Interblock flow and transmissibility

Quick refresher

The flow of a phase through a face is the product of three factors: the transmissibility of the connection $T_{ij}$ (geometry and permeability, calculated once at initialisation), the mobility of the phase at the face (it changes at every step), and the difference between the phase potentials at the cell centres.
The flow passes through half of one cell and then through half of the other, so the resistances add: $T_{ij} = 1/(1/t_i + 1/t_j)$ . The half that conducts less decides the connection, and an impermeable cell closes it completely. For a common face this is the harmonic mean of the permeabilities (an average in which the smaller values count the most), weighted by the lengths.
The pressure at the face is the average of the centre pressures, weighted by the transmissibilities of the halves. It is closer to the pressure of the cell that conducts better, and most of the pressure drop is lost in the one that conducts less.
The vertical connection between thin layers is very strong: for equal cells $T_z/T_x = (k_v/k_h)(\Delta x/\Delta z)^2$ , usually tens or hundreds. Only a very tight thin layer that continues over the whole area, or a zero transmissibility multiplier, separates the layers.
The mobility at a face is taken from the cell the phase flows out of, separately for each phase. The arithmetic mean of the two cells gives a wrong solution, and making the cells smaller (grid refinement) does not repair it. The harmonic mean shuts the face ahead of the front.
The two-point formula is exact when the line between the centres is perpendicular to the face and the axes of anisotropy (the directions of the highest and lowest permeability) match the grid axes. Otherwise its error does not go away when the cells are made smaller.

Key relationships

F_{\alpha} = 0.0853\,T_{ij}\,\lambda_{\alpha\uparrow}\left(\Phi_{\alpha,j} - \Phi_{\alpha,i}\right)

flow of the phase from cell $j$ into cell $i$ ; $\lambda_{\alpha\uparrow}$ is the mobility from the upstream cell

T_{ij} = \left(\frac{1}{t_i} + \frac{1}{t_j}\right)^{-1},\qquad t_i = \frac{k_iA}{\Delta x_i/2}

transmissibility of the connection: two half-cells in series, mD·m

p_f = \frac{t_i\,p_i + t_j\,p_j}{t_i + t_j}

pressure at the common face

\frac{T_z}{T_x} = \frac{k_v}{k_h}\left(\frac{\Delta x}{\Delta z}\right)^2

vertical and horizontal connections of equal cells

Common pitfalls

Averaging the permeabilities of neighbouring cells arithmetically. When one permeability is 100 times the other, the transmissibility comes out about 25 times too high.
Treating a thin tight layer as a barrier without setting a transmissibility multiplier for it. Over the area of a cell, it still passes a noticeable flow.
Carrying the transmissibility multiplier of a fault from one model to another. It depends on the cell sizes.
Using at a face the average mobility of the two neighbouring cells, or the mobility of the downstream cell (the cell the phase flows into).

After this lesson you can

derive the transmissibility of a connection from two half-cells that the flow crosses one after the other (in series), find the pressure at the face, and explain why permeabilities are averaged harmonically and not arithmetically
calculate transmissibilities with the geometric factors of Cartesian (rectangular), radial and corner-point cells, include the net-to-gross ratio (the share of the cell that is reservoir rock), and estimate the vertical connection between thin layers
find the connections between cells across a fault that offsets the layers (non-neighbour connections) and their transmissibilities
explain upstream weighting of the mobility (taking it from the cell the phase flows out of), find the upstream cell for each phase with gravity taken into account, and show what goes wrong when the mobility is averaged arithmetically or harmonically
explain when the two-point flow formula is inaccurate, and link this to the grid orientation effect

What is calculated at the face between cells

A simulator stores the properties in cells: pressure, saturations (the shares of the pore space filled by each fluid), permeability and porosity. There is one value per cell, and it belongs to the cell centre (Grids in reservoir simulation models). The fluid, however, moves from cell to cell through the common faces, and the balance of each cell (mass balance of a cell) is made up of the flows through its faces. So for every pair of neighbours we must write the flow through the face in terms of the values at the two centres. In other words, we must decide which “average” properties of the two cells act at the face.

We write Darcy's law for a phase $\alpha$ between the centres of cells $i$ and $j$ in a way that separates two kinds of quantities: those that depend only on the grid and the rock, and those that change during the calculation:

Flow of a phase through the common face of two cells

F_{\alpha} = 0.0853\,T_{ij}\,\lambda_{\alpha\uparrow}\left(\Phi_{\alpha,j} - \Phi_{\alpha,i}\right),\qquad \lambda_\alpha = \frac{k_{r\alpha}}{\mu_\alpha B_\alpha},\qquad \Phi_\alpha = p_\alpha - \gamma_\alpha D

where:

$F_{\alpha}$ flow of phase $\alpha$ from cell $j$ into cell $i$ ; negative if the phase flows from $i$ to $j$ , m³/d (standard conditions)
$T_{ij}$ transmissibility of the connection: the geometry of the cells and their absolute permeability, without the fluid properties, mD·m
$\lambda_{\alpha\uparrow}$ mobility of the phase at the face, taken from the cell upstream of this phase (see the section below), 1/(mPa·s)
$k_{r\alpha},\ \mu_\alpha,\ B_\alpha$ relative permeability (how well the phase flows when other phases are present), viscosity and formation volume factor (the volume the phase takes at reservoir conditions per unit volume at the surface) of the phase, —, mPa·s, m³/m³
$\Phi_\alpha$ phase potential at the cell centre: the phase pressure converted to zero depth, which accounts for the weight of the fluid, MPa
$\gamma_\alpha,\ D$ specific weight of the phase $\rho_\alpha g$ (in MPa/m it equals $0.00981\,\rho_\alpha$ with $\rho_\alpha$ in g/cm³) and the depth of the cell centre, measured downwards, MPa/m, m

Units: metric oilfield units: $k$ in mD (as part of $T_{ij}$ ), lengths in m, $\mu$ in mPa·s, pressures in MPa

Holds when

Darcy's law applies between the cell centres: the flow through a face depends only on the potential difference of these two cells (the two-point formula; its limits are in the last section).
The properties inside each cell are uniform.
The water pressure in a cell is $p_w = p_o - p_c(S_w)$ , so the capillary pressure enters the potential difference of the water.

The transmissibility $T_{ij}$ depends only on the grid and the rock. The simulator calculates it when it initialises the model and changes it only with multipliers. The mobility depends on saturation and pressure, so the simulator recalculates it at every time step and at every iteration (every repeat of the calculation within a step). For one phase in reservoir volumes the formula reduces to $q = 0.0853\,T_{ij}\,\Delta p/\mu$ from the lesson on the reservoir simulation model.

In other unit systems (1)

Oilfield units

F_{\alpha} = 1.127\cdot10^{-3}\,T_{ij}\,\lambda_{\alpha\uparrow}\left(\Phi_{\alpha,j} - \Phi_{\alpha,i}\right)

$T_{ij}$ in mD·ft ( $k$ in mD, area in ft², distance in ft), $\mu$ in cP, $B$ in rb/STB, potentials in psi; $F$ in STB/d.

