Exact solution of electroosmotic flow in generalized Burgers fluid

Appl. Math. Mech. -Engl. Ed., 32(9), 1119–1126 (2011) DOI 10.1007/s10483-011-1486-6 c Shanghai University and Springer-Verlag Berlin Heidelberg 2011

Applied Mathematics and Mechanics (English Edition)

Exact solution of electroosmotic flow in generalized Burgers fluid∗ T. HAYAT1,2 , S. AFZAL1 ,

A. HENDI3

(1. Department of Mathematics, Quaid-i-Azam University, Islamabad 45320, Pakistan; 2. Department of Mathematics, King Saud University, Riyadh 11451, Saudi Arabia; 3. Department of Physics, Faculty of Science, King Saud University, Riyadh 11321, Saudi Arabia) (Communicated by Jian-zhong LIN)

Abstract An exact solution is developed for the time periodic electroosmotic flow of a non-Newtonian fluid between the micro-parallel plates. The constitutive equations of a generalized Burgers fluid are used in the mathematical formulation. The resulting problem is solved by a Fourier transform technique. Graphs are plotted and discussed for various emerging parameters of interest. Key words

generalized Burgers fluid, periodic flow, Fourier transform

Chinese Library Classification O373 2010 Mathematics Subject Classification

1

74A05

Introduction

Electroosmosis has received substantial interest by many researchers recently because of its applications in biochemistry and medicine. In particular, microelectro mechanical systems (MEMs) are encountered in micro-heat exchangers, micro-pumps, micro-turbines, and biochemical analysis instruments[1–3] . With such facts in mind, Ngoma and Erchiqui[4] examined the effects of the heat flux and slip on the viscous flow in a microchannel. Zhang et al.[5] studied the electroosmotic flow in an irregular shape microchannel. Duan and Muzychka[6] presented the analysis of the slip effect on the viscous flow in elliptic microchannels. Dutta and Beskok[7] provided the analytic solution for time-periodic electroosmotic flows in two-dimensionsal (2D) straight channels. Wang and Wu[8] constructed the analytic solution for the periodic electroosmosis in a microchannel for biochips. Although electroosmotic flows occur in microscales where the electrorheological flow appears on bulk scales, scarce information is available to the electroosmotically driven flow of non-Newtonian fluids[9–11] . Das and Chakraborty[12] and Akg¨ ul and Pakdemirli[13] discussed the electroosmotic flow of the power law and the third-grade fluids, respectively. These fluid models are the subclasses of differential type fluids in non-Newtonian fluid mechanics. However, no attempt has been made so far for the electroosmotic flow in rate type fluids. In the present paper, we provide an attempt for the periodic electroosmotic flow of rate type fluids. ∗ Received Jan. 19, 2011 / Revised Jun. 9, 2011 Project supported by the Visiting Professor Programming of King Saud University (No. KSU-VPP117) Corresponding author T. HAYAT, Professor, Ph. D., E-mail: pensy [email protected]

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T. HAYAT, S. AFZAL, and A. HENDI

A subclass of rate type non-Newtonian fluids, i.e., the generalized Burgers fluid, is used. The relevant equations are first constructed, and the closed form solution is derived. The physical interpretation to the obtained solution is assigned by the graphs.

2

Governing equations

We consider an incompressible generalized Burgers fluid bounded by the micro-parallel plates with the distance 2h. The length of the plate is L. The geometry of the 2D uniform microchannel with an isolated wall is shown in Fig. 1. In this figure, the distance from the channel center to the slipping plane of the electric double layer is h. Considering the symmetry in the electric potential and velocity field, the domain is reduced to a half. Note that the electrical potential and pressure gradient are parallel to the x-axis of the channel. The governing equations are div V ∗ = 0, ρf

(1)

∗

dV = div T + ρe b, dt∗

ρe b = ρe E sin(Ωt∗ ),

(2) (3)

where V ∗ is the velocity vector, ρf is the fluid density, T is the Cauchy stress tensor, b is the body force consisting of the sinusoidal electrical field E sin(Ωt∗ ) only in which the gravity is neglected, ρe is the net electric charge density, and Ω is the frequency of the unsteady external electric field. The electrical field is assumed to be in the x-direction only.

