Two-phase upscaling: pseudo-functions

Quick refresher

Pseudo relative permeability is defined through phase permeabilities: $\hat k_{rf} = \hat k_f/k^{*}$ . This is the equivalent permeability of the coarse cell for phase $f$ , divided by the single-phase equivalent permeability in the same direction. Even when the core relative permeabilities are the same, pseudo relative permeabilities depend on the direction.
Steady-state methods reduce the two-phase problem to single-phase ones. In the capillary equilibrium limit (slow flow, small scales) the capillary pressure is the same in the whole cell. In the viscous limit the water fraction in the flow (the fractional flow) is the same. If the relative permeability curves of all the fine cells are the same, the viscous limit gives back the core curves.
Dynamic methods take the saturation, the phase rates and the pressures from a displacement calculation on a fine grid. The Kyte–Berry method averages the pressures of the central column with the weight $k\,k_{rf}h$ and takes the rates at the outlet face of the block. The pore-volume weighted method (PVW) averages the pressure over the whole block.
The rate at the outlet face compensates for numerical dispersion: until the water reaches the face, $\hat k_{rw} = 0$ , however much water there is in the block. In a layered cross-section something else is stronger. The water passes the block along the best layer, and at the same average saturation $\hat k_{rw}$ is many times larger than from the core curves.
Different weights for the oil and water pressures create a false pseudo capillary pressure even where there is no capillary pressure on the fine grid. Pore-volume weighted averaging does not do this.
Pseudo functions are tied to the balance of forces (the injection rate), the boundary conditions of the calculation, the well positions, the flow direction and the coarse grid itself. We check them by comparing the fine and the coarse model in conditions different from those in which they were found.

Key relationships

\hat k_{rf} = \dfrac{\hat k_f}{k^{*}},\qquad \hat S_w = \dfrac{\sum V_i\phi_iS_{w,i}}{\sum V_i\phi_i}

pseudo relative permeability of phase $f$ , and the saturation of the coarse cell

\hat k_{rf} = \dfrac{\mu_f\,q_f\,\Delta x}{0.0853\,k^{*}A\,\Delta\bar p_f}

dynamic pseudo relative permeabilities: Darcy's law between block centres, metric oilfield units

\bar p_f = \dfrac{\sum_j k_jk_{rf,j}h_j\,p_{f,j}}{\sum_j k_jk_{rf,j}h_j}

pressure of a phase in the central column of the block by Kyte–Berry (horizontal reservoir)

N_{cv} = \dfrac{k_v\,k_{ro}'\,\Delta p_c\,L}{\mu_o\,u\,H^2}

ratio of the time to cross the cell to the time of capillary equalisation across the layers

Common pitfalls

Thinking that a correct single-phase permeability guarantees a correct displacement calculation.
Carrying pseudo functions over to a different injection rate, grid, wells or flow direction without a check.
Using Kyte–Berry pseudo relative permeabilities without their pseudo capillary pressure, and not checking the tables for negative and infinite values.
Accounting for the numerical dispersion of one grid in the pseudo functions, then making the cells smaller but keeping the same tables.
Applying the capillary equilibrium limit to cells with layers metres thick at a field injection rate.

After this lesson you can

explain why the equivalent permeability and the core relative permeability curves are not enough for a coarse cell: heterogeneity inside the cell, numerical dispersion (the smearing of the front by the grid), gravity and capillary forces
write the definition of pseudo relative permeability through phase permeabilities, and calculate steady-state pseudo functions (for flow that does not change in time) in the capillary equilibrium limit (very slow flow: the capillary pressure is the same everywhere in the cell) and in the viscous limit (capillary forces and gravity do not matter)
calculate dynamic pseudo relative permeabilities (found from a displacement calculation on a fine grid) by the Kyte–Berry method and by pore-volume weighted averaging, and explain where the false pseudo capillary pressure comes from
compare a fine-grid calculation with coarse-grid calculations that use core curves, different pseudo functions, and a non-uniform grid (one that keeps the important layers separate), and estimate the error of each approach
list the conditions in which pseudo functions stop working (the injection rate, the boundary conditions, the well positions, the flow direction, the grid), and check them before use

Why equivalent permeability is not enough for a coarse cell

Single-phase upscaling gives each coarse cell an equivalent permeability $k^{*}$ , so that the cell passes the same single-phase flow as the fine cells inside it (Permeability upscaling for single-phase flow). In waterflooding this is not enough. The coarse model knows only the average saturation of the cell. But the share of water that the cell gives to its neighbour depends on where inside the cell the water is. Three causes act on this distribution.

Heterogeneity inside the cell. Water moves faster along permeable layers and leaves the cell when most of its volume is still oil-saturated. With core curves, a cell with such an average saturation would hardly pass any water (Displacement in heterogeneous reservoirs).
Numerical dispersion. A coarse cell “mixes” the incoming water over its whole volume, and passes it on to the neighbour as soon as the average saturation has risen. The front is smeared over one or two cells, that is, over hundreds of metres (Numerical dispersion).
Gravity and capillary forces. Water sinks to the base (Segregated flow and gravity displacement), and capillary imbibition (the rock soaking up water by capillary forces) draws it into tight layers and into laminae (very thin layers of rock) (Capillary pressure and the transition zone). How the phases are distributed depends on the ratio of these forces to the viscous forces, that is, also on the injection rate.

All three causes change the same relation: the relation between the average saturation of the coarse cell and the water fraction in the flow through its face. This relation is what the tables of pseudo functions give to the coarse model. For a layered cross-section at vertical equilibrium (the fluids have separated by gravity over the thickness) we get pseudo functions analytically (pseudo functions of a cross-section). In the general case we get them by a displacement calculation on a fine grid.

The model of the examples of the lesson

What: Cross-section
Value: the eight layers of the lesson Displacement in heterogeneous reservoirs, thickness 16 m; inside each layer $k_v/k_h = 0.01$

What: Element
Value: 500 m between the rows of wells, width 300 m, a horizontal reservoir

What: Relative permeability and fluids
Value: the curves of the lesson Relative permeability in all the layers; oil 3.0 mPa·s, water 0.5 mPa·s (mobility ratio $M'$ = 1.765: at the end points of the curves, the water moves 1.765 times more easily than the oil)

What: Injection
Value: 340 m³/d at reservoir conditions: a Darcy velocity of 0.0708 m/d, as in the lesson Buckley–Leverett theory and the Welge method

What: Fine grid
Value: 8 layers × 45 columns of 11.1 m; the IMPES scheme (implicit pressure, explicit saturation), upstream weighting (the flow takes the water fraction from the cell it comes from)

What: Coarse grid
Value: one layer × 5 blocks of 100 m (9 columns of the fine grid in each)

What	Value
Cross-section	the eight layers of the lesson Displacement in heterogeneous reservoirs, thickness 16 m; inside each layer $k_v/k_h = 0.01$
Element	500 m between the rows of wells, width 300 m, a horizontal reservoir
Relative permeability and fluids	the curves of the lesson Relative permeability in all the layers; oil 3.0 mPa·s, water 0.5 mPa·s (mobility ratio $M'$ = 1.765: at the end points of the curves, the water moves 1.765 times more easily than the oil)
Injection	340 m³/d at reservoir conditions: a Darcy velocity of 0.0708 m/d, as in the lesson Buckley–Leverett theory and the Welge method
Fine grid	8 layers × 45 columns of 11.1 m; the IMPES scheme (implicit pressure, explicit saturation), upstream weighting (the flow takes the water fraction from the cell it comes from)
Coarse grid	one layer × 5 blocks of 100 m (9 columns of the fine grid in each)

Gravity and capillary pressure are not included in the grid calculations, so that the influence of the heterogeneity and the grid can be seen in pure form. The data are assumed for illustration.

