Question 30.3: 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 $300 \mathrm{~K}$. The properties of air may be obtained from Appendix I, with $\rho=1.1769 \mathrm{~kg} / \mathrm{m}^{3}$ and $v=1.5689 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}$ at $300 \mathrm{~K}$ and $1 \mathrm{~atm}$. The Reynolds number is

$\operatorname{Re}_{D}=\frac{D v_{\infty}}{v_{\text {air }}}=\frac{(0.076 \mathrm{~m})(4.6 \mathrm{~m} / \mathrm{s})}{1.5689 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}}=22283$

From Appendix Table J.1, the diffusivity of water in air at $298 \mathrm{~K}$ and $1 \mathrm{~atm}$ is $2.60 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}$, which corrected for temperature becomes

$D_{A B}=\left(2.60 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\left(\frac{300 \mathrm{~K}}{298 \mathrm{~K}}\right)^{3 / 2}=2.63 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}$

The Schmidt number is

$\mathrm{Sc}=\frac{\nu_{\text {air }}}{D_{A B}}=\frac{1.5689 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}}{2.63 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}}=0.6$

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

$G_{M}=\frac{v_{\infty} \rho_{\text {air }}}{M_{\text {air }}}=\frac{(4.6 \mathrm{~m} / \mathrm{s})\left(1.1769 \mathrm{~kg} / \mathrm{m}^{3}\right)}{29 \mathrm{~kg} / \mathrm{kg} \mathrm{mol}}=0.187 \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s}}$

Upon substitution of the known values into equation (30-16),

$\frac{k_{G}P(S c)^{0.56}}{G_{M}}=\frac{k_{c}(S c)^{0.56}}{v_{\infty}}=0.281(\mathrm{Re}_{D})^{-0.4}$        (30-16)

we can solve for the gas-phase film mass-transfer coefficient

$\frac{k_{G} P \mathrm{Sc}^{0.56}}{G_{M}}=0.281\left(\operatorname{Re}_{D}\right)^{-0.4}$

or

$\frac{k_{G}\left(1.013 \times 10^{5} \mathrm{~Pa}\right)(0.60)^{0.56}}{0.187 \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s}}}=\frac{0.281}{(22283)^{0.4}}$

Finally

$k_{G}=1.26 \times 10^{-8} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{Pa}}$

The flux of water can be evaluated by

$N_{A}=k_{G}\left(p_{A, i}-p_{A \infty}\right)$

The vapor pressure of water at $290 \mathrm{~K}$ is $1.73 \times 10^{3} \mathrm{~Pa}$, and the partial pressure of the dry air $\left(p_{A \infty}\right)$ is zero, as the surrounding air stream is assumed to be an infinite sink for mass transfer. Consequently,

$N_{A}=\left(1.26 \times 10^{-8} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{Pa}}\right)\left(1.73 \times 10^{3} \mathrm{~Pa}-0\right)=2.18 \times 10^{-5} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s}}$

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

$\begin{aligned} W_{A}=N_{A} M_{A}(\pi D L) & =\left(2.18 \times 10^{-5} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s}}\right)\left(18 \frac{\mathrm{kg}}{\mathrm{kg} \mathrm{mol}}\right)(\pi)(0.076 \mathrm{~m} \cdot 1.22 \mathrm{~m}) \\ & =1.14 \times 10^{-4} \mathrm{~kg} / \mathrm{s} \end{aligned}$