Heat Transfer · Chapter 3 of 10 · Intermediate
Fins and Extended Surfaces
When convection is the bottleneck, add area. Fins are the cheapest way to move more heat from a surface, and the fin equation tells you how long is long enough.
Readiness check
From Chapter 2. Tick only what you can do closed-notes.
- Use the convection resistance 1/(hA).
- See that low h means a large surface resistance.
- Work with hyperbolic tangent on a calculator.
- Compute a cross-section area and a perimeter.
- Read a temperature difference as a driving potential.
The core idea
A fin trades a little conduction along its length for a lot of extra convecting area.
m = √(hP/(kAc))qfin = √(hPkAc) · θb · tanh(mL)Heat enters the fin at its base and leaks off its sides as it travels outward, so the temperature decays from the base value θb = Tb − T∞ toward the fluid. The fin parameter m sets how fast it decays. A good fin is made of high-k metal, kept thin, and used where h is small.
The skills, taught in order
A fin is a tiny conduction-plus-convection problem with a tidy answer. Learn the fin parameter, the heat rate, and the two ratios (efficiency and effectiveness) that tell you whether a fin earns its place.
3.1 Why fins work
Convection off a bare surface is limited by hA. You cannot easily raise h, but you can raise A by growing fingers of metal into the fluid. Fins are worth most exactly where convection is poor: gas cooling, natural convection, and electronics in still air.
3.2 The fin equation
Along a fin, conduction in balances convection off, giving d²θ/dx² = m²θ with m = √(hP/(kAc)), where P is the perimeter, Ac the cross-section, and θ = T − T∞. The solution decays from the base; m is the inverse decay length.
3.3 The fin heat rate
For a fin with an effectively insulated tip, the heat it carries is qfin = √(hPkAc) · θb · tanh(mL). The tanh saturates near 1 once mL passes about 2, so beyond that length the fin adds almost nothing.
Micro-example. Doubling a fin from mL = 1 to mL = 2 raises tanh from 0.76 to 0.96, only a 26% gain in heat for twice the metal.
3.4 Fin efficiency
Fin efficiency compares the real heat to the heat an ideal fin (all at the base temperature) would carry: ηf = tanh(mL)/(mL). A short, stubby, high-k fin runs near 100% efficient; a long one wastes its tip.
| Quantity | Formula | What it tells you |
|---|---|---|
| Fin parameter | m = √(hP/(kAc)) | how fast temperature decays |
| Heat rate (insulated tip) | √(hPkAc) θb tanh(mL) | watts the fin moves |
| Efficiency | ηf = tanh(mL)/(mL) | fraction of ideal heat |
| Effectiveness | εf = qfin/(hAcθb) | gain over the bare base |
3.5 Fin effectiveness
Effectiveness asks the blunt question: does the fin beat the bare spot it covers? εf = qfin/(hAcθb). A fin is only worth adding if εf is comfortably above 2, which favours high k, small h, and thin fins (large P/Ac).
3.6 When more fin stops helping
Because tanh saturates, length has diminishing returns; past mL ≈ 2 to 3 the tip is at the fluid temperature and contributes nothing. Real designs therefore use many short fins (an array) rather than a few long ones, packing area into a fixed volume. Chapter 6 supplies the h that sets m.
Engineering connection: heat sinks, air-cooled engines, radiators, and any surface fighting a low convection coefficient. The h needed here comes from Chapter 6.
Worked example 1: a single pin fin
An aluminium pin fin (k = 180 W/m·K) is 5 mm in diameter and 30 mm long. Its base is 75 °C above the surrounding air, which gives h = 40 W/m²·K. Find the heat the fin carries, its efficiency, and its effectiveness.
- ProblemFind qfin, ηf, and εf for the pin fin in Figure 1.
- Given / findk = 180, D = 0.005 m, L = 0.030 m, θb = 75 K, h = 40 W/m²·K. Find the heat rate, efficiency, and effectiveness.
- AssumptionsSteady state, one-dimensional fin, constant properties, tip effectively insulated (short fin).
- ModelCompute Ac = πD²/4 and P = πD, then m, then the heat rate and the two ratios.
- Equationsm = √(hP/(kAc)) qfin = √(hPkAc) θb tanh(mL) ηf = tanh(mL)/(mL)
- SolveAc = 1.96×10⁻⁵ m², P = 0.0157 m. m = √(40 × 0.0157 / (180 × 1.96×10⁻⁵)) = 13.3 /m, so mL = 0.40. qfin = √(hPkAc)(75)tanh(0.40) = 0.0471 × 75 × 0.380 = 1.34 W. ηf = 0.380/0.40 = 0.95. εf = 1.34/(40 × 1.96×10⁻⁵ × 75) = 22.8.
- CheckmL = 0.40 is small, so the fin is nearly isothermal and efficiency is high (95%), exactly as a short high-k fin should be. Effectiveness far above 2 means the fin is well worth adding.
- ConclusionThis fin moves 23 times the heat of the base spot it occupies, the whole reason heat sinks bristle with fins. Aluminium's high k and the small diameter (large P/Ac) are what make it so effective.
Worked example 2: a heat-sink array
An aluminium heat sink (k = 180) carries 12 rectangular fins, each 20 mm long, 1.5 mm thick, and 40 mm deep. A fan gives h = 50 W/m²·K, and the base runs 35 °C above the air. Estimate the heat the fin array removes, using the efficiency method.
