Pseudo-steady-state flow and the shape factor

Quick refresher

In a closed area without recharge, the only thing that replaces the withdrawal is the expansion of rocks and fluids. So the average pressure falls linearly with time. After the transient period the pressure at every point falls at the same rate.
Through a circle of radius $r$ passes not the whole flow rate, but only the fluid released outside the circle: the fraction $1 - r^2/r_e^2$ . So far from the well the pressure gradient is smaller than in steady-state inflow.
For a well at the centre of a circle, the drawdown $\bar p - p_{wf}$ is proportional to $\ln(r_e/r_w) - 3/4 + S$ , and the difference $p_e - p_{wf}$ is proportional to $\ln(r_e/r_w) - 1/2 + S$ . The logarithm is the natural one.
For an area of any shape, $\tfrac12\ln\big(4A/(\gamma C_Ar_w^2)\big)$ replaces $\ln(r_e/r_w) - 3/4$ . For a circle $C_A$ = 31.62. For elongated areas, and for wells shifted towards a boundary, $C_A$ is smaller and productivity is lower.
In a circular area with the well at the centre, the regime sets in when $kt/(\phi\mu c_tA) \approx 0.1$ . In elongated and asymmetric areas it sets in several times later.
In a block with several wells, each well drains a pore volume proportional to its flow rate. The average pressure of the block is the average of the pressures of the areas, weighted by flow rate.

Key relationships

\bar p - p_{wf} = \dfrac{qB\mu}{2\pi kh}\left(\ln\dfrac{r_e}{r_w} - \dfrac34 + S\right)

well at the centre of a circular area, consistent units

\bar p - p_{wf} = \dfrac{qB\mu}{2\pi kh}\left(\dfrac12\ln\dfrac{4A}{\gamma C_A r_w^2} + S\right)

area of any shape, $\gamma = 1.781$

\dfrac{\dd\bar p}{\dd t} = -\dfrac{qB}{c_t V_p}

rate of pressure decline in a closed area

S_A = \dfrac12\ln\dfrac{4\pi}{\gamma C_A} + \dfrac34

shape pseudo-skin

t_{DA} = \dfrac{k\,t}{\phi\mu c_t A}

dimensionless time based on area (SI)

Common pitfalls

Putting the term $\ln(r_e/r_w) - 1/2$ into the pseudo-steady-state equation written with the average pressure. In this regime that term belongs to the pressure at the outer boundary. It is steady-state inflow that uses $-1/2$ with the average pressure.
Applying the shape factor before the pseudo-steady-state regime has set in. While $t_{DA}$ is below the tabulated value, the equation is wrong.
Converting drainage volumes into areas without allowing for faults, or for differences in thickness, porosity and saturation.
Expecting a constant productivity index below the bubble point pressure (the pressure below which gas comes out of the oil). Free gas lowers the index with every step of pressure decline.
Dividing volumes in proportion to flow rates among wells that lie in different blocks, separated by a sealing fault.

After this lesson you can

explain why the pressure in a closed area falls at the same rate everywhere after the transient period, and link this rate to the material balance
derive the pseudo-steady-state inflow equation and explain the terms $-1/2$ and $-3/4$
calculate the productivity index, the average reservoir pressure and the bottomhole pressure in the pseudo-steady-state regime
allow for the shape of the drainage area and the position of the well through the shape factor and the shape pseudo-skin (an extra, skin-like term that comes from the shape of the area)
estimate the time the regime takes to set in, and find the average pressure of a block with several wells

A closed area: the pressure falls at the same rate everywhere

Take a well at the centre of a circular area of radius $r_e$ . Its outer boundary is impermeable. Such a boundary may be a fault (a break in the rock layers), a pinch-out of the formation (where it thins out to nothing) or a flow divide between neighbouring wells (a no-flow boundary). The well starts to produce at a constant flow rate.

At first the drawdown cone (the funnel of lowered pressure around the well) expands as in an infinite reservoir. The regime is transient (Flow regimes). When the cone reaches the boundary, there is no recharge. The withdrawal is now replaced only by the expansion of rocks and fluids throughout the area. Soon the pressure distribution takes a constant shape and moves down as a whole: the pressure at every point falls at the same rate. This is the pseudo-steady-state regime. It is also called quasi-steady state, and British literature calls it semi-steady state.

CalculatedCalculated profiles of pressure decline in a closed circular area of radius 300 m (data of the examples in the lesson Flow regimes, flow rate 100 m³/d, skin 0). After 1 and 10 h (dashed lines) the cone is still expanding. After 100, 250 and 400 h (solid lines) the profile keeps its shape and moves down parallel to itself. The distance scale is logarithmic, starting from 1 m.

