Electrical streaming potential generated by 2-phase flow

J.D. Sherwood Department of Applied Mathematics & Theoretical Physics University of Cambridge

IMA: 10 December 2009

E. Lac, Schlumberger Doll Research

1 Electrical double layers: equilibrium

Electrical potential φ + − + − + Thermal equilibrium: + y + − − + − + − i eziφ + − n = n∞ exp − − + − − − i kT − − + − − − − −   Poisson’s equation: + + + +++++++++++ ∇2φ = −ρ/ǫ Ions above positively charged ezin∞ − surface = e eziφ/kT ǫ i X Ion valence z , charge ez i i 1–1 electrolyte Ion number density ni i At infinity ni → n∞ eφ 2e2 eφ ∇2 = sinh Charge density ρ = i ezini + − kT ǫkT kT 1–1 electrolyte: n∞ = n∞ = n∞     P

2 Electrical double layers: equilibrium (2)

1D solution: φ = ζe−κy If φˆ = eφ/kT ≪ 1, linearize: 2 2 2 i ρ = −ǫκ φ e z n∞ 2 i 2 eζ − y ∇ φ = φ = κ φ n+ = n∞ 1 − e κ ǫkT kT i X −  eζ −κy n = n∞ 1 + kT e −1 where κ is Debye length   Wall kT/e ≈ 25 mV σ = −ǫn.∇φ = ǫκζ

Wall potential ζ Total charge in cloud: ∞ Typically eζ/kT < 5 − 0 ρ dy = ǫκζ R

3 Standard electrokinetic problems

Electro-osmosis + + + + + + + + + + wall −1 charge cloud κ

E u E +Q

+ + + + + + + + + + negative charge cloud Stokes eqn:

0 = µ∇2u −∇p + ρE −1 Plane 2-D channel, width 2h ≫ κ Hence At wall y = 0 ǫζE −κy 2 2 u = e − 1 ∂ u ǫκ ζE − µ = e κy ∂y2 µ 

4 Electrokinetic problems (2)

Electrophoresis + + + + + + + + + + wall −1 charge cloud κ

E +Q u

+Q + + + + + + + + + +

E Sphere (ζp) in capillary (ζc)

Thin double layer: apparent slip at wall BC: utan = Etanǫζ/µ Electric field: irrotational velocity ∇2 ∇ φ = .E = 0 ǫζ u = E with E.n = 0 on surface of particle µ

5 Electrokinetic problems (3)

Streaming Potential Sedimentation potential

+ + + + + + + + + + wall negative −1 charge cloud charge cloud κ

u E +Q

+ + + + + + + + + + mg

Charge cloud convected to right Dipole field Return current: ∆φ between top & bottom via Ohmic conduction in liquid of column of sedimenting or some other route particles

6 Governing equations: the deformed charge cloud

Ions move under electric & thermodynamic forces

i vi = u + ωi(−ezi∇φ − kT∇ ln n )

where u is fluid velocity, 2 2 i ωi is ionic mobility, conductivity σ = i e zi n∞ωi No reactions. Conservation of ions P

i i i i ∇.(n vi)= ∇. n u − ωi(ezin ∇φ + kT∇n ) = 0

For typical fluid velocity U and ion mobility ω  P´eclet number measures ratio U convective forces Pe = ∼ ωkTκ diffusive forces

7 The deformed charge cloud (2)

0.1 moles/litre NaCl in water Debye length

−1/2 2 2 i − e z n∞ κ 1 = i ≈ 1 nm ǫkT i ! X Mobility ω = 4 × 1011 mN−1s−1 Hence ωkTκ ≈ 1.7 m s−1 P´eclet number Pe = U/(ωkTκ) ≪ 1

In non-aqueous fluids: Debye length larger, e.g κ−1 = 600 nm (Prieve 2008) mobility smaller, e.g. ω = 4 × 1010 mN−1s−1 Set U =Γκ−1 Pe =Γ/(ωkTκ2), with ωkTκ2 = 460 s−1

8 The deformed charge cloud (3) Look for O(Pe) perturbations

i i i Ion density n = n0 + n1 + · · · φ = φ0 + φ1 + · · · Charge density ρ = ρ0 + ρ1 + · · ·

Stokes equation: 0 = −∇P + µ∇2u − ρ∇φ

2 At O(Pe) µ∇ u1 = ∇p1 + ρ1∇φ0 + ρ0∇φ1 if potentials small

N 2 i 2 µ∇ u1 = ∇ p1 − kTn1 + ρ0∇(φ1 + ρ1/ǫκ ) ! Xi=1 Perturbed velocity field of magnitude

