Modelling a droplet moving in an electric field

Mathematics and Computers in Simulation 68 (2005) 157–169

Modelling a droplet moving in an electric field Dirk Langemann∗ Department of Mathematics, University of Rostock, D-18051 Rostock, Germany Received 4 August 2004; accepted 16 December 2004 Available online 22 January 2005

Abstract Droplets on insulators in outdoor high-voltage equipment move and leave water films on the insulating material. These films further the development of undesirable electric currents or even flash-overs. The paper deals with the behaviour of a single droplet laying on a solid support in a strong electric field. Inside the solid support two electrodes generate the electric field. Although the experiment is designed so that the electric field is nearly homogeneous in the absence of a droplet, the remaining inhomogeneity motivates the discussion of its influence on the droplet. Therefore, water droplets at general positions on the experimental set-up are considered. The deformation of the droplet is calculated together with the total force acting on whole the droplet. The total horizontal force initiates the droplet to move. An analytical proof of the existence of a non-vanishing total force acting on an extended uncharged body in a general electric field is given. © 2004 IMACS. Published by Elsevier B.V. All rights reserved. Keywords: Stationary electric field; Insulators; Droplets; Free boundary value problem; Numerical analysis

1. Introduction The behaviour of a droplet put into a homogeneous electric field is well-understood [5,8,9,12]. The droplet disturbs the field, the field effects a concentration of charge at the droplet surface, and the droplet deforms due to the force density which the disturbed electric field causes on the electric charge. In this case, the droplet suffers a pure deformation without any translation.

∗

Tel.: +49 381 4986644; fax: +49 381 4986553. E-mail address: [email protected].

0378-4754/$30.00 © 2004 IMACS. Published by Elsevier B.V. All rights reserved. doi:10.1016/j.matcom.2004.12.001

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If, however, the uncharged droplet is put into a slightly inhomogeneous electric field, a translation of whole the droplet may occur. The main topic of the present paper is the calculation of the total force and the caused translation of a droplet in a general electric field. The consideration is motivated by droplets and water films on insulators of outdoor high-voltage equipment. The moisture on insulating material furthers undesirable electric currents or even flash-overs. In particular, water films are critical. Since water droplets influence negatively the insulating and hydrophobic material properties and the aging process of the insulating material, several experiments [5] and simulations [7,11,12] have been done to analyze the droplet behaviour. The differences between dielectric and conductive droplet fluids are discussed in [8]. But realistic rainwater is sufficiently conductive cause of its additives. Here, a conductive two-dimensional (2D) model of a droplet on a solid support is regarded. Two electrodes with the potentials +U and −U generate a time-constant electric field. In opposite to preceding investigations [7,8], the droplet position is not restricted to the symmetric centre between the electrodes. The electric field in the experimental set-up is nearly homogeneous close to the symmetric centre in the absence of a droplet. Considerable inhomogeneities exist outside the symmetric centre. The occurrence of inhomogeneous electric fields in realistic high-voltage equipment seems to be convincing. The actual investigations can be extended in a straightforward manner to the three-dimensional (3D) case. This has been done in [9] for droplets at the symmetric position. Section 2 introduces the basic notations and splits the problem into two sub-problems. The first, socalled electric sub-problem consists in the determination of the force density pe on a given droplet surface. The second sub-problem is the determination of the droplet shape for a given outer force density pe . A parametrization of the droplet surface is introduced, and the incompressibility of the droplet fluid is handled as an algebraic constraint. An academic example of a circular droplet in an electric field is discussed in Section 3. The material outside the droplet is completely homogeneous, and the electric potential on the boundary of a large circular domain is given by its Fourier series. The domain is arbitrarily large and so the feed-back of the droplet on the electric field is arbitrarily small. The total force F can be given as a series in terms of the Fourier coefficients. It is shown that the total force is non-vanishing in general. It vanishes in some particular cases, e.g. in the case that the droplet was put into a completely homogeneous field. The discussion of the academic example ends by a point-wise indicator of the inhomogeneity of the undisturbed electric field. The indicator behaves asymptotically like the total force for small droplets. The numerical efforts to solve both sub-problems are discussed in Section 4. Starting with a more detailed view on the feed-back problem between the electric and the mechanical sub-problems, the calculation of the electric potential Φ by finite elements including a non-standard boundary condition is described. A next critical point is the transcription of the one-dimensional non-linear boundary value problem of the droplet shape with algebraic conditions into a parabolic problem with algebraic constraints in an auxiliary time τ. Its solution converges to the time-constant solution of the original mechanical subproblem. Section 5 presents the droplet shapes for different voltages U and at different droplet positions on the support. The resulting total force is given. The separation of the droplet deformation and its motion as a whole leads to a first attempt for the simulation of a droplet moving in the electric field. Trajectories of the translation of whole the droplet are computed. The paper is finished by the concluding remarks and the motivation for a closer look to the hydrodynamic processes inside the droplet in future.

