Permeability upscaling for single-phase flow

Quick refresher

Porosity, net-to-gross ratio (the share of the rock that is reservoir) and saturation are transferred exactly, with the weights of rock volume and pore volume. Permeability does not add up. The equivalent permeability of a block depends on where the patches of different rock are, on the flow direction and on the boundary conditions of the calculation.
The equivalent permeability found for the block alone (with no buffer zone, that is, with no extra cells added around it) lies, for any arrangement of cells, between the harmonic and the arithmetic mean (the averages for layers across the flow and along it). These are the Wiener bounds. The Cardwell–Parsons bounds are narrower. The lower one assumes that lines of cells along the flow do not exchange fluid. The upper one assumes that the pressure evens out in every cross-section across the flow.
A flow calculation in the block solves the equations of the cells of the detailed grid inside the block. From the total rate it finds $k^{*}$ with Darcy's law for the block.
Boundary conditions are an assumption about the neighbours of the block. Sealed side faces give the diagonal values. Periodic and linear conditions give a tensor. Linear conditions give the largest $k^{*}$ . The spread between the conditions measures the uncertainty of the upscaling itself.
The coarse model calculates flows between cells through transmissibilities (numbers that say how easily fluid flows between two cells). Transmissibility upscaling finds them directly from the detailed model. It does not “smear” a thin layer at a cell boundary over both faces of the cell.
A cell with a well is upscaled separately. Near the wellbore the flow is radial (it converges on the well from all sides). The fine cells along the wellbore, where there are measurements, matter far more for the inflow than their share of the volume.

Key relationships

k^{*} = \dfrac{Q\,\mu\,L}{A\,\Delta p}

equivalent permeability of a block: Darcy's law for a coarse cell

\bar S_w = \dfrac{\sum V_i\phi_iS_{w,i}}{\sum V_i\phi_i}

saturation of a coarse cell, weighted by pore volume

\bar k_{\text{H}} \le \underline{k} \le k^{*} \le \overline{k} \le \bar k_{\text{A}}

the Wiener bounds (outer) and the Cardwell–Parsons bounds (inner), for sealed side faces and periodic conditions

T^{*} = \dfrac{\mu\,Q}{0.0853\,(\bar p_I - \bar p_{II})}

transmissibility of the link between coarse cells from a fine-grid calculation, mD·m

Common pitfalls

Averaging saturation by the volume of the cells instead of by the pore volume.
Taking the geometric mean as a universal recipe. For elongated bodies it is wrong by tens of per cent.
Taking “average each cross-section arithmetically, then the sections harmonically” for a good estimate. It is the upper bound.
Taking the result of a block flow calculation as the truth, without checking how sensitive it is to the boundary conditions.
Giving a cell with a well a well index (the factor that links the well rate to the pressure drop) by Peaceman's formula, with a permeability averaged for linear flow.
Counting the net-to-gross ratio twice: in the upscaled permeability, and again in the transmissibility multiplier of the simulator.

After this lesson you can

explain why the grid is coarsened and by what factor, and why porosity and saturation are transferred exactly but permeability is not
estimate the equivalent permeability of a block by averages, and give its strict bounds (limits that the answer cannot cross): the Wiener bounds and the Cardwell–Parsons bounds
calculate the equivalent permeability of a block by a flow calculation with sealed side faces (no flow through the sides), with periodic conditions (the block is repeated in all directions) and with linear boundary conditions (the pressure changes in a straight line along every face), and explain why the answers differ
tell permeability upscaling from transmissibility upscaling (upscaling the flow link between two cells), and explain why a cell with a well is upscaled separately
check a coarsened model: the volumes, the bounds, the sensitivity to boundary conditions, and a comparison of flow on the fine and the coarse grid

Why coarsen the model: the gap between scales

The geological model is built to keep everything that is known about the reservoir: the top and the base, thin layers and facies (rock bodies of one origin), and the porosity and permeability from well logs and core. Its cells are tens of metres across horizontally and a fraction of a metre thick. In a large accumulation there are millions or tens of millions of them. The simulation model needs something else. It must run the flow calculation many times: for development options, for history matching and for estimating uncertainty. The calculation time grows faster than the number of cells (how many cells are needed). So the simulation grid is made tens or hundreds of times coarser, and the properties are transferred to it. This is upscaling.

Scales at which the reservoir is described

Scale: Core plug (a small cylinder cut from the core)
Typical size: 3–4 cm × 5 cm
Volume: tens of cm³
Where the properties come from: laboratory measurements

Scale: Cell of the geological model
Typical size: 25–50 m horizontally, 0.3–1 m thick
Volume: hundreds to thousands of m³
Where the properties come from: well logs, core, geostatistics (statistical methods that fill the space between wells)

Scale: Cell of the simulation model
Typical size: 50–200 m horizontally, 1–10 m thick
Volume: thousands to hundreds of thousands of m³
Where the properties come from: upscaling

Scale: Distance between wells
Typical size: hundreds of metres
Volume: —
Where the properties come from: well tests, production history

Scale	Typical size	Volume	Where the properties come from
Core plug (a small cylinder cut from the core)	3–4 cm × 5 cm	tens of cm³	laboratory measurements
Cell of the geological model	25–50 m horizontally, 0.3–1 m thick	hundreds to thousands of m³	well logs, core, geostatistics (statistical methods that fill the space between wells)
Cell of the simulation model	50–200 m horizontally, 1–10 m thick	thousands to hundreds of thousands of m³	upscaling
Distance between wells	hundreds of metres	—	well tests, production history

The orders of magnitude are a guide, not a standard. Cell sizes are chosen for the task and the computing resources, and the cells get smaller as the resources grow.

Worked exampleHow many core plugs and geological-model cells fit in one simulation cell

Given (assumed data)

Core plug: diameter 3.8 cm, length 5 cm
Cell of the geological model: 25 × 25 m, thickness 0.5 m
Cell of the simulation grid: 100 × 100 m, thickness 4 m
Reservoir: 12 × 6 km, total thickness 60 m

Solution

Volume of the plug: $V = \pi d^2L/4 = \pi\cdot3.8^2\cdot5/4 = 56.7$ cm³.
Volume of a simulation cell: $100\cdot100\cdot4 = 40{,}000$ m³. It holds $4\cdot4\cdot8 = 128$ cells of the geological model and $40{,}000\cdot10^6/56.7 \approx 705$ million plug volumes.
The number of cells of the geological model of the reservoir is $480\cdot240\cdot120 = 13{,}824{,}000$ . The number of simulation cells is $120\cdot60\cdot15 = 108{,}000$ , which is 128 times fewer.

Answer: One simulation cell holds 128 cells of the geological model and about 705 million core plug volumes. The model of the reservoir is coarsened from 13.8 million to 108 thousand cells.

What it means. Nobody has measured the property that the model assigns to a coarse cell. We can only calculate it from the fine cells, which themselves came from transferring core and log data. How to calculate it depends on the property: porosity adds up, permeability does not. The data in this example are assumed for illustration.

Equivalent permeability of a block: Darcy's law for a coarse cell

Q = \frac{k^{*}A\,\Delta p}{\mu\,L}\qquad\Longrightarrow\qquad k^{*} = \frac{Q\,\mu\,L}{A\,\Delta p}

where:

$Q$ total rate through the block at reservoir conditions, m³/s
$k^{*}$ equivalent permeability of the block along the flow direction, m²
$A$ area of the block face perpendicular to the flow, m²
$\Delta p$ pressure difference between the inlet and outlet faces, Pa
$L$ length of the block along the flow, m
$\mu$ viscosity of the fluid, Pa·s

Units: SI (any consistent system)

Holds when

The flow is single-phase and steady, and the fluid is incompressible. Darcy's law holds in the fine cells and for the block as a whole.
The rate and the pressure difference belong to one block under definite boundary conditions. $k^{*}$ depends on them (see the section on boundary conditions).
If a pressure difference along one axis causes flow along another axis too, one number is not enough. We need the tensor $\mathbf{K}^{*}$ .