2 Un∞ eζ Uǫζ2 u ∼ ∼ 1 ωκ2µ kT ωµkT   9 The deformed charge cloud (4)

Perturbed velocity u1/U = O(H) where electric Hartmann number

ǫζ2 eζ 2 ǫkT H = = ωµkT kT µωe2   In water, ǫkT/µωe2 ≈ 0.28

10 Streaming potential: single phase

y In pipe of radius R, pressure u gradient G = −8µQ/πR4 charge κ du dp R cloud = dr dz 2µ ∞ Convected current Ic = 0 ρu dy Convected current Approximate u by y ∂u ∂y R dp πR2 ∞ 2πRIc = −ζǫ ∂u dz µ Ic = yρ dy ∂y 0 Streaming potential gradient for Z ∞ ∂u d2φ return current through fluid = −ǫ y dy ∂y dy2 (conductivity σ) Z0 ∂u = −ζǫ dφ ζǫ dp ∂y s = dz µσ dz

11 Streaming potential: single phase (2)

Integrate:

ζǫ φ = p (1) s µσ   Non-uniform shear rate Ic charge cloud Is ∂u u ∇ . −ζǫ = −σn.∇φ s ∂y s  

Suppose ζ potential uniform All surfaces rigid, no-slip Impose velocity u and pressure p Can show (non-trivial) (1) holds for arbitrary geometry

12 Streaming potential: two-phase

Rock pump u Water Oil surface E w Eo

Rock charge cloud

Presence of 2nd phase changes well a) wall shear rate b) electrical conductance oil c) streaming potential water

2R 2d V U Potential difference between base z 1 z2 of well and infinity changed by arrival of water Need to understand: Pore-scale Need very stable electrodes Darcy scale Reservoir scale Morgan et al. (1989)

13 Tight-fitting rigid sphere

r Integral form of momentum eqn

2R c z over z1 < z < z2

y Rp h h0 in frame of particle, leakage flux

z1 z 2 4 5h0 Rigid sphere, radius Rp, velocity U Q = − πR h U 1 − + · · · p 3 c 0 3R Capillary radius Rc = Rp + h0  c  Pressure drop across sphere Potential ζp on particle ζc on capillary wall 4πµdU p1 − p2 = p(z1) − p(z2)= 1/2 h0Rc lubrication theory d = (2Rph0)

14 Tight-fitting rigid sphere (2)

sphere Lubrication theory for streaming potential u(y) 1/2 ǫπU (2R h ) capillary wall − p 0 − z φ1 φ2 = 2 (ζc ζp) 2σ h0

If ζc = ζp expect 2R 2d V U

z 1 z2 ǫζ ǫπUζ 16 φ −φ = c (p −p )= c 1 2 1 2 1/2 Spherical bubble: σµ 2σ (2h0Rc) 4πµU For uncharged stress-free bubble p − p = 1 2 1/2 (spherical or Bretherton) (2h0Rp)

ǫζ Bretherton bubble φ − φ = c (p − p ) 1 2 σµ 1 2 γ 3µU 2/3 p −p = 4.81 1 2 R γ c   15 BCs at liquid drop/bubble

Charge at mobile interface? Ionic surfactants V capillary wall

Baygents & Saville (1991) −1 κ charge cloud

Charged mercury drop (Ohshima et al. u h external fluid 1984) 0

−1 Perfect conductor, uniform potential κ charge cloud

No ion discharge at interface drop interface us Tangential motion of interface enhances convective flux of ions

E Q Gas — similar problems Spinning drop interfacial tensiometer (JDS, JFM 162)

16 Small drop in tube

2R Charged drop, potential ζd z 0 viscosity µi = λµ radius a

∆φ between tube ends: z z r charge cloud dipole perturbation θ

Velocity field: Hetsroni et al. (1970) a R0 b unperturbed Poiseuille flow (0) 2 2 v = ˆz(1 − R /R0)U0

Need no-slip on drop (velocity V) r sin θ R Φ Φ (1) c s add perturbation velocity v a R b v(1) fails to satisfy no-slip on capillary wall 0 add v(2) etc

