Question

A material $\text{P}$ of thickness $1\text{mm}$ is sandwiched between two steel slabs, as shown in the figure below. A heat flux $10{\text{kW/m}}^2$ is supplied to one of the steel slabs as shown. The boundary temperatures of the slabs are indicated in the figure. Assume thermal conductivity of this steel as $10\text{W/mK}$. Considering one-dimensional steady state heat conduction for the configuration, the thermal conductivity $k\left(\text{in W/mK}\right)$ of material $\text{P}$ is

• A. $0.1\text{W/mK}$
• B. $0.2\text{W/mK}$
• C. $0.3\text{W/mK}$
• D. $0.4\text{W/mK}$

Answer 1

The correct option is A $0.1\text{W/mK}$

From the data given in the figure:

$x=1\text{mm}=0.001\text{m}$

$q^{\prime \prime }=10{\text{kW/m}}^2=10000{\text{W/m}}^2$

$K_s=10\text{W/mK}$

$\delta =20\text{mm}=0.02\text{m}$

$T_1=500K$

$T_2=360K$


We know that heat flow through the combinations of the slab:

$Q=\frac{T_1-T_2}{\frac{\delta }{k_{s}A}+\frac{x}{kA}+\frac{\delta }{k_{s}A}}$

$\Rightarrow Q=\frac{T_1-T_2}{\frac{2\delta }{k_{s}A}+\frac{x}{kA}}$

$\Rightarrow \frac{Q}{A}=\frac{T_1-T_2}{\frac{2\delta }{k_s}+\frac{x}{k}}$

Substituting $\frac{Q}{A}=q^{\prime \prime }$, which is a heat flux passing through the front surface in unit time,

$q^{\prime \prime }=\frac{T_1-T_2}{\frac{2\delta }{k_s}+\frac{x}{k}}$

$\Rightarrow q^{\prime \prime }(\frac{2\delta }{k_s}+\frac{x}{k})=T_1-T_2$

Substituting the values from figure we get

$10000(\frac{2\times 0.02}{10}+\frac{0.001}{k})=500-360$

$\Rightarrow 40+\frac{10}{k}=140$

$\Rightarrow \frac{10}{k}=140-40=100$

$\Rightarrow k=\frac{10}{100}$

$\Rightarrow k=0.1\text{W/mK}$

Hence, Option (a) is the correct answer.