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Fig. 1. (Left) Experimental set-up and discretization principle, d is distance of the droplet’s centre of gravity from the symmetric centre between the electrodes. (Right) Electric potential in the neighbourhood of the droplet in the case U = 5 kV and d = 15 mm.

In the computed examples, we use the surface tension, σ = 0.072 N m−1 ; the mass density, δ = 1000 kg m−3 and the gravitational acceleration, g = 9.81 m s−2 . Concordant with the experimental set-up [5] the permittivity is ε(x) = 4 for x2 ≤ 0 in the solid support made of resin and ε(x) ≈ 1 for x2 > 0 in the air, Fig. 1. The two-dimensional droplet model has a fixed area of V = 1 × 10−5 m2 what corresponds to a 50 ␮l 3D droplet. Section 5.2 refers the relation between the 2D model and the three-dimensional reality. Different droplet positions with respect to the symmetric centre between the electrodes are modelled by different displacements d, Fig. 1.

2. Basic notations and relations 2.1. The electric sub-problem The support Ω ⊂ R2 of the electric potential Φ(x) is whole the two-dimensional plane except the electrodes and the conductive droplet itself. The points are called x ∈ R2 with x = (x1 , x2 )T . The boundary of the electrodes is denoted by Γe . The boundary of the droplet is divided into the upper part Γu , where the air and the rainwater touch, and the lower part Γs , where the rainwater meets the solid support. The normal n is the outer normal with respect to the droplet, i.e. the inner normal with respect to Ω. The electric potential [6] fulfils ∇ · [ε(x)∇Φ(x)] = 0

in x ∈ Ω

(1)

with the permittivity ε(x) depending on the position x. The boundary conditions on the electrodes and at infinity are Φ(x) = ±U on x ∈ Γe

and

lim Φ(x) = 0.

x →∞

The conductive droplet itself is free of electric charge and thus
∂ Φ(x) = c on x ∈ Γu ∪ Γs with ε(x) Φ(x) dx = 0, ∂n Γu ∪Γs

(2)

(3)

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where the second expression determines the constant c. Finally, the force density pe (x) acting in normal direction on the droplet boundary [6,7,12] is found by  2 1 ∂ pe (x) = ε0 ε(x) Φ(x) n, for x ∈ Γu ∪ Γs . (4) 2 ∂n An example of the electric potential Φ close to a droplet outside the symmetric centre is given in the right plot of Fig. 1. 2.2. The sub-problem of the droplet shape The droplet shape is completely described by its upper boundary Γu = {(x1 (s), x2 (s)), s ∈ [0, sfin ]}

with x2 (0) = x2 (sfin ) = 0,

where s is the line parameter growing from the left triple point to the right one. The parametrization of the upper boundary is not at all unique. The lower boundary is Γs = {(y, 0), y ∈ [x1 (0), x1 (sfin )]}. In opposite to [7,8], the description of the upper boundary Γu by a parameterized curve allows us to consider droplets at general positions on the solid support. Additionally to the electric force density pe (x), the capillary pressure pk (x) and the hydrostatic pressure ph (x) [3] are acting on the upper droplet boundary. They are given for all x(s) ∈ Γu with s ∈ (0, sfin ) by pk (x) = σκ(x)n

with x˙ i = dxi (s)/ds  ph (x) = δg

x˙ 1 x¨ 2 − x¨ 1 x˙ 2 with κ(x) =  3 x˙ 12 + x˙ 22 and

x¨ i = d2 xi (s)/ds2

and by



 max x2 (s) − x2

s∈[0,sfin ]

n.

