1 Molecular Diffusion
Notion of concentration
Molecular diffusion, Fick’s Law
Mass balance
Transport analogies; salt-gradient solar
ponds
Simple solutions
Random walk analogy to diffusion
Examples of sources and sinks
Motivation
Molecular diffusion is often negligible in
environmental problems
Exceptions: near interfaces, boundaries
Responsible for removing gradients at
smallest scales
Analytical framework for turbulent and
dispersive transport
Concentration
Contaminant => mixture
„
„
„
Carrier fluid (B) and contaminant/tracer (A)
If dissolved, then solvent and solute
If suspended, then continuous and dispersed phase
Concentration (c or ρA) commonly based on
mass/volume (e.g. mg/l); also
„
„
mol/vol (chemical reactions)
Mass fraction (salinity): ρA/ρ
Note: 1 mg/l ~ 1 mg/kg (water) = 1ppm
Molecular Diffusion
One-way flux (M/L2-T)
z
= ρ A wA
(2)
Net flux
= wA ( ρ A1 − ρ A 2 )
lm
(1)
wA
c or ρΑ
= − wAl m ∂ρ A / ∂z
DAB
r
J A = − DAB ∇ρ A
Ficks Law and Diffusivities
r
J A = − DAB ∇ρ A
r
r ∂
r ∂
∂
∇( ) = ( ) i +
() j+ ()k
∂x
∂y
∂z
DAB is isotropic and essentially uniform
(temperature dependent), but depends on A, B
Table 1.1 summarizes some values of DAB
Roughly: Dair ~ 10-1 cm2/s; Dwater ~ 10-5 cm2/s
Diffusivities, cont’d
Diffusivities often expressed through Schmidt no.
Sc = ν/D
Roughly: νair ~ 10-1 cm2/s; νwater ~ 10-2 cm2/s
Scair ~ 1
Scwater ~ 103
Also: Prandlt no. Pr = ν/κ (κ = thermal cond.)
Add advection; total flux of A is:
r
r
N A = ρ A q − DAB ∇ρ A
macroscopic velocity vector
Conservation of Mass
z
Like a bank account except
expressed as rates:
∆z
(NA)x
∆y
∆x
y
(rate of) change in account =
(rate of) (inflow – out) +/(NA)x+∆x
(rate of) prod/consumption
x
Example for x-direction
in = ( N A ) x ∆y∆z
⎡
⎛ ∂N ⎞ ⎤
out = ( N A ) x + ∆x ∆y∆z = ⎢( N A ) x + ⎜ A ⎟∆x ⎥ ∆y∆z
⎝ ∂x ⎠ ⎦
⎣
⎡ ⎛ ∂N ⎞ ⎤
net in = ⎢− ⎜ A ⎟∆x ⎥ ∆y∆z
⎣ ⎝ ∂x ⎠ ⎦
Conservation of Mass, cont’d
z
Account balance
= ρ A ∆x∆y∆z
Rate of change of account balance
x
y
∂
= ( ρ A )∆x∆y∆z
∂t
Rate of production
= rA ∆x∆y∆z
Conservation of Mass, cont’d
z
Sum all terms (incl. advection in 3D)
x
y
∂ρ A ∂
∂
∂
+ (N A )x + (N A ) y + (N A )z
∂t ∂x
∂y
∂z
r
∂ρ A
+ ∇ ⋅ N A = rA
∂t
Flux divergence (dot product of two
vectors is scalar)
For carrier fluid B
r
∂ρ B
∂t
+ ∇ ⋅ N B = rB
Conservation of Mass, mixture
rA = − rB
r
r
r
N A + N B = ρq
r
∂ρ
+ ∇ ⋅ ( ρq ) = 0
∂t
∂ρ
≅ ∇ρ ≅ 0
∂t
r
∇⋅q = 0
Conservation of total mass
ρA +ρB =ρ
Liquids are nearly incompressible
Divergence = 0; Continuity
Conservation of Mass, contaminant
ρA = c
drop subscript A
r
∂c
+ ∇ ⋅ (cq ) = ∇ ⋅ ( D∇c) + r Conservative form of mass cons.
