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Mechanical EngineeringQ&A Library3-28E A wall is constructed of two layers of 0.6-in-thick sheetrock (k = 0.10 Btu/h-ft-°F), which is a plasterboard made of two layers of heavy paper separated by a layer of gypsum, placed 7 in apart. Thespace between the sheetrocks is filled with fiberglass insulation (k = 0.020 Btu/h-ft F). Determine (a) the thermal resistance of the wall and (b) its R-value of insulation in English units.FIGURE P3-28EFiberglassinsulationSheetrock0.6 in7 inI 0.6 inQuestion

Asked Sep 20, 2019

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Step 1

Thickness of fiber glass insulation, L = 7 in.

Thermal conductivity of sheetrock K_{s} = 0.10 Btu/h. ft. °F

Thermal conductivity of insulation K_{i} = 0.020 Btu/h. ft. °F

Thickness of sheet rock L_{1} = 0.6 in.

Step 2

Part (a): The thermal resistance of the wall:

According to the diagram the thermal resistance of the wall is calculated as:

Here, L_{1} and L_{3} is the thickness of the sheet rock and L_{2} is the thickness of the fiber glass. And, k_{1} and k_{3} is the thermal conductivity of the sheet rock and k_{2} is the thermal conductivity of fiber glass.

Step 3

Thus, the total resistance of the wall is 30...

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