25x105 rf206 b samerf cthemaximumheatrateqc16qfqbqi

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Unformatted text preview: transfer rate? a) Rf = 1/hAfηf ηf = tanh mLc/mLc Lc = L + D/4 = .0154 m = (4h/kD)1/2 = (4 *1000/400*.0015)1/2 = 81.7 mLc = 1.26 tanh mLc = .84 ηf = .67 h = 1000 Af = πDLc = 7.25 X 10‐5 Rf = 20.6 b) Same Rf. c) The maximum heat rate qc = 16qf + qb + qi Now to figure out some of our constants: From p. 144: The heat rate from the chip top is by convection: The heat rate from the board is by conduction and convection: qi = Tc − T∞ ,i 1 (1 / hi + Lb / k b ) Ac We use Ac =( 0.0127 m) 2 because we’re looking at the bottom of the board. qi = 0.29 W qc = 16(2.703) + 7.32 + 0.03 = 50.9 W When we add a contact resistance, the qi changes insignificantly: qi = Tc − T∞ ,i 1 (1 / hi + Lb / k b + R,c ) Ac = 0.29 W as well. Same heat transfer rate. d) W/out the fins, qc = 1000 W/m2‐K (0.0127 m)2 55o C + 0.29 = 9.16 W. 50.9/9/16 = 555 % enhancement. e) The infinite fin assumption is only for L > 2.65/m. In this case, = .032 won’t work. If we us...
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