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20) Me Steady-state temperature distribution in the s figure. The heat flow is one-dimensional. ibution in the sandwich of th
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Fourier's law of steady state 1D conduction is used to find the solutions.

90 30 A130 0.03 0.08 0.1 a = 64000 W/m² KA - 3.6w/mk a at heat By flux at left side w Fouriers law a = - KADT ax -- K dT. da

5) Net heat flux from Bi aur = Å LB Our = 64000 (0.08 -0.03) = 3200 w/m2 (right side) = whe So, heat flux from c a = VB - WA

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