Question

with an array of circular cross-section fins is used to cool a central processing the to ensure that the temperature of the CPU surface remains at Te-1000. is placed directly on the CPU with a special thermal paste. The associated contact resistance 8. (15 Marks) A heat sink unit (CPU) which dissipates a maximum of 1680 [W] in heat. The heat sink is l-pase T and is square in shape (WcrUry32 mm x 32mm). The temperature 25 C. The circular pin fins are arranged in a structured pattern (see Fig. 1). The spacing 10 [mm] thick of the fluid is To between The diameter of each of the fins is d 0.5 [mm), and the height of each fin heat sink is Ra-2x10-6 [ ]. WcPU Lf Lpase Rt.c 1708133 Fig. 1: (Left) Heat sink top view; (Right) Zoomed in heat sink top and side views Assume a convective heat transfer coefficient of h 2000 [mr] and a thermal conductivity of the metal of k - 500 Assume heat transfer only occurs through the heat sink (i.e, adiabatic boundary conditions at bottom and side of the CPU). For a circular fin efficiency, use the following tanh(mlc) mlc Where, Lc = Lf + d/4 and Af = PLC (e) (3 Marks) Draw a thermal circuit from the CPU surface to the fluid (include parallel resistances). (f) (7 Marks) Calculate each of the thermal resistances in the circuit in terms of the fin spacing parameter 's (g) (5 Marks) Calculate the required fin spacing based on the given maximum heat transfer rate. PLEASE NOTE PARTS

Answer 1

F) Given that R_base = L base/A base = R base = 5 times 10^- 3/500 times 32 times 10^- 3 times 32 times 10^- 3 R_base = 9.76 timse 10^- 3 k/w (l_+ c)_total = l_+ f/A_base = 1 times 10^- 6/(32 times 10^- 3)^2 Resistance of f is L_c = -L_f + w/A = L_c = 10 + 0.5/4 = l_c = 10.125 mm A F = p L_c = 4 w L_c = A f = 20.25 mm^2 r_f = tan f_1 (m L_c)/m L_c r_f = 0.948/1.811 r_f = 0.523 \r\n R_fin = 1/f_1 eta f_in A f_in R_fin = 1/2000 times 0.523 times (20.25)^2 times (10^- 3)^2 R_fin = 2.331 k/w (R_fin)_total = 2.28 times 10^- 3 k/w R_c = 1/F_1 A_base A_base = (N - 1) (rho - w) L_f R_c = 1/2000 times (024 - 1) (1 - 0.5) times 10^- 3 times 10 times 10^- 3 R_c = 9.78 times 10^- 2 k/w g) Total rate of heat transfer Q = T_c - T_alpha/R_total = T_c - T_alpha/(R_f a C)_total + R_base + (R_fin)_total + R C