Fig. 1

3

Time periodic electroosmotic flow in straight channel

Basic equations The Cauchy stress tensor T in a generalized Burgers fluid is given by T = −P I + S,

(4)

δS δ2 S δA δ2A (5) + λ2 2 = μ A + λ3 + λ4 2 , δt δt δt δt where P is the pressure, λ1 and λ2 are the relaxation time, λ3 (< λ1 ) and λ4 are the retardation time, A is the first Rivlin-Ericksen tensor, S is the extra stress tensor, μ is the dynamic viscosity, and δ dS δ δS δ2S = − LS − SLT , = (6) 2 δt δt δt δt dt S + λ1

A = grad V + (grad V )T ,

(7)

d denotes the material derivative. It should be pointed out that the model (4) includes in which dt the Burgers fluid when λ4 = 0. Equation (4) represents an Oldroyd-B fluid when λ2 = λ4 = 0. This expression, respectively, reduces to the Maxwell and viscous fluids when λ3 = λ4 = 0.

Exact solution of electroosmotic flow in generalized Burgers fluid

4

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Poisson-Boltzmann equation

For the body force, one needs the Poisson-Boltzmann equation and the Debye–Huckel approximation. The Poisson-Boltzmann equation is related to the potential distribution within the electric double layer. In the y-direction, it is expressed in the following form: d2 ψ ∗ ρe =− , dy ∗2

(8)

where ψ ∗ is the electrical potential, is the dielectric constant or the permitivity of the fluid, and ρe is the net electric charge density. Assuming that the equilibrium Boltzmann equation is applicable, which implies a uniform dielectric constant. The number concentration of the type-i ions are given by ni± = no exp

∓Zeψ ∗ , kB θ∗

(9)

where no is the bulk ionic concentration, Z is the valance of the type-i ions, e is the elementary charge, kB is the Boltzmann constant, and θ∗ is the absolute temperature. The net electric charge density ρe can be expressed by assuming a symmetric electrolyte as follows: ρe = Ze(ni+ − ni− ).

(10)

From Eqs. (9) and (10), we can write ρe = −2Zeno sinh

Zeψ ∗ . kB θ∗

(11)

Substituting the above expression in Eq. (7) yields Zeψ ∗ d2 ψ ∗ 2Zeno . sinh = dy ∗2 kB θ∗

(12)

The above equation is called the Poisson-Boltzmann equation. The Debye-Huckel linear approximation gives sinh() = (). Therefore, Eq. (12) reduces to d2 ψ ∗ ∼ 2Z 2 e2 no ψ ∗ . = dy ∗2 kB θ∗

(13)

This approximation holds when the electric potential is small in comparison with the thermal energy of the ions. Equation (13) can be put into the form d2 ψ ∗ = k2 ψ∗ dy ∗2

(14)

with k = Ze

2no kB θ∗

(15)

as the Debye-Huckel parameter, in which 1/k is used for the thickness of the electric double layer. The boundary conditions for Eq. (14) are taken as follows: dψ ∗ (0) = 0, dy ∗

ψ ∗ (h) = ξ ∗ ,

(16)

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where ξ ∗ is the zeta potential (the electrical potential at the shear plane). The dimensionless problem is now expressed as d2 ψ = K 2 ψ, dy 2 dψ(0) = 0, dy

(17) ψ(1) = ξ,

(18)

where y is dimensionless coordinate, ψ is the dimensionless electric potential, and K is the electro-kinetic separation distance based on the plate height. The dimensionless parameters here are Zeψ ∗ y∗ Zeξ ∗ y= , ψ= , ξ= , K = kh. (19) ∗ h kB θ kB θ∗ Equation (17) is satisfied by the following expression: ψ(y) = A cosh(Ky) + B sinh(Ky),

(20)

which, after using Eq. (18), reduces to ψ(y) =

ξ cosh(Ky) . cosh y

(21)

The one-dimensional velocity field V ∗ satisfying the continuity equation is expressed as V ∗ = (v ∗ (y ∗ , t∗ ), 0, 0).

(22)

S = S(y ∗ , t∗ ).

