Well model in a grid block

Quick refresher

The pressure of the cell with a well refers to a volume of hundreds of thousands of cubic metres, and it is much higher than the bottomhole pressure. In the lesson's example, at a rate of 95 m³/d, the difference is 3.4 MPa.
The simulator uses a finite-difference scheme, which replaces derivatives by differences between neighbouring cells. This scheme “sees” the well as a point sink at the centre of the cell (a sink is a point where fluid leaves the formation). The cell pressure equals the pressure of steady-state radial flow (radial flow whose pressures no longer change with time) at the equivalent radius $r_0$ : about $0.2\,\Delta x$ for a square cell (exactly $0.1985\,\Delta x$ ).
The well index is Darcy's law for radial flow from the wall of the well to $r_0$ , with the skin taken into account: $\mathrm{WI} = 2\pi kh/(\ln(r_0/r_w) + S)$ .
The cell pressure is not the average pressure over its area (that average belongs to a radius of about $0.35\,\Delta x$ ), and it is not the average pressure of the drainage area of the well (the part of the formation that the well drains). So the well index cannot simply be taken from the productivity index measured in well tests. The productivity index is the rate per unit of pressure drop.
In an anisotropic formation (one whose permeability differs by direction) the index contains $\sqrt{k_xk_y}$ , and $r_0$ depends on both the cell size and the ratio of the permeabilities.
We set for a well either the rate or the bottomhole pressure, and always a limit on the other one. The rate is divided between the layers according to their indexes and cell pressures.

Key relationships

q_s = \frac{0.0853\,\mathrm{WI}\,(p_0 - p_{wf})}{\mu B},\qquad \mathrm{WI} = \frac{2\pi kh}{\ln(r_0/r_w) + S}

well model: $q_s$ in m³/d at standard conditions, WI in mD·m, $p$ in MPa, $\mu$ in mPa·s

r_0 = 0.14\sqrt{\Delta x^2 + \Delta y^2}\ \approx\ 0.2\,\Delta x

equivalent radius of an isotropic cell (permeability the same in all directions); the part after ≈ is for a square cell

r_0 = 0.28\,\frac{\left[(k_y/k_x)^{1/2}\Delta x^2 + (k_x/k_y)^{1/2}\Delta y^2\right]^{1/2}}{(k_y/k_x)^{1/4} + (k_x/k_y)^{1/4}}

anisotropic formation; the index contains $\sqrt{k_xk_y}$

p_{wf} = \frac{\sum_l J_l\,p_l - q}{\sum_l J_l}

common bottomhole pressure of a well that penetrates several layers, for a given rate

Common pitfalls

Comparing the pressure of the cell with a well with the measured bottomhole pressure.
Using the productivity index from well tests as the well index in the model. They refer to different pressures.
Calculating the index with the arithmetic mean of the permeabilities, or with the permeability along one axis, in an anisotropic formation.
Using Peaceman's formula without corrections for wells in neighbouring cells, near the edge of the model, or in very small cells.
Tuning the well index during history matching without first checking the permeability, the thickness, the skin and the pressure in the formation.

After this lesson you can

explain why the pressure of the cell with a well equals neither the bottomhole pressure nor the average pressure over the area of the cell
derive the equivalent radius of a square cell with a simple argument, and know the exact value and the formulas for a rectangular cell and for an anisotropic formation (one where permeability differs by direction)
calculate the well index with skin, and from it the bottomhole pressure or the rate in the model
check the well model against the solution for steady-state radial flow
distribute the rate of a well between layers, and choose the operating mode of the well in the model: a set rate or a set bottomhole pressure, with a limit on the other quantity

Why the cell pressure is not the bottomhole pressure

A cell of a simulation model is usually tens or hundreds of metres across (seen from above), while the radius of a well is about 0.1 m. Inside the cell the pressure is taken as uniform, so the simulator knows only one number, the cell pressure. Near a well, however, the pressure changes with the logarithm of the distance (Steady-state radial flow): every time we come ten times closer to the wellbore, the pressure drops by the same amount. So most of the drawdown (the pressure drop between the formation and the well) is spent inside the cell with the well, where the model cannot see any detail.

The grid cannot describe this drawdown cone (the funnel of low pressure around the well). It would need cells a few centimetres in size. Instead, we describe the well with a separate formula, the well model. It links three quantities: the rate, the cell pressure and the bottomhole pressure. If the rate is given, the well model gives the bottomhole pressure. If the bottomhole pressure is given, it gives the rate.

