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Using the method of Laplace transform, analytical expressions are derived for the time periodic pulse electroosmotic flow (EOF) velocity of the triangle and sawtooth of Maxwell fluid in circular microchannel. The solution involves analytically solving the linearized Poisson-Boltzmann (P-B) equation, together with the Cauchy momentum equation and the general Maxwell constitutive equation. By numerical computations of inverse Laplace transform, the effects of electrokinetic width
*K*, relaxation time
and pulse width
*a* on the above several pulse EOF velocities are investigated. In addition, we focused on the comparison and analysis of the formulas and graphs between the triangle and sawtooth pulse EOF with the rectangle pulse EOF. The study found that there are obvious differences in formulas and graphs between triangle and sawtooth pulse EOF with rectangle pulse EOF, and the difference mainly depends on the different definitions of the three kinds of time periodic pulse waves. Finally, we also studied the stability of the above three kinds of pulse EOF and the influence of relaxation time on pulse EOF velocity under different pulse widths is discussed. We find that the rectangle pulse EOF is more stable than the triangle and sawtooth pulse EOF. For any pulse, as the pulse width
*a* increases, the influence of the relaxation time on the pulse EOF velocity will be weakened.

Microfluidic devices have become increasingly more important because of their application in micro-electro-mechanical systems (MEMS) and microbiological sensors (such as lab-on-a-chip) [

In previous research, a large number of theoretical and experimental studies [

All of the above-mentioned studies are related to Newtonian fluids. But most of solutions of industry and biopharmaceutical are fluid that has the structural characteristic of non-Newtonian fluids, such as biological fluid and other solutions of long-chain molecules, which structural characteristics include strain force, normal shear stress, hysteresis effect, variable viscosity, memory effect and so on [

Although some basic characteristics of EOF of non-Newtonian fluids have been reported in the above studies, its rich properties still need to be examined. Recent study have shown that Maxwell fluid model simulation of blood in narrow conical vessels has achieved an ideal effect, and it is completely possible to analyze the blood-based microfluidics and other microbial fluid transmission systems by means of electric mechanism [

However, from the current research situation, there is almost no research on pulse EOF, and it has not attracted enough attention from the majority of researchers. Therefore, based on the rectangle pulse EOF, we have re-selected several common pulses (such as triangle pulse and sawtooth pulse) to study the time periodic pulse EOF of Maxwell fluid through a circular microchannel in present work. The target of this article is to derive the semi-analytical solutions of the above two time periodic pulse EOF for viscoelastic fluid. Meanwhile, we analyzed the effect of several parameters such as the electrokinetic width, the relaxation time and the pulse width on the pulse EOF of Maxwell fluid. Moreover, we also discussed the different effects of relaxation time on the EOF velocity of different pulses. It is mainly compared with the rectangle pulse EOF [

The time periodic pulse EOF of an incompressible Maxwell fluid through a circular microchannel is sketched in _{0} with pulse amplitude of 1, pulse repetition period of 2a and pulse width of a, respectively.

Due to the symmetry of the geometry, we only study the semi-section of the microchannel. Provided that the pressure gradient along z direction is ignored, then the one-dimensional Cauchy momentum equation can be written as

ρ ∂ u ( r , t ) ∂ t = − 1 r ∂ ∂ r ( r τ r z ) + ρ e ( r ) E 0 f ( t ) (1)

where u ( r , t ) is the velocity along z axial direction, ρ is the fluid density, t is the time, τ r z is the stress tensor and ρ e ( r ) is the volume charge density, E 0 f ( t ) is the ideal pulsed electric field of strength E_{0}.

Provided that the boundary condition of Equation (1) is no slip, and it can be given as [

u ( r , t ) | r = R = 0 , ∂ u ( r , t ) ∂ r | r = 0 = 0 (2)

For the Maxwell fluid, the constitutive equation satisfies [

τ r z + λ 1 ∂ ∂ t τ r z = − η 0 ∂ u ( r , t ) ∂ r (3)

where λ 1 is the relaxation time, η 0 is zero shear rate viscosity.

