Novel procedure for the estimation of swelling pressures of compacted bentonites based on diffuse double layer theory

Abstract

Bentonite clays are proven to be attractive as buffer and backfill material in high-level nuclear waste repositories around the world. A quick estimation of swelling pressures of the compacted bentonites for different clay–water–electrolyte interactions is essential in the design of buffer and backfill materials. The theoretical studies on the swelling behavior of bentonites are based on diffuse double layer (DDL) theory. To establish theoretical relationship between void ratio and swelling pressure (e versus P), evaluation of elliptic integral and inverse analysis are unavoidable. In this paper, a novel procedure is presented to establish theoretical relationship of e versus P based on the Gouy–Chapman method. The proposed procedure establishes a unique relationship between electric potentials of interacting and non-interacting diffuse clay–water–electrolyte systems. A procedure is, thus, proposed to deduce the relation between swelling pressures and void ratio from the established relation between electric potentials. This approach is simple and alleviates the need for elliptic integral evaluation and also the inverse analysis. Further, application of the proposed approach to estimate swelling pressures of four compacted bentonites, for example, MX 80, Febex, Montigel and Kunigel V1, at different dry densities, shows that the method is very simple and predicts solutions with very good accuracy. Moreover, the proposed procedure provides continuous distributions of e versus P and thus it is computationally efficient when compared with the existing techniques.

Appendix 1

Calculation of swelling pressure from half-space distance by the proposed approach:

Known

Specific surface, S = 800 m²/g; base exchange capacity, B = 100 meq/100 g (P′ = 0.125 meq/100 m²); n = 0.001 M; v = 1; ε = 78.54 T = 298 K; and d = 50 Å

Calculation

• Step 1. Surface potential of single clay particle (Eq. 9),

  $$ \phi_{0} = 0.1725\frac{T}{\nu }\sinh^{ - 1} \left( {\frac{{1256.81 \times P^{'} }}{{\sqrt {n\varepsilon T} }}} \right) = 214.554\,{\text{mV}} $$

• Step 2. Scaled surface potential,

  $$ z = \frac{{\nu F\varphi_{0} }}{RT} = \frac{{1 \times 9.6487 \times 10^{7} \times 0.214554}}{{8.314 \times 10^{3} \times 298}} = 8.3556 $$

• Step 3. The dimensionless parameter, ξ, at x, which equals to the half-space distance d, where the mid-plane potential is to be calculated

  $$ \xi = \kappa \times x = \kappa \times d = 10^{6} \times 50 \times 10^{ - 8} = 0.5 $$

• Step 4. Scaled potential at any distance from the particle surface,

  $$ y = 2 \times \ln \left( {\frac{{\exp \left( \xi \right) + \tanh (\frac{z}{4})}}{{\exp \left( \xi \right) - \tanh (\frac{z}{4})}}} \right) = 2 \times \ln \left( {\frac{{\exp \left( {0.5} \right) + \tanh (\frac{8.3556}{4})}}{{\exp \left( {0.5} \right) - \tanh (\frac{8.3556}{4})}}} \right) = 2.6997 $$

• Step 5. Electric potential of a single non-interacting plate at distance, x, which equals to half-space distance d, where the midway potential is to be calculated,

  $$ \phi_{x = d} = y\frac{RT}{\nu F} = 2.6997 \times \frac{{8.314 \times 10^{3} \times 298}}{{1 \times 9.6487 \times 10^{7} }} \times 1000 = 69.319\,{\text{mV}} $$

• Step 6. The mid-plane potential is calculated using (12),

  $$ \phi_{d} = - 6.24 \times 10^{ - 4} \phi_{x = d}^{2} + 1.205\phi_{x = d} + 8.582 = { 89}. 1 1 3 {\text{ mV}} $$

• Step 7. Scaled mid-plane potential

  $$ u = \frac{{\nu F\phi_{d} }}{RT} = \frac{{1 \times 9.6487 \times 10^{7} \times 89.113}}{{8.314 \times 10^{3} \times 298}} \times 10^{ - 3} = 3.47 $$

• Step 8. The swell pressure corresponding to the half-space distance 50 Å is calculated using the Langmuir’s formula (5).

  $$ P = 2nRT(\cosh u - 1) = 74.77\,{\text{kPa}} $$