MASS 3.5 TRANSFER RATES WITHIN AND BETWEEN PHASES
Below is a conceptual view of different mass transfer environment
Aqueous and gas diffusion occurs within a phase
Interphase mass transfer occurs between phases
stagnant
air/water film
soil or
adsorbent
grain
1) air-air
mass transfer
2) water-air
mass transfer
3) water-water
mass transfer
moving water or air phase
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1
Mass transfer occurs when there is fugacity gradient within a phase or between
phases.
At equilibrium fi(air) = fi(water) = fi(pure org.) = fi(solid) = fi(octanol)
If fi(water) > fi(air); mass travels from the water to the air
If fi(water) > fi(solid); mass travels from the water to the solid
If fi(water at point 1) > fi(water at point 2); mass travels from point 1 to point 2
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2
These relationships can be restated in terms of water phase equivalence:
Ci,water* = concentration of i in the water phase that would be in
equilibrium with Ci,air
Ci,water* =
Ci,air/Hi,cc
Ci,water** = concentration of i in the water phase that would be in
equilibrium with qi,solid
Ci,water** = (qi,solid/KF)-nF
If Ci,water > Ci,water*; mass travels from the water to the air
If Ci,water > Ci,water**; mass travels from the water to the solid
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3.5.1 Molecular Diffusion
Molecular diffusion is a transport process driven by the kinetic energy of
individual molecules (i.e. Thermal motion). The rate of transport is governed
by the concentration gradient and the molecular diffusivity, a property of the
solute-solvent system.
Fick's first law defines this relationship at a given point in time (i.e.
independent of time):
F = -D C/x = -D C/x
(3.77)
F = mass flux of chemical [g m-2 s-1]
D = molecular diffusion coefficient [m2/s]
C/x = change in concentration in one dimensional space
C = concentration [g/m3]
x = space coordinate [m]
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3.5.1.1 Molecular diffusion coefficients
The molecular diffusion coefficient depends on:
- solvent temperature
- solvent viscosity
- molecular size & shape
Derivation of the diffusion coefficient from first principles (i.e. kinetic gas
theory) is not practical because molecular interactions are poorly
understood (especially in the liquid phase) or difficult to measure.
Hence, we rely on empirical correlations.
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5
Diffusion in Water
The most commonly used empirical correlation for predicting the liquid-phase
diffusivity is that of Wilke and Chang (A.I.Ch.E.J., 1, 264, 1955) given below:
D=
7.4 *10 -8 (M)0. 5 T
accuracy 15% (3.78)
Vb 0.6
where:
D = diffusion coefficient [cm2s-1]
NOTE UNITS!!!!!
= association parameter for solvent
= 2.6 for water (=1 for nonpolar solvents)
Vb = molar volume of solute at its normal boiling pt. [cm3mol-1]
= liquid viscosity [centipoise] = 1 centipoise for water at 22C)
T = absolute temperature (i.e. in Kelvin)
M = molecular wt of solvent [g/mol]
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For simple, non dissociating organic molecules the following Schroeder
Increments may be used to estimate Vb (Reid and Sherwood, 'The Properties
of Liquids and Gases', McGraw-Hill, 1958):
Increment [cm3mol-1]
Carbon
Hydrogen
Oxygen
Nitrogen
Fluorine
Chlorine
Bromine
Iodine
Sulfur
Carbon Double Bond, C=C
Carbon Triple Bond, C C
Aromatic Ring
CEE 440
7
7
7
7
10.5
24.5
31.5
38.5
21
7
14
-7
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7
Example:
Estimate the diffusivity of TCE in water at 288K
Vb = 2*7(two Cs)+7(double bond)+7(one H)+3*24.5(three Cls) = 101.5
cm3mol-1
M = 18
= 2.6 for water
D=
7.4 *10 -8 ( 2.6 * 18)0. 5 288
1(101.5)0.6
D = 9.12x10-6 cm2sec-1
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Diffusion in Vapor
A commonly used empirical correlation for calculating vapor phase
diffusivity was developed by Arnold and is defined as follows:
D12 =
0.00837 T
P
(
5
2
0.5
[(M1 + M 2 ) (M1M 2 )]
0.33
Vb
1
0.33
+ Vb
2
2
) (T + S12 )
Accuracy 15% (3.79)
where:
1=solvent
2=solute
D12 = diffusivity [cm2s-1]
T = temperature [K]
P = pressure
[atm]
M = molecular weight [gmol-1]
Vb = molar volume at normal boiling point [cm3mol-1], calculated
from Schroeder increments
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S12 = Sutherland Constant = 1.47 F (Tb1*Tb2)0.5
Tb = boiling points [K]
F = Function of ratio (Vb2/Vb1), determine F by interpolating between the data
given below:
Vb2 / Vb1
F
CEE 440
1
2
3
4
6
8
10
1.00
0.98
0.95
0.92
0.88
0.84
0.81
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10
Example:
Estimate the diffusivity of CHCl3 in air at 298K and P = 1 atm.