CalculatedWater saturation of the cross-section on the fine grid after 0.20 pore volumes have been injected. Water has passed the whole element along the 900 mD layer. Flowing across the layers out of it, the water has partly filled the neighbouring layers of 120 and 18 mD. In the layers of 310 and 200 mD it has reached about 390 and 280 m. In the tight layers of 75 and 8 mD it has reached only 120 and 45 m (the pale strip further along the 8 mD layer is water that flowed in from the neighbouring layers). The vertical lines are the boundaries of the coarse blocks. The average saturation of block 3 is only 0.37, but 62% of water already flows through its outlet face.

What pseudo relative permeability is

The coarse model calculates the flow of each phase between neighbouring cells by Darcy's law, with the permeability $k^{*}$ and a relative permeability taken at the saturation of the upstream cell (the cell the flow comes from). Pseudo relative permeabilities are the values $\hat k_{rf}$ at which this flow agrees with the flow of the phase on the fine grid. The pseudo saturation $\hat S_w$ is the average saturation of the coarse cell, weighted by pore volume. It is averaged exactly, as in the single-phase case (porosity and saturation).

It is convenient to think of each phase as its own single-phase flow in a rock with the permeability $k_f = k\,k_{rf}(S_w)$ . This is the phase permeability (the effective permeability for the phase, relative permeability). If we average it over the coarse cell in the same way as the absolute permeability, we get the equivalent phase permeability $\hat k_f$ . Dividing by $k^{*}$ gives the pseudo relative permeability.

Pseudo relative permeability and pseudo saturation of a coarse cell

\hat k_{rf,x} = \frac{\hat k_{f,x}}{k^{*}_x},\qquad \hat S_w = \frac{\sum_i V_i\,\phi_i\,S_{w,i}}{\sum_i V_i\,\phi_i}

where:

$\hat k_{rf,x}$ pseudo relative permeability of phase $f$ (oil or water) for flow along $x$ , —
$\hat k_{f,x}$ equivalent phase permeability of the coarse cell along $x$ : the permeability at which the cell passes the same rate of the phase as the fine cells, mD
$k^{*}_x$ single-phase equivalent permeability of the same cell along $x$ , mD
$V_i,\ \phi_i,\ S_{w,i}$ volume, porosity and water saturation of the $i$ -th fine cell, m³, fraction, fraction

Units: permeabilities in equal units; saturation as a fraction

Holds when

$k^{*}$ and $\hat k_f$ are calculated for one direction and under the same boundary conditions. The directions $y$ and $z$ have their own values.
The sums run over the fine cells inside the coarse one.
How to get $\hat k_f$ depends on the method. Steady-state methods solve single-phase problems, and dynamic methods take the rates and pressures from a displacement calculation.

Pseudo relative permeability describes not the rock but the cell. Even with the same core curves in all the fine cells, it depends on how the water and the oil are placed in the cell. So we calculate it separately for each direction. Along the layers and across them the pseudo relative permeabilities can differ by several times.

Steady-state methods: capillary equilibrium and the viscous limit

Suppose the flow inside the coarse cell has settled, so that the saturation does not change in time. Then the balance of each phase reduces to the equation of single-phase steady flow with the permeability $k\,k_{rf}(S_w)$ (the derivation is in the advanced section). If we know the saturation distribution in the cell, we find $\hat k_f$ by the same calculations as $k^{*}$ : by the averages for layers, or by a flow calculation in the block (flow calculation in the block). This is the basis of steady-state upscaling. In it, the saturation distribution is set by the balance of forces.

Capillary equilibrium limit (capillary limit): The flow is very slow, and gravity does not matter: the capillary pressure is the same in the whole cell. We find the saturation of each fine cell from its own capillary pressure curve. For rocks of one type we use the Leverett J-function (a way to bring the capillary pressure curves of similar rocks to one curve). The method at once gives the pseudo capillary pressure: $\hat p_c(\hat S_w)$ is equal to this common pressure.
Viscous limit: Capillary forces and gravity are negligible, and in steady flow the water fraction is the same in the whole cell. We find the saturation of each fine cell from its $f_w$ curve. If the relative permeability curves of all the fine cells are the same, the saturation comes out the same, and the pseudo relative permeabilities coincide with the core curves. The viscous limit changes nothing.
Gravity equilibrium: The phases are separated by gravity in the vertical direction. This is vertical equilibrium. For a sharp boundary, the pseudo relative permeabilities are straight lines between the end points (the lesson on segregated flow). For a layered cross-section they are sums over the flooded layers (the lesson on layered reservoirs). With a capillary transition zone they are integrals over the thickness (see the advanced section).

Pseudo functions in the capillary equilibrium limit

1
Set the capillary pressure
Take a level $P_c$ , from the largest one, at which all the fine cells are at irreducible water (the water that cannot be pushed out), down to zero.
2
Saturations of the fine cells
For each fine cell, find $S_{w,i}$ from its curve: $p_{c,i}(S_{w,i}) = P_c$ . At the same pressure, tight rocks have more water.
3
Average saturation
$\hat S_w$ is weighted by pore volume.
4
Phase permeabilities
$k_{w,i} = k_ik_{rw}(S_{w,i})$ and $k_{o,i} = k_ik_{ro}(S_{w,i})$ for each fine cell.
5
Single-phase upscaling of each phase
Average the field $k_w$ and the field $k_o$ in the same way as the field $k$ : for layers along the flow arithmetically, across the flow harmonically, and for any heterogeneity by a flow calculation in the block. This gives $\hat k_w$ and $\hat k_o$ for each direction.
6
Pseudo relative permeabilities
Divide by $k^{*}$ of the same direction. Repeat for other values of $P_c$ . The points $(\hat S_w, \hat k_{rw}, \hat k_{ro}, P_c)$ form a table.

Pseudo relative permeabilities of a layered cell in the capillary equilibrium limit

p_{c,j}(S_{w,j}) = P_c,\qquad \hat k_{rf}^{\parallel} = \frac{\sum_j k_jh_j\,k_{rf}(S_{w,j})}{\sum_j k_jh_j},\qquad \hat k_{rf}^{\perp} = \frac{\left(\sum_j h_j/\bigl(k_jk_{rf}(S_{w,j})\bigr)\right)^{-1}}{\left(\sum_j h_j/k_j\right)^{-1}}

where:

$p_{c,j}(S_w)$ capillary pressure curve of layer (lamina) $j$ . For one rock type it is $p_{c,j} = J(S_w)\,\sigma\cos\theta\sqrt{\phi_j/k_j}$ in consistent units (Leverett J-function), MPa
$P_c$ common capillary pressure of the cell, MPa
$k_j,\ h_j$ permeability and thickness of the layer, mD, m
$\hat k_{rf}^{\parallel},\ \hat k_{rf}^{\perp}$ pseudo relative permeability of phase $f$ along the layers and across them, —

Units: permeabilities in mD, thicknesses in m; pseudo relative permeabilities are dimensionless

Holds when

The layers are continuous over the whole cell, and each one is uniform. Along the layers they work in parallel, and across them in series (averaging over layers).
The flow is slow: capillary forces have time to equalise the pressure between the layers (the criterion is below).
The relative permeability and capillary pressure curves are those of the process being modelled (for waterflooding of a water-wet rock, imbibition).

