ch18-p059

ch18-p059 - 59. (a) We use Pcond = kA TH TC L with the...

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where the sum in the denominator is over the layers. If L g is the thickness of a glass layer, L a is the thickness of the air layer, k g is the thermal conductivity of glass, and k a is the thermal conductivity of air, then the denominator is 22 . g ga ag a ga a g LL k L k L L kk k k k + =+ = ¦ Therefore, the heat conducted per unit area occurs at the following rate: () () () () () () ( ) () cond 3 2 51.1 C 0.026W m K 1.0W m K 2 2 3.0 10 m 0.026W m K 0.075m 1.0W m K 18W m . HC a g ga ag TT k k P AL k L k °⋅ == + ×⋅ +⋅ = 59. (a) We use cond H C TT Pk A L = with the conductivity of glass given in Table 18-6 as 1.0 W/m·K. We choose to use the Celsius scale for the temperature: a temperature difference of () 72 F 20 F 92 F HC TT −=° ° = ° is equivalent to 5 9 (92) 51.1C
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