Insulating power of a wall (Fig. 10A.6). The \"insulating power\" of a wall can

Question

Insulating power of a wall (Fig. 10A.6). The "insulating power" of a wall can be measured by means of the arrangement shown in the figure. One places a plastic panel against the wall. In the panel two thermocouples are mounted flush with the panel surfaces. The thermal con ductivity and thickness of the plastic panel are known. From the measured steady-state tem peratures shown in the figure, calculate: The steady-state heat flux through the wall (and panel). The "thermal resistance" (wall thickness divided by thermal conductivity).

Explanation / Answer

(a) Heat flux = Heat current /Area = k dT/L

= [ (0.075 Btu.hr-1.ft-1.oF-1) * (69 - 61 )oF ] / [(0.052/12)]ft

= 14.3 Btu/ft2.hr

.

(b) In the steady state flow of heat through wall and panel will be same.

[dTwall / (Rwall/A)] = [dTpanel / (Rpanel/A)]                     ...R = resistance, A = Area of surface

or, [dTwall / (Rwall)] = [dTpanel / (Rpanel)]

or, Rwall = (dTwall / dTpanel)* Rpanel

= (dTwall / dTpanel)* (Lpanel/k)                          ...As R = L/k

= (61/8) * (0.502/12)ft * [ 1/(0.075 Btu.hr-1.ft-1.oF-1) ]

= 4.25 ft2.hr.oF/Btu

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