Training Cooling Fins Rectangular Fin — Temperature Distribution and Heat Transfer
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Rectangular Fin — Temperature Distribution and Heat Transfer

24 min Cooling Fins

Rectangular Fin — Temperature Distribution and Heat Transfer

The fin equation is a second-order ODE. Its solution depends on the tip boundary condition. The most common case assumes an insulated (adiabatic) tip, which is surprisingly accurate because the tip area is small.

Fin Equation (Uniform Cross-Section)

$$\frac{d^2\theta}{dx^2} - m^2\theta = 0$$

where $\theta(x) = T(x) - T_\infty$ and $m = \sqrt{hP/(kA_c)}$.

Solution — Insulated Tip

$$\frac{\theta(x)}{\theta_b} = \frac{\cosh[m(L-x)]}{\cosh(mL)}$$

$$\dot{Q}_{\text{fin}} = \sqrt{hPkA_c}\,\theta_b\,\tanh(mL)$$

$\theta_b = T_b - T_\infty$.

Other Tip Conditions

Convection at tip: $\dot{Q} = \sqrt{hPkA_c}\,\theta_b\,\frac{\sinh(mL) + (h/mk)\cosh(mL)}{\cosh(mL) + (h/mk)\sinh(mL)}$

Corrected length (simpler): Use $L_c = L + t/2$ in the insulated-tip formula to approximate convection from the tip. Works well for thin fins.

Infinite fin ($mL > 2.5$): $\dot{Q} = \sqrt{hPkA_c}\,\theta_b$ (tanh ≈ 1)

Example 1 — Rectangular Fin Heat Transfer

Aluminum fin: $k = 200$, $w = 8$ cm, $t = 2$ mm, $L = 5$ cm. $h = 30$, $T_b = 100$°C, $T_\infty = 25$°C. Find $\dot{Q}_{\text{fin}}$.

$A_c = 0.08 \times 0.002 = 1.6 \times 10^{-4}$ m²

$P = 2(0.08 + 0.002) = 0.164$ m

$m = \sqrt{30 \times 0.164/(200 \times 1.6 \times 10^{-4})} = \sqrt{4.92/0.032} = \sqrt{153.75} = 12.4$ m⁻¹

$mL = 12.4 \times 0.05 = 0.620$

$\tanh(0.620) = 0.551$

$\sqrt{hPkA_c} = \sqrt{30 \times 0.164 \times 200 \times 1.6 \times 10^{-4}} = \sqrt{0.1574} = 0.397$ W/K

$\dot{Q}_{\text{fin}} = 0.397 \times 75 \times 0.551 = 16.4$ W

Example 2 — Tip Temperature

In Example 1, find the temperature at the fin tip.

$$\frac{\theta(L)}{\theta_b} = \frac{\cosh(0)}{\cosh(0.620)} = \frac{1}{1.198} = 0.835$$

$T_{\text{tip}} = T_\infty + 0.835 \times \theta_b = 25 + 0.835 \times 75 = 87.6$°C

The tip is 12.4°C cooler than the base — quite effective since most of the 75°C $\Delta T$ is preserved.

Example 3 — Infinite Fin Approximation

A steel fin ($k = 50$) with $m = 20$ m⁻¹ and $L = 15$ cm. $mL = 3.0$. Can we use the infinite fin formula?

$mL = 3.0 > 2.5$, so $\tanh(3.0) = 0.9951 \approx 1$.

Yes — the infinite fin approximation is accurate to within 0.5%. The tip is essentially at $T_\infty$.

Practice Problems

1. $\sqrt{hPkA_c} = 0.5$ W/K, $\theta_b = 60$°C, $mL = 1.0$. Find $\dot{Q}_{\text{fin}}$.
2. $mL = 2.0$, $\theta_b = 80$°C. What fraction of $\theta_b$ remains at the tip?
3. Compare $\tanh(0.5)$, $\tanh(1)$, $\tanh(2)$, $\tanh(3)$.
4. Corrected length: $L = 4$ cm, $t = 3$ mm. Find $L_c$.
5. $m = 10$, $L = 10$ cm, $k = 180$, $A_c = 2 \times 10^{-4}$, $\theta_b = 50$. $\dot{Q}$?
6. A fin has $mL = 0.3$ vs. $mL = 3$. Which has a more uniform temperature?
Show Answer Key

1. $\dot{Q} = 0.5 \times 60 \times \tanh(1.0) = 0.5 \times 60 \times 0.762 = 22.9$ W

2. $\theta(L)/\theta_b = 1/\cosh(2) = 1/3.762 = 0.266$. Only 26.6% of base excess temperature remains.

3. $\tanh(0.5) = 0.462$, $\tanh(1) = 0.762$, $\tanh(2) = 0.964$, $\tanh(3) = 0.995$

4. $L_c = 0.04 + 0.003/2 = 0.0415$ m

5. $mL = 1.0$. Need $P$ and $h$: from $m^2 = hP/(kA_c)$, $hP = 100 \times 180 \times 2 \times 10^{-4} = 3.6$. $\sqrt{hPkA_c} = \sqrt{3.6 \times 180 \times 2 \times 10^{-4}} = \sqrt{0.1296} = 0.360$. $\dot{Q} = 0.360 \times 50 \times 0.762 = 13.7$ W

6. $mL = 0.3$. Smaller $mL$ = slower temperature decay = more uniform. $1/\cosh(0.3) = 0.956$ vs. $1/\cosh(3) = 0.0993$.

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