Question 30.2: In a humidification apparatus, liquid water flows in a thin ......

The liquid film on the outside of the cylinder represents the source for mass transfer, and the air stream flowing normal to the cylinder represents an infinite sink. The properties of the air stream are evaluated at the film-average temperature of 300 K. The properties of air may be obtained from Appendix I, with \rho=1.1769\;{\mathrm{kg/m}}^{3}\;{\mathrm{and}}\;\nu=1.5689\;\times\;10^{-5}\;{\mathrm{m}}^{2}/s at 300 K and 1 atm. The Reynolds number is

\mathrm{Re}_{D}={\frac{\mathrm{v_{\infty }.D}}{\nu_{\mathrm{air}}}}={\frac{(4.6\,\mathrm{m/s})(0.076\,\mathrm{m})}{1.5689\times10^{-5}\,\mathrm{m/\!/s}}}=22.283

From Appendix J, Table J.1, the diffusion coefficient of water vapor in air at 298 K and 1.0 atm is 2.6\times10^{-5}\,\mathrm{m}^{2}/s, which, corrected for temperature, becomes

D_{A B}=\left(2.60\times10^{-5}\,\mathrm{m}^{2}/s\right)\left(\frac{300\,\mathrm{K}}{298\,\mathrm{K}}\right)^{3/2}=2.63\times10^{-5}\,\mathrm{m}^{2}/s

The Schmidt number is

\mathrm{Sc}={\frac{\nu_{\mathrm{air}}}{D_{A B}}}={\frac{1.5689\times10^{-5}~\mathrm{m^{2}/s}}{2.63\times10^{-5}~\mathrm{m^{2}/s}}}=0.60

The superficial molar velocity of the air normal to the cylinder is

G_{M}=\frac{\mathrm{v_{\infty }.}\rho_{\mathrm{air}}}{M_{\mathrm{air}}}=\frac{(4.6\ \mathrm{m/s})(1.1769\,\mathrm{kg/m^{3}})}{(29\,\mathrm{kg/kg\,mole})}= 0.187{\frac{\mathrm{kgmole}}{{\mathrm{m^{2}}{\mathrm{s}}}}}

Upon substitution of the known values into equation (30-16), we can solve for the gas-phase film mass-transfer coefficient:

k_{G}=\frac{G_{M}}{P S c^{0.56}}0.281(\mathrm{Re}_{D})^{-0.4}

or

k_{G}={\frac{0.281(0.187\,{\mathrm{kgmole}}/m^{2}\cdot{\mathrm{s}})(22.283)^{-0.4}}{\left(1.013\times10^{5}\,{\mathrm{Pa}}\right)(0.60)^{0.56}}} = 26\times10^{-8}{\frac{\mathrm{~kgmole}}{{\mathrm{~m}}^{2}\cdot{\mathrm{s}}\cdot{\mathrm{Pa}}}}

The flux of water can be evaluated by

N_{A}=k_{G}{\bigl(}p_{A s}-p_{A,\infty }{\bigr)}

The vapor pressure of water at 290 K is 1.73\times10^{3}\,\mathrm{Pa}\,(p_{A s}=P_{A}), and the partial pressure of the dry air (p_{A∞}) is zero, as the surrounding air stream is assumed to be an infinite sink for mass transfer. Consequently,

N_{A}=\left(1.26\times10^{-8}{\frac{{\mathrm{kgmole}}}{{\mathrm{m}}^{2}\cdot{\mathrm{s}}\cdot{\mathrm{Pa}}}}\right)\left(1.73\times10^{3}\,{\mathrm{Pa}}-0\right)=2.18\times10^{-5}{\frac{{\mathrm{kgmole}}}{{\mathrm{m}}^{2}\cdot{\mathrm{s}}}}

Finally, the mass-feed rate of water for a single cylinder is the product of the flux, and the external surface area of the cylinder is

W_{A}=N_{A}M_{A}(\pi D L)\,=\,\left(2.18\times10^{-5}{\frac{{\mathrm{kgmole}}}{{\mathrm{m^{2}}}\cdot{\mathrm{s}}}}\right)\left({\frac{18\,\mathrm{kg}}{{\mathrm{kgmole}}}}\right)(\pi)(0.076\,\mathrm{m})(1.22\,\mathrm{m})

=\;1.14\times10^{-4}\,\mathrm{kg/s}