A 15-cm × 20-cm integrated circuit board is to be cooled by attaching 4-cm long aluminum (k = 237 W/m∙K) fins on one side of it (Fig. 3–50). Each fin has a 2-mm × 2-mm square cross section. The surrounding ambient temperature is 25°C and the convection heat transfer coefficient on each fin surface is 20 W / m ^{2} \cdot K To prevent the circuit board from overheating, the upper surface of the circuit board needs to be at 85°C or cooler. Design a finned surface having the appropriate number of fins, with an overall effectiveness of 3 that can keep the circuit board surface from overheating.
Question : A 15-cm × 20-cm integrated circuit board is to be cooled by ...
SOLUTION In this example, the concepts of Prevention through Design (PtD) are applied in conjunction with the fin analysis. An integrated circuit board is to be cooled by attaching aluminum fins of square cross section on one side. The number of fins needed to keep the circuit board surface cooler than 85°C, while having an overall effectiveness of 3, is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat conduction is one dimensional. 3 Heat transfer from the fin tips is negligible. 4 Fins are very long. 5 The heat transfer coefficient is constant and uniform over the entire fin surface. 6 The thermal properties of the fins are constant.
Properties The thermal conductivity of the aluminum fins is given to be k = 237 W/m·K.
Analysis Noting that the cross-sectional areas of the fins are constant, the efficiency of the square cross-section fins that are assumed to be very long with adiabatic tips can be determined to be
m=\sqrt{\frac{h p}{k A_{c}}}=\sqrt{\frac{4 h a}{k a^{2}}}=\sqrt{\frac{4\left(20 W / m ^{2} \cdot K \right)(0.02 m )}{(237 W / m \cdot K )(0.02 m )^{2}}}=12.99 m ^{-1}where a is the length of each side of the square fin.
\eta_{\text {fin }}=\frac{\tanh m L}{m L}=\frac{\tanh \left(12.99 m ^{-1} \times 0.04 m \right)}{12.99 m ^{-1} \times 0.04 m }=0.919The finned and unfinned surface areas, and heat transfer rates from these areas are
A_{\text {fin }}=n_{\text {fin }} \times 4 \times(0.002 m )(0.04 m )=0.00032 n_{\text {fin }} m ^{2}A_{\text {unfinnat }}=(0.15 m )(0.20 m )-n_{ fin }(0.002 m )(0.002 m )
=0.03-0.000004 n_{ fin } m ^{2}
\dot{Q}_{\text {finnad }}=\eta_{\text {fin }} \dot{Q}_{\text {fin, } \max }=\eta_{\text {fin }} h A_{\text {fin }}\left(T_{b}-T_{\infty}\right)
=0.919\left(20 W / m ^{2} \cdot K \right)\left(0.00032 n_{\text {fin }} m ^{2}\right)(85-25)^{\circ} C=0.3529 n_{\text {fin }} W
\dot{Q}_{\text {untinnad }} =h A_{\text {untinad }}\left(T_{b}-T_{\infty}\right)=\left(20 W / m ^{2} \cdot K \right)\left(0.03-0.000004 n_{ fin } m ^{2}\right)(85-25)^{\circ} C =36-0.0048 n_{ fin } W
Then the total heat transfer from the finned surface (circuit board) becomes
\dot{Q}_{\text {total } \text { fin }}=\dot{Q}_{\text {finnat }}+\dot{Q}_{\text {untinnat }}=0.3529 n_{\text {fin }}+36-0.0048 n_{\text {fin }} WThe rate of heat transfer if there were no fins attached to the plate would be
A_{ mfin }=(0.15 m )(0.20 m )=0.03 m ^{2}\dot{Q}_{\text {rofin }}=h A_{\text {nofin }}\left(T_{b}-T_{\infty}\right)=\left(20 W / m ^{2} \cdot K \right)\left(0.03 m ^{2}\right)(85-25)^{\circ} C =36 W
The number of fins can be determined from the overall fin effectiveness equation
\varepsilon_{ fin }=\frac{\dot{Q}_{ fin }}{\dot{Q}_{ nofin }}3=\frac{0.3529 n_{ fin }+36-0.0048 n_{ fin }}{36}
n_{ fin }=207
Discussion To keep the circuit board surface from heating above 85°C, the finned surface having an overall effectiveness of 3 needs to have at least 207 fins. Number of fins on the circuit board may be reduced by using different fin material and geometry.