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Page Title: Electrical Analogy
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Equivalent Resistance Method
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Thermodynamics Heat Transfer and Fluid Flow Volume 2 of 3
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Conduction-Cylindrical Coordinates

CONDUCTION HEAT TRANSFER Heat Transfer Electrical Analogy Figure 2    Equivalent Resistance Example: A composite protective wall is formed of a 1 in. copper plate, a 1/8 in. layer of asbestos, and  a  2  in.  layer  of  fiberglass.   The  thermal  conductivities  of  the  materials  in  units  of Btu/hr-ft-oF  are  as  follows:  kCu   =  240,  kasb   =  0.048,  and  kfib   =  0.022.  The  overall temperature difference across the wall is 500°F.   Calculate the thermal resistance of each layer of the wall and the heat transfer rate per unit area (heat flux) through the composite structure. Solution: RCu DxCu kCu 1  inæ ç è ö ÷ ø 1  ft 12  in 240 Btu hr   ft   °F 0.000347hr   ft2°F Btu Rasb Dxasb kasb 0.125  inæ ç è ö ÷ ø 1  ft 12  in 0.048 Btu hr   ft   °F 0.2170hr   ft2°F Btu Rfib Dxfib kfib 2  inæ ç è ö ÷ ø 1  ft 12  in 0.022 Btu hr   ft   °F 7.5758hr   ft2°F Btu HT-02 Page 10 Rev. 0

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