The rate of pressure decline follows from the material balance. In a time $\dd t$ the well withdraws a reservoir volume $qB\,\dd t$ . The expansion of the fluids and pores throughout the area replaces it: $c_tV_p$ for each unit of fall in the average pressure.

Rate of pressure decline in a closed area

\frac{\dd \bar p}{\dd t} = -\frac{qB}{c_t V_p} = -\frac{qB}{c_t\,\phi h A}

where:

$\bar p$ average reservoir pressure: the pressure averaged over the pore volume of the area, Pa
$q$ flow rate at standard conditions, m³/s
$B$ fluid formation volume factor, m³/m³
$c_t$ total compressibility of the rocks and fluids (how much their volume changes when the pressure changes), 1/Pa
$V_p = \phi hA$ pore volume of the drainage area, m³
$\phi,\ h,\ A$ porosity, net thickness and area of the region, —, m, m²

Units: any consistent units: for example, m³/d, 1/MPa and m³ give MPa/d

Holds when

A closed area: no fluid enters across the boundaries.
Constant flow rate; single-phase fluid, constant compressibility.
For the average pressure the equality holds at any time. For the pressure at each point it holds only after the pseudo-steady-state regime has set in.

The rate of decline depends only on the withdrawal, the compressibility and the pore volume. It does not depend on the permeability or the skin. Integrating gives $\bar p = p_i - qBt/(c_tV_p)$ : the average pressure falls linearly with time. We find the pore volume of the area from the slope of this line (Well testing: purposes and types of test).

The inflow equation

In the pseudo-steady-state regime every element of the reservoir expands at the same rate. So only the fluid released in the ring between $r$ and $r_e$ passes through a circle of radius $r$ . This is the share of the withdrawal that equals the share of the area of the ring:

Flow rate through a circle of radius r

q(r) = q\left(1 - \frac{r^2}{r_e^2}\right)

where:

$q(r)$ flow rate through a cylindrical surface of radius $r$ , m³/s
$q$ well flow rate at the same conditions, m³/s
$r,\ r_e$ distance from the well axis and radius of the area, m

Units: any consistent units

Holds when

Pseudo-steady-state regime, well at the centre of a circular area, homogeneous reservoir.
$r_w \ll r_e$ .

Close to the well, almost the whole flow rate passes through the circle. At the boundary, nothing does. In steady-state inflow, by comparison, the whole flow rate passes through any circle (Steady-state radial flow). So near the well the pressure gradients in the two regimes are almost the same. Far from the well they are smaller in the pseudo-steady-state regime.

We put this flow rate into Darcy's law for a cylindrical surface and integrate from the wall of the well. (The derivation is in the advanced section.) This gives the pressure distribution:

Pressure distribution in pseudo-steady-state inflow

p(r) - p_{wf} = \frac{qB\mu}{2\pi kh}\left(\ln\frac{r}{r_w} - \frac{r^2}{2r_e^2} + S\right)

where:

$p(r)$ pressure at distance $r$ from the well axis, outside the altered near-wellbore zone, Pa
$p_{wf}$ bottomhole pressure, Pa
$q$ flow rate at standard conditions, m³/s
$B,\ \mu$ formation volume factor and viscosity of the fluid, m³/m³, Pa·s
$k,\ h$ permeability and net thickness of the reservoir, m², m
$S$ skin factor, —

Units: SI; $\ln$ is the natural logarithm

Holds when

Pseudo-steady-state regime, circular area with the well at the centre.
$r_w \ll r_e$ : terms of the order of $(r_w/r_e)^2$ are left out.
The skin is an additional pressure drop in a thin zone at the wellbore (Skin factor).

Near the well the term $r^2/(2r_e^2)$ is negligible, so the profile is the same as the logarithmic profile of steady-state inflow. The difference appears at distances close to $r_e$ . There the profile flattens, and at the no-flow boundary the pressure gradient is zero.

