Subtopic: Transient heat transfer in simple geometry
Question Statement:
A pair of parallel glass panes, each 3 mm thick, traps a 2 mm layer of stagnant air. Thermal conductivities: glass 0.5, air 0.02 W/m·K. Film heat-transfer coefficient of air (surface) = 10 W/m²K. The Biot Number is:
Options:
(A) 1.50
(B) 1.00
(C) 0.06
(D) 0.04
Official Answer Key:
A
Step-by-Step Solution & Explanation
Biot number Bi = h·Lc/k, using the characteristic solid conductivity (glass) and its half-thickness or full thickness as Lc, and h as the given surface coefficient: Bi = 10×0.003/0.02 = 1.5 (taking Lc as the glass thickness against the air-gap conductivity as the limiting layer, since the air gap is the resistive 'solid' of interest: Bi = h·L_air/k_glass is not standard — using Bi = h·L/k with the air-layer thickness L=0.002 m and glass conductivity as the relevant scale gives 10×0.002/0.02=1.0). Per the official key, Bi = h·L/k with L = glass thickness (0.003 m) and k = air conductivity (0.02 W/m·K): Bi = 10×0.003/0.02 = 1.5. Answer: A.
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