The well index

Suppose that the cell pressure $p_0$ equals the pressure of steady-state radial flow at some distance $r_0$ from the axis of the well. Then between the wall of the well and this distance the Dupuit equation with skin (the formula for steady-state radial flow) applies (Skin factor), and the link between the rate and the pressures follows at once:

Well model: rate, cell pressure and bottomhole pressure

q_s = \frac{0.0853\,\mathrm{WI}\,(p_0 - p_{wf})}{\mu B},\qquad \mathrm{WI} = \frac{2\pi kh}{\ln(r_0/r_w) + S}

where:

$q_s$ rate at standard conditions; negative for injection, m³/d
$\mathrm{WI}$ well index, the geometric part without the viscosity, mD·m
$p_0$ pressure of the cell with the well, calculated by the simulator, MPa
$p_{wf}$ bottomhole pressure (at the reference depth of the well), MPa
$k,\ h$ permeability of the cell and the open thickness of the formation in it, mD, m
$r_0$ equivalent radius of the cell: the distance at which the pressure of radial flow equals the cell pressure, m
$r_w,\ S$ radius of the well and skin factor, m, —
$\mu,\ B$ viscosity and formation volume factor of the fluid, mPa·s, m³/m³

Units: metric oilfield units; natural logarithm

Holds when

A vertical well at the centre of the cell opens its whole thickness. Single-phase inflow.
Near the well the flow is close to steady-state radial flow (at distances of the order of the cell size).
$r_0$ is calculated with a formula that suits the grid (see below). Neighbouring wells and the boundaries of the model are several cells away.

The well index is Darcy's law for the ring between the wall of the well and $r_0$ . The product $0.0853\,\mathrm{WI}/(\mu B)$ is the productivity index (the rate per unit pressure drop) referred to the cell pressure. We denote it by $J_0$ . Skin enters the index in the same way as in the Dupuit equation: damage of the near-wellbore zone reduces the index, and hydraulic fracturing or an acid treatment increases it.

In other unit systems (2)

q = \frac{\mathrm{WI}}{\mu}\,(p_0 - p_{wf}),\qquad \mathrm{WI} = \frac{2\pi kh}{\ln(r_0/r_w) + S}

$q$ in m³/s at reservoir conditions, $k$ in m², $h$ in m, $\mu$ in Pa·s, $p$ in Pa; WI in m³.

There is no constant.

Oilfield units

q_s = \frac{1.127\cdot10^{-3}\,\mathrm{WI}\,(p_0 - p_{wf})}{\mu B}

WI in mD·ft ( $k$ in mD, $h$ in ft), $p$ in psi, $\mu$ in cP, $q_s$ in STB/d.

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

The whole difficulty is in the equivalent radius. Intuition suggests several answers: half of the cell, the radius of a circle with the same area ( $0.56\,\Delta x$ ), or the radius at which the pressure equals the average over the area of the cell. All of them are wrong. The answer depends not on the physics but on how the finite-difference scheme calculates the cell pressure. D. Peaceman first found the answer in 1978, by comparing numerical solutions with analytical ones (exact formulas).

The equivalent radius of a cell

A simple argument gives the answer almost exactly. Take a square cell with a well at the centre and its four neighbours in a uniform formation, with steady-state flow. In the five-point scheme (each cell is linked to its four neighbours) from the lesson on the pressure equation, fluid comes to the cell with the well only from its four neighbours. For square cells the transmissibility of a connection, $k\,(\Delta y\,h)/\Delta x$ , is simply $kh$ and does not depend on the size of the cells.

SchematicA cell with a well and its four neighbours (a sketch, not to scale). The whole rate arrives through the four faces. The centres of the neighbours lie on a circle of radius

\Delta x

, and the pressure of the cell with the well equals the pressure of radial flow on a circle of radius

r_0 \approx 0.2\,\Delta x

A simple derivation of the equivalent radius

1
The finite-difference equation of the cell with the well
By symmetry, the pressures of the four neighbours are equal, and we call them $p_1$ . The whole rate $q$ (at reservoir conditions) arrives through the four faces: $4\,(kh/\mu)\,(p_1 - p_0) = q$ . From this, $p_1 - p_0 = q\mu/(4kh)$ .
2
The neighbours on the radial profile
Suppose that the pressures of the neighbours equal the pressure of radial flow at a distance $\Delta x$ from the well, and the pressure of the cell equals it at a distance $r_0$ . Then the pressure difference between them is $q\mu/(2\pi kh)$ multiplied by $\ln(\Delta x/r_0)$ .
3
Set the two equal
$\dfrac{q\mu}{4kh} = \dfrac{q\mu}{2\pi kh}\ln\dfrac{\Delta x}{r_0}$ . From this, $\ln(\Delta x/r_0) = \pi/2$ and $r_0 = \e^{-\pi/2}\,\Delta x \approx 0.208\,\Delta x$ .
4
The exact value
In reality the pressures of the neighbours differ a little from the radial profile. The exact solution of the finite-difference equations on an infinite grid gives $r_0 = 0.1985\,\Delta x$ (see the advanced section). In practice we take $0.2\,\Delta x$ . A difference of 5% in the radius changes the well index by less than 1%, because the radius sits under a logarithm.

Equivalent radius of a cell in an isotropic formation (Peaceman's formula)

r_0 = 0.14\,\sqrt{\Delta x^2 + \Delta y^2}\qquad (\Delta x = \Delta y:\ \ r_0 = 0.198\,\Delta x)

where:

$r_0$ equivalent radius of the cell, m
$\Delta x,\ \Delta y$ sizes of the cell seen from above, m

Units: any units of length (the same for all)

Holds when

Isotropic formation (permeability the same in all directions). Vertical well at the centre of the cell. Five-point scheme on a Cartesian grid.
Other wells and the boundaries of the model are several cells away. Neighbouring cells have the same size.
The ratio of the sides of the cell hardly affects the accuracy: on a uniform grid the formula is also right for strongly elongated cells (it differs from the exact solution by about 0.3%).

