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

(2) = ρ AwA Net flux l m = wA (ρ A1 − ρ A2 )

wA (1) = −wAl m∂ρ A / ∂z D r AB J = −D ∇ρ c or ρΑ A AB 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 -1 2 -5 2 Roughly: Dair ~ 10 cm /s; Dwater ~ 10 cm /s Diffusivities, cont’d

Diffusivities often expressed through Schmidt no. Sc = ν/D

-1 2 -2 2 Roughly: νair ~ 10 cm /s; νwater ~ 10 cm /s

Scair ~ 1 3 Scwater ~ 10 Also: Prandlt no. Pr = ν/κ (κ = thermal cond.) Add advection; total flux of A is: r r N A = ρ Aq − DAB∇ρ A macroscopic velocity vector Conservation of Mass

z Like a bank account except expressed as rates: (rate of) change in account = ∆z (N ) (rate of) (inflow – out) +/- (NA)x A x+∆x x (rate of) prod/consumption ∆y Example for x-direction ∆x in = (N ) ∆y∆z y A x ⎡ ⎛ ∂N A ⎞ ⎤ out = (N A ) x+∆x ∆y∆z = ⎢(N A ) x + ⎜ ⎟∆x⎥∆y∆z ⎣ ⎝ ∂x ⎠ ⎦

⎡ ⎛ ∂N A ⎞ ⎤ net in = ⎢− ⎜ ⎟∆x⎥∆y∆z ⎣ ⎝ ∂x ⎠ ⎦ Conservation of Mass, cont’d z Account balance

= ρ A∆x∆y∆z Rate of change of account balance ∂ x = (ρ )∆x∆y∆z ∂t A Rate of production y = rA∆x∆y∆z Conservation of Mass, cont’d z Sum all terms (incl. advection in 3D) ∂ρ ∂ ∂ ∂ A + (N ) + (N ) + (N ) ∂t ∂x A x ∂y A y ∂z A z ∂ρ r A + ∇ ⋅ N = r ∂t A A x Flux divergence (dot product of two vectors is scalar)

y For carrier fluid B ∂ρ r B + ∇ ⋅ N = r ∂t B B Conservation of Mass, mixture

rA = −rB Conservation of total mass r r r N A + N B = ρq ρA +ρB =ρ ∂ρ + ∇ ⋅(ρqr) = 0 ∂t ∂ρ ≅ ∇ρ ≅ 0 Liquids are nearly incompressible ∂t ∇ ⋅qr = 0 Divergence = 0; Continuity Conservation of Mass, contaminant

ρ A = c drop subscript A ∂c + ∇ ⋅(cqr) = ∇ ⋅(D∇c) + r Conservative form of mass cons. ∂t ∂c + c∇ ⋅qr + qr ⋅∇c = D∇2c + (∇D)(∇c) + r ∂t 00 ∂c + qr ⋅∇c = D∇2c∇ + r N.C. form ∂t

∂c 2 = D∇ c∇ If qr = r = 0 => Ficks Law of Diffusion ∂t Heuristic interpretation of Advection and diffusion

c t-1 t Advection U u∆t Flux ~ negative gradient ∆c

x c Diffusion t Difference in fluxes (divergence) ~ curvature t-1 JJ x Analogs

∂c = D∇2c Ficks Law (mass transfer) ∂t ∂T = κ∇2T Fourier’s Law (heat transfer) ∂t ∂qr rr + (qr ⋅∇)qr =υ∇2qr + m Newton’s Law (mom. Transfer) ∂t ρ Air Water ν/DSc ~1 ~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 ST BCZ

dense brine Solar Pond, diffusive salt flux to UCZ S = 50 o oo Area = 10,000 m2 UCZ Z1=0.3m GZ Z2=1.8m ST BCZ

S = 250 o oo c = ρS 3 -3 3 cUCZ = (1033 Kg/m )(50x10 Kg/Kg) = 52 Kg/m 3 -3 3 cBCZ = (1165 Kg/m )(250x10 Kg/Kg) = 291 Kg/m

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

2 o φsn = 250 W/m T1 = 30 C z 1 T

z2 ηz φs(z) = φsn(1-β)e-

o T2 = 80 C β= fraction of φsn absorbed at surface 2 d T −ηz η = extinction coefficient 0 = ρC pκ 2 +η(1− β )φsn e o dz Cp = heat capacity (4180 J/Kg C); −φ * κ = thermal diffusivity (1.5x10-7 m2/s) T = e −ηz + c z + c η 2 1 2 φ* = η(1− β )φsn / ρC pκ

c1, c2 from T=T1 at z1, T=T2 at z2 Solar Pond: thermal flux

2 o φsn = 250 W/m T1 = 30 C z 1 T

z2 ηz φs(z) = φsn(1-β)e-

o T2 = 80 C J = ρC κ (dT / dz) t p z=z2 ⎡ −ηz2 −ηz1 ⎤ T2 −T1 −ηz2 (e − e ) = ρC pκ + (1− β )φsn ⎢e + ⎥ z2 − z1 ⎣ η(z2 − z1) ⎦

