Chapter 10
Transfer between the gas-liquid interface
Develop a physical picture of the air-water boundary and how it may control mass transfer
Calculate mass transfer rates for water
Extend these rates to other compounds
Figure 10.1 page 217
bubbles, oil films, aerosol droplets, etc. which perturb the interface are not included in this model
From the velocity of a molecule colliding with a surface normal to the direction of movement let us calculate an upper limit of the velocity at which molecules enter the liquid phase from the gas phase.
R = 8.31x107 g/cm2 K-1sec-2 mol-1
At 298oC and for a compound of 100g/mol
If only one out of 1000 gas phase molecules ends up in the liquid below the surface the transfer velocity is still 5 kilometers/day
This is much higher than what is observed and requires additional treatment (1cm/sec)
The Two film model
stagnant air layer
Flux = -DC/z
stagnant water layer
Assumes that concentrations at the boundary layers are constant long enough for a concentration profile to reach steady state
No chemical reactions in the boundary phases
Figure 10.2 page 220
There may be different resistivities in the different boundary layers
If we first look at the interface; immediately above is an air concentration, and immediately below a water conc. that are probably in equilibrium
air
Water
KH’= Ca/w/Cw/a
The flux at the interface is the same whether is coming from the direction of the water or the air
substituting for Ca/w from Henry’s law
and solving for Cw/a
recalling the expression for flux in the liquid boundary
and substituting we get an expression which includes mass transfer through both boundary layers
If the bulk water conc. equals Ca/K’H; which means
K’H= Ca/Cw ie Ca and Cw are in equilibrium
there will be no net flux
If the bulk water conc. > Ca/K’H; ie.Cw > Ca/K’H then F will be positive and there will be a net flow from the water to the air phase and vice versa
moles/(cm2 sec)cm/secmoles/cm3
where this net velocity flux, vtot, is called a mass transfer coef
note that zw/Dw has the units of sec/cm or inverse velocity
1/vtot= 1/vw + 1/(vaK’H )
1/vtot= 1/vw+ 1/(vaK’H)
This has the “look and feel” of resistors in parallel
1/Rtot= 1/R1 + 1/R2
Rtot R1 R2
if vw is < vaKH
vtot ~ vw
and transfer will be dominated by the water boundary layer or the resistance will be in the water layer
if vw > vaKH
resistance will be in air boundary layer and vtot~ vaKH
Boundary layer theory assumptions
1. We don’t know much about z thicknesses
as air velocities increase we would expect the thickness of za and zw to decrease, and this is observed in that as volatilization increases so does transfer from the water to that air phase
2. It is assumed that no reactions take place in the boundary layer
What is the time needed to traverse the boundary layer compared to reaction times?
x = (2Dt)1/2
dropping the 2
in the water layerw= zw2/Dw and zw/Dw = 1/vw
so w= zw/vw
in the air boundarya= za/va
Typical lengths for zw are 5x10-2 to 5x10-3 cm
and Dw = 10-5cm2/sec;
this gives w= zw2/Dw=
and for a=
For PAH reacting in sunlight in the air or on particles
dPAH/PAH = -k PAH; k = ~0.02 min-1
r ~ 1/k= 3000 sec
3. It is assumed that the bulk concentrations in air and water “above and below” the boundary layers are constant long enough to establish a steady state gradient. What if they are not?
Surface Renewal Model
This model attempts to describe a continual turnover at the air water interface.
- Eddies in the water and air immediately above and below the air water interface transport material from the bulk phase to an interface boundary
- Henry’s law equilibrium at the interface is rapidly re-established
- Mass transported across the interface does not remain there and is mixed into the bulk phase on the other side.
Figure 10.4 page 225
{
Cw
water
The concentration in an element just below the interface line will have will have a diffusion length that is some function of (D t)1/2
s = const(D t)1/2
the concentration in this element Cw - Cw/a
the product of this concentration and the diffusion length is the mass/unit area that moves across the boundary or fw
fw = const.x (Dw t)1/2 (Cw - Cw/a) = fa
fa = const.x(Da t)1/2(Ca/w - Ca)
const.x (Dw t)1/2 (Cw - Cw/a) = const.x(Da t)1/2(Ca/w - Ca)
substituting K’H x Cw/a = Ca/w and solving for Cw/a
fw =const.x (Dw t)1/2 (Cw - Cw/a)
substituting for Cw/a
fw = = fa
fa and fw have the units of mass/area, so if we divide bytime, which can be thought of the transfer time across the interface, we have the units
mass/(area time) or flux, F
if we call 1/t the surface renewal rate , r
fw/t = = flux= F
we can even decouple the renewal rates from different turn over rates in the air and water layers
fw/t = =F
This looks like the stagnant two film expression
so by analogyvtot; with partial transfer velocities
vw= (rwDw)1/2 & va=(raDa)1/2 KH’
a. quiescent conditions ----> stagnant film model
b. streams, etc.------> renewal model
flux will vary D; where ranges from 0.5 to 1
Example: What is the Flux of SO2 to the ocean’s surface?
