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

24 min Cooling Fins

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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