Air at 27°C and a velocity of 5 m/s passes over the small region As (20 mm × 20 mm) on a large surface, which is maintained at Ts = 127°C. For these conditions, 0.5 W is removed from the surface As. To increase the heat removal rate, a stainless steel (AISI 304) pin fin of diameter 5 mm is affixed to As, which is assumed to remain at Ts = 127°C. (a) Determine the maximum possible heat removal rate through the fin. (Hint: look back to Chapter 3 to see which fin case produces the maximum heat transfer.) (b) What fin length would provide a close approximation to the heat rate found in part (a)? (Hint: refer to Example 3.9.) (c) Determine the fin effectiveness, εf. (d) What is the percentage increase in the heat rate from As due to installation of the fin? In other words, what is the percentage increase in q obtained by adding the fin as compared to not having the fin?

Answers 2

Answer:

a) The maximum possible heat removal rate = 2.20w

b) Fin length = 37.4 mm

c) Fin effectiveness = 89.6

d) Percentage increase = 435%

Explanation:

See the attached file for the explanation.

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AI generated Answer

a) The maximum possible heat removal rate through the fin corresponds to the case of an externally finned surface, which is given by q_max = \alpha_f h_f A_s(T_s - T_{\infty} ) where \alpha_f is the fin-to-base thermal conductivity ratio, h_f is the fin heat transfer coefficient, and A_s is the total area of the base surface. b) The fin length that would provide a close approximation to the maximum heat rate is given by L_f \simeq \frac{\alpha_f h_f A_s }{A_c (T_s - T_{\infty} )} where A_c is the cross-sectional area of the fin. c) The fin effectiveness is given by \varepsilon_f = \frac{q_f}{q_max} where q_f is the actual heat rate through the fin. d) The percentage increase in the heat rate from As due to installation of the fin can be calculated as \Delta q = \frac{ q_f - q_0 }{q_0} \times 100 \% where q_0 is the heat rate from As with no fin.
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