Confined viscoplastic flows with heterogeneous wall slip

Abstract

The steady, pressure-driven flow of a Herschel-Bulkley fluid in a microchannel is considered, assuming that different power-law slip equations apply at the two walls due to slip heterogeneities, allowing the velocity profile to be asymmetric. Three different flow regimes are observed as the pressure gradient is increased. Below a first critical pressure gradient G ₁, the fluid moves unyielded with a uniform velocity, and thus, the two slip velocities are equal. In an intermediate regime between G ₁ and a second critical pressure gradient G ₂, the fluid yields in a zone near the weak-slip wall and flows with uniform velocity near the stronger-slip wall. Beyond this regime, the fluid yields near both walls and the velocity are uniform only in the central unyielded core. It is demonstrated that the central unyielded region tends towards the midplane only if the power-law exponent is less than unity; otherwise, this region rends towards the weak-slip wall and asymmetry is enhanced. The extension of the different flow regimes depends on the channel gap; in particular, the intermediate asymmetric flow regime dominates when the gap becomes smaller than a characteristic length which incorporates the wall slip coefficients and the fluid properties. The theoretical results compare well with available experimental data on soft glassy suspensions. These results open new routes in manipulating the flow of viscoplastic materials in applications where the flow behavior depends not only on the bulk rheology of the material but also on the wall properties.

Appendices

Appendix 1. General solution in Regime I

In the general case with different slip exponents at the two walls

$$ \begin{array}{ll}{\tau}_{wi}={\beta}_i\;{u}_{wi}^{s_i},\hfill & i=1,2\hfill \end{array} $$

(A1)

and the two slip velocities satisfy

$$ {\beta}_1\;{u}_{w1}^{s_1}+{\beta}_2\;{u}_{w2}^{s_2}= GH $$

(A2)

In Regime I, u _w1 = u _w2 = u _w and thus

$$ {\beta}_1\;{u}_w^{s_1}+{\beta}_2\;{u}_w^{s_2}= GH $$

(A3)

After solving the above equation for u _w , we can calculate the two wall shear stresses by means of Eq. (A1). The first critical pressure gradient can be then found by setting the hydrophilic wall shear stress equal to the yield stress, τ _w1 = τ ₀.

Independent experiments of Vayssade et al. (2014) on soft glassy suspensions showed that s ₁ = 1 and s ₂ = 1/2. From Eq. (A3), we get

$$ {u}_w=\frac{\beta_2^2}{4{\beta}_1^2}{\left(\sqrt{1+\frac{4{\beta}_1 GH}{\beta_2^2}}-1\right)}^2 $$

(A4)

The first critical pressure gradient is given by

$$ {G}_1=\left(1+\frac{\sqrt{\beta_1{\tau}_0}}{\beta_2}\right)\frac{\beta_2\sqrt{\tau_0/{\beta}_1}}{H} $$

(A5)

Appendix 2. Bingham-plastic flow with Navier slip

It is clear from Eq. (13) that the first critical pressure G ₁ required for the material to yield at the lower wall is independent of the consistency index and the power-law exponent. In the case of Bingham plastic flow with Navier slip (n = s = 1), Eq. (19) for the second critical pressure gradient (the pressure gradient at which the material adjacent to the upper wall yields) is simplified as follows:

$$ \frac{1}{2}{\left( GH-2{\tau}_0\right)}^2-\left[\left({B}_1+{B}_2\right){\tau}_0-{B}_1 GH\right] GH=0 $$

(B1)

and thus G ₂ is given by

$$ {G}_2=\frac{2{\tau}_0/ H}{1+\frac{B_1+{B}_2}{2}\left[1\pm \sqrt{1+4\left({B}_2-{B}_1\right)/{\left({B}_1+{B}_2\right)}^2}\right]} $$

(B2)

The lower root is chosen if it is greater than G ₁ and the higher one otherwise.

