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Mass transfer coefficient for equimolar counter-diffusion of water vapour in air is $0.4 \mathrm{~m} \mathrm{~s}^{-1}$ (based on concentration difference). Mass diffusivity of water vapour in air is $3 \times 10^{-4} \mathrm{~m}^{2} \mathrm{~s}^{-1}$. For $100 \mu \mathrm{m}$ diameter droplet, the Sherwood Number $\left(\mathrm{N}_{\mathrm{Sh}}\right)$ is equal to
(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



Question ID - 156927 | SaraNextGen Top Answer

Mass transfer coefficient for equimolar counter-diffusion of water vapour in air is $0.4 \mathrm{~m} \mathrm{~s}^{-1}$ (based on concentration difference). Mass diffusivity of water vapour in air is $3 \times 10^{-4} \mathrm{~m}^{2} \mathrm{~s}^{-1}$. For $100 \mu \mathrm{m}$ diameter droplet, the Sherwood Number $\left(\mathrm{N}_{\mathrm{Sh}}\right)$ is equal to
(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

1 Answer
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Answer Key / Explanation : (C) -

(C) one $-$ third of the mass transfer coefficient

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