Question

Heat flows through a composite slab, as shown. The depth of the slab is $1m$. The $k$ values are in $W/mK$, then the overall thermal resistance in $K/W$ is

• A. $17$
• B. $21.9$
• C. $28.6$
• D. $39.2$

Answer 1

Answer: (C)

$28.6$

Given: Heat flows through a composite slab, as shown. The depth of the slab is 1 m. The k values are in $W/mK$,

To find the overall thermal resistance in $K/W$

Solution:

There is no variation in the horizontal direction. Therefore, we consider a 1 m

deep and 1 m high portion of the slab, since it representative of the entire wall. Assuming any cross-section of the slab normal to the x - direction to be isothermal, the thermal resistance network for the slab is shown in the figure.

The corresponding thermal resistance network is as shown in above fig(ii).

The individual thermal resistance is, (substitute values from above fig)

$R_1=\frac{L_1}{K_1{A}_1}⟹R_1=\frac{0.5}{0.02(1\times 1)}⟹R_1=25$

Similarly we will calculate the other two thermal resistance value

$R_2=\frac{L_2}{K_2{A}_2}⟹R_2=\frac{0.25}{0.10(0.5\times 1)}⟹R_2=5$

$R_3=\frac{L_3}{K_3{A}_3}⟹R_3=\frac{0.25}{0.04(0.5\times 1)}⟹R_3=12.5$

Now $R_2$ and $R_3$ are in parallel.

So the equivalent resistance becomes.

$R_{eq}=R_1+\frac{R_2{R}_3}{R_2+R_3}$

by substituting the values, we get

$R_{eq}=25+\frac{5\times 12.5}{5+12.5}⟹R_{eq}=25+3.57=28.6K/W$