homework and exercises - 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?

Answer

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.