Question 30.2: Estimate the distance a spherical drop of liquid water, orig...

The physical system requires a combined analysis of momentum and mass transport. The liquid water droplet is the source for mass transfer, the surrounding air serves as an infinite sink, and water vapor (species A ) is the transferring species. The rate of evaporation is sufficiently small so that the water droplet is considered isothermal at 293 \mathrm{~K}; otherwise, a combined analysis of momentum, mass, and heat transport would be required! By considering a force balance on a spherical particle falling in a fluid medium, we can show that the terminal velocity of the particle is

v_{o}=\sqrt{\frac{4 d_{p}\left(\rho_{w}-p_{\mathrm{air}}\right) g}{3 C_{D} \rho_{\mathrm{air}}}}

where d_{p} is the diameter of the particle, \rho_{w} is the density of the water droplet, \rho_{\text {air }} is the density of the surrounding fluid (air), g is the acceleration due to gravity, and C_{D} is the drag coefficient, which is a function of the Reynolds number of the spherical particle as illustrated in Figure 12.4. The arithmetic mean droplet diameter is evaluated by

\begin{aligned} \bar{d}_{p} & =\frac{d_{p \mid t_{1}}+d_{p \mid t_{2}}}{2}=\frac{d_{p \mid t_{1}}+\left(\frac{1}{2}\right)^{1 / 3} \cdot d_{p \mid t_{1}}}{2} \\ & =0.897 d_{p \mid t_{1}}=(0.897)\left(1 \times 10^{-3} \mathrm{~m}\right)=8.97 \times 10^{-4} \mathrm{~m} \end{aligned}

Hence, the arithmetic mean radius is equal to 4.48 \times 10^{-4} \mathrm{~m}. At 293 \mathrm{~K}, the density of the water droplet \left(\rho_{w}\right) is 9.95 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3}. At 308 \mathrm{~K}, the density of the air is 1.14 \mathrm{~kg} / \mathrm{m}^{3} and the viscosity of air is 1.91 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}. Substitution of these values into the terminal velocity equation yields

v_{o}=\sqrt{\frac{(4)\left(8.97 \times 10^{-4} \mathrm{~m}\right)\left(9.95 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3}-1.14 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)}{(3)\left(1.14 \mathrm{~kg} / \mathrm{m}^{3}\right) C_{D}}}=\sqrt{\frac{10.22 \mathrm{~m}^{2} / \mathrm{s}^{2}}{C_{D}}}

By trial and error, guess a value for ν_{o}, calculate a Reynolds number, and read C_{D} from Figure 12.4. Then, check the guessed value of ν_{o} by the above equation. Guess ν_{o}=3.62 \mathrm{~m} / \mathrm{s}. The Reynolds number is

\operatorname{Re}=\frac{d_{p} v_{o} \rho_{\text {air }}}{v_{\text {air }}}=\frac{\left(8.97 \times 10^{-4} \mathrm{~m}\right)(3.62 \mathrm{~m} / \mathrm{s})\left(1.14 \mathrm{~kg} / \mathrm{m}^{3}\right)}{1.19 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}\left(\frac{\mathrm{kg} / \mathrm{m} \cdot \mathrm{s}}{\mathrm{Pa} \cdot \mathrm{s}}\right)}=194

and Figure 12.4, C_{D}=0.78. Now recalculate v_{o}

v_{o}=\sqrt{\frac{10.22 \mathrm{~m}^{2} / \mathrm{s}^{2}}{C_{D}}}=\sqrt{\frac{10.22 \mathrm{~m}^{2} / \mathrm{s}^{2}}{0.78}}=3.62 \mathrm{~m} / \mathrm{s}

Therefore, the guessed value for ν_{o} is correct. The Schmidt number must now be calculated. From Appendix J.1, the gas diffusivity \left(D_{A B}\right) for water vapor in air at 298 \mathrm{~K} is 2.60 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, which is corrected to the desired temperature by

