A fin of triangular section is made of an aluminum alloy of thermal conductivity equal to k = 185W/m^◦C. It has a base thickness of 2yb = 3mm and a length of L = 56 mm. The lateral surfaces of the fin are cooled by air at Tamb = 25^◦C with a heat transfer coefficient of h = 47W/m^2^◦C. Howmuch heat

Question

A fin of triangular section is made of an aluminum alloy of thermal conductivity equal to k = 185W/m°C. It has a base thickness of [latex] 2y_{b} = 3mm[/latex] and a length of L = 56 mm. The lateral surfaces of the fin are cooled by air at T_{amb} = 25 °C with a heat transfer coefficient of [latex] h = 47W/m^{2} °C[/latex] . Howmuch heat does the fin dissipate if its width is W = 0.3m and the base is maintained at [latex] T_{b} = 86 °C[/latex] ? Use fin efficiency from Fig. 4.9b to solve this problem.

Accepted Answer

From the given data [latex]y_{b} = 1.5mm = 0.0015 m[/latex]. The fin parameter is calculated using definition 4.37 as

[latex] ho ^{2} = \frac{hL}{ky_{b} }\sqrt{1+\left(\frac{y_{b}}{L} ight)^{2} }[/latex] (4.37)

[latex] ho = \sqrt{ \frac{hL}{ky_{b} }\sqrt{1+\left(\frac{y_{b}}{L} ight)^{2} }}=\sqrt{\frac{47 imes 0.056}{185 imes 0.0015 } \sqrt{1+\left(\frac{0.0015}{0.056} ight)^{2} }} =3.08m^{-\frac{1}{2} }[/latex]

Hence the non-dimensional fin parameter is

[latex]\mu = ho \sqrt{L} =3.08 imes \sqrt{0.056} =0.729[/latex]

From Fig. 4.9b, the fin efficiency is read off as, η = 0.77. Hence the total heat loss from the fin is calculated as

[latex]Q_{b}=2L\sqrt{1+\left(\frac{y_{b}}{L} ight)^{2} }Wh \eta (T_{b} -T_{amb})[/latex]

[latex]=2 imes 0.056 \sqrt{1+\left(\frac{0.0015}{0.056} ight)^{2} } imes 0.3 imes 47 imes 0.77(86-25)=74.2W[/latex]

Question 4.3: A fin of triangular section is made of an aluminum alloy of ...

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