Heat Transfer CONDUCTION HEAT TRANSFERwhere:= rate of heat transfer (Btu/hr)QA = cross-sectional area of heat transfer (ft^{2})Dx = thickness of slab (ft)Dr = thickness of cylindrical wall (ft)DT = temperature difference (°F)k = 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 conductionis demonstrated in the following examples.Conduction-RectangularCoordinatesExample:1000 Btu/hr is conducted through a section of insulating material shown in Figure 1 thatmeasures 1 ft^{2} in cross-sectional area. The thickness is 1 in. and the thermal conductivityis 0.12 Btu/hr-ft-°F. Compute the temperature difference across the material.Figure 1 Conduction Through a SlabRev. 0 Page 7 HT-02