The chemical interaction between the electrolyte liquid and the solid wall produces an electric double layer (EDL), a very thin layer of charged liquid at the solid-liquid interface. A cylindrical coordinate system ( r , θ , z ) is adopted. In this theoretical model, it is assumed that the channel wall is uniformly charged, so that the electrical potential in the EDL only varies in this r direction and does not depend on θ [

1 r d d r ( r d ψ ( r ) d r ) = − ρ e ( r ) ε (4)

ρ e ( r ) = − 2 n 0 z ν e 0 sinh [ z ν e 0 ψ ( r ) k b T ] (5)

where ε is the dielectric constant of the electrolyte liquid, ψ ( r ) is the electrical potential of the EDL, n_{0} is the ion density of the bulk liquid, z ν is the valence, e_{0} is the electron charge, k_{b} is the Boltzmann constant, T is the absolute temperature, and sinh is the sine function.

Combing Equation (4) and Equation (5) gives

1 r d d r ( r d ψ ( r ) d r ) = 2 n 0 z ν e 0 ε sinh [ z ν e 0 ψ ( r ) k b T ] (6)

which is subject to the following boundary conditions

ψ ( r ) | r = R = ψ 0 , d ψ ( r ) d r | r = 0 = 0 (7)

where ψ 0 is wall zeta potential, r is radial coordinate and R is radius of the circular microchannel.

Assuming that the electrical potential is small enough, the Debye-Hückel linearization approximation can be applied to the hyperbolic sine function appearing on the right hand side of Equation (6), which means that the electrical potential is physically small compared to the thermal energy of the charged species [

1 r d d r ( r d ψ ( r ) d r ) = κ 2 ψ ( r ) , and κ = ( 2 n 0 z ν 2 e 0 2 ε k b T ) 1 / 2 (8)

where κ is the Debye-Hückel parameter, which 1 / κ usually represents the thickness of the EDL in physical.

By solving Equation (7) and Equation (8), the net charge density distribution for circular microchannel can be express as

ρ e ( r ) = − ε κ 2 ψ 0 I 0 ( κ r ) I 0 ( κ R ) (9)

where I 0 is the first kind modified Bessel function of order zero.

In order to obtain the solution of the velocity field of the triangle pulse EOF and the sawtooth pulse EOF, let us first briefly review the process of solving the velocity field of the rectangle pulse EOF, and then analyze the difference among the three formulas to obtain the corresponding velocity field solution above.

The ideal rectangle pulse can be expressed as the following form

f ( t ) = { 1 , t ∈ [ 0 , a ) , − 1 , t ∈ ( a , 2 a ] . (10)

For simplicity, the following dimensionless groups are introduced:

r ¯ = r R , K = κ R , ( t ¯ , λ ¯ 1 ) = ( t , λ 1 ) ρ R 2 / η 0 , u ¯ ( r ¯ , t ¯ ) = u ( r , t ) U e o , τ ¯ r z ¯ = τ r z η 0 U e o / R , U e o = − ε ψ 0 E 0 η 0 (11)

where U e o denotes steady Helmholtz-Smoluchowshi EOF velocity of Newtonian fluids, K is the ratio of the characteristic width of the microchannel to Debye length.

Using Equation (11), Equations of (1) and (3) and boundary conditions (2) are normalized as

∂ u ¯ ( r ¯ , t ¯ ) ∂ t ¯ = − 1 r ¯ ∂ ∂ r ¯ ( r ¯ τ r z ¯ ¯ ) + f ( t ¯ ) K 2 I 0 ( K r ¯ ) I 0 ( K ) (12)

τ r z ¯ ¯ + λ ¯ 1 ∂ ∂ t ¯ τ r z ¯ ¯ = − ∂ u ¯ ( r ¯ , t ¯ ) ∂ r ¯ (13)

u ¯ ( r ¯ , t ¯ ) | r ¯ = 1 = 0 , ∂ u ¯ ( r ¯ , t ¯ ) ∂ r ¯ | r ¯ = 0 = 0 (14)