2 z1 = 0.3m; z2 = 1.5m, β=0.5, η = 0.6 => Jt = 7 W/m 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

Feed pump G

Cooling water

Condenser

Figure by MIT OCW. Solar Pond: total energy extraction

2 o φsn = 250 W/m T1 = 30 C z 1 T

z2 ηz φs(z) = φsn(1-β)e-

o T2 = 80 C % of φsn Energy Flux at surface 250 W/m2 100 Energy Flux reaching BCZ 51 20 Energy Flux extracted 34 14 Carnot efficiency η = (T -T )/(T +273) Electricity extracted (theoretical) 4.8 2 c 2 1 2 Electricity extracted (net actual) 2.4 1 [24 KWe for 1 ha] Simple Solutions

Inst. injection of mass M

A x ∂c ∂ 2c = D ∂t ∂x2 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 Simple Solutions, cont’d

Inst. injection of mass M

A x Solution by similarity transform (Crank, 1975) or inspection

2 x2 x B − M − c = e 4Dt c(x,t) = e 4Dt t1/ 2 2A πDt ∞ Add a current A cdx = M ( x−ut )2 ∫ − −∞ M 4 Dt c(x,t) = e M = 2AB πD 2A πDt Gaussian Solution c

t

x

σx x Spatial Moments interpretation ∞ m = c(x,t)dx M = m A Mass; indep of t o ∫ o −∞ m ∞ x = 1 = ut Center of mass m = cxdx c 1 ∫ mo −∞ 2 ∞ 2 m ⎛ m ⎞ 2 2 1 Plume m = cx dx σ x = − ⎜ ⎟ = 2Dt 2 ∫ m ⎜ m ⎟ variance −∞ o ⎝ 2 ⎠ Spatial Moments, cont’d

Relationship of moments to equation parameters

2 ∞ M − x m = e 4Dt x2dt 2 ∫ −∞2A πDt M = 2 Dt A

2 m2 2MDt / A σ x = = = 2Dt mo M / A

Without current, odd moments are 0 Spatial Moments, cont’d

Rewrite in terms of σ or in 3-D (isotropic) 2 2 2 x2 (x + y +z ) M − M − c(x,t) = e 4Dt c(x,t) = e 4Dt 2A πDt 8(πDt)3 / 2 x2 2 2 2 − (x + y +z ) M 2 − = e 2σ M 2σ 2 = 3 / 2 3 e A 2πσ x (2π ) σ

Plume dilutes by spreading:

-1/2 -1 In 1-D, c ~ t ~ σx In 3-D, c ~ t-3/2 ~ σ-3 Moment generating equation

∂c ∂ 2c ∞ = D m = x i cdx 2 i ∫ ∂t ∂x −∞

Approach 1: moments of c(x,t) => σ2 = m2/mo = 2Dt Approach 2: moments of ge => moment generation eq. ∞ ∫ xi (each term) dx −∞ Moment generating eq., cont’d ∂c ∂ 2c ∞ = D m = xicdx 2 i ∫ ∂t ∂x −∞ ∞ ∂c ∂ ∞ dm 0th ∫ dx = ∫ cdx = o −∞ ∂t ∂t −∞ dt ∞ ∞ ∂2c ⎛ ∂c ⎞ moment D dx = D = 0 ∫ 2 ⎜ ⎟ −∞ ∂x ⎝ ∂x ⎠−∞ ∞ 0 ∞ ∂c ∂ dm ∫ x2 dx = ∫ x2cdx = 2 2nd −∞ ∂t ∂t −∞ dt ∞ 2 ∞ ∞ ∞ 2 ∂ c 2 ⎛ ∂c ⎞ ∂c ∞ moment Dx dx = Dx ⎜ ⎟ − 2 xD dx = −2xD(c)−∞ + 2 Dcdx = 2Dmo ∫ ∂x2 ⎝ ∂x ⎠ ∫ ∂x ∫ −∞ 0 −∞ −∞ 0 −∞ Moment generating eq., cont’d ∂c ∂ 2c ∞ = D m = x i cdx 2 i ∫ ∂t ∂x −∞

th 0 dmo = 0 => mo = const = M/A moment dt dm 2 = 2Dm 2nd o dt => dσ2/dt = 2D or σ2 = 2Dt dσ 2 moment m = 2Dm o dt o How fast is molecular diffusion?

z Creating linear salinity distribution from initial step profile Assume 80 cm tank; 40 2cm steps

S Table 1-1 Time to diffuse: σ2 = 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

