air stripping
DESCRIPTION
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 PresentationTRANSCRIPT
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
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
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?
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
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
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
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
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)
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