Question

A wall of 0.5m thickness is to be constructed from a material that has an average thermal conductivity of 1.4 W/mK. The wall is to be insulated with a material having an average thermal conductivity of 0.35 W/mK so that heat loss per square meter will not exceed 1450 W. Assuming that the inner and outer wall surface temperatures are 1200°C and 15°C respectively, the thickness of insulation required is

• A. 0.20 m
• B. 0.08 m
• C. 0.16 m
• D. 0.12 m

Answer 1

The correct option is C 0.16 m

Given, $L_1=0.5m,$

$k_1=1.4W/mK,$ $k_2=0.35W/mK,$

$Q=1450W$

$T_1={1200}^{\circ }C$, $T_2={15}^{\circ }C$


For heat transfer by conduction through the wall and insulation, we have

$Q=\frac{T_1-T_2}{\frac{L_1}{k_{1}A}+\frac{L_2}{k_{2}A}}$

$\frac{Q}{A}=\frac{1200-15}{\frac{0.5}{1.4}+\frac{L_2}{0.35}}=\frac{1185}{\frac{0.5}{1.4}+\frac{L_2}{0.35}}$

$\Rightarrow L_2=0.161m$