Heat Transfer
CONDUCTION HEAT TRANSFER
Using Equation 2-3:
Q
k A
DT
Dx
Q A
24
Btu
hr ft2
(1200 ft2)
28,800Btu
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
Rth
where:
= Heat Flux ( /A) (Btu/hr-ft2)
Q
Q
DT = Temperature Difference (oF)
Rth
= Thermal Resistance (Dx/k) (hr-ft2-oF/Btu)
Rev. 0
Page 9
HT-02