This is not new physics but a definition: $k^{*}$ is the permeability of a uniform block that passes the same rate as the non-uniform one (equivalent permeability). It is also called the effective permeability of the block. It has nothing to do with the effective (phase) permeability in multiphase flow.

In other unit systems (2)

Metric oilfield units

Q\,[\text{m}^3/\text{d}] = 0.0853\,\frac{k^{*}\,[\text{mD}]\;A\,[\text{m}^2]\;\Delta p\,[\text{MPa}]}{\mu\,[\text{mPa·s}]\;L\,[\text{m}]}

$k^{*}$ in mD, $A$ in m², $\Delta p$ in MPa, $\mu$ in mPa·s, $L$ in m, $Q$ in m³/d at reservoir conditions.

$0.0853$ is the same constant as in Darcy's law (Darcy's law). It converts mD to m², MPa to Pa, mPa·s to Pa·s, and seconds to days.

Oilfield units

q\,[\text{bbl/d}] = 1.127\cdot10^{-3}\,\frac{k^{*}\,[\text{mD}]\;A\,[\text{ft}^2]\;\Delta p\,[\text{psi}]}{\mu\,[\text{cP}]\;L\,[\text{ft}]}

$k^{*}$ in mD, $A$ in square feet, $\Delta p$ in psi, $\mu$ in cP, $L$ in feet, $q$ in barrels per day at reservoir conditions.

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

Porosity, net-to-gross ratio and saturation

Porosity and saturation describe volumes, and volumes add up. The pore volume of a coarse cell is the sum of the pore volumes of the fine cells. The volume of water is the sum of the volumes of water. So the rule is this: we average porosity with the weight of rock volume, and saturation with the weight of pore volume. If the model uses the net-to-gross ratio (the share of the rock that is reservoir), the porosity refers to the volume of reservoir rock.

Porosity and saturation of a coarse cell

\bar\phi = \frac{\sum_i V_i\,\phi_i}{\sum_i V_i},\qquad \bar S_w = \frac{\sum_i V_i\,\phi_i\,S_{w,i}}{\sum_i V_i\,\phi_i}

where:

$V_i$ rock volume of the $i$ -th fine cell, m³
$\phi_i,\ S_{w,i}$ porosity and water saturation of the fine cell, fraction
$\bar\phi,\ \bar S_w$ porosity and water saturation of the coarse cell, fraction

Units: any equal units of volume; porosity and saturation as fractions

Holds when

The sums run over all the fine cells inside the coarse one.
With a net-to-gross ratio $N_i$ : $\bar N = \sum V_iN_i/\sum V_i$ , the porosity of the reservoir rock is $\bar\phi = \sum V_iN_i\phi_i/\sum V_iN_i$ , and the weight in the saturation is $V_iN_i\phi_i$ . In this way the pore volume of the reservoir rock, $\bar N\bar\phi\sum V_i$ , is preserved.
Oil and gas saturations are averaged in the same way. If the formation volume factor differs from cell to cell, we add it to the weight to get the volumes in place at standard conditions.

The formulas are exact: a coarse cell holds exactly the same pore volume and the same volumes of fluids as the fine cells. An error appears only if we take the wrong weight, for example if we average saturation by rock volume.

Worked exampleWater saturation of a cell made of three rock types

Given (assumed data)

Shares of the volume of the simulation cell: channel sandstone 0.30; silty sandstone 0.45; shale 0.25
Porosity: 0.24; 0.17; 0.06
Water saturation: 0.22; 0.45; 1.00

Solution

Porosity of the cell: $\bar\phi = 0.30\cdot0.24 + 0.45\cdot0.17 + 0.25\cdot0.06 = 0.072 + 0.0765 + 0.015 = 0.1635$ . The terms are the pore volumes of the rocks per 1 m³ of the cell.
Water saturation weighted by pore volume: $\bar S_w = (0.072\cdot0.22 + 0.0765\cdot0.45 + 0.015\cdot1.00)/0.1635 = 0.06527/0.1635 = 0.3992$ .
The average by rock volume would give $0.30\cdot0.22 + 0.45\cdot0.45 + 0.25\cdot1.00 = 0.5185$ .
Oil per 1 m³ of the cell: the correct value is $0.1635\cdot(1 - 0.3992) = 0.0982$ m³. This is the same as in the rocks themselves: $0.072\cdot0.78 + 0.0765\cdot0.55 = 0.0982$ m³. With the wrong average we get $0.1635\cdot(1 - 0.5185) = 0.0787$ m³, which is 20% less.

Answer: The porosity of the cell is 0.1635 and the water saturation is 0.3992. The average by rock volume would raise the saturation to 0.5185 and underestimate the oil in place by 20%.

What it means. Shale takes a quarter of the volume but only 9% of the pore volume, so its full water saturation weighs little. The average by rock volume gives it the weight of a quarter and “pours” extra water into the cell. If shale is not counted as reservoir, its pores are not part of the pore volume of the model at all. Then the net-to-gross ratio is 0.75 and the porosity of the reservoir rock is 0.198. The oil per cubic metre of the cell is the same, 0.0982 m³, while the water and the storage capacity are smaller. The data in this example are assumed for illustration.

Permeability does not work like this. It describes not a volume but conductivity (how easily the rock passes fluid), and conductivities add up in different ways, depending on how the patches are connected. Take two cells of equal size, 10 and 1000 mD. Placed side by side, so that the flow runs through both at once (in parallel), they pass it like one cell of 505 mD. Placed one after the other, so that the flow runs through one and then the other (in series), they pass it like one cell of 19.8 mD, which is 25.5 times less.

Averages and bounds of the equivalent permeability

For two simple cases the answer comes from the averages of the lesson Reservoir heterogeneity and permeability averaging. For continuous layers along the flow it is the arithmetic mean. Across the flow it is the harmonic mean. For an isotropic random mosaic in plane flow it is the geometric mean (random mosaic). (A random mosaic is a patchwork of cells with different permeabilities, placed at random. Isotropic means the same in all directions. Plane flow means flow in two dimensions.) A power average with a fitted exponent includes all three as special cases (power average). A real block has neither perfect layers nor a perfect mosaic, and the average turns into a guess. But this guess has strict limits that it cannot cross.

Wiener bounds

\bar k_{\text{H}} = \left(\sum_i \frac{v_i}{k_i}\right)^{-1} \;\le\; k^{*} \;\le\; \bar k_{\text{A}} = \sum_i v_i\,k_i

where:

$v_i$ share of the block volume occupied by the $i$ -th fine cell ( $\sum v_i = 1$ ), —
$k_i$ permeability of the fine cell, mD
$k^{*}$ equivalent permeability of the block in any direction, mD

Units: permeabilities in any equal units

Holds when

The flow is single-phase and steady. The fine cells are isotropic (for anisotropic cells, we take the permeability of each direction separately).
The bounds hold for any arrangement of the cells and for all three kinds of boundary conditions of the lesson, if $k^{*}$ is calculated from the flow in the block itself, with no buffer zone.

The lower bound is reached only when all the patches are connected in series (layers across the flow). The upper bound is reached only when they are connected in parallel (layers along the flow). These Wiener bounds are the first check of any upscaling: a value outside them means an error.

The Wiener bounds do not know how the cells are arranged, and for a non-uniform block they are very wide. Narrower bounds appear if we imagine thin partitions inserted in the block and see how the flow changes. These are the Cardwell–Parsons bounds.

Impermeable partitions along the flow cut the block into tubes: lines of cells from the inlet face to the outlet face. Inside a tube the cells are in series (harmonic mean), and the tubes are in parallel with each other (arithmetic mean). The partitions only stop the flow from going round the tight cells, so this gives the lower bound.
Perfectly conducting partitions across the flow make the pressure equal in every cross-section. The cells of one cross-section work in parallel (arithmetic mean), and the cross-sections are in series with each other (harmonic mean). The partitions only help the flow to redistribute itself, so this gives the upper bound.

