A Stochastic Method for Modelling the Geometry of a Single Fracture: Spatially Controlled Distributions of Aperture, Roughness and Anisotropy

Abstract

We describe a simple but effective stochastic method to model the void structure of a single fracture in a form of voxel representation. A fracture void is delineated by two bounding wall surfaces that are separated by some distance (i.e. the local aperture) at each location on the medial surface that lies within the fracture void and serves as a model reference frame. The three surface height fields are generated based on four parameters (mean, standard deviation and two spatial correlation lengths) for each field and two parameters (coefficient and synergistic length) for the spatial correlation between the fracture walls. Testing of generated models demonstrates that not only are the model fracture apertures spatially correlated and characterized as a Gaussian field, but also the two fracture walls are closely correlated, with a similar shape and/or height. With respect to fracture apertures, three quantities, i.e. the mean aperture, roughness and anisotropy, can be derived from the fracture models to describe fracture morphology. The effect of model fractures on fluid flow is investigated in order to establish the relationship between fracture permeability and the three morphological quantities, revealing a way to avoid the significant estimation error associated with the use of the cubic law.

Appendix

The spatial correlation of the elements in I is quantified by semi-variogram and has the following analytical relation with the filtering template S₀:

$$ \gamma_{I} (\varvec{h}) = \left\{ {\begin{array}{*{20}l} {0,} \hfill & {h = 0;} \hfill \\ {\frac{{N - N_{2} (\varvec{h})}}{N}\sigma_{I}^{2} ,} \hfill & {h \in (0,D_{\varvec{h}} );} \hfill \\ {\sigma_{I}^{2} ,} \hfill & {h \ge D_{\varvec{h}} .} \hfill \\ \end{array} } \right. $$

(25)

where $h = \sqrt {x^{2} + y^{2} }$ is the norm of h =(x, y), see Fig. 6, representing the distance between two locations at ζ and ζ + h. D_h is the geometrical length (see the distance between point C and point D in Fig. 6) of the template S₀ for fracture model generation, which is related to the direction vector h = (x, y); for example, D_h is the major axis (length) of S₀ if h =(x, 0) or the minor axis (length) if h =(0, y); N₂(h) is the overlapped area from the two filter templates, as shown in Fig. 6 and calculated by Eq. (15).

Proof

According to Eq. (6), the semi-variogram function of I with regard to h = (x, y) is

$$ \gamma_{I} \left( \varvec{h} \right) = \frac{1}{2M}\mathop \sum \limits_{k = 1}^{M} \left( {I\left( {\zeta_{k} } \right) - I\left( {\zeta_{k} + \varvec{h}} \right)} \right)^{2} $$

(26)

$I(\zeta_{k} ) = \frac{1}{N}\sum\nolimits_{j = 1}^{N} {I_{0} (\zeta_{kj} )}$, where $I_{0} (\zeta_{k1} ),I_{0} (\zeta_{k2} ),I_{0} (\zeta_{k3} ), \ldots ,I_{0} (\zeta_{kN} )$ are N pixel values in image I₀ specified by the template S₀, centred at ζ_k.

$I\left( {\zeta_{k} + \varvec{h}} \right) = \frac{1}{N}\sum\nolimits_{j = 1}^{N} {I_{0} \left( {\zeta_{kj} + \varvec{h}} \right)}$, where $I_{0} (\zeta_{k1} + \varvec{h}),I_{0} (\zeta_{k2} + \varvec{h}),I_{0} (\zeta_{k3} + \varvec{h}), \ldots ,I_{0} (\zeta_{kN} + \varvec{h})$ are N pixel values in image I₀ specified by the template S₀, centred at ζ_k + h.

Therefore,

$$ \gamma_{I} \left( \varvec{h} \right) = \frac{1}{2M}\mathop \sum \limits_{k = 1}^{M} \left( {\frac{1}{N}\mathop \sum \limits_{j = 1}^{N} I_{0} \left( {\zeta_{kj} } \right) - \frac{1}{N}\mathop \sum \limits_{j = 1}^{N} I_{0} \left( {\zeta_{kj} + \varvec{h}} \right)} \right)^{2} $$

(27)

For h < D_h, there are overlapped squares of the two filter templates, which have centres of ζ_k and ζ_k+ h, as illustrated in Fig. 6. For h ≥ D_h, there are no overlapped squares. Let N₁ be the number of no-overlap squares in one template and N₂ be the number of overlapped squares; thus, N₁+ N₂= N.

