Question 30.1: 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 K. The terminal velocity of the droplet is given by

\mathrm{v}_{o}={\sqrt{\frac{4\,d_{p}(\rho_{w}-\rho_{\mathrm{air}})g}{3\,C_{D}\,\rho_{\mathrm{air}}}}}

where d_{d} is the diameter of the particle, \rho _{w} is the density of the water droplet, \rho _{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 particle diameter is taken at the arithmetic mean of the evaporating droplet diameter:

\overline{{{d}}}_{p}=\frac{d_{p,o}+d_{p}}{2}=\frac{d_{p,o}+\left(0.5\right)^{1/3}\!d_{p,o}}{2}=8.97\times10^{-4}\,\mathrm{m}

At 293 K, the density of the water droplet (\rho _{w}) is 995\;{\mathrm{kg/m}}^{3}. At 308 K, the density of the air is 1.14{\mathrm{~kg/m}}^{3}, and the viscosity of air is 1.91\,\times\,10^{-5}\ \mathrm{~k g/m}\cdot{\mathrm{s}}. Substitution of these values into the terminal velocity equation yields

\mathrm{v}_{o}=\sqrt{\frac{(4)(8.97\times10^{-4}\,\mathrm{m})\bigl(9.95\times10^{2}\,\mathrm{kg/m^{3}-1.14\,\mathrm{kg/m^{3}}\bigl)(9.8\,\mathrm{m/s^{2}}\bigr)}}{(3)(1.14\,\mathrm{kg/m^{3}})C_{D}}}\,\mathrm{\large~} =\sqrt{\frac{10.22\,\mathrm{m}^{2}/s^{2}}{C_{D}}}

To obtain \mathrm{v}_{o} , trial and error is required. First, a value for \mathrm{v}_{o} is guessed, the Reynolds number is calculated, and \mathrm{C}_{D} is read from Figure 12.4. To start, guess \mathrm{v}_{o}=3.62~\mathrm{m}/s The Reynolds number is

{\mathrm{Re}}={\frac{d_{p}\mathrm{v}_{o}\rho_{\mathrm{air}}}{\nu_{\mathrm{air}}}} ={\frac{(8.97\times10^{-4}\,{\mathrm{m}})(3.62\,{\mathrm{m}}/{\mathrm{s}})(1.14\,{\mathrm{kg/m}}^{3})}{1.19\times10^{-5}\,{\mathrm{kg/m}}\cdot{\mathrm{s}}}}

and from Figure 12.4, \mathrm{C}_{D} = 0.78 Now recalculate \mathrm{v}_{o} :

\mathrm{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 \mathrm{v}_{o} is correct. The Schmidt number must now be calculated. From Appendix J.1, the gas diffusivity (D _{AB}) for water vapor in air at 298 K is 2.60\times10^{-5}\mathrm{~m}^{2}/s, which is corrected to the desired temperature by

D_{A B}=\left(2.60\times10^{-5}\,\mathrm{m}^{2}/s\right)\left(\frac{308\,\mathrm{K}}{298\,\mathrm{K}}\right)^{3/2}=2.73\times10^{-5}\,\mathrm{m}^{2}/s

The Schmidt number is

\mathrm{Sc}={\frac{\mu_{\mathrm{air}}}{\rho_{\mathrm{ai}}D_{A B}}}={\frac{\left(1.91\times10^{-5}\mathrm{~kg/m}\cdot\mathrm{s}\right)}{\left(1.14\mathrm{Kg/m^{3}}\right)\left(0.273\times10^{-4}\,\mathrm{m^{2}/s}\right)}}=0.61

The Fröessling equation (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:

\mathrm{Sh}={\frac{k_{c}D}{D_{A B}}}=2.0+0.552\,\mathrm{Re}^{1/2}\,\mathrm{Sc}^{1/3}        (30-9)

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

or

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

={\frac{\left(0.273\times10^{-4}\,{\mathrm{m}}^{2}/{\mathrm{s}}\right)}{8.97\times10^{-4}\,{\mathrm{m}}}}\left(2.0+0.552(194)^{1/2}(0.61)^{1/3}\right)=0.276\,{\mathrm{m}}/{\mathrm{s}}

The average rate of water evaporation from the droplet is

W_{A}=4\pi r_{p}^{2}\,N_{A}=4\pi r_{p}^{2}\,k_{c}(c_{A s}-c_{A\infty })

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

c_{A s}={\frac{P_{A}}{R T}} = \frac{2.33\times10^{3}\,{\mathrm{Pa}}}{\left(8.314\frac{\mathrm{Pa}\cdot\mathrm{m^{3}}}{\mathrm{gmole}\cdot\mathrm{K}}\right)(293\,{\mathrm{K}})} =0.956{\frac{\mathrm{gmole}}{\mathrm{m^{3}}}}

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

W_{A}=4\pi\bigl(4.48\times10^{-4}\,{ m}\bigr)^{2} (0.276\,\mathrm{m/s})\bigl(0.956\,\mathrm{gmole/m^{3}\,-\,0}\bigr)=6.65\times10^{-7}\,\mathrm{gmole/s}

or 1.2\times10^{-8}\ \mathrm{kg/s} on a mass basis. The amount of water evaporated is

m_{A}=\rho_{w}\Delta V=\rho_{w}(V(t_{1})-V(t_{2}))={\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\times10^{2}\,{\mathrm{kg/m}}^{3}\right)\left(4.48\times10^{-4}\,{\mathrm{m}}\right)^{3}=1.87\times10^{-7}\,{\mathrm{kg}}

The time necessary to reduce the volume by 50% is

t={\frac{m_{A}}{W_{A}}}={\frac{1.87\times10^{-7}\,\mathrm{kg}}{1.20\times10^{-8}\,\mathrm{kg/s}}}=15.6\,\mathrm{s}

Finally, the distance of the fall is equal to \mathrm{v}_{o}t or 56.5 m.