At $r = r_e$ the profile gives the pressure at the boundary, and averaging the profile over the pore volume gives the average pressure of the area:

Pseudo-steady-state inflow to a well at the centre of a circular area

\begin{aligned} p_e - p_{wf} &= \frac{qB\mu}{2\pi kh}\left(\ln\frac{r_e}{r_w} - \frac12 + S\right)\\ \bar p - p_{wf} &= \frac{qB\mu}{2\pi kh}\left(\ln\frac{r_e}{r_w} - \frac34 + S\right)\end{aligned}

where:

$p_e$ pressure at the impermeable outer boundary, Pa
$\bar p$ pressure averaged over the pore volume of the area, Pa
$r_e,\ r_w$ radius of the area and well radius, m
$q,\ B,\ \mu,\ k,\ h,\ S$ as in the pressure distribution, m³/s, m³/m³, Pa·s, m², m, —

Units: SI (any consistent system of units); $\ln$ is the natural logarithm

Holds when

The same as for the pressure distribution.
The fluid is single-phase and slightly compressible, Darcy's law holds, and the skin does not depend on the flow rate.

The pressure at the boundary is higher than the average by a quarter of a “unit” of radial pressure drop, $qB\mu/(2\pi kh)$ . Most of the pore volume lies far from the well, where the pressure is close to $p_e$ . The profile reaches the value $\bar p$ at a distance of about $0.55\,r_e$ from the well.

In other unit systems (2)

Metric oilfield units (productivity index)

J = \frac{q}{\bar p - p_{wf}} = \frac{0.536\,k\,h}{\mu\,B\left(\ln\dfrac{r_e}{r_w} - 0.75 + S\right)}

$J$ in m³/(d·MPa), $k$ in mD, $h$ in m, $\mu$ in mPa·s, $B$ in m³/m³; flow rate at standard conditions.

$0.536 = 2\pi\cdot0.0853$ , as in the Dupuit equation (Steady-state radial flow).

Oilfield units

J\,[\text{STB/(d·psi)}] = \frac{7.08\cdot10^{-3}\,k\,h}{\mu\,B\left(\ln\dfrac{r_e}{r_w} - 0.75 + S\right)}

$k$ in mD, $h$ in ft, $\mu$ in cP, $B$ in rb/STB.

$7.08\cdot10^{-3} = 2\pi\cdot1.127\cdot10^{-3}$ .

Worked exampleProductivity and bottomhole pressure during the depletion of a closed area

Given (assumed data)

Permeability $k$ ; thickness $h$: 50 mD; 15 m
Viscosity $\mu$ ; formation volume factor $B$: 1.2 mPa·s; 1.3 m³/m³
Porosity; connate water; total compressibility $c_t$: 0.20; 0.25; $2.11\cdot10^{-3}$ MPa⁻¹
Drainage area: circle of radius 300 m, no-flow boundary; $r_w$ = 0.1 m; skin +2
Oil flow rate $q$ ; initial pressure $p_i$: 100 m³/d; 25.0 MPa (bubble point pressure 18 MPa)

Solution

Logarithmic term: $\ln(300/0.1) - 0.75 + 2 = 9.256$ .
Productivity index: $J = \dfrac{0.536\cdot50\cdot15}{1.2\cdot1.3\cdot9.256} = 27.8$ m³/(d·MPa). Drawdown: $q/J = 3.59$ MPa.
Pore volume of the area: $V_p = \pi\cdot300^2\cdot15\cdot0.20 = 848$ thousand m³. Rate of pressure decline: $qB/(c_tV_p) = 0.0726$ MPa/d.
After 30 days the average pressure is $\bar p = 25.0 - 0.0726\cdot30 = 22.82$ MPa, and the bottomhole pressure is $p_{wf} = 22.82 - 3.59 = 19.23$ MPa.
The pressure at the boundary is above the average by a quarter of the “unit” $qB\mu/(0.536\,kh) = 0.388$ MPa. That is 0.097 MPa.
The average pressure falls to the bubble point pressure after $(25.0 - 18)/0.0726 = 96$ days. In that time the well produces 9.6 thousand m³ of oil. This is 2.0% of the oil in place in the area ( $V_p\cdot0.75/1.3 = 489$ thousand m³).

Answer: The productivity index is about 27.8 m³/(d·MPa). Without recharge the pressure in the area falls by 0.0726 MPa per day, and after a month the bottomhole pressure is about 19.23 MPa.

What it means. In the pseudo-steady-state regime the productivity stays constant as long as the inflow is single-phase. But the pressure itself falls fast. In three months a small closed area loses all of its pressure margin above the bubble point pressure, after producing only about 2.0% of its oil. Gas starts to come out of the oil at the well even earlier: the bottomhole pressure drops below 18 MPa after about 47 days, and from then on the productivity declines (Inflow performance relationship (IPR)). After 96 days gas comes out of the oil throughout the area (Reservoir drive mechanisms). To hold the flow rate, we maintain the pressure by water injection. The data in this example are assumed for illustration; the reservoir and oil are the same as in the lesson Flow regimes.