$$ \begin{aligned} I\left( {\zeta_{k} } \right) - I\left( {\zeta_{k} + \varvec{h}} \right) & = \frac{1}{N}\mathop \sum \limits_{j = 1}^{N} I_{0} \left( {\zeta_{kj} } \right) - \frac{1}{N}\mathop \sum \limits_{j = 1}^{N} I_{0} \left( {\zeta_{kj} + \varvec{h}} \right) \\ & = \frac{1}{N}\left[ {\mathop \sum \limits_{i = 1}^{{N_{1} }} \left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right) - I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right)} \right) + \mathop \sum \limits_{i = 1}^{{N_{2} }} \left( {I_{0}^{\prime \prime } \left( {\zeta_{ki} } \right) - I_{0}^{\prime \prime } \left( {\zeta_{ki} + \varvec{h}} \right)} \right)} \right] \\ & = \frac{1}{N}\mathop \sum \limits_{i = 1}^{{N_{1} }} \left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right) - I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right)} \right) \\ \end{aligned} $$

(28)

where $I_{0}^{\prime } \left( {\zeta_{ki} } \right),I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right),\,i = 1,2, \ldots ,N_{1}$ are the no-overlap pixel values in image I₀ specified by two templates, centred at ζ_k and ζ_k + h, so that $I_{0}^{\prime } \left( {\zeta_{ki} } \right) \ne I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right)$, while $I_{0}^{\prime \prime } \left( {\zeta_{ki} } \right), I_{0}^{\prime \prime } \left( {\zeta_{ki} + \varvec{h}} \right),\,i = 1,2, \ldots ,N_{2}$ are the overlapped pixel values in image I₀ specified by the two templates, so $I_{0}^{\prime \prime } \left( {\zeta_{ki} } \right) = I_{0}^{\prime \prime } \left( {\zeta_{ki} + \varvec{h}} \right)$.

Thus, Eq. (26) can be further arranged as:

$$ \begin{aligned} 2 \cdot \gamma_{I} \left( \varvec{h} \right) & = \frac{1}{M}\frac{1}{{N^{2} }}\mathop \sum \limits_{k = 1}^{M} \left[ {\mathop \sum \limits_{i = 1}^{{N_{1} }} \left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right) - I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right)} \right)} \right]^{2} = \frac{1}{{N^{2} }}\frac{{\mathop \sum \nolimits_{k = 1}^{M} \left[ {\mathop \sum \nolimits_{i = 1}^{{N_{1} }} \left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right) - I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right)} \right)} \right]^{2} }}{M} \\ & = \frac{1}{{N^{2} }}\frac{{\left[ {\mathop \sum \nolimits_{i = 1}^{{N_{1} }} \left( {I_{0}^{\prime } \left( {\zeta_{1i} } \right) - I_{0}^{\prime } \left( {\zeta_{1i} + \varvec{h}} \right)} \right)} \right]^{2} + \left[ {\mathop \sum \nolimits_{i = 1}^{{N_{1} }} \left( {I_{0}^{\prime } \left( {\zeta_{2i} } \right) - I_{0}^{\prime } \left( {\zeta_{2i} + \varvec{h}} \right)} \right)} \right]^{2} + \cdots + \left[ {\mathop \sum \nolimits_{i = 1}^{{N_{1} }} \left( {I_{0}^{\prime } \left( {\zeta_{Mi} } \right) - I_{0}^{\prime } \left( {\zeta_{Mi} + \varvec{h}} \right)} \right)} \right]^{2} }}{M} \\ \end{aligned} $$

(29)

Because

$$ \begin{aligned}& E\left( {\mathop \sum \limits_{i = 1}^{{N_{1} }} \left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right) - I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right)} \right)} \right)^{2} \\ &\quad = \frac{{\left[ {\mathop \sum \nolimits_{i = 1}^{{N_{1} }} \left( {I_{0}^{\prime } \left( {\zeta_{1i} } \right) - I_{0}^{\prime } \left( {\zeta_{1i} + \varvec{h}} \right)} \right)} \right]^{2} + \left[ {\mathop \sum \nolimits_{i = 1}^{{N_{1} }} \left( {I_{0}^{\prime } \left( {\zeta_{2i} } \right) - I_{0}^{\prime } \left( {\zeta_{2i} + \varvec{h}} \right)} \right)} \right]^{2} + \cdots + \left[ {\mathop \sum \nolimits_{i = 1}^{{N_{1} }} \left( {I_{0}^{\prime } \left( {\zeta_{Mi} } \right) - I_{0}^{\prime } \left( {\zeta_{Mi} + \varvec{h}} \right)} \right)} \right]^{2} }}{M} \end{aligned} $$

(30)

then:

$$ 2\gamma_{I} \left( \varvec{h} \right) = \frac{1}{{N^{2} }}E\left( {\mathop \sum \limits_{i = 1}^{{N_{1} }} \left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right) - I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right)} \right)} \right)^{2} $$

(31)

Because $I_{0}^{\prime } \left( {\zeta_{k1} } \right),I_{0}^{\prime } \left( {\zeta_{k1} + \varvec{h}} \right),I_{0}^{\prime } \left( {\zeta_{k2} } \right),I_{0}^{\prime } \left( {\zeta_{k2} + \varvec{h}} \right), \ldots ,I_{0}^{\prime } \left( {\zeta_{{kN_{1} }} } \right),I_{0}^{\prime } \left( {\zeta_{{kN_{1} }} + \varvec{h}} \right)$ are the pixel values of I₀ and are a sequence of independent and identically distributed random variables with $E(I_{0}^{\prime } (\zeta_{ki} )) = E(I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right) = \mu_{0}$ and $E\big[ \left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right) - I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right)} \right) \cdot \left( {I_{0}^{\prime } \left( {\zeta_{kj} } \right) - I_{0}^{\prime } \left( {\zeta_{kj} + \varvec{h}} \right)} \right) \big] = 0, \quad i \ne j$, then:

$$ E\left( {\mathop \sum \limits_{i = 1}^{{N_{1} }} \left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right) - I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right)} \right)} \right)^{2} = N_{1} E\left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right) - I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right)} \right)^{2} $$

(32)

Because $I_{0}^{\prime } \left( {\zeta_{1i} } \right),I_{0}^{\prime } \left( {\zeta_{1i} + \varvec{h}} \right),I_{0}^{\prime } \left( {\zeta_{2i} } \right),I_{0}^{\prime } \left( {\zeta_{2i} + \varvec{h}} \right), \ldots ,I_{0}^{\prime } \left( {\zeta_{Mi} } \right),I_{0}^{\prime } \left( {\zeta_{Mi} + \varvec{h}} \right)$ are the pixel values of I₀ and are a sequence of independent and identically distributed random variables with $E\left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right)} \right) = \mu_{0}$ and ${\text{Var}}\left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right)} \right) = \sigma_{0}^{2}$, then:

$$ \begin{aligned} E\left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right) - I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right)} \right)^{2} & = E\left( {I_{0}^{\prime 2} \left( {\zeta_{ki} } \right) - 2I_{0}^{\prime } \left( {\zeta_{ki} } \right)I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right) + I_{0}^{\prime 2} \left( {\zeta_{ki} + \varvec{h}} \right)} \right) \\ & = 2E\left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right)} \right)^{2} - 2E^{2} \left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right)} \right) = 2{\text{Var}}\left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right)} \right) = 2\sigma_{0}^{2} \\ \end{aligned} $$

(33)

Substituting Eqs. (32) and (33) into Eq. (31), we can get:

$$ 2\gamma_{I} \left( \varvec{h} \right) = \frac{1}{{N^{2} }}N_{1} E\left( {I_{0}^{\prime } \left( {\zeta_{ki} } \right) - I_{0}^{\prime } \left( {\zeta_{ki} + \varvec{h}} \right)} \right)^{2} = \frac{1}{{N^{2} }}N_{1} 2\sigma_{0}^{2} = \frac{1}{{N^{2} }}\left( {N - N_{2} } \right)2\sigma_{0}^{2} $$

(34)

Because $\sigma_{I}^{2} = \frac{{\sigma_{0}^{2} }}{N}$, then:

$$ \gamma_{I} \left( \varvec{h} \right) = \frac{{N - N_{2} }}{N}\sigma_{I}^{2} $$

(35)

Noting that, as shown in Fig. 6, if h =0, then N₂ = N, so γ_I(h) = 0; if 0< h < D_h, then N₂< N, which increases with smaller values of h, so $\gamma_{I} \left( \varvec{h} \right) = \sigma_{I}^{2} \cdot (N - N_{2} )/N$ is an increase function in the interval (0, D_h); if h ≥ D_h, then N₂ = 0, so $\gamma_{I} \left( \varvec{h} \right) = \sigma_{I}^{2} \cdot (N - 0)/N = \sigma_{I}^{2}$.

From the above analysis, we can obtain:

$$ \gamma_{I} (\varvec{h}) = \left\{ {\begin{array}{*{20}l} {0,} \hfill & {h = 0;} \hfill \\ {\frac{{N - N_{2} (\varvec{h})}}{N}\sigma_{I}^{2} ,} \hfill & {h \in (0,D_{\varvec{h}} );} \hfill \\ {\sigma_{I}^{2} ,} \hfill & {h \ge D_{\varvec{h}} .} \hfill \\ \end{array} } \right. $$