Equivalent Resistance Method

Heat Transfer CONDUCTION HEAT TRANSFER Using Equation 2-3: Q k A � � � � � � DT Dx Q A � � � � � � 24 Btu hr ft² (1200 ft²) 28,800^Btu hr Equivalent Resistance Method It is possible to compare heat transfer to current flow in electrical circuits. The heat transfer rate may be considered as a current flow and the combination of thermal conductivity, thickness of material, and area as a resistance to this flow. The temperature difference is the potential or driving function for the heat flow, resulting in the Fourier equation being written in a form similar to Ohm’s Law of Electrical Circuit Theory. If the thermal resistance term Dx/k is written as a resistance term where the resistance is the reciprocal of the thermal conductivity divided by the thickness of the material, the result is the conduction equation being analogous to electrical systems or networks. The electrical analogy may be used to solve complex problems involving both series and parallel thermal resistances. The student is referred to Figure 2, showing the equivalent resistance circuit. A typical conduction problem in its analogous electrical form is given in the following example, where the "electrical" Fourier equation may be written as follows. = (2-6) Q DT R_th where: = Heat Flux ( /A) (Btu/hr-ft²) Q Q DT = Temperature Difference (^oF) R_th = Thermal Resistance (Dx/k) (hr-ft²-^oF/Btu) Rev. 0 Page 9 HT-02