An extruded aluminum alloy heat sink consists of 4 parallel flat fins integral on a flat base. Air is blown parallel to fin flats and the convective heat transfer coefficient has been estimated from experiments to be h = 28W/m^2^◦C. Aluminum alloy has a thermal conductivity of k = 175W/m^◦C. The ba

Question

An extruded aluminum alloy heat sink consists of 4 parallel flat fins integral on a flat base. Air is blown parallel to fin flats and the convective heat transfer coefficient has been estimated from experiments to be [latex] h = 28W/m^{2} °C[/latex] . Aluminum alloy has a thermal conductivity of k = 175W/m°C. The base of heat sink is W = 30mm, H = 50mmwith the fins being parallel to the 50mmedge. All the fins as well as the base have a thickness of 2t = 0.4mm. Choosing optimal proportioned fins determine the heat loss from the heat sink when the base is at a temperature of [latex] T_{b} = 65°C[/latex] and ambient air is at [latex] T_{amb} = 30°C.[/latex]

Accepted Answer

Step 1 With the data specified in the problem, we assume the material thickness to be [latex] 2t^{*} = 0.4mm[/latex] or [latex] t^{*} = 0.2mm[/latex] .We choose μ = 1.4192 to obtain the fin length as

[latex] L^{*} =\frac{\mu ^{*} }{m} =\frac{\mu^{*} }{\sqrt{\frac{h}{kt^{*}} } } =\frac{1.4192}{\sqrt{\frac{28}{175 imes 0.0002} } }=0.05018m\approx 50.2mm[/latex]

Step 2 With n = 4 and W = 30 mm, fin spacing is given by

[latex]S=\frac{W-2nt^{*}}{(n-1)} =\frac{30-2 imes 4 imes 0.2}{(4-1)} = 9.47mm[/latex]

Heat sink cross section will appear as shown in Fig. 4.19.

Step 3 Fin efficiency is then calculated using Eq. 4.22.

[latex]\eta =\frac{Q_{b} }{Q_{max} } =\frac{mkA heta _{b}tanh(mL)}{PLh heta _{b}} =\frac{tanh(mL)}{mL}=\frac{tanh(\mu )}{\mu }[/latex] (4.22)

[latex]\eta =\frac{tanh(1.4192)}{1.4192} =0.627[/latex]

The effectiveness of fin array may now be calculated using Eq. 4.71 as

[latex]\varepsilon =\frac{Q_{t} }{Q_{p} }=\frac{A_{e}+A_{f} \eta }{A_{p}}[/latex] (4.71)

[latex]\varepsilon =\frac{2 imes 4 imes 0.0502+(4-1) imes 0.0947}{0.03} =9.332[/latex]

Step 4 Heat transfer in the absence of fins is given by

[latex]Q_{no fins} = WHhθ_{b} = 0.03 × 0.05 × 28 × (65 − 30) = 1.47W[/latex]

Hence the heat transfer from heat sink is given by

[latex]Q_{heat sink }= ε × Q_{no fins} = 9.332 × 1.47 = 13.72W[/latex]

Question 4.10: An extruded aluminum alloy heat sink consists of 4 parallel ...

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