# HW3Soln - Problem 17.4 Assume a unit depth of 1m Length of...

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Problem 17.4 Assume a unit depth of 1m Length of cell includes 0.3m of Insulation 0.06m of wood Total length = 0.36m Resistance calculation through each material () ( ) W K m m R plaster 068 . 0 36 . 0 1 mK W 0.814 0.02m kA x = = = ( ) W K m m R wood 7 . 16 06 . 0 1 mK W 0.15 0.15m kA x = = = ( ) W K m m R insulation 3 . 14 3 . 0 1 mK W 0.035 0.15m kA x = = = W K .7 7 W K 3 . 4 1 W K 6.7 1 W K 3 . 14 W K 6.7 1 = + = + = insulation wood insulation wood eq R R R R R W K m m K A h R in in 28 . 0 36 . 0 1 m W 10 1 1 2 = = = W K m m K A h R out out 14 . 0 36 . 0 1 m W 20 1 1 2 = = = W W K W K W K W K C C R q 3 . 4 28 . 0 14 . 0 7 . 7 068 . 0 10 - 5 2 T T out in = + + + = = in plaster R R q + = inter in T T C W K W K W C T er 5 . 23 28 . 0 068 . 0 3 . 4 5 2 int = + = eq in plaster R R R q + + = out wall in T T C W K W K W K W 5 . 9 7 . 7 28 . 0 068 . 0 3 . 4 25C T out wall = + + = wood wood R q out wall inter T T = ( ) W W K C C q wood 0 . 2 7 . 16 .5 9 - .5 3 2 = = ( ) W W K C C q insulation 3 . 2 3 . 14 .5 9 - .5 3 2 = = Problem 17.9 Apply heat to different sides of plastic/aluminum layered problem Resistance calculation through each material Assume 1m 2 of area W K m R plastic 0103 . 0 . 0

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## This note was uploaded on 09/07/2009 for the course CBME Kinetics & taught by Professor Lobban during the Spring '09 term at The University of Oklahoma.

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HW3Soln - Problem 17.4 Assume a unit depth of 1m Length of...

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