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Straight Fin Efficiency Calculator

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Understanding the straight fin efficiency calculator — concept and principle

Fins are thin extensions added to a surface to increase its area and therefore its heat transfer, as on heat sinks, radiators, motor housings and air-conditioner coils. Fin efficiency tells you how well a fin performs compared with an ideal fin whose whole surface stays at the base temperature.

Straight aluminium fin on a hot baseA 20 mm long, 1.5 mm thick aluminium fin with h = 40 W/m²K has m = 16.33 1/m, mLc = 0.339 and efficiency 96.34%.A fin pushes heat out into more airhotbasealuminium k = 200h = 40L = 20 mmt = 1.5mm(thickness drawn larger)m = √(2h ÷ kt) = √(80 ÷ 0.3) = 16.33mLc = 16.33 × 0.02075 = 0.339Fin efficiency η = 96.34%tanh(0.339) ÷ 0.339 · nearly ideal
A short, thick aluminium fin stays almost as hot as its base, so it works at 96.34% efficiency.

How it is calculated

η = tanh(m·L_c) / (m·L_c), with m = √(2h / (k·t)) and L_c = L + t/2.

The calculator shows η as a percentage and the dimensionless parameter mL_c.

The principle

Heat enters the fin at the base and is conducted along it while convection removes heat from both faces. Because heat leaks away along the way, the temperature falls from base to tip, so outer parts of the fin are less effective. Solving the one-dimensional fin equation, d²θ/dx² = m²θ, gives a hyperbolic temperature profile and the efficiency above. The parameter m compares convection loss with conduction along the fin: a thin, poorly conducting fin in a strong flow has a large m and a low efficiency.

Worked example

  1. Inputs: h = 40 W/m²K, k = 200 W/m·K (aluminium), t = 1.5 mm, L = 20 mm.
  2. m = √(2 × 40 / (200 × 0.0015)) = √266.67 ≈ 16.330 m⁻¹.
  3. L_c = 20 + 0.75 = 20.75 mm = 0.02075 m, so mL_c ≈ 0.33885.
  4. tanh(0.33885) ≈ 0.32645, and η = 0.32645 / 0.33885 ≈ 96.34%.
How Fin efficiency changes with Fin length (mm)How Fin efficiency changes with Fin leng…8587.59092.59597.51005101520253040Example: 96.34%Fin length (mm)
Each point is the calculator re-run with only “Fin length (mm)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.
Fin length (mm)Fin efficiency
599.71%
1098.99%
1597.85%
2096.34%
2594.5%
3092.36%
4087.46%

How to read the result

η near 100% means almost the whole fin is close to base temperature, so it uses its material well. At mL_c = 1, η ≈ 76%; at mL_c = 2, η ≈ 48%. Values of mL_c much beyond 2 mean extra length adds little heat transfer while adding weight and cost. Higher efficiency is not the only goal: a short fin can be very efficient yet add little area, so designers balance η against total heat removed.

Common mistakes

Keep in mind: The model assumes a straight rectangular fin that is wide compared with its thickness, constant h and k, and one-dimensional conduction. Real heat sinks need testing or detailed simulation by a qualified engineer.

Frequently asked questions

What is the difference between fin efficiency and fin effectiveness?

Efficiency compares the fin with an ideal isothermal fin; effectiveness compares heat transfer with the fin to the heat the bare base area would transfer without it.

Why use copper instead of aluminium?

Copper's higher k (about 400 W/m·K) lowers m and raises efficiency, but it is heavier and more expensive.

When is the one-dimensional assumption valid?

When the Biot number h·t/(2k) is much smaller than 0.1; here it is about 0.00015.

Sources: Incropera, DeWitt, Bergman and Lavine, Fundamentals of Heat and Mass Transfer, Wiley · Kraus, Aziz and Welty, Extended Surface Heat Transfer, Wiley

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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