For the fin and conditions shown, compute the temperature at the distances from the base shown in the table. x (m) 0.05 0.25 0.5 5cm 40°C h=20 w/m²°C 20°C air 1mm 0.5m Kfin = 200 W/m.k $\theta/\theta_b$ T(x) mx
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The image shows a rectangular fin. Length of the fin, $L = 0.5$ m Width of the fin, $W = 5$ cm $= 0.05$ m Thickness of the fin, $t = 1$ mm $= 0.001$ m 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.
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A triangular straight fin of 0.1 m in length, 0.02 m thick at the base, and 0.2 m in depth is used to extend the surface of a wall at 200°C. The wall and the fin are made of mild steel (k = 54 W/m°C). Air at 10°C (h = 200 W/m²°C) flows over the surface of the fin. Evaluate the temperature at 0.05 m from the base and at the tip of the fin. Determine the rate of heat removal from the fin and the fin efficiency. [Use MS-Excel to solve the problem and plot temperature profile].
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