Training Cooling Fins Why Fins Work — The Extended Surface Concept
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Why Fins Work — The Extended Surface Concept

24 min Cooling Fins

Why Fins Work — The Extended Surface Concept

When convection-side resistance is much larger than conduction resistance (low $h$, high $k$), adding area on the convection side dramatically improves heat transfer. Fins are extended surfaces attached to a base, trading a small conduction penalty for a large area gain.

When Are Fins Effective?

Fins help when the Biot number of the fin is small: $Bi = hL_c/k \ll 1$. This means the fin material conducts so well internally that the temperature is nearly uniform across its thickness. Fins on the air side of a heat exchanger (low $h$) are very effective. Fins on the water side (high $h$) rarely help.

Fin Terminology

$L$ = fin length (from base to tip) • $t$ = fin thickness • $w$ = fin width (into the page) • $P$ = fin perimeter $\approx 2(w + t) \approx 2w$ for thin fins • $A_c$ = fin cross-sectional area $= wt$ • $A_f$ = fin surface area $= PL$ (lateral) • $T_b$ = fin base temperature • $T_\infty$ = surrounding fluid temperature

Fin Parameter $m$

$$m = \sqrt{\frac{hP}{kA_c}}$$

Units: m⁻¹. $m$ controls how quickly the fin temperature drops from base to tip. Large $m$ (thin fin, low $k$, high $h$) = rapid decay = less effective fin.

Example 1 — Why Not Fin the Water Side?

Compare: Air side ($h = 20$, $k = 200$, $t = 2$ mm) vs. water side ($h = 5000$, same $k$ and $t$). Compute $Bi$ for each.

Air side: $Bi = hL_c/k = h(t/2)/k = 20 \times 0.001/200 = 10^{-4}$ → excellent for fins!

Water side: $Bi = 5000 \times 0.001/200 = 0.025$ → still < 1, but fins add cost and pressure drop for minimal gain since the bottleneck isn’t area but fluid-side resistance.

Example 2 — Computing $m$

Aluminum fin: $k = 200$ W/(m·K), $t = 3$ mm, $w = 10$ cm, $h = 25$ W/(m²·K). Find $m$.

$A_c = 0.10 \times 0.003 = 3 \times 10^{-4}$ m²

$P \approx 2 \times 0.10 = 0.20$ m (thin fin, width ≫ thickness)

$$m = \sqrt{\frac{25 \times 0.20}{200 \times 3 \times 10^{-4}}} = \sqrt{\frac{5.0}{0.06}} = \sqrt{83.3} = 9.13 \text{ m}^{-1}$$

Example 3 — Bare vs. Finned

A pipe has $A_{\text{bare}} = 0.1$ m² with $h = 20$. $\Delta T = 50$°C. Heat loss bare: $\dot{Q}_{\text{bare}}$? If fins increase effective area to 1.5 m² with average $h = 18$, $\dot{Q}_{\text{finned}}$?

$\dot{Q}_{\text{bare}} = 20 \times 0.1 \times 50 = 100$ W

$\dot{Q}_{\text{finned}} = 18 \times 1.5 \times 50 = 1350$ W

Fins increased heat transfer by 13.5×! (Even though $h$ dropped slightly due to fin geometry effects.)

Practice Problems

1. Compute $m$ for a copper fin ($k = 385$): $t = 1$ mm, $w = 5$ cm, $h = 30$.
2. $m = 15$ m⁻¹, fin length $L = 5$ cm. What is $mL$?
3. Why is $mL$ important? (Hint: it determines how much of the fin is “active.”)
4. $Bi_{\text{fin}} = h(t/2)/k = 0.001$. Is this fin thermally thin?
5. If $k$ doubles, $m$ changes by what factor?
6. Bare area = 0.05 m², 40 fins each adding 0.01 m². Total surface area?
Show Answer Key

1. $A_c = 0.05 \times 0.001 = 5 \times 10^{-5}$. $P = 0.10$. $m = \sqrt{30 \times 0.10/(385 \times 5 \times 10^{-5})} = \sqrt{3/0.01925} = \sqrt{155.8} = 12.5$ m⁻¹

2. $mL = 15 \times 0.05 = 0.75$

3. $mL$ is the dimensionless fin length. If $mL > 2.5$, the tip is essentially at $T_\infty$ and making the fin longer adds little benefit.

4. Yes — temperature is nearly uniform across the fin thickness.

5. $m \propto 1/\sqrt{k}$, so $m$ decreases by $1/\sqrt{2} = 0.707$.

6. $0.05 + 40 \times 0.01 = 0.45$ m² (but subtract the base area covered by fin roots if specified).

Rectangular Fin — Temperature Distribution and Heat Transfer