We start with an average atmospheric conc. of 3 g/m3 and the fact that SO2 is rapidly oxidized in the slightly alkaline environments of sea water.
Hence what is the Cw value at the water surface?
From tables we can obtain values for vw and va.; we will later learn how to calculate these.
va(SO2)= 4.4 x 10-3 ms-1
vw(SO2)= 9.6 x 10-2 ms-1
KH’ = 0.02
vtot = 9x10-5 ms-1
F = vtot(0 - 3g/m3/KH) = -0.013 g/m-2s-1 (-sign;air-->water)
Total ocean surf= 3.6x1014 m2; Ftot oceans= 1.5x1014g/year
This is about equal to the annual emissions of SO2
How do we obtain air and water transfer velocites?
let us 1st use evaporating water as an example
The transfer velocity for water in either the stagnant or renewal film model is probably very high
this gives
F = va (K’H Cw-Ca)
The relative humidity of air is the existing conc of water in air at a given temperature, Ca,compared to the saturated conc of water in air at that temperature.
h = Ca /Casatand hCasat = Ca
substituting h Casat for Ca in the flux expression and solving for va
so by measuring Fwater,h, va can be known
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Can we calculate the film thickness z, from va = Da/z in the stagnant model or r from va= (raDa)1/2 in the renewal model?
Mass transfer velocities (mass transfer coef) over water (va) have been related to wind speed (page 229 Figure 10.5)
page 229 Fig 10.5
The wind speed measured at one height,z, is reference to a standard height like 10 meters
(MacKay and Yeun)
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What emerges from the above figure is that surface mass transfer coef. for water are in the range of 0.3 to 3 cm/sec when surface winds at 10 m range from 0 -10 m/sec.
Many of the lines have non zero intercepts suggesting a stagnant film
page 231
A general relationship of Va(H2O) ~ 0.2 10(m/sec)+0.3
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As an example, let’s calculate the flux of water from a natural lake at 20oC.
Flux = va (CwK’H - Ca )
at 20oC the relative humidity over a lake is say 80%
Ca= 0.8 K’H Cw
Flux(H2O) = va (CwK’H - 0.8 K’H Cw )
K’H(H2O) at 20oC = 2.3x10-5
va= 0.3 to 3 cm/sec
Flux = 10-6 to 10-5 g cm-2sec-1
Flux = 0.1 to 1 cm/day, and this agrees
with observed evaporation rates
Air mass transfer rates for other compounds
recall
va ~ D/z in the stagnant model
to (Dr)1/2 in the renewal model
va(1) / va(H2O)= {Da(i) /Da(H2O)}
so
va(i) = va(H2O) {Da(i) /Da(H2O)}
***Di /Dj= [Mwj /Mwi]1/2for both air and water
In summary
the mass transfer velocity across the air boundary layer is related to wind speed over a surface
Va(H2O) ~ 0.2 10(m/sec)+0.3
Water trans. velocities vw generally slower than in air, va
page 233
page 233 figure 10.6
wind speed and shear stress
wave motion
surface contamination (surface tension)
PCBs were banned in the early 1970s
In 1980 Eisenreich and co-workers estimated that still 85% of the PCBs in the US great lakes came from atmospheric sources.
For Lake Superior
SourcesKg/yearSinksKg/year
atmos7,500River 140
Municipal1,200Sed 1,400
Indust 2H2O->Air ?
For the next 15 years sources to the lakes had declined because of the PCB ban.
In 1990 Eisenreich and co-workers reported that ambient measurements over the great lakes were generally constant for the past 10 years.
RT lnfair /fiopure liquid = RT lnf H2O/fiopure liquid
fair = f H2O
Based on mass transfer calculations it was proposed that during the summer months the lakes were actually a source of atmospheric PCBs.