∂t
r r
∂c
+ c∇ ⋅ q + q ⋅ ∇c = D∇ 2 c + (∇D)(∇c) + r
∂t
0
0
∂c r
N.C. form
+ q ⋅ ∇c = D∇ 2 c∇ + r
∂t
r
∂c
2
= D∇ c∇
If q = r = 0 => Ficks Law of Diffusion
∂t
Heuristic interpretation of Advection and diffusion
c
t-1
U
t
u∆t
Advection
Flux ~ negative gradient
∆c
x
c
Diffusion
t
Difference in fluxes
(divergence) ~ curvature
t-1
J
J
x
Analogs
∂c
Ficks Law (mass transfer)
= D∇ 2 c
∂t
∂T
Fourier’s Law (heat transfer)
= κ∇ 2T
∂t
r
r
r
rm
∂q r
2r
+ ( q ⋅ ∇ ) q = υ∇ q +
Newton’s Law (mom. Transfer)
ρ
∂t
ν/D
Sc
Air
~1
Water
~103
ν/κ
Pr
~0.7
~8
D~κ∼ν
D<<κ<ν
Example: Salt Gradient Solar
Ponds (WE 1-1)
like El Paso Solar Pond
UCZ
GZ
~3m
S
T
dense brine
BCZ
Solar Pond, diffusive salt flux to UCZ
S = 50 o
Area = 10,000 m2
oo
Z1=0.3m
Z2=1.8m
UCZ
GZ
S
c = ρS
S = 250 o
T
BCZ
oo
cUCZ = (1033 Kg/m3)(50x10-3 Kg/Kg) = 52 Kg/m3
cBCZ = (1165 Kg/m3)(250x10-3 Kg/Kg) = 291 Kg/m3
J s = Ddc / dz
=(2x10-9)(239 Kg/m3)/1.2m = 4.0x10-7 Kg/m2-s
= 344 Kg/day
Solar Pond: diffusive thermal flux to UCZ
T1 =
z1
z2
30oC
T
φs(z) = φsn(1-β)e-ηz
T2 = 80oC
d 2T
0 = ρC pκ 2 + η (1 − β )φ sn e −ηz
dz
−φ *
T = 2 e −ηz + c1 z + c 2
η
φsn = 250 W/m2
β= fraction of φsn absorbed at surface
η = extinction coefficient
Cp = heat capacity (4180 J/KgoC);
κ = thermal diffusivity (1.5x10-7 m2/s)
φ * = η (1 − β )φ sn / ρC pκ
c1, c2 from T=T1 at z1, T=T2 at z2
Solar Pond: thermal flux
T1 =
z1
z2
30oC
φsn = 250 W/m2
T
φs(z) = φsn(1-β)e-ηz
T2 = 80oC
J t = ρC pκ (dT / dz ) z = z2
⎡ −ηz2 (e −ηz2 − e −ηz1 ) ⎤
T2 − T1
= ρC pκ
+ (1 − β )φsn ⎢e +
⎥
−
z 2 − z1
η
(
z
z
)
2
1
⎣
⎦
z1 = 0.3m; z2 = 1.5m, β=0.5, η = 0.6 => Jt = 7 W/m2
Compare with (1-0.5)(250)exp(-0.6*1.5) = 51 W/m2
reaching BCZ (~13% lost)
Rankine Cycle Heat Engine
Evaporator
Brine
Turbine
generator
G
Feed pump
Cooling water
Condenser
Figure by MIT OCW.