The shape of the area and the position of the well: the shape factor

A circular area with the well at the centre is an idealisation (a simplified picture). Real drainage areas are polygons bounded by faults and by flow divides with neighbouring wells, and the well does not always stand in the middle. Suppose the well is closer to one of the boundaries. Then the paths from the distant parts of the area are longer, and the streamlines (the paths the fluid follows) crowd together at the nearer boundary. So a larger drawdown is needed for the same flow rate.

The Dietz shape factor $C_A$ accounts for this effect. It writes the logarithmic term with the area $A$ of the region:

Pseudo-steady-state inflow in an area of arbitrary shape

\bar p - p_{wf} = \frac{qB\mu}{2\pi kh}\left(\frac12\ln\frac{4A}{\gamma C_A r_w^2} + S\right)

where:

$A$ area of the drainage region, m²
$C_A$ shape factor: depends on the shape of the area and the position of the well, —
$\gamma$ $\e^{0.5772} = 1.781$ , that is, e raised to the power of Euler's constant, —
$\bar p,\ p_{wf},\ q,\ B,\ \mu,\ k,\ h,\ S,\ r_w$ as in the equation for a circular area, Pa, Pa, m³/s, m³/m³, Pa·s, m², m, —, m

Units: SI; $A$ and $r_w^2$ in the same units of area; $\ln$ is the natural logarithm

Holds when

Pseudo-steady-state regime: the dimensionless time $t_{DA}$ is greater than the tabulated value for the given shape.
A homogeneous isotropic reservoir, the well penetrates the whole thickness, and the boundary is impermeable along its whole perimeter.

For a circle, $A = \pi r_e^2$ and $C_A = 4\pi\e^{3/2}/\gamma = 31.62$ . The equation then turns into the term $\ln(r_e/r_w) - 3/4$ . The smaller $C_A$ is, the greater the drawdown at the same flow rate. The constant $\gamma$ comes from the solutions for transient inflow, where it appears naturally (Well testing: purposes and types of test).

In other unit systems (1)

Metric oilfield units (productivity index)

J = \frac{0.536\,k\,h}{\mu\,B\left(\dfrac12\ln\dfrac{4A}{1.781\,C_A\,r_w^2} + S\right)}

$J$ in m³/(d·MPa), $A$ in m², $r_w$ in m, $k$ in mD, $h$ in m, $\mu$ in mPa·s.

$0.536$ is the same constant as for a circular area; in oilfield units it is $7.08\cdot10^{-3}$ with $A$ in square feet.

It is often handier to compare with an ideal circular area of the same size. Then the shape pseudo-skin $S_A$ expresses the effect of shape. It is added to the skin of the well:

Shape pseudo-skin

\frac12\ln\frac{4A}{\gamma C_A r_w^2} = \ln\frac{r_e}{r_w} - \frac34 + S_A,\qquad S_A = \frac12\ln\frac{4\pi}{\gamma C_A} + \frac34

where:

$r_e = \sqrt{A/\pi}$ radius of the circle of the same area, m
$S_A$ shape pseudo-skin, —
$C_A,\ \gamma$ shape factor and the constant 1.781, —

Units: dimensionless

Holds when

The same as for the equation with the shape factor.

For a circle $S_A = 0$ . For a well at the centre of a square it is 0.01, and at the centre of a 4:1 rectangle it is 0.89. A well right at a boundary or in a corner of the area gets a pseudo-skin of 2–3 or more. This loses as much productivity as clear damage to the near-wellbore zone. But an acid treatment cannot remove it (pseudo-skin); only the placement of wells can.

Shape factors of some closed areas

Shape and position of the well: Circle, at the centre
$C_A$: 31.62
$S_A$: 0
Equation valid for $t_{DA} >$: 0.1
Infinite reservoir for $t_{DA} <$: 0.1

Shape and position of the well: Square, at the centre
$C_A$: 30.88
$S_A$: 0.01
Equation valid for $t_{DA} >$: 0.1
Infinite reservoir for $t_{DA} <$: 0.09

Shape and position of the well: Square, midway between the centre and the middle of a side
$C_A$: 12.99
$S_A$: 0.45
Equation valid for $t_{DA} >$: 0.7
Infinite reservoir for $t_{DA} <$: 0.03