$1.127\cdot10^{-3}$ is the constant of Darcy's law in oilfield units (lesson Rock permeability).

Transmissibility: two half-cells in series

Take two neighbouring cells along the $x$ axis. They have sizes $\Delta x_i$ and $\Delta x_j$ and permeabilities $k_i$ and $k_j$ , and they share a face of area $A$ . The permeability inside each cell is constant, so the pressure changes in a straight line between the centre and the face. At the face itself the pressure has one value, $p_f$ . All the fluid that left the centre of cell $i$ towards the face arrives from the face at the centre of cell $j$ , because no fluid builds up at the face. We write Darcy's law for each half:

Darcy's law for the halves of two cells

q = \frac{0.0853\,t_i}{\mu}\,(p_i - p_f) = \frac{0.0853\,t_j}{\mu}\,(p_f - p_j),\qquad t_i = \frac{k_i A}{\Delta x_i/2},\quad t_j = \frac{k_j A}{\Delta x_j/2}

where:

$q$ flow from cell $i$ into cell $j$ at reservoir conditions, m³/d
$t_i,\ t_j$ transmissibilities of the half-cells, from the centre to the common face, mD·m
$p_i,\ p_j,\ p_f$ pressures at the cell centres and at the common face, MPa
$k_i,\ k_j$ permeabilities of the cells in the direction of the connection, mD
$A,\ \Delta x_i,\ \Delta x_j$ area of the common face and sizes of the cells along the connection, m², m
$\mu$ viscosity of the fluid, mPa·s

Units: metric oilfield units

Holds when

Single-phase flow through the pair of cells. The permeability inside each cell is uniform.
The flow is perpendicular to the face, and the cell centres lie on the line perpendicular to it.

Each half is a stretch of the path with its own resistance $1/t$ . The pressure at the face is unknown. We remove it by adding the pressure drops over the two halves.

The pressure drops over the halves are $q\mu/(0.0853\,t_i)$ and $q\mu/(0.0853\,t_j)$ , and their sum is the full drop $p_i - p_j$ . From this we get Darcy's law between the centres, with the transmissibility of the connection, and the pressure at the face:

Transmissibility of the connection between two cells

T_{ij} = \left(\frac{1}{t_i} + \frac{1}{t_j}\right)^{-1} = \frac{A}{\dfrac{\Delta x_i}{2k_i} + \dfrac{\Delta x_j}{2k_j}},\qquad \bar k = \frac{\Delta x_i + \Delta x_j}{\Delta x_i/k_i + \Delta x_j/k_j}

where:

$T_{ij}$ transmissibility of the connection: the flow between the centres equals $0.0853\,T_{ij}(p_i - p_j)/\mu$ , mD·m
$t_i,\ t_j$ transmissibilities of the half-cells, mD·m
$\bar k$ the permeability that gives the same transmissibility over the distance between the centres $(\Delta x_i + \Delta x_j)/2$ : $T_{ij} = \bar k A/[(\Delta x_i + \Delta x_j)/2]$ , mD
$A,\ \Delta x,\ k$ area of the common face, sizes and permeabilities of the cells, m², m, mD

Units: any consistent units; in this lesson $k$ is in mD, $A$ in m² and lengths in m, so $T$ is in mD·m

Holds when

The same as for the half-cells. If the cross-sections of the halves differ (a one-dimensional model of a formation of variable thickness or width), each of $t_i$ and $t_j$ gets its own cross-section.
For equal cells $\bar k$ is the simple harmonic mean $2k_ik_j/(k_i + k_j)$ .

The resistances add up, so the half that conducts less decides the connection, and an impermeable cell ( $k = 0$ ) closes the connection completely. The average permeability of the connection is the harmonic mean, weighted by the lengths of the halves. This is the same law as for layers that the flow crosses at right angles to the bedding (layers in series).

Pressure at the common face

p_f = \frac{t_i\,p_i + t_j\,p_j}{t_i + t_j}

where:

$p_f$ pressure at the common face of the cells, MPa
$p_i,\ p_j$ pressures at the cell centres, MPa
$t_i,\ t_j$ transmissibilities of the half-cells, mD·m

Units: any consistent units

Holds when

Single-phase flow. The pressure changes linearly inside each half.

The pressure at the face is the average of the centre pressures, weighted by the transmissibilities of the halves. It lies closer to the pressure of the cell that conducts better, because most of the pressure drop is lost in the half that conducts less. The simulator does not need $p_f$ , since the calculation eliminated it, but it shows where the pressure drop is “spent”.

Worked exampleA large cell next to a short tight one: transmissibility, face pressure and averaging errors

Given (assumed data)

Cell 1: $\Delta x_1$ = 240 m, $k_1$ = 180 mD
Cell 2: $\Delta x_2$ = 60 m, $k_2$ = 15 mD, a tight zone divided into smaller cells
Common face: 200 × 8 m
Pressures at the centres: $p_1$ = 18.6 MPa, $p_2$ = 18.2 MPa
Oil viscosity: 0.9 mPa·s

Solution

Face area $A = 200\cdot8 = 1600$ m². Transmissibilities of the half-cells: $t_1 = 180\cdot1600/120 = 2400$ mD·m, $t_2 = 15\cdot1600/30 = 800$ mD·m.
Transmissibility of the connection: $T_{12} = 1/(1/2400 + 1/800) = 600$ mD·m. The average permeability of the connection over the distance between the centres, 150 m, is $\bar k = 300/(240/180 + 60/15) = 56.25$ mD.
Pressure at the face: $p_f = (2400\cdot18.6 + 800\cdot18.2)/(2400 + 800) = 18.50$ MPa. The half of cell 1, 120 m long, takes 0.1 MPa of the pressure drop. The half of cell 2, 30 m long, takes 0.3 MPa, which is 75% of the drop.
Flow: $q = 0.0853\cdot600\cdot0.4/0.9 = 22.7$ m³/d at reservoir conditions.
For comparison, here is the transmissibility with other “averages” of the permeability over the same distance of 150 m. The arithmetic mean, 97.5 mD, gives 1040 mD·m, which is 1.73 times too large. The mean weighted by the cell lengths, 147 mD, gives 1568 mD·m, which is 2.61 times too large. The geometric mean, 51.96 mD, gives 554 mD·m, which is 0.92 of the correct value.

Answer: The transmissibility of the connection is 600 mD·m. At a pressure drop of 0.4 MPa the flow is 22.7 m³/d, and 75% of the pressure drop is lost in the short tight half.

What it means. The arithmetic mean overestimates the flow 1.73 times, and the length-weighted mean overestimates it 2.61 times. The long cell of high permeability almost alone decides the length-weighted mean, although the flow must also pass through the tight cell. Here the geometric mean came close to the correct value by chance. With other cell lengths and another contrast it is clearly wrong (see the figure below). The pressure at the face shows where the pressure drop is spent. If we “dissolve” the tight zone into the permeability of the large cell, the model also puts the drawdown (the pressure drop that drives the flow) in the wrong place. The data in this example are assumed for illustration.