SchematicImagined partitions that give the Cardwell–Parsons bounds for the block of the examples of the lesson (flow from west to east). On the left, impermeable partitions along the flow give the lower bound. On the right, conducting partitions across the flow give the upper bound. The numbers are the permeabilities of the cells, mD.

Cardwell–Parsons bounds for flow along the

x

axis

\begin{aligned} \underline{k}_x &= \frac{1}{n_y}\sum_{j=1}^{n_y}\frac{n_x}{\sum_{i=1}^{n_x} 1/k_{ij}},\\ \overline{k}_x &= n_x\left[\sum_{i=1}^{n_x}\left(\frac{1}{n_y}\sum_{j=1}^{n_y} k_{ij}\right)^{-1}\right]^{-1},\\ \underline{k}_x &\le k^{*}_x \le \overline{k}_x \end{aligned}

where:

$k_{ij}$ permeability of a fine cell: $i$ is the number along the flow (the column), $j$ is the number of the line of cells along the flow (the row), mD
$n_x,\ n_y$ number of cells along the flow and across it, —
$\underline{k}_x,\ \overline{k}_x$ lower and upper bounds of the equivalent permeability along $x$ , mD

Units: permeabilities in any equal units; cells of equal size (otherwise the means get weights of lengths and areas)

Holds when

The flow is single-phase and steady. The block has sealed side faces, or it is periodic. For linear boundary conditions the upper bound is not guaranteed (see the advanced section).
In a three-dimensional block $j$ runs over all the cells of the $y$ – $z$ cross-section.
For flow along $y$ , rows and columns swap roles.

The lower bound is the arithmetic mean of the harmonic means along the lines of cells in the flow direction. The upper bound is the harmonic mean of the arithmetic means over the cross-sections. A common quick recipe, “average each cross-section arithmetically, and the sections harmonically”, is exactly the upper bound. It overestimates $k^{*}$ , or at best guesses it. For layers along or across the flow the two bounds coincide and give the exact answer.

Permeability of the cells of the block, mD (rows from north to south, columns from west to east)

Row: 1
1: 40
2: 25
3: 280
4: 350

Row: 2
1: 30
2: 260
3: 300
4: 15

Row: 3
1: 220
2: 320
3: 8
4: 35

Row: 4
1: 300
2: 45
3: 20
4: 60

Row	1	2	3	4
1	40	25	280	350
2	30	260	300	15
3	220	320	8	35
4	300	45	20	60

A simulation cell of 100 × 100 m made of 16 cells of the geological model of 25 m, one layer 1 m thick. Channel sandstone (220–350 mD) crosses the block as a stepped band from the south-western corner to the north-eastern corner. 8 mD is a cemented patch next to the channel, and the rest is silty rock. The data are assumed for illustration.

Worked exampleAverages and bounds for a block of 100 × 100 m

Given (assumed data)

Block: 16 cells of 25 m (the table above)
Flow: from west to east, along $x$ ; then from south to north, along $y$

Solution

Wiener bounds: $\bar k_{\text{A}} = 2308/16 = 144.25$ mD and $\bar k_{\text{H}} = 16/\sum(1/k_i) = 16/0.43207 = 37.0$ mD. The geometric mean is $k_G = 75.4$ mD.
Lower Cardwell–Parsons bound along $x$ : the harmonic means of the rows are 56.0, 37.3, 24.8 and 43.4 mD. Their mean is $\underline{k}_x = 40.4$ mD.
Upper bound: the arithmetic means of the columns are 147.5, 162.5, 152.0 and 115.0 mD. Their harmonic mean is $\overline{k}_x = 4/(1/147.5 + 1/162.5 + 1/152.0 + 1/115.0) = 141.8$ mD.
Along $y$ rows and columns swap roles: $\underline{k}_y = 43.8$ mD and $\overline{k}_y = 139.7$ mD.

Answer:

k^{*}_x

lies between 40.4 and 141.8 mD, and

k^{*}_y

lies between 43.8 and 139.7 mD. The Wiener bounds are from 37.0 to 144.25 mD.

What it means. For this block the Cardwell–Parsons bounds are hardly narrower than the Wiener bounds: the upper one is below the arithmetic mean by only 1.7%. The reason is the structure. The channel runs along the diagonal, so every row and every column holds both good and bad cells. The bounds get narrower when the block looks like layers along or across the flow. They are almost useless when conducting bodies cross the block at an angle. Then we have to calculate the equivalent permeability. The data in this example are assumed for illustration.

Flow calculation in the block

If averages do not work, we calculate the equivalent permeability the way it was defined. We solve the flow problem inside the block on the detailed grid and measure the rate. The method is called flow-based upscaling, or the pressure-solver method.

How to calculate

k^{*}_x

of a block

1
Cut out the block
Take the fine cells that make up the coarse cell, sometimes together with a border of neighbouring cells (see the section on boundary conditions).
2
Set the boundary conditions
For example, set the pressure $p_1$ on the west face and $p_2$ on the east face. The north and south faces are impermeable.
3
Solve the cell equations
For each fine cell, write the balance of steady flow: the sum of the flows from the neighbours is zero. The transmissibilities are as in any model: the halves of the two neighbouring cells in series (cell equation).
4
Find the rate
Add up the flows through the outlet face. As a check, add them up through the inlet face too. In steady flow with no sources the two sums are equal.
5
Calculate $k^{*}_x$
Use Darcy's law for the block: $k^{*}_x = Q\mu L/(A\Delta p)$ . Repeat for $y$ and, in a three-dimensional model, for $z$ .

Cell equations of a block in steady flow

\sum_{m} T_{lm}\,(p_m - p_l) + \sum_{b} T_{b}\,(p_b - p_l) = 0,\qquad T_{lm} = \frac{A_f}{\frac{\Delta x_l}{2k_l} + \frac{\Delta x_m}{2k_m}},\qquad T_b = \frac{2k_lA_f}{\Delta x_l}

where:

$p_l,\ p_m$ pressure in the fine cell $l$ and in the neighbouring cell $m$ , MPa
$T_{lm}$ transmissibility of the link between neighbouring cells: two half-cells in series, mD·m
$A_f$ area of the common face, m²
$\Delta x_l,\ \Delta x_m$ sizes of the cells across this face, m
$k_l,\ k_m$ permeability of the cells in the direction perpendicular to the face, mD
$T_b,\ p_b$ transmissibility from the cell centre to the face of the block, and the pressure set on that face, mD·m, MPa

Units: metric oilfield: $k$ in mD, lengths in m, pressures in MPa. The viscosity and the constant 0.0853 are the same in all terms and cancel

Holds when

The flow is steady and the fluid is incompressible: there is no accumulation. There are no wells inside the block.
The scheme has five points (seven in three dimensions): the cells are linked only through common faces, and the permeability tensor of the fine cells is diagonal (one value for each axis, with no cross terms).
The equations are linear: the pressures can be written as fractions of the applied pressure difference, and $k^{*}$ does not depend on the size of the difference.

This is the cell equation of the lesson Pressure equation for single-phase flow, without accumulation and without wells. For a block of $n$ fine cells it is a system of $n$ linear equations. In a model with millions of cells such systems are solved for each of hundreds of thousands of blocks. So programs do the calculation, but it is built in the same way as in the example below.