Along the layers each phase goes along its best paths in parallel, so the pseudo relative permeability is decided by the permeable layer. Across the layers a phase has to cross every layer, so the pseudo relative permeability is decided by the layer where that phase is least mobile.

Worked exampleLaminae at capillary equilibrium: pseudo relative permeabilities along and across

Given (assumed data)

Laminae (fractions of the thickness): A: 400 mD, $\phi$ = 0.25 (0.45); B: 60 mD, $\phi$ = 0.21 (0.35); C: 6 mD, $\phi$ = 0.15 (0.20)
Relative permeability: the curves of the lesson Relative permeability in all the laminae
Imbibition capillary pressure: $0.04\,(1 - S_n)^2$ MPa, the assumed curve of the lesson Two-phase flow equations. Here we take it as the curve of a rock of 50 mD with $\phi$ = 0.20. For the other rocks we multiply it by $\sqrt{(\phi/k)/(0.20/50)}$ , following the Leverett J-function
Level of the capillary pressure: $P_c$ = 0.012 MPa

Solution

The largest capillary pressure of the laminae (at the irreducible water saturation $S_{wi}$ ): for A it is $0.04\sqrt{(0.25/400)/0.004} = 0.01581$ MPa, for B it is 0.03742, and for C it is 0.100 MPa.
Saturation from $(1 - S_n)^2 = P_c/p_{c,\max}$ : for A, $1 - S_n = \sqrt{0.012/0.01581} = 0.8712$ and $S_w = 0.22 + 0.53\cdot0.1288 = 0.2883$ . For B, $S_w = 0.4499$ . For C, $S_w = 0.5664$ .
Relative permeabilities of the laminae: $k_{rw}$ = 0.000534, 0.02040, 0.06980 and $k_{ro}$ = 0.6022, 0.2051, 0.0600.
Average saturation, weighted by $\phi h$ : $\hat S_w = (0.1125\cdot0.2883 + 0.0735\cdot0.4499 + 0.030\cdot0.5664)/0.216 = 0.382$ .
Along the laminae, $k^{*} = 400\cdot0.45 + 60\cdot0.35 + 6\cdot0.20 = 202.2$ mD: $\hat k_{rw} = (180\cdot0.000534 + 21\cdot0.02040 + 1.2\cdot0.06980)/202.2 = 0.0030$ and $\hat k_{ro} = 0.558$ .
Across, $k^{*} = 1/(0.45/400 + 0.35/60 + 0.20/6) = 24.82$ mD. The phase permeability of water is $1/\bigl(0.45/(400\cdot0.000534) + 0.35/(60\cdot0.02040) + 0.20/(6\cdot0.06980)\bigr) = 0.348$ mD, and $\hat k_{rw} = 0.0140$ . For oil it is 1.707 mD, and $\hat k_{ro} = 0.0688$ .
Core curves at $S_w = 0.382$ : $k_{rw}$ = 0.0071 and $k_{ro}$ = 0.341.

Answer: At the average saturation 0.382, along the laminae

\hat k_{rw}

= 0.0030 and

\hat k_{ro}

= 0.558. Across them these are 0.0140 and 0.0688. From the core curves they are 0.0071 and 0.341.

What it means. Capillary forces have drawn the water into the tight laminae C and B, and the oil has stayed in the permeable lamina A. Along the laminae the oil flows through A, and the cell behaves more favourably than a rock with the same saturation: it passes 1.6 times more oil and 2.4 times less water. Across the laminae the oil has to cross the water-saturated lamina C, where its relative permeability is only 0.0600. So the pseudo relative permeability of oil is 8.1 times lower than along the laminae: for flow across them the oil in A is almost trapped. The ratio of the vertical and horizontal phase permeabilities of the cell is 0.015 for oil and 0.574 for water, while the single-phase $k_v/k_h$ is 0.123. So the anisotropy depends on the phase. The data in this example are assumed for illustration.

CalculatedPseudo relative permeabilities of the laminae of the example at capillary equilibrium, along and across the laminae, compared with the core curves (dashed). Along the laminae the oil flows more easily than by the core curves, and across them much less easily.

When can we use the capillary equilibrium limit? Capillary imbibition equalises the saturation across the layers the faster, the thinner the layers and the higher the vertical permeability. The viscous flow carries water through the cell the faster, the shorter the cell and the higher the velocity. We compare the two times, as in the lesson on layered reservoirs we compared the viscous forces with gravity.

Ratio of capillary and viscous forces in a cell

N_{cv} = \frac{t_x}{t_c} = \frac{k_v\,k_{ro}'\,\Delta p_c\,L}{\mu_o\,u\,H^2}

where:

$t_x$ time for the flow to cross the cell length $L$ , about $\phi\,\Delta S\,L/u$ , s
$t_c$ time for capillary forces to even out the saturation over the thickness $H$ , about $\phi\,\Delta S\,\mu_oH^2/(k_vk_{ro}'\Delta p_c)$ , s
$k_v$ vertical permeability of the unit of layers (a group of neighbouring layers), m²
$\Delta p_c$ typical capillary pressure difference between the layers, Pa
$L,\ H$ length of the cell along the flow, and thickness of the layers between which the saturation equalises, m
$\mu_o,\ u,\ k_{ro}'$ oil viscosity, Darcy velocity, end-point relative permeability of oil, Pa·s, m/s, —

Units: SI

Holds when

This is an estimate of the order of magnitude, like the ratio of gravity to viscous forces in the lesson on layered reservoirs. The pressure difference $\Delta p_c/H$ over the thickness of a layer takes the place of the gravity gradient.
$N_{cv} \gg 1$ is the capillary equilibrium limit. $N_{cv} \ll 1$ means that capillary crossflows (flow of water between layers, driven by capillary forces) can be neglected. In between, a calculation is needed.

The thickness enters squared. In laminae a few centimetres thick, capillary equilibrium is reached almost always. In a unit of layers that is metres thick, it is reached only with good vertical permeability (in the cross-section of the lesson, with its shale layers, $N_{cv}$ is a few thousandths). Do not confuse $N_{cv}$ with the capillary number $N_c = \mu v/\sigma$ . That one compares the viscous and capillary forces at the pore scale and grows with the rate. $N_{cv}$ is the opposite: it is large when the flow is slow.

In other unit systems (1)

Metric oilfield units

N_{cv} = \frac{0.0853\,k_v\,k_{ro}'\,\Delta p_c\,L}{\mu_o\,u\,H^2}

$k_v$ in mD, $\Delta p_c$ in MPa, $L$ and $H$ in m, $\mu_o$ in mPa·s, $u$ in m/d.

0.0853 is the constant of Darcy's law in metric oilfield units (Darcy's law).