Shape and position of the well: Square, at the centre of a quarter
$C_A$: 4.51
$S_A$: 0.97
Equation valid for $t_{DA} >$: 0.6
Infinite reservoir for $t_{DA} <$: 0.025

Shape and position of the well: 2:1 rectangle, at the centre
$C_A$: 21.84
$S_A$: 0.19
Equation valid for $t_{DA} >$: 0.3
Infinite reservoir for $t_{DA} <$: 0.025

Shape and position of the well: 2:1 rectangle, on the axis at 1/8 of the length from the short side
$C_A$: 0.58
$S_A$: 2.00
Equation valid for $t_{DA} >$: 2.0
Infinite reservoir for $t_{DA} <$: 0.02

Shape and position of the well: 4:1 rectangle, at the centre
$C_A$: 5.38
$S_A$: 0.89
Equation valid for $t_{DA} >$: 0.8
Infinite reservoir for $t_{DA} <$: 0.01

Shape and position of the well: 5:1 rectangle, at the centre
$C_A$: 2.36
$S_A$: 1.30
Equation valid for $t_{DA} >$: 1.0
Infinite reservoir for $t_{DA} <$: 0.025

Shape and position of the well	$C_A$	$S_A$	Equation valid for $t_{DA} >$	Infinite reservoir for $t_{DA} <$
Circle, at the centre	31.62	0	0.1	0.1
Square, at the centre	30.88	0.01	0.1	0.09
Square, midway between the centre and the middle of a side	12.99	0.45	0.7	0.03
Square, at the centre of a quarter	4.51	0.97	0.6	0.025
2:1 rectangle, at the centre	21.84	0.19	0.3	0.025
2:1 rectangle, on the axis at 1/8 of the length from the short side	0.58	2.00	2.0	0.02
4:1 rectangle, at the centre	5.38	0.89	0.8	0.01
5:1 rectangle, at the centre	2.36	1.30	1.0	0.025

Classical calculated values for a homogeneous reservoir (Dietz, 1965). $S_A$ is calculated from the equation above. The pseudo-steady-state equation is valid when $t_{DA}$ is greater than the value in the fourth column. The infinite-reservoir solution gives the bottomhole pressure with an error below 1% when $t_{DA}$ is less than the value in the fifth column. Between the two lies a transition period.

CalculatedCalculated shape pseudo-skin of a rectangular area against the length-to-width ratio: a well at the centre and a well on the long axis at a quarter of the length from the short side. The lines show the approximate equation from the advanced section, and the dots show the tabulated values.

Worked exampleAn elongated area: productivity and time to reach the regime

Given (assumed data)

Area of the region $A$: 282,700 m² — the same as a circle of radius 300 m
Shape: a circle with the well at the centre ( $C_A$ = 31.62, the regime sets in at $t_{DA}$ = 0.1) or a 4:1 rectangle with the well at the centre ( $C_A$ = 5.38, the regime sets in at $t_{DA}$ = 0.8)
Reservoir and well: as in the previous example: $k$ = 50 mD, $\phi\mu c_t = 5.07\cdot10^{-4}$ MPa⁻¹·mPa·s, $r_w$ = 0.1 m, skin +2

Solution

Circle: $\tfrac12\ln\dfrac{4\cdot282{,}700}{1.781\cdot31.62\cdot0.1^2} = 7.256$ . This is $\ln(r_e/r_w) - 3/4$ .
4:1 rectangle: $\tfrac12\ln\dfrac{4\cdot282{,}700}{1.781\cdot5.38\cdot0.1^2} = 8.142$ . The shape pseudo-skin is $S_A = 0.89$ . The dimensions of the rectangle are 266 by 1063 m.
Ratio of productivities at a skin of +2: $(7.256 + 2)/(8.142 + 2) = 0.913$ . In the rectangular area the well delivers 8.7% less.
Time for the pseudo-steady-state regime to set in: $t = t_{DA}\,\phi\mu c_tA/(3.553\cdot10^{-3}\,k)$ . For the circle it is $0.1\cdot5.07\cdot10^{-4}\cdot282{,}700/(3.553\cdot10^{-3}\cdot50) = 81$ h. For the rectangle it is 645 h, or 27 days.

Answer: The elongated area takes away about 8.7% of the productivity. It reaches the pseudo-steady-state regime eight times later: after 27 days instead of 3.4.

What it means. As long as the well stands in the middle, the loss of productivity is small: a pseudo-skin of 0.9 is less than a typical damage skin. The time matters much more. Until the regime has set in, the ratio of flow rate to pressure drop is higher than the stabilised value. So the productivity measured in the first weeks of a well in an elongated block comes out too high. The data in this example are assumed for illustration.

