Training Cooling Fins Practice Test — Cooling Fins
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Practice Test — Cooling Fins

24 min Cooling Fins

This practice test covers cooling fins: the extended-surface concept, rectangular fin temperature distributions and heat transfer with different tip conditions, fin efficiency and effectiveness metrics, and the analysis of pin fins, annular fins, and fin arrays. Use this test to confirm your ability to design and evaluate finned surfaces for thermal management applications.

Practice Test — Cooling Fins

Practice Test — 20 Questions

1. Why do fins enhance heat transfer?
2. Define the fin parameter $m$.
3. Aluminum fin: $k = 200$, $P = 0.10$ m, $A_c = 10^{-4}$ m², $h = 20$. Find $m$.
4. $m = 10$, $L = 8$ cm. What is $mL$?
5. Heat transfer from a fin (insulated tip): formula?
6. $mL = 0.8$. Find $\eta_f$.
7. $mL = 5$. Find $\eta_f$. Is this fin too long?
8. Define fin effectiveness $\varepsilon_f$.
9. $\varepsilon_f = 50$. The fin is __ times better than no fin.
10. Pin fin: $D = 2$ mm, $k = 385$ (copper), $h = 30$. $m$?
11. Annular fin: $r_1 = 1$ cm, $r_2 = 3$ cm. Fin area (both sides)?
12. 30 fins with $\eta_f = 0.88$, $A_f = 5 \times 10^{-4}$ each, $A_b = 0.01$ m². $\eta_o$?
13. $\eta_o = 0.95$, $h = 40$, $A_t = 0.2$ m², $\theta_b = 30$°C. $\dot{Q}$?
14. A fin with $mL = 2.5$: $\tanh(2.5)/(2.5) = $ ?
15. When does adding a fin decrease heat transfer?
16. What is the corrected length $L_c$?
17. Infinite fin approximation is valid when $mL >$ ?
18. Doubling $h$ does what to $m$?
19. $\sqrt{hPkA_c} = 0.3$ W/K, $\theta_b = 100$°C, $mL = 1.5$. $\dot{Q}$?
20. For maximum heat transfer per unit fin mass, should $mL$ be large or small?
Show Answer Key

1. They increase surface area on the low-$h$ side, reducing convection resistance.

2. $m = \sqrt{hP/(kA_c)}$

3. $m = \sqrt{20 \times 0.10/(200 \times 10^{-4})} = \sqrt{2/0.02} = \sqrt{100} = 10$ m⁻¹

4. $mL = 10 \times 0.08 = 0.80$

5. $\dot{Q} = \sqrt{hPkA_c}\,\theta_b\,\tanh(mL)$

6. $\eta_f = \tanh(0.8)/0.8 = 0.664/0.8 = 0.830$

7. $\eta_f = \tanh(5)/5 = 0.9999/5 = 0.200$. Yes — 80% of the fin area is wasted.

8. $\varepsilon_f = \dot{Q}_{\text{fin}}/(hA_c\theta_b)$ = ratio of fin heat transfer to bare-base heat transfer

9. 50 times

10. $m = \sqrt{4 \times 30/(385 \times 0.002)} = \sqrt{120/0.77} = \sqrt{155.8} = 12.5$ m⁻¹

11. $A_f = 2\pi(0.03^2 - 0.01^2) = 2\pi \times 8 \times 10^{-4} = 5.03 \times 10^{-3}$ m²

12. $A_t = 30 \times 5 \times 10^{-4} + 0.01 = 0.025$ m². $\eta_o = 1 - (0.015/0.025)(0.12) = 1 - 0.072 = 0.928$

13. $\dot{Q} = 0.95 \times 40 \times 0.2 \times 30 = 228$ W

14. $\tanh(2.5)/2.5 = 0.9866/2.5 = 0.395$

15. When $\varepsilon_f < 1$ — practically, only with very low-$k$ material (an insulator, not a metal).

16. $L_c = L + t/2$ (or $L + D/4$ for pin fins) — accounts for tip convection approximately.

17. $mL > 2.5$

18. $m \propto \sqrt{h}$, so $m$ increases by $\sqrt{2} \approx 1.41$.

19. $\dot{Q} = 0.3 \times 100 \times \tanh(1.5) = 30 \times 0.905 = 27.2$ W

20. Small ($mL \approx 1$). Beyond that, the extra mass adds little heat transfer (law of diminishing returns).

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