17 Small drop in tube (2) Thin double layer aκ ≫ 1 Drop viscosity λµ

3 8U ζ ǫ a aκ 1 + 3λ − ∆φ = − 0 d + + O(aκ) 1 R σ R 2 + 3λ 2 + 3λ 0  0   

Uncharged drop, capillary at ζc 5 2 4 Shear rate at wall O(U0a /R0r ) Drop force free: Stokeslet = 0 Convective ion flux 5 2 4 on centreline: stresslet = 0 O(ǫζU0a /R0r ) Current normal to wall Far-field perturbation 5 2 5 O(ǫζU0a /R0r ) 5 5 U0a a Integrate once to get change in u ∼ ∼ U 2 3 σ µ 0 6 R0r R0 axial current Integrate again decays on lengthscale R0 5 ǫζU0 a Additional pressure drop ∆φ ∼ 5 6 σR0 R0 ∆p ∼ µU0a /R0   18 Small drop in tube (3)

2 Electric field when no drop: E = 4U0ǫζc/R σ Dipole perturbation due to insulating drop

a3 φ = − 1 + E.r 2r3   Streaming potential change

8U ζ a 3 16Uζ a 3 ∆φ = 0 c = c σR R σR R 0  0  0  0  where U = U/2 is average flow rate

19 Viscous Bretherton drop

2R 2d V U Fluid (mean) velocity U Drop velocity V z 1 z2 Drop cylindrical radius d = δR = R − h V Drop volume 4πa3/3 = 4π(αR)3/3 u s Drop length L ≈ 4a3/3d2

Bretherton: h = 1.337R Ca2/3 valid λ ≪ Ca−1/3 Park & Homsy (1984) Hodges et al. (2004) for higher viscosity Drop viscosity λµ Continuity of stress & velocity Interfacial tension γ R2 dp 8U Capillary number = − 4 −1 − Ca = µU/γ µ dz 1 + δ (λ 1)

20 Viscous Bretherton drop (2)

2 7 10 6 5 λ = 0.01 2 0 1 5 0.5

0.2 U 4 / p V ∆ 0.1 -2 3 0.1 0.2 2 0.5

λ = 0.01 10 -4 1

0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 h h

∆p = pressure increase due to drop V = drop velocity

− 3R4 δ2(λ 1 − 1) V 2 + δ(λ−1 − 2) ∆p = − = 32Ua3µ 1 + δ4(λ−1 − 1) U 1 + δ4(λ−1 − 1)  

21 Viscous Bretherton drop (3)

Change in streaming potential caused by uncharged drop

32Uǫζ a3 1 + (δ4 − δ2)(λ−1 − 1) ∆φ = c 3R4σ (1 − δ2)[1 + δ4(λ−1 − 1)]

∆φ = 0 when 4 1/2 2δ2 = 1 ± 1 − λ−1 − 1   real roots if λ ≤ 1/5 Change in potential above that expected for pressure change

ǫζ ∆p 32Uǫa3ζ ∆φ − c = c µσ 3R4σ(1 − δ2)[1 + δ4(λ−1 − 1)]

22 Viscous Bretherton drop (4)

V 2 + δ2(λ−1 − 2) = 4 −1 100 − a U 1 + δ (λ 1) 80 ( )

w 60 fr µU/R Expand for δ ≪ 1 40 20 − 0 V − 2U = −δ2(2 − λ 1)U 4 −1 6 −2δ (λ − 1)U + O(δ U) (b) -2 w fx -4 µU/R V < 2U if λ> 1/2 -6 (c) -8 Normal stress balance e -4 -2 0 2 4 V − 2U γ a) wall normal stress µ ∼ R Rδ b) steady drop Hence c) wall shear stress α = 1.1, Ca = 0.05 − − − δ ∼ (2 − λ 1) 1/3Ca 1/3 λ> 1/2 λ = 10 − ∼ Ca 1/5 λ = 1/2

23 Boundary integral computations

λ=10

λ=1

λ=0.1

Steady drop profiles Ca=0.05 λ = 0.1, 1, 10 4 3 Drop volume 3 πa α = a/R = 0.6, 0.8, [0.1], 1.3

24 Boundary integral computations (2)

λ = 10 λ = 0.1 (a) (c)

(b) (d)

Steady drop profile at increasing capillary number (a) λ = 10, α = 0.8, Ca = 0.05, 0.1, [0.1], 0.5 (b) λ = 10, α = 1.1, Ca = 0.05, 0.1, 0.2, 0.3 (c) λ = 0.1, α = 0.8, Ca = 0.05, 0.2, 0.5, 1, 2 (d) λ = 0.1, α = 1.1, Ca = 0.05, 0.2, 0.5, 2.35

