ch18-p058 - 58. (a) As in Sample Problem 18-6, we take the...

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to the rate across layer 2 ( P 2 ). Using Eq. 18-37 and canceling out the common factor of area A , we obtain T H - T c ( L 1 / k 1 + L 2 / k 2 + L 3 / k 3 ) = Δ T 2 ( L 2 / k 2 ) ¡ 45 C° (1 + 7/9 + 35/80) = Δ T 2 (7/9) which leads to Δ T 2 = 15.8 °C. (b) We expect (and this is supported by the result in the next part) that greater conductivity should mean a larger rate of conductive heat transfer. (c) Repeating the calculation above with the new value for k 2 , we have 45 C° (1 + 7/11 + 35/80) = Δ T 2 (7/11) which leads to Δ T 2 = 13.8 °C. This is less than our part (a) result which implies that the
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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