PROBLEM 4.28 KNOWN:Dimensions and surface temperatures of a square channel. Number of chips mounted on outer surface and chip thermal contact resistance. FIND:Heat dissipation per chip and chip temperature. SCHEMATIC:ASSUMPTIONS:(1) Steady state, (2) Approximately uniform channel inner and outer surface temperatures, (3) Two-dimensional conduction through channel wall (negligible end-wall effects), (4) Constant thermal conductivity. ANALYSIS:The total heat rate is determined by the two-dimensional conduction resistance of the channel wall, q = (T2– T1)/Rt,cond(2D), with the resistance determined by using Equation 4.21 with Case 1 of Table 4.1. For W/w = 1.6 > 1.4 1()t,cond(2D)0.930 ln W / w0.0500.387R0.00160 K / W2L k20.160m 240 W / m Kππ−===⋅The heat rate per chip is then ( )21ct,cond 2D50 20 CTTq156.3 WN R120 0.0016 K / W−°−=<and, with q
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This note was uploaded on 12/07/2010 for the course MAE Heat Trans taught by Professor Lee,j.s. during the Spring '10 term at Seoul National.