Transcript AIR STRIPPING

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:
CA =
KH pA
where,
CA
KH
pA
= concentration of A, [mol/L] or [mg/L]
= equilibrium constant (Henry's Law
constant), [mol/L-atm] or [mg/L-atm]
= 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 30°C 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 5°C and
40.2 mg/L-atm at 25°C. What is the solubility of
oxygen at 5°C and at 25°C?
Solution O2 Solubility Example
At 50C the solubility is:
C O 2 (5 C ) = K H ,O 2 P O 2 = 61.2
C O 2 (5 C ) = 12.8
mg
L- atm
x 0.209 atm
mg
L
At 250C the solubility is:
C O 2 (25 C ) = K H ,O 2 P O 2 = 40.2
C O 2 (25 C ) = 8.40
mg
L- atm
mg
L
x 0.209 atm
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:
C CO 2 = K H, CO 2 p CO 2 = 10
C C O 2 = 10
-5
M = 10
C CO 2 = 0.44 mg/L
-5
-1 . 3
m ole
L
mole
L- atm
x
10
44 g
m ole
- 3 .7
atm
3
x
10 m g
g
Answer = 0.9 mg/L CO2
Two-film partitioning in gas-liquid
ys*
Bulk
gas
ys
Bulk
liquid
Cs*
Bulk
liquid
Cb
Cs
Cb
yb
yb
Cs
Cs*
Stripping
Bulk
gas
ys
ys*
Absorption
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
Another approach is that all the resistance
to transfer is in the liquid phase, so
yb = HC*s
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