CalculatedHow many times a “wrong” average of the permeabilities overstates the transmissibility of the connection between two equal cells, compared with the correct, harmonic mean. It is shown against the contrast

r = k_2/k_1

. The arithmetic mean overstates it

(1 + r)^2/(4r)

times, and the geometric mean

(1 + r)/(2\sqrt r)

times. Both scales are logarithmic.

Geometric factors: directions, radial and corner-point grids

The formula for a connection is always the same: two halves in series. Only the geometry of a half changes: the area that the flow passes through and the length of the path from the centre to the face. In a Cartesian cell (a rectangular block with straight sides) this is $A/(\Delta x/2)$ . If the cross-section changes along the path, we find the resistance of the half by integration (adding up the small pieces of the path): $1/t = \int \dd s/(k\,A(s))$ from the centre to the face.

Transmissibility of a half-cell in different cases

Connection: Along $x$ in a Cartesian grid
Half-cell transmissibility $t_i$: $k_x\,\Delta y\,\Delta z\,\mathrm{NTG}/(\Delta x/2)$
Comment: the net-to-gross ratio NTG reduces the horizontal connection

Connection: Along $z$ (vertical)
Half-cell transmissibility $t_i$: $k_z\,\Delta x\,\Delta y/(\Delta z/2)$
Comment: the area is large and the path is short

Connection: Between the rings of a radial grid
Half-cell transmissibility $t_i$: $\theta h k/\ln(r_f/r_i)$
Comment: $r_f$ is the boundary between the rings, $r_i$ is the calculation point of the ring

Connection: Skewed cell of a corner-point grid
Half-cell transmissibility $t_i$: $k\,(\mathbf{A}\cdot\mathbf{c})/(\mathbf{c}\cdot\mathbf{c})$
Comment: $\mathbf{A}$ is the area vector of the face, $\mathbf{c}$ goes from the cell centre to the face centre

Connection: Cell to well
Half-cell transmissibility $t_i$: $2\pi kh/(\ln(r_0/r_w) + S)$
Comment: the well index (Well model in a grid block)

Connection: Cell to a constant-pressure boundary
Half-cell transmissibility $t_i$: $kA/(\Delta x/2)$
Comment: one half only: the pressure is set at the face

Connection	Half-cell transmissibility $t_i$	Comment
Along $x$ in a Cartesian grid	$k_x\,\Delta y\,\Delta z\,\mathrm{NTG}/(\Delta x/2)$	the net-to-gross ratio NTG reduces the horizontal connection
Along $z$ (vertical)	$k_z\,\Delta x\,\Delta y/(\Delta z/2)$	the area is large and the path is short
Between the rings of a radial grid	$\theta h k/\ln(r_f/r_i)$	$r_f$ is the boundary between the rings, $r_i$ is the calculation point of the ring
Skewed cell of a corner-point grid	$k\,(\mathbf{A}\cdot\mathbf{c})/(\mathbf{c}\cdot\mathbf{c})$	$\mathbf{A}$ is the area vector of the face, $\mathbf{c}$ goes from the cell centre to the face centre
Cell to well	$2\pi kh/(\ln(r_0/r_w) + S)$	the well index (Well model in a grid block)
Cell to a constant-pressure boundary	$kA/(\Delta x/2)$	one half only: the pressure is set at the face

In the common input file format, the net-to-gross ratio is applied to the horizontal transmissibilities but not to the vertical one. The vertical permeability is entered already averaged over the whole thickness of the cell (cell properties).

Now we compare the vertical and horizontal connections of equal cells. The area of the horizontal face of a cell is $\Delta x\Delta y$ , and the path between the centres in the vertical direction is $\Delta z$ . For a vertical face the area is $\Delta y\Delta z$ and the path is $\Delta x$ . Model cells are usually flat: tens or hundreds of metres across and a few metres thick.

Vertical and horizontal transmissibility of equal cells

\frac{T_z}{T_x} = \frac{k_v}{k_h}\left(\frac{\Delta x}{\Delta z}\right)^2

where:

$T_z,\ T_x$ transmissibilities of the connection with the neighbour above (below) and with the neighbour at the side, mD·m
$k_v,\ k_h$ vertical and horizontal permeability, mD
$\Delta x,\ \Delta z$ cell size in plan ( $\Delta x = \Delta y$ ) and in thickness, m

Units: dimensionless

Holds when

Equal, uniform neighbouring cells with $\Delta x = \Delta y$ . The net-to-gross ratio is one.

The ratio grows as the square of the “flatness” of the cells. Even for $k_v/k_h$ = 0.1, cells of 100 × 100 × 3 m are connected vertically more than a hundred times more strongly than horizontally. So gravity quickly separates the phases between the thin layers of the model. The vertical connections usually set the largest time step of an explicit scheme, a scheme that calculates the new values from the old ones only (why pressure is calculated implicitly).

Worked exampleVertical connection of two layers and a thin tight layer between them

Given (assumed data)

Cells: 100 × 100 × 3 m, one above the other
Permeability: $k_h$ = 150 mD, $k_v$ = 15 mD
Thin tight layer: 0.2 m thick, permeability 0.001 mD, lying between the cell centres, continuous over the whole area

Solution

Horizontal connection with the neighbouring cell: $T_x = \dfrac{100\cdot3}{50/150 + 50/150} = 450$ mD·m.
Vertical connection without the thin layer: $T_z = \dfrac{100\cdot100}{1.5/15 + 1.5/15} = 50{,}000$ mD·m. This is 111 times larger. By the formula, $0.1\cdot(100/3)^2 = 111$ .
Resistance per 1 m² of area without the thin layer: $1.5/15 + 1.5/15 = 0.20$ m/mD. The thin layer replaces 0.2 m of rock: minus $0.2/15 = 0.0133$ , plus $0.2/0.001 = 200$ m/mD, which gives 200.19 m/mD in total.
The vertical transmissibility multiplier is $0.20/200.19 = 0.000999$ . Then $T_z = 50.0$ mD·m, which is 11% of the horizontal connection.
At a pressure difference of 1 MPa, the thin layer passes $0.0853\cdot50.0\cdot1/1 = 4.3$ m³/d of a fluid with a viscosity of 1 mPa·s over the area of the cell (1 hectare).

Answer: Without the thin layer, the vertical connection is 111 times stronger than the horizontal one. A tight layer 0.2 m thick with a permeability of 0.001 mD weakens it about 1000 times but does not cut it: 50.0 mD·m remains.

What it means. A thin, very tight layer over the area of a whole cell still passes a few cubic metres per day for each megapascal of pressure difference. Over ten years at a pressure difference of 1 MPa this adds up to about 16,000 cubic metres. If the geology says that the layer is continuous and sealing, we set it with a zero vertical transmissibility multiplier. If the layer has breaks, the breaks decide the connection, not the permeability of the layer. We must represent the breaks in the grid or in the transmissibilities. The permeability of the layer measured on a core does not settle this question. The data in this example are assumed for illustration.