(23)

The stress field S is taken as

From Eqs. (5), (6), (22), and (23), we have Sx∗ x∗ + λ1 (∂t∗ Sx∗ x∗ − 2Sx∗ y∗ ∂y∗ v ∗ ) + λ2 (∂t∗ (∂t∗ Sx∗ x∗ − 2Sx∗y∗ ∂y∗ v ∗ ) − 2(∂t∗ Sx∗ y∗ − Sy∗ y∗ ∂y∗ v ∗ )∂y∗ v ∗ ) = −2μλ3 (∂y∗ v ∗ )2 − 6μλ4 ∂y∗ v ∗ (∂t2∗ y∗ v ∗ ),

(24)

Sx∗ y∗ + λ1 (∂t∗ Sx∗ y∗ − 2Sy∗ y∗ ∂y∗ v ∗ ) + λ2 (∂t∗ (∂t∗ Sx∗ y∗ − Sy∗ y∗ ∂y∗ v ∗ ) − ∂t∗ Sy∗ y∗ ∂y∗ v ∗ ) = μ(1 + λ3 ∂t∗ + λ4 ∂t2∗ )∂y∗ v ∗ ,

(25)

Sx∗ z∗ + λ1 (∂t∗ Sx∗ z∗ − 2Sy∗ z∗ ∂y∗ v ∗ ) + λ2 (∂t∗ (∂t∗ Sx∗ z∗ − Sy∗ z∗ ∂y∗ v ∗ ) − ∂t∗ Sy∗ z∗ ∂y∗ v ∗ ) = 0,

(26)

(1 + λ1 ∂t∗ + λ2 ∂t2∗ )(Sy∗ y∗ , Sy∗ z∗ , Sz∗ z∗ ) = 0.

(27)

The above equations give Sy∗ y∗ = Sy∗ z∗ = Sz∗ z∗ = 0. Therefore, Eqs. (24)–(27) yield Sx∗ x∗ + λ1 (∂t∗ Sx∗ x∗ − 2Sx∗ y∗ ∂y∗ v ∗ ) + λ2 (∂t2∗ Sx∗ x∗ − 4∂t∗ Sx∗ y∗ ∂y∗ v ∗ − 2Sx∗ y∗ ∂t2∗ y∗ v ∗ ) = −2μλ3 (∂y∗ v ∗ )2 − 6μλ4 ∂y∗ v ∗ (∂t2∗ y∗ v ∗ ),

(28)

(1 + λ1 ∂t∗ + λ2 ∂t2∗ )Sx∗ y∗ = μ(1 + λ3 ∂t∗ + λ4 ∂t2∗ )∂y∗ v ∗ ,

(29)

(1 + λ1 ∂t∗ + λ2 ∂t2∗ )Sx∗ z∗ = 0.

(30)

Equation (30) further gives Sx∗ z∗ = 0.


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The scalar forms of Eq. (2) are ρf ∂t∗ v ∗ = −∂x∗ P − ∂y∗ Sx∗ y∗ + ρe Ex sin(Ωt∗ ),

(31)

∂y∗ P = ∂z∗ P = 0.

(32)

Equation (32) implies P = P (y, z). Then, eliminating Sx∗ y∗ between Eqs. (29) and (31) yields ρf (1 + λ1 ∂t∗ + λ2 ∂t2∗ )∂t∗ v ∗ + (1 + λ1 ∂t∗ + λ2 ∂t2∗ )∂x∗ P = μ(1 + λ3 ∂t∗ + λ4 ∂t2∗ )∂y2∗ v ∗ + ρe Ex (1 + λ1 ∂t∗ + λ2 ∂t2∗ ) sin(Ωt∗ ),

(33)

which, in the absence of the pressure gradient, takes the form ρf (1 + λ1 ∂t∗ + λ2 ∂t2∗ )∂t∗ v ∗ = μ(1 + λ3 ∂t∗ + λ4 ∂t2∗ )∂y2∗ v ∗ + ρe Ex (1 + λ1 ∂t∗ + λ2 ∂t2∗ ) sin(Ωt∗ ).