There are two constrained conditions. First, the water is incompressible and the volume
sfin V = x2 (s)˙x1 (s) ds

(5)

0

stays constant. Second, we assume here that the droplet’s centre of gravity does not move in horizontal direction. The horizontal motion of the whole droplet is considered separately in Section 5.2. So, the x1 -component of the centre of gravity
1 sfin x2 (s)x1 (s)˙x1 (s) ds = 0 (6) x¯ 1 = V 0

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does not change. The equilibrium of forces at x(s) ∈ Γu with s ∈ (0, sfin ) is pe (x) + pk (x) + ph (x) + λ(1) n + λ(2) n = 0,

(7)

where λ(1) and λ(2) are the Lagrangian multipliers to the conditions (5) and (6), respectively. The boundary conditions in the triple points [3,7] are x˙ 2 (0) = x˙ 1 (0) tan ϑ

and

x˙ 2 (sfin ) = −˙x1 (sfin ) tan ϑ.

(8)

The upper boundary is determined by the boundary value problem (7) with the boundary conditions (8) and the constraints (5) and (6). The Lagrangian multiplier to Eq. (5) is the constant part of the hydrostatic pressure λ(1) = po . The Lagrangian multiplier to the condition x¯˙ 1 = 0, Eq. (6), can be given explicitly. The only outer force is generated by the electric field, and thus the total force acting on the droplet as a whole is found by  F=

F1 F2




=

Γu ∪Γs

pe (x) dx.

The force density pe (x) is orthogonal to the boundary at x ∈ Γs . The constant force density f1 with F1 = |Γu |f1 is the horizontal component of the total force F applied on whole the upper boundary with the length |Γu |. Now, the Lagrangian multiplier is λ(2) = −f1 n1 . Implicitly, we allow a unrealistic tangential pressure on Γu by the introduction of the multiplier λ(2) . The existence of a total force F = 0 acting on an electrically uncharged body may astonish. The following Section 3 discusses an academic sample where the total force can be given analytically.

3. An academic sample problem A conductive, electrically uncharged body takes up a disk with the radius rΓ . The area around has constant permittivity. The support of the electric field in the sample problem is the domain Ω = {(r, ϕ) : rΓ < r < 1, 0 ≤ ϕ < 2π}. The boundary of the domain Ω is ∂Ω = Γ ∪ H where H is the circle with the outer radius 1 and Γ is the circle with the inner radius rΓ . The calculation of the potential u in Ω leads to the boundary value problem "u(x) = 0 u(x) = g(x)

in x ∈ Ω, on x ∈ H,

u(x) = c on x ∈ Γ,
∂ u(x) dx = 0 Γ ∂n

(9) (10) (11) (12)

with the Dirichlet boundary condition g(x) on H in Eq. (10) and a constant potential c on the boundary of the conductive body in Eq. (11). The constant c is determined by Eq. (12) which expresses the absence of electric charge on the body. The outer normal to the body is denoted by n. It is identical with the inner normal of Ω on Γ .

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We want to give an expression for the total force 2
 1 ∂ F = ε0 u(x) n dx 2 ∂n Γ

(13)

acting on the uncharged body. Eq. (13) is the analogue in the academic sample to Eq. (4) like Eq. (9) reduces Eq. (1), Eq. (10) is analogous to Eq. (2) and Eqs. (11) and (12) correspond to Eq. (3). We use polar co-ordinates and denote the points by x = x(r, ϕ). The Fourier expansion of the Dirichlet condition is g(x) = g(ϕ) = a0 +

∞ 

ak cos kϕ + bk sin kϕ.

