Fin efficiency
A fin works by adding surface area, but its own temperature falls along the length — so the average temperature difference is smaller than at the base, and the fin delivers less than its area suggests. Fin efficiency is that shortfall, and it depends on a single group, mL: a long, thin, poorly conducting fin has a low efficiency, and beyond mL ≈ 2 the extra length buys almost nothing.
| Quantity | Value | Note |
|---|---|---|
| Fin length | 25.0 mm | as entered |
| Thickness | 1.50 mm | drives m |
| Conductivity | 200 W/m·K | aluminium 200, copper 400 |
| Coefficient h | 50 W/m²·K | from the convection correlation |
| Fin parameter m | 18.3 /m | √(2h/kt) |
| Corrected length | 25.75 mm | L + t/2 for the tip |
| mL | 0.470 | the governing group |
| Fin efficiency | 93.2% | tanh(mL)/mL |
| Fin area each | 51.5 cm² | two sides plus the tip |
| Total area | 1230 cm² | 20 fins plus the base |
| Surface efficiency | 94.3% | weighted by area |
| Heat transfer | 232 W | at ΔT 40 K |
| Useful length | 110 mm | 2/m |
mL is 0.47, so the fin efficiency is 93%. A useful rule is that a fin longer than 2/m adds little; here that is 110 mm against 25 mm actual.
m = √(2h/kt) · Rectangular fin.
η_f = tanh(mL_c)/(mL_c) · Convective tip.
η_o = 1 − (A_f/A_t)(1 − η_f) · Area weighted.
q = η_o h A_t θ_b · The whole array.
A clear path from inputs to results.
A fin works by adding surface area, but its own temperature falls along the length — so the average temperature difference is smaller than at the base, and the fin delivers less than its area suggests.
How to use
- Enter the fin length, thickness and width.
- Choose the material and the convection coefficient.
- Set the number of fins and the temperature difference.
- Read the fin efficiency, the surface efficiency and the useful length.
The formula
Reference: The hyperbolic solution for a straight rectangular fin with a convective tip (Incropera & DeWitt).
Worked example
A 25 mm aluminium fin at h = 50
- 25 mm long, 1.5 mm thick, 100 mm wide, aluminium at 200 W/m·K.
- h = 50 W/m²·K, 20 fins, 40 K difference.
- 1
m = √(2 × 50/(200 × 0.0015)) = 18.3 /m.
- 2
The corrected length is 25.75 mm, so mL = 0.47.
- 3
The fin efficiency is tanh(0.47)/0.47 = 0.932.
- 4
With 20 fins the total area is 0.123 m² and the surface efficiency is 0.943.
- 5
The array transfers 232 W, against 246 W if every fin were isothermal.
93.2% fin efficiency · 94.3% surface efficiency · 232 W.
Interpreting the result
The useful length is 2/m, which here is 110 mm — four times the actual fin. That is why a heat sink with short dense fins beats one with long sparse ones: past 2/m the extra length is at almost the ambient temperature and adds area without adding heat.
Assumptions & limitations
- A straight rectangular fin with a convective tip.
- Uniform base temperature and a constant convection coefficient.
- No radiation from the fin and no bond resistance.
- Steady state, one-dimensional conduction along the fin.
Frequently asked questions
Why does a longer fin not help?
Because its tip approaches the ambient temperature, so the extra area has almost no temperature difference across it. The efficiency falls faster than the area grows.
Does thickness matter as much as length?
Yes — m depends on thickness through √(1/t). Doubling the thickness reduces m by 30% and raises the efficiency substantially.
What about a pin fin?
A pin fin uses a different m and a different efficiency expression, and is usually less efficient per unit volume than a plate fin at the same h.
What is the difference between fin and surface efficiency?
Fin efficiency is per fin; surface efficiency weights it by the fraction of total area that is finned, because the unfinned base is at 100% efficiency.
Why are heat sinks for fast fans so dense?
Because a high h gives a large m and therefore a short useful length. The fins must be short and closely spaced to stay efficient, which is exactly what a fast fan allows.
Methodology last reviewed 2026-10-22. Results are informational and not financial or engineering advice — see our disclaimer.