Question
Asked Sep 20, 2019
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3-28E 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. The
space 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 P3-28E
Fiberglass
insulation
Sheetrock
0.6 in
7 in
I 0.6 in
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3-28E 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. The space 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 P3-28E Fiberglass insulation Sheetrock 0.6 in 7 in I 0.6 in

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Expert Answer

Step 1

Thickness of fiber glass insulation, L = 7 in.

Thermal conductivity of sheetrock Ks = 0.10 Btu/h. ft. °F

Thermal conductivity of insulation Ki = 0.020 Btu/h. ft. °F

Thickness of sheet rock L1 = 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, L1 and L3 is the thickness of the sheet rock and L2 is the thickness of the fiber glass. And, k1 and k3 is the thermal conductivity of the sheet rock and k2 is the thermal conductivity of fiber glass.

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R2 RI WMWMWML Sheet rock Sheet rock Fiberglass RwRRR or, walf + walf

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Step 3

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

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0.6 0.6 0.10 0.10 0.020 0.6/12 7/120.6/12 R 0.10 0.020 0.10 R 0.5+29.1667+0.5 R30.1667 ft'-h-'F/Btu

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