Question

Three different type of glasses are used to make a large glass slab. The thermal conductivity of glass slabs are $k_1$, $k_2$ and $k_3$. The width of each glass slab is '$\omega$' and cross-sectional area is $A$. The glass slabs are kept in a pattern as shown in the figure. The upper surface of glass slab is maintained at temperature $T_1$ and lower surface at $T_2$. It is given that $k_1 = 1.10\,\mathrm{W/mK}$, $k_2 = 2.82\,\mathrm{W/mK}$, $k_3 = 2.08\,\mathrm{W/mK}$, $\omega = 10\mathrm{\,cm}$, $A = 1.3\,\mathrm{m^2}$, $T_1 = 100^\circ\mathrm{C}$, $T_2 = 0^\circ\mathrm{C}$. Find equivalent thermal conductivity of large glass slab in W/mK.

Answer 1

$$k_{\text{eq}} = \frac{3}{\frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3}} = \frac{3}{\frac{1}{1.10} + \frac{1}{2.82} + \frac{1}{2.08}} \approx 1.72\ \mathrm{W/mK}$$

Answer 2

Step‑1: Identify series combination
All three slabs have the same thickness $\omega$ and are stacked one above the other, so they conduct heat in series.

Step‑2: Thermal resistance of each slab

$$R_i = \frac{\omega}{k_i A} \quad\Longrightarrow\quad R_{\text{total}} = \sum_{i=1}^3 R_i = \frac{\omega}{A}\Bigl(\frac1{k_1} + \frac1{k_2} + \frac1{k_3}\Bigr)$$

Step‑3: Equivalent conductivity definition
For an equivalent slab of total thickness $3\omega$:

$$R_{\text{total}} = \frac{3\omega}{k_{\text{eq}}\,A}$$

Equate resistances:

$$\frac{3\omega}{k_{\text{eq}}\,A} = \frac{\omega}{A}\Bigl(\frac1{k_1} + \frac1{k_2} + \frac1{k_3}\Bigr) \;\Longrightarrow\; k_{\text{eq}} = \frac{3}{\frac1{k_1} + \frac1{k_2} + \frac1{k_3}}$$

Step‑4: Numerical substitution

$$k_{\text{eq}} = \frac{3}{\frac{1}{1.10} + \frac{1}{2.82} + \frac{1}{2.08}} \approx 1.72\ \mathrm{W/mK}$$