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The steady-state mass transfer coefficient $\left(\mathrm{k}_{\mathrm{g}}\right)$ based on water vapour pressure differential (VPD) operating across stagnant, non-diffusing air was estimated to be $0.05 \mathrm{~g}$ mole $\mathrm{s}^{-1} \mathrm{~m}^{-2} \mathrm{kPa}^{-1}$. If VPD varies from $12 \mathrm{kPa}$ to $7 \mathrm{kPa}$ over a distance of $2 \mathrm{~mm}$, then the mass transfer coefficient $\left(\mathrm{k}_{\mathrm{y}}^{\prime}\right)$ based on equimolar counter-diffusion in $\mathrm{g}$ mole $\mathrm{s}^{-1} \mathrm{~m}^{-2}$ (mole fraction) $^{-1}$ ( rounded off to one decimal place) is________________



Question ID - 156458 | SaraNextGen Top Answer

The steady-state mass transfer coefficient $\left(\mathrm{k}_{\mathrm{g}}\right)$ based on water vapour pressure differential (VPD) operating across stagnant, non-diffusing air was estimated to be $0.05 \mathrm{~g}$ mole $\mathrm{s}^{-1} \mathrm{~m}^{-2} \mathrm{kPa}^{-1}$. If VPD varies from $12 \mathrm{kPa}$ to $7 \mathrm{kPa}$ over a distance of $2 \mathrm{~mm}$, then the mass transfer coefficient $\left(\mathrm{k}_{\mathrm{y}}^{\prime}\right)$ based on equimolar counter-diffusion in $\mathrm{g}$ mole $\mathrm{s}^{-1} \mathrm{~m}^{-2}$ (mole fraction) $^{-1}$ ( rounded off to one decimal place) is________________

1 Answer
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Answer Key / Explanation : (4.5 to 4.7) -

4.5 to 4.7

127 votes


127