Worked example

k^{*}

of the block with sealed side faces

Given (assumed data)

Block: from the previous example: 4 × 4 cells of 25 m, thickness $h$ = 1 m
Boundary conditions: pressure 1 on the west face and 0 on the east face (as fractions of the pressure difference); the north and south faces are impermeable

Solution

Transmissibilities: for square cells $A_f/\Delta x = h$ , so $T = h\cdot2k_1k_2/(k_1 + k_2)$ , and at the faces of the block $T_b = 2kh$ . The link between a channel cell of 320 mD and the cemented cell of 8 mD is $2\cdot320\cdot8/328 = 15.6$ mD·m. The link between two channel cells of 220 and 320 mD is $2\cdot220\cdot320/540 = 260.7$ mD·m.
The sixteen cell equations are solved on a computer. The pressures are in the table after the example. In the east column (from north to south) they are 0.0842, 0.1498, 0.0849 and 0.0851.
The rate through the east face, written through the transmissibilities (the pressure difference is 1, and the viscosity and the constant 0.0853 cancel): $Q = 2h\,(350\cdot0.0842 + 15\cdot0.1498 + 35\cdot0.0849 + 60\cdot0.0851) = 79.6$ mD·m.
Check: through the west face the inflow is $2h\,(40\cdot(1 - 0.8682) + 30\cdot(1 - 0.8506) + 220\cdot(1 - 0.9178) + 300\cdot(1 - 0.9601)) = 79.6$ mD·m, the same.
Darcy's law for the block in these units is $Q = k^{*}Wh\,\Delta p/L$ , where $W$ is the width of the block. With $L = W$ = 100 m, $h$ = 1 m and $\Delta p = 1$ we get $k^{*}_x = Q = 79.6$ mD. The same calculation from south to north gives $k^{*}_y = 81.0$ mD.

Answer:

k^{*}_x \approx 79.6

mD and

k^{*}_y \approx 81.0

mD. They are inside the Cardwell–Parsons bounds and almost twice below the arithmetic mean.

What it means. Three quarters of the flow (76%) enter the block through the two southern cells of the west face. And 74% leave through the north-eastern cell of 350 mD (see the figure). The fluid follows the channel and not a straight line. The geometric mean (75.4 mD) came close to the answer only because the channel crosses the block along the diagonal and helps the flow equally along both axes. Suppose the same 16 values were arranged so that the channel ran from west to east. The geometric mean would not change, but the calculation would give $k^{*}_x$ of about 115 mD and $k^{*}_y$ of about 49 mD. The data in this example are assumed for illustration.

Pressures in the cells of the block for flow from west to east, fractions of the pressure difference

Row: 1
1: 0.8682
2: 0.5453
3: 0.2677
4: 0.0842

Row: 2
1: 0.8506
2: 0.6068
3: 0.4208
4: 0.1498

Row: 3
1: 0.9178
2: 0.7516
3: 0.4360
4: 0.0849

Row: 4
1: 0.9601
2: 0.7911
3: 0.4260
4: 0.0851

Row	1	2	3	4
1	0.8682	0.5453	0.2677	0.0842
2	0.8506	0.6068	0.4208	0.1498
3	0.9178	0.7516	0.4360	0.0849
4	0.9601	0.7911	0.4260	0.0851

The pressure is 1 on the west face and 0 on the east face. The north and south faces are impermeable. Rows are from north to south, and columns are from west to east.

CalculatedCalculated flows between the cells of the block for flow from west to east, with the north and south faces impermeable. The thickness of an arrow is proportional to the rate through the face. The numbers are the permeabilities of the cells, mD.

The channel in the example is deliberately built from cells that share faces. In a five-point scheme, cells that touch only at the corners do not exchange fluid. A “diagonal” channel made of such cells would look broken to the fine model. The fine grid is also an approximation (Finite differences: from equations to computation), and its shortcomings pass into the coarse model.

Boundary conditions: what the block knows about its neighbours

The block is cut out of the reservoir. How fluid really passes through its faces depends on the neighbours, which the calculation does not see. Upscaling boundary conditions are an assumption about these neighbours, and the answer depends on the assumption. There are three kinds of conditions, and one technique that weakens their influence.

Sealed side faces (no-flow): The inlet and outlet faces hold given pressures, and the other faces are closed. This assumes that the fluid moves along the axis and does not exchange with the neighbours at the sides. It is reasonable for nearly horizontal layering and for a fine mosaic with no elongated bodies. In practice these conditions are used more often than the others. They give only the diagonal of the tensor, $k^{*}_x$ , $k^{*}_y$ , $k^{*}_z$ . They underestimate the conductivity of bodies that leave the block through the side faces.
Periodic: The block is one tile of an endless tiled floor made of copies of itself. What leaves through the east face enters through the west face. Along the given gradient the pressure drops by $\Delta p$ over one period (the length of one block). The flows in all the copies are the same, and the pressure in the neighbouring copy is only shifted by $\Delta p$ . Flow goes through all the faces, and two calculations, with the gradient along $x$ and along $y$ , give the full tensor. The conditions are exact for repeating structures. If a body in the block does not continue through the opposite face, the conditions “stitch” it to foreign rock.
Linear pressure: The pressure on all the faces of the block changes linearly, as in a uniform reservoir with the same average gradient. Flow goes through the side faces too, and we get a tensor. The conditions impose uniform behaviour on the boundary of the block. They give the largest values of the three.
Buffer zone (flow jacket, border region): A border of fine cells from the neighbouring blocks is added to the block. The boundary conditions are set on the outer boundary of the border, and $k^{*}$ is calculated only for the block itself, from the average velocity and gradient in it. The influence of the conditions weakens. The price is a larger calculation and the question of how wide the border should be.

SchematicBoundary conditions of the flow calculation in a block (schematic): a) sealed side faces; b) periodic, a block among its copies; c) linear pressure on all the faces; d) a buffer zone of neighbouring cells around the block.

Darcy's law for a block with a permeability tensor

\langle\mathbf{u}\rangle = -\frac{1}{\mu}\,\mathbf{K}^{*}\langle\nabla p\rangle,\qquad \mathbf{K}^{*} = \begin{pmatrix} k^{*}_{xx} & k^{*}_{xy}\\ k^{*}_{yx} & k^{*}_{yy}\end{pmatrix},\qquad k^{*}_{xx} = \frac{\mu\langle u_x\rangle}{G},\quad k^{*}_{yx} = \frac{\mu\langle u_y\rangle}{G}

where:

$\langle\mathbf{u}\rangle$ velocity of flow (Darcy velocity) averaged over the volume of the block, m/s
$\langle\nabla p\rangle$ pressure gradient averaged over the volume of the block, Pa/m
$\mathbf{K}^{*}$ tensor of the equivalent permeability of the block, m²
$G$ magnitude of the average gradient in the calculation where the pressure falls along $x$ , Pa/m
$k^{*}_{yx}$ flow along $y$ caused by a gradient along $x$ , written as a permeability, m²

Units: SI (any consistent system)

Holds when

The averages are over the volume of the block. With periodic and linear conditions the average gradient equals the given one, and the average velocity is found from the rates through the faces (see the advanced section).
The second calculation, with the gradient along $y$ , gives $k^{*}_{yy}$ and $k^{*}_{xy}$ . With these conditions a correctly calculated tensor is symmetric, $k^{*}_{xy} = k^{*}_{yx}$ , and $k^{*}_{xx}k^{*}_{yy} > (k^{*}_{xy})^2$ .
With sealed side faces the average gradient acquires a transverse component that the calculation does not set. So from such calculations we take only the diagonal values.

An off-diagonal term (such as $k^{*}_{xy}$ , which is not on the main diagonal of the table of numbers) says that a pressure difference along one axis drives fluid along the other axis too. This happens when conducting bodies lie at an angle to the grid. The usual five- and seven-point schemes of simulators use only the diagonal. How the full tensor is handled is in the lesson Special topics in upscaling.

Worked exampleThe same block with periodic and linear conditions

Given (assumed data)

Block: from the previous examples
Calculations: two for each kind of condition, with the pressure difference along $x$ and along $y$

Solution

Periodic conditions: $k^{*}_{xx} = 62.4$ , $k^{*}_{yy} = 64.2$ , $k^{*}_{xy} = k^{*}_{yx} = 20.0$ mD.
Linear: $k^{*}_{xx} = 86.3$ , $k^{*}_{yy} = 87.0$ , $k^{*}_{xy} = k^{*}_{yx} = 10.8$ mD.
Sealed side faces (the previous example): $k^{*}_x = 79.6$ and $k^{*}_y = 81.0$ mD.
The periodic tensor in axes turned by 46° from $x$ towards $y$ , that is, along the channel (these are the axes in which flow and pressure drop point the same way): $\tfrac{62.4 + 64.2}{2} \pm \sqrt{0.9^2 + 20.0^2} = 63.3 \pm 20.02$ . This is 83.3 mD along the channel and 43.3 mD across it.

