A sheet of aluminum alloy 1.5mm thick is available for providing fins in a heat transfer application. Determine the length of the fin you would suggest such that it is an optimum fin. The thermal conductivity of the material of the sheet is 200W/m^◦C while the heat transfer coefficient is 67W/m^2^◦

Question

A sheet of aluminum alloy 1.5mm thick is available for providing fins in a heat transfer application. Determine the length of the fin you would suggest such that it is an optimum fin. The thermal conductivity of the material of the sheet is 200W/m°C while the heat transfer coefficient is [latex] 67W/m^{2} °C[/latex] . What is the heat loss per unit temperature difference between the fin base and the ambient? Discuss a few non-optimal cases.

Accepted Answer

Step 1 Since the fin material is of specified thickness we assume that this corresponds to the optimum thickness. Thus

[latex] t=t^{*} =\frac{0.0015}{2} =0.00075m[/latex]

Step 2 The length of the fin is chosen such that [latex] μ = μ_{0} = 1.4192[/latex] . Since [latex] μ_{0} = m^{*} L^{*}[/latex] , we have

[latex] L^{*} = \frac{\mu ^{*}}{m^{*}} =\frac{\mu _{0} }{\sqrt{\frac{h}{kt^{*}}}}=\frac{1.4192}{\sqrt{\frac{67}{200 imes 0.00075} } }=0.067m[/latex]

Step 3 Figure 4.14 compares non-optimal proportioned fins with a fin that is proportioned to obtain optimal behavior. The figures are to scale. All the cases have a profile area [latex]A_{p}[/latex] equal to [latex]2t^{*}L^{*} = 2 × 0.00075 × 0.067 = 0.0001m^{2}[/latex] .

Step 4 Heat transfer per unit temperature difference may then be calculated as

[latex]\frac{Q_{b}}{ heta _{b}} = 2hL^{*} η ^{*} = 2 × 67 × 0.067 × 0.6267 = 5.639W/m°C[/latex]

Heat transfer from the fin per unit volume and temperature difference is thus given by

[latex]\frac{Q_{b}}{ heta _{b}A_{p}} =\frac{5.639}{0.0001}=56390W/m^{3}°C[/latex]

Step 5 The calculations have been performed also for a number of non-optimal fins and the resulting data has been shown plotted, in Fig. 4.15, as a graph of ratio heat transfer from the fin per unit volume to the maximum possible heat transfer per unit volume and temperature difference, as a function of fin length. The optimum value is shown by the black circle labeled (b) in this figure.

Question 4.7: A sheet of aluminum alloy 1.5mm thick is available for provi...

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