A 12-cm-wide and 18 -cm-high vertical hot surface in 30^{\circ} C air is to be cooled by a heat sink with equally spaced fins of rectangular profile Fig. (9-23) . The fins are 0.1 cm thick and 18 cm long in the vertical direction and have a height of 2.4 cm from the base. Determine the optimum fin spacing and the rate of heat transfer by natural convection from the heat sink if the base temperature is 80^{\circ} C.
SOLUTION A heat sink with equally spaced rectangular fins is to be used to cool a hot surface. The optimum fin spacing and the rate of heat transfer are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 The atmospheric pressure at that location is 1 atm. 4 The thickness t of the fins is very small relative to the fin spacing s so that Eq. (9-32)
T_{s}=\text { constant: } \quad S_{ opt }=2.714\left(\frac{S^{3} L}{ Ra _{S}}\right)^{0.25}=2.714 \frac{L}{ Ra _{L}^{0.25}}for optimum fin spacing is applicable. 5 All fin surfaces are isothermal at base temperature.
Properties The properties of air at the film temperature of T_{f}=\left(T_{s}+T_{\infty}\right) / 2=(80+30) / 2=55^{\circ} C and 1 atm pressure are (Table A -15)
k=0.02772 W / m \cdot K\operatorname{Pr}=0.7215
\nu=1.847 \times 10^{-5} m ^{2} / s
\beta=1 / T_{f}=1 / 328 K
Analysis We take the characteristic length to be the length of the fins in the vertical direction (since we do not know the fin spacing). Then the Rayleigh number becomes
Ra _{L} =\frac{g \beta\left(T_{s}-T_{\infty}\right) L^{3}}{\nu^{2}} \operatorname{Pr}=\frac{\left(9.81 m / s ^{2}\right)[1 /(328 K )](80-30 K )(0.18 m )^{3}}{\left(1.847 \times 10^{-5} m ^{2} / s \right)^{2}}(0.7215) =1.845 \times 10^{7}The optimum fin spacing is determined from Eq. (9-32) to be
S_{ opt }=2.714 \frac{L}{ Ra _{L}^{0.25}}=2.714 \frac{0.18 m }{\left(1.845 \times 10^{7}\right)^{0.25}}=7.45 \times 10^{-3} m =7.45 mmwhich is about seven times the thickness of the fins. Therefore, the assumption of negligible fin thickness in this case is acceptable. The number of fins and the heat transfer coefficient for this optimum fin spacing case are
n=\frac{W}{S+t}=\frac{0.12 m }{(0.00745+0.001) m } \approx 14 \text { fins }The convection coefficient for this optimum fin spacing case is, from Eq. (9-33),
S=S_{ opt }: \quad Nu =\frac{h S_{ opt }}{k}=1.307 h= Nu _{ opt } \frac{k}{S_{ opt }}=1.307 \frac{0.02772 W / m \cdot K }{0.00745 m }=4.863 W / m ^{2} \cdot KThen the rate of natural convection heat transfer becomes
\dot{Q} =h A_{s}\left(T_{s}-T_{\infty}\right)=h(2 n L H)\left(T_{s}-T_{\infty}\right) =\left(4.863 W / m ^{2} \cdot K \right)[2 \times 14(0.18 m )(0.024 m )](80-30)^{\circ} C =29.4 WTherefore, this heat sink can dissipate heat by natural convection at a rate of 29.4 W.