Answer: The three kinds of conditions gave, for one block,

k^{*}_x

from 62.4 to 86.3 mD. The periodic and linear conditions showed an off-diagonal term of 20.0 and 10.8 mD.

What it means. Sealed side faces let the flow follow the channel from one face with constant pressure to the other. Along these faces the fluid spreads freely. Periodic conditions put a copy of the block next to it. The north-eastern end of the channel runs into the cell of 40 mD of the neighbouring copy, and the south-western end into the cell of 60 mD. So the channel turns out to be broken, and this gives the smallest $k^{*}_{xx}$ . But the periodic conditions do see that the channel runs at an angle: $k^{*}_{xy} > 0$ , and a pressure difference from west to east drives fluid to the north too. Linear conditions impose on the north and south faces the pressure of a uniform reservoir, and let fluid enter and leave through them. Their value is the largest. Which one is closer to the truth depends on the neighbours of the block: does the channel continue in the neighbouring cells? The block “does not know” this. The spread between the conditions measures the uncertainty of the upscaling itself. The data in this example are assumed for illustration.

The block of the examples: equivalent permeability by different methods, mD

Method: Arithmetic mean: the upper Wiener bound
$k^{*}_x$: 144.3
$k^{*}_y$: 144.3
$k^{*}_{xy}$: —

Method: Upper Cardwell–Parsons bound
$k^{*}_x$: 141.8
$k^{*}_y$: 139.7
$k^{*}_{xy}$: —

Method: Linear conditions
$k^{*}_x$: 86.3
$k^{*}_y$: 87.0
$k^{*}_{xy}$: 10.8

Method: Sealed side faces
$k^{*}_x$: 79.6
$k^{*}_y$: 81.0
$k^{*}_{xy}$: —

Method: Geometric mean
$k^{*}_x$: 75.4
$k^{*}_y$: 75.4
$k^{*}_{xy}$: —

Method: Periodic conditions
$k^{*}_x$: 62.4
$k^{*}_y$: 64.2
$k^{*}_{xy}$: 20.0

Method: Lower Cardwell–Parsons bound
$k^{*}_x$: 40.4
$k^{*}_y$: 43.8
$k^{*}_{xy}$: —

Method: Harmonic mean: the lower Wiener bound
$k^{*}_x$: 37.0
$k^{*}_y$: 37.0
$k^{*}_{xy}$: —

Method	$k^{*}_x$	$k^{*}_y$	$k^{*}_{xy}$
Arithmetic mean: the upper Wiener bound	144.3	144.3	—
Upper Cardwell–Parsons bound	141.8	139.7	—
Linear conditions	86.3	87.0	10.8
Sealed side faces	79.6	81.0	—
Geometric mean	75.4	75.4	—
Periodic conditions	62.4	64.2	20.0
Lower Cardwell–Parsons bound	40.4	43.8	—
Harmonic mean: the lower Wiener bound	37.0	37.0	—

The rows are in decreasing order of $k^{*}_x$ . The flow calculations for all three kinds of conditions lie inside the Cardwell–Parsons bounds, but they differ from each other by almost 1.4 times.

Sealed side faces: when the layers are almost parallel to the grid axes and there is little cross-flow through the side faces. We usually start with them.
Periodic: for statistically uniform fields and repeating structures, when a tensor is needed.
Linear: when an upper estimate or a tensor is needed and the heterogeneity is finer than the block. They are used more often on the outer boundary of a buffer zone than directly on the faces of the block.
A buffer zone, a different cell size or transmissibility upscaling: when different conditions give noticeably different answers for many blocks.

The conditions matter most for blocks of the size of the correlation length (the distance over which neighbouring values of permeability are still alike). Such a block holds both conducting bodies and barriers, but too few of them to average out. In a block much smaller than the correlation length, the properties are almost uniform. In a block much larger, the heterogeneity averages out. In both cases the answer hardly depends on the conditions. The error also grows with the permeability contrast. The hardest things to represent are thin barriers and narrow conducting paths, which decide whether neighbouring cells communicate: continuous shale layers, non-conducting faults, isolated channels and open fractures.

Permeability or transmissibility

The simulator does not use the permeability of a cell directly. It calculates the flow between neighbouring cells through the transmissibility, joining the halves of two cells in series (cell equation, Interblock flow and transmissibility). After permeability upscaling this is a second averaging: first the fine cells into a coarse one, and then two halves of coarse cells into a link. The resistance of a coarse cell is shared equally between its halves. If a tight thin layer lies at one face of the coarse cell, this is wrong.

Transmissibility upscaling skips the middle step. The link between two coarse cells is calculated directly from the fine grid. We solve the flow in the region from the centre of one coarse cell to the centre of the other, and divide the rate by the pressure difference.

Transmissibility between coarse cells from the detailed model

T^{*}_{I,II} = \frac{\mu\,Q}{0.0853\,\left(\bar p_{I} - \bar p_{II}\right)}

where:

$T^{*}_{I,II}$ transmissibility of the link between the coarse cells $I$ and $II$ , mD·m
$Q$ rate between them in the fine-grid calculation, at reservoir conditions, m³/d
$\bar p_I,\ \bar p_{II}$ pressures of the coarse cells in the same calculation: the pressures at their centres, or the averages by pore volume of their fine cells, MPa
$\mu$ viscosity, mPa·s

Units: metric oilfield: $Q$ in m³/d, $p$ in MPa, $\mu$ in mPa·s, $T^{*}$ in mD·m

Holds when

The flow is single-phase and steady. The problem is solved for the region between the centres of two cells (a local calculation) or for the whole model with wells (a global one).
Like the equivalent permeability, the result depends on the boundary conditions of the local calculation.
In the one-dimensional case the formula reduces to the sum of the resistances of the fine cells between the centres of the coarse ones.

Transmissibility is a property of a link, not of a cell. It remembers where inside the coarse cells the thin layers and channels lie. The price is that it cannot be transferred to another grid, and each link needs its own calculation.

In other unit systems (1)

T^{*}_{I,II} = \frac{\mu\,Q}{\bar p_{I} - \bar p_{II}}

$Q$ in m³/s, $\mu$ in Pa·s, $p$ in Pa, $T^{*}$ in m³ (permeability in m² multiplied by a length).

There is no constant: all the quantities are in consistent units.

A column of the reservoir under a simulation cell of 100 × 100 m

Layer (1 m each): 1
$k_z$ , mD: 40
Coarse layer: upper

Layer (1 m each): 2
$k_z$ , mD: 60
Coarse layer: upper

Layer (1 m each): 3
$k_z$ , mD: 50
Coarse layer: upper

Layer (1 m each): 4
$k_z$ , mD: 0.01
Coarse layer: middle

Layer (1 m each): 5
$k_z$ , mD: 80
Coarse layer: middle

Layer (1 m each): 6
$k_z$ , mD: 70
Coarse layer: middle

Layer (1 m each): 7
$k_z$ , mD: 90
Coarse layer: lower

Layer (1 m each): 8
$k_z$ , mD: 60
Coarse layer: lower

Layer (1 m each): 9
$k_z$ , mD: 45
Coarse layer: lower

Layer (1 m each)	$k_z$ , mD	Coarse layer
1	40	upper
2	60	upper
3	50	upper
4	0.01	middle
5	80	middle
6	70	middle
7	90	lower
8	60	lower
9	45	lower

Nine layers of the geological model are combined into three layers of the simulation grid, 3 m each. Layer 4 is a thin shale layer. The data are assumed for illustration.