Solar Pond: total energy extraction
T1 =
z1
z2
φsn = 250 W/m2
30oC
T
φs(z) = φsn(1-β)e-ηz
T2 = 80oC
Energy Flux at surface
% of φsn
250 W/m2
100
Energy Flux reaching BCZ
51
20
Energy Flux extracted
34
14
Electricity extracted (theoretical)
4.8
2
Electricity extracted (net actual)
2.4
1
Carnot efficiency
ηc = (T2-T1)/(T2+273)
[24 KWe for 1 ha]
Simple Solutions
Inst. injection of mass M
A
x
∂c
∂ c
=D 2
∂t
∂x
bc : c = 0 at x = ±∞
M
ic : c = δ ( x) at t = 0 +
alternative
A
M
r = δ ( x)δ (t ) with c = 0 at t = 0
A
2
Simple Solutions, cont’d
Inst. injection of mass M
A
x
Solution by similarity transform (Crank, 1975) or inspection
c=
B
t
1/ 2
e
x2
−
4 Dt
∞
A ∫ cdx = M
−∞
M = 2 AB πD
c ( x, t ) =
M
e
x2
−
4 Dt
2 A πDt
Add a current
( x −ut )2
−
M
4 Dt
c ( x, t ) =
e
2 A πDt
Gaussian Solution
c
t
x
σx
x
Spatial Moments
interpretation
∞
mo = ∫ c( x, t )dx
−∞
∞
m1 = ∫ cxdx
M = mo A
m1
xc =
= ut
mo
−∞
Center of mass
2
∞
m2 = ∫ cx dx
2
−∞
Mass; indep of t
σx
2
m2 ⎛ m1 ⎞
=
− ⎜⎜ ⎟⎟ = 2 Dt
mo ⎝ m2 ⎠
Plume
variance
Spatial Moments, cont’d
Relationship of moments to equation parameters
∞
m2 =
∫ 2A
−∞
M
πDt
e
− x2
4 Dt
x 2 dt
M
= 2 Dt
A
m2 2 MDt / A
2
σx =
=
= 2 Dt
mo
M/A
Without current, odd moments are 0
Spatial Moments, cont’d
Rewrite in terms of σ
c ( x, t ) =
=
M
2 A πDt
e
x2
−
4 Dt
M
e
A 2π σ x
−
or in 3-D (isotropic)
M
c ( x, t ) =
e
3/ 2
8(πDt )
x2
2σ 2
Plume dilutes by spreading:
In 1-D, c ~ t-1/2 ~ σx-1
In 3-D, c ~ t-3/2 ~ σ-3
( x2 + y2 + z 2 )
−
4 Dt
−
M
=
e
3/ 2
3
(2π ) σ
( x2 + y 2 + z 2 )
2σ 2
Moment generating equation
∂c
∂ c
=D 2
∂t
∂x
2
∞
mi =
i
x
∫ cdx
−∞
Approach 1: moments of c(x,t) => σ2 = m2/mo = 2Dt
Approach 2: moments of ge => moment generation eq.
∞
i
x
∫ (each term) dx
−∞
Moment generating eq., cont’d
∞
∂c
∂ c
=D 2
∂t
∂x
2
∞
0th
moment
moment
i
x
∫ cdx
−∞
∞
dmo
∂c
∂
dx
c
dx
=
=
∫ ∂t ∂t −∫∞
dt
−∞
∞
∞
∂ 2c
⎛ ∂c ⎞
dx
D
D
=
⎜ ⎟ =0
∫−∞ ∂x 2
⎝ ∂x ⎠ −∞
0
∞
2nd
mi =
∞
dm2
∂
∂c
2
dx
x
c
dx
x
=
=
∫ ∂t ∂t −∫∞
dt
−∞
2
∞
∞
∞
∞
2
∂c
2 ⎛ ∂c ⎞
2 ∂ c
2
dx = −2 xD(c) ∞−∞ + 2 ∫ Dcdx = 2 Dmo
−
xD
dx
Dx
Dx
=
⎟
⎜
∫−∞ ∂x 2
∫
∂x
⎝ ∂x ⎠ −∞
−∞
−∞
0
0
Moment generating eq., cont’d
∂c
∂ c
=D 2
∂t
∂x
2
0th
moment
2nd
moment
dmo
=0
dt
∞
mi =
i
x
∫ cdx
−∞
=> mo = const = M/A
dm2
= 2 Dmo
dt
=> dσ2/dt = 2D or σ2 = 2Dt
dσ 2
mo
= 2 Dmo
dt
How fast is molecular diffusion?
