# In what situations do I use the characteristic length of a fin to find the surface area?

So I'm learning about fins in heat transfer and it seems that there are two separate formulas for the surface area of a rectangular fin of length L, width w and thickness t. The fin is attached to a wall at one of its L x t face.

The formula for the surface area for a normal rectangular fin would be: $$A_{f}=2Lw+2Lt+tw$$

Now there is also a formula using the characteristic length $L_{c}$ $$A_{f}=2wL_{c}$$

Now my question is Under what conditions do I use the characteristic length to determine the surface area of the fin?

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The first expression is the actual exposed area of the fin. The second uses $L_c=L+(t/2)\$ to get $A_f=2Lw+tw,\$ which seemingly ignores the fin's sides. This is because $L_c\$ and $A_f\$ are the corrected length and corrected area, to make the fin fit into the standard efficiency relation:

\begin{align}\eta_f&\equiv\frac{q_f}{q_\text{max}}=\frac{q_f}{hA_f\,\theta_b}=\frac{\tanh mL_c}{mL_c}\\\vphantom{\Large \frac{Q}{Q}}{\text{where}}\qquad m&=\sqrt{\frac{2h}{kt}}\end{align}

We need the corrections to fix two problems:

1. The equation is intended for fins with adiabatic tips. Pointy tips are adiabatic since they have zero surface area. But rectangular fins' tips have area and transfer heat. So we fudge the length a bit to pretend that the extra heat transfer is coming from the fin's body. See National Advisory Committee for Aeronautics Report 158 (1929).
2. The equation implies $q_\text{max}=hA_f\,\theta_b,\$ which in turn implies that the entire surface area of the fin "sees" a uniform heat transfer coefficient $h.\$ Obviously that isn't true. So for rectangular fins we also fudge the area by only counting those faces which are parallel to the air flow – not the sides. It just so happens we get the rather neat expression $A_f=2wL_c\$ doing this.
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Here's why I ask. I just had an exam the other day where there was a rectangular fin attached to the wall the protruded a distance L=8mm and exposed ta fluid. It's width was 10mm and it's thickness was 1mm. We had to determine the resistance and efficiency of the fin. He said $$A_{f}=PL+A_{c}=2(w+t)L+wt=186mm^{2}$$ whereas I said $$A_{f}=2wL_{c}=2w(L+\frac{t}{2})=170mm^{2}$$ Now there is a $16mm^{2}$ difference between the two areas. So which way is correct? –  Greg Harrington Feb 25 '12 at 20:23
@Greg Well, in the report I linked, that 16 mm² is ignored simply because "The fin is so long that the effect of the exposed ends... is negligible." The efficiency relation $\eta_f=\tanh(mL_c)/(mL_c)\$ comes from that report, so if you use the relation, you automatically make the assumption. It's inconsistent to apply the assumption for one equation, but not for another. And if the fluid flows, then – like I said – the exposed ends "see" a quite different heat transfer coefficient than the rest of the fin, so you shouldn't lump all the sides together in any case. $A_f=2wL_c\$ is correct. –  rdhs Feb 26 '12 at 5:16
If your professor wanted to be really rigorous, I'd suggest the correction $$L_c^\prime=L+\frac{t}{2}+\frac{h_l+h_t}{2hw}tL,\qquad A_f=2wL_c^\prime.$$ But nobody does that. –  rdhs Feb 26 '12 at 5:25
He didnt use that rigorous one. I looked at the solution and he used $$\eta_{f}=\frac{tanh(mL_{c})}{mL_{c}}$$ and $$A_{f}=2(w+t)L+wt$$ Essentially, he is using the corrected length for one equation but not for the other which, like you said, is inconsistent. But what if he used the area equation without the corrected length but use the other equation for the efficiency by using the temperature distribution. Would that theoretically yield the same result or would it be different? Pretty much does it matter which method you use as long as you stay consistent with which 'type' of length you use? –  Greg Harrington Feb 26 '12 at 6:13
@Greg If you mean this equation:$$q_f=\sqrt{hPkA_c}\;\theta_b\,\frac{\sinh mL+\frac{h}{mk}\cosh mL}{\cosh mL+\frac{h}{mk}\sinh mL},$$then that doesn't use $A_f\$ at all. And it'd give the most accurate answer, *if* you had the right value for $h.$ –  rdhs Feb 26 '12 at 6:22