(34)

The dimensionless problem is ˜1 ∂t + λ ˜ 2 ∂ 2 )∂t v (1 + λ t =

2 1 ˜ 3 ∂t + λ ˜ 4 ∂ 2 )∂ 2 v + ∧1 d ψ (1 + λ ˜1 ∂t + λ ˜ 2 ∂ 2 ) sin t, (1 + λ t y t Re Re dy 2

∂v (0, t) = 0, ∂y

v(1, t) = 0, where

⎧ ∗ ⎪ ⎨ t = Ωt ,

˜ 1 = λ1 Ω, λ

v∗ ⎪ ⎩v = , vo

y=

y∗ , h

(36)

˜ 2 = λ2 Ω2 , λ

Re =

(35)

h2 Ω , ν

˜3 = λ3 Ω, λ

Λ1 =

˜ 4 = λ4 Ω2 , λ

kB θ∗ Ex . Zevo μ

(37)

In the above equations, vo is the reference velociy, Re is the Reynold number, and Λ1 is the dimentionless parameter for the electro-kinetic effect. Differentiating Eq. (21) yields d2 ψ ξK 2 cosh(Ky) . = dy 2 cosh y

(38)

Inserting Eq. (38) into Eq. (35) yields ˜ 1 ∂t + λ ˜ 2 ∂ 2 )∂t v (1 + λ t =

5

2 1 ˜ 3 ∂t + λ ˜ 1 ∂t + λ ˜ 4 ∂ 2 )∂ 2 v − Λ1 ξK cosh(Ky) (1 + λ ˜ 2 ∂ 2 ) sin t. (1 + λ t y t Re Re cosh y

(39)

Solution procedure The temporal Fourier transform pair is expressed as ∞ v(y, t)e−iωt dt, v¯(y, ω) =

(40)

−∞

v(y, t) =

1 2π

∞

−∞

v¯(y, ω)eiωt dω,

(41)

1124


where ω is the temporal frequency. The transformed problem is ˜ 1 − iω 3 λ ˜2 ) d2 v¯ Re(iω − ω 2 λ v¯ − ˜3 − ω2λ ˜4 dy 2 1 + iω λ =

˜1 − λ ˜ 2 )δ(ω − 1) − (1 − iλ ˜1 − λ ˜ 2 )δ(ω + 1)) cosh(Ky) 2πΛ1 K 2 ξ((1 + iλ , ˜3 − ω2λ ˜ 4 ) cosh K 2i(1 + iω λ v¯(1, ω) = 0,

∂¯ v (0, ω) = 0. ∂y

(42)

(43)

The solution to Eq. (42) subjected to the boundary conditions (43) is v¯(y, ω) =

˜1 − λ ˜ 2 )δ(ω − 1) −2πΛ1 K 2 ξ(1 + iλ

˜3 − ω2λ ˜4 ) K 2 − Re(iω−ω2 λ˜ 1 −iω3 λ˜ 2 ) 2i(1 + iω λ ·

cosh

˜4 ˜ 3 −ω 2 λ 1+iω λ

˜ 1 −iω 3 λ ˜2) Re(iω−ω 2 λ y ˜4 ˜ 3 −ω 2 λ 1+iω λ

cosh +

cosh(Ky) cosh K

−

˜ 1 −iω 3 λ ˜2 Re(iω−ω 2 λ ˜4 ˜ 3 −ω 2 λ 1+iω λ

˜1 − λ ˜ 2 )δ(ω + 1) 2πΛ1 K 2 ξ(1 − iλ

˜ 4 ) K 2 − Re(iω−ω2 λ˜1 −iω3 λ˜ 2 ) ˜3 − ω2λ 2i(1 + iω λ ˜4 ˜ 3 −ω 2 λ 1+iω λ

2˜ 3λ ˜2 ) cosh ( Re(iω−ω˜ λ1 −iω )y cosh(Ky) ˜4 1+iω λ3 −ω 2 λ · . − 2λ ˜ 1 −iω 3 λ ˜2 ) cosh K cosh Re(iω−ω 2 ˜ ˜ −ω λ 1+iω λ 3

(44)