Eliminating τ r z ¯ ¯ from Equation (12) and Equation (13) yields

∂ u ¯ ( r ¯ , t ¯ ) ∂ t ¯ + λ ¯ 1 ∂ 2 u ¯ ( r ¯ , t ¯ ) ∂ t ¯ 2 = 1 r ¯ ∂ ∂ r ¯ ( r ¯ ∂ u ¯ ( r ¯ , t ¯ ) ∂ r ¯ ) + ( 1 + λ ¯ 1 ∂ ∂ t ¯ ) f ( t ¯ ) K 2 I 0 ( K r ¯ ) I 0 ( K ) (15)

Let us employ the method of Laplace transforms defined by

U ( r ¯ , s ) = L [ u ¯ ( r ¯ , t ¯ ) ] = ∫ 0 ∞ u ¯ ( r ¯ , t ¯ ) e − s t ¯ d t ¯ (16)

Obviously ∂ ∂ t ¯ f ( t ¯ ) = 0 in Equation (15), and the Laplace transform of f ( t ¯ ) is given by the Appendix A.

From the literature [

U ( r ¯ , s ) = tanh ( a s 2 ) K 2 s ( K 2 − β 2 ) ( I 0 ( β r ¯ ) I 0 ( β ) − I 0 ( K r ¯ ) I 0 ( K ) ) (17)

where β = λ ¯ 1 s 2 + s , tanh is a hyperbolic tangent function.

The inverse Laplace transform is defined by

u ¯ ( r ¯ , t ¯ ) = L − 1 [ U ( r ¯ , s ) ] = 1 2 π i ∫ Γ U ( r ¯ , s ) e s t ¯ d s (18)

where Γ is a vertical line to the right of all singularities of U ( r ¯ , s ) in the complex s plane. The exact solution of the EOF velocity cannot be obtained analytically due to the complexity of the express of U ( r ¯ , s ) . Therefore, the numerical computation must be performed by numerical inverse Laplace transform [

The ideal triangle pulse can be expressed as the following form

f ( t ) = { 2 a t , t ∈ [ 0 , a 2 ) , 2 − 2 a t , t ∈ [ a 2 , 3 a 2 ] , 2 a t − 4 , t ∈ ( 3 a 2 , 2 a ] . (19)

The difference from the rectangle pulse wave is ∂ ∂ t ¯ f ( t ¯ ) ≠ 0 in Equation (15), the others are the same, and the Laplace transform of ( 1+ ∂ ∂ t ¯ ) f ( t ¯ ) is given by the Appendix B.

If initial condition satisfies u ¯ ( r ¯ , 0 ) = 0 , then the transforms of Equation (15) and Equation (14) can be written as

λ ¯ 1 s 2 U ( r ¯ , s ) + s U ( r ¯ , s ) = ∂ 2 U ( r ¯ , s ) ∂ r ¯ 2 + 1 r ¯ ∂ U ( r ¯ , s ) ∂ r ¯ + 2 ( 1 + λ ¯ 1 s ) ( 1 − sech ( a s 2 ) ) K 2 a s 2 I 0 ( K r ¯ ) I 0 ( K ) (20)

U ( r ¯ , s ) | r ¯ = 1 = 0 , ∂ U ( r ¯ , s ) ∂ r ¯ | r ¯ = 0 = 0 (21)

Equation (20) can be simplified as

∂ 2 U ( r ¯ , s ) ∂ r ¯ 2 + 1 r ¯ ∂ U ( r ¯ , s ) ∂ r ¯ − β 2 U ( r ¯ , s ) = − 2 ( 1 + λ ¯ 1 s ) ( 1 − sech ( a s 2 ) ) K 2 a s 2 I 0 ( K r ¯ ) I 0 ( K ) (22)

where β = λ ¯ 1 s 2 + s , sech is a hyperbolic secant function.

Equation (22) is a linear and inhomogeneous ordinary differential equation, and its solution can be written as the sum of a general solution U h ( r ¯ , s ) corresponding to homogeneous equation and a special solution U s ( r ¯ , s ) .

U ( r ¯ , s ) = U h ( r ¯ , s ) + U s ( r ¯ , s ) (23)

The homogeneous solution of Equation (22) is expressed as

U h ( r ¯ , s ¯ ) = A I 0 ( β r ¯ ) + B K 0 ( β r ¯ ) (24)

where I 0 and K 0 are modified Bessel functions of first and second kinds of order zero, respectively.