Worked exampleA shale layer at the boundary of a coarse layer

Given (assumed data)

Column: 9 layers of 1 m each (the table above), cross-section area 10,000 m²; the coarse layers are three fine ones each
Thin layer: 0.01 mD, the upper fine layer of the middle coarse layer

Solution

The vertical permeability of the coarse layers is harmonic. Upper: $3/(1/40 + 1/60 + 1/50) = 48.65$ mD. Middle: $3/(1/0.01 + 1/80 + 1/70) = 3/100.027 = 0.02999$ mD. Lower: $3/(1/90 + 1/60 + 1/45) = 60$ mD.
Transmissibilities from the permeabilities of the coarse layers, with 1.5 m from the centres to the common face: $T_{12} = 10{,}000/(1.5/48.65 + 1.5/0.02999) = 199.8$ mD·m and $T_{23} = 10{,}000/(1.5/0.02999 + 1.5/60) = 199.8$ mD·m.
Transmissibilities directly from the fine layers between the centres of the coarse ones, from the middle of layer 2 to the middle of layer 5, and from the middle of layer 5 to the middle of layer 8: $T^{*}_{12} = 10{,}000/(0.5/60 + 1/50 + 1/0.01 + 0.5/80) = 100.0$ mD·m and $T^{*}_{23} = 10{,}000/(0.5/80 + 1/70 + 1/90 + 0.5/60) = 250{,}124$ mD·m.

Answer: From the permeabilities of the coarse layers, both links get about 199.8 mD·m. The direct calculation gives 100.0 mD·m for the upper link and 250,124 mD·m for the lower one, which is 1252 times more.

What it means. Permeability upscaling “smeared” the thin layer over both faces of the middle coarse layer. The link with the upper layer became twice as conductive as it really is. The link with the lower layer became 1252 times weaker, although there is no barrier between the middle and the lower layers. The resistance of the whole path from the upper coarse layer to the lower one hardly changed. What differs is which link it is assigned to. Suppose there is bottom water (water lying under the oil) below, and a well opens the middle layer. A model with upscaled permeability will hardly show water coning (the water rising towards the well in a cone), and in history matching this coning will have to be “created” by other parameters. Transmissibility upscaling puts the barrier where it lies. The data in this example are assumed for illustration.

Transmissibility upscaling matters where the heterogeneity is tied to cell boundaries. Examples are barriers and high-permeability thin layers at the faces; faults and non-neighbour connections (links between cells that are not neighbours in the grid) of corner-point grids (Grids in reservoir simulation models; these are grids whose cells are bent to follow layers and faults); and neighbouring cells of very different sizes. In the horizontal plane the task is solved in the same way as for a block, with boundary conditions, a buffer zone and the same doubts. Some methods take the boundary conditions of each link from a calculation of the whole model with wells. Others use non-uniform grids that keep the main channels. They are discussed together with displacement in the lesson Two-phase upscaling: pseudo-functions.

Cells with wells

Everything said so far refers to linear flow through a block. Near a well the flow is radial: it converges on the wellbore, and the resistance of each zone is proportional to the logarithm of the ratio of its outer and inner radii (ring zones). The fine cells that the wellbore passes through matter far more for the inflow than their share of the volume. It is in them that the geological model keeps what was measured in the well: the well logs and the core. Upscaling for linear flow dissolves them in the average permeability of the cell.

So a cell with a well is upscaled separately (near-well upscaling). We solve the inflow to the well on a fine grid, in the coarse cell itself and its neighbours. Then we choose the well index and the transmissibilities to the neighbouring cells so that the coarse model gives the same rate at the same drawdown (the pressure difference that drives the flow into the well) (Well model in a grid block).

Well index of a coarse cell

\mathrm{WI}^{*} = \frac{\mu\,q}{0.0853\,\left(\bar p_0 - p_{wf}\right)}

where:

$\mathrm{WI}^{*}$ well index of the coarse cell, mD·m
$q$ well rate at reservoir conditions, from the fine-grid calculation, m³/d
$\bar p_0$ pressure of the coarse cell in the same calculation, defined in the same way as the pressures of the neighbouring cells when the transmissibilities to them were calculated, MPa
$p_{wf}$ bottomhole flowing pressure, MPa
$\mu$ viscosity, mPa·s

Units: metric oilfield, as in the lesson Well model in a grid block

Holds when

The flow to the well is single-phase and steady. The boundary conditions of the local calculation are, for example, the pressure of radial inflow on the outer boundary of the region.
The index and the transmissibilities around the cell with the well are calculated together, from one calculation. If the cell pressure is taken as the average over its volume, the index of a uniform cell corresponds to a radius of about $0.35\,\Delta x$ , not Peaceman's $0.2\,\Delta x$ . The transmissibilities to the neighbours must be consistent with this.
The skin factor of the wellbore is added separately, if it is not in the fine model.

Heterogeneity near the wellbore enters the index in the same way as the skin factor. Better rock at the well increases the index, and worse rock reduces it. More on upscaling near wells is in the lesson Special topics in upscaling.

Worked exampleA well in the best cell of the block: an estimate

Given (assumed data)

Simulation cell: 100 × 100 m, a vertical well in the centre, $r_w$ = 0.1 m, wellbore skin 0; upscaling for linear flow gave 60 mD
Cell of the geological model with the wellbore: 25 × 25 m, 300 mD, from the logs and core of this well
Assumption: we replace this cell by a circle of the same area around the wellbore, and beyond it the rock is 60 mD; the calculation is for 1 m of thickness

Solution

Peaceman's equivalent radius (the radius at which the pressure of the radial flow equals the pressure of the cell): $r_0 = 0.14\sqrt{100^2 + 100^2} = 19.8$ m. The radius of a circle with the area of 25 × 25 m is $r_s = 25/\sqrt{\pi} = 14.1$ m. This is less than $r_0$ , so the zone is wholly inside it.
Index from the permeability of the cell: $\mathrm{WI} = 2\pi\cdot60/\ln(19.8/0.1) = 2\pi\cdot60/5.288 = 71.3$ mD·m.
The zone at the wellbore is like a Hawkins skin (Skin factor: a number that shows how much the zone around the well helps or hinders the flow): $S = (60/300 - 1)\ln(14.1/0.1) = -0.8\cdot4.949 = -3.959$ .
Index with the zone: $\mathrm{WI}^{*} = 2\pi\cdot60/(5.288 - 3.959) = 283.7$ mD·m. The average permeability of the ring zones from $r_w$ to $r_0$ is $60\cdot5.288/(5.288 - 3.959) = 238.7$ mD.

Answer: The model with the permeability of the cell underestimates the well index by a factor of 4.0: 71.3 instead of 283.7 mD·m per metre of thickness.

What it means. The best rock takes 1/16 of the volume of the cell and hardly changes its permeability for linear flow. But for the inflow to the well it is the main thing. At the same cell pressure and the same bottomhole pressure, the coarse model without a correction will show a rate four times lower. In history matching this difference will be “found” in a negative skin or in a multiplier of the index, and nobody will understand where it came from. The estimate is rough. The zone is not round, and if $r_s$ is close to $r_0$ or larger than it, we need a calculation on a fine grid. The data in this example are assumed for illustration.

How to check a coarsened model

Upscaling is done by programs, and errors in it are not visible on property maps. The coarse model looks plausible until it is compared with the fine one. The checks go from simple to time-consuming.

Volumes. The pore volume and the fluids in place of each region, of the reservoir and of a sector must be the same before and after coarsening, up to rounding. A difference means an error in the averaging weights or in the net-to-gross ratio.
Bounds. Each $k^{*}$ calculated for the block alone, with no buffer zone, must lie between the harmonic and the arithmetic mean of its fine cells. With sealed side faces and periodic conditions it must also lie between the Cardwell–Parsons bounds. A value outside the bounds means an error of direction, of units, or a double count of the net-to-gross ratio.
Sensitivity to the boundary conditions. If two kinds of conditions give noticeably different values for many blocks, the heterogeneity is comparable to the size of the block. We need a buffer zone, a different cell size or transmissibility upscaling.
Flow on the fine and the coarse grid. On the whole model, or on a typical sector, we calculate single-phase steady flow, from face to face or between wells. Then we compare the rates, the productivity of wells and the pressure field. This is the main check (example below).
Coarsening in two stages. If the method is exact, the result does not depend on the path: coarsening by two and then by two again gives the same as coarsening by four at once. A difference points to an error of the method for this heterogeneity. For layered blocks there is none, because the arithmetic and harmonic means can be calculated in parts. But agreement does not prove accuracy: the arithmetic, geometric and harmonic means pass this check for any heterogeneity.
Barriers and channels. Discontinuous shale layers (layers that do not run through the whole block) are tricky. A layer that crosses a coarse cell from edge to edge makes its vertical permeability zero, although in the fine model the fluid goes round the layers by their edges. The danger is greater when the length of the layers is close to the cell size. The opposite error is a barrier “dissolved” by averaging. It opens a communication between layers that does not exist. We check the communication between layers separately, by calculating the vertical flow on the fine and the coarse grid.
Well data. We compare the $kh$ of the model near wells (permeability times thickness) with the results of well tests (Pressure build-up analysis), and the inflow profiles with field measurements (a check against the behaviour of the reservoir).