M1 = 29;
M2 = 119
Vb1 = 14;
Vb2 = 87.5
Vb2/Vb1 = 6;
F = 0.88
Tb1 = 79K; Tb2=334K [from Handbook of Physics and
Chemistry]
S12=1.47 F (Tb1*Tb2)0.5 = 1.47 * 0.88 (79*333)0.5 = 210
5
D12 =
8.37 *10 -3 * 298
(
2
(29 + 119) (29 * 119)]0.5
[
1 * 140. 33 + 87.5 0.33
2
) (298 + 210)
D12=0.11 cm2s-1 = 1.1e-5 m2s-1
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NOTE: the diffusivity of a compound in air is three to five orders of
magnitude greater than the diffusivity of the same compound in water. As
a general rule, mixing of the gas phase by diffusion is faster than mixing of
the liquid phase by diffusion.
Temperature dependence of diffusion coefficients
liquid
gas
CEE 440
Note: the temperature
dependence of diffusivity in
5
the gas phase ( D12 T 2 )
is much weaker than in the
liquid phase ( D T ).
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12
3.5.1.2 Diffusion in Porous Media
The diffusion path is often altered by the presence of boundaries. In the
subsurface organic chemicals must diffuse around soil and sediment grains.
Within soil and sediment grains, organic chemicals must diffuse inside narrow
and possibly undulating pores.
To account for diffusion around grains and small pores the effective diffusion
coefficient is modified by a restrictivity factor, Kr, and a tortuosity factor, , as
follows.
Deff = Dmol Kr /
CEE 440
(3.80)
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Kr accounts for the effects of steric hindrance on diffusion in small pores.
Kr is a strong function of the adsorbate to pore size ratio, and
exponentially approaches unity as this ratio decreases (Chantong and
Massoth, 1983; Satterfield and Colton, 1973).
The effects of Kr become small (i.e., the value is close to unity) when the
adsorbate size is less than 1/10 of the pore size.
Kr is typically equal to 1 when considering diffusion between grains, or
within intragranular pore spaces greater than 5nm
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The parameter accounts for the deviation of the diffusion path from a
straight line. Grathwohl (1992) demonstrated that is inversely related to
the porosity, n, raised to an exponent, m, as shown in eqn. 67.
= 1 / nm
(3.81)
For diffusion between soil and sediment particles, n represents the
interstitial porosity. For diffusion within soil and sediment particles, n
represents the intraparticle porosity. Values of m close to 1 are common for
diffusion in porous media (Werth et al., 1997). In low porosity materials this
value can increase (Wong et al., 1984).
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3.5.1.3 Transient diffusion
Ficks first law defined the rate of diffusion at steady state or at a given point
in time. In order to describe how diffusion drives mass transfer over time we
need to derive Fick's second law. We can derive Fick's second law from
Fick's first law and the equation of continuity for a solute:
Consider an element of volume in the form of a rectangular parallelpiped
whose sides are parallel to the axes of coordinates and are of lengths 2dx,
2dy, 2dz. Assume that the flux, Fx, at the center of the volume element is
known.