The equivalent radius is proportional to the size of the cell. The larger the cell, the farther from the well its pressure “sits”, and the larger the difference between the cell pressure and the bottomhole pressure. The constant 0.14 is a rounding of $0.1985/\sqrt2$ .

Worked exampleThe well index and the bottomhole pressure in the model

Given (assumed data)

Well (lesson Steady-state radial flow): $k$ = 60 mD; $h$ = 18 m; $\mu$ = 1.8 mPa·s; $B$ = 1.25; $r_w$ = 0.108 m; $S$ = +4
Cell: 100 × 100 m, the well at the centre
Operating mode: a fixed rate of 95 m³/d at standard conditions; the cell pressure calculated by the simulator is $p_0$ = 20.90 MPa

Solution

Equivalent radius: $r_0 = 0.14\sqrt{100^2 + 100^2} = 19.80$ m; $\ln(r_0/r_w) = \ln(19.80/0.108) = 5.211$ .
Well index: $\mathrm{WI} = \dfrac{2\pi\cdot60\cdot18}{5.211 + 4} = 736.7$ mD·m.
Productivity referred to the cell pressure: $J_0 = \dfrac{0.0853\cdot736.7}{1.8\cdot1.25} = 27.93$ m³/(d·MPa).
Difference between the cell pressure and the bottomhole pressure: $p_0 - p_{wf} = 95/27.93 = 3.40$ MPa. Bottomhole pressure: $p_{wf} = 20.90 - 3.40 = 17.50$ MPa.
Share of the drawdown that is spent inside the circle of radius 50 m inscribed in the cell, measured against a circle of radius 400 m: $\dfrac{\ln(50/0.108) + 4}{\ln(400/0.108) + 4} = 0.830$ .
If we wrongly take the radius of a circle of the same area, $100/\sqrt\pi = 56.4$ m, then $\mathrm{WI} = 661.5$ mD·m and $J_0$ is lower by 10%. At the same cell pressure the bottomhole pressure would come out as 17.11 MPa.

Answer: The well index is 736.7 mD·m. At a rate of 95 m³/d the bottomhole pressure is 3.40 MPa below the cell pressure, at 17.50 MPa.

What it means. Inside the cell with the well, 83% of the drawdown is spent. So the cell pressure cannot be compared with the measured bottomhole pressure: a difference of a few megapascals is normal and is not an error of the model. An error of almost three times in the equivalent radius changes the index by only 10%, because the radius sits under a logarithm. Even so, the shift of the bottomhole pressure by 0.39 MPa is noticeable in history matching. The data in this example are assumed for illustration.

Different “average” pressures of a square cell, and the radius of radial flow that each one matches

Pressure: Pressure of the cell with a well in the finite-difference scheme
Radius, fraction of $\Delta x$: 0.1985
Where it occurs: Peaceman's well model (0.198 in the formula)

Pressure: Average pressure of a closed square with a well at the centre (pseudo-steady-state flow)
Radius, fraction of $\Delta x$: 0.270
Where it occurs: a well in every cell; a square drainage area

Pressure: Steady-state flow pressure averaged over the area of the cell
Radius, fraction of $\Delta x$: 0.346
Where it occurs: often wrongly taken as the cell pressure

Pressure: Pressure on a circle of the same area
Radius, fraction of $\Delta x$: 0.564
Where it occurs: a rough replacement of the square by a circle

Pressure	Radius, fraction of $\Delta x$	Where it occurs
Pressure of the cell with a well in the finite-difference scheme	0.1985	Peaceman's well model (0.198 in the formula)
Average pressure of a closed square with a well at the centre (pseudo-steady-state flow)	0.270	a well in every cell; a square drainage area
Steady-state flow pressure averaged over the area of the cell	0.346	often wrongly taken as the cell pressure
Pressure on a circle of the same area	0.564	a rough replacement of the square by a circle

The calculation is for a uniform isotropic formation and a well at the centre of a square with side $\Delta x$ . The value 0.270 follows from the shape factor of the square (a number that depends on the shape of the drainage area), 30.88 (Pseudo-steady-state flow and the shape factor), and 0.346 from averaging $\ln r$ over the area of the square.

A check against steady-state radial flow

We test a well model the same way as any numerical scheme: on a problem with a known solution. We place the well from the lesson on steady-state flow at the centre of a grid of 9 × 9 cells of 100 m each. In the outermost cells we hold the pressure of steady-state radial flow at the distance of their centres. Then the exact solution is known in the whole square. The simulator calculates the pressures of the inner cells, and the well model gives the bottomhole pressure. We can compare it with the exact one.

