10A.6 Insulating power of a wall. The "insulating power" of a wall can be measured by means of the arrangement shown in Fig. 10A.6. One places a plastic panel against the wall. In the panel two thermocouples are mounted flush with the panel surfaces. The thermal conductivity and thickness of the plastic panel are known. From the measured steady-state temperatures shown in the figure, calculate: T3=0°F Wall T2 61°F T₁ = 69°F Surface temperatures of plastic panel Plastic panel has thermal conductivity k = 0.075 Btu/hr. ft. °F (average value between T₁ and T₂) -0.502" Fig. 10A.6 Determination of the thermal resistance of a wall. Problems 315 (a) The steady-state heat flux through the wall (and panel). (b) The "thermal resistance" (wall thickness divided by thermal conductivity). Answers: (a) 14.3 Btu/hr ft2; (b) 4.2 ft² hr "F/Btu