CalculatedA sector of 600 × 600 m from the example below: the permeability of the fine grid of 24 × 24 cells, and the equivalent permeabilities of the coarse grid of 6 × 6 along

x

and along

y

(flow calculation in blocks of 4 × 4 with sealed side faces). Bodies elongated from west to east give the coarse cells anisotropy:

k^{*}_x

is larger than

k^{*}_y

in all 36 cells.

Worked exampleA sector of 600 × 600 m: the fine and the coarse grid

Given (assumed data)

Fine model: one layer, 24 × 24 cells of 25 m; lognormal permeability (a few high values and many low ones), median 50 mD, $\sigma_{\ln k}$ = 1.0 (the standard deviation of the logarithm of permeability, a measure of how widely it varies); the bodies are elongated from west to east, with a correlation length of about 600 m along $x$ and 75 m along $y$ (the figure)
Coarse model: 6 × 6 cells of 100 m, that is, blocks of 4 × 4; the permeability of the blocks by four methods
Check: steady flow through the whole sector from west to east, then from south to north; the side faces of the sector are impermeable

Solution

Fine model (576 equations): the equivalent permeability of the sector is 78.8 mD along $x$ and 50.2 mD along $y$ . The averages of all the cells are: arithmetic 97.6, geometric 61.6, harmonic 38.4 mD.
Coarse model with the geometric mean of each block: 62.1 and 60.2 mD. The errors are −21.2% and +19.9%.
Flow calculation in blocks with sealed side faces: 77.2 and 49.3 mD, which is −2.0% and −1.8%.
With periodic conditions (the diagonal of the tensor): 75.5 and 48.7 mD, which is −4.2% and −3.0%. With linear conditions: 79.0 and 57.7 mD, which is +0.3% and +14.9%.

Answer: The flow calculation in blocks with sealed side faces reproduced the permeability of the sector with an error of about 2%. The geometric mean was wrong by −21.2% along the bodies and by +19.9% across them.

What it means. The geometric mean does not know that the bodies are elongated. Along them it underestimates the conductivity, and across them it overestimates it. The flow calculation in blocks sees the elongation: in all 36 coarse cells $k^{*}_x > k^{*}_y$ , on average by a factor of 1.5 (the figure). Linear conditions overestimated the flow across the bodies. At the faces of the block they let the fluid go round the tight thin layers, which in the reservoir run further than the block. For the two directions together, the best here were the sealed side faces, and the check problem, flow from face to face, is similar to them. For flow between wells or at an angle to the grid the order may be different. So the check is repeated for your own tasks. The data in this example are assumed for illustration: one random version (realisation) of a random field.

Terms and symbols

Upscaling: The transfer of properties from the fine grid of a geological model to the coarse grid of a simulation model.
Equivalent permeability of a block $k^{*}$: The permeability of a uniform block that passes the same rate as the real block, at the same pressure difference and with the same boundary conditions.
Wiener bounds: The harmonic and arithmetic means. They are the limits of the equivalent permeability for any arrangement of cells.
Cardwell–Parsons bounds: The arithmetic mean of the harmonic means along the lines in the flow direction, and the harmonic mean of the arithmetic means over the cross-sections. They are narrower than the Wiener bounds.
Flow calculation in a block: Finding $k^{*}$ by solving the equations of the cells of the detailed grid inside the block, with given boundary conditions.
Upscaling boundary conditions: Sealed side faces, periodic and linear conditions. A buffer zone weakens their influence.
Equivalent permeability tensor: A $2\times2$ or $3\times3$ matrix. The off-diagonal terms describe flow across the applied gradient.
Transmissibility upscaling: Calculating the transmissibility of the link between two coarse cells directly from the detailed model.
Well index of a coarse cell $\mathrm{WI}^{*}$: An index chosen from a fine-grid inflow calculation so that the coarse model gives the same rate at the same drawdown.

Common mistakes and misconceptions

Wrong

“The saturation of a coarse cell is the average of the saturations of the fine cells by their volume.”

In fact

The weight is the pore volume. In the example of the lesson, the average by rock volume raised the water saturation from 0.40 to 0.52 and underestimated the oil in place by 20%.

Wrong

“The geometric mean is the standard way to upscale permeability.”

In fact

It is exact only for an isotropic random mosaic in plane flow. For elongated bodies in the example of the lesson it was wrong by −21% along the bodies and by +20% across them, while the flow calculation in blocks was wrong by about 2%.

Wrong

“We average each cross-section arithmetically, and the sections harmonically, and we get a good estimate.”

In fact

This is the upper Cardwell–Parsons bound. For the block of the lesson it gave 141.8 mD, while the flow calculations gave from 62.4 to 86.3 mD.

Wrong

“A flow calculation in a block gives the true permeability of the cell.”

In fact

It answers a question with given boundary conditions. For the block of the lesson the three kinds of conditions gave from 62.4 to 86.3 mD. We reduce this spread by a buffer zone, by the cell size or by transmissibility upscaling, and we check it by comparison with the fine model.

Wrong

“The well index of a coarse cell is calculated by Peaceman's formula with its permeability.”

In fact

Near a well the flow is radial, and the fine cells along the wellbore matter far more than their volume. In the example of the lesson the best rock, which fills 1/16 of the volume of the cell, increased the well index by a factor of 4.0.

In depth: advanced detail

Why the bounds are strict and how the boundary conditions are ordered

Steady flow is “economical”. Of all the pressure distributions with the given pressures on the faces, the one that is realised dissipates the least energy. The dissipated power at a unit pressure difference is proportional to $k^{*}$ . So any other admissible pressure distribution gives an estimate of $k^{*}$ from above. A pressure that is linear along the flow gives the arithmetic mean, the upper Wiener bound. A pressure that is constant in each cross-section but changes along the flow in any way gives the best of such estimates. It is the harmonic mean of the arithmetic means of the cross-sections, the upper Cardwell–Parsons bound.

The matching statement is about rates. Of all the flow distributions without sources and with a given total rate, the one that is realised has the smallest energy losses. So any such distribution gives an estimate from below. A uniform flow over the whole cross-section gives the harmonic mean, the lower Wiener bound. A flow locked in tubes along the lines of cells gives the arithmetic mean of the harmonic means, the lower Cardwell–Parsons bound. Both statements hold also for the difference equations of the cells, so the calculation on a grid does not go outside the bounds.

The same argument puts the boundary conditions in order. Linear conditions fix the pressure on the side faces, and sealed faces leave it free. The second problem has more room to reduce the dissipation. So $k^{*}$ under linear conditions is not less than under sealed faces. For the same reason it is not less than under periodic conditions. Between sealed faces and periodic conditions there is no general order: in the example of the lesson the periodic ones gave less. The Cardwell–Parsons bounds hold for sealed faces and periodic conditions, but not for linear ones. In a block of stripes across the flow, both bounds are equal to the harmonic mean, and sealed faces give exactly it, while linear conditions give more.