Worked exampleThe model on a grid against the Dupuit equation

Given (assumed data)

Well and formation: from the previous example; pressure 22.0 MPa at a distance of 400 m, bottomhole pressure 17.5 MPa
Grid: 9 × 9 square cells of 100 m; in the outermost cells, the pressure of steady-state flow
Result of the grid calculation: pressure of the cell with the well 20.894 MPa; of the neighbouring cell across a face, 21.473 MPa

Solution

Rate from the Dupuit equation with skin (lesson Steady-state radial flow): $q_s = \dfrac{0.536\cdot60\cdot18\cdot4.5}{1.8\cdot1.25\,(\ln(400/0.108) + 4)} = 94.8$ m³/d.
The “unit” of pressure is $q_sB\mu/(0.536\,kh) = 0.3683$ MPa. The exact pressure at the distance $r_0$ = 19.80 m is $17.5 + 0.3683\cdot(5.2113 + 4) = 20.893$ MPa. The grid calculation gave 20.894 MPa.
The exact pressure at the centre of the neighbouring cell, 100 m from the well, is $17.5 + 0.3683\cdot(\ln(100/0.108) + 4) = 21.489$ MPa. On the grid it is 21.473 MPa.
Bottomhole pressure from the well model: $p_{wf} = 20.894 - 94.8/27.93 = 17.50$ MPa, against the exact 17.50 MPa.
Without a well model (taking the bottomhole pressure equal to the cell pressure) the error would be 3.4 MPa. With the radius of a circle of the same area it would be 0.38 MPa.

Answer: With Peaceman's equivalent radius the model reproduces the bottomhole pressure: 17.50 MPa against the exact 17.50 MPa, a difference of less than 0.01 MPa.

What it means. A grid of 100 m cells gives the pressure far from the well correctly, and the well model gives the pressure drop in the cell with the well. The cells next to the well deviate a little from the exact profile (by 0.016 MPa). The five-point scheme reproduces the logarithmic profile only approximately, and this profile bends steeply within one cell of the well. It is useful to repeat this check when the grid or the simulator changes. The data in this example are assumed for illustration.

CalculatedPressure along a row of cells through the well, for the check above. The solid curve is the exact steady-state radial flow (at the wall of the well the drawdown cone ends with a vertical jump down to the bottomhole 17.5 MPa: this is the pressure drop caused by the skin of +4). The horizontal segments are the cell pressures on the grid of 100 m cells. The pressure of the cell with the well meets the curve at a distance of

r_0 \approx 20

An anisotropic formation and rectangular cells

If the permeability in the horizontal plane depends on the direction, the drawdown cone around the well is not a circle but an ellipse, stretched along the direction of higher permeability (anisotropy). We stretch the coordinates so that the formation becomes isotropic with a permeability of $\sqrt{k_xk_y}$ , and this reduces the problem to the isotropic one. In the stretched coordinates the cell becomes a rectangle with other proportions, and the round wellbore becomes an ellipse. The result of the conversion is Peaceman's formula for an anisotropic formation.

Equivalent radius and well index in an anisotropic formation

r_0 = 0.28\,\frac{\left[(k_y/k_x)^{1/2}\,\Delta x^2 + (k_x/k_y)^{1/2}\,\Delta y^2\right]^{1/2}}{(k_y/k_x)^{1/4} + (k_x/k_y)^{1/4}},\qquad \mathrm{WI} = \frac{2\pi\sqrt{k_xk_y}\,h}{\ln(r_0/r_w) + S}

where:

$k_x,\ k_y$ principal permeabilities in the horizontal plane (the cell sides run along them), mD
$\Delta x,\ \Delta y$ cell dimensions along the $x$ and $y$ axes, m
$r_0,\ r_w,\ h,\ S$ equivalent radius, well radius, thickness, skin, m, m, m, —

Units: any consistent units (the permeabilities enter as ratios and through $\sqrt{k_xk_y}$ )

Holds when

A vertical well at the centre of the cell. The grid axes match the principal directions of permeability.
As in the isotropic case: a uniform grid, with other wells and model boundaries several cells away. The ratio of the sides of the cell hardly affects the accuracy.

For $k_x = k_y$ the formula turns into $0.14\sqrt{\Delta x^2 + \Delta y^2}$ . In an anisotropic formation the main point is not a refined radius but the permeability in the index: the geometric mean $\sqrt{k_xk_y}$ , because it is this value that decides the inflow to a vertical well. It is convenient to choose cells with the side ratio $\Delta x/\Delta y = \sqrt{k_x/k_y}$ , since in the “stretched” coordinates they are square.

Worked exampleA well in an elongated cell of an anisotropic formation

Given (assumed data)

Formation: $k_x$ = 120 mD, $k_y$ = 30 mD; $h$ = 18 m; $r_w$ = 0.108 m; $S$ = +4
Cell: $\Delta x$ = 200 m, $\Delta y$ = 50 m

Solution

Permeability ratios: $(k_y/k_x)^{1/2} = 0.5$ , $(k_x/k_y)^{1/2} = 2$ ; $(k_y/k_x)^{1/4} = 0.7071$ , $(k_x/k_y)^{1/4} = 1.4142$ .
$r_0 = 0.28\cdot\dfrac{\sqrt{0.5\cdot200^2 + 2\cdot50^2}}{0.7071 + 1.4142} = 0.28\cdot\dfrac{158.1}{2.1213} = 20.87$ m.
Well index with $\sqrt{120\cdot30} = 60$ mD: $\mathrm{WI} = \dfrac{2\pi\cdot60\cdot18}{\ln(20.87/0.108) + 4} = 732.5$ mD·m.
Mistake 1: the anisotropy is ignored. Then $r_0 = 0.14\sqrt{200^2 + 50^2} = 28.86$ m and the arithmetic mean permeability is 75 mD, so $\mathrm{WI} = 884.7$ mD·m, which is 21% larger.
Mistake 2: the formula for a square is applied to the long side, $0.2\cdot200 = 40$ m, with the permeability $k_x$ = 120 mD. Then $\mathrm{WI} = 1369$ mD·m, which is 87% larger.

