# 25x105 rf206 b samerf cthemaximumheatrateqc16qfqbqi

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online