Due to the finite of U ( r ¯ , s ) at r ¯ = 0 , the constant B equal to zero from the boundary condition Equation (21). Therefore, the homogeneous solution of Equation (22) is rewritten as

U h ( r ¯ , s ¯ ) = A I 0 ( β r ¯ ) (25)

here A is constant, which can be determined from boundary conditions of Equation (21).

Considering the formation of the right hand side of Equation (22), the special solution can be given as

U s ( r ¯ , s ) = C I 0 ( K r ¯ ) (26)

Inserting Equation (26) into Equation (22) gives

C [ d 2 I 0 ( K r ¯ ) d r ¯ 2 + 1 r ¯ d I 0 ( K r ¯ ) d r ¯ − β 2 I 0 ( K r ¯ ) ] = − 2 ( 1 + λ ¯ 1 s ) ( 1 − sech ( a s 2 ) ) K 2 a s 2 I 0 ( K r ¯ ) I 0 ( K ) (27)

From Equation (8) and Equation (11), we can get

d 2 I 0 ( K r ¯ ) d r ¯ 2 + 1 r ¯ d I 0 ( K r ¯ ) d r ¯ = K 2 I 0 ( K r ¯ ) (28)

Substituting Equation (28) into Equation (27) and equalizing the coefficients in front of the modified Bessel functions I 0 ( K r ¯ ) at the two sides of the equation yields

C = − 2 ( 1 + λ ¯ 1 s ) ( 1 − sech ( a s 2 ) ) K 2 a s 2 ( K 2 − β 2 ) I 0 ( K ) (29)

Thus, the solution of velocity U ( r ¯ , s ¯ ) can be written as

U ( r ¯ , s ) = A I 0 ( β r ¯ ) − 2 ( 1 + λ ¯ 1 s ) ( 1 − sech ( a s 2 ) ) K 2 a s 2 ( K 2 − β 2 ) I 0 ( K ) I 0 ( K r ¯ ) (30)

The coefficient A with boundary condition of Equation (21) can be determined as

A = 2 ( 1 + λ ¯ 1 s ) ( 1 − sech ( a s 2 ) ) K 2 a s 2 ( K 2 − β 2 ) I 0 ( β ) (31)

Inserting Equation (31) into Equation (30), we have

U ( r ¯ , s ) = 2 ( 1 + λ ¯ 1 s ) ( 1 − sech ( a s 2 ) ) K 2 a s 2 ( K 2 − β 2 ) ( I 0 ( β r ¯ ) I 0 ( β ) − I 0 ( K r ¯ ) I 0 ( K ) ) (32)

where β = λ ¯ 1 s 2 + s , sech is a hyperbolic secant function.

As with rectangle pulse wave, the numerical computation must be performed by numerical inverse Laplace transform of Equation (32).

The ideal sawtooth pulse can be expressed as the following form

f ( t ) = { t a , t ∈ [ 0 , a ) , t a − 2 , t ∈ ( a , 2 a ] . (33)

The Laplace transform of ( 1 + ∂ ∂ t ¯ ) f ( t ¯ ) is given by the Appendix C.

Making the Laplace transform for Equation (15), we have

∂ 2 U ( r ¯ , s ) ∂ r ¯ 2 + 1 r ¯ ∂ U ( r ¯ , s ) ∂ r ¯ − β 2 U ( r ¯ , s ) = − ( 1 + λ ¯ 1 s − a s csch ( a s ) ) K 2 a s 2 I 0 ( K r ¯ ) I 0 ( K ) (34)

where β = λ ¯ 1 s 2 + s , csch is a hyperbolic cosecant function.

Therefore, the solution of Equation (34) with boundary condition (21) can be given as

U ( r ¯ , s ) = ( 1 + λ ¯ 1 s − a s csch ( a s ) ) K 2 a s 2 ( K 2 − β 2 ) ( I 0 ( β r ¯ ) I 0 ( β ) − I 0 ( K r ¯ ) I 0 ( K ) ) (35)

Similar to the above several pulse waves, the numerical calculation must also be performed by the numerical inverse Laplace transform of Equation (35).