In a radial grid the cross-section grows with the radius. Integrating, instead of dividing the length by the area, gives a logarithm. This is the Dupuit equation (the formula for steady-state radial flow) for each half of a ring (Steady-state radial flow):

Transmissibility between the rings of a radial grid

t_i = \frac{\theta\,h\,k_i}{\ln(r_f/r_i)},\qquad t_{i+1} = \frac{\theta\,h\,k_{i+1}}{\ln(r_{i+1}/r_f)},\qquad T = \left(\frac{1}{t_i} + \frac{1}{t_{i+1}}\right)^{-1}

where:

$\theta$ angle of the sector; a full circle is $2\pi$ , rad
$h$ thickness of the layer of rings, m
$k_i,\ k_{i+1}$ permeability of the rings, mD
$r_i,\ r_{i+1}$ radii of the calculation points of the rings, m
$r_f$ radius of the boundary between the rings, m

Units: $k$ in mD and lengths in m, so $T$ is in mD·m (natural logarithm)

Holds when

Inside each half-ring the flow is steady radial flow in uniform rock.
If the rings have the same permeability, $T = \theta h k/\ln(r_{i+1}/r_i)$ , and the position of the boundary $r_f$ does not matter.

The well index is the same kind of half-cell transmissibility, from the wall of the well to the equivalent radius of the cell $r_0$ (well index). A logarithmic grid (radial grid) makes the transmissibilities of all the rings equal.

In corner-point geometry (corner-point grid) a cell is a skewed hexahedron (a block with six faces that are not exactly rectangles), and the path from the cell centre to the centre of the face is not perpendicular to the face. A common formula allows for this with a projection (it uses the part of the path that is perpendicular to the face):

Transmissibility of half of a skewed cell

t_i = k_i\,\frac{\mathbf{A}\cdot\mathbf{c}_i}{\mathbf{c}_i\cdot\mathbf{c}_i}

where:

$\mathbf{A}$ area vector of the common face: it points perpendicular to the face, and its length equals the area of the face, m²
$\mathbf{c}_i$ vector from the centre of cell $i$ to the centre of the face, m
$k_i$ permeability of the cell in the direction of the connection, mD

Units: $k$ in mD and lengths in m, so $t$ is in mD·m

Holds when

The two-point formula: the flow follows from the pressures of two cells only.
The net-to-gross ratio and the transmissibility multipliers enter as extra factors.

For a rectangular cell $\mathbf{A}\cdot\mathbf{c} = A\,\Delta x/2$ and $\mathbf{c}\cdot\mathbf{c} = (\Delta x/2)^2$ , and we get $kA/(\Delta x/2)$ . If the vector $\mathbf{c}$ is tilted away from the perpendicular to the face by an angle $\varphi$ , at the same distance to the plane of the face, the transmissibility is multiplied by $\cos^2\varphi$ , so it becomes smaller. A flow formula that uses the pressures of only two cells is called the two-point flux approximation (TPFA). The last section shows when it stops being accurate.

Faults, barriers and non-neighbour connections

A fault affects the flows in two ways. The first is geometry: the displacement of the two sides of the fault changes which layers touch and over what area. The second is the properties of the fault zone itself: crushed rock and clay smear (a smear of clay dragged along the fault plane) add their own resistance (fault sealing). A model with corner-point geometry accounts for the first effect by itself. We set the second with a transmissibility multiplier.

When the sides of a fault are displaced, a cell of layer 2 on one side can touch layers 1 and 2 on the other side. Connections between cells that are not neighbours by their index numbers $(i, j, k)$ are called non-neighbour connections (NNC). We calculate their transmissibility with the same two-half formula. The only difference is that the area is the overlap area of the cells on the plane of the fault. Other connections work in the same way. One kind goes through a layer that pinches out (thins to zero thickness), so that a cell of the upper layer connects directly to the lower one. Another joins a coarse grid to a finer grid placed inside it (local grid refinement). A third joins the model to an aquifer (a water-bearing layer next to the reservoir). For the simulator all these are rows of one list of connections: a pair of cells and a transmissibility.

Fault transmissibility multiplier for cells of different sizes

F = \frac{\dfrac{d_i}{k_i} + \dfrac{d_j}{k_j}}{\dfrac{d_i}{k_i} + \dfrac{d_j}{k_j} + \dfrac{l_f}{k_f}}

where:

$F$ multiplier of the transmissibility of the connection without the fault, from 0 to 1, —
$d_i,\ d_j$ distances from the cell centres to the plane of the fault, m
$k_i,\ k_j$ permeabilities of the cells across the fault, mD
$l_f,\ k_f$ width of the fault zone and its permeability; only the ratio $l_f/k_f$ enters the formula, m, mD

Units: dimensionless; all lengths in one unit and all permeabilities in one unit

Holds when

The fault zone is thin compared with the cells. Single-phase flow across the fault.
For equal cells the formula turns into the multiplier of the lesson on partially communicating faults.

The fault zone is a third “resistor” in the chain. The multiplier depends on the cell sizes. When the cells are made smaller, the resistance of the halves falls, but the resistance of the zone does not, so the same fault needs a smaller multiplier. That is why we carry $l_f/k_f$ (the resistance of the fault zone per unit area) from one model to another, and not the multiplier.

SchematicCross-section at the fault of the example below (a sketch: the vertical scale is true, the horizontal scale is not). The right side has dropped by 5 m: this is the throw of the fault. Layer 2 of the left side borders layers 1 and 2 of the right side, and layer 3 of the left side borders layer 2 of the right side. The arrows are the connections across the fault, with their transmissibilities. Layer 1 of the left side and layer 3 of the right side face shale.

Worked exampleConnections across a fault that offsets the layers

Given (assumed data)

Unit (a group of neighbouring layers) at the fault, top to bottom: layer 1: 4 m, 300 mD; layer 2: 6 m, 80 mD; layer 3: 5 m, 150 mD; impermeable shale above and below the unit
Fault: vertical, with a throw of 5 m (the right side is lower); we ignore the resistance of the fault zone (multiplier 1)
Cells: 100 m across the fault and 100 m along it

Solution

We measure depths from the top of layer 1 on the left. On the left the layers occupy 0–4, 4–10 and 10–15 m. On the right they occupy 5–9, 9–15 and 15–20 m.
Overlaps: layer 2 on the left with layer 1 on the right, 4 m (from 5 to 9 m); layer 2 on the left with layer 2 on the right, 1 m (from 9 to 10 m); layer 3 on the left with layer 2 on the right, 5 m (from 10 to 15 m). Layer 1 on the left faces the shale above the right side, and layer 3 on the right faces the shale below the left side, so they have no connections across the fault.
Transmissibilities from the overlap areas (100 m along the fault, 50 m from each centre to the fault): $T_{2\text{–}1} = 400/(50/80 + 50/300) = 505$ ; $T_{2\text{–}2} = 100/(50/80 + 50/80) = 80$ ; $T_{3\text{–}2} = 500/(50/150 + 50/80) = 522$ mD·m. Connections 2–1 and 3–2 are non-neighbour connections: the layer numbers differ.
The sum is 1107 mD·m. Without the offset, each layer would touch the same layer on the other side: $1200 + 480 + 750 = 2430$ mD·m. The offset left 46% of this connection.