Worked exampleLaminae and the cross-section: where capillary equilibrium works

Given (assumed data)

Laminae: thickness about 1 cm; a cell of the detailed model 0.3 m long; $k_v$ = 24.8 mD (across the laminae of the previous example)
Cross-section of the examples of the lesson: thickness 16 m, block 100 m; $k_v$ of the unit = $16/\sum(h_j/0.01k_j)$ = 0.258 mD
Common data: $k_{ro}'$ = 0.85; $\Delta p_c$ = 0.04 MPa; oil 3.0 mPa·s; $u$ = 0.0708 m/d

Solution

Laminae: $N_{cv} = 0.0853\cdot24.8\cdot0.85\cdot0.04\cdot0.3/(3.0\cdot0.0708\cdot0.01^2) = 1016$ .
Cross-section: $N_{cv} = 0.0853\cdot0.258\cdot0.85\cdot0.04\cdot100/(3.0\cdot0.0708\cdot16^2) = 0.0014$ .
The thickness at which $N_{cv} = 1$ in a block of 100 m with $k_v$ = 0.258 mD: $H^2 = 0.0853\cdot0.258\cdot0.85\cdot0.04\cdot100/(3.0\cdot0.0708) = 0.352$ m², so $H = 0.59$ m.

Answer: In the laminae

N_{cv}

≈ 1016, which is capillary equilibrium. In the cross-section it is 0.0014: a block 16 m thick is very far from capillary equilibrium.

What it means. The capillary equilibrium limit is good for upscaling in several steps. First laminae and layers a centimetre thick are turned into a cell of the detailed geological model. Then that cell is upscaled further, by other methods. For a coarse cell that a front crosses at a field injection rate, the limit usually does not apply. In the cross-section of the lesson the layers do not have time to exchange water, and the water runs ahead along the permeable ones. The data in this example are assumed for illustration.

Dynamic pseudo functions: the Kyte–Berry and pore-volume weighted methods

When a front crosses a coarse cell, the flow in it is not steady, and the steady-state assumptions do not hold. Then we take the pseudo functions from a displacement calculation on a fine grid. The calculation covers a whole sector with wells, or each coarse cell with a border of neighbouring cells (boundary conditions). During the calculation we save the saturations, pressures and flows. Afterwards, for each coarse cell and each direction, we build a table. The best-known way is the Kyte–Berry method. It is called dynamic, and the curves it gives are called dynamic pseudo functions (dynamic pseudos).

Pseudo relative permeabilities of block I by the Kyte–Berry method

1
Permeability between the centres
Calculate the single-phase equivalent permeability $k^{*}$ of the region from the centre of block I to the centre of the neighbouring block II, in the same way as in single-phase upscaling (Permeability upscaling for single-phase flow). For layers along the flow it is the arithmetic mean.
2
Average saturation
Find $\hat S_w$ of block I, weighted by pore volume, for each saved moment of time.
3
Phase rates
Add up the flows of water and oil of the fine cells through the face between blocks I and II, the outlet face of block I.
4
Phase pressures
Average the pressure of each phase in the central column of fine cells of block I and of block II, with the weight $k\,k_{rf}h$ , and correct it to the depth of the block centre.
5
Pseudo relative permeabilities
By Darcy's law between the block centres, find $\hat k_{rw}$ and $\hat k_{ro}$ . Assign them to the $\hat S_w$ of block I, the upstream block, whose saturation the coarse model will use for the flow.
6
Pseudo capillary pressure
This is the difference of the averaged oil and water pressures at the centre of block I.

Pseudo relative permeabilities and pseudo capillary pressure by the Kyte–Berry method

\begin{aligned} \bar p_{f,I} &= \frac{\sum_j k_jk_{rf,j}h_j\bigl(p_{f,j} - \rho_fg(D_j - \bar D_I)\bigr)}{\sum_j k_jk_{rf,j}h_j},\qquad \Delta\bar\Phi_f = \bar p_{f,I} - \bar p_{f,II} - \rho_fg\,(\bar D_I - \bar D_{II}),\\ \hat k_{rf} &= \frac{\mu_f\,q_f\,\Delta x}{0.0853\,k^{*}A\,\Delta\bar\Phi_f},\qquad \hat p_c = \bar p_{o,I} - \bar p_{w,I} \end{aligned}

where:

$\bar p_{f,I}$ pressure of phase $f$ in block I: the average over the central column of fine cells $j$ , corrected to the depth of the block centre $\bar D_I$ , MPa
$k_j,\ k_{rf,j},\ h_j$ permeability, relative permeability of the phase and thickness of a fine cell of the column: the averaging weights, mD, —, m
$D_j,\ \bar D_I$ depth of a fine cell and of the centre of the coarse block (measured downwards), m
$\rho_fg$ hydrostatic gradient of the phase, $0.00981\rho_f$ with the density in g/cm³, MPa/m
$\Delta\bar\Phi_f$ difference of the phase potential (the phase pressure corrected for depth) between the centres of blocks I and II, MPa
$q_f$ rate of the phase through the face between the blocks: the sum of the flows of the fine cells of this face, at reservoir conditions, m³/d
$\Delta x,\ A$ distance between the block centres and the area of their common face, m, m²
$k^{*}$ single-phase equivalent permeability between the block centres, mD
$\mu_f$ viscosity of the phase, mPa·s

Units: metric oilfield: $k$ in mD, $q$ in m³/d, $p$ in MPa, $\mu$ in mPa·s, lengths in m

Holds when

The reservoir and the flow are the same as in the fine-grid calculation from which the pressures and rates were taken. The pseudo relative permeabilities are assigned to the saturation of the upstream block.
If a phase is immobile in the whole column (all $k_{rf,j} = 0$ ), the weight has no meaning. In the examples of the lesson, the pressure of such a phase is averaged with the weight $kh$ .
$0.0853$ is the constant of Darcy's law (Darcy's law). In a horizontal reservoir the gravity terms vanish.

This is Darcy's law solved for the relative permeability. The rate, the potential difference and the permeability are known, and the relative permeability is unknown. All that the fine model knows about the placement of the phases in the block reaches $\hat k_{rf}$ through the rate at the face and the averaged pressure.

Why the rate is taken at the outlet face. The coarse model decides how much water to give to the neighbour only from the average saturation of the block. While the front on the fine grid moves inside a uniform block, this saturation grows. With the core curves the block would begin to give water earlier than it reaches the block boundary, and this is numerical dispersion. The rate at the outlet face is zero at this time: $q_w = 0$ , so $\hat k_{rw} = 0$ at all such average saturations (see the example with a homogeneous reservoir below). The pseudo curve of water is shifted to the right, and the coarse model holds the water back until it reaches the face. In this way the method compensates for numerical dispersion and for heterogeneity at the same time.

Pore-volume weighted averaging (PVW). The weight $k\,k_{rf}h$ stresses the zones where the phase flows. But this is a choice that we make, not a consequence of physics, and the weights for oil and water come out different. So the averaged pressures of the phases differ even where there is no capillary pressure on the fine grid. This gives a false pseudo capillary pressure, and it depends on which column was averaged, that is, on the direction. The pore-volume weighted method takes the pressure of the whole block with the weight $\phi V$ of the fine cells, the same for both phases. The rest is as in the Kyte–Berry method.