4

The inverse Fourier transform yields the following expression: v(y, t) =

˜ 2 + iλ ˜ 1 )eit −Λ1 K 2 ξ(1 − λ

˜ 4 + iλ ˜ 3 ) K 2 − Re(i(1−λ˜ 2 )−λ˜ 1 ) 2i(1 − λ ·

cosh

·

6

˜ 2 )−λ ˜1) Re(i(1−λ y ˜ 4 +iλ ˜3 1−λ

cosh +

˜ 4 +iλ ˜3 1−λ

˜ 2 )−λ ˜1 ) Re(i(1−λ ˜ 4 +iλ ˜3 1−λ

−

cosh(Ky) cosh K

˜ 2 − iλ ˜ 1 )e−it Λ1 K 2 ξ(1 − λ

˜ 4 − iλ ˜ 3 ) K 2 − Re(−i(1−λ˜ 2 )−λ˜ 1 ) 2i(1 − λ

˜ 2 )−λ ˜1 ) Re(−i(1−λ y ˜ 4 −iλ ˜3 1−λ ˜ 2 )−λ ˜1 ) λ cosh Re(−i(1− ˜ 4 −iλ ˜3 1−λ

cosh

˜ 4 −iλ ˜3 1−λ

−

cosh(Ky) . cosh K

(45)

Results and discussion

This section deals with the effects of different parameters on the velocity profile of the time periodic electroosmotically driven flow across the half of the channel for the fixed zeta potential ξ. The results are shown in Figs. 2–7. Figure 2 discusses the influence of Re on the velocity field. It shows that the velocity is uniform for small Re. Moreover, the velocity has a wavelike structure when Re increases, and this phenomenon is more pronounced when Re = 50. The effects of Λ1 on the velocity is plotted in Fig. 3. It is found that the velocity is an increasing function of the electrokinetic parameter.


1125

Figure 4 shows the influence of K on the dimensionless velocity profile. It shows that the magnitude of the velocity increases while the electric double layer thickness decreases when K increases. The effect of t on the velocity is displayed in Fig. 5. It is noticed that the velocities are positive for t = 2π and t = 1.5π and are negative for t = 0.5π and t = π. The magnitudes of the velocities for t = 1.5π and t = 0.5π are similar, while the magnitudes of the velocities for t = π and t = 2π are the same. The magnitudes of the velocities for t = π and t = 2π are greater than those for t = 1.5π and t = 0.5π.

Fig. 2

Influence of Re on dimensionless velocity profile with Λ1 = 1, K = 30, ˜ 2 = 5, ˜ 1 = 7, λ ζ = 0.5, t = π, λ ˜4 = 4 ˜ 3 = 1, and λ λ

Fig. 3

Influence of Λ1 on dimensionless velocity profile with Re = 50, K = 30, ˜ 2 = 5, ˜ 1 = 7, λ ζ = 0.5, t = π, λ ˜4 = 4 ˜ 3 = 1, and λ λ

Fig. 4

Influence of K on dimensionless velocity profile with Re = 50, Λ1 = 1, ˜ 2 = 5, ˜ 1 = 7, λ ζ = 0.5, t = π, λ ˜4 = 4 ˜ 3 = 1, and λ λ

Fig. 5

Influence of t on dimensionless velocity profile with Re = 50, Λ1 = 1, ˜ 2 = 5, ˜ 1 = 7, λ ζ = 0.5, K = 30, λ ˜4 = 4 ˜ 3 = 1, and λ λ

Figures 6 and 7 depict the effect of the dimensionless relaxation time and the dimensionless retardation time on the velocity field, respectively. The magnitude of the velocity increases with an increase in the relaxation time, and decreases with an increase in the retardation time. The velocity field has an opposite behavior for the relaxation time and the retardation times. From the above figures, we can see that the magnitude of the velocity is constant across the half of the channel, and the magnitude of the velocity is zero at the channel wall. Acknowledgements The first author, as a visiting professor, thanks the support of Global Research Network for Computational Mathematics and King Saud University for this research.

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Fig. 6


˜ 1 on dimensionless veInfluence of λ locity profile with Re = 50, K = 30, ˜ 2 = 5, ζ = 0.5, t = π, Λ1 = 1, λ ˜4 = 4 ˜ 3 = 1, and λ λ

Fig. 7

˜ 3 on dimensionless veInfluence of λ locity profile with Re = 50, K = 30, ˜ 1 = 7, ζ = 0.5, t = π, Λ1 = 1, λ ˜4 = 4 ˜ 2 = 5, and λ λ

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