- ProblemEstimate the heat removed by the 12-fin array in Figure 2.
- Given / findN = 12, k = 180, L = 0.020 m, t = 0.0015 m, depth w = 0.040 m, h = 50 W/m²·K, θb = 35 K. Find the array heat rate.
- AssumptionsIdentical fins, one-dimensional, insulated tip handled by the corrected length Lc = L + t/2, base at uniform θb.
- ModelUse the efficiency method per fin: q = ηf h Afin θb, with Afin = 2wLc, then multiply by N.
- Equationsm = √(hP/(kAc)), Lc = L + t/2 qfin = ηf h (2wLc) θb
- SolveAc = wt = 6.0×10⁻⁵ m², P = 2(w+t) = 0.083 m, m = 19.6 /m. Lc = 0.0208 m, mLc = 0.41, so ηf = tanh(0.41)/0.41 = 0.95. Afin = 2 × 0.040 × 0.0208 = 1.66×10⁻³ m². q per fin = 0.95 × 50 × 1.66×10⁻³ × 35 = 2.75 W, so the array removes 12 × 2.75 ≈ 33 W.
- CheckEach fin is again 95% efficient (short and high-k). Thirty-three watts from a palm-sized sink in forced air is the right order for cooling a small processor. Halving h (no fan) would cut the heat roughly in half and lower η only slightly.
- ConclusionAn array of short, efficient fins is how real heat sinks work: not a few long fins, but many stubby ones that each run near the base temperature. The same per-fin efficiency method scales to any array.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Longer fin always better | Fins drawn very long for more heat | "Is mL already past about 2?" | tanh saturates; beyond mL ≈ 2 the tip is at the fluid temperature and adds nothing. Use more short fins. |
| Fins on the high-h side | Fins placed in fast liquid flow | "Which side has the poor convection?" | Fins help where h is small (gases, still air), not where convection is already strong. |
| Efficiency and effectiveness mixed | Calling a 95% fin "barely worth it" | "Am I comparing to an ideal fin or to no fin?" | Efficiency compares to an ideal fin; effectiveness compares to no fin. Both can be high at once. |
| Low-k material for fins | Plastic or steel fins underperform | "Does conduction reach the tip easily?" | Fins need high k so the whole length stays near the base temperature; aluminium and copper dominate. |
Practice ladder
For the Worked Example 1 fin, find m and mL if the air flow doubles h to 80 W/m²·K. Does efficiency rise or fall?
Show answer
m = √(80 × 0.0157 / (180 × 1.96×10⁻⁵)) = 18.9 /m, so mL = 0.566. tanh(0.566)/0.566 = 0.90: efficiency falls slightly, because stronger convection makes the fin less isothermal, even though it now carries more heat.
Compare the Worked Example 1 aluminium fin (k = 180) with an identical stainless-steel one (k = 15). Which moves more heat, and why?
Show answer
For steel, m = √(40 × 0.0157/(15 × 1.96×10⁻⁵)) = 46 /m, mL = 1.39, and √(hPkAc) is smaller, giving q ≈ 0.90 W versus 1.34 W. Lower k means heat cannot reach the tip, so efficiency drops to about 64%. Material matters: high k is the first requirement of a good fin.
A long aluminium fin has mL = 2.5. Find its efficiency and the fraction of the long-fin-limit heat it delivers. Comment on whether the last centimetres earn their weight.
Show answer
η = tanh(2.5)/2.5 = 0.987/2.5 = 0.39, only 39% efficient. tanh(2.5) = 0.987, so it delivers 99% of the long-fin limit already; the extra length past mL ≈ 2 is essentially dead metal. A shorter fin would carry nearly the same heat for far less material.
Measure a real heat sink (CPU cooler, amplifier, motor housing). Estimate fin count, size, and likely h, then compute the array heat rate and the per-fin efficiency, and judge whether the fins are well proportioned.
What good work looks like
Measured fin geometry, a justified h, m and η per fin, the array total, and a verdict on whether the fins are too long (low η) or about right.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Design a fin or small heat sink for a stated heat load: pick material, fin size, and count, then report m, efficiency, effectiveness, and the total heat, with a sentence defending the length you chose.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the fin parameter and what it measures.
m = √(hP/(kAc)); the inverse decay length of temperature along the fin.
2. Write the fin heat rate for an insulated tip.
qfin = √(hPkAc) θb tanh(mL).
3. Distinguish fin efficiency and effectiveness.
Efficiency η = tanh(mL)/(mL) compares to an ideal isothermal fin; effectiveness ε = q/(hAcθb) compares to no fin at all.
4. When is a fin worth adding?
When effectiveness exceeds about 2, favoured by high k, small h, and thin fins (large P/Ac).
5. Why use many short fins instead of a few long ones?
tanh saturates past mL ≈ 2, so long fins waste their tips; short, efficient fins pack more useful area into a volume.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Lienhard and Lienhard, A Heat Transfer Textbook (6th ed), Section 4.5 (fin design) |
| Cross-reference | Incropera, Ch. 3 (extended surfaces) · Çengel and Ghajar, Ch. 3 |
| Core topics | 3.1 Why fins · 3.2 Fin equation · 3.3 Heat rate · 3.4 Efficiency · 3.5 Effectiveness · 3.6 Diminishing returns and arrays |
| Engineering connection | Heat sinks, radiators, and air-cooled machinery; the h comes from Chapter 6. |
| Read next | Chapter 4: Transient Conduction. |