Part 2 1.a < www Effective fin surface area: A_f = 2N_finsl_fW_f = 2600.0315 m0.00835 m = 3.006 m^2 < Fin efficiency: m = (hP_f)/k_A/A_f = (25 W/m^2-K0.0315 m)/(237 W/m-K0.00835 m) = 0.6684 = Convection coefficient: Re_L = (U_infL_f)/nu = (4 m/s0.0315 m)/(1.57e-5 m^2/s) = 8.0e4 -> Turbulent flow h = 0.037k_air/L_Re_L^0.8)(Pr^1/3) = 0.037(0.0257 W/m- K)/(0.0315 m)(8.0e4^0.8) (0.707^1/3) = 25.3 W/m^2-K < worm Base conduction resistance: R_base = t_base/(k_AIA_base) = (0.0035 m)/(237 W/m- KO.0125^2 m^2) = 0.061 K/W < Fin resistance: R_fin = 1/(hA_fm) = 1/(25.3 W/m^2-K3.006 m^20.668) = 0.064 K/W < Spreading resistance: R sp = 0.25*(t_base/k_Al)^0.5/(A_base^0.5h) = 0.25(0.0035 m/237 W/m-K)^0.5/(0.0125 m^0.5*25.3 W/m^2-K) = 0.008 K/W < Total resistance: R_total = R base + R fin + R sp = 0.061 + 0.064 + 0.008 = 0.133K/W < www mommmmmm mattum manhw madm This matches well with the manufacturer's given value of R_ja = 0.13 K/W at this air velocity. < b. < utom Shear stress on fins: tau_w= 0.023 rhoU_inf^2 = 0.023*(1.16 kg/m^3)*(4 m/s)^2 = 0.882 Pa Pressure drop: F_p = 2N_finsL_fW_ftau_w= 2600.0315 m0.00835 m0.882 Pa = 3.24 N < Delta P = F p/A_duct = 3.24 N/(0.035 m*0.05 m) = 186 Pa C. < untamw inochi initium At U_inf = 4 m/s, the fin spacing of 8.35 mm is still large compared to the expected boundary layer thickness. Treating each fin as an individual flat plate with external flow is reasonable. < At a lower U_inf = 0.5 m/s, the boundary layer growth would be larger and the fins would begin to act more like a channel flow with the boundary layers merging. The flat plate assumption would no longer be appropriate. 1. (25 Pts) This problem focuses on the performance of the Alpha Novatech LPD50-35B finned heat sink. Assume the following information about this heat sink and operating conditions: The whole unit is constructed from 6063 aluminum alloy The heat sink has a 50 50 mm square base of thickness tbase = 3.5 mm A 12.7 x 12.7 mm square heater is installed on the center of the base to simulate an electronic deviser that would be cooled by this heat sink The heat sink has 12 evenly spaced rows of 5 fins (approximate as rectangular). The fins are L = 31.5 mm tall, t=0.85 mm thick, and W=8.35 mm long in the flow direction. Air at 25C is supplied to the heat sink through a close-fitting duct of height 35 mm and width 50 mm (i.e., the duct fits perfectly over the heat sink). The air flows parallel to the fins. 1.60 TYP 50.0 SQ -05 0.25 12X 0.85 0.25 T 3.45 4X1.50 a. (10 pts) Estimate the total thermal resistance from the 12.7 x 12.7 mm square heater to the air stream for an upstream average air velocity in the duct of U = 4 m s. Model the fins as individual flat plates, and assume that parallel flow flat plate convection boundary layers develop along each. Account for spreading resistance into the base of the heat sink using the method suggested in HW2 question 3D. Hint: the manufacturer provides data for the thermal performance of this heat sink. You should find close agreement between your model and their resistance data at U = 4 m s. b. (8 pts.) Estimate the total frictional pressure drop for the air flow through this finned assembly at U 4 m s. To do this, first find the average wall shear stress (Tw) on the individual fins based on the boundary layer flow model. Then perform a force balance accounting for the pressures acting on the inlet and outlet faces of the close-fitting duct (35 mm 50 mm) and the frictional drag assuming Tw acts on all surfaces of the heat sink and points against the flow direction. c. (7 pts) The above analysis assumes that the flow over the individual fins is more like external flow over individual flat plates rather than channel flow between closely spaced walls. Is this a reasonable approximation at U = 4 m s? What about at U = 0.5 m s?