solution1 - KNOWN Dimensions and thermal conductivity of a...

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Unformatted text preview: KNOWN: Dimensions and thermal conductivity of a chip. Power dissipated on one surface. FIND: Temperature drop across the chip. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform heat dissipation, (4) Negligible heat loss from back and sides, (5) One-dimensional conduction in chip. ANALYSIS: All of the electrical power dissipated at the back surface of the chip is transferred by conduction through the chip. Hence, from Fourier’s law, T P = q = kA AT 01' AT: t-P 2 0.001mX4W 2 kW 150W/n1-K(0.005n1) AT = 1.1" c. < COMMENTS: For fixed P, the temperature drop across the chip decreases with increasing k and W, as well as with decreasing t. 2. KNOWN: Chip width and maximum allowable temperature. Coolant conditions. FIND: Maximum allowable chip power for air and liquid coolants. SCHEMATIC: —|> 7;, =150C —[> —|> A Er, h =200 W/m‘ 'K Dielecfnc fluid, In 3000 W/m3°/< ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from sides and bottom, (3) Chip is at a uniform temperature (isothermal), (4) Negligible heat transfer by radiation in air. ChlP) me = 85°C ANALYSIS: All of the electrical power dissipated in the chip is transferred by convection to the coolant. Hence, P=q and from Newton’s law of cooling, P = hA(T — Tm) = h W2(T — T0,). In air, Pmax = 200 mez-K(0.005 m)2(85 — 15) ° c = 0.35 w. < In the diefectric liquid Pm,x = 3000 Wlmz-K(0.005 m)2(85-15) ° c = 5.25 w. < COMMENTS: Relative to liquids, air is a poor heat transfer fluid. Hence, in air the chip can dissipate far less energy than in the dielectric liquid ...
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