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

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

127 votes

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

127 votes

127