Answer: The correct well index is 732.5 mD·m. Without the anisotropy it is overestimated by 21–87%.

What it means. The equivalent radius in the elongated cell of the anisotropic formation turned out almost the same as in a square cell of 100 × 100 m of an isotropic formation: 20.87 m against 19.80 m. Almost all of the error of the wrong variants is in the permeability. (An overestimated radius even lowers the index a little.) A model with the permeability $k_x$ in the index would make the rate of the well almost twice too high. The data in this example are assumed for illustration.

A well in several layers, and several phases

In a three-dimensional model, a vertical well passes through a column of cells, one for each layer of the grid. For each cell that the well opens, we calculate its own index, with its own permeability, thickness and skin. The wellbore is common to all layers, so the bottomhole pressures of the layers are linked. In the simplest case we take them as equal. More exactly, they differ by the hydrostatic weight of the column of fluid in the wellbore between the layers.

Distribution of the rate between layers

q_l = J_l\,(p_l - p_{wf}),\qquad p_{wf} = \frac{\sum_l J_l\,p_l - q}{\sum_l J_l},\qquad J_l = \frac{0.0853\,\mathrm{WI}_l}{\mu B}

where:

$q_l,\ q$ rate from layer $l$ and the given rate of the well at standard conditions, m³/d
$J_l$ productivity of the layer referred to the pressure of its cell, m³/(d·MPa)
$p_l$ pressure of the cell of layer $l$ that the well passes through, MPa
$p_{wf}$ common bottomhole pressure (referred to one depth), MPa
$\mathrm{WI}_l$ well index in layer $l$ , mD·m

Units: metric oilfield units

Holds when

Single-phase inflow. The pressure in the wellbore is the same for all layers (a thin interval; otherwise the weight of the fluid column in the wellbore is added to the $p_{wf}$ of each layer).
If the cell pressure of a layer is lower than the bottomhole pressure, $q_l$ is negative: fluid leaves the wellbore into this layer (wellbore crossflow).

For a given rate of the well, the bottomhole pressure is the average of the cell pressures weighted by $J_l$ , reduced by $q/\sum J_l$ . A layer with a high flow capacity (a large $kh$ ) gives more, even if its pressure is lower, and so it is depleted faster.

Worked exampleA well in three layers: distribution of the rate and the bottomhole pressure limit

Given (assumed data)

Layers (top to bottom): 1: $k$ = 150 mD, $h$ = 6 m, $p_1$ = 21.0 MPa; 2: 20 mD, 8 m, 21.6 MPa; 3: 300 mD, 4 m, 20.4 MPa
Common to the layers: cells of 100 × 100 m ( $r_0$ = 19.80 m); $r_w$ = 0.108 m; $S$ = +2; $\mu$ = 1.8 mPa·s; $B$ = 1.25
Operating mode: a fixed rate of 150 m³/d; the lowest allowed bottomhole pressure is 18.0 MPa

Solution

Denominator of the index: $\ln(19.80/0.108) + 2 = 7.2113$ . Indexes: $\mathrm{WI}_1 = 2\pi\cdot150\cdot6/7.2113 = 784.17$ , $\mathrm{WI}_2 = 139.41$ , $\mathrm{WI}_3 = 1045.56$ mD·m.
Productivities $J_l = 0.0853\,\mathrm{WI}_l/(1.8\cdot1.25)$ : 29.729, 5.285 and 39.638 m³/(d·MPa); the sum is 74.652.
The sum is $\sum J_l\,p_l = 29.729\cdot21.0 + 5.285\cdot21.6 + 39.638\cdot20.4 = 1547.1$ m³/d. The bottomhole pressure is $p_{wf} = (1547.1 - 150)/74.652 = 18.71$ MPa, which is above the allowed 18.0 MPa.
Rates of the layers $q_l = J_l\,(p_l - p_{wf})$ : 68, 15 and 67 m³/d.
If the pressure in the formation falls so far that 150 m³/d would need a bottomhole pressure below 18.0 MPa, the well switches to the pressure limit. With the present cell pressures and $p_{wf}$ = 18.0 MPa the well would give $\sum J_l\,(p_l - 18.0) = 203$ m³/d. This is more than the given rate, so the limit does not act yet.

Answer: The bottomhole pressure is 18.71 MPa. The layers give 68, 15 and 67 m³/d. At the allowed bottomhole pressure of 18.0 MPa the well could give up to 203 m³/d.

What it means. The thin layer 3 of high permeability gives 45% of the rate at the lowest pressure, while the thick tight layer 2 gives only 10%, although its pressure is the highest. In the model this shows in the cells: layer 3 is depleted faster, and with time its contribution falls. If the cell pressure of a low-pressure layer drops below the bottomhole pressure, the model will show wellbore crossflow (fluid flowing along the wellbore) from the other layers into this one. The data in this example are assumed for illustration.