(14)

k=1

The potential u [10] in the disk Ω is given by ∞  u(x) = u(r, ϕ) = a0 + (ak+ rk + ak− r −k ) cos kϕ + (bk+ rk + bk− r −k ) sin kϕ. k=1

Eq. (10) yields ak+ + ak− = ak and bk+ + bk− = bk . Due to Eq. (11) the potential u is constant on Γ and thus it holds u(rΓ , ϕ) = a0 = c. With ak+ rΓk + ak− rΓ−k = 0 and bk+ rΓk + bk− rΓ−k = 0 we get ak+ =

ak , 1 − rΓ2k

ak− =

−rΓ2k ak , 1 − rΓ2k

bk+ =

bk 1 − rΓ2k

and

bk− =

−rΓ2k bk . 1 − rΓ2k

(15)

The normal derivative of u on Γ is ∞

 k k ∂ u(x) = (ak+ rΓk − ak− rΓ−k ) cos kϕ + (bk+ rΓk − bk− rΓ−k ) sin kϕ ∂n r r Γ k=1 Γ and thus condition (12) is fulfilled. The Fourier coefficients of the normal derivative are abbreviated by Ak =

2krΓk−1 ak 1 − rΓ2k

and

Bk =

2krΓk−1 bk 1 − rΓ2k

and the total force is  
∞ ∞ cos ϕ ε0 2π   (Ak cos kϕ + Bk sin kϕ)(Aj cos jϕ + Bj sin jϕ) rΓ dϕ. F= 2 0 k=1 j=1 sin ϕ

(16)

After the evaluation of the trigonometric integrals in Eq. (16) we find ∞  ε0 Ak Ak+1 + Bk Bk+1 F1 = πrΓ 2 k=1

and

∞  ε0 F2 = πrΓ Ak Bk+1 − Bk Ak+1 . 2 k=1

(17)

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The components of the total force in Eq. (17) can be expressed in terms of the Fourier expansion (14) of the Dirichlet boundary condition g(x) and that yields F1 = ε0 π

∞  k=1

2k(k + 1)rΓ2k (1 − rΓ2k )(1 − rΓ2k+2 )

(ak ak+1 + bk bk+1 )

(18)

(ak bk+1 − bk ak+1 ).

(19)

and F2 = ε0 π

∞  k=1

2k(k + 1)rΓ2k (1 − rΓ2k )(1 − rΓ2k+2 )

Eqs. (18) and (19) are the sought expressions of F depending on g. Example 1. The total force is not vanishing in general. For instance, may be g(ϕ) = cos ϕ + cos 2ϕ, i.e. a1 = a2 = 1 and all other Fourier coefficients vanish. Then, F2 = 0 but F1 = 4ε0 π

rΓ2 (1 − rΓ2 )(1 − rΓ4 )

is not vanishing for all rΓ = 0. Corollary 2. The series (18) and (19) converge if the function g(ϕ) is piece-wise continuous. Proof. If the function g(ϕ) is piece-wise continuous then the Fourier expansion (14) exists and it holds ak < O(1/k) and bk < O(1/k). Hence, the terms k(k + 1)ak ak+1 and the respective terms in Eqs. (18) and (19) are bounded. The series ∞  k=1

rΓ2k (1 − rΓ2k )(1 − rΓ2k+2 )

=

∞ 1  1 1 rΓ2 − = 1 − rΓ2 k=1 1 − rΓ2k (1 − rΓ2 )2 1 − rΓ2k+2

has positive terms only and it is absolutely convergent. Hence, the sums (18) and (19) exist.
Corollary 3. It holds lim F = 0. rΓ →0

Proof. Let M an upper bound for |k(k + 1)ak ak+1 | and the respective terms in the series (18) and (19). Then, √

F 2 ≤ 4 2ε0 π

rΓ2 M, (1 − rΓ2 )2

which tends to zero for rΓ → 0. Thus, the known statement that a dimensionless uncharged particle does not suffer any force effected by an electric field, is not touched by the present investigation and by Example 1.
Corollary 4. If g(x) is a cut-off of the potential of an homogeneous electric field then F = 0. Proof. In this case, only a1 = 0 and b1 = 0 and all other Fourier coefficients are vanishing. With Eqs. (18) and (19) the assertion holds.


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Corollary 5. May be g(ϕ) = g(2ψ − ϕ) = −g(π + 2ψ − ϕ), i.e. there is a symmetric axis at ψ and an odd-symmetric axis at ψ + π/2. Then, holds F = 0. Proof. Without loss of generality we set ψ = 0 and then bk = 0 for all k and ak = 0 for all even k. The assertion follows immediately.
Finishing the academic sample, we regard Eq. (9) with the boundary condition (10) in the domain ˜ ˜ i.e. the potential u˜ in the absence of a body with Ω = {(r, ϕ) : 0 ≤ r < 1, 0 ≤ ϕ < 2π} with H = ∂Ω, the radius rΓ . It holds [10] u˜ (x) = u˜ (r, ϕ) = a0 +

∞ 

ak rk cos kϕ + bk r k sin kϕ.

