Heat Transfer CONDUCTION HEAT TRANSFERUsing Equation 2-3:Qk ADTDxQ A24Btuhr ft2(1200 ft2)28,800BtuhrEquivalentResistanceMethodIt is possible to compare heat transfer to current flow in electrical circuits. The heat transfer ratemay be considered as a current flow and the combination of thermal conductivity, thickness ofmaterial, and area as a resistance to this flow. The temperature difference is the potential ordriving function for the heat flow, resulting in the Fourier equation being written in a formsimilar to Ohm’s Law of Electrical Circuit Theory. If the thermal resistance term Dx/k is writtenas a resistance term where the resistance is the reciprocal of the thermal conductivity divided bythe thickness of the material, the result is the conduction equation being analogous to electricalsystems or networks. The electrical analogy may be used to solve complex problems involvingboth series and parallel thermal resistances. The student is referred to Figure 2, showing theequivalent resistance circuit. A typical conduction problem in its analogous electrical form isgiven in the following example, where the "electrical" Fourier equation may be written asfollows.= (2-6)QDTRthwhere:= Heat Flux ( /A) (Btu/hr-ft2)QQDT = Temperature Difference (oF)Rth= Thermal Resistance (Dx/k) (hr-ft2-oF/Btu)Rev. 0 Page 9 HT-02
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