Q. A house wall may be approximated as two 1.2 cm layers of fiber insulating board, an 8.0 cm layer of loosely packed asbestos, and a 10 cm layer of common brick, the high and width of the wall are 3 and 5m respectively. The wall contain a window with single glass plate 5 mm thick, the high and width of the window are 1 and 1.5m respectively. If the inside and outside of wall temperature are 15 and 45 C° respectively, calculate the heat-transfer for this arrangement. Note: for fiberglass k = 0.038 W/m.Co, asbestos k = 0.154 W/m.Co, brick k-0.69 W/m.Co, glass k=0.78 W/m.Co

Q. A house wall may be approximated as two 1.2 cm layers of fiber insulating board, an 8.0 cm layer of loosely packed asbestos, and a 10 cm layer of common brick, the high and width of the wall are 3 and 5m respectively. The wall contain a window with single glass plate 5 mm thick, the high and width of the window are 1 and 1.5m respectively. If the inside and outside of wall temperature are 15 and 45 C° respectively, calculate the heat-transfer for this arrangement. Note: for fiberglass k = 0.038 W/m.Co, asbestos k = 0.154 W/m.Co, brick k-0.69 W/m.Co, glass k=0.78 W/m.Co

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Q. A house wall may be approximated as two 1.2 cm layers of fiber
insulating board, an 8.0 cm layer of loosely packed asbestos, and a 10 cm
layer of common brick, the high and width of the wall are 3 and 5m
respectively. The wall contain a window with single glass plate 5 mm thick,
the high and width of the window are 1 and 1.5m respectively. If the inside
and outside of wall temperature are 15 and 45 C° respectively, calculate the
heat-transfer for this arrangement. Note: for fiberglass k = 0.038 W/m.Cº,
asbestos k = 0.154 W/m.Co, brick k-0.69 W/m.Co, glass k-0.78 W/m.Co

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I have a problem similar to this in my case the problem is from the outside in. the heat is entering the interior and I don't know what temperature it will have inside since part of the heat will be lost in the thickness of the cylinder, I want to know how to fix the formula so that I can obtain the temperature inside the cylinder.
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