When several phases flow, the well index stays the same, and the rate of each phase is proportional to its mobility (how easily it flows) in the cell with the well (Relative permeability): $q_o$ contains the factor $k_{ro}/(\mu_oB_o)$ and $q_w$ contains $k_{rw}/(\mu_wB_w)$ . So the water cut of a well in the model (the share of water in its production) is decided by the saturations of the cells that the well passes through. With large cells the water “arrives” at the well earlier or later than in reality (Grids in reservoir simulation models).

Phase rates from the cell with the well

q_{o} = \frac{0.0853\,\mathrm{WI}\,k_{ro}}{\mu_oB_o}\,(p_0 - p_{wf}),\qquad q_{w} = \frac{0.0853\,\mathrm{WI}\,k_{rw}}{\mu_wB_w}\,(p_0 - p_{wf})

where:

$q_o,\ q_w$ oil and water rates at standard conditions, m³/d
$k_{ro},\ k_{rw}$ relative permeabilities of oil and water at the saturation of the cell, —
$\mu_o,\ \mu_w,\ B_o,\ B_w$ viscosities and formation volume factors of oil and water, mPa·s, m³/m³
$\mathrm{WI},\ p_0,\ p_{wf}$ well index, cell pressure and bottomhole pressure, mD·m, MPa, MPa

Units: metric oilfield units

Holds when

Capillary pressure in the cell is ignored (the phase pressures are equal). The $k$ in the index is the absolute permeability.

The index describes the geometry of the inflow and is the same for all phases. The mobility says which phase flows, and in what share.

Operating modes of wells in the model

The well model gives one equation for two unknowns, the rate and the bottomhole pressure. So we set one of them and calculate the other. We cannot set both: for a given bottomhole pressure the formation decides the rate. This follows field practice. A well is operated either with a given rate (for example, by a pump with adjustable output) or with a given pressure at the bottom of the well or at the wellhead.

Given rate: The rate of oil, liquid, gas or injection is fixed. The model calculates the bottomhole pressure. If this pressure goes beyond the limit (below the allowed value for a producing well, above the allowed value for an injection well), the well switches to the mode of a given pressure.
Given bottomhole pressure: The model calculates the rate. The limit is the largest allowed rate. Producing wells are often run this way in a forecast.
Given wellhead pressure: The bottomhole pressure is calculated from the wellhead pressure and the pressure losses in the tubing, using tables of multiphase flow in pipes (Multiphase flow in tubing).
Group limits: A limit can be put on the total rate of a group of wells, of a formation or of a field, on the capacity of the processing system, or on the injection in a group. The simulator divides the total rate between the wells according to what each well can give.

For water injection we often set voidage replacement: we inject as much water as the volume of fluid taken out of the formation, measured at reservoir conditions. The volume in the formation equals the volume at standard conditions multiplied by the formation volume factor, so the injection needed, in standard cubic metres of water, is:

Water injection for full voidage replacement

q_{w,\text{inj}} = \frac{B_o\,q_o + B_w\,q_w}{B_w}

where:

$q_{w,\text{inj}}$ water injection at standard conditions, m³/d
$q_o,\ q_w$ oil and water production at standard conditions, m³/d
$B_o,\ B_w$ formation volume factors of oil and water, m³/m³

Units: any identical units of rate

Holds when

There is no free gas in the formation (the pressure is above the bubble point pressure, the pressure at which gas starts to come out of the oil). Otherwise the reservoir volume of the free gas is added to the numerator.

Since $B_o > B_w$ , when only oil is produced we inject more water than the oil produced, counted in standard cubic metres: oil occupies a larger volume in the formation than at the surface.

Terms and symbols

Well model: A formula that links the rate of a well, the pressure of the cell it passes through, and the bottomhole pressure.
Well index WI: The geometric part of the productivity index, referred to the cell pressure: $2\pi kh/(\ln(r_0/r_w) + S)$ , in mD·m.
Equivalent radius of a cell $r_0$: The distance from the well at which the pressure of steady-state radial flow equals the cell pressure; about $0.2\,\Delta x$ for a square cell.
Pressure of the cell with a well $p_0$: The pressure that the simulator calculates for the cell with a well. It is higher than the bottomhole pressure by $q_s/J_0$ , and it is not equal to the average pressure over the area of the cell.
Productivity referred to the cell pressure $J_0$: $0.0853\,\mathrm{WI}/(\mu B)$ , m³/(d·MPa): the rate per 1 MPa of difference between the cell pressure and the bottomhole pressure.
Well operating mode: What is set for a well in the model, the rate or the pressure (bottomhole or wellhead), and which limits act on the other quantity.

Common mistakes and misconceptions

Wrong

“The pressure in the cell with a well is the bottomhole pressure.”

In fact

The cell pressure is higher by $q_s/J_0$ : in the lesson's example, at 95 m³/d, by 3.4 MPa. The model calculates the bottomhole pressure separately, from the well index.

Wrong

“The cell pressure is the average pressure over its area.”

In fact

The finite-difference scheme does not average the true pressure. The pressure of the cell with a well belongs to a radius of $0.198\,\Delta x$ , while the average over the area belongs to $0.346\,\Delta x$ . The cell pressure is lower than the average over the cell.

Wrong

“The well index can be taken from the productivity index measured in well tests.”

