Question

Heat flows through a composite slab, across a temperature difference $\Delta T$ as shown below. The depth of the slab is $1\text{m}$. The conductivity ($K$) values are in $\text{W/mK}$. The overall thermal resistance in $\text{K/W}$ is
(rounded to the nearest integer)

Answer 1

$R_1=\frac{l_1}{k_1{A}_1},R_2=\frac{l_2}{k_2{A}_2},R_3=\frac{l_3}{k_3{A}_3}$
$R_1=\frac{0.5}{0.02\times (1\times 1)}=25\text{K/W}$
$R_2=\frac{0.25}{0.1\times (0.5\times 1)}=5\text{K/W}$
$R_3=\frac{0.25}{0.15\times (0.5\times 1)}=3.33\text{K/W}$
So, for $R_2\&R_3$,
$\frac{1}{R_{e{q}_1}}=\frac{1}{R_2}+\frac{1}{R_3}$
$\Rightarrow R_{e{q}_1}=\frac{R_2{R}_3}{R_2+R_3}=\frac{5\times 3.33}{8.33}=2$
Now for $R_1\&R_{e{q}_1}$,
$R_{eq}=R_1+R_{e{q}_1}$
$=25+2=27\text{K/W}$