Project ForgeA brick wall of 20 cm thickness has thermal conductivity of 0.7 W m⁻¹ K⁻¹. An insulation of thermal conductivity 0.2 W m⁻¹ K⁻¹ is to be applied on one side of the wall, so that the heat transfer through the wall is reduced by 75%. The same temperature difference is maintained across the wall before and after applying the insulation. The required thickness (in cm) of the insulation is _____
Explanation
Original resistance (per unit area) R1=L/k=0.20/0.7=0.2857 m²K/W. A 75% reduction in Q (same ΔT) means the new resistance must be 4× the original: R_new=4×0.2857=1.1429. Insulation resistance=R_new−R1=1.1429−0.2857=0.8571=L_ins/0.2, so L_ins=0.8571×0.2=0.1714 m=17.14 cm.
Source: GATE CH 2014 (2014)
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