Mass transfer coefficient for equimolar counter-diffusion of water vapour in air is 0.4 m s⁻¹ (based on concentration difference). Mass diffusivity of water vapour in air is 3×10⁻⁴ m² s⁻¹. For 100 µm diameter droplet, the Sherwood Number (N_Sh) is equal to
Options:
(A) the mass transfer coefficient
(B) diameter divided by mass diffusivity
(C) one-third of the mass transfer coefficient
(D) three times of the mass diffusivity
Official Answer Key:
C
Step-by-Step Solution & Explanation
Sherwood number is defined as N_Sh = k_c×d/D_AB, where k_c is the mass transfer coefficient, d the characteristic (droplet) diameter and D_AB the mass diffusivity. Substituting: N_Sh = 0.4×(100×10⁻⁶)/(3×10⁻⁴) = 0.4×0.0001/0.0003 = 0.4×0.3333 = 0.1333, which numerically equals one-third of the mass transfer coefficient (0.4/3 = 0.1333) → option (C). This matches because d/D_AB itself evaluates to exactly 1/3 s/m for the given d and D_AB, making N_Sh = k_c×(1/3) numerically.
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