Effects of Stirring(shallow streams) page 234
page 234 Fig 10.7
O’connor and Dobbins (1958) proposed a relationship of the renewal rate of the film surface with the velocity of water and water depth r w/dw
r w/dw
vw= (Dw rw)1/2 = (Dw w/ dw)1/2
for a stream of 50 cm depth and a water velocity of 1m/sec and with a benzene contamination(Dw= 1.02x10-5 cm2/sec)
vw= (1.02x10-5 x100/50)1/2= 4.5x10-3cm/sec
typically vw is greater than 5x10-4 cm/sec; so if we take this as a reference value and use a typical water diffusion coef., and estimate the importance of turbulence
This is done by asking at what critical velocity to depth ratio will vw be less than 5x10-4 cm/sec
5x10-4 cm/sec < (Dw w/ dw)1/2
this means that if w/ dw < 0.03 sec-1, va will be very low
Measurements of the vw from streams using tracers or from spills are generally with in a factor of 2 from predictions by
vw= (Dw w/ dw)1/2
page 236
page 236, Figure 10.8
Schwartzenbach suggests vw(O2) = 4 x10-4+ 4x10-5210
from vw(O2) we can estimate vw(i) = vw(O2) {Dw(i) /Dw(O2)}
lab data = 0.57 and = 0.67
Total Transfer Velocity, ratios of vw /vaand how these compare with Henry’s law values for different compounds
1/vtot= 1/vw+ 1/(vaK’H)
1/vtot= 1/vw+ 1/(va’)
where va’= vaK’H = vaKH/RT
If we call the inverse transfer velocities layered resistances, a resistance ratio would measure the importance of the air film resistance to the water film resistance
if Ra/w < 0.1 mass transfer is controlled by the water film
Ra/w > 10 mass transfer is controlled by the air film
in watervw(i) = vw(O2) {Dw(i) /Dw(O2)}
in airva(i) = va(H2O) {Da(i) /Da(H2O)}0.67
we can calculate va(O2) from
vw(O2) = 4 x10-4+ 4x10-5210
va(H2O) from
Va(H2O) ~ 0.2 10(m/sec)+0.3
wind speed measured at one level can be translated to10
Diffusion coef.
Dw= 2.3x10-4/ V0.71
Da= 2.35/ V0.73
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Table 10.3 page 238
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Transfer velocities vs. Henry’s law values
page 240, Figure 10.9
Surface Films
Oily layers (amphiphilic or surface active)
How do they alter the exchange of volatile chemicals through an air water boundary?
1.. Create another layer through with diffusion must take place.
2. dampen mixing and turbulence -lengthens diffusion distances and slows renewal rate
page 252 Figure 10.13
air
oil
water
Fluxtotal=fluxwater boundary= fluxoil boundary =fluxair boundary
flux = velocity x conc
vw(Cw-Cw/o) = vw(Co/w-Co/a) = va(Ca/w-Ca)
At the interface we previously assumed that Henry’s law was operating
Ka/o= Ca/CoCo= conci in oil
Ko/w= Co/Co
Ca/Cw= Ka/o Ko/w =K’H
Fluxtotal =
Fluxtotal =
for O2 , K’H is high an we can neglect 1/ (K’H va )
for organics Ko/w is probably high
page 253 Figure 10.14
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Rate constants and fluxes
If we look at the flux coming off a water surface, it has the units of
flux = moles/(cm2 sec)
depth
conci
if we divide the flux by the conci (moles/cm3)xdepth(cm) we come up with 1/sec, which is really the rate constant in rate of loss of conci from the entire body of water
dconci/dt = -krate x concI
If the conc = Cw
dCw/dt = flux/(Cw x depth) x Cw = flux/depth
dCw/dt = flux/depth = -krate x Cw
flux x surface/volume = -krate x Cw
The Influence of Chemical Reactions
1. Reaction rates are much slower then transport rates across the air water boundary
2. Reaction is extremely fast compared to transport time. Newly formed species equilibrate immediately
3. Reactions and transport are of the same order.
on the air side
charged species can not volatilize in air
on the water side of the boundary
substituting
for A-w above
if we define Dw (1+Ka/H+) as an enhanced diffusivity D’w
it can ultimately be shown that the total flux
a similar expression can be derived for the surface renewal model
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