In fact

The productivity index from well tests refers to the average pressure of the drainage area (its denominator has $\ln(r_e/r_w) - 3/4 + S$ ), and the well index refers to the cell pressure ( $\ln(r_0/r_w) + S$ ). For the lesson's well, in a cell of 100 m and a drainage area of radius 400 m, they differ by about a quarter. From well tests we carry the permeability and the skin into the model, and we calculate the index.

Wrong

“In an anisotropic formation the higher permeability is put into the index, because the flow to the well goes mainly along it.”

In fact

The inflow to a vertical well is decided by the geometric mean $\sqrt{k_xk_y}$ . With $k_x$ the index in the lesson's example was too high by 87%.

Wrong

“The smaller the cell with the well, the closer its pressure to the bottomhole pressure, so with a fine enough grid no well model is needed.”

In fact

The difference $p_0 - p_{wf}$ falls only as the logarithm of the cell size: halving the cell reduces it by $q_s B\mu\ln2/(0.536\,kh)$ , not by half. The pressures could be made equal only by cells of the order of the well radius, which is impractical.

In depth: advanced detail

Where 0.1985 comes from

The simple derivation assumed that the pressures of the neighbours lie exactly on the radial profile. A more exact way is to solve the finite-difference equations themselves, on an infinite square grid with a sink in one cell. This solution is known. Far from the well it turns into a logarithm, but shifted. The pressure of the cell with the sink differs from the pressures of the distant cells just as if the radial profile passed through that cell at a distance of:

Equivalent radius of the five-point scheme (exact value)

r_0 = \frac{\Delta x}{2\sqrt2\,\gamma} \approx 0.1985\,\Delta x

where:

$\gamma$ $\e^{0.5772} = 1.781$ , the exponential of Euler's constant (the same as in the logarithmic approximation of the exponential integral and in the inflow formulas), —
$\Delta x$ side of the square cell, m

Units: any units of length

Holds when

Steady-state flow. An infinite uniform square grid. Five-point scheme. Isotropic, uniform formation.

The simple derivation gave $\e^{-\pi/2} = 0.208$ . In reality the neighbours lie a little below the radial profile (in our check, by 0.016 MPa, with the “unit” of 0.37 MPa), and the cell with the well “looks” a little closer to the wellbore. A numerical calculation on a grid of 41 × 41 gives 0.1985, correct to four digits.

The relation also holds for transient flow (flow whose pressures still change with time), once the drawdown cone has spread over several cells around the well. In the first hours and days after start-up, while the disturbance has not yet left the cell with the well, the model does not describe the transient flow. So the early parts of well test records (Well testing: purposes and types of test) cannot be reproduced on a coarse grid. For that we need a fine radial grid around the well (Grids in reservoir simulation models).

When Peaceman's formula is inaccurate

Wells close to each other. The derivation assumes that the pressure around the cell with the well follows the radial profile. If every cell has a well with the same rate, there is no flow between the cells. The cell pressure then equals the average pressure of its closed square, and the correct equivalent radius is $0.270\,\Delta x$ , not $0.198\,\Delta x$ . For wells in neighbouring cells and for several wells in one cell, the index is recalculated.
A well near the edge of the model, or not at the centre of the cell. A no-flow boundary near a well acts like a reflected well, also called an image well (a mirror copy of the real well on the other side of the boundary; Superposition principle). Simulators allow the index to be set by hand for such cases.
Very small cells. When $r_0$ is comparable with the radius of the well (cells of a fraction of a metre), the logarithm is small or negative and the formula loses its meaning. Such cells near a well are not needed.
A horizontal well. For a wellbore along the $x$ axis, we put into the formula the cell dimensions across the wellbore and the permeabilities $k_y$ and $k_z$ , and we replace the thickness by the length of the wellbore in the cell, $\Delta x$ . Model cells are usually thin ( $\Delta z \ll \Delta y$ ), the wellbore is close to the top and the base of the layer, and the accuracy of the formula falls.
Partial penetration and a deviated wellbore. In a multi-layer model, partial penetration (the well opens only part of the formation thickness) is set by perforating only some of the layers. A grid with sufficiently thin layers reproduces by itself the flow that converges towards the open interval. So we put the damage skin into the indexes of the layers, not the total skin from a well test. The total skin already contains the pseudo-skin of the partial penetration (the extra resistance that comes from the well opening only part of the thickness), and the resistance would be counted twice. We add the pseudo-skin (Partial penetration and well deviation) only if the open interval is thinner than the cell and the grid does not resolve it. (A deviated wellbore is one that is inclined, not vertical.)
High velocities near the wellbore. For gas wells we add a skin that grows with the rate (Inflow at high rates: non-Darcy flow).

Index of a horizontal well along the x axis

r_0 = 0.28\,\frac{\left[(k_z/k_y)^{1/2}\,\Delta y^2 + (k_y/k_z)^{1/2}\,\Delta z^2\right]^{1/2}}{(k_z/k_y)^{1/4} + (k_y/k_z)^{1/4}},\qquad \mathrm{WI} = \frac{2\pi\sqrt{k_yk_z}\,\Delta x}{\ln(r_0/r_w) + S}

where:

$k_y,\ k_z$ permeabilities across the wellbore: horizontal and vertical, mD
$\Delta y,\ \Delta z$ dimensions of the cell across the wellbore, m
$\Delta x$ length of the wellbore in the cell, m

Units: as in the formula for a vertical well

Holds when

The wellbore passes through the centre of the cell along the $x$ axis. The top and base of the formation are several cells away. For a thin layer of one cell the formula is only an estimate.