Block pressure with pore-volume weighted averaging

\bar p_{f,I} = \frac{\sum_{i\in I} V_i\phi_i\,\bigl(p_{f,i} - \rho_fg(D_i - \bar D_I)\bigr)}{\sum_{i\in I} V_i\phi_i}

where:

$i \in I$ all the fine cells of block I, —
$V_i\phi_i$ pore volume of the fine cell, m³
$p_{f,i}$ pressure of the phase in the fine cell, MPa

Units: metric oilfield, as above

Holds when

The rest is the same potential difference and Darcy's law as in the Kyte–Berry method. For the last block before a well, we use the difference with the pressure at its face and the distance $\Delta x/2$ .

With zero capillary pressure on the fine grid, the PVW pseudo capillary pressure is also zero, and the water fraction from the pseudo relative permeabilities is exactly equal to the water fraction in the flow through the face.

Worked exampleOne point of the pseudo curves of block 3

Given (assumed data)

Moment: 0.20 pore volumes injected; block 3 is from 200 to 300 m, and block 4 from 300 to 400 m
Fine grid: average saturation of block 3 $\hat S_w$ = 0.369; through the face at 300 m pass 209.2 m³/d of water and 130.8 m³/d of oil
Kyte–Berry pressures (central columns at 250 and 350 m): oil 6.763 and 3.836 MPa; water 6.837 and 3.959 MPa
PVW pressures (whole block): 6.749 and 3.882 MPa
Blocks: $k^{*}$ = 139.6 mD (the arithmetic mean of the layers), $A = 300\cdot16 = 4800$ m², $\Delta x$ = 100 m

Solution

Darcy's law factor: $G = 0.0853\,k^{*}A/\Delta x = 0.0853\cdot139.6\cdot4800/100 = 571.6$ m³·mPa·s/(d·MPa).
Kyte–Berry: $\Delta\bar p_o = 6.763 - 3.836 = 2.927$ MPa and $\hat k_{ro} = 3.0\cdot130.8/(571.6\cdot2.927) = 0.2345$ . Also $\Delta\bar p_w = 2.878$ MPa and $\hat k_{rw} = 0.5\cdot209.2/(571.6\cdot2.878) = 0.0636$ .
The Kyte–Berry pseudo capillary pressure: $6.763 - 6.837 = -0.074$ MPa, although there is no capillary pressure on the fine grid.
PVW: $\Delta\bar p = 6.749 - 3.882 = 2.867$ MPa, $\hat k_{ro} = 0.2394$ and $\hat k_{rw} = 0.0638$ .
The water fraction of the coarse block from the PVW pseudo relative permeabilities: $(0.0638/0.5)/(0.0638/0.5 + 0.2394/3.0) = 0.615$ . This is exactly the water fraction in the flow through the face, $209.2/340 = 0.615$ . From the Kyte–Berry pseudo relative permeabilities without their pseudo capillary pressure it is 0.619.
Core curves at $S_w$ = 0.369: $k_{rw}$ = 0.0056 and $k_{ro}$ = 0.372, and the water fraction is 0.083.

Answer: At

\hat S_w

= 0.369 the pseudo relative permeabilities of block 3 are

\hat k_{rw}

≈ 0.0638 and

\hat k_{ro}

≈ 0.2394. That is 11.4 times more for water and 1.6 times less for oil than from the core curves.

What it means. A coarse block with the saturation 0.369 should give its neighbour 62% of water, while the core curves at this saturation would give 8%. The water passed the block along the 900 mD layer when the tight layers were still almost untouched (see the figure at the beginning of the lesson). The Kyte–Berry and PVW pseudo relative permeabilities are close here, but the pseudo capillary pressures differ. In the Kyte–Berry method it is false: −0.074 MPa, with zero capillary pressure on the fine grid. In PVW it is zero. If we take the Kyte–Berry pseudo relative permeabilities without their pseudo capillary pressure, the water fraction of the block (0.619) is slightly different from the water fraction in the flow through the face. The data in this example are assumed for illustration.

Negative and infinite pseudo relative permeabilities

If the averaged potential difference of a phase between the block centres is close to zero, the formula gives huge values. If it has the opposite sign to the flow, the values are negative. This happens when a phase flows through a face by crossflows that the averaged pressures do not see, when gravity separates the phases, or in countercurrent flow (the phases moving in opposite directions). We check the tables. We discard the “strange” points or replace them by smoothing, and we make the curves monotonic (the water curve only rises, the oil curve only falls). In the example of the lesson, among 16,905 points of the five blocks there are no negative values. The reservoir is horizontal, there is no capillary pressure, and both phases pass the block faces only from the injection row to the production row.

There are other ways to average the pressure: for example with the weight of transmissibilities, which works better with strong segregation of gas and oil. A separate family is the total mobility methods. They do not average the phase pressures. They upscale the total mobility and the water fraction in the flow, and from them they get the pseudo relative permeabilities. All dynamic methods need a calculation on a fine grid for each region with its boundary conditions. So they are expensive and sensitive to how they are set up.

Fine grid, coarse grid with core curves, coarse grid with pseudo functions

Now we check what all this gives. We calculate the cross-section of the lesson on the fine grid up to two pore volumes injected, and we build the pseudo functions of each of the five blocks. Then we calculate the same element on a coarse grid of five blocks, with different curves. On a one-dimensional coarse grid at a constant rate, the saturation is decided only by the water fraction in the flow, and without capillary pressure this is decided by the ratio of the mobilities of the phases. So the coarse calculations use the pseudo relative permeabilities alone. For PVW this is exact: their pseudo capillary pressure is zero here. For Kyte–Berry the false pseudo capillary pressure is left out in this way, and the small difference of their result from the fine grid is the price of this simplification. As in the lesson Numerical dispersion, we take 1% of water in the outlet flow as the first water.

Worked exampleOne waterflood on the fine and the coarse grids

Given (assumed data)

Reference: fine grid of 8 × 45; the cross-section and the data are in the table at the beginning of the lesson
Coarse grid: 5 blocks of 100 m; step 0.0005 pore volumes: the Courant number (the fraction of a cell that the front crosses in one step) is small, and the smearing of the front is decided only by the grid itself
Curves of the coarse blocks: core curves; the vertical equilibrium pseudo functions of the lesson Displacement in heterogeneous reservoirs; the capillary equilibrium limit for the layers of the cross-section, with the same assumed imbibition curve as in the example with laminae; Kyte–Berry and PVW pseudo relative permeabilities

Solution

Fine grid: first water after 0.109 pore volumes; oil production 0.292 after 0.5 and 0.360 after 1 pore volume.
Core curves: 0.285, 0.379, 0.419. The first water is 2.6 times later, and the production after 1 pore volume is 16% higher.
Vertical equilibrium pseudo functions: 0.022, 0.213, 0.298. The exact solution with them gives the first water after 0.083 pore volumes. But on five blocks their steep initial part is smeared even more.
Capillary equilibrium limit: 0.346, 0.424, 0.457. This is even more optimistic than the core curves.
Kyte–Berry and PVW pseudo relative permeabilities, own for each block: 0.112 and 0.109. The production after 1 pore volume is 0.359 and 0.360, against 0.360.
PVW pseudo relative permeabilities of block 3 used in all five blocks: 0.095, 0.285, 0.353.

