Thermodynamics HW Solutions 167

# Thermodynamics HW Solutions 167 - a unit surface area A = 1...

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Chapter 3 Steady Heat Conduction 3-28E A wall is constructed of two layers of sheetrock with fiberglass insulation in between. The thermal resistance of the wall and its R-value of insulation are to be determined. Assumptions 1 Heat transfer through the wall is one-dimensional. 2 Thermal conductivities are constant. Properties The thermal conductivities are given to be k sheetrock = 0.10 Btu/h ft °F and k insulation = 0.020 Btu/h ft °F. R 1 R 2 R 3 L 1 L 2 L 3 Analysis ( a ) The surface area of the wall is not given and thus we consider
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Unformatted text preview: a unit surface area ( A = 1 ft 2 ). Then the R-value of insulation of the wall becomes equivalent to its thermal resistance, which is determined from. F.h/Btu . ft 21.66 2 ° = + × = + = ° = ° = = = ° = ° = = = = 83 . 20 417 . 2 2 F.h/Btu . ft 83 . 20 F) Btu/h.ft. 020 . ( ft 12 / 5 F.h/Btu . ft 417 . F) Btu/h.ft. 10 . ( ft 12 / 5 . 2 1 2 2 2 2 2 1 1 3 1 R R R k L R R k L R R R total fiberglass sheetrock ( b ) Therefore, this is approximately a R-22 wall in English units. 3-10...
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