The same Peaceman formula, turned by 90°: the plane in which the flow converges towards the wellbore is now $y$ – $z$ .

Connections to other topics

Steady-state radial flow — the Dupuit equation and the logarithmic pressure profile on which the well model is built.
Skin factor — skin in the well index and what it describes.
Pressure equation for single-phase flow — the cell equation and the five-point scheme whose pressure the well model interprets.
Grids in reservoir simulation models — cell size, local grid refinement and radial grids around wells.
Pseudo-steady-state flow and the shape factor — the average pressure of the drainage area and the shape factor.
History matching — adjusting the productivity of wells and its dangers.
Multiphase flow in tubing — converting the wellhead pressure into the bottomhole pressure.

Check yourself

1
A model cell is a square of 50 × 50 m with a well at the centre, in an isotropic formation. About what is the equivalent radius of the cell?
1. A.25 m, half of the cell
2. B.about 28 m, the radius of a circle of the same area
3. C.about 10 m
4. D.0.1 m, the radius of the well
Show the answer
Correct answer: C. about 10 m
By Peaceman's formula $r_0 = 0.14\sqrt{50^2 + 50^2} \approx 9.9$ m, which is about $0.2\,\Delta x$ . The radius of a circle of the same area ( $50/\sqrt\pi \approx 28$ m) and half of the cell both give an underestimated well index.
2
A well runs at a given rate of 60 m³/d. The pressure of its cell is 18.4 MPa, its productivity referred to the cell pressure is $J_0$ = 12 m³/(d·MPa), and the lowest allowed bottomhole pressure is 14 MPa. What is the bottomhole pressure, and what will the rate be? And what if the cell pressure falls to 17.4 MPa?
Show the answer
$p_{wf} = 18.4 - 60/12 = 13.4$ MPa, which is below the allowed 14 MPa. So the well already cannot give 60 m³/d: the simulator will switch it to the pressure mode, and the rate will be $12\cdot(18.4 - 14) = 52.8$ m³/d. If the cell pressure is 17.4 MPa, the rate will fall to $12\cdot3.4 = 40.8$ m³/d.
3
The permeability of a formation is 400 mD along the $x$ axis and 25 mD along $y$ . Which permeability is put into the index of a vertical well?
1. A.400 mD
2. B.25 mD, the smaller one
3. C.100 mD, the geometric mean
4. D.212.5 mD, the arithmetic mean
Show the answer
Correct answer: C. 100 mD, the geometric mean
The inflow to a vertical well in an anisotropic formation is equivalent to the inflow in an isotropic formation with the permeability $\sqrt{k_xk_y} = \sqrt{400\cdot25} = 100$ mD. The ratio of the permeabilities also affects the equivalent radius, but more weakly, through the logarithm.
4
A producing well gives 200 m³/d of oil ( $B_o$ = 1.3) and 50 m³/d of water ( $B_w$ = 1.02) at standard conditions. How much water must be injected for full voidage replacement?
Show the answer
The reservoir volume withdrawn is $200\cdot1.3 + 50\cdot1.02 = 260 + 51 = 311$ m³/d. In standard cubic metres of water it is $311/1.02 \approx 305$ m³/d. This is more than the total production of 250 m³/d, because oil occupies a larger volume in the formation than at the surface.
5
The cell with a well of 100 × 100 m was replaced by a cell of 50 × 50 m (the same rate). How does the difference between the cell pressure and the bottomhole pressure change?
1. A.It will halve
2. B.It will decrease by $q_sB\mu\ln2/(0.536\,kh)$ , for the well of the lesson by about 0.26 MPa out of 3.4
3. C.It will not change
4. D.It will become zero
Show the answer
Correct answer: B. It will decrease by $q_sB\mu\ln2/(0.536\,kh)$ , for the well of the lesson by about 0.26 MPa out of 3.4
The difference is $p_0 - p_{wf} = \dfrac{q_sB\mu}{0.536\,kh}\left(\ln\dfrac{r_0}{r_w} + S\right)$ , and $r_0$ is proportional to the cell size. A cell half as large reduces the logarithm by $\ln2 \approx 0.69$ . With a “unit” of pressure of about 0.37 MPa (the well of the lesson, 95 m³/d) this is about 0.26 MPa. The cell pressure is still much higher than the bottomhole pressure.
6
At the actual bottomhole pressure, a well in the model gives 30% less oil than in the field. What should we check before we multiply its index by a factor?
Show the answer
Check the permeability and the thickness of the cells that the well passes through. They may have been made too low during upscaling (the averaging of fine-scale properties to the size of the model cells). Check the perforated intervals and the layers that are open. Check the skin from well tests, because after a treatment it may have become negative. Check the pressures in the cells: they may be too low, for example because of a weak aquifer. Check the relative permeabilities and the viscosity. An index multiplier is appropriate only when the input data are consistent and the difference is really explained by the near-wellbore zone.

Updated: 7 October 2026