Estimate the distance a spherical drop of liquid water, originally 1 mm in diameter, must fall in quiet, dry air at 323 K in order to reduce its volume by 50%. Assume that the velocity of the drop is its terminal velocity evaluated at its mean diameter and that the water temperature remains at

Question

Estimate the distance a spherical drop of liquid water, originally [latex]1 \mathrm{~mm}[/latex] in diameter, must fall in quiet, dry air at [latex]323 \mathrm{~K}[/latex] in order to reduce its volume by [latex]50 \%[/latex]. Assume that the velocity of the drop is its terminal velocity evaluated at its mean diameter and that the water temperature remains at [latex]293 \mathrm{~K}[/latex]. Evaluate all gas properties at the average gas film temperature of [latex]308 \mathrm{~K}[/latex].

Accepted Answer

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 [latex]A[/latex] ) is the transferring species. The rate of evaporation is sufficiently small so that the water droplet is considered isothermal at [latex]293 \mathrm{~K}[/latex]; 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

[latex]v_{o}=\sqrt{\frac{4 d_{p}\left( ho_{w}-p_{\mathrm{air}} ight) g}{3 C_{D} ho_{\mathrm{air}}}}[/latex]

where [latex]d_{p}[/latex] is the diameter of the particle, [latex] ho_{w}[/latex] is the density of the water droplet, [latex] ho_{ ext {air }}[/latex] is the density of the surrounding fluid (air), [latex]g[/latex] is the acceleration due to gravity, and [latex]C_{D}[/latex] 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

[latex] \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} ight)^{1 / 3} \cdot d_{p \mid t_{1}}}{2} \ & =0.897 d_{p \mid t_{1}}=(0.897)\left(1 imes 10^{-3} \mathrm{~m} ight)=8.97 imes 10^{-4} \mathrm{~m} \end{aligned} [/latex]

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

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

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

[latex]\operatorname{Re}=\frac{d_{p} v_{o} ho_{ ext {air }}}{v_{ ext {air }}}=\frac{\left(8.97 imes 10^{-4} \mathrm{~m} ight)(3.62 \mathrm{~m} / \mathrm{s})\left(1.14 \mathrm{~kg} / \mathrm{m}^{3} ight)}{1.19 imes 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}\left(\frac{\mathrm{kg} / \mathrm{m} \cdot \mathrm{s}}{\mathrm{Pa} \cdot \mathrm{s}} ight)}=194[/latex]

and Figure 12.4, [latex]C_{D}=0.78[/latex]. Now recalculate [latex]v_{o}[/latex]

[latex]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}[/latex]

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

[latex]D_{A B}=\left(2.60 imes 10^{-5} \mathrm{~m}^{2} / \mathrm{s} ight)\left(\frac{308 \mathrm{~K}}{298 \mathrm{~K}} ight)^{3 / 2}=2.73 imes 10^{-5} \mathrm{~m}^{2} / \mathrm{s}[/latex]

The Schmidt number is

[latex]\mathrm{Sc}=\frac{\mu_{\mathrm{air}}}{ ho_{\mathrm{air}} D_{A B}}=\frac{\left(1.91 imes 10^{-5} \mathrm{~Pa} \cdot \mathrm{s} ight)\left(\frac{\mathrm{kg} / \mathrm{m} \cdot \mathrm{s}}{\mathrm{Pa} \cdot \mathrm{s}} ight)}{\left(1.14 \mathrm{~kg} / \mathrm{m}^{3} ight)\left(0.273 imes 10^{-4} \mathrm{~m}^{2} / \mathrm{s} ight)}=0.61[/latex]

The Fröessling equation (30-9)

[latex]\mathrm{Sh}={\frac{k_{\mathrm{c}}D}{D_{A B}}}=2+0.552\,\mathrm{Re}^{1/2}\,\mathrm{Sc}^{1/3}[/latex] (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

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

or

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

The average rate of water evaporation from the droplet is

[latex]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} ight)[/latex]

The dry-air concentration, [latex]c_{A \infty}[/latex], 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 [latex]293 \mathrm{~K}[/latex]

[latex]c_{A s}=\frac{P_{A}}{R T}=\frac{2.33 imes 10^{3} \mathrm{~Pa}}{\left(8.314 \frac{\mathrm{Pa} \cdot \mathrm{m}^{3}}{\mathrm{~mol} \cdot \mathrm{K}} ight)(293 \mathrm{~K})}=0.956 \frac{\mathrm{mol}}{\mathrm{m}^{3}}[/latex]

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

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

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

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

The time necessary to reduce the volume by [latex]50 \%[/latex] is

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

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

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

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