25 BI computations: gap width h = 1 − d

1 1 (a) Film thickness hˆ = h/R vs. Ca λ=1 λ=10 λ=0.5 λ=0.2 λ=0.1 λ dotted: 2/3 hˆ λ=0.01 δ Bretherton hˆ ≈ 1.3375 Ca

0.1 (b) drop breadth δ = 1 − hˆ vs. Ca same λ as in (a) (a) (b) dashed ∼Ca−1/3 0.01 0.1 −1/5 0.01 0.1 1 10 0.01 0.1 1 10 dotted ∼Ca Ca Ca • experiments Taylor (1961)

26 BI computations: velocity

0 0 10 10 • α=0.8 1 ˆ V − −

ˆ 10 V λ= -1 -1 10  α=1.1 2 10 λ=1 λ=0.1 λ=0.5 λ=1 (a) λ=10 (b)

-2 -2 10 10 0.001 0.01 0.1 1 0.001 0.01 0.1 1 Ca Ca (a) Drop velocity Vˆ = V/U vs. Ca for λ = 0.1, 1, 10 • α = 0.8  α = 1.1 Dotted: Bretherton bubble, Vˆ ≈ 1 + 1.29 (3 Ca)2/3

(b) 2U − V for α = 1.1 and λ > 0.5 dashed ∼Ca−2/3 dotted ∼Ca−4/5

27 BI computations: velocity (2)

V/U vs. film thickness h Dotted: approximate limit values λ=0.01 2.5 λ=0.1 of h and V for λ< 0.5 λ=0 λ=0.2 V λ=0.3 − U 2 2 2 1 2λ δ∞ = 5 1 − λ λ=0.5   1.5 λ=1 λ=10 Ca ◦ last point for drop λ 6 1 λ→∞ 1 0 0.2 0.4 0.6 0.8 1 • Soares et al. (2005) h/R = 1 − δ infinite finger, Ca ≫ 1 λ−1 = 103, 12, 4

28 BI computations: instability

ˆt=1 7

2 8

4 9

5 10

6 11.6

Breakup of low-viscosity drop, λ = 0.1 α = 1.1 Ca increased from 1 to 2 at ˆt = tU/R = 0 dashed : steady drop profile at Ca = 1 Experiments: Olbricht (λ = 0.0013?) Annual Rev. Fluid Mech. 28 (1996) 187

29 BI computations: instability (2)

-1 10 max ] n time · ′ u [

-2 10

(b)

-3 10 (a) 0 200 400 600 800 1000 ˆt = Ut/R Travelling wave instability, in centre-of-mass frame λ = 10, α = 1.1, Ca = 0.5 Capillary wall shown on first (bottom) profile ′ (b) maximum normal velocity (u .n)max on drop surface solid line: N = 100, ∆ˆt = 0.005 dashed: N = 160, ∆ˆt = 0.002

30 BI computations: instability (3)

Ca=0.3

-1 10 Ca=0.50 0.4 -2 10 Ca=0.40

max -3 0 6 n 10 . · ′

u -4 10 Ca=0.30 0.8

-5 10 Ca=0.25 1.0 -6 10 Ca=0.20

-7 1.2 10 0 200 400 600 800 1000 Ut/R Maximum normal velocity, Drop profiles λ = 10, α = 1.1 λ = 10, α = 1.1 Only Ca = 0.3 is steady Various Ca, started from steady Kuramoto-Sivashinsky eqn state at lower Ca Papageorgiou et al. (1990)

31 BI computations: instability (4)

2 100 L/R 1 h/R = 1 − δ −3/2 10 2Rδ/L 2 Cac 0.5 1 −1/2

0.1 (a) (b)

0.1 0.01 0.5 1 2 0.1 0.5 1 2 α α

Critical capillary number Cac vs drop size α = a/R drop length L λ = 10 film thickness h error bars: slenderness ratio 2Rδ/L last stable and first unstable Ca at last stable Ca

32 BI computations: pressure

0.8 0 Additional pressure drop ∆p vs. Ca

∆p 1.1  µU/R ◦ α = 0.8 α = 1.1 ⋄ α = 2 -50 (a) λ = 0.1 (dashed)

2 -100 α= (b) λ = 1 (dot-dash) λ = 10 solid (a) λ = 0.1

-150 Drop length (spherocylinder) 0.001 0.01 0.1 1 Ca 4 3 3 2 l = 3 (a − d )/d 2 10 Increase in pressure due to drop

