Heat dissipation from a pin fin

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; $\eta \approx 96\%$; $\varepsilon \approx 19$.