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Let the origin be centered in the box and let the concentration of diffusing
substance be C. The rate at which diffusing substance enters the element
through the face ABCD is given by:
4 dy dz (Fx - Fx/ x dx)
(3.82)
Similarly, the rate of diffusing substance leaving the element through A'B'C'D' is
given by:
(3.83)
4 dy dz (Fx + Fx/ x dx)
Performing a mass balance on the rectangular parallelpiped in one dimension
yields:
Mass Accumulated = Mass In - Mass Out
(3.84)
Dividing by the time interval yields
Rate of Accumulation = Rate In - Rate Out
CEE 440
(3.85)
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Substituting in from above yields:
Rate In Rate Out =
4 dy dz (Fx -
Fx/
x dx) - 4 dy dz (Fx + Fx/ x dx)
Rate In Rate Out= -8 dx dy dz
Fx/
x
(3.86)
(3.87)
Now we can define the rate of accumulation as follows:
Rate of Accum. = 2dx 2dy 2dz C/ t = 8 dx dy dz C/ t (3.89)
Setting these two equations equal to each other yields:
C/ t = -
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Fx/
x
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(3.90)
18
Substituting in for Fx from Fick's first law yields:
C/ t
= - / x (-D C/ x)
(3.91)
If the diffusion coefficient is constants (i.e. not a function of concentration
or distance):
C
t
2 C
=D 2
x
(3.92)
(Note: this argument could easily be extended to two or three
dimensions. However, this is as complicated as we need to get in
CEE440).
Solving this partial differential equation (p.d.e.) and any others is beyond
the scope of Haz. and Solid Waste Management. Analytical solutions will
be provided to all p.d.e.s.
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3.5.1.3.1. Application of Ficks 2nd Law Defining Initial & Boundary Conditions
Example #1 1D diffusion
Let's apply the gas phase diffusion coefficient to a pulse input of contaminant
in the vadose zone. Let's assume that a small amount of a pure phase
chemical has just spilled into the vadose zone and become immobilized. The
pure phase chemical volatilizes and begins to diffuse away from its
immobilized position.
Let's also assume that:
i) there is no sorption
ii) at time zero the pure phase chemical has just vaporized and is represented
by a source of vapor at initial concentration Co that is 2 meters in width
iii) the chemical diffuses away from the source in only the x direction
iv) the vapor in the vadose zone is stagnant so only diffusion controls mass
transfer
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Formally we can represent these assumptions as:
i)
C = Co at t = 0, -h<x<h
ii)
C = 0 at t = 0, x<-h, x>h
iii) dC/dx = 0, x = 0
iv) C=0 as xinf & -inf
The first two are called initial conditions and the second two are called
boundary conditions
A solution to Fick's second Law and the above boundary and initial conditions
can be obtained from Crank (The Mathematics of Diffusion, Oxford University
Press, 1975)
C
CEE 440
= 1 C0
2
h - x
h + x
e
+erf
rf
2 Dt
2 Dt
(3.93)
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21
What is the error function?
2z
(z ) = 0.5 exp - 2 d
erf
0
()
+1
-4
-2
erf(z)
1
3
5
z
-1
The error function is a function that we treat similar to the exponential or log
function. We can obtain values for the error function from tables or we can use
Excel to calculate the error function for us. The error function has the
following properties:
erf(-z) = -erf(z); erf(0) = 0; erf(infinity) = 1; 1-erf(z) = erfc(z)
(3.95)
Often times Excel does not recognize one or more of these properties and you
must define everything in terms of the positive error function or you must
substitute in 1 or 0.