When the pseudo-steady-state regime sets in

It is convenient to measure the approach to the regime with the dimensionless time based on area:

Dimensionless time based on the area of the region

t_{DA} = \frac{k\,t}{\phi\mu c_t A}

where:

$t_{DA}$ dimensionless time based on the area of the region, —
$t$ time of production at a constant flow rate since start-up or the last change of flow rate, s
$A$ area of the drainage region, m²
$k,\ \phi,\ \mu,\ c_t$ permeability, porosity, viscosity, total compressibility, m², —, Pa·s, 1/Pa

Units: SI

Holds when

A homogeneous reservoir; time is counted from the start-up of the well or from the last change of flow rate.

The pseudo-steady-state inflow equations are valid when $t_{DA}$ is greater than the tabulated value for the given shape. This value is 0.1 for a circle and for a square with the well at the centre, 0.8 for a 4:1 rectangle, and 2 or more for a well near the boundary of an elongated area. Asymmetric areas take longer to settle. The disturbance must reach the most distant boundary and “rearrange” the flow throughout the area.

In other unit systems (2)

Metric oilfield units

t\,[\text{h}] = \frac{t_{DA}\;\phi\;\mu\,[\text{mPa·s}]\;c_t\,[\text{MPa}^{-1}]\;A\,[\text{m}^2]}{3.553\cdot10^{-3}\,k\,[\text{mD}]}

$\mu$ in mPa·s, $c_t$ in 1/MPa, $A$ in m², $k$ in mD; time in hours.

$3.553\cdot10^{-3}$ converts mD to m², hours to seconds, mPa·s to Pa·s and 1/MPa to 1/Pa (Flow regimes).

Oilfield units

t\,[\text{h}] = \frac{t_{DA}\;\phi\;\mu\,[\text{cP}]\;c_t\,[\text{psi}^{-1}]\;A\,[\text{ft}^2]}{2.637\cdot10^{-4}\,k\,[\text{mD}]}

$\mu$ in cP, $c_t$ in 1/psi, $A$ in square feet, $k$ in mD; time in hours.

$2.637\cdot10^{-4}$ is the dimensionless-time factor in oilfield units.

Until the regime has set in, the ratio of flow rate to pressure drop is higher than in the pseudo-steady-state regime, and it slowly decreases. So we measure the productivity used for forecasting after stabilisation. In tight reservoirs (low-permeability rock), where stabilisation takes months, we calculate it from the permeability and skin obtained from well tests (Well testing: purposes and types of test).

If we add the depletion from the material balance to the drawdown, we get the fall of the bottomhole pressure from its initial value at any moment of the pseudo-steady-state regime:

Bottomhole pressure in the pseudo-steady-state regime in terms of the initial pressure

p_i - p_{wf} = \frac{qB\mu}{2\pi kh}\left(\frac12\ln\frac{4A}{\gamma C_A r_w^2} + S + 2\pi t_{DA}\right)

where:

$p_i$ initial reservoir pressure, Pa
$t_{DA}$ dimensionless time based on area, —
$q,\ B,\ \mu,\ k,\ h,\ A,\ C_A,\ r_w,\ S$ as in the equation with the shape factor, m³/s, m³/m³, Pa·s, m², m, m², —, m, —

Units: SI; $\ln$ is the natural logarithm

Holds when

Constant flow rate from the initial state; the pseudo-steady-state regime has already set in.
A closed area, single-phase slightly compressible fluid.

The term $2\pi t_{DA}$ is the fall of the average pressure from the material balance, expressed in “units” of $qB\mu/(2\pi kh)$ . It grows linearly with time, while the other terms are constant. So on an ordinary plot of bottomhole pressure against time (both axes linear), the late points lie on a straight line. Its slope gives the pore volume. Its intercept on the pressure axis gives the shape factor, if the permeability and skin are known from the early part of the record (reservoir limit test, Reservoir limit test).

Several wells: drainage volumes and the average pressure

Suppose several wells produce from a closed block, and all of them have reached the pseudo-steady-state regime. Then the pressure throughout the block falls at the same rate. Flow divides form between the wells. Fluid does not cross them, so for each well they work as a no-flow boundary, just like a fault.

SchematicDrainage areas of wells in a block bounded by a fault (schematic, not to scale). In the pseudo-steady-state regime the flow divides between the wells behave like no-flow boundaries. In a homogeneous block with equal flow rates the areas are equal; with different flow rates they are proportional to the flow rates.