Answer: The dynamic pseudo functions calculated for each block reproduced the production of the fine grid to within thousandths. Its first water was reproduced exactly (PVW) or with a deviation of 0.003 pore volumes (Kyte–Berry without their pseudo capillary pressure). The core curves gave the first water 2.6 times later, and the vertical equilibrium pseudo functions gave it 5 times earlier.

What it means. The core curves do not see the 900 mD layer. The vertical equilibrium pseudo functions assume that crossflows at once equalise the pressure between the layers. But with $k_v/k_h = 0.01$ inside each layer, the $k_v$ of the unit is only 0.258 mD, and $R_L \approx 1.3$ (the effective length-to-thickness ratio, which shows how well the pressure equalises across the thickness). This is an intermediate case (communication between layers). With $M' > 1$ , full crossflows make the displacement worse and speed up breakthrough. The dynamic pseudo functions agreed with the fine grid because they come from the same calculation. So this is a check of consistency, not of forecasting. A single curve of block 3 for all the blocks is already wrong: a block next to the injection row is flooded differently from a block next to the production row. The data in this example are assumed for illustration.

First water and production on the fine and the coarse grids, pore volumes

Calculation: Fine grid 8 × 45, the reference
First water: 0.109
$N_{pD}$ at 0.5: 0.292
$N_{pD}$ at 1.0: 0.360

Calculation: Coarse, core curves
First water: 0.285
$N_{pD}$ at 0.5: 0.379
$N_{pD}$ at 1.0: 0.419

Calculation: Coarse, vertical equilibrium pseudo functions (VE)
First water: 0.022
$N_{pD}$ at 0.5: 0.213
$N_{pD}$ at 1.0: 0.298

Calculation: Coarse, capillary limit
First water: 0.346
$N_{pD}$ at 0.5: 0.424
$N_{pD}$ at 1.0: 0.457

Calculation: Coarse, Kyte–Berry, own for each block
First water: 0.112
$N_{pD}$ at 0.5: 0.292
$N_{pD}$ at 1.0: 0.359

Calculation: Coarse, PVW, own for each block
First water: 0.109
$N_{pD}$ at 0.5: 0.292
$N_{pD}$ at 1.0: 0.360

Calculation: Coarse, PVW of block 3 for all blocks
First water: 0.095
$N_{pD}$ at 0.5: 0.285
$N_{pD}$ at 1.0: 0.353

Calculation: Non-uniform 3 × 5, core curves
First water: 0.086
$N_{pD}$ at 0.5: 0.311
$N_{pD}$ at 1.0: 0.381

Calculation	First water	$N_{pD}$ at 0.5	$N_{pD}$ at 1.0
Fine grid 8 × 45, the reference	0.109	0.292	0.360
Coarse, core curves	0.285	0.379	0.419
Coarse, vertical equilibrium pseudo functions (VE)	0.022	0.213	0.298
Coarse, capillary limit	0.346	0.424	0.457
Coarse, Kyte–Berry, own for each block	0.112	0.292	0.359
Coarse, PVW, own for each block	0.109	0.292	0.360
Coarse, PVW of block 3 for all blocks	0.095	0.285	0.353
Non-uniform 3 × 5, core curves	0.086	0.311	0.381

First water means 1% of water in the outlet flow. $N_{pD}$ is the oil produced, in pore volumes of the cross-section, after 0.5 and 1.0 pore volumes have been injected. “Non-uniform 3 × 5” has three layers on the coarse grid: layers 1–3, the 900 mD layer on its own, and layers 5–8. The Kyte–Berry pseudo relative permeabilities are taken without their pseudo capillary pressure. The data are assumed for illustration.

CalculatedWater fraction in the outlet flow of the element: the fine grid, and a coarse grid of five blocks with core curves, with vertical equilibrium pseudo functions, at the capillary equilibrium limit, and with dynamic pseudo relative permeabilities (PVW). The curve with the dynamic pseudo relative permeabilities almost coincides with the fine grid.

CalculatedDynamic pseudo relative permeabilities of block 3 (PVW) and the core curves (dashed). While the average saturation of the block is low, the pseudo relative permeability of water is tens or hundreds of times larger than the core one: the water passes the block along the 900 mD layer. The curves end at about

\hat S_w

= 0.6: after that the block was not flooded within two pore volumes.

Worked exampleA homogeneous reservoir: pseudo functions against numerical dispersion

Given (assumed data)

Reservoir: the same element, but homogeneous: 139.6 mD, with the average porosity of the cross-section; fine grid 1 × 45, coarse grid 1 × 5
Exact solution: breakthrough after 0.396 pore volumes (Buckley–Leverett theory and the Welge method)

Solution

First water: the fine grid 0.382, the coarse grid with core curves 0.285, and the coarse grid with Kyte–Berry pseudo relative permeabilities 0.381 pore volumes.
After 0.20 pore volumes the average saturation of block 3 is 0.420, but the front is still inside the block. No water passes through its outlet face, and $\hat k_{rw}$ = 0, although by the core curves at this saturation $k_{rw}$ = 0.0134.
After 0.5 pore volumes: $\hat S_w$ = 0.617, and $\hat k_{rw}$ = 0.0983 against 0.1051 by the core curve.

Answer: The Kyte–Berry pseudo relative permeabilities gave the coarse grid the first water of the fine grid (0.381 against 0.382, instead of 0.285 with the core curves).

What it means. In a homogeneous reservoir there is no heterogeneity, and the pseudo curve of water is shifted to the right: the block does not give water until it has reached its outlet face. The pseudo functions reproduce the fine grid, not the exact solution. The fine grid itself smears the front, and its first water is 3.5% earlier than the exact breakthrough. So the cells of the reference grid must be much smaller than the width of the front. The data in this example are assumed for illustration.

First, the best single-phase upscaling

Dynamic pseudo functions are expensive. For each coarse cell and direction we need a displacement calculation on a fine grid. A model has tens of thousands of cells, and then the curves must be grouped. So in practice we more often improve the single-phase upscaling so that the main paths of the water stay visible to the coarse grid.

Non-uniform grid (non-uniform upscaling). The layers along which water escapes are kept as separate layers of the coarse grid, and the uniform units are combined. In the example of the lesson, three layers instead of one (the 900 mD layer on its own) reduced the overestimate of production after 1 pore volume from 16 to 6% with no pseudo functions at all (the table above). The principle is the same: make the cells smaller where the single-phase flow calculation shows the main flow.
Global boundary conditions. The transmissibilities between coarse cells are calculated not for each block separately but from a single-phase calculation of the whole detailed model with wells. The rate between regions is divided by the difference of their average pressures (transmissibility upscaling). In this way the coarse model “knows” the paths along which water flows between its wells.
A smaller coarsening factor where fronts move and where the wells are, and a check of convergence with the grid (how many cells are needed).

We keep pseudo functions for the cases where this is not enough: strong heterogeneity inside the cells that is comparable to their size, an unfavourable mobility ratio, and an important role of capillary or gravity forces. In the example of the lesson the non-uniform grid still gives the first water earlier than the reference. The 900 mD layer on cells of 100 m is smeared by numerical dispersion.

Where pseudo functions stop working

Pseudo functions are the “memory” of one fine-grid calculation. They are correct while the coarse model works in the same conditions.