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Let's solve the analytical solution to Fick's second law that we derived using
CHCl3 (trichloromethane), where h=1 meter (i.e. the vapor source is centered
at x=0 and extends for 1 meter on either side)
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Example Spherical #2 Diffusion
When working with spheres it is best to use spherical rather than
rectangular coordinates
Fick's second law in spherical coordinates is:
2 C 2 C
C
= D
r2 + r r
t
r
a
(3.96)
If a sphere of radius a is initially at uniform concentration Co and the
concentration outside the sphere is uniform and constant at C1, the
solution to the radial diffusion equation is:
22
C - Co
n r
2a ( - 1)n
=1 +
exp - Dn t 2
sin
r n =1 n
C1 - Co
a
a
CEE 440
(3.97)
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24
and the fractional uptake is:
Mt
61
- Dn2 2 t
f=
=1 exp
2
2= 2
M
a
n 1n
(3.98)
where:
a = radius of sphere [cm]
r = radial distance from the center of the sphere [cm]
D = diffusion coefficient [cm2/s]
t = time [sec]
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, 20,000 sec
10,000 sec
5,000 sec
1/s
1/s
1/s
1/s
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3.5.2 Interphase Mass Transfer
What determines the rate of interphase mass transfer if the phases are not in
equilibrium?
water
water
air
NAPL water
solid
There are several theories which attempt to describe the mechanisms
controlling this rate. Some of these theories are :
film theory - will cover in CEE440
boundary layer theory - will cover in CEE440
penetration theory - will NOT cover in CEE440
surface renewal theory - will NOT cover in CEE440
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3.5.2.1 Film Theory
This is the oldest approach, first proposed by Lewis (Ind. Eng. Chem., 8,
825, 1916) and Whitman (Chem. Met. Eng., 29, 147, 1923).
Formally, this theory assumes that:
1) the entire resistance to mass
transfer resides in a stagnant film
at the phase interface
2) bulk fluid phase is well-mixed so
that concentration gradients are
negligible
3) concentration gradient in the
stagnant boundary layer (or film)
is linear (i.e., steady state
diffusion)
4) equilibrium obtains at the
interface
CEE 440
interface
CbL
liquid
lm
gas
lm
CintG
bulk
liquid
CintL
L
Hcc=CintG/CintL
CbG
G
bulk
gas
(3.99)
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28
If we convert the gas phase concentrations to equivalent liquid-phase
concentrations:
interface
L
Cb
liquid
film
gas
film
CintL = CintG /Hcc
bulk
liquid
CL* =CbG /Hcc
L
CEE 440
G
bulk
gas
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The mass flux is given by Fick's first law as follows:
F = (1/A) dM/dt = -(DL/L)(CbL-CintL) = -(DG/G)(CintG-CbG)
F = (1/A) dM/dt = -(DL/L)(CbL-CintL) = -(DG/G)(CintL-CL*)Hcc
(3.100a)
(3.100b)
kL=DL/L = specific liquid phase mass transfer coefficient [ms-1]
kG=DG/G = specific gas phase mass transfer coefficient [ms-1]
dM/dt = -kL A(CbL-CintL)=-kG A Hcc(CintL-CL*)
(3.101)
where:
F = mass flux [gm-2s-1]
DG = diffusivity in the gas phase [m2s-1]
dM = change in mass [g]
CintL = conc. at the interface [gm-3]
dt = change in time [t]
L = thickness of stagnant film [m]
A = interfacial area [m2]
DL = diffusivity in the liquid phase [m2s-1]
CbL = bulk liquid phase concentration [gm-3]
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We could also define the mass transfer rate between the two phases in terms
of an overall mass transfer coefficient:
dM/dt = -KL A(CbL-CL*) =-KG A(CG*-CbG)
KL = overall liquid phase mass transfer coefficient [ms-1]
KG = overall gas phase mass transfer coefficient [ms-1]
(3.102)
Film theory predicts that k is proportional to D and inversely proportional do
. The parameter D can be calculated from either the Wilke-Chang or the
Arnold equations. However, is impossible to predict and impractical to
measure. Hence, we rely on empirical correlations (coupled with Fick's first
law) to define mass transfer across an interface.
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3.5.2.2 Boundary Layer Theory
The boundary layer approach to mass transfer between phases is based on
the analogy between mass- and momentum transport in the fluid boundary
layer at the interface. This approach has been successfully used in heat
transfer (see Bennett and Myers, Momentum, Heat, and Mass Transfer,
McGraw-Hill, 3rd ed., 1982).
The boundary layer theory can be applied to the flow of fluids with high
viscosity (i.e. water) at low velocity (laminar flow) through flow channels
characterized by small characteristic dimensions (i.e. groundwater flow,
fixed bed adsorption processes, packed-bed contactors).
