Pin Fins, Annular Fins, and Fin Arrays
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.
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.
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.
$N$ 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.
Brass pin fin: $D = 3$ mm, $L = 3$ cm, $k = 110$, $h = 40$, $\theta_b = 70$°C. Find $\dot{Q}$.
$m = \sqrt{4 \times 40/(110 \times 0.003)} = \sqrt{160/0.33} = \sqrt{484.8} = 22.0$ m⁻¹
$mL = 22.0 \times 0.03 = 0.660$
$A_c = \pi(0.003)^2/4 = 7.07 \times 10^{-6}$ m²
$P = \pi \times 0.003 = 9.42 \times 10^{-3}$ m
$\sqrt{hPkA_c} = \sqrt{40 \times 9.42 \times 10^{-3} \times 110 \times 7.07 \times 10^{-6}} = \sqrt{2.93 \times 10^{-4}} = 0.01712$
$\dot{Q} = 0.01712 \times 70 \times \tanh(0.660) = 0.01712 \times 70 \times 0.577 = 0.692$ W
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 $\dot{Q}$.
$A_b = 0.02 - 20 \times 5 \times 10^{-5} = 0.02 - 0.001 = 0.019$ m²
$A_{\text{total}} = 20 \times 8 \times 10^{-4} + 0.019 = 0.016 + 0.019 = 0.035$ m²
$\eta_o = 1 - \frac{20 \times 8 \times 10^{-4}}{0.035}(1 - 0.90) = 1 - \frac{0.016}{0.035} \times 0.10 = 1 - 0.0457 = 0.954$
$\dot{Q} = 0.954 \times 35 \times 0.035 \times 45 = 52.6$ W
Annular fin on a pipe: $r_1 = 2$ cm, $r_2 = 5$ cm, $t = 2$ mm, $k = 180$, $h = 50$. Using a chart, find $\eta_f$.
$m = \sqrt{2h/(kt)} = \sqrt{2 \times 50/(180 \times 0.002)} = \sqrt{100/0.36} = \sqrt{277.8} = 16.67$ m⁻¹
$L_c = r_2 - r_1 + t/2 = 0.03 + 0.001 = 0.031$ m
Chart parameter: $L_c \sqrt{2h/(kt)} = 0.031 \times 16.67 = 0.517$
$r_{2c}/r_1 = (r_2 + t/2)/r_1 = 0.051/0.02 = 2.55$
From the annular fin chart at $x = 0.517$, $r_{2c}/r_1 = 2.55$: $\eta_f \approx 0.87$
Practice Problems
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 $h$.