(20)

k=1

The next corollary deals with the feed-back of the droplet to the electric field. Corollary 6. The smaller rΓ the smaller is the disturbance of the electric field by the droplet. Proof. Due to Eq. (15) it holds ak+ → ak and ak− → 0 for rΓ → 0 and thus lim u(x) = u˜ (x).


rΓ →0

A body with fixed small radius rΓ compared to the measurements of the support of the electric field reduces the feed-back effect of the droplet. Rescaling the problem yields a low feed-back remote of the droplet. Corollary 7. Let be J(x) = ∇|∇ u˜ (x)|2 = 2∇E(x) · E(x) an indicator of the inhomogeneity of E(x) = ∇ u˜ (x). Then, holds F = ε0 J(0). rΓ →0 πr 2 Γ lim

Proof. From Eqs. (18) and (19), follows F1 = 4ε0 (a1 a2 + b1 b2 ) rΓ →0 πr 2 Γ lim

and

F2 = 4ε0 (a1 b2 − b1 a2 ). rΓ →0 πr 2 Γ lim

The components of the indicator J(x) = (J1 , J2 )T are Jj = 2u˜ 1 u˜ j1 + 2u˜ 2 u˜ j2 with j ∈ {1, 2} and u˜ j = ∂u˜ /∂xj and u˜ ij = ∂2 u˜ /∂xi ∂xj . Eq. (20) reads in Cartesian co-ordinates x1 = r cos ϕ and x2 = r sin ϕ as u˜ (x) = a0 + a1 x1 + b1 x2 + a2 (x12 − x22 ) + 2b2 x1 x2 +



O(x1α1 x2α2 )

α1 +α2 =3

where the last term means a sum of products of at least three xj . We find u˜ 1 (0) = a1 , u˜ 2 (0) = b1 , u˜ 11 (0) = −˜u22 (0) = 2a2 and u˜ 12 (0) = 2b2 , and the assertion follows immediately.


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4. Numerical handling 4.1. Separation of the sub-problems The computation of the electric potential Φ for a given droplet shape Γu and thus Γs by Eq. (1) with the boundary conditions (2) and (3) is called P-problem. The other sub-problem consists in the determination of Γu via (x1 (s), x2 (s)). It is named R-problem and contains the boundary value problem (7) with the boundary conditions (8) and the constraints (5) and (6). The operators P and R map P : Γu → pe

and

R : pe → Γu .

We are searching for a fixed point of the combined operator RP by using a Banach-like iteration Γui+1 = RPΓui , [7,8]. Since the parametrization of Γu is not unique, R does not map to a particular (x1 (s), x2 (s)). 4.2. Numerics of the electric sub-problem The solution of the electric P-problem for finding the outer force density pe is found in two steps. The critical point of the numerical solution is the unboundedness of the domain Ω. We introduce two sub-domains Ω2 ⊂ Ω1 ⊂ Ω whose outer boundaries are the rectangles [x1i− , x1i+ ] × [x2i− , x2i+ ]. The sub-domain Ω1 is large enough to contain the electrodes and the droplet and to leave a sufficiently large margin outside. Since a concentration of charge has local influence to the potential, Corollary 6, it is suitable to replace the boundary condition (2) at infinity by ∂ Φ1 (x) = 0 ∂nΩ1

on x ∈ ∂Ω1 .