1 10 λ = 10 ∆p 3 −1 −1 −2 3 −2 3 ∆p ∼ α (1−λ )(2−λ ) / Ca / µU/R U R 0 µ / 10 λ = 1

−2/3 -1 10 dotted lines ∼ Ca − (b) λ = 1: end caps R∆p/µU ∼ Ca 5/3 -2 10 −5/3 0.001 0.01 0.1 1 thick dashed line ∼ Ca Ca

33 BI computations: pressure (2)

200 λ→∞ Additional pressure drop ∆p vs. film

10 100 λ= thickness h (α = 2) ∆p µU/R λ=1 Thin lines: asymptote 0

λ=0.5 λ=0.2 (cyl) 2 -100 ∆p 32 (1 − λ) δ λ=0.1 ≈− 3− 3 Ca 4 (α δ ) -200 λ=0.01 U R 3 1 − λ=0 µ / λ + ( λ) δ 0 0.2 0.4 0.6 0.8 1 h/R ◦ numerical results Dotted: approximate limit film thickness (λ< 1/2)

2 2 1 − 2λ δ∞ = λ = 0.01, α = 1.1, Ca = 3.5 5 1 − λ  

34 Streaming potential ∆Φˆ = ∆Φ(σR/ǫζcU) vs. drop size

Ca=0 Ca=0 0.05 0.10 20 0.20 (a) λ = 0.1 0.05 2 10 (c) λ = 10 0.10 0 0.30 ˆ ∆Φ 2 ˆ 10 ∆Φ 0.50 0.20 1 1 10 -20 10

0 10 0.40

-1 10 -40 -2 0 10 1 10

-3 10

-4 -60 10 0.01 0.1 1 -1 0 0.5 1 1.5 2 10 0.1 0.5 1 2 α = a/R α = a/R

0.05 Ca=0 Wall potential ζc 2 10 (b) λ = 1 0.10 λ = 0.1, 1, 10 ∆Φˆ 0.20 ˆ 3 1 Dotted line: 16 for small 10 0.50 ∆φ = α spherical drops

0 10 thick dashed line in (b) and (c): asymptote 32 α3 for Ca ≫ 1 -1 3 10 0.1 0.5 1 2 when > 1 2 (long drops) α = a/R λ /

35 ∆Φˆ − ∆ˆp = ∆Φ(σR/ǫζcU) − ∆p(R/µU)

Ca=0 0.05 0.10 Ca=0 2 2 10 0.05 10 p p ∆ˆ ∆ˆ 0.20 0 10

. − − 1 1 ˆ ˆ Φ Φ 10 10 0.50 ∆ ∆ 0.20

0 1 0 10 10 (a) λ = 0.1 (c) λ = 10

-1 -1 10 10 0.1 0.5 1 2 0.1 0.5 1 2 α = a/R α = a/R

0.05 Ca=0 Dotted: ∆φˆ = 16α3 for small 2 10

p 0.10 spherical drops ∆ˆ

− 0.20 1 ˆ Φ 10

∆ 0.50 thick dashed line in (b) and (c): 32 3 ≫ 0 asymptote α for Ca 1 10 3 (b) λ = 1 when λ > 1/2 (long drops)

-1 10 0.1 0.5 1 2 α = a/R

36 Polygonal capillary

Modern network simulators for 2-phase flow in porous media use polygonal capillaries Wetting meniscus Apparent contact line (e.g. Blunt) Wetting fluid film Films of wetting fluid (Wong, Morris & Radke 1995) y Thickness h ∼ O(Ca2/3) varies downstream

r and across capillary

−1 x If h ≫ κ R convected ionic current computed from integral of shear rate π−2β Ohmic return current mainly through corners

37 Future

Theory now ahead of experiment

Have not discussed: high potentials Hartmann number surface conduction at wall

Yet to understand: 1. Limiting film thickness for λ< 1/2 2. Onset of capillary wave instability 3. Electrokinetics at oil/rock interface 4. Electrokinetics at water/oil interface 5. Flow through constrictions 6. Scale up from pore to Darcy scale

38 Jiang, Shan, Jin, Zhou & Sheng, Geophys. Res. Lett. 25 (1998) 1581 Effect of conductivity σ: ∆φ = (ǫζ/σµ)∆p

39 Lorne, Perrier & Avouac, Bol`eve, Crespy, Revil, Janod & J. Geophys. Res. B104 (1999) Mattiuzzo, J. Geophys. Res. B112 17857 (2007) B08204

40 Lorne, Perrier & Avouac, J. Geophys. Res. B104 (1999) 17857

41