfm per pound of product), the static pressure that airflow has to fight through the stack, and the fan power needed to deliver both. Total airflow is simply the airflow per pound times the load mass; fan power is the airflow against the static pressure, divided by fan efficiency.
This tool draws the tunnel cross-section with cold air pulled through two pallet stacks into a central return plenum. The airflow arrows thicken as you raise the cfm/lb, the cooling curve steepens toward the target, and if the design misses your target time the tool solves the airflow you actually need — or warns you when no fan can beat the product's own heat conduction. That is the gap left by every cooling-curve calculator: this one specifies the fan.
How to use it in five steps
- 1Pick the commodity
The tool sets the target pulp temperature, design airflow and how slowly the fruit cools.
- 2Enter the load and airflow
Enter the load mass and your chosen airflow per pound (start near 1 cfm/lb).
- 3Enter temperatures and static
Add the field heat, supply-air temperature and the stack static pressure.
- 4Set the target cooling time
Enter the 7/8 (commercially cool) time you want to achieve.
- 5Read and right-size the fan
Read the total cfm, static pressure and fan power; raise airflow if the design misses the target.
Frequently Asked Questions
How much airflow does a forced-air cooler need per pound of produce?+
Good practice is roughly 1 to 2 cubic feet per minute (cfm) per pound of product for most fruits and vegetables, with about 1 cfm/lb a common design target. Below about 0.5 cfm/lb cooling is slow and uneven; above about 2 cfm/lb you hit diminishing returns because the limit becomes the product's own internal heat conduction, not the air. Total airflow = (cfm/lb) × the load mass in pounds.
How do I calculate the fan power for forced-air precooling?+
Fan power follows air power divided by fan efficiency: air power (kW) = airflow Q (m³/s) × static pressure ΔP (Pa) ÷ 1000, then fan electrical power = air power ÷ efficiency (a typical wire-to-air efficiency is about 0.55). The tool converts your cfm to m³/s, applies the stack static pressure you enter, and reports the fan in both kW and horsepower.
What is seven-eighths cooling time and why size for it?+
Seven-eighths (7/8) cooling is when 87.5% of the original temperature difference between the produce and the supply air has been removed — the produce is then 'commercially cool.' It equals three half-cooling times. Coolers are sized to the 7/8 point because chasing the last eighth takes as long again as the first seven-eighths and rarely earns the extra cooler time.
What static pressure should I expect through the stack?+
Typical forced-air pallet stacks run around 0.4 to 0.8 inches of water gauge (about 100 to 200 Pa), depending on stacking depth, package vent area and product size. Denser stacks and poorly aligned vents raise the static pressure, which raises fan power for the same airflow — so vent alignment is as important as fan size.
Why does more airflow stop helping above about 2 cfm/lb?+
Forced-air cooling moves heat from the surface of the produce to the air. Once the air is removing surface heat faster than the product can conduct heat from its centre to its surface, adding airflow only spends more fan energy without cooling the centre any faster. That conduction floor is why this tool caps the achievable 7/8 time and flags targets that are physically too fast.
Can I cool produce below the supply-air temperature?+
No. Cooling is driven by the gap between the pulp and the supply air; as the pulp nears the air temperature the gap shrinks toward zero and cooling stalls. If your target pulp temperature is at or below the supply-air temperature it is unreachable — you must supply colder air. The tool flags this case.
How is the fan size to hit a target cooling time worked out?+
The tool inverts its airflow-versus-time model: starting from the 7/8 time you want, it solves for the cfm/lb that delivers it, then multiplies by the load mass for total cfm. If your target time is below the product's conduction floor, no airflow can reach it and the tool tells you to allow more time instead of buying a bigger fan.
Is forced-air cooling right for my crop?+
Forced-air suits most fruits, berries, peppers, broccoli and cut flowers in vented packages. Very high-respiration crops like sweet corn are often hydrocooled or iced instead, and leafy greens like lettuce are usually vacuum-cooled — though forced-air can work for all of them with enough airflow. The commodity note in the tool flags the typical choice.
What field heat and supply-air temperatures should I enter?+
Field heat is the pulp temperature of the produce as it arrives — push a probe thermometer into the centre of a representative item. Supply air is the temperature of the cold air the fan delivers (the evaporator off-coil temperature). The larger the gap, the faster the early cooling, but the 7/8 and half-cooling times depend on the airflow and commodity, not the gap.
How much faster is forced-air than room cooling?+
Typically four to ten times faster. Room cooling lets cold air drift past packages, giving a half-cooling time near seven hours; forced-air pulls cold air through the vented packages and reaches a half-cooling time under two hours at about 1 cfm/lb. For fast-respiring crops that difference decides shelf life.
Does the load mass change the fan size or just the airflow?+
Both. Total airflow is cfm/lb times the load mass, so a bigger load needs proportionally more total cfm — and because fan power scales with airflow at a given static pressure, the fan grows with the load too. The cfm/lb and the 7/8 cooling time, however, stay the same for a given commodity and airflow density.
Is anything uploaded?+
No. The whole calculation runs in your browser using the forced-air design relations and the built-in commodity table. Nothing you enter is sent anywhere.