The rate of pressure decline $qB/(c_tV_p)$ is the same in all areas. So the pore volume that each well drains is proportional to its flow rate. A well with a higher flow rate “pushes” the flow divides towards its neighbours.

Drainage volumes and the average pressure of the block

V_i = V_t\,\frac{q_i}{q_t},\qquad \bar p_{\text{blk}} = \frac{\sum_i \bar p_i V_i}{\sum_i V_i} = \frac{\sum_i \bar p_i q_i}{\sum_i q_i}

where:

$V_i$ pore volume drained by the $i$ -th well, m³
$V_t$ pore volume of the block, m³
$q_i,\ q_t$ flow rate of the $i$ -th well and total flow rate of the block, m³/d
$\bar p_i$ average pressure in the area of the $i$ -th well (for example, from the pressure build-up curve), MPa
$\bar p_{\text{blk}}$ average pressure of the block, MPa

Units: any consistent units

Holds when

All wells of the block operate in the pseudo-steady-state regime at constant flow rates.
Single-phase fluid; the compressibility and formation volume factor are the same throughout the block.
Wells of different blocks separated by sealing faults are considered separately.

For the material balance the form with flow rates is handier. We know the flow rate of each well from daily measurements, but we cannot measure the pore volume of its area directly. The volumes we get are pore volumes. To plot the areas on a map, we divide the volume of each by the local product of porosity and thickness. If the saturation varies a lot across the block, we allow for that too.

Worked exampleAverage pressure of a block with three wells

Given (assumed data)

Block between faults: pore volume 5.0 million m³; three wells in the pseudo-steady-state regime
Oil flow rates: 120, 80 and 50 m³/d
Average pressures in the areas of the wells (from build-up curves): 21.6, 22.0 and 22.6 MPa
$B$ ; $c_t$: 1.3 m³/m³; $2.11\cdot10^{-3}$ MPa⁻¹
Fourth well: in the neighbouring isolated block: 60 m³/d, 18.9 MPa

Solution

The total flow rate of the block is 250 m³/d. Drainage volumes: $5.0\cdot120/250 = 2.4$ , $5.0\cdot80/250 = 1.6$ and $5.0\cdot50/250 = 1.0$ million m³.
Average pressure of the block: $\bar p = \dfrac{21.6\cdot120 + 22.0\cdot80 + 22.6\cdot50}{250} = 21.93$ MPa. The simple average of the three pressures is 22.07 MPa.
Rate of pressure decline in the block: $250\cdot1.3/(2.11\cdot10^{-3}\cdot5.0\cdot10^6) = 0.031$ MPa/d. It is the same in all three areas.
We leave the fourth well out of the calculation: its area is its whole isolated block. If we counted it by mistake as part of the block, the weighted average would come out as 21.34 MPa.

Answer: The average pressure of the block is 21.93 MPa. The first well drains 2.4 million m³, almost half of the block.

What it means. The weighted average is lower than the simple average by 0.14 MPa. The most productive well drains a larger volume, so its lower pressure carries more weight. The mistake with the fourth well would shift the average by a further 0.59 MPa and distort the material balance of the block. The data in this example are assumed for illustration.

When productivity stops being constant

In the ideal pseudo-steady-state regime the productivity index is constant. The flow rate is proportional to the drawdown, and the fall of the average pressure does not change $J$ . In practice productivity decreases with time and with a growing drawdown. The causes are:

Free gas. Below the bubble point pressure, gas comes out of the oil at the well. It lowers the relative permeability to oil (Relative permeability), which is how well oil flows when gas takes up part of the pore space. The inflow performance curve then bends (Inflow performance relationship (IPR)). Below the bubble point pressure, a constant productivity index loses its meaning.
Oil viscosity. Oil that has lost its gas is more viscous. Above the bubble point pressure the opposite happens: the viscosity of oil decreases slightly as the pressure falls.
Inertial losses. At high velocities near the wellbore, an inertial resistance is added to the viscous resistance. It grows faster than the flow rate (Inflow at high rates: non-Darcy flow).
Compaction of the reservoir. As the effective stress (the stress that presses the rock grains together) increases, pores and fractures narrow and the permeability falls. Fractured and poorly cemented rocks are especially sensitive.
Water and gas near the well. When a water front or a gas cone arrives, the phase permeabilities near the wellbore change (Water and gas coning).

