# A (2 mathrm{~cm})-diameter, (19 mathrm{~cm})-long tube is placed touching a pool of liquid. The end away from

## Question:

A \(2 \mathrm{~cm}\)-diameter, \(19 \mathrm{~cm}\)-long tube is placed touching a pool of liquid. The end away from the liquid pool \((\mathrm{z}=0.19 \mathrm{~m})\) is in an air stream (component C) so that it is pure air, \(\mathrm{y}_{\mathrm{C}}(\mathrm{z}=0.19 \mathrm{~m})=1.0\). The liquid is \(80 \mathrm{~mol} \%\) component \(\mathrm{A}\) and \(20 \mathrm{~mol} \%\) component \(\mathrm{B}\). The ratio of these two components in the gas at \(\mathrm{z}=0\) is the same as in the liquid, \(\mathrm{y}_{\mathrm{A}, \mathrm{i}} / \mathrm{y}_{\mathrm{B}, \mathrm{i}}=4\). The total vapor pressure of the liquid is \(13,005 \mathrm{~Pa}\). Thus \(13,005 / \mathrm{P}_{\text {total }}=\left(\mathrm{y}_{\mathrm{A}, \mathrm{i}}+\mathrm{y}_{\mathrm{B}, i}\right)\). Total pressure is \(107,404 \mathrm{~Pa}\), and temperature is \(40^{\circ} \mathrm{C}\). Assume ideal gas. At \(40^{\circ} \mathrm{C}\) and 107,404 Pa, \(D_{\mathrm{AB}}=2.45 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}(\mathrm{C}=\) air \(), D_{\mathrm{BC}}=\) \(2.69 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), and \(D_{\mathrm{AC}}=1.01 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s} . \mathrm{MW} \mathrm{B}=18, \mathrm{MW}\) air \(=28.9, \mathrm{MW} \mathrm{A}=\) 60.1. A partial solution is \(\mathrm{N}_{\mathrm{A}}=0.000228074 \mathrm{~mol} /\left(\mathrm{m}^{2} \mathrm{~s}\right)\). What are the molar fluxes, \(\mathrm{mol} /\left(\mathrm{s}^{2}\right)\), of B and C (air) at steady state?

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**Related Book For**

## Separation Process Engineering Includes Mass Transfer Analysis

**ISBN:** 9780137468041

5th Edition

**Authors:** Phillip Wankat