Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The diameter of the fin is 3 cm. The thermal conductivity of the fin is 150 W/m·K. The heat transfer coefficient is 123 W/m2·K. Estimate the fin temperature in °C at a distance of 10 cm from the base

98°C

Explanation:

Total surface area of cylindrical fin = πr² + 2πrl , r = 0.015m; l= 0.1m; π =22/7

22/7*(0.015)² + 22/7*0.015*0.1 = 7.07 X 10∧-4 + 47.1 X 10∧-4 = (54.17 X 10∧-4)m²

Temperature change, t = (50 - 25)°C = 25°C = 298K

Hence, Temperature =  150 X (54.17 X 10∧-4) X 298/123 = 242.14/124 = 2.00K =

∴ Temperature change = 2.00K

But temperature, T= (373 - 2)K = 371 K

In °C = (371 - 273)K = 98°C         

answer:

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