In the first step, an approximation Φ1 of the potential Φ is calculated on a coarse rectangular grid in Ω1 by finite differences [1]. In Ω2 , the boundary value problem (1) is solved by finite elements with the boundary condition Φ(x) = Φ1 (x) on x ∈ ∂Ω2 . Here, we used an adaptive triangular mesh with 6000 points and about 12,000 triangular elements. The mesh is refined near the droplet surface Γu ∪ Γs . The finite elements are constructed so that each element is either completely above the surface of the support or completely beneath it. Fig. 1 illustrates the two steps of discretization. It shows a detail of Ω1 and the smaller sub-domain Ω2 where the finite-element approximation is done, and it presents the potential Φ(x) in Ω2 . The droplet lays asymmetrically between the electrodes and thus the potential inside the droplet and on its boundary is not vanishing, c = 0. 4.3. Numerics of the mechanical sub-problem The solution of the non-linear elliptic boundary value problem (7) [2] with the boundary conditions (8) and the constrained conditions (5) and (6) yields the droplet shape described by Γu . We introduce an

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auxiliary time τ and abbreviate   x1 x= = x(s, τ) and x2

p(s, τ) = pe (x) + pk (x) + ph (x) − (f1 n1 )n.

Instead of solving p(s, τ) + λ(1) n = 0, we consider the corresponding parabolic system [7] ∂ x(s, τ) = p(s, τ) + λ(1) n ∂τ with the boundary conditions x2 (0, τ) = x2 (sfin , τ) = 0 and xτ (s, τ) =

(21)

∂ ∂ ∂ ∂ x2 (0, τ) = x1 (0, τ) tan ϑ, x2 (sfin , τ) = − x1 (sfin , τ) tan ϑ (22) ∂s ∂s ∂s ∂s and the algebraic constraint (5). Let us mention that a tangential motion xτ (s, t)⊥n does not influence Γu . The limit xlim (s) = lim x(s, τ) τ→∞

is the stable solution xlim (s) ∈ Γu of the boundary value problem (7). The initial condition x(s, 0) in Eq. (21) is arbitrary as long as it describes a connected domain in the positive half-plane x2 ≥ 0 with a smooth upper boundary and the boundary conditions (22) resp. (8). We discretize the parameter s ∈ [0, sfin ] in equidistant grid points si , i = 0, . . . , N and without loss of generality, we can set s0 = 0, si = i and sN = sfin = N. We introduce xi (τ) = (x1i , x2i )T = x(si , τ) and pi (τ) = p(si , τ) and we get xi (τ) = pi (τ) + λ(1) ∇GT ,

for i = 1, . . . , N − 1

(23)

with the algebraic constraint form Eq. (5) N−1 1 G(xi ) = x2i (x1i+1 − x1i−1 ) − V 2 i=1

(24)

and the discretized boundary conditions x21 = (x11 − x10 ) tan ϑ and x2N−1 = (x1N − x1N−1 ) tan ϑ from Eq. (22). After having expressed the components x10 and x1N via the boundary conditions, we get a differential-algebraic system in xi (τ) for i = 1, . . . , N − 1 which can be solved by standard methods [4] or alternatively by deriving Eq. (24) and giving an expression for λ(1) . In the second case, the differential algebraic system transforms into a ordinary differential equation which is not particularly stiff. Finally, Eq. (23) is extended to xi (τ) = pi (τ) + λ(1) ∇GT + k( xi+1 − xi − xi − xi−1 )t,

i = 1, . . . , N − 1

by an artificial force in tangential direction t avoiding discretization points getting numerically into contact [13]. Here, k is a suitable spring constant. Furthermore, the tangential component −f1 t1 of the Lagrangian multiplier to Eq. (6) is eliminated by an additional horizontal motion of whole the droplet. The numerical solution has been checked with different tolerances, and it has been compared with the analytical solution of the droplet shape which can be given implicitly, e.g. in the case pe = 0.

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Fig. 2. (Left) Comparison of the droplet shapes for U = 0, 4, 6, 8, 10 kV with d = 0 mm (symmetric position, left) and d = 11 mm (out of centre, right). (Right) Comparison of the droplet shapes for U = 10 kV and d = 0, 4, . . . , 24 mm (from left to right, d = 4, 12, 20 mm dotted lines).

