CONDUCTION HEAT TRANSFER
= rate of heat transfer (Btu/hr)
A = cross-sectional area of heat transfer (ft2)
Dx = thickness of slab (ft)
Dr = thickness of cylindrical wall (ft)
DT = temperature difference (°F)
= thermal conductivity of slab (Btu/ft-hr-°F)
The use of Equations 2-4 and 2-5 in determining the amount of heat transferred by conduction
is demonstrated in the following examples.
1000 Btu/hr is conducted through a section of insulating material shown in Figure 1 that
measures 1 ft2 in cross-sectional area. The thickness is 1 in. and the thermal conductivity
is 0.12 Btu/hr-ft-°F. Compute the temperature difference across the material.
Figure 1 Conduction Through a Slab