Answer: Three connections remain across the fault, with a total transmissibility of 1107 mD·m, which is 46% of the connection without the displacement. The best layer, layer 1 of the left side (300 mD), has no connection across the fault at all.

What it means. Even a fully conducting fault can cut the layers off from each other by geometry alone. Oil of layer 1 on the left side can reach the right side only after it moves down into layer 2 inside the left side. An error of a couple of metres in the throw changes the overlap areas, and so the transmissibilities. That is why faults whose throw is comparable with the thickness of the layers matter so much in history matching. We did not take into account the vertical offsets of the cell centres from the middle of the overlap. At these sizes, including them would reduce the transmissibilities by less than 0.3%. The data in this example are assumed for illustration.

Mobility at the face: upstream weighting

The transmissibility is usually constant in time, but the mobility changes. At the displacement front (the moving edge of the zone swept by the injected fluid), the mobility of a phase in two neighbouring cells can differ many times over. Ahead of the front the water cannot move ( $k_{rw} = 0$ ), and behind it the water moves. Which value should we take at the face? For permeability, a conservation law gave the answer: what flows in must flow out, so the same flow passes through both halves one after the other. For mobility the reasoning is different. A phase that passes through the face comes from the cell it flows out of, and it carries the saturation of that cell. Saturation is carried by the flow, and the “information” about it moves together with the fluid. So we take the mobility of a phase at the face from the cell that is upstream of its flow. This is called upstream weighting.

Mobility of a phase at the face (upstream weighting)

\lambda_{\alpha\uparrow} = \begin{cases} \lambda_\alpha(S_j,\ p_j), & \Phi_{\alpha,j} \ge \Phi_{\alpha,i}\\ \lambda_\alpha(S_i,\ p_i), & \Phi_{\alpha,j} < \Phi_{\alpha,i}\end{cases}

where:

$\lambda_{\alpha\uparrow}$ mobility of phase $\alpha$ at the face between cells $i$ and $j$ , 1/(mPa·s)
$S,\ p$ saturations and pressure of the cell from which the mobility is calculated, —, MPa
$\Phi_{\alpha,i},\ \Phi_{\alpha,j}$ potentials of the phase at the cell centres, MPa

Units: as in the flow formula

Holds when

The direction is found for each phase separately, from the difference of its own potentials. With gravity and capillary pressure, the phases can flow in opposite directions through one face.
The viscosity and the formation volume factor change little from cell to cell. We take them together with the relative permeability from the upstream cell, or as averages of the two cells. This hardly affects the result.
The specific weight of the phase in the potential difference is taken as the average of the two cells, if the phase is present in both.

The displacing phase enters a cell where it is still immobile, with the mobility of the cell it came from. A phase that is immobile in the upstream cell does not flow through the face. So the saturations stay within physical limits. The price is first-order accuracy (the error falls only in proportion to the cell size) and a numerical smearing of the front over a few cells (Numerical dispersion).

The lesson on the two-phase flow equations shows the main point. With the mobility of the downstream cell, water would not enter a cell at irreducible water saturation (the water that stays in the pores and cannot flow), and the front would stop. Now we look at what other rules give at the front, first at a single face and then in a whole one-dimensional element of the formation (a straight row of cells).

Worked exampleOne face at a waterflood front: three rules

Given (assumed data)

Upstream cell (behind the front): $S_w$ = 0.60
Downstream cell (ahead of the front): $S_w$ = 0.22, irreducible, so the water cannot move
Relative permeabilities and fluids: Corey model of the lesson Relative permeability; oil 3.0 mPa·s, water 0.5 mPa·s (the base case of the lesson Buckley–Leverett theory and the Welge method)
Face: $T$ = 1000 mD·m; the total flow through it is 20 m³/d at reservoir conditions

Solution

Behind the front $k_{rw}$ = 0.09214 and $k_{ro}$ = 0.03622. The mobilities per unit permeability are $\lambda_w = 0.09214/0.5 = 0.184$ and $\lambda_o = 0.03622/3.0 = 0.0121$ 1/(mPa·s). Ahead of the front $\lambda_w = 0$ and $\lambda_o = 0.85/3.0 = 0.2833$ .
Upstream weighting (both phases flow out of the cell behind the front): $\lambda_w = 0.184$ , $\lambda_o = 0.0121$ , and their sum is $\lambda_t = 0.196$ . The fraction of water at the face is $f_w = 0.184/0.196 = 0.94$ .
Arithmetic mean: $\lambda_w = 0.184/2 = 0.092$ , $\lambda_o = (0.0121 + 0.2833)/2 = 0.1477$ , $\lambda_t = 0.240$ . Then $f_w = 0.092/0.240 = 0.38$ .
Harmonic mean: $\lambda_w = 0$ , because one of the two mobilities is zero. $\lambda_o = 2\cdot0.0121\cdot0.2833/(0.0121 + 0.2833) = 0.0232$ and $f_w = 0$ .
Of the 20 m³/d that pass through the face, upstream weighting gives 19 m³/d of water and 1 of oil. The arithmetic mean gives 8 and 12. The harmonic mean gives only oil. The pressure drop at the face is $20/(0.0853\cdot1000\cdot\lambda_t)$ : 1.2, 1.0 and 10.1 MPa.

Answer: With upstream weighting, the share of water in the flow through the face is 94%, the same as in the flow of the cell behind the front. With the arithmetic mean it is 38%. With the harmonic mean no water passes at all, and it needs a pressure drop about 8 times larger.

What it means. Only upstream weighting passes through the face the mixture that really comes from the cell behind the front. The arithmetic mean adds the mobility of the untouched cell to the oil mobility, and so it drives mostly oil through the face. The cell behind the front then loses oil faster than it does in reality. The harmonic mean locks the water in: the face passes only oil, and only under a huge pressure drop. The calculation below shows what this means for the whole waterflood. The data in this example are assumed for illustration.

CalculatedWater saturation of a one-dimensional element after 0.30 pore volumes of water have been injected: the exact Buckley–Leverett solution, and a calculation on a grid of 50 cells with the mobilities at the faces taken upstream and taken as the arithmetic mean (explicit scheme, time step half of the limit). The base case of the lesson Buckley–Leverett theory and the Welge method; see the example below.

