Training Cooling Fins Pin Fins, Annular Fins, and Fin Arrays
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Pin Fins, Annular Fins, and Fin Arrays

24 min Cooling Fins

Real thermal systems use pin fins, annular (circumferential) fins, and arrays of many fins rather than single rectangular fins. Pin fins offer high surface area in compact spaces, annular fins appear on cylinders and pipes, and fin arrays require analysis of inter-fin spacing to avoid choking the air flow. The overall surface efficiency ηo = 1 − (Af/At)(1 − ηf) lets you compute the effective heat-transfer rate for an entire finned surface in a single step.

Pin Fins, Annular Fins, and Fin Arrays

Beyond rectangular fins, two other geometries are common: pin (cylindrical) fins and annular (circular) fins. Real heat sinks use arrays of many fins with exposed base area between them.

Pin (Cylindrical) Fin

Circular cross-section with diameter $D$.

$A_c = \pi D^2/4$, $P = \pi D$

$$m = \sqrt{\frac{h \cdot \pi D}{k \cdot \pi D^2/4}} = \sqrt{\frac{4h}{kD}}$$

Same $\tanh(mL)/(mL)$ efficiency formula applies.

Annular (Circular) Fin

A disk of outer radius $r_2$ attached to a tube of radius $r_1$.

$$\eta_f = \frac{2r_1}{m(r_2^2 - r_1^2)} \cdot \frac{K_1(mr_1)I_1(mr_2) - I_1(mr_1)K_1(mr_2)}{K_0(mr_1)I_1(mr_2) + I_0(mr_1)K_1(mr_2)}$$

$I_0, I_1, K_0, K_1$ = modified Bessel functions. In practice, use efficiency charts: $\eta_f$ vs. $L_c\sqrt{2h/(kt)}$ with curves for $r_2/r_1$.

For annular fins: $m = \sqrt{2h/(kt)}$ where $t$ = fin thickness.

Fin Array (Overall Surface)

$N$ fins on a base of total area $A_{\text{total}}$:

$$\dot{Q}_{\text{total}} = \eta_o h A_{\text{total}} \theta_b$$

Overall surface efficiency:

$$\eta_o = 1 - \frac{NA_f}{A_{\text{total}}}(1 - \eta_f)$$

$A_{\text{total}} = NA_f + A_b$ where $A_b$ = unfinned (exposed) base area.

Example 1 — Pin Fin

Brass pin fin: $D = 3$ mm, $L = 3$ cm, $k = 110$, $h = 40$, $\theta_b = 70$°C. Find $\dot{Q}$.

  1. $m = \sqrt{4 \times 40/(110 \times 0.003)} = \sqrt{160/0.33} = \sqrt{484.8} = 22.0$ m⁻¹
  2. $mL = 22.0 \times 0.03 = 0.660$
  3. $A_c = \pi(0.003)^2/4 = 7.07 \times 10^{-6}$ m²
  4. $P = \pi \times 0.003 = 9.42 \times 10^{-3}$ m
  5. $\sqrt{hPkA_c} = \sqrt{40 \times 9.42 \times 10^{-3} \times 110 \times 7.07 \times 10^{-6}} = \sqrt{2.93 \times 10^{-4}} = 0.01712$
  6. $\dot{Q} = 0.01712 \times 70 \times \tanh(0.660) = 0.01712 \times 70 \times 0.577 = 0.692$ W
Example 2 — Fin Array

A heat sink has 20 rectangular aluminum fins ($\eta_f = 0.90$). Each fin: $A_f = 8 \times 10^{-4}$ m². Total base area (including under fins) = 0.02 m². Fin footprint each = $5 \times 10^{-5}$ m². $h = 35$, $\theta_b = 45$°C. Find $\dot{Q}$.

  1. $A_b = 0.02 - 20 \times 5 \times 10^{-5} = 0.02 - 0.001 = 0.019$ m²
  2. $A_{\text{total}} = 20 \times 8 \times 10^{-4} + 0.019 = 0.016 + 0.019 = 0.035$ m²
  3. $\eta_o = 1 - \frac{20 \times 8 \times 10^{-4}}{0.035}(1 - 0.90) = 1 - \frac{0.016}{0.035} \times 0.10 = 1 - 0.0457 = 0.954$
  4. $\dot{Q} = 0.954 \times 35 \times 0.035 \times 45 = 52.6$ W
Example 3 — Annular Fin

Annular fin on a pipe: $r_1 = 2$ cm, $r_2 = 5$ cm, $t = 2$ mm, $k = 180$, $h = 50$. Using a chart, find $\eta_f$.

  1. $m = \sqrt{2h/(kt)} = \sqrt{2 \times 50/(180 \times 0.002)} = \sqrt{100/0.36} = \sqrt{277.8} = 16.67$ m⁻¹
  2. $L_c = r_2 - r_1 + t/2 = 0.03 + 0.001 = 0.031$ m
  3. Chart parameter: $L_c \sqrt{2h/(kt)} = 0.031 \times 16.67 = 0.517$
  4. $r_{2c}/r_1 = (r_2 + t/2)/r_1 = 0.051/0.02 = 2.55$
  5. From the annular fin chart at $x = 0.517$, $r_{2c}/r_1 = 2.55$: $\eta_f \approx 0.87$

Practice Problems

1. Pin fin: $D = 5$ mm, $k = 200$, $h = 25$. Find $m$.
2. 50 fins, each $A_f = 10^{-3}$ m², $\eta_f = 0.85$, $A_b = 0.04$ m². Find $\eta_o$.
3. With $\eta_o = 0.92$, $A_t = 0.1$ m², $h = 20$, $\theta_b = 50$°C. $\dot{Q}$?
4. Annular fin: $r_1 = 1.5$ cm, $r_2 = 4$ cm. $A_f = $ ?
5. Why do pin fins give more surface area per unit base area than plate fins?
6. If you double the number of fins, does $\dot{Q}$ double? Why or why not?
Show Answer Key

1. $m = \sqrt{4 \times 25/(200 \times 0.005)} = \sqrt{100} = 10$ m⁻¹

2. $A_t = 50 \times 10^{-3} + 0.04 = 0.09$ m². $\eta_o = 1 - (0.05/0.09)(0.15) = 1 - 0.0833 = 0.917$

3. $\dot{Q} = 0.92 \times 20 \times 0.1 \times 50 = 92$ W

4. $A_f = 2\pi(r_2^2 - r_1^2) = 2\pi(0.04^2 - 0.015^2) = 2\pi(0.001375) = 8.64 \times 10^{-3}$ m² (both sides)

5. Cylindrical geometry provides circumferential area that rectangular fins lack. Pin fins radiate heat in all directions from the cross-section.

6. Not quite — more fins reduce the exposed base area $A_b$ and can cause boundary layer interference, reducing effective $h$.

🧮 Fin Array / Overall Surface Efficiency
Total fin area NAᶠ
Total area Aₜ
Overall efficiency ηₒ
q per K of ΔT
Fin Efficiency and Effectiveness Practice Test — Cooling Fins