Average velocity and the tensor from a block calculation

The velocity averaged over the volume is expressed through the rates on the faces of the block. There are no sources inside the block, so the integral of $u_x$ over the volume equals the integral of $x\,(\mathbf{u}\cdot\mathbf{n})$ over its surface (the divergence theorem for the product $x\mathbf{u}$ ). For the difference equations of the cells the same equality holds exactly:

Average Darcy velocity from the rates through the faces of the block

\langle u_x\rangle = \frac{1}{V}\sum_{f} x_f\,q_f,\qquad \langle u_y\rangle = \frac{1}{V}\sum_{f} y_f\,q_f

where:

$q_f$ rate outward through the boundary face $f$ of a fine cell (an inflow has a minus sign), m³/s
$x_f,\ y_f$ coordinates of the centre of this face, m
$V$ volume of the block, m³

Units: SI (any consistent system)

Holds when

The flow is steady, and there are no sources inside the block.
The sum is over all the faces of fine cells on the boundary of the block.

With sealed side faces the formula gives $\langle u_x\rangle = Q/(Wh)$ and agrees with Darcy's law for the block. With periodic conditions it reduces to the rate through one cross-section: $\langle u_x\rangle = Q_x/(Wh)$ and $\langle u_y\rangle = Q_y/(Lh)$ , where $Q_y$ is the total flow to the north through any horizontal line of faces.

A tensor calculated correctly with periodic or linear conditions on the faces of the block itself has two signs. It is symmetric ( $k^{*}_{xy} = k^{*}_{yx}$ , up to the error of the solution). And it is positive definite ( $k^{*}_{xx}k^{*}_{yy} > (k^{*}_{xy})^2$ , which means that the flow always has a component down the pressure gradient). In the example of the lesson the two off-diagonal terms agreed to hundredths. An asymmetric tensor with such conditions most often means that the velocity was averaged not over the volume but, for example, over one outlet face. With a buffer zone, symmetry is not guaranteed even for a correct calculation.

Quick approximate methods

Besides a flow calculation in every block, there are methods that are orders of magnitude faster. Renormalisation combines cells in groups of 2 × 2 (2 × 2 × 2 in a volume). It replaces each group by a small network of resistances with a ready formula. Then it combines the resulting blocks again, until it reaches the size of the coarse cell. The number of operations is proportional to the number of cells, but at each step some information about the arrangement is lost, and with a high permeability contrast the error accumulates.

The power average has an exponent fitted by flow calculations for several typical blocks. The fitted exponent is then transferred to the other blocks of the same structure (power average). The exponent depends on the direction: for bodies elongated along the flow it is closer to one, and across the flow it is closer to minus one. Both methods are checked by comparison with a flow calculation, at least for a part of the blocks.

Connections to other topics

Reservoir heterogeneity and permeability averaging — averages, lognormal permeability, correlation length and anisotropy are the basis of upscaling.
Grids in reservoir simulation models — what simulation grids there are and how the cell size affects the result.
Pressure equation for single-phase flow — the cell equations and transmissibilities that the flow calculation in a block solves.
Interblock flow and transmissibility — how the simulator calculates the flows between cells.
Well model in a grid block — the well index and the pressure of the cell with a well.
Two-phase upscaling: pseudo-functions — pseudo functions and the check of upscaling on displacement.
Special topics in upscaling — the full tensor, upscaling near wells, small-scale heterogeneity.
Displacement in heterogeneous reservoirs — pseudo functions of a layered cross-section at vertical equilibrium.
Reservoir architecture and performance — sizes of sand bodies and barriers, which decide the choice of cells.
History matching — how upscaling errors are disguised during history matching.

Check yourself

1
A simulation cell consists of two parts of equal volume: porosity 0.25 and water saturation 0.20; porosity 0.05 and water saturation 1.00. What is the water saturation of the cell?
1. A.0.50
2. B.0.20
3. C.0.60
4. D.0.33
Show the answer
Correct answer: D. 0.33
The weight is the pore volume: $\bar S_w = (0.25\cdot0.20 + 0.05\cdot1.00)/(0.25 + 0.05) = 0.10/0.30 \approx 0.33$ . The simple average, 0.60, would give the tight part, which has almost no pores, the weight of a half. It would overestimate the water in the cell by almost two times.
2
A block of 3 × 2 equal cells: along the flow, the first line has 200, 150 and 250 mD, and the second has 20, 30 and 10 mD. Find the Wiener and Cardwell–Parsons bounds for flow along the lines. Is a flow calculation needed here?
Show the answer
Wiener: the arithmetic mean is $660/6 = 110$ mD, and the harmonic mean is $6/0.199 \approx 30.2$ mD. Cardwell–Parsons: the harmonic means of the lines are $3/0.015667 \approx 191.5$ and $3/0.1833 \approx 16.4$ mD, and the lower bound is their mean, about 104 mD. The averages of the cross-sections are 110, 90 and 130 mD, and the upper bound is $3/(1/110 + 1/90 + 1/130) \approx 107.5$ mD. The block is almost layered along the flow, and the bounds differ by less than 4%. Any value between them will do. A flow calculation (it gives about 104.7 mD) adds almost nothing.
3
Which boundary conditions give the largest equivalent permeability of a block along the applied gradient?
1. A.Periodic
2. B.Sealed side faces
3. C.Linear
4. D.Any of the three, depending on the block
Show the answer
Correct answer: C. Linear
Linear conditions fix the pressure on the side faces as in a uniform reservoir, and let the fluid enter and leave through them. Because of this, the dissipation of energy is not less than in problems with freer faces, and $k^{*}$ under linear conditions is not less than under sealed faces and under periodic conditions. Between the last two there is no general order.
4
Two coarse layers of 4 m each are made of fine layers of 1 m each. In the upper one the vertical permeabilities of the fine layers are 50, 50, 50 and 0.02 mD (a thin layer at the base). In the lower one there are four layers of 50 mD each. Find the vertical transmissibility between the centres of the coarse layers per 1 m² of cross-section: from the permeabilities of the coarse layers, and directly from the fine layers.
Show the answer
The permeability of the upper coarse layer is $4/(3/50 + 1/0.02) = 4/50.06 \approx 0.0799$ mD, and that of the lower one is 50 mD. From these, $T/A = 1/(2/0.0799 + 2/50) \approx 1/25.07 \approx 0.040$ mD/m. Directly from the fine layers between the centres, taking the lower two layers of the upper coarse layer and the upper two layers of the lower one: $T/A = 1/(1/50 + 1/0.02 + 1/50 + 1/50) = 1/50.06 \approx 0.020$ mD/m. Permeability upscaling overestimated the transmissibility through the thin layer by a factor of two. It assigned half of its resistance to the upper face of the upper coarse layer, where there is no thin layer.
5
The horizontal correlation length of the permeability is about 100 m. At what horizontal size of the simulation cells does the upscaling result depend most strongly on the boundary conditions?
1. A.2000 m
2. B.It does not depend on the cell size
3. C.10 m
4. D.100–200 m
Show the answer
Correct answer: D. 100–200 m
In a cell of the size of about the correlation length there are both conducting bodies and barriers, but too few of them to average out. The answer depends most strongly on the assumption about the neighbours. In a cell of 10 m the properties are almost uniform, and in a cell of 2000 m the heterogeneity averages out.
6
A simulation cell of 200 × 200 m, 80 mD; the wellbore ( $r_w$ = 0.1 m) passes through a compacted zone, roughly a circle of radius 12 m with a permeability of 15 mD. By how many times does this zone reduce the well index compared with a uniform cell?
Show the answer
Peaceman's equivalent radius is $r_0 = 0.14\sqrt{200^2 + 200^2} \approx 39.6$ m, and the zone is inside it. The skin of the zone is $S = (80/15 - 1)\ln(12/0.1) \approx 4.33\cdot4.79 \approx 20.7$ . The index is proportional to $1/(\ln(r_0/r_w) + S)$ : $\ln 396 \approx 5.98$ , and the ratio is $5.98/(5.98 + 20.7) \approx 0.22$ . The index is about 4.5 times smaller. Upscaling of the cell for linear flow will hardly notice the zone.

Updated: 16 October 2026