Eight uniform area fins of thickness 1.2mm and length 35mm are integral with a base in a heat sink used in electronic cooling application. The thermal conductivity of the material (an aluminum alloy) is 187W/m^◦C. All the exposed surfaces are convectively cooled by ambient air at 22^◦C subject to

Question

Eight uniform area fins of thickness 1.2mm and length 35mm are integral with a base in a heat sink used in electronic cooling application. The thermal conductivity of the material (an aluminum alloy) is 187W/m°C. All the exposed surfaces are convectively cooled by ambient air at 22°C subject to a heat transfer coefficient of [latex] 16.5W/m^{2} °C[/latex] . The base temperature has been measured at 64°C. Determine (a) the overall surface efficiency and (b) effectiveness of the fin array. Also determine the heat loss from the heat sink. The heat sink is 98mm wide and the spacing between the fins is 4 mm.

Accepted Answer

Step 1 The nomenclature follows that given in Fig. 4.18. The given data may be written down as under.

Fin length: [latex] L = 35 mm or 0.035 m[/latex]
Fin semi-thickness: [latex] t = 0.6 mm or 0.0006 m[/latex]
Width of fin array: [latex] W = 98 mm or 0.098 m[/latex]
Fin spacing: [latex] S = 4 mm or 0.004 m[/latex]
Fin material thermal conductivity: [latex] k = 187 W/m°C[/latex]
Fin side heat transfer coefficient: [latex] h = 16.5 W/m^{2} °C[/latex]
Parent structure temperature: [latex] T_{b} = 64 °C[/latex]
Ambient temperature: [latex] T_{amb} = 22 °C[/latex]

Step 2 The fin parameter is then calculated as

[latex]m=\sqrt{\frac{h}{kt} } =\sqrt{\frac{16.5}{187 imes 0.0006} }=12.13m^{-1}[/latex]

The non-dimensional fin parameter is then obtained as

[latex]μ = mL = 12.13 × 0.035 = 0.424[/latex]

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

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

[latex]\eta =\frac{tanh(0.424)}{0.424}=0.944[/latex]

Step 4 Various areas needed are calculated now. Number of fins is specified as n = 8. The breadth of the base is hence given by

[latex]B = 2nt + (n − 1)S = 2 × 8 × 0.0006 + (8 − 1) × 0.004 = 0.0376m[/latex]

The total exposed area for convection may now be calculated as

[latex]A _{t} =2LWn + SW(n − 1) = 2 × 0.035 × 0.098 × 8+[/latex]

[latex]0.004 × 0.098 × (8 − 1) = 0.0576m^{2}[/latex]

Area of fins may be calculated as

[latex]A _{f} = 2LWn = 2 × 0.035 × 0.098 × 8 = 0.0549m^{2}[/latex]

Step 5 (a) Overall surface efficiency may now be calculated using Eq. 4.70.

[latex]\eta_{0} =\frac{Q_{t} }{Q_{max}}=\left[1-\frac{A_{f} }{A_{t} }(1-\eta ) ight][/latex] (4.70)

[latex]\eta_{0} =\left[1-\frac{0.0549 }{0.0576 }(1-0.944 ) ight]=0.9466[/latex]

Area of parent surface is calculated as

[latex]A_{p} = BW = 0.0376 × 0.098 = 0.00369m^{2}[/latex]

Exposed parent area is given by

[latex]A_{e} = (n − 1)SW = (8 − 1) × 0.004 × 0.098 = 0.00274m^{2}[/latex]

Step 6 (b) The effectiveness 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{0.00274 +0.0549 imes 0.944 }{0.00369} =14.78[/latex]

Step 7 Heat loss from the heat sink may now be calculated using two different methods as follows.

Based on the overall surface efficiency concept, we have

[latex]Q_{b} = A_{t} h(T_{b} − T_{amb} )η_{0} = 0.0576 × 16.5 × (64 − 22) × 0.9466 = 37.79W[/latex]

Based on the effectiveness concept, we have

[latex]Q_{b} = A_{p} h(T_{b} − T_{amb} )ε = 0.00369 × 16.5 × (64 − 22) × 14.78 = 37.79W[/latex]

Question 4.9: Eight uniform area fins of thickness 1.2mm and length 35mm a...

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