Injection rate and the balance of forces: Where the water is in the cell depends on the ratio of capillary forces to viscous forces, and of gravity to viscous forces. In the example of the lesson, the capillary equilibrium pseudo functions (very slow injection) give on the coarse grid the first water after 0.346 pore volumes, while the fine grid at the field rate gives it after 0.109. If the rate differs in different parts of the reservoir, near wells and between them, one table is not enough.
Boundary conditions of the calculation: For the calculation of a single cell we must set what enters it, and with what water fraction. An error in this assumption passes into the pseudo functions. A border of neighbouring cells, and a calculation of the whole sector with wells, weaken the influence of the assumption. But they make the work more expensive.
Well positions and the path of the front: The pseudo functions of a block depend on how far it is from the injection well and how the front approached it. In the example the five blocks have five different curves. After two pore volumes the average saturation is from 0.68 in the first block to 0.58 in the last one. New wells, or the conversion of producers to injection, change the path of the front, and the tables must be recalculated.
Flow direction: Pseudo relative permeabilities are separate for $x$ , $y$ and $z$ , and even for flow in one direction and in the opposite one. In a cell where a permeable layer reaches only one face, water leaves through that face earlier. The example with laminae showed a difference of several times along and across the layers.
Grid and time step: Dynamic pseudo functions compensate for the numerical dispersion of the particular coarse grid they were calculated for. If the cells are made smaller, the compensation is no longer needed. If the step of the fully implicit scheme is made larger, smearing is added that was not in the calculation (choice of the step).

How pseudo functions are checked

1
Reject points
Reject negative and abnormally large values, non-monotonic parts, and points with an almost zero pressure difference. Extend the pseudo curves to the end points that the calculation did not reach.
2
Group
Combine similar curves by rock type and by region. Check what the grouping costs (in the example, one curve for all the blocks moved the first water from 0.109 to 0.095 pore volumes).
3
Compare the fine and the coarse model
Do this on a typical sector, with a different injection rate, different well positions and a different flow direction, and not only in the conditions in which the curves were found.
4
Do not confuse with history matching
Pseudo functions are found from the detailed model before the comparison with history. Adjusting the relative permeability during history matching (adjusting the model until it reproduces the past production) can hide in them the errors of the grid and of the geology (the grid first).

How the structure of heterogeneity, anisotropy and scale affect upscaling, and in which cases it does not work at all, is the subject of the lesson Special topics in upscaling.

Terms and symbols

Pseudo functions: The pseudo relative permeabilities and the pseudo capillary pressure of a coarse cell, as functions of its average saturation.
Rock curves (core curves): The relative permeability and capillary pressure curves that describe the rock at the scale of a sample.
Phase permeability $k_f = k\,k_{rf}$: The permeability of the rock for one phase at a given saturation. It is upscaled as a single-phase property.
Steady-state methods: Pseudo functions found from the assumption of steady flow in the cell: the capillary equilibrium limit, the viscous limit and gravity equilibrium.
Dynamic methods: Pseudo functions found from a displacement calculation on a fine grid: Kyte–Berry, pore-volume weighted, transmissibility-weighted and total mobility methods.
Kyte–Berry method: Pseudo relative permeabilities found from the rate through the outlet face of the block and the pressures of the central columns, averaged with the weight $k\,k_{rf}h$ .
PVW: Averaging of the pressure over the whole block with the weight of pore volume. It does not create a false pseudo capillary pressure.
$N_{cv}$: The ratio of the time to cross the cell to the time for capillary forces to even out the saturation across the layers.

Common mistakes and misconceptions

Wrong

“The permeability is transferred to the coarse grid correctly, so the coarse model will also calculate the waterflood correctly.”

In fact

Single-phase upscaling does not see where the water is in the cell. In the example of the lesson the coarse grid with core curves gave the first water 2.6 times later than the fine grid, and 16% more oil after one pore volume was injected.

Wrong

“Pseudo relative permeabilities are a property of the rock: calculate them once and use them everywhere.”

In fact

They depend on the injection rate, the boundary conditions, the wells, the flow direction and the grid. We carry them over only after a check in the new conditions.

Wrong

“The Kyte–Berry pseudo capillary pressure shows the capillary forces in the cell.”

In fact

It comes from the different weights of averaging of the phase pressures, and it can be non-zero when the capillary pressure on the fine grid is zero. In the example of the lesson it is −0.074 MPa. Pore-volume weighted averaging does not create it.

Wrong

“The pseudo curve of water is always shifted to the right of the core curve, because it compensates for numerical dispersion.”

In fact

In a uniform cell, yes. In a layered one the heterogeneity is usually stronger. In the example of the lesson, at an average saturation of 0.369 the pseudo relative permeability of water is 11.4 times larger than the core one.

Wrong

“Capillary equilibrium is a handy way to get pseudo functions for any cell.”

In fact

Only for thin layers and slow flow. For laminae a centimetre thick $N_{cv}$ is about a thousand. For the cross-section of the lesson, 16 m thick, at a field rate it is a few thousandths. Otherwise the pseudo functions come out too optimistic.

In depth: advanced detail

Why steady flow reduces to single-phase problems

The balance of phase $f$ in an incompressible system is $\phi\,\partial S_f/\partial t + \nabla\cdot\mathbf u_f = 0$ (Two-phase flow equations). In steady flow $\partial S_f/\partial t = 0$ , and what remains is $\nabla\cdot\mathbf u_f = 0$ . With Darcy's law without gravity, $\mathbf u_f = -(k\,k_{rf}/\mu_f)\nabla p_f$ , this is $\nabla\cdot\bigl(k\,k_{rf}(S_w)\nabla p_f\bigr) = 0$ . This is the equation of single-phase flow with the permeability $k\,k_{rf}$ . If the saturation field is known, for each phase we are left with the same problem as in the calculation of $k^{*}$ .

The limits give the saturation field. It is convenient to write the water flow through the total velocity $\mathbf u_t$ : $\mathbf u_w = f_w\mathbf u_t + k\lambda_of_w\nabla p_c$ , where $\lambda_o = k_{ro}/\mu_o$ (Fractional flow function and mobility ratio). In steady flow $\nabla\cdot\mathbf u_t = 0$ , and the water balance takes the form $\mathbf u_t\cdot\nabla f_w + \nabla\cdot(k\lambda_of_w\nabla p_c) = 0$ . As $\mathbf u_t \to 0$ (the capillary limit), what remains is $\nabla\cdot(k\lambda_of_w\nabla p_c) = 0$ . With zero capillary flows through the boundaries, its solution is a constant $p_c$ . When $p_c$ is negligible (the viscous limit), what remains is $\mathbf u_t\cdot\nabla f_w = 0$ . The water fraction is constant along the streamlines, and with a constant water fraction at the inlet, it is constant in the whole cell.

