---
title: "Heat dissipation from a pin fin"
subject: "Heat Transfer"
chapter: 6
key: "semester_5/heat_transfer:pin-fin-aluminium-heat-dissipation"
source: https://www.bctengineer.com/q/semester_5/heat_transfer/pin-fin-aluminium-heat-dissipation
site: BCT Engineering
---

# Heat dissipation from a pin fin

Heat Transfer · Chapter 6: Applications of Heat Transfer

- Asked in: year not given, 6 marks
- Video: [PROBLEMS ON FINS|HEAT TRANSFER THROUGH FINS|FINDING TEMPRATURE DISTRIBUTION AND HEAT LOSS ON PLATE](https://www.youtube.com/watch?v=5UPoFI_DrKg) — mech.solutions.garage
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/heat_transfer/pin-fin-aluminium-heat-dissipation
- Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-6

## Question

A cylindrical aluminium pin fin (k = 200 W/m K) of diameter 10 mm and length 50 mm projects from a wall at 120 °C into air at 30 °C with h = 25 W/m^2 K. Assuming an insulated tip, find (a) the heat dissipated by the fin, (b) the temperature at the tip, (c) the fin efficiency, and (d) the fin effectiveness.

## Answer

Data: $D = 0.01$ m, $L = 0.05$ m, $\theta_b = 120 - 30 = 90$ K.

$$
P = \pi D = 0.031416\ \text{m},\qquad A_c = \frac{\pi D^2}{4}=7.854\times10^{-5}\ \text{m}^2
$$

$$
m = \sqrt{\frac{hP}{kA_c}} = \sqrt{\frac{4h}{kD}} = \sqrt{\frac{4\times25}{200\times0.01}} = 7.071\ \text{m}^{-1}
$$

$$
mL = 7.071\times0.05 = 0.3536,\qquad \tanh(mL)=0.3395,\qquad \cosh(mL)=1.0630
$$

#### (a) Heat dissipated

$$
\begin{aligned}
\sqrt{hPkA_c} &= \sqrt{25\times0.031416\times200\times7.854\times10^{-5}} = 0.1111\ \text{W/K}\\
Q &= 0.1111\times90\times0.3395 = 3.39\ \text{W}
\end{aligned}
$$

#### (b) Tip temperature

$$
T_{tip} = T_\infty + \frac{\theta_b}{\cosh mL} = 30 + \frac{90}{1.0630} = 114.7\ ^\circ\text{C}
$$

#### (c) Efficiency

$$
\eta = \frac{\tanh mL}{mL} = \frac{0.3395}{0.3536} = 0.960 = 96\%
$$

(Check: $Q_{max} = hPL\theta_b = 25\times0.031416\times0.05\times90 = 3.534$ W, and $3.394/3.534 = 0.960$.)

#### (d) Effectiveness

$$
\varepsilon = \frac{Q}{hA_c\theta_b} = \frac{3.394}{25\times7.854\times10^{-5}\times90} = 19.2
$$

An effectiveness of 19 shows the fin is well worth using.

**Answer:** $Q \approx 3.39$ W; $T_{tip} \approx 114.7\ ^\circ$C; $\eta \approx 96\%$; $\varepsilon \approx 19$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a pin fin, m = sqrt(hP/(k Ac)) simplifies to
   A. sqrt(4h/(kD))
   B. sqrt(hD/k)
   C. sqrt(kD/(4h))
   D. sqrt(h/(kD))
   Answer: A. P/Ac = pi D / (pi D^2/4) = 4/D, so m = sqrt(4h/(kD)) = 7.07 per m. (Key point: Fin parameter m)

2. How is the insulated tip temperature found?
   A. T_tip = T_inf + theta_b x tanh(mL)
   B. T_tip = T_inf + theta_b / cosh(mL)
   C. T_tip = T_b
   D. T_tip = T_inf + theta_b x cosh(mL)
   Answer: B. At x = L, theta/theta_b = 1/cosh(mL), giving about 114.7 C. (Key point: Tip temperature)

3. The fin efficiency is tanh(mL)/(mL) = 0.96. What does this mean?
   A. 96 percent of the base heat is lost
   B. The fin is 96 percent effective compared with no fin
   C. The fin dissipates 96 percent of the heat it would if the whole fin were at base temperature
   D. The tip is at 96 percent of the air temperature
   Answer: C. Efficiency is actual loss over the maximum possible loss with the entire fin at base temperature. (Key point: Fin efficiency)

4. Effectiveness of 19 for this fin is obtained from
   A. Q / (h Ac theta_b), heat with fin over heat without fin
   B. Q / (h P L theta_b)
   C. tanh(mL)/mL
   D. Q x h Ac theta_b
   Answer: A. Effectiveness compares the fin heat with the heat that the bare base area would lose; 3.39 divided by about 0.1767 is about 19. (Key point: Fin effectiveness)

## Related in Heat Transfer

- [Fin heat dissipation, efficiency and effectiveness](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Afin-equation-tip-conditions-efficiency) — Heat Transfer, chapter 6
- [Heat loss from insulated steam pipe](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Ainsulated-steam-pipe-heat-loss) — Heat Transfer, chapter 2
- [Composite furnace wall heat loss](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Acomposite-furnace-wall-interface-temperatures) — Heat Transfer, chapter 2

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