Discussion
A pair of parallel glass panes, each of 3 mm thickness traps 2 mm layer of stagnant air. Thermal conductivities of glass and air are 0.5 and 0.02 W m−1 K−1, respectively. If the film heat transfer coefficient of air is 10 W m−2 K−1, then Biot Number is
Answer: C
The Biot number is 0.06.
The Biot number compares conduction resistance inside a solid with convection resistance at its surface: \(Bi = hL/k\), where k is the conductivity of the solid. Here the solid is a glass pane, 3 mm thick, with the air film coefficient on its surface:
\[ \begin{aligned} Bi &= \frac{hL}{k_{glass}} \\ &= \frac{10 \times 0.003}{0.5} = 0.06 \end{aligned} \]
Using the conductivity of air in place of glass gives 1.50 with the 3 mm pane, or 1.00 with the 2 mm air layer. Using the 2 mm air gap with the glass conductivity gives 0.04. None of these is a Biot number for the pane.
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