D_{A B}=\left(2.60 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\left(\frac{308 \mathrm{~K}}{298 \mathrm{~K}}\right)^{3 / 2}=2.73 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}

The Schmidt number is

\mathrm{Sc}=\frac{\mu_{\mathrm{air}}}{\rho_{\mathrm{air}} D_{A B}}=\frac{\left(1.91 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}\right)\left(\frac{\mathrm{kg} / \mathrm{m} \cdot \mathrm{s}}{\mathrm{Pa} \cdot \mathrm{s}}\right)}{\left(1.14 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(0.273 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\right)}=0.61

The Fröessling equation (30-9)

\mathrm{Sh}={\frac{k_{\mathrm{c}}D}{D_{A B}}}=2+0.552\,\mathrm{Re}^{1/2}\,\mathrm{Sc}^{1/3}        (30-9)

can now be used to evaluate the mass-transfer coefficient for transfer of water vapor from the surface of the droplet to the surrounding air

\frac{k_{c} d_{p}}{D_{A B}}=2+0.552 \operatorname{Re}^{1 / 2} \mathrm{Sc}^{1 / 3}

or

\begin{aligned} k_{c} & =\frac{D_{A B}}{d_{p}}\left(2+0.552 \mathrm{Re}^{1 / 2} \mathrm{Sc}^{1 / 3}\right) \\ & =\frac{\left(0.273 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\right)}{8.97 \times 10^{-4} \mathrm{~m}}\left(2.0+0.552(194)^{1 / 2}(0.61)^{1 / 3}\right)=0.276 \mathrm{~m} / \mathrm{s} \end{aligned}

The average rate of water evaporation from the droplet is

W_{A}=4 \pi \bar{r}_{p}^{2} N_{A}=4 \pi \bar{r}_{p}^{2} k_{c}\left(c_{A s}-c_{A \infty}\right)

The dry-air concentration, c_{A \infty}, is zero, and the surrounding is assumed to be an infinite sink for mass transfer. The surface concentration is evaluated from the vapor pressure of water at 293 \mathrm{~K}

c_{A s}=\frac{P_{A}}{R T}=\frac{2.33 \times 10^{3} \mathrm{~Pa}}{\left(8.314 \frac{\mathrm{Pa} \cdot \mathrm{m}^{3}}{\mathrm{~mol} \cdot \mathrm{K}}\right)(293 \mathrm{~K})}=0.956 \frac{\mathrm{mol}}{\mathrm{m}^{3}}

When we substitute the known values into the rate of evaporation equation, we obtain

W_{A}=4 \pi\left(4.48 \times 10^{-4} \mathrm{~m}\right)^{2}(0.276 \mathrm{~m} / \mathrm{s})\left(0.956 \mathrm{~mol} / \mathrm{m}^{3}-0\right)=6.65 \times 10^{-7} \mathrm{~mol} / \mathrm{s}

or 1.2 \times 10^{-8} \mathrm{~kg} / \mathrm{s} on a mass basis. The amount of water evaporated is

\begin{aligned} m_{A} & =\rho_{w} \Delta V=\rho_{w}\left(V_{t, 1}-V_{t, 2}\right)=\rho_{w}\left(V_{t, 1}-0.5 V_{t, 1}\right)=\frac{\rho_{w} V_{t, 1}}{2} \\ & =\frac{\rho_{w}}{2} \frac{4 \pi}{3} r_{p}^{-3}=\frac{4 \pi}{6}\left(9.95 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3}\right)\left(4.48 \times 10^{-4} \mathrm{~m}\right)^{3}=1.87 \times 10^{-7} \mathrm{~kg} \end{aligned}

The time necessary to reduce the volume by 50 \% is

t=\frac{m_{A}}{W_{A}}=\frac{1.87 \times 10^{-7} \mathrm{~kg}}{1.20 \times 10^{-8} \mathrm{~kg} / \mathrm{s}}=15.6 \mathrm{~s}

and the distance of the fall is equal to v_{o} t or 56.5 \mathrm{~m}.