Question
The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip?
Step 1
First, we need to understand the purpose of a fin. A fin is used to enhance heat transfer from a surface by increasing the surface area exposed to the surrounding medium. Show more…
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Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially that of the surrounding air, i.e., 20°C. Its width is 5.0 cm; thickness is 1.0 mm; thermal conductivity is 200 W/m·K; and base temperature is 40°C. The heat transfer coefficient is 20 W/m^2·K. Estimate the fin temperature at a distance of 5.0 cm from the base and the rate of heat loss from the entire fin.
A 1-cm-diameter, 30-cm-long fin made of aluminum $(k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ is attached to a surface at $80^{\circ} \mathrm{C}$. The surface is exposed to ambient air at $22^{\circ} \mathrm{C}$ with a heat transfer coefficient of $18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. If the fin can be assumed to be very long, the rate of heat transfer from the fin is (a) $2.0 \mathrm{~W}$ (b) $3.2 \mathrm{~W}$ (c) $4.4 \mathrm{~W}$ (d) $5.5 \mathrm{~W}$ (e) $6.0 \mathrm{~W}$
A plane wall surface at $200^{\circ} \mathrm{C}$ is to be cooled with aluminum pin fins of parabolic profile with blunt tips. Each fin has a length of $25 \mathrm{~mm}$ and a base diameter of $4 \mathrm{~mm}$. The fins are exposed to ambient air at $25^{\circ} \mathrm{C}$, and the heat transfer coefficient is $45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. If the thermal conductivity of the fins is 230 $\mathrm{W} / \mathrm{m} \cdot \mathrm{K}$, determine the heat transfer rate from a single fin and the increase in the rate of heat transfer per $\mathrm{m}^{2}$ surface area as a result of attaching fins. Assume there are 100 fins per $\mathrm{m}^{2}$ surface area.
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