Air Stripping of Radon from Groundwater Groundwater from some geological formations may contain radon, a gas that has been implicated in lung disease, so its concentration should be reduced. Air stripping is one method of doing this. Groundwater from a well is found to contain 10 parts per million

Question

Air Stripping of Radon from Groundwater

Groundwater from some geological formations may contain radon, a gas that has been implicated in lung disease, so its concentration should be reduced. Air stripping is one method of doing this. Groundwater from a well is found to contain 10 parts per million by weight of radon, and it is desired to reduce its concentration to 0.1 parts per million. This is to be done by air stripping in the device shown here, which is open to the atmosphere, using previously humidifed air so that water does not evaporate in the air stripping process. Assuming that the air leaving the stripper is at 20°C and in equilibrium with the liquid, how many kilograms of air must be supplied per kilogram of water to reduce the radon content to the desired level? Radon has a molecular weight of 222, and its Henry’s constant is [latex]K=P_{ R } / x_{ R }=5.2 imes 10^{3}[/latex] bar/mole fraction, where [latex]P_{ R }[/latex] is the partial pressure of radon.

Accepted Answer

Since the air has been prehumidified, no water should evaporate. Also, since the concentration of radon in the water is so low, even if all of it was removed, there would be very little change (indeed, only one part in 100,000) in the original mass of liquid in the air stripper, so that the small change in total mass (or number of moles) can be neglected. Therefore, in the equations that follow, the total number of moles N is replaced by the number of moles of water [latex]N_{ W }[/latex]. The mass (mole) balance on the radon in the liquid in the air stripper is

[latex]\frac{d N_{ R }}{d t}=\frac{d\left(N_{ W } x_{ R } ight)}{d t}=N_{ W } \frac{d x_{ R }}{d t}=-\frac{\dot{N}_{ air } K_{ R } x_{ R }}{P_{ atm }}[/latex]

or

[latex]N_{ W } \frac{d x_{ R }}{d t}=-\frac{\dot{N}_{ air } K_{ R } x_{ R }}{P_{ atm }}[/latex]

On integration from t = 0 we obtain

[latex]\ln \left[\frac{N_{ R }(t)}{N_{ R }(t=0)} ight]=-\frac{\dot{N}_{ air } \cdot t \cdot K_{ R }}{N_{ W } \cdot P_{ atm }}=\frac{N_{ air } \cdot K_{ R }}{N_{ W } \cdot P_{ atm }}[/latex]

where [latex]N_{ air }=\dot{N}_{ air } \cdot t[/latex] is the number of moles of air that have passed through the radon-containing water in the stripping device. Note that this equation gives the amount of radon remaining in the water as a function of the number of moles of air that have been used. Consequently,

[latex]\ln (0.01)=-4.605=\frac{N_{ air } \cdot 0.52 imes 10^{4}}{N_{ W } \cdot 1.103}[/latex]

or

[latex]\frac{N_{ air }}{N_{ W }}=8.97 imes 10^{-4}[/latex]

so that [latex]8.97 imes 10^{-4}[/latex] moles of air are needed for each mole of radon-contaminated water, or equivalently, [latex]1.45 imes 10^{-3}[/latex] kg of air are needed for each kg of radon-contaminated water in the air stripper to reduce the radon content of the water to one-hundredth of its initial concentration.

Question 11.1.3: Air Stripping of Radon from Groundwater Groundwater from som...

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