In the previous section, we have obtained the semi-analytical solutions of the time periodic pulse EOF velocity of the triangle and sawtooth of Maxwell fluid through a circular microchannel, which rely mainly on electrokinetic width K, relaxation time λ ¯ 1 and pulse width a. Then, we also compare with the rectangle pulse EOF on the graph, and discuss the influence of the above several parameters on the velocity distribution of different pulse EOF. Among them, it is important to analyze the influence of relaxation time on the three kinds of EOF under the condition of a fixed pulse width, that is, the stability. Additionally, the effect of relaxation time on pulse EOF under different pulse widths is discussed.

and r ¯ = 0.5 . We can clearly find that for different pulses, the variations of velocity are relatively significant. Therefore, it is very necessary to study different time periodic pulse EOF. At the same time, it can be seen from

that the amplitude of velocity increases with relaxation time λ ¯ 1 . At the same time, we can also be seen from both

variations in the amplitude of the three kinds of pulse EOF velocity, we can see that the relaxation time has different effects on the velocity amplitude for different pulses. In particular, it has a greater impact on triangle pulse and sawtooth pulse than rectangle pulse (see

In this article, the semi-analytical solutions for both triangle and sawtooth time periodic pulse EOF of Maxwell fluid through a circular microchannel under the Debye-Hückel approximation are presented. Based on the results obtained in this work, it can be concluded that with the electrokinetic width K increases, the velocity variations are mainly limited to the narrow area close to the EDL for small relaxation time λ ¯ 1 . However, as the increase of relaxation time λ ¯ 1 , the elasticity of the fluid becomes significant and the velocity variations can be extended to the entire flow field. At the same time, the velocity amplitude will significantly larger, and the flow needs longer time to attain steady status. Moreover, the time it takes for the fluid to change from a static state to a flowing state increases with relaxation time λ ¯ 1 . For given pulse widtha, the effect of relaxation time λ ¯ 1 on triangle pulse EOF and sawtooth pulse EOF is greater than rectangle pulse EOF, which implies that the rectangle pulse EOF is more stable. With the increase of pulse width a, the effect of relaxation time λ ¯ 1 on the velocity will be weakened, the change period of the velocity profiles becomes larger, and the time required for the velocity profiles to reach a steady state also becomes longer.

This work was supported by the Scientific Research Project of Inner Mongolia University of Technology (Grant No. ZZ201813).

The authors declare no conflicts of interest regarding the publication of this paper.

Li, D.S. and Li, K. (2021) Talk about Several Time Periodic Pulse Electroosmotic Flow of Maxwell Fluid in a Circular Microchannel. Journal of Applied Mathematics and Physics, 9, 617-634. https://doi.org/10.4236/jamp.2021.94045

The Laplace transform of ideal rectangle pulse is expressed as follows [

∫ 0 ∞ f ( t ¯ ) e − s t ¯ d t ¯ = ∫ 0 T f ( t ¯ ) e − s t ¯ d t ¯ + ∫ T ∞ f ( t ¯ ) e − s t ¯ d t ¯ = ∫ 0 T f ( t ¯ ) e − s t ¯ d t ¯ + e − s T ∫ 0 ∞ f ( τ ) e − s τ d τ ( Let t ¯ = T + τ ) (A.1)

here T = 2 a . It needs to point out that all of the following pulses have a period of 2a.

By shifting the term of Equation (A.1) and using the Equation (10), we can get

F ( s ) = ∫ 0 ∞ f ( t ¯ ) e − s t ¯ d t ¯ = 1 1 − e − s T ∫ 0 T f ( t ¯ ) e − s t ¯ d t ¯ = 1 1 − e − 2 a s F 1 ( s ) = 1 s tanh ( a s 2 ) (A.2)

where

F 1 ( s ) = ∫ 0 a f ( t ¯ ) e − s t ¯ d t ¯ + ∫ a 2 a f ( t ¯ ) e − s t ¯ d t ¯ = 1 − e − a s s − e − 2 a s ( e a s − 1 ) s = e − 2 a s ( e a s − 1 ) 2 s (A.3)

With the aid of Equation (A.1) and Equation (A.2), the Laplace transform of ideal triangle pulse can be given as