Creating linear salinity distribution
from initial step profile
z
Assume 80 cm tank; 40 2cm steps
S
Time to diffuse:
σ2
Table 1-1
= 2Dt
t = σ2/2D ~(2cm)2/(2)(1.3x10-5 cm2/s)
= 1.5x105s ~ 2 days
slow!
If thermal diffusion (100 x faster), t < 1 hr
Spatially distributed sources
c
co
t
∂ 2c
∂c
0
=D 2
∂x
∂t
bc c = 0 at x = ∞
c = co at x = −∞
ic c = co
c=0
x
or c = co/2 at x = 0
for
x < 0 at t = 0
for
x > 0 at t = 0
Spatially distributed sources
co
ξ
dξ
x
0
dc(ξ , t ) =
∞
c ( x, t ) = ∫
x
dM
2 A πDt
c o dξ
2 πDt
e
e
−
−ξ 2
4 Dt
ξ2
4 Dt
=
co
=
2
c o dξ
2 πDt
e
−
x
ξ2
4 Dt
⎡
⎛ x ⎞⎤ co
⎛ x ⎞
⎟⎟⎥ = erfc⎜⎜
⎟⎟
⎢1 − erf ⎜⎜
⎝ 2 Dt ⎠⎦ 2
⎝ 2 Dt ⎠
⎣
Error Function
2
π
e
−α 2
erf(ω)
erfc(ω)
ω
erf (ω ) =
erfc(ω ) =
2
π
2
ω
∫e
−α 2
dα
erf (∞) = 1
0
∞
e
∫
π ω
erf (0) = 0
−α 2
dα
erfc( x) = 1 − erf ( x)
α
Example: DO in Fish aquarium (WE 1-4)
2(c2-c1)
C1
C2
t=0: c=c1=8 mg/l (csat at 27oC)
2C2 – C1
c
t
t>0: c(0)=c2=10 mg/l (csat at 16oC)
(
1
c( z ,t ) = c1 + 2(c 2 − c1 ) erfc z / 2 Dt
2
(
c − c1
= erfc z / 2 Dt
c 2 − c1
z
)
)
Evaluate at z =15 cm, D = 2x10-5 cm2/s (Table 1-1)
t
1hr
z/(4Dt)0.5
erfc[(z/(4Dt)0.5]
28
0
1d
5.7
10-15
1 mo
1.0
0.15
Again,
very slow!
Diffusion as correlated
movements
t
x=0 at t=0
x (t ) =
∫ u ( t ' ) dt '
0
p(x,t)
x
Analogy between p(x) and c(x); ergotic assumption
For many particles, both distributions become Normal
(Gaussian) through Central Limit Theorem
Statistics of velocity
u =0
mean velocity
u 2 = const.