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Basically, the boundary layer analogy presumes that the velocity and
concentration profiles in the boundary layer are similar in shape, and that the
mass transport rate is proportional to the momentum transfer rate.
uo
moving bulk fluid
y
uo
uo
uo
velocity
profile next
to stationary
surface
x
y /
u/uo or c/co 1
As shown above, friction/shear between the moving fluid and the surface cause
the velocity to decrease as we approach the surface.
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Unfortunately, we do not have time in CEE440 to derive the momentum
transfer equation and equate it to a mass transfer equation (see the book
by Bennett and Myers if you are interested in this). Instead, we will
examine the form of the derived equation and look at its application
a
b
du
kd
= consto + const1 o
D
D
Sh
Re Sc
(3.103)
where k = mass transfer coefficient [m s-1]
D = diffusivity [m2 s-1]
d = characteristic length scale (e.g., diameter of grain) [m]
uo = superficial velocity of the moving fluid (air or water) [m s-1]
v = kinematic viscosity of fluid (air or water) (i.e. viscosity/density) [m2 s-1]
Sh = Sherwood Number [-]
Re = Reynold's Number[-]
Sc = Schmidt Number [-]
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Sh can be thought of as the ratio of a mass transfer velocity to a diffusion
velocity
Re represents the ratio of inertial to viscous forces and it is a quantitative
indicator of the amount of turbulence in the boundary layer. The boundary
layer theory approximation is valid for 0<Re<3x105
Sc represents the ratio of kinematic viscosity to molecular diffusivity. This ratio
accounts for differences between the diffusivities of solutes, as well as
adjusting for the differences between the thickness of the velocity and the
concentration profiles.
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Aside: The constants can be derived for simple systems like the flat plate.
However, for more complex real systems, the constants are fit by designing
experiments such that one parameter from each dimensionless group is varied
and k is measured. By varying each dimensionless group in turn and
measuring k the dependence on k of each dimensionless group can be
determined.
Fitting parameters are specific for the experimental conditions investigated. In
the paper by Powers et al. (see reader) there are several correlations that
describe dissolution from a packed bed containing DNAPL at the residual
saturation. These correlations do not adhere to the structure determined via
Boundary Layer Theory. Hence, they are more empirical in nature. Some of
these expressions are presented on the next page.
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36
Sh' = 18.4 Re0.736
Sh' = 57.7 Re0.611 d500.643 Ui0.413
Sh' = 340 Re0.621 nd-1.16 (i/d)-0.796 d0.186
(3.104)
(3.105)
(3.106)
where:
Sh' = k (d50)2 / Daq
d50 = mean grain size (i.e. sieve size through which 50% of grains
by mass pass through)
nd, i, d - fitting parameters from the van Genuchten equation
k = kf a
kf = this is the equivalent of KL with units of length/time
a = surface area between NAPL and aqueous phase per unit
volume of matrix
Re = w q d50 / w
q = Darcy velocity
Ui = grain uniformity index (=d60/d10)
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With these correlations we can try to predict mass transfer across a NAPL
water boundary layer by assuming that Fick's first law (i.e. boundary layer
theory) describes mass transfer across an interface. For instance, if we
wanted to calculate mass transfer across a NAPL-water interface we could
assume the following:
dM/dt = -KL A(CbL-Csol)
(3.107)
CbL
where KL = k /a = kf
CbL
We could also substitute in other water equivalent concentrations and other
Sh expressions to calculate mass transfer between other phases.
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Example:
Approximately 100 L of perchloroethylene (PCE) escaped from a leaking
underground storage tank (LUST) and migrated below the water table.
PCE blobs are trapped in pore spaces (n=0.33) below the water table at a
residual saturation of 15% (15% of the pore space is occupied by the
NAPL). The Darcy velocity of flowing groundwater is approximately 3x10-9
m/s (this flow rate is typical of that found in silty clays). The average size
of these blobs is on the order of the grain size; where the grain size of silt
is approximately 75 m with a uniformity index of 1.21. Assuming mass
transfer resistance in the water phase controls dissolution, how long will it
take for 99.99% of the PCE to dissolve in the flowing groundwater if the
bulk aqueous phase concentration of PCE around the perimeter of each
grain remains at Cbulk,aq/Csol,aq = 0.10000 (T=293K) (use Sh' = 57.7 Re0.611
d500.643 Ui0.413).
