Subtopic: Steady state heat transfer in conduction, convection and radiation
Question Statement:
If thermal conductivity, mass diffusivity, equimolar mass transfer coefficient based on concentration gradient, density and specific heat capacity of air are 0.03 W m⁻¹ K⁻¹, 2.4 × 10⁻⁵ m² s⁻¹, 0.3 m s⁻¹, 1.0 kg m⁻³ and 1.0 kJ kg⁻¹ K⁻¹ respectively, then convective heat transfer coefficient of air is
Options:
(A) 7.43 W m⁻² K⁻¹
(B) 74.27 W m⁻² K⁻¹
(C) 348.12 W m⁻² K⁻¹
(D) 794.39 W m⁻² K⁻¹
Official Answer Key:
C
Step-by-Step Solution & Explanation
Heat–mass transfer (Chilton–Colburn/Lewis relation) analogy: h = k_c·ρ·c_p·Le^(2/3), where Le = α/D_AB (thermal diffusivity / mass diffusivity). α = k/(ρc_p) = 0.03/(1.0×1000) = 3.0×10⁻⁵ m²/s. Le = α/D_AB = 3.0×10⁻⁵/2.4×10⁻⁵ = 1.25. h = 0.3×1.0×1000×(1.25)^(2/3) = 300×1.160 = ≈348.1 W m⁻²K⁻¹.
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