Worked exampleWaterflooding an element with different rules for the mobility

Given (assumed data)

Element: one-dimensional and horizontal; the rock and the fluids are incompressible; relative permeabilities and fluids as in the previous example
Grids: 50 and 200 equal cells; explicit scheme with a time step of half the stability limit
Comparison: after 0.30 pore volumes injected (the front is where the water saturation falls below 0.40), and at breakthrough (1% of water in the production)

Solution

Exact solution: the water saturation jumps between 0.22 and 0.564 at 0.76 of the length of the element. Breakthrough comes after 0.40 pore volumes.
Upstream weighting: the front is at 0.77 of the length on 50 cells and at 0.76 on 200 cells. Breakthrough comes after 0.38 and 0.39 pore volumes. On 50 cells the front, smeared by numerical dispersion, is slightly ahead of the exact one. As the cells get smaller, the calculation approaches the exact solution.
Arithmetic mean: behind the front the water saturation is almost at its limit, 0.74 against $1 - S_{or}$ = 0.75, and the front lags behind: it is at 0.58 of the length on 50 cells and at 0.57 on 200 cells. Breakthrough comes after 0.51 and 0.52 pore volumes.
Harmonic mean: water does not pass the face between cells 1 and 2 while cell 2 is at irreducible water saturation. Cell 1 fills with water up to $1 - S_{or}$ after $(1 - 0.25 - 0.22)/50 = 0.0106$ pore volumes are injected. Then both mobilities at the face become zero, and the model cannot inject any more water into the formation.

Answer: With upstream weighting the calculation gets closer to the exact solution as the cells get smaller (breakthrough after 0.39 pore volumes on 200 cells, against the exact 0.40). With the arithmetic mean it settles on a wrong, almost piston-like displacement (the water pushes the oil ahead of it like a piston), with breakthrough after 0.52 pore volumes, 30% later. With the harmonic mean the injection stops.

What it means. The error of the arithmetic mean is hard to notice. Before breakthrough the result looks plausible: a smooth front and believable saturations. Yet the water breaks through almost a third too late, and a finer grid does not reveal this. A smaller time step does not help either: with a step five times smaller, breakthrough on 200 cells comes after 0.49 pore volumes. After breakthrough the result also becomes physically impossible. With the time step of the example, the water saturation at the outlet rises to 0.83 by one pore volume injected on 50 cells, which is above the limit $1 - S_{or}$ = 0.75. The rule carries oil out of a cell where the oil is already immobile. If we matched such a model to the production history, we would have to compensate the “late” breakthrough with false changes of the relative permeabilities or the permeability. Upstream weighting smears the front over a few cells, which is numerical dispersion, but its error falls as the cells get smaller. The data in this example are assumed for illustration.

We find the direction of flow for each phase separately. In a vertical connection, gravity can make water and oil flow towards each other (in opposite directions), and then the “upstream” cells of the two phases are different.

Worked exampleCountercurrent flow in a vertical connection: each phase has its own upstream cell

Given (assumed data)

Cells: two, one above the other, 100 × 100 × 4 m; $k_v$ = 25 mD, porosity 0.20; no wells
Saturations: water saturation in the upper cell $S_w$ = 0.60 (the water came along the upper layer), in the lower cell 0.22
Fluids: water density 1.04 g/cm³, oil density 0.80 g/cm³ (lesson Fractional flow function and mobility ratio); relative permeabilities and viscosities as above; capillary pressure not taken into account

Solution

Vertical transmissibility: $T_z = \dfrac{100\cdot100}{2/25 + 2/25} = 62{,}500$ mD·m. Specific weights: $\gamma_w = 0.00981\cdot1.04 = 0.01020$ and $\gamma_o = 0.00981\cdot0.80 = 0.00785$ MPa/m. The distance between the centres is 4 m.
There are no wells and everything is incompressible, so the total flow through the face is zero: as much oil rises as water sinks. Water flows down, so we take its mobility from the upper cell, $\lambda_w = 0.1843$ . Oil flows up, so we take its mobility from the lower cell, $\lambda_o = 0.2833$ 1/(mPa·s).
From the condition of zero total flow we find the pressure difference between the centres (lower minus upper). It is the mean hydrostatic pressure difference (the difference caused by the weight of the fluid columns), weighted by the mobilities: $\delta p = 4\,(\lambda_w\gamma_w + \lambda_o\gamma_o)/(\lambda_w + \lambda_o) = 0.0351$ MPa. It lies between the pressure difference of a column of water, 0.0408 MPa, and that of a column of oil, 0.0314 MPa.
Potential differences (upper minus lower): for water $0.0408 - 0.0351 = +0.0057$ MPa, so it flows down. For oil $0.0314 - 0.0351 = -0.0037$ MPa, so it flows up. The directions agree with what we assumed, so the upstream cells were chosen correctly.
Flow: $0.0853\cdot62{,}500\cdot0.1843\cdot0.0057 = 5.6$ m³/d of water down and the same amount of oil up. The water saturation of the lower cell (pore volume 8000 m³) grows by 0.00070 per day.
With mobilities averaged over the two cells, the arithmetic mean gives 2.8 m³/d, half as much. The harmonic mean gives zero: the water never sinks into the lower cell.

Answer: Water sinks into the lower cell at 5.6 m³/d, and oil rises to meet it. The mobility of the water is taken from the upper cell, and that of the oil from the lower cell.

What it means. The model reproduces the separation of the phases by gravity (countercurrent flow of the phases, Segregated flow and gravity displacement) only with upstream weighting taken separately for each phase. The water saturation of the lower cell grows by about 0.26 per year. This is comparable with the pace of a waterflood, which is why in thick formations the injected water “sinks” to the base. Capillary pressure in a water-wet rock (a rock that water clings to more than oil) pulls water into the drier lower cell, in the same direction. From the imbibition curve of the lesson Two-phase flow equations (imbibition is the uptake of water by the rock), the capillary pressure difference between these two saturations is about 37 kPa, against 9.4 kPa for the difference between the hydrostatic pressure drops of the water and the oil. So imbibition would speed up the exchange several times over. The data in this example are assumed for illustration.

The direction can change during the iterations

In the fully implicit method (a method that solves for all the unknowns together at the end of the step), the mobility is taken from the upstream cell. The simulator chooses that cell by the potentials at the end of the step, and the iterations keep improving these potentials. If at some face the potential difference of a phase is close to zero, the direction can flip from one iteration to the next. Then the derivatives in the Jacobian matrix (the table of derivatives that Newton's method uses) jump. Simulators damp such flipping, for example by fixing the direction for the duration of the iterations (Newton iterations). Schemes that take the mobility from two upstream cells, or that use limiters (extra rules that stop the numerical solution from overshooting), reduce the smearing of the front (ways to reduce dispersion).

The two-point formula, the stencil and the grid orientation effect

The two-point formula connects a cell only with the neighbours that share a face with it. In a two-dimensional Cartesian grid there are four such neighbours. The cell and its four neighbours form a pattern of five linked cells, called the five-point stencil. In a three-dimensional grid there are six neighbours (a seven-point stencil). This is economical: each row of the system of equations has only seven non-zero numbers (in three dimensions), and such systems are solved faster (Solving systems of linear equations in simulators). But this scheme has two known limitations.