Vertical equilibrium with a capillary transition zone

Suppose the reservoir is thin compared with the capillary transition zone (the zone where water saturation changes gradually with height). Then there is no sharp boundary between water and oil, but vertical equilibrium can remain (the lesson on segregated flow). In every cross-section the phase pressures are then distributed hydrostatically over the height. The capillary pressure grows with the height $z$ above the base as $p_c(z) = p_c(0) + \Delta\rho\,g\,z$ (water is heavier than oil), and the saturation of each point is given by its capillary pressure curve. The pseudo functions are integrals over the thickness:

Pseudo functions of a cross-section at capillary–gravity equilibrium

\hat S_w = \frac{\int_0^H\phi\,S_w(z)\,\dd z}{\int_0^H\phi\,\dd z},\qquad \hat k_{rf} = \frac{\int_0^H k\,k_{rf}\bigl(S_w(z)\bigr)\dd z}{\int_0^H k\,\dd z},\qquad p_c\bigl(S_w(z)\bigr) = p_c(0) + \Delta\rho\,g\,z

where:

$z$ height above the base of the reservoir, m
$H$ thickness of the reservoir, m
$p_c(0)$ capillary pressure at the base: the parameter that sets the average saturation of the cross-section, Pa
$\Delta\rho\,g$ difference of the hydrostatic gradients of water and oil, Pa/m

Units: SI

Holds when

Vertical equilibrium: the phase pressures equalise over the thickness faster than the front advances along the reservoir.
The flow is along the reservoir: the pseudo relative permeabilities are the conductivities of the layers $k\,k_{rf}$ , added in parallel.
Going through $p_c(0)$ from the largest to the smallest, we get a table $(\hat S_w, \hat k_{rw}, \hat k_{ro})$ . The capillary pressure is that of imbibition or of drainage, according to the process.

This is the capillary equilibrium limit with the weight of the phases added. Without capillary forces the integrals give the straight lines of segregated flow, and without gravity they give the pseudo functions of the capillary limit.

Connections to other topics

Permeability upscaling for single-phase flow — equivalent permeability, boundary conditions and transmissibility upscaling are the basis of pseudo functions.
Displacement in heterogeneous reservoirs — pseudo functions of a layered cross-section at vertical equilibrium, and communication between layers.
Segregated flow and gravity displacement — pseudo relative permeabilities for a sharp boundary between phases and vertical equilibrium.
Numerical dispersion — smearing of the front on a coarse grid, which dynamic pseudo functions compensate for.
Relative permeability — the core relative permeability curves from which pseudo curves are built.
Capillary pressure and the transition zone — the Leverett J-function and the imbibition curves for the capillary equilibrium limit.
Petrophysical input for simulation: relative permeability and capillary pressure — how core curves are prepared for the model.
Special topics in upscaling — the structure of heterogeneity, anisotropy, and the cases when upscaling does not work.
History matching — why pseudo functions do not replace history matching, and the reverse.

Check yourself

1
A coarse cell consists of two layers of the same porosity: 2 m with a permeability of 300 mD, and 1 m with 30 mD. The relative permeability curves are the same. At capillary equilibrium, at some $P_c$ the water saturations of the layers are 0.30 and 0.55, and the oil relative permeabilities are 0.62 and 0.12. Find the average saturation and the pseudo relative permeability of oil along and across the layers.
Show the answer
The average saturation is $(2\cdot0.30 + 1\cdot0.55)/3 \approx 0.383$ . Along the layers, $k^{*} = (300\cdot2 + 30\cdot1)/3 = 210$ mD, and the phase permeability of oil is $(300\cdot2\cdot0.62 + 30\cdot1\cdot0.12)/3 = (372 + 3.6)/3 \approx 125.2$ mD. So $\hat k_{ro} \approx 125.2/210 \approx 0.596$ . Across the layers, $k^{*} = 3/(2/300 + 1/30) = 75$ mD, and the phase permeabilities of the layers are 186 and 3.6 mD. Then $3/(2/186 + 1/3.6) = 3/0.2885 \approx 10.4$ mD, and $\hat k_{ro} \approx 10.4/75 \approx 0.139$ . Across the layers the oil is 4.3 times less mobile: its path is blocked by the water-saturated tight layer.
2
In the homogeneous reservoir of the lesson, after 0.20 pore volumes have been injected, the average water saturation of block 3 is 0.420, but the front on the fine grid has not yet reached the outlet face of the block. What value of $\hat k_{rw}$ will the Kyte–Berry method give at this point?
1. A.Zero: water does not yet flow through the outlet face
2. B.Negative: the pressure difference of the water is directed against the flow
3. C.0.0134, like the core curve at the same saturation
4. D.More than the core value: the water passed the block along the best layer
Show the answer
Correct answer: A. Zero: water does not yet flow through the outlet face
The method takes the phase rate at the outlet face of the block, and there $q_w = 0$ for now. So $\hat k_{rw} = 0$ at any average saturation while the front is inside the block. With the core curve (0.0134) the coarse model would begin to give water to the neighbour too early. This is numerical dispersion, and the pseudo curve compensates for it. The answer “more than the core value” is correct for a layered cross-section, where water passes the block along the best layer, but not for a homogeneous reservoir.
3
There is no capillary pressure on the fine grid, but the Kyte–Berry method gave a pseudo capillary pressure of −0.1 MPa. What does this mean, and what should we do?
Show the answer
These are not capillary forces. They come from the different averaging weights: the oil pressure is averaged with the weight $k\,k_{ro}h$ and the water pressure with $k\,k_{rw}h$ , and with different pressures in the layers the averages differ. We cannot use the Kyte–Berry pseudo relative permeabilities without this pressure, because the water fraction of the coarse cell would be distorted. It is better to move to pore-volume weighted averaging, which with zero capillary pressure gives zero pseudo capillary pressure.
4
A unit of layers 4 m thick, $k_v$ = 2 mD, a block of 200 m, $\Delta p_c$ = 0.05 MPa, oil 2.0 mPa·s, $k_{ro}'$ = 0.8, Darcy velocity 0.05 m/d. Does the capillary equilibrium limit apply?
1. A. $N_{cv}$ is about one: a calculation is needed
2. B.No, $N_{cv}$ is much smaller than one
3. C.Yes, $N_{cv}$ is much larger than one
Show the answer
Correct answer: A. $N_{cv}$ is about one: a calculation is needed
$N_{cv} = 0.0853\cdot2\cdot0.8\cdot0.05\cdot200/(2.0\cdot0.05\cdot4^2) = 1.365/1.6 \approx 0.85$ . The capillary crossflows and the viscous flow are comparable. Neither the capillary equilibrium limit nor neglecting capillary forces is accurate, and the pseudo functions are found by a fine-grid calculation with capillary pressure.
5
The pseudo relative permeabilities were calculated on a fine grid for the scheme “an injection well in the corner of the element”. Then it was decided to add a well in the centre. Can we keep the tables?
Show the answer
Not without a check. The pseudo functions of each block describe how the front approaches it and how it passes through the block. A new well changes the directions and speeds of the flow, the length of the path of the front, and even the direction of flow through some blocks. We recalculate the tables, or at least compare the fine and the coarse model for the new scheme.
6
Why, in the example of the lesson, did the vertical equilibrium pseudo functions give the first water much earlier than the fine grid?
Show the answer
For two reasons. They assume that crossflows at once equalise the pressure between the layers ( $R_L \gg 1$ ). But with $M' > 1$ , crossflows make the displacement worse and speed up breakthrough along the fast layer. In the cross-section of the lesson, with $k_v/k_h = 0.01$ inside the layers, $R_L \approx 1.3$ : the crossflows are limited, and the fine model is closer to the intermediate case. And they do not compensate for numerical dispersion. Their steep initial part is smeared on the coarse blocks, and water leaves the element even earlier than by their own exact solution.

Updated: 16 October 2026