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AIR STRIPPING The removal of volatile contaminants from water and contaminated soils

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AIR STRIPPING. The removal of volatile contaminants from water and contaminated soils. Air Stripping Tower Use. Air Stripping Towers. Some column internals. Clockwise from top left: Packing, bubble caps, mist eliminator, sieve tray. Case Study: TCE Contaminated Site Remediation. - PowerPoint PPT Presentation

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Page 1: AIR STRIPPING

AIR STRIPPING

The removal of volatile contaminants from water

and contaminated soils

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Air Stripping Tower Use

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Air Stripping Towers

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Some column internals

Clockwise from top left: Packing, bubble caps, mist eliminator, sieve tray

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Case Study:TCE Contaminated Site Remediation

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Henry’s Law

Henry's Law states that the amount of a gas that dissolves into a liquid is proportional to the partial pressure that gas exerts on the surface of the liquid. In equation form, that is:

A H AC = K pwhere,

CA = concentration of A, [mol/L] or [mg/L]

KH =equilibrium constant (Henry's Law constant), [mol/L-atm] or [mg/L-atm]

pA = partial pressure of A, [atm]

See also http://en.wikipedia.org/wiki/Henry's_law

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Example: Solubility of O2 in Water

Although the atmosphere we breathe is comprised of approximately 20.9 percent oxygen, oxygen is only slightly soluble in water. In addition, the solubility decreases as the temperature increases. Thus, oxygen availability to aquatic life decreases during the summer months when the biological processes which consume oxygen are most active. Summer water temperatures of 25 to 30C are typical for many surface waters in the U.S. Henry's Law constant for oxygen in water is 61.2 mg/L-atm at 5C and 40.2 mg/L-atm at 25C. What is the solubility of oxygen at 5C and at 25C?

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Solution O2 Solubility Example

2 2 2O H,O OC (5 C) = K P = 61.2 mg

L- atm x 0.209 atm

At 50C the solubility is:

2OC (5 C) = 12.8 mg

L

At 250C the solubility is:

2 2 2O H,O OC (25 C) = K P = 40.2 mg

L- atm x 0.209 atm

2OC (25 C) = 8.40 mg

L

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Air Stripping Example

An air stripping tower, similar to that shown, is to be used to remove dissolved carbon dioxide gas from a groundwater supply. If the tower lowers the level to twice the equilibrium concentration, what amount of dissolved gas will remain in the water after treatment? The partial pressure of carbon dioxide in the atmosphere is 10-3.7 atm.

Example 4.2 from Ray

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Solution Air Stripping Example

Henry's Law constant for carbon dioxide = 1.14L/L Divide by RT, i.e. 1.14/(0.083x288) = 0.048 =10-1.3 mol L-1atm-1 The equilibrium solubility is then:

atm10atmL-

mole10 = pK = C 7.3-3.1-

COCOH,CO 222

2CO-5 -5

3

C = 10 M = 10mole

L x

44 g

mole x

10 mg

g

mg/L 0.44 = CCO2

Answer = 0.9 mg/L CO2

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Two-film partitioning in gas-liquid

ys*

ys

Cs*

Cs

ys*

ys

Cs*

Cs

yb

Cb

ybCb

Stripping Absorption

Bulk gas

Bulk liquid

Bulk liquid

Bulk gas

Page 38: AIR STRIPPING

Mass transfer calculations

Henry’s law applies because of equilibrium at interface:

ys = HCs

Fick’s law applies: flux is proportional to the driving force

JA = kL(Cb – Cs) – kg(ys – yb)

Page 39: AIR STRIPPING

Relationship of transfer coefficients

Alternatively all the resistance to transfer could be in the gas phase, so

ys* = HCb

Mass flux based on water phase, substitutions, and applying Henry’s law to liquid concentrations leads to

1/KLa = 1/kLa + 1/Hkga

Similarly 1/KGa = H/kLa + 1/kga

Another approach is that all the resistance to transfer is in the liquid phase, so

yb = HC*s