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39
From the paper by Powers et al. we know that:
2
K L a (d 50 )
D aq
2
K L a (d 50 )
D aq
0. 643
= 57.7 Re 0. 611 (d 50 )
qd
= 57 .7 w 50
w
0.611
(Ui )0.41
(d 50 )0.643 (U i )0.41
First let's calculate each parameter required in SI units. Daq,PCE can be
calculated from the Wilke-Chang equation and it is:
Daq,PCE = 8.58x10-10 m2/s
We can look up the viscosity and density of water in the Handbook of Physics
and Chemistry:
vw = w/w
vw = 10-3 Pa s / (998 kg/m3) = 9.98x10-7 m2/s
We also dened the following
d50 = 75x10-6 m
q = 3x10-9 m/s
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Now lets calculate KLa
(
K L a 75 x 10
-6
8.58 x 10 10
)
2
3x 10 - 9 * 75 x 10 -6
= 57 .7
9.98 x 10 7
0. 611
0. 643
( 5 x 10 )
7
-6
(1.15 )0.41
2
0 .643
0 .611
-6
K L a 75 x 10 6 = 57.7 2.25 x10 -7
(1.21)0 .41
75 x10
8.58 x 10 10
(
)
(
)(
)
K L a =1.08x10 7 1 / s
Now we can rearrange Fick's rst law:
dM/dt = -KL A (Csol,aq-Cbulk,aq)
where:
Csol,aq = concentration of PCE at the solubility limit (g/ml)
Cbulk,vapor = concentration of PCE in the bulk aqueous phase (g/ml)
A = interfacial area of blobs (cm2) = a * VREV
CEE 440
2011 Charles J. Werth, University of Illinois at Urbana-Champaign. All rights reserved.
41
We can calculate VREV by assuming the NAPL is uniformly distributed at
15% residual saturation.
VREV = VNAPL/(porosity*0.15) = 100 L / (0.33 * 0.15) = 2020 L = 2.02 m3
To obtain a solution in terms of the mass remaining we integrate between Mo
and Mt and between t=0 and t.
dM/dt = -KL A (Csol,aq-Cbulk,aq)
t = - (Mt-Mo) / KL a VREV (Csol,aq-Cbulk,aq)
CEE 440
2011 Charles J. Werth, University of Illinois at Urbana-Champaign. All rights reserved.
42
To obtain a solution in terms of the mass remaining we integrate between Mo and
Mt and between t=0 and t.
t = - (Mt-Mo) / KL a VREV (Csol,aq-Cbulk,aq)
We can calculate the initial mass of PCE from the density of PCE, we can calculate
the mass at time t, and we can calculate the concentration driving force from the
solubility limit of PCE.
Mo = 100 L * 1.6 g/ml * 103 ml/L = 1.6x105 g
Mt = 0.0001 * 1.6x105 = 16 g (mass left after 99.99% of mass dissolved)
Mt-Mo = 16 - 1.6x105 = - 1.6x105 g
Csol,aq = 0.15000 g/L = 150.00 g/m3
Cbulk,aq = 0.1000 * Csol,aq = 0.10000 * 0.15000 g/L = 0.015 g/L = 15 g/m3
CEE 440
2011 Charles J. Werth, University of Illinois at Urbana-Champaign. All rights reserved.
43
Now we can calculate the time to vaporize this mass:
t = - (Mt-Mo) / KL a VREV (Cs,aq-Cbulk,aq)
= - (-1.6x105 g) / [1.08x10-7 1/s * 2.02 m3 * (150.00 - 15 g/m3)]
= 5.43x109 seconds = 172 years
WOW, this is really slow. Is this realistic? Are our assumptions correct?
CEE 440
2011 Charles J. Werth, University of Illinois at Urbana-Champaign. All rights reserved.
44