The lower-wall shear stress in the three regimes is given by

$$ \frac{\tau_{w1}}{ G H}=\left\{\begin{array}{c}\hfill \begin{array}{cc}\hfill \frac{B_2}{B_1+{B}_2},\hfill & \hfill \kern17em 0\le G\le {G}_1\hfill \end{array}\hfill \\ {}\hfill \begin{array}{cc}\hfill \frac{\tau_0}{ G H}+\sqrt{{\left({B}_1+{B}_2\right)}^2+2{B}_2-2\left({B}_1+{B}_2\right)\frac{\tau_0}{ G H}}-{B}_1-{B}_2,\hfill & \hfill {G}_1\le G\le {G}_2\hfill \end{array}\hfill \\ {}\hfill \begin{array}{cc}\hfill \frac{1+2{B}_2-2{\tau}_0/ GH}{2\left(1+{B}_1+{B}_2-2{\tau}_0/ GH\right)},\kern11em \hfill & \hfill {G}_2\le G\kern1em \hfill \end{array}\hfill \end{array}\right. $$

(B3)

The two slip velocities can be calculated by means of

$$ \begin{array}{lll}{u}_{w1}=\frac{\tau_{w1}}{\beta_1}\hfill & \mathrm{and}\hfill & {u}_{w2}=\frac{GH-{\tau}_{w1}}{\beta_2}\hfill \end{array} $$

(B4)

and the positions of the yield points by:

$$ \frac{y_1}{H}=\frac{\tau_{w1}-{\tau}_0}{ G H}=\left\{\begin{array}{cc}\hfill \sqrt{{\left({B}_1+{B}_2\right)}^2+2{B}_2-2\left({B}_1+{B}_2\right)\frac{\tau_0}{ G H}}-{B}_1-{B}_2,\hfill & \hfill {G}_1\le G\le {G}_2\hfill \\ {}\hfill \frac{1+2{B}_2-2{\tau}_0/ GH}{2\left(1+{B}_1+{B}_2-2{\tau}_0/ GH\right)}-\frac{\tau_0}{ G H},\kern5em \hfill & \hfill {G}_2\le G\hfill \end{array}\right. $$

(B5)

and

$$ \frac{y_2}{H}=\frac{\tau_{w1}+{\tau}_0}{G H}=\frac{1+2{B}_2-2{\tau}_0/ GH}{2\left(1+{B}_1+{B}_2-2{\tau}_0/ GH\right)}+\frac{\tau_0}{G H},\kern1em {G}_2\le G $$

(B6)

Finally, the velocity in Regimes I–III is given respectively by

$$ {u}_x^I(y)=\frac{GH}{\beta_1+{\beta}_2}, $$

(B7)

$$ {u}_x^{II}(y)=\left\{\begin{array}{cc}\hfill {u}_{w1}+\frac{G}{2\mu}\left[{y}_1^2-{\left({y}_1- y\right)}^2\right],\hfill & \hfill 0\le y\le {y}_1\hfill \\ {}\hfill {u}_{w1}+\frac{G}{2\mu}{y}_1^2,\kern6em \hfill & \hfill {y}_1\le y\le H\hfill \end{array}\right. $$

(B8)

and

$$ {u}_x^{III}(y)=\left\{\begin{array}{c}\hfill \begin{array}{cc}\hfill {u}_{w1}+\frac{G}{2\mu}\left[{y}_1^2-{\left({y}_1- y\right)}^2\right],\kern1.12em \hfill & \hfill \kern3em 0\le y\le {y}_1\hfill \end{array}\hfill \\ {}\hfill \begin{array}{cc}\hfill {u}_{w1}+\frac{G}{2\mu}{y}_1^2,\hfill & \hfill \kern9.5em {y}_1\le y\le {y}_2\hfill \end{array}\hfill \\ {}\hfill \begin{array}{cc}\hfill {u}_{w2}+\frac{G}{2\mu}\left[{\left( H-{y}_2\right)}^2-{\left( y-{y}_2\right)}^2\right],\kern1em \hfill & \hfill {y}_2\le y\le H\hfill \end{array}\hfill \end{array}\right. $$

(B9)

The solution for the symmetric problem is obtained by setting β ₁ = β ₂. The two critical pressure gradients are then equal, G ₁ = G ₂ = 2τ ₀/H, so that the intermediate Regime II disappears. Moreover, τ _w1 = τ _w2 = GH/2 and the positions of the yield points in Regime III are given by:

$$ \begin{array}{ll}{y}_1=\frac{H}{2}-\frac{\tau_0}{G},\hfill & {y}_2=\frac{H}{2}+\frac{\tau_0}{G}\hfill \end{array} $$

(B10)

Hence, ${u}_x^I= GH/\left(2\beta \right)$ while ${u}_x^{III}$ is given by Eq. (B9).