G ( s ) = ∫ 0 ∞ ( 1 + λ ¯ 1 ∂ ∂ t ) f ( t ¯ ) e − s t ¯ d t ¯ = 1 1 − e − s T ∫ 0 T ( 1 + λ ¯ 1 ∂ ∂ t ) f ( t ¯ ) e − s t ¯ d t ¯ = 1 1 − e − s T ( G 1 ( s ) + λ ¯ 1 G 2 ( s ) ) = 1 1 − e − 2 a s { 2 e − 2 a s ( − 1 + e a s 2 ) 3 ( 1 + e a s 2 ) a s 2 + λ ¯ 1 e − a s ( − 8 sinh ( a s 2 ) + 4 sinh ( a s ) ) a s ( 1 − e − 2 a s ) } = 2 − 2 sech ( a s 2 ) a s 2 + λ ¯ 1 2 − 2 sech ( a s 2 ) a s = 2 ( 1 + λ ¯ 1 s ) ( 1 − sech ( a s 2 ) ) a s 2 (B.1)

where

G 1 ( s ) = ∫ 0 a 2 f ( t ¯ ) e − s t ¯ d t ¯ + ∫ a 2 3 a 2 f ( t ¯ ) e − s t ¯ d t ¯ + ∫ 3 a 2 2 a f ( t ¯ ) e − s t ¯ d t ¯ = 2 ( 1 − 1 2 e − a s 2 ( 2 + a s ) ) a s 2 + e − 3 a s 2 ( 2 + a s + e a s ( − 2 + a s ) ) a s 2 + e − 2 a s ( − 2 + e a s 2 ( 2 − a s ) ) a s 2 = 2 e − 2 a s ( − 1 + e a s 2 ) 3 ( 1 + e a s 2 ) a s 2 (B.2a)

G 2 ( s ) = ∫ 0 a 2 ∂ ∂ t f ( t ¯ ) e − s t ¯ d t ¯ + ∫ a 2 3 a 2 ∂ ∂ t f ( t ¯ ) e − s t ¯ d t ¯ + ∫ 3 a 2 2 a ∂ ∂ t f ( t ¯ ) e − s t ¯ d t ¯ = 2 − 2 e − a s 2 a s 2 − 2 e − 3 a s 2 ( − 1 + e a s ) a s 2 + 2 e − 2 a s ( − 1 + e a s 2 ) a s 2 = e − a s ( − 8 sinh ( a s 2 ) + 4 sinh ( a s ) ) a s ( 1 − e − 2 a s ) (B.2b)

Similarly, the Laplace transform of ideal sawtooth pulse with Equation (A.1) and Equation (A.2) can be written as

H ( s ) = ∫ 0 ∞ ( 1 + λ ¯ 1 ∂ ∂ t ) f ( t ¯ ) e − s t ¯ d t ¯ = 1 1 − e − s T ∫ 0 T ( 1 + λ ¯ 1 ∂ ∂ t ) f ( t ¯ ) e − s t ¯ d t ¯ = 1 1 − e − s T ( H 1 ( s ) + λ ¯ 1 H 2 ( s ) ) = 1 1 − e − 2 a s ( 1 − e − 2 a s − 2 a s e − a s a s 2 + λ ¯ 1 1 − e − 2 a s a s ) = 1 − a s csch ( a s ) a s 2 + λ ¯ 1 1 a s = 1 + λ ¯ 1 s − a s csch ( a s ) a s 2 (C.1)

where

H 1 ( s ) = ∫ 0 a f ( t ¯ ) e − s t ¯ d t ¯ + ∫ a 2 a f ( t ¯ ) e − s t ¯ d t ¯ = 1 − e − a s ( 1 + a s ) a s 2 + e − 2 a s ( − 1 + e a s ( 1 − a s ) ) a s 2 = 1 − e − 2 a s − 2 a s e − a s a s 2 (C.2a)

H 2 ( s ) = ∫ 0 a ∂ ∂ t f ( t ¯ ) e − s t ¯ d t ¯ + ∫ a 2 a ∂ ∂ t f ( t ¯ ) e − s t ¯ d t ¯ = 1 − e − a s 2 a s 2 + e − 2 a s ( − 1 + e a s ) a s = 1 − e − 2 a s a s (C.2b)