Analysis of capillary filling in nanochannels with electroviscous effects

Abstract

Capillary filling is the key phenomenon in planar chromatography techniques such as paper chromatography and thin layer chromatography. Recent advances in micro/nanotechnologies allow the fabrication of nanoscale structures that can replace the traditional stationary phases such as paper, silica gel, alumina, or cellulose. Thus, understanding capillary filling in a nanochannel helps to advance the development of planar chromatography based on fabricated nanochannels. This paper reports an analysis of the capillary filling process in a nanochannel with consideration of electroviscous effect. In larger scale channels, where the thickness of electrical double layer (EDL) is much smaller than the characteristic length, the formation of the EDL plays an insignificant role in fluid flow. However, in nanochannels, where the EDL thickness is comparable to the characteristic length, its formation contributes to the increase in apparent viscosity of the flow. The results show that the filling process follows the Washburn’s equation, where the filled column is proportional to the square root of time, but with a higher apparent viscosity. It is shown that the electroviscous effect is most significant if the ratio between the channel height (h) and the Debye length (κ ⁻¹) reaches an optimum value (i.e. κh ≈ 4). The apparent viscosity is higher with higher zeta potential and lower ion mobility.

Appendix 1

To analyze the stability of the critical point found in Eq. (30), it is necessary to rewrite equation Eq. (29) as below

$$ \left\{ {\begin{array}{*{20}l} {\dot{\chi } = \alpha (\chi ,\xi ,\bar{u}) = \frac{{\hat{F}_{\text{s}} }}{C} - \frac{A}{C}\chi - \frac{B}{C}\xi } \\ {\dot{\xi } = \beta (\chi ,\xi ,\bar{u}) = - D\xi + E\chi + \frac{\xi }{\chi }\bar{u}^{2} } \\ {\dot{\bar{u}} = \gamma (\chi ,\xi ,\bar{u}) = \left( {\frac{{\hat{F}_{\text{s}} }}{C} - \frac{A}{C}\chi - \frac{B}{C}\xi } \right)\frac{{\bar{u}}}{\chi } - \frac{{\bar{u}^{3} }}{\chi }} \\ \end{array} } \right. $$

(38)

These non-linear equations cannot be linearized because γ _x , γ _ξ , $\gamma_{{\bar{u}}}$ reduce to 0 when $\bar{u} = 0.$

However, the stability still can be discovered by considering some experimental facts. First, the capillary filling length x increases with time. Therefore, $\bar{u} = \dot{x}$ is a non-negative function. Second, the viscous force is proportional to $\bar{u}x.$ Because the surface tension, as the capillary filling driving force, is a constant, the viscous force must be bound. Therefore, the average velocity $\bar{u}$ must approach 0 when t → ∞. So, the problem reduces to determining the stability of the critical point in $\bar{u} = 0$ hyperplane. With this condition, the Eq. (38) can be rewritten as.

$$ \left\{ {\begin{array}{*{20}l} {\alpha (\chi ,\xi ,0) = \frac{{\hat{F}_{\text{s}} }}{C} - \frac{A}{C}\chi - \frac{B}{C}\xi } \\ {\beta (\chi ,\xi ,0) = - D\xi + E\chi } \\ \end{array} } \right. $$

(39)

To linearize this equation in $\bar{u} = 0$ hyperplane, at the critical point, the Jacobian matrix J is evaluated at that point.

$$ J = \left[ {\begin{array}{*{20}c} {\alpha_{\chi } } & {\alpha_{\xi } } \\ {\beta_{\chi } } & {\beta_{\xi } } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} { - \frac{A}{C}} & { - \frac{B}{C}} \\ E & { - D} \\ \end{array} } \right] $$

(40)

The eigenvalues λ are calculated by

$$ \left| {\begin{array}{*{20}c} { - \frac{A}{C} - \lambda } & { - \frac{B}{C}} \\ E & { - D - \lambda } \\ \end{array} } \right| = \left( { - \frac{A}{C} - \lambda } \right)( - D - \lambda ) + \frac{\text{BE}}{C} = 0 $$

(41)

Expand the variable

$$ \lambda^{2} + \left( {\frac{A}{C} + D} \right)\lambda + \frac{{{\text{AD}} + {\text{BE}}}}{C} = 0 $$

(42)

Discriminant Δ

$$ \begin{aligned} \Updelta & = \left( {\frac{A}{C} + D} \right)^{2} - 4\frac{{{\text{AD}} + {\text{BE}}}}{C} \\ & = \frac{{A^{2} }}{{C^{2} }} - 2\frac{\text{AD}}{C} + D^{2} - \frac{{4{\text{BE}}}}{C} \\ & = \left( {\frac{A}{C} - D} \right)^{2} - \frac{{4{\text{BE}}}}{C} \\ \end{aligned} $$

(43)

It is necessary to know the sign of the reduced variables. By their definition, the variables A, C and D are positive. Both variables B and E are opposite in sign to ζ; therefore, the product BE is positive.

As a result, if Δ ≥ 0, two eigenvalues are

$$ \lambda_{1} = \frac{1}{2}\left( { - \frac{A}{C} - D - \sqrt \Updelta } \right) $$

(44)

$$ \lambda_{1} = \frac{1}{2}\left( { - \frac{A}{C} - D + \sqrt \Updelta } \right) $$

(45)

With A, C and D are positive, λ ₁ is negative obviously. In quadratic equation (42)

$$ \lambda_{1} \lambda_{2} = 4\frac{{{\text{AD}} + {\text{BE}}}}{C} > 0 $$

(46)

Then, λ ₂ is also negative. The critical point is a nodal sink (stable).

If Δ < 0, two eigenvalues are complex numbers, with the real parts are

$$ {\text{Re(}}\lambda_{1} )= {\text{Re}}(\lambda_{2} ) = \frac{ - A - D}{2C} < 0 $$

(47)

The critical point is a spiral sink (stable).

Because there is only one finite critical point, and three variables χ, ξ, and $\bar{u}$are bound as t → ∞, the critical point as in Eq. (30) describes the asymptotic solution of the system.