SchematicStencils on a two-dimensional grid (a sketch): (a) five-point: a cell is linked to its four neighbours that share a face; (b) nine-point: links to the diagonal neighbours are added; (c) with a five-point stencil the flow to a well on the diagonal goes in “steps” across the faces, and the flow to a well in the same row goes straight.

The first is the grid orientation effect. Flow along a row of cells goes along a straight chain, and a diagonal flow goes in “steps” through the neighbours that share a face. The mobility at each face is taken from the upstream cell, and the displacing phase advances more easily along the grid lines than along the diagonal. When the mobility ratio is unfavourable, that is, when the displacing agent is more mobile than the oil, the front is unstable and the difference grows. Tongues of the agent grow along the grid lines, and a production well on a grid line gets breakthrough earlier than a well on the diagonal at the same distance (orientation effect, numerical diffusion along the grid directions). What helps:

nine-point stencils: part of the transmissibility is given to the diagonal connections, and the stencil becomes almost the same in all directions
orienting the grid along the main direction of flow, and unstructured grids
a check: the same calculation on a grid rotated by 45°, and a comparison of the breakthroughs

The second is accuracy on skewed cells and with rotated anisotropy. The two-point formula assumes that the flow through a face is decided by the pressure difference along the line between the centres. This is right if the line between the centres is perpendicular to the face and the axes of permeability anisotropy (the directions in which the permeability differs) match the grid axes. In the general case it is right if the vector $\mathbf{K}\mathbf{n}$ points along the line between the centres. Here $\mathbf{K}$ is the permeability tensor (it describes how permeability depends on direction) and $\mathbf{n}$ is the direction perpendicular to the face. Such a grid is called K-orthogonal (orthogonal in the sense of the permeability tensor $\mathbf{K}$ ). Otherwise, part of the flow through the face is driven by the pressure gradient along the face, which the formula does not see, and the error does not fall when the cells are made smaller. Then we use multipoint flux approximations (MPFA): the flow through a face is worked out from the pressures of several cells around it. An example is in the advanced section.

Terms and symbols

Transmissibility of a connection $T_{ij}$: The part of Darcy's law between the centres of two neighbouring cells that comes from their geometry and permeability. It is measured in mD·m and is found from two half-cells in series.
Half-cell transmissibility $t_i$: $k_iA/d_i$ , from the cell centre to the common face; $T_{ij} = 1/(1/t_i + 1/t_j)$ .
Pressure at the face $p_f$: The average of the centre pressures, weighted by $t_i$ and $t_j$ . It shows in which half the pressure drop is lost.
Transmissibility multiplier: A factor applied to the calculated transmissibility: barriers, faults, history matching parameters.
Non-neighbour connection (NNC): A connection between cells that are not neighbours by their index numbers. It runs across a fault that offsets the layers, through a layer that pinches out, or to a finer grid placed inside the model. Its area is the overlap area.
Phase potential $\Phi_\alpha$: $p_\alpha - \gamma_\alpha D$ . The potential difference decides where the phase flows.
Upstream weighting: The mobility of a phase at the face is taken from the cell that the phase flows out of.
Two-point formula (TPFA): The flow through a face worked out from the pressures of two cells only. It is exact on a K-orthogonal grid.

Common mistakes and misconceptions

Wrong

“The permeability of a connection between two cells is the average of their permeabilities.”

In fact

The flow passes through the halves of the cells one after the other, and the resistances add up. The permeability of the connection is the harmonic mean, weighted by the lengths. In the lesson's example the arithmetic mean made the transmissibility 1.73 times too high. For two equal cells whose permeabilities differ 100 times, it makes it about 25 times too high.

Wrong

“A thin shale layer with a permeability of 0.001 mD cuts the layers off from each other.”

In fact

Over the area of a cell it weakens the vertical connection about a thousand times, but in the lesson's example about 50 mD·m remains, a tenth of the horizontal connection. A thin layer that is continuous and sealing is set with a zero vertical transmissibility multiplier.

Wrong

“A fault transmissibility multiplier tuned in one model will suit another model as well.”

In fact

The multiplier is the ratio of the transmissibilities with and without the fault, and it depends on the cell sizes. From model to model we carry the ratio of the width of the fault zone to its permeability, and we recalculate the multiplier.

Wrong

“It is more accurate to take the mobility at the face as the average of the two cells.”

In fact

In the lesson's calculation the arithmetic mean settled on an almost piston-like displacement, with breakthrough almost a third later than the exact one, and smaller cells did not repair it. The harmonic mean shut the face ahead of the front. We take the mobility from the upstream cell, separately for each phase.

In depth: advanced detail

When the two-point formula is wrong: rotated anisotropy

Suppose the principal axes of anisotropy (the directions of the highest and lowest permeability) are rotated against the grid axes by an angle $\theta$ . This happens, for example, when fractures run along the diagonal of the grid. Then the permeability tensor in the grid axes gets an off-diagonal component (a term that links the flow in one direction to the pressure gradient in the other):

Permeability tensor in the grid axes for rotated principal axes

k_{xx} = k_1\cos^2\theta + k_2\sin^2\theta,\qquad k_{yy} = k_1\sin^2\theta + k_2\cos^2\theta,\qquad k_{xy} = (k_1 - k_2)\sin\theta\cos\theta

where:

$k_1,\ k_2$ principal permeabilities (along and across the dominant direction), mD
$\theta$ angle between the first principal axis and the $x$ axis of the grid, °
$k_{xx},\ k_{yy},\ k_{xy}$ components of the tensor in the grid axes, mD

Units: mD

Holds when

A two-dimensional problem. Darcy's law $\mathbf{u} = -(\mathbf{K}/\mu)\nabla p$ with a symmetric tensor.

The flow along $x$ equals $-(k_{xx}\,\partial p/\partial x + k_{xy}\,\partial p/\partial y)/\mu$ . When $k_{xy} \ne 0$ , a gradient along $y$ also drives the fluid along $x$ . For faces perpendicular to $x$ , the two-point formula uses only the pressure difference of the neighbours along $x$ , and it loses the term with $k_{xy}$ .

Worked exampleWhat the two-point formula does not see

Given (assumed data)

Principal permeabilities: $k_1$ = 100 mD, $k_2$ = 10 mD
Rotation: 45° to the $x$ axis of the grid
Pressure gradient: only along the $y$ axis

Solution

Tensor components: $k_{xx} = k_{yy} = 100\cdot0.5 + 10\cdot0.5 = 55$ mD, $k_{xy} = (100 - 10)\cdot0.5 = 45$ mD.
The true flow through the faces perpendicular to $x$ is $k_{xy}/k_{yy} = 0.82$ of the flow along $y$ : the fluid goes sideways, towards the direction of high permeability.
For these faces the two-point formula sees a zero pressure difference between the neighbours along $x$ and gives zero flow, however fine the grid.

