Consider an annular fin of uniform thickness equal to 1.6mm. The base radius of the fin is 20mm while the tip radius is 40 mm. The base structure to which the fin is attached is at 77^°C while the air flowing over the fin is at 15^°C. The average heat transfer coefficient is given to be 15 W/m^2^◦C

Question

Consider an annular fin of uniform thickness equal to 1.6mm. The base radius of the fin is 20mm while the tip radius is 40 mm. The base structure to which the fin is attached is at 77 °C while the air flowing over the fin is at 15 °C. The average heat transfer coefficient is given to be [latex] 15W/m^{2}°C[/latex] while the fin has a thermal conductivity of 16W/m°C. Determine the heat loss from a single fin.

Accepted Answer

The given data is recorded below, in the notation of the text.

[latex]r_{1} = 0.02 m; r_{2} = 0.04 m; t = 0.0008 m; h = 15 W/m^{2}°C;[/latex]

[latex]k = 16 W/m°C; T_{b}= 77 °C; T_{amb}= 15°C[/latex]

The two parameters that characterize the fin are calculated now.

[latex]\alpha =\frac{r_{2} }{r_{1} }=\frac{0.04}{0.02}=2 ;\mu =r_{2}\sqrt{\frac{h}{kt} } =0.04 imes \sqrt{\frac{15}{16 imes 0.0008} } =1.369[/latex]

The fin efficiency from is read off Fig. 4.13 as η = 0.82. The heat loss from a single fin is then calculated as

[latex]Q_{b} =2\pi (r_{2}^{2}-r_{1}^{2})h\eta (T_{b}-T_{amb} )[/latex]

[latex]= 2 × π × (0.04^{2} − 0.02^{2}) × 15 × 0.82 × (77 − 15)[/latex]

= 5.75W

Question 4.6: Consider an annular fin of uniform thickness equal to 1.6mm....

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