In a humidification apparatus, liquid water flows in a thin film down the outside of a vertical, circular cylinder. Dry air at 310 K and 1.013 × 10^5 Pa (1 atm) flows at right angles to the 0.076-m diameter, 1.22-m-long vertically aligned cylinder at a velocity of 4.6 m/s. The liquid film

Question

In a humidification apparatus, liquid water flows in a thin film down the outside of a vertical, circular cylinder. Dry air at [latex]310 \mathrm{~K}[/latex] and [latex]1.013 imes 10^{5} \mathrm{~Pa}(1 \mathrm{~atm})[/latex] flows at right angles to the [latex]0.076-\mathrm{m}[/latex] diameter, [latex]1.22-\mathrm{m}[/latex]-long vertically aligned cylinder at a velocity of [latex]4.6 \mathrm{~m} / \mathrm{s}[/latex]. The liquid film temperature is [latex]290 \mathrm{~K}[/latex]. Calculate the rate at which liquid must be supplied to the top of the cylinder if the entire surface of the cylinder is to be used for the evaporating process and no water may drip off from the bottom of the cylinder.

Accepted Answer

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 [latex]300 \mathrm{~K}[/latex]. The properties of air may be obtained from Appendix I, with [latex] ho=1.1769 \mathrm{~kg} / \mathrm{m}^{3}[/latex] and [latex]v=1.5689 imes 10^{-5} \mathrm{~m}^{2} / \mathrm{s}[/latex] at [latex]300 \mathrm{~K}[/latex] and [latex]1 \mathrm{~atm}[/latex]. The Reynolds number is

[latex]\operatorname{Re}_{D}=\frac{D v_{\infty}}{v_{ ext {air }}}=\frac{(0.076 \mathrm{~m})(4.6 \mathrm{~m} / \mathrm{s})}{1.5689 imes 10^{-5} \mathrm{~m}^{2} / \mathrm{s}}=22283[/latex]

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

[latex]D_{A B}=\left(2.60 imes 10^{-5} \mathrm{~m}^{2} / \mathrm{s} ight)\left(\frac{300 \mathrm{~K}}{298 \mathrm{~K}} ight)^{3 / 2}=2.63 imes 10^{-5} \mathrm{~m}^{2} / \mathrm{s}[/latex]

The Schmidt number is

[latex]\mathrm{Sc}=\frac{ u_{ ext {air }}}{D_{A B}}=\frac{1.5689 imes 10^{-5} \mathrm{~m}^{2} / \mathrm{s}}{2.63 imes 10^{-5} \mathrm{~m}^{2} / \mathrm{s}}=0.6[/latex]

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

[latex]G_{M}=\frac{v_{\infty} ho_{ ext {air }}}{M_{ ext {air }}}=\frac{(4.6 \mathrm{~m} / \mathrm{s})\left(1.1769 \mathrm{~kg} / \mathrm{m}^{3} ight)}{29 \mathrm{~kg} / \mathrm{kg} \mathrm{mol}}=0.187 \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s}}[/latex]

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

[latex]\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}[/latex] (30-16)

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

[latex]\frac{k_{G} P \mathrm{Sc}^{0.56}}{G_{M}}=0.281\left(\operatorname{Re}_{D} ight)^{-0.4}[/latex]

or

[latex]\frac{k_{G}\left(1.013 imes 10^{5} \mathrm{~Pa} ight)(0.60)^{0.56}}{0.187 \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s}}}=\frac{0.281}{(22283)^{0.4}}[/latex]

Finally

[latex]k_{G}=1.26 imes 10^{-8} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{Pa}}[/latex]

The flux of water can be evaluated by

[latex]N_{A}=k_{G}\left(p_{A, i}-p_{A \infty} ight)[/latex]

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

[latex]N_{A}=\left(1.26 imes 10^{-8} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{Pa}} ight)\left(1.73 imes 10^{3} \mathrm{~Pa}-0 ight)=2.18 imes 10^{-5} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s}}[/latex]

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

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

Question 30.3: In a humidification apparatus, liquid water flows in a thin ...

Related Answered Questions

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 ...

Verified Answer:

Consider the remediation trench shown in Figure 30.7, a very simple process to treat contaminated wastewater before discharge to a lake or river. The remediation trench consists of a narrow outdoor open channel with an air sparger aligned along the bottom of the trench. Wastewater containing a ...

Verified Answer:

Verified Answer:

Verified Answer:

The dissolution of lead from lead-soldered joints in household piping is a possible health problem. If the water flowing through the pipe is mildly acidic, then some of the lead will dissolve from the inner surface of the lead-soldered pipe section into the water. New drinking water requirements ...

Verified Answer:

Verified Answer: