lecture 5 reservoir sim practical

Reservoir Simulator Practical Lecture 5: Effective Permeability Calculations

By: Dr. Julian E. Mindel

Lecture Milestones Permeability

Calculation of effective values

Isotropy vs. Anisotropy

Heterogeneities

Example: Diagonal permeability tensor calculation

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Permeability: Calculation of effective values Up until today we have always addressed porous medium properties as isotropic and

homogeneous, which allowed us to simplify modelling approaches of our governing equations.

The truth is that, while most of us enjoy writing theory behind simplified isotropic and homogeneous domains, real porous media are in general heterogeneous and anisotropic.

It is normal to "up-scale" heterogeneous fields of permeability or porosity to, at least, the level of our smallest mesh cell or region of interest, which will allow us to continue applying our Darcy + Mass Conservation laws for low Reynolds numbers1. It also allows us to homogenize our material, at least, to the cell-size level.

Anisotropy, however, is not something that can be overlooked while upscaling. Anisotropy establishes that flow entering a control volume through only one face will not necessarily remain in that direction while within the control volume, and may exit through the other faces. This is normally due to the internal lower resistance flow paths established via the heterogeneities within that control volume.

In this lecture we will be addressing two important aspects of our attempts at adding further porous-media-realism to our simulations : 1. How to model heterogeneous media without incurring into extreme levels of resolution

(typically, less cells = less computational expense). 2. How to address anisotropy.

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1. If we would actually geometrically discretize (i.e. generate a mesh!) our domain to the level of the void space, our macroscopic modelling laws might break down and we would need to generate new ones! Remember that the Reynolds number depends on a reference length or size.

Direction shown of this vector is merely illustrative and not necessarily the actual direction of the pressure gradient!

Imagine a portion of an incompressible porous medium, with an unknown permeability tensor in the absence of gravity.

This "control volume" has a prismatic shape, with sides of length , , and , which are aligned with the Cartesian axes. As we saw in our derivation of the mass conservation equation, the total volume of the domain is given by,

=

Given our incompressibility assumption, an assumed constant viscosity, and uniform (within the cube) permeability tensor we know that there will likely be a resulting pressure gradient in each direction linked to fluid flow in each direction through .

We also know that the Darcy velocity relationship yields the relationship between the fluxes and pressure gradients (in the absence of gravity),

= 1

,

This velocity may be integrated (or in our simplified cube case, multiplied by the face area) over each face of to obtain the mass fluxes (when multiplied by the constant density), as shown in the figure to the left. (final result of integration)

We also know that the mass flow

is related to the

volumetric flow rate via, =

Permeability: Isotropy vs anisotropy

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x y

z

=

+ =

=

+ =

+ =

=

=

Suppose we are told that our medium is isotropic. In that case we know that our tensor looks like this,

= 0 00 00 0

To find out the value of k, either through numerical simulation (also applicable to physical experiments), we only need to apply a pressure differential across two opposing faces of the cube (say, x direction), while preventing flow in the others1.

= 1

0 00 00 0

, where =

00

And so we know that ,

=

= 1

00

We can thus determine , after measuring volumetric flow (hint: surface flow in Paraview!) and dividing by the face area ( in our case) to calculate velocity,

=

=

=


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x y

z

= 0

+ = 0

= 0

+ = 0

+ =

=

=

00

1. Applying no-flow BC's is not necessary if we know that the material is truly isotropic. Even if we do not prevent flow across the remaining faces, no flow should appear!

Suppose now that we are told that the tensor is diagonal, however it is no longer isotropic. The diagonal-anisotropic permeability tensor is given by,

=

0 00 0

0 0

Applying the strategy from the previous slide, we can obtain each component of the tensor via applying pressure drops in each respective direction, =

, =

x

, =

Where we would determine , , and in three separate numerical experiments (simulations), applying the same pressure drop in each direction.

Notice how although we have applied the same pressure differential in each direction, the flow rates will be different since the permeability tensor has different diagonal values (while also assuming = = ).

This case shows anisotropy, although the absence of off-diagonal values in the tensor prevent pressure drops in one direction to potentially cause flow in another.


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x y

z

=

+ =

=

+ =

+ =

=

1 =

00

or 2 =

0

0

or 3 =

00

These are the gradients used in rach independent simulation to determine the three tensor diagonal components.

Imposed pressure gradient is oriented in the x direction as stated in the example. (if is positive, otherwise it points in the opposite direction)

Isotropy and diagonal anisotropy are particular cases of the otherwise more general full anisotropic case.. For any our risk assessment modelling purposes, it would not be prudent to assume isotropy in general.

Let us now suppose that our cube is filled with an anisotropic medium. Thus, our tensor within Darcy's law would be,

= 1

,

In contrast to the isotropic and diagonal anisotropy cases, we are not able to simply "close" the boundaries and apply independent pressure differentials. This would only get us the values of the diagonal components.

Calculating the off diagonal components in practice involves imposing fluxes in one direction, and with all boundaries free, measuring resulting pressure drops and fluxes in the other two. By doing this in three directions, you obtain a system of equations, where the coefficients to solve for are the components of the tensor.

Remember, pressure drop or flow imposed in one direction (x in this example) will cause fluid flow and pressure drops in all three others. The directional "coupling" is caused by the off-diagonal terms.

Remember that, in general, the permeability tensor is symmetric.


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x y

z

= +

+ =

+

+

= +

+ =

+

+

+ =

+

+

= +

Permeability: Isotropy vs anisotropy The case of diagonal anisotropy is a special case that happens rarely in nature, particularly

while assuming the orthogonal coordinate system , , and . It is important to note at this point that a generic permeability tensor1 can be "diagonalized"

by changing/rotating the reference local coordinate system. In short, given,

=

typically possesses "eigenvectors" 1 and 2 and 3, also known as the principal components or directions of a particular tensor. Each one of those vectors corresponds to a particular "eigenvalue", also charateristic of .

In this particular case, the mentioned eigenvectors can be used to construct a new orthogonal coordinate system (a basis like x, y, and z), in which there exists a new matrix . This new matrix is the diagonalized version of generic ,

=

11 0 0

0 22 0

0 0 33

Note that unless special care is taken (i.e. a specific pre calculation is in place) matrix may not be used directly in a simulator.

Also note that 1 and 2 and 3 coupled to the eigenvalues give rise to the elipsoidal representation of tensor . (see next slide)

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1. With particular characteristics (e.g. Symmetric positive definite or semi-definite)


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= 0 00 00 0

=

=

11 0 0

0 22 0

0 0 33

Isotropic Permeability Tensor Representation (one point in space)

Anisotropic Permeability Tensor Representation (one point in space)

1

3

2

Note: Principal directions 1, 2, and 3 are interchangeable (i.e. axis 3 is not always the smallest axis)

Sphere

Permeability: Heterogeneities While the homogeneous isotropic case is relatively simple to model and useful

to formulate and understand theories, heterogeneities play a very important role in defining an overall anisotropy within our control volume, and thus the complexity of the fluid flow. This anisotropy must be honoured correctly to model its effects appropriately.

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Figure 1: Representative plot of a cross section of our control volume . Different colours represent different permeabilities.

Permeability: Heterogeneities Taking a rather macroscopic view of the rock, one

could formulate a permeability tensor within an REV (Representative Elementary Volume).

This permeability tensor, while quite possibly anisotropic, can be assumed homogeneous within this representative volume for all modelling purposes. Note that this does not mean that your whole model is homogeneous, but it means that certain regions have been homogenized for the sake of modelling simplicity.

While the detailed studies regarding the statistics of how to obtain such values will remain outside the scope of this course, we will perform a series of simulations that will allow us to calculate flow within an REV, to obtain effective values of permeability.

Note that we will be assuming that those models have been created with such detail that they are true or statistically true models of reality.

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Example: Diagonal permeability tensor calculation Use the model ThreeSectors with the following parameters:

= 1 1012 for SECTOR1 (matrix rock)

= 1 1011 for SECTOR2

= 1 1013 for SECTOR3

= 0.001

The model is a 10x10x10 m cube containing heterogeneities, and we will calculate their effects on an otherwise diagonal permeability tensor (we will assume no flow on the side walls during the individual directional tests).

Using the SteadyStatePressure simulator (use the launcher provided!), perform the required independent directional flow tests to obtain the effective permeability tensor characteristic of this model if it was to be used as an REV. Set a pressure differential in the x-direction (i.e. try 3 bar), and measure the

volumetric flow rate using Paraview. (Make sure to remain slightly away from the domain boundary when measuring!!!).

Perform identical measurements in the y and z directions.

Using the theory from class, and assuming a diagonal tensor (off diagonal terms neglected), formulate the permeability tensor.

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lecture 5 reservoir sim practical

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