Fin Efficiency and Effectiveness
Fin Efficiency and Effectiveness
Two metrics quantify how well a fin performs: efficiency (how much of the fin area is actually useful) and effectiveness (how much better is the fin vs. no fin at all).
$$\eta_f = \frac{\dot{Q}_{\text{fin}}}{\dot{Q}_{\text{max}}} = \frac{\dot{Q}_{\text{fin}}}{hA_f\theta_b}$$
$\dot{Q}_{\text{max}}$ = heat transfer if the entire fin were at $T_b$. Since the fin temperature drops toward the tip, $\eta_f < 1$.
For a rectangular fin (insulated tip):
$$\eta_f = \frac{\tanh(mL)}{mL}$$
$$\varepsilon_f = \frac{\dot{Q}_{\text{with fin}}}{\dot{Q}_{\text{without fin}}} = \frac{\dot{Q}_{\text{fin}}}{hA_b\theta_b}$$
$A_b$ = fin base (footprint) area $= A_c$. If $\varepsilon_f < 1$, the fin actually hurts (acts as insulation)!
Design rule: $\varepsilon_f > 2$ for a fin to be cost-effective. High $k$, low $h$ gives high $\varepsilon_f$.
$$\varepsilon_f = \frac{\dot{Q}_{\text{fin}}}{hA_c\theta_b} = \sqrt{\frac{kP}{hA_c}}\tanh(mL)$$
$mL = 1.0$. Find $\eta_f$. What does it mean?
$$\eta_f = \frac{\tanh(1.0)}{1.0} = \frac{0.762}{1.0} = 0.762$$
The fin transfers 76.2% of the heat it would if it were entirely at the base temperature. The remaining 23.8% is “lost” because the fin temperature drops along its length.
$mL = 3.0$. Find $\eta_f$.
$$\eta_f = \frac{\tanh(3.0)}{3.0} = \frac{0.995}{3.0} = 0.332$$
Only 33.2%! The outer 2/3 of the fin is essentially at $T_\infty$ and contributes almost nothing. This fin is too long — material is wasted.
Aluminum fin in air: $k = 200$, $h = 20$, $P = 0.20$ m, $A_c = 2 \times 10^{-4}$ m², $mL = 0.8$. Find $\varepsilon_f$.
$$\varepsilon_f = \sqrt{\frac{kP}{hA_c}}\tanh(mL) = \sqrt{\frac{200 \times 0.20}{20 \times 2 \times 10^{-4}}} \times \tanh(0.8)$$
$= \sqrt{\frac{40}{0.004}} \times 0.664 = \sqrt{10{,}000} \times 0.664 = 100 \times 0.664 = 66.4$
Effectiveness of 66 — the fin transfers 66× more heat than the bare base area. This is why finned heat sinks are so effective in air cooling!
What $mL$ gives 90% efficiency?
$\tanh(mL)/(mL) = 0.90$. By trial: $mL \approx 0.72$ gives $\tanh(0.72)/0.72 = 0.618/0.72 = 0.858$. Try $mL = 0.55$: $0.500/0.55 = 0.909$. So $mL \approx 0.55$ for 90% efficiency. Short, fat fins are most efficient per unit material!
Practice Problems
Show Answer Key
1. $\eta_f = \tanh(0.5)/0.5 = 0.462/0.5 = 0.924$
2. $\eta_f = \tanh(2)/2 = 0.964/2 = 0.482$
3. $\dot{Q} = \eta_f \cdot h A_f \theta_b = 0.85 \times 25 \times 0.01 \times 60 = 12.75$ W
4. $5 \times 2 = 10$ W
5. When a very low-$k$ fin material actually insulates the surface. Practically never happens with metals.
6. Both give the same factor: $\varepsilon_f \propto \sqrt{k/h}$, so doubling $k$ or halving $h$ each multiply $\varepsilon_f$ by $\sqrt{2} \approx 1.41$.