Prediction of inter-particle capillary forces for non-perfectly wettable granular assemblies

Abstract

At a moisture content that corresponds to the so-called pendular regime, granular assemblies are subjected to the development of inter-particle capillary forces. These forces provide a tensile resistance at the particle level, which results into a cohesion shear strength at the macroscopic scale. Granular assemblies with non-perfectly wettable particles show a non-zero contact angle between the liquid bridge and the particle surface. It is worth mentioning that such an angle is an intrinsic property of the particle pair in contact and the liquid bridge. Its value has a significant effect on the magnitude of the capillary force and its behaviour as a function of the particle separation distance. This study is mainly motivated by the large range of values of the contact angle observed experimentally. In this paper, the governing equations for non-perfectly wettable granular assemblies in the pendular regime are first developed using the toroidal approximation. A robust numerical procedure is then proposed to solve these equations. Experimental validation of the numerical model shows that the capillary forces are predicted with a very good accuracy. The influence of the contact angle on the predicted inter-particle capillary force is also discussed.

Appendices

Appendix 1

Expressions of functions $f_a$, $a=1,\ldots ,5$, involved in Eqs.  7–10 and 14.

$$\begin{aligned}&f_1 (\varvec{X})=(x_A -\xi )^{2}+(y_A -\eta )^{2}-\rho ^{2}\\&f_2 (\varvec{X})=(x_B -\xi )^{2}+(y_B -\eta )^{2}-\rho ^{2}\\&f_3 (\varvec{X})=(s\xi -c\eta )x_A +(c\xi +s\eta )y_A -s\\&f_4 (\varvec{X})=(sL+s\xi +c\eta )x_B \\&\qquad \qquad \quad +\,(cL-c\xi +s\eta )y_B -L(s\xi +c\eta )-s\\&f_5 (\varvec{X})=\pi \left[ {F_v (x_B )-F_v (x_A )} \right] -V_A^*-V_B^*-V_w^*\end{aligned}$$

In the expressions above, $F_v (x)$, $V_I^*,I=A,B$ and $V_w^*$ are defined by (15), (12) and (2), respectively. The parameters c and s correspond to $\cos (\theta )$ and $\sin (\theta )$, respectively.

Appendix 2

Expression of the components of the Jacobian matrix defined by Eq. 18.

$$\begin{aligned}&J_{11} =2\frac{x_A }{y_A }\eta -2\xi ; \quad J_{12} =0; \quad J_{13} =2\left( {\xi -x_A } \right) ; \\&J_{14} =2\left( {\eta -y_A } \right) ; \quad J_{15} =-2\rho ;\\&vJ_{21} =0; \quad J_{22} =-2\eta \left( {\frac{L-x_B }{y_B }} \right) -2\xi +2L; \\&J_{23} =2\left( {\xi -x_B } \right) ; \quad J_{24} =2\left( {\eta -y_B } \right) ; \quad J_{25} =-2\rho ;\\&J_{31} =\left( {s\xi -c\eta } \right) -\left( {c\xi +s\eta } \right) \frac{x_A }{y_A }; \\&J_{32} =0; \quad J_{33} =sx_A +cy_A ; \quad J_{34} =-cx_A +sy_A ;\\&J_{35} =0;\\&J_{41} =0; \quad J_{42} =\left( {cL-c\xi +s\eta } \right) \left( {\frac{L-x_B }{y_B }} \right) \\&\qquad \quad +\,\left( {sL+s\xi +c\eta } \right) ; \quad J_{43} =sx_B -cy_B -Ls;\\&J_{44} =cx_B +sy_B -Lc; \quad J_{45} =0;\\&J_{51} =0; \quad J_{52} =0; \quad J_{53} =\pi \left( {y_B -y_A } \right) \left( {y_B +y_A -4\eta } \right) ;\\&J_{54} =\pi \left[ {x_B \left( {\eta +y_B } \right) -x_A \left( {\eta +y_A } \right) -\xi \left( {y_B -y_A } \right) } \right] \\&\qquad \quad +\,\pi \rho ^{2}\left[ {\arctan \left( {\frac{x_B -\xi }{y_B -\eta }} \right) -\arctan \left( {\frac{x_A -\xi }{y_A -\eta }} \right) } \right] \\&J_{55} =2\pi \rho \left[ {x_B -x_A } \right] \\&\qquad \quad +\,2\pi \rho \eta \left[ {\arctan \left( {\frac{x_B -\xi }{y_B -\eta }} \right) -\arctan \left( {\frac{x_A -\xi }{y_A -\eta }} \right) } \right] . \end{aligned}$$

Appendix 3

Algorithm for the numerical procedure to solve Eqs.  7–10 and 14.

• (1) Obtain $\varvec{X}^{(1)}$using the secant procedure proposed in Harireche et al. [11] and set $\varvec{X}^{(0)}=\varvec{X}^{(1)}$

• (2) Set increment counter to zero: $\textit{inc}=0$

• (3) Perform a new increment: $\textit{inc}\leftarrow \textit{inc}+1$

• (4) Set iteration counter to zero: $i=0$

• (5) Perform a new iteration: $i\leftarrow i+1$

• (6) Obtain $\Delta \varvec{X}$by solving: $\sum \nolimits _{b=1}^5 {J_{ab} } (\varvec{X}^{(i-1)})\Delta X_b =f_a (\varvec{X}^{(i-1)})$, $a=1,\ldots ,5$

• (7) Obtain $f_a (\varvec{X}^{(i)})=f_a (\varvec{X}^{(i-1)}+\Delta \varvec{X})$, $a=1,\ldots ,5$

• (8) Check convergence: $\left\| \varvec{F} \right\| <Tol$; where $\varvec{F}^{T}=(f_1 (\varvec{X}^{(i)}),\ldots ,f_5 (\varvec{X}^{(i)}))$ and Tol is typically $10^{-6}$.

• (9) If (convergence) Then set $\varvec{X}^{(0)}=\varvec{X}^{(i)}$ and Go to (3) Else Goto (5)

• (10) END