∂c ∂2c x = D 0 ∂t ∂x2 bc c = 0 at x = ∞

c = co at x = −∞ or c = co/2 at x = 0 ic c = co for x < 0 at t = 0 c = 0 for x > 0 at t = 0 Spatially distributed sources

c o ξ dξ

0 x x ξ 2 ξ 2 dM − c dξ − dc(ξ,t) = e 4Dt = o e 4Dt 2A πDt 2 πDt 2 ∞ −ξ ⎡ ⎤ co dξ 4Dt co ⎛ x ⎞ co ⎛ x ⎞ c(x,t) = ∫ e = ⎢1− erf ⎜ ⎟⎥ = erfc⎜ ⎟ x 2 πDt 2 ⎣ ⎝ 2 Dt ⎠⎦ 2 ⎝ 2 Dt ⎠ 2 2 Error Function e−α π

erf(ω) erfc(ω)

α ω

ω 2 2 erf (ω) = e−α dα erf (0) = 0 π ∫ 0 erf (∞) =1 ∞ 2 2 erfc(ω) = ∫ e−α dα erfc(x) =1− erf (x) π ω Example: DO in Fish aquarium (WE 1-4)

2(c2-c1) o t=0: c=c1=8 mg/l (csat at 27 C) C C 2C – C o 1 2 2 1 t>0: c(0)=c2=10 mg/l (csat at 16 C) c 1 t c( z,t ) = c + 2()c − c erfc(z / 2 Dt ) 1 2 2 1 c − c 1 = erfc(z / 2 Dt ) c2 − c1 z Evaluate at z =15 cm, D = 2x10-5 cm2/s (Table 1-1) tz/(4Dt)0.5 erfc[(z/(4Dt)0.5] 1hr 28 0 Again, -15 1d 5.7 10 very slow! 1 mo 1.0 0.15 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 ) = R(τ ) auto correlation u(t)2 R(τ)

1

0 τ Statisticst of position x(t) = ∫ u(t')dt' 0 t t x = ∫ u(t')dt' = ∫ u(t)dt = 0 0 0 x 2 (t) increases with time, as follows dx 2 (t) dx ⎡ t ⎤ t = 2x(t) = 2⎢∫ u(t')dt'⎥u(t) = 2∫ u(t)u(t')dt' dt dt ⎣ 0 ⎦ 0 d x 2 (t) t t = 2u 2 (t)∫ R(t − t')dt' = 2u 2 (t)∫ R(τ )dτ dt 0 0 d x 2 dσ 2 t D = = = u 2 ∫ R(τ )dτ Taylor’s Theorem (1921); classic 2dt 2dt 0 [D] = [V2T] σ 2 Earlier, D = w [VL] or D = [L2/T] Al m 2t Random Walk (WE 1-3)

Special case: u(t) = U or −U direction changes randomly after ∆t N Walker’s position at time t=N∆t x = ∑u∆t 1 Probability distribution N ! ⎛ 1 ⎞ p()χ, N = ⎜ ⎟ ⎛ N + χ ⎞ ⎛ N − χ ⎞ ⎝ 2N ⎠ ⎜ ⎟ ⎜ ⎟ 2 ! 2 ! 0.5 ⎝ ⎠ ⎝ ⎠

3/8 0.375 Bernoulli Distribution

2/8 0.25 Approaches Gaussian 1/8 0.125 for large N

0 Example for N=3 -3 -1χ = x/Ut 1 3 -3 -1 1 3 Statistics of position

N x()t = ∑u( t ) = 0 i=1

N 2 2 ⎛ ⎞ 2 σ = ⎜∑u∆t⎟ = ∆t []()u1 + u2 Kui +Ku N (u1 + u2 +Ku N ) ⎝ i=1 ⎠ = N∆t 2U 2 = tU 2 ∆t = t∆x 2 / ∆t

σ 2 U 2∆t ∆x2 D = = = [L2/T ] 2t 2 2∆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

Example: c/co radioactive decay 1

dc = −kc 0.5 dt 0.37 −kt 0.1 c / co = e 0 t -1 t1/2 k 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 0th Order

Example: silica uptake S/So by diatoms (high diatom 1 conc)

dS = −B dt S = S − Bt o 0 t S /B S=substrate (silica) concentration o B=rate (depends on diatom population, but assume large) 2nd Order

Example: particle- c/co particle 1 collisions/reactions; flocculant settling) dc = −Bc2 dt co c 1 0 = t co 1+ Btco -kt Behavior depends on co; slower than e . Can be confused with multiple species undergoing 1st order removal Coupled Reactions

Example: Nitrogen oxidation

dN 1 = −K N N = NH -N dt 12 1 1 3 dN 2 = K N − K N N = NO -N dt 12 1 23 2 2 2 dN 3 = K N N = NO -N dt 23 2 3 3 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

r dc kS rmax = dt S + So Rmax/2

c = algal concentration S = substrate concentration So S So = half-saturation const dc kS ≅ = k'S (1st Order) S << So => dt So S >> S => dc th o ≅ k (0 Order) dt 1 Wrap-up Molecular diffusivities

D = wl m Molecular motion; Eulerian frame dσ 2 D = 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