5. Results 5.1. Droplet shapes at fixed positions Both sub-problems can be solved by use of the given algorithms. The restrictions included in the R-problem bring a droplet shape for a fixed position of its centre of gravity. By this algorithm, the feedback adaption between droplet and electric field was separated from the total translation of the droplet as a whole. As we have seen in Section 3, the case of a motionless centre of gravity is a rather rare one. Fig. 2 compares droplet shapes for different applied voltages U for droplets at different positions d. It can be seen that a droplet in the symmetric centre between the electrodes suffers a larger deformation than a droplet near one of the electrodes. Droplets between the electrodes become flattened and lengthened in a reinforced manner. Furthermore, the out-centred droplet stays not completely symmetric, but the deviation from symmetry is tiny. The right plot in the same figure shows seven droplet shapes for droplets at different positions for the same voltage. If the droplet is not displaced between the electrodes but outside them, i.e. |d| > x1e with half the distance between the centres of the electrodes, then the observed effects are smaller and they are opposite. The droplets become heightened and pressed slightly. Fig. 3 presents the heights and widths of droplets at different positions d and under different applied voltages U.

Fig. 3. Heights (left) and widths (right) of the droplets depending on U and d.

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Fig. 4. (Left) Total horizontal force F1 depending on U and d. (Right) Motion of the droplet centre for different initial positions d(0).

5.2. Total motion of the droplet The left plot of Fig. 4 shows the total horizontal force F1 = F1 (d, U) depending on the voltage U and on the distance d of the droplet centre from the symmetric centre between the electrodes. The unit of the force in Fig. 4 is N m−1 due to the lacking third dimension. For comparison, we give the mass m = δV = 0.01 kg m−1 . The present 2D model of a droplet can be interpreted as a sectional view of an infinitely long cylindrical droplet line. Then, the force F1 and the mass m are a force resp. a mass per length unit. We remark that the F1 > 0 if d > 0 and that the total force F1 has its maximum at d ≈ 11 mm. For comparison, the electrodes have a distance of 2x1e = 35 mm to each other. A positive d means a displacement of the droplet out of the symmetric centre of the electrodes in negative direction, Fig. 1. So, the total force moves the droplet into the symmetric centre. Only there, that is for d = 0, the force F1 vanishes. A droplet which is not held by adhesion would move into the symmetric centre and leave a water film. A first attempt to simulate the droplet translation consists in handling it as one body sliding over the support. The co-ordinate of the droplet’s centre of gravity with respect to the symmetric centre is −d. The voltage U is fixed. It is assumed that the frictional force is proportional to the normal force and to the sliding velocity with the friction coefficient µ. This behaviour is known from rolling. The droplet is pressed onto the support by the normal force F2 − mg < 0, and we get the equation of motion ˙ 2 − mg) − md¨ = F1 − µd(F

˙ with d(0) = d0 , d(0) = 0.

(25)

The trajectories of the droplet’s horizontal mass centre x¯ 1 are shown in the right plot of Fig. 4 for different initial positions d0 . It is visible that the droplets move indeed into the symmetric centre. If the displacement out of the centre is small, then the trajectory is approximately a part of a damped harmonic oscillation. The droplets move very fast because the electric field is strongly inhomogeneous due to the low distance of the electrodes. They move much slowlier in realistic applications. Of course, the trajectories depend on the coefficient µ, which is completely hypothetical at this simulation level. Furthermore, modelling the droplet as a sliding or rolling rigid body like done in Eq. (25) can be regarded only as a very first step to describe the motion of whole the droplet.

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6. Concluding remarks The procedure presented here to calculate the shape of a droplet in a general electric field contains two new features. First, the parametrization of the droplet surface allows us to handle the incompressibility of the fluid as an algebraic constraint in an easy way. Second, the droplet may be at a general position on the support. We get slightly asymmetrical droplet shapes, and we get a total horizontal force acting on the droplet as a whole. A sample problem is used to express the total force acting on an uncharged body in terms of the electric potential remote from the body. It has been shown that the total force is non-vanishing in general. In addition, a function was given which indicates the inhomogeneity of the undisturbed electric field and which is proportional to the leading term of the series expansion of the total force. The total force is computed for a droplet together with its shape. The total force vanishes only if the droplet lays symmetrically between the electrodes. In all other cases, the droplet tends to moving into the symmetric centre. This motion is modelled in a first attempt by separating the feed-back adaptation of droplet shape and electric field from the translation of the droplet. This procedure is neglecting many effects like the flux of the fluid, the adhesion forces between the droplet and the support and so on. The observation of the described motion strictly motivates a forthcoming paper extending ansatzes like [14] to modern computational opportunities and modelling the droplet fluid in detail as a hydrodynamical boundary value problem with free surface.

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