Answer: The two-point formula misses the sideways flow completely. This flow is 82% of the flow along the pressure gradient.

What it means. The error does not come from the size of the cells. It comes from the fact that $k_{xy}$ is missing from the formula, so smaller cells do not reduce it. We avoid it by aligning the grid axes with the principal axes of anisotropy, or we use multipoint schemes. They work out the flow through a face from the pressures of several cells and take the off-diagonal components into account. Their price is more connections in each row of the system of equations. With strong anisotropy and skewed cells they can also give non-monotonic solutions (results with false ups and downs that the physics does not allow). The data in this example are assumed for illustration.

Gravity and capillary pressure in the potential difference

We write the difference of phase potentials between cell centres at different depths as $\Phi_{\alpha,j} - \Phi_{\alpha,i} = (p_{\alpha,j} - p_{\alpha,i}) - \bar\gamma_\alpha\,(D_j - D_i)$ . Here $\bar\gamma_\alpha$ is the specific weight of the phase at the face: the average of the two cells (sometimes weighted by the saturations), or, if the phase is absent in one of the cells, the value of the other.

This matters already at initialisation. We calculate the initial pressures from hydrostatic equilibrium (the balance of the weights of the fluid columns; initial state). Suppose they were calculated with a different averaging of the density than the one the model uses for the flows. Then in a formation at rest the potential differences are not zero, and false flows start from the first step, without any wells. The check is simple: run the model without wells for several years. The pressures and saturations must not change.

Capillary pressure enters through the water pressure $p_w = p_o - p_c(S_w)$ . In a water-wet rock, $p_c$ is higher where there is less water, so water is drawn into the drier cell even against the oil pressure difference. This is how the model describes capillary imbibition and the smearing of the front. On large cells this contribution is usually small compared with the viscous forces (scale of the capillary forces). But in vertical countercurrent flow without injection, as in the example of the lesson, it can dominate.

Connections to other topics

Grids in reservoir simulation models — grids, non-neighbour connections and the orientation effect.
Pressure equation for single-phase flow — the equations of the cells with the connection coefficients $0.0853\,T_{ij}/\mu$ .
Two-phase flow equations — the cell equations for two phases, and why the mobility is taken upstream.
Well model in a grid block — the connection between a cell and a well: the well index.
Reservoir heterogeneity and permeability averaging — average permeabilities of layers and what they mean.
Permeability upscaling for single-phase flow — transmissibilities of large cells directly from a detailed model.
Special topics in upscaling — the full permeability tensor after upscaling, and what a two-point scheme loses with it.
Partially communicating faults — transmissibility of a fault zone from a well test.
Numerical dispersion — smearing of the front by the upstream scheme, and high-order schemes.
History matching — transmissibility multipliers as history matching parameters.
Solving systems of linear equations in simulators — how the systems of equations built from these transmissibilities are solved.

Check yourself

1
Two equal neighbouring cells have permeabilities of 120 and 3 mD. Which permeability decides the flow between their centres?
1. A.61.5 mD, the arithmetic mean
2. B.≈ 5.9 mD, the harmonic mean
3. C.3 mD, the smaller of the two
4. D.19 mD, the geometric mean
Show the answer
Correct answer: B. ≈ 5.9 mD, the harmonic mean
The flow passes through the halves of the cells one after the other, and the resistances add up: $\bar k = 2\cdot120\cdot3/(120 + 3) \approx 5.85$ mD. The arithmetic mean would overestimate the flow $(1 + r)^2/(4r) \approx 10.5$ times ( $r$ = 0.025), and the geometric mean more than three times. The smaller permeability would underestimate the flow by almost half: the high-permeability half also has some resistance, even if it is small.
2
The model cells are 50 × 50 × 1 m, with $k_v/k_h$ = 0.05. How many times stronger is the vertical connection between neighbouring layers than the horizontal connection between neighbouring cells, and what follows from this for the calculation?
Show the answer
$T_z/T_x = 0.05\cdot(50/1)^2 = 125$ . The phases quickly redistribute between thin layers under gravity, so the layers of the model exchange fluid much more easily than neighbouring cells side by side. An explicit scheme would be limited exactly by the vertical connections, which is one of the reasons to calculate pressure implicitly. Only a continuous, sealing barrier set with a zero multiplier can separate such layers.
3
In a vertical connection with no wells, the upper cell holds a lot of water, and the lower cell holds oil at irreducible water saturation. Which cell supplies the mobility for each phase?
1. A.The upper cell for both phases
2. B.The lower cell for both phases
3. C.The upper cell for water and the lower cell for oil
4. D.The average of the two cells for both phases
Show the answer
Correct answer: C. The upper cell for water and the lower cell for oil
The heavy water sinks, and the light oil rises to meet it. Each phase has its own upstream cell, the one the phase flows out of. The water mobility of the lower cell is zero, so with it (or with the harmonic mean) the water would never sink. The arithmetic mean would give only half of the true exchange.
4
A unit of two layers, each 6 m thick, with shale above and below, is offset by a fault with a throw of 6 m. Which layers are connected across the fault?
Show the answer
The throw equals the thickness of a layer. On the higher side the layers occupy 0–6 and 6–12 m, and on the lower side 6–12 and 12–18 m. Only the lower layer of the higher side and the upper layer of the lower side overlap, over the full thickness of 6 m. This is a non-neighbour connection, because the layer numbers differ. The upper layer of the higher side faces the shale above the lower side, and the lower layer of the lower side faces the shale below the higher side.
5
A waterflood calculation with the arithmetic mean of the mobilities was repeated on a grid four times finer. What happened?
1. A.Almost nothing: the calculation settled on a wrong solution, and breakthrough is still at about 0.52 pore volumes instead of 0.40
2. B.The calculation became unstable
3. C.The front stopped at the inlet
4. D.The front came closer to the exact solution
Show the answer
Correct answer: A. Almost nothing: the calculation settled on a wrong solution, and breakthrough is still at about 0.52 pore volumes instead of 0.40
In the example of the lesson, breakthrough with the arithmetic mean comes at 0.51 pore volumes on 50 cells and 0.52 on 200 cells, and the exact solution is 0.40. The error does not come from the cell size but from the rule at the face. The rule drives oil through the face from an almost fully swept cell, and makes the displacement almost piston-like. A finer grid does not cure such an error.
6
Two cells have half-cell transmissibilities $t_1$ = 900 and $t_2$ = 300 mD·m, and the pressures at their centres are 21.0 and 20.2 MPa. Find the transmissibility of the connection and the pressure at the face.
Show the answer
$T = 1/(1/900 + 1/300) = 225$ mD·m. $p_f = (900\cdot21.0 + 300\cdot20.2)/1200 = 24{,}960/1200 = 20.8$ MPa. The half of cell 1 takes 0.2 MPa, and the half of cell 2 takes 0.6 MPa, which is three quarters of the pressure drop.

Updated: 28 October 2026