Question

Heat Sinks 3.25 ure, find (a) the fin efficiency, (b) the surface efficiency, For the heat sink shown in the accompanying fig- and (c) the heat transfer from the heat sink. T,, = 40°C 5 mm h = 16 W/m·K 4 cm 6061 aluminum 12 cm L= 25 cm To 90°C

3.25 For the heat sink shown in the accomping figure, find (a) the fin efficiency, (b) the surface efficiency, and (c) the heat transfer from the heat sink.

Answer 1

efficiency of fin: - n_fin = tan h (m.H_f) /m.H_f H_f rightarrow height of fin. m = squareroot 2h/K_fin middot t_fin K_fin rightarrow thermal conductivitu of f_in. t_fin rightarrow thickness of fin. k_fin for A1 6061 is K_fin = 167 w/m-k. t_fin 5 mm = 0.005 m therefore m = squareroot 2 times 16 (Given) /167 times 0.005 m = 6.190 \r\n therefore n_fin = tan 4 (6.19 times 0..04) /6.19 times 0.04 = 0.98 n_fin = 98% b) Surface efficiency: - n_0 = 1-A_f/A (1-n_f) A rightarrow Area A_f rightarrow F_in area. nf rightarrow f_in efficiency A_f = 2 times H_f times L = 2 times 0.04 times 0.25 A_f = 0.02 m^2 n_0 = 1-0.02/0.021251 (1- 0.98) therefore n_0 = 981.11 % A = A_f + A_p F_p = t times L = 0.001251 m^2 therefore A = 0.02 + 0.001251 A = 0.021251 m^2 c) Heat transfer: - Q = N[KA delta Tn] times [x_1/x_2]