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I am looking at the solutions that my professor put up and I feel that he did something wrong. Here is the question and I will give my stab at the solution so you can see why I think that it is wrong.

A straight fin is fabricated of aluminum alloy with conductivity of 185 $\frac{W}{mK}$ with a base thickness of t = 3mm and L = 15mm. Its base temperature is $T_{b}=100 C$ and it is exposed to a fluid for which $T_{\infty}=20C$ and $h=50 \frac{W}{m^{2}k}$. Compare the efficiencies of rectangular, triangular and parabolic profiles assuming a fin of unit width.

Solution for rectangular:

$$m=\sqrt{\frac{hP}{kA_{c}}}=\sqrt{\frac{2h(w+t)}{kwt}}=13.4432m^{-1}$$ $$mL=(13.4432)(.015)=.201649$$ $$mL_{c}=13.4432(.015+\frac{.003}{2})=.221813$$

This is where the problem lies. My professor then uses the $mL_{c}$ value in the efficiency for the rectangular fin which is fine. But he then goes on to use the above mL value for the triangular and parabolic fins. I would think that the mL values for these shapes would be different than for a rectangular fin.

I assumed that for fins of non-uniform surface area like the triangular and parabolic fins then you would use $$mL_{c}=\sqrt{\frac{2h}{kA_{p}}}L_{c}^{\frac{3}{2}}$$ $$m=\sqrt{\frac{2hL_{c}}{kA_{p}}}$$ to find the m value for non-uniform surface areas. Then you would multiply that by L to get the mL values used in the efficiency equations for the triangular and parabolic shapes. So my question is, Should the mL values for the rectangular and triangular fins be the same value or should they be different? And when should I use the equation that uses the area profile if it cannot be used here?

Clarification of variables:

P: Perimeter

w: fin width (unit width)

$A_{c}$: cross-sectional area

t: fin thickness

$A_{p}$: profile area

L: Fin length

$L_{c}$: characteristic length

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here,perimeter is not constant along the length,it varies along the length,so u can't use his formulae –  user27300 Jul 20 '13 at 13:11

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