variance
u (t )u (t − τ ) = u (0)u (τ ) = u (−τ )u (0) auto co-variance
u (t )u (t − τ )
u (t )
2
= R(τ )
auto correlation
R(τ)
1
0
τ
Statistics of position
t
x(t ) = ∫ u (t ' )dt '
0
t
t
0
0
x = ∫ u (t ' )dt ' = ∫ u (t )dt = 0
x 2 (t )
increases with time, as follows
t
⎡t
⎤
dx 2 (t )
dx
= 2 x(t )
= 2 ⎢ ∫ u (t ' )dt '⎥u (t ) = 2 ∫ u (t )u (t ' )dt '
dt
dt
0
⎣0
⎦
t
t
d x 2 (t )
= 2u 2 (t ) ∫ R (t − t ' )dt ' = 2u 2 (t ) ∫ R (τ )dτ
dt
0
0
d x 2 dσ 2
D=
=
= u 2 ∫ R(τ )dτ Taylor’s Theorem (1921); classic
2dt
2dt
0
t
[D] = [V2T]
Earlier, D = wAl m [VL] or
D=
σ2
2t
[L2/T]
Random Walk (WE 1-3)
Special case: u (t ) = U or − U direction changes randomly after ∆t
Walker’s position at time t=N∆t
Probability distribution
N
x = ∑ u∆t
1
p (χ , N ) =
0.5
N!
⎛ 1 ⎞
⎜ N⎟
+
N
χ
N
−
χ
⎛
⎞ ⎛
⎞ ⎝2 ⎠
⎜
⎟ ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
!
!
3/8 0.375
Bernoulli Distribution
2/8
Approaches Gaussian
for large N
0.25
1/8 0.125
0
-3
-3
χ = x/Ut 1
-1
1
-1
3
3
Example for N=3
Statistics of position
N
x (t ) = ∑ u ( t ) = 0
i =1
2
⎛
⎞
σ 2 = ⎜ ∑ u∆t ⎟ = ∆t 2 [(u 1 + u 2 K u i + K u N )(u 1 + u 2 + K u N )]
⎝ i =1
⎠
= N∆t 2U 2 = tU 2 ∆t = t∆x 2 / ∆t
N
σ2
U 2 ∆t ∆x 2
D=
=
=
2t
2
2∆t
[L2/T ]
Alternatively, derive D from Taylor’s Theorem
Examples of Sources and
Sinks (r terms)
1st order
0th order
2nd order
Coupled reactions
Mixed order
1st Order
c/co
Example:
radioactive decay
dc
= − kc
dt
c / co = e − kt
1
0.5
0.37
0.1
0
t1/2
k-1
t90
Half life e-folding time
Linearity => 1st O decay multiplies simple sol’n by e-kt; e.g.
Also very convenient in particle tracking models
t
0th Order
S/So
Example: silica uptake
by diatoms (high diatom 1
conc)
dS
= −B
dt
S = S o − Bt
0
S=substrate (silica) concentration
B=rate (depends on diatom
population, but assume large)
So/B
t
2nd Order
Example: particleparticle
collisions/reactions;
flocculant settling)
dc
= − Bc 2
dt
1
c
=
co 1 + Btco
c/co
1
0
co
Behavior depends on co; slower than e-kt.
Can be confused with multiple species undergoing
1st order removal
t
Coupled Reactions
Example: Nitrogen oxidation
dN1
= − K12 N1
dt
dN 2
= K12 N1 − K 23 N 2
dt
dN 3
= K 23 N 2
dt
N1 = NH3-N
N2 = NO2-N
N3 = NO3-N
If N’s are measured as molar quantities, or atomic
mass, then successive K’s are equal and opposite
Mixed Order—Saturation Kinetics
(Menod kinetics)
Example: algal uptake of nutrients—focus on algae
rmax
dc
kS
=
dt S + S o
Rmax/2
c = algal concentration
S = substrate concentration
So = half-saturation const
S << So =>
S >> So =>
r
So
dc kS
≅
= k'S
dt S o
(1st Order)
dc
≅k
dt
(0th Order)
S
1 Wrap-up
Molecular diffusivities
Molecular motion; Eulerian frame
D = wl m
dσ
D=
2
2dt
2
Method of moments
∞
D = u 2 ∫ R (τ )dτ
Molecular motion; Lagrangian frame
0
D is “small” ~ 1x10-5 cm2/s for water
Inst. point source solutions are Gaussian; other
solutions built from
„
Spatial and temporal integration, coordinate translation,
linear source/sink terms