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Unformatted text preview: rate? R f = θ b /16 q f Look on p. 144 to find q f for Case A. (The θ b ’s cancel out). Same R f . The maximum heat rate q c = 16q f + q b + q i 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 also by convection: We use A c =( 0.0127 m) 2 because we’re looking at the bottom of the board. q i = 0.29 W q c = 16(2.703) + 7.32 + 0.03 = 50.9 W W/out the fins, q c = 1000 W/m 2-K (0.0127 m) 2 55 o C + 0.29 = 9.16 W. 50.9/9/16 = 555 % enhancement. When we add a contact resistance, the q i changes insignificantly: = 0.29 W as well. Same heat transfer rate. The infinite fin assumption is only for L > 2.65/m. In this case, = .032 won’t work. If we use it, the actual heat transfer rate will be overestimated since q f = M = 3.17W and our q f = 2.7 W....
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This note was uploaded on 08/13/2011 for the course CHEM 3210 taught by Professor Degrazia during the Fall '10 term at Colorado.
- Fall '10