For gas wells we describe the pseudo-steady-state regime with pseudo-pressures. The compressibility of gas is close to $1/p$ and changes a lot during depletion, so the pressure does not fall linearly with time. What is linear is the material balance written with $p/Z$ (Material balance for gas reservoirs, Gas formation volume factor, compressibility and viscosity).

Terms and symbols

Pseudo-steady-state flow: A regime in which the pressure in a closed area falls at the same rate everywhere. The difference between the average and the bottomhole pressure is constant.
Average reservoir pressure $\bar p$: The pressure averaged over the pore volume of the drainage area.
Pressure at the boundary $p_e$: The pressure at the impermeable outer boundary. In the pseudo-steady-state regime it is higher than the average by a quarter of a “unit” of $qB\mu/(2\pi kh)$ .
Shape factor $C_A$: A number that accounts for the shape of the drainage area and the position of the well. It is 31.62 for a well at the centre of a circle.
Shape pseudo-skin $S_A$: The additional resistance that is equivalent to the difference between the shape of the area and a circle of the same area.
Dimensionless time $t_{DA}$: The quantity $kt/(\phi\mu c_tA)$ . It tells us whether the pseudo-steady-state regime has set in.
Flow divide: The boundary between the drainage areas of neighbouring wells. Fluid does not flow across it.

Common mistakes and misconceptions

Wrong

“In the pseudo-steady-state regime the pressure has stabilised.”

In fact

The pressure keeps falling, at the same rate at all points. Only the pressure differences stay constant, including the drawdown $\bar p - p_{wf}$ .

Wrong

“The equation with −3/4 is suitable for any drainage area.”

In fact

Only for a well at the centre of a circular or nearly circular area. Other shapes and well positions need the shape factor $C_A$ or the pseudo-skin $S_A$ .

Wrong

“The volume that a well drains depends on the distance to its neighbours.”

In fact

In the pseudo-steady-state regime the flow rate decides it. The drainage volume is proportional to the well's share of the withdrawal from the block.

Wrong

“The productivity measured a week after start-up is good for forecasting.”

In fact

Only if the regime has set in by then, that is, $t_{DA}$ is greater than the tabulated value. In tight or elongated blocks the first week may fall in the transient period, and the measurement then overestimates the productivity.

Wrong

“Skin and shape pseudo-skin are the same thing.”

In fact

Skin describes the near-wellbore zone and is reduced by stimulation. The shape pseudo-skin describes the geometry of the area and the position of the well in it; it can be changed only by the placement of wells.

In depth: advanced detail

Derivation of the inflow equation

We write Darcy's law for a cylindrical surface of radius $r$ and put into it the flow rate $q(r)$ at reservoir conditions:

Darcy's law for a cylindrical surface

q_{\text{res}}\left(1 - \frac{r^2}{r_e^2}\right) = \frac{2\pi kh\,r}{\mu}\,\frac{\dd p}{\dd r}\quad\Rightarrow\quad \dd p = \frac{q_{\text{res}}\mu}{2\pi kh}\left(\frac{1}{r} - \frac{r}{r_e^2}\right)\dd r

where:

$q_{\text{res}}$ flow rate at reservoir conditions, $qB$ , m³/s
$p,\ r$ pressure and distance from the well axis, Pa, m
$k,\ h,\ \mu,\ r_e$ permeability, thickness, viscosity, radius of the area, m², m, Pa·s, m

Units: SI

Holds when

Pressure increases with distance from the well, and the flow is directed towards it.

Integrating from the wall of the well to $r$ gives the profile of the main section (without skin): $\ln(r/r_w)$ minus $(r^2 - r_w^2)/(2r_e^2)$ , where $r_w^2$ is left out. At $r = r_e$ we get the term $\ln(r_e/r_w) - 1/2$ .

The average pressure is the integral of the profile weighted by the volume of the ring, $2\pi rh\phi\,\dd r$ , divided by the volume of the area, $\pi r_e^2h\phi$ :

Average pressure of the area

\bar p - p_{wf} = \frac{2}{r_e^2}\int_{r_w}^{r_e}(p - p_{wf})\,r\,\dd r = \frac{q_{\text{res}}\mu}{2\pi kh}\left(\ln\frac{r_e}{r_w} - \frac12 - \frac14\right)

where:

$\bar p$ pressure averaged over the pore volume, Pa
$p - p_{wf}$ pressure profile relative to the bottomhole, Pa

Units: SI

Holds when

$r_w \ll r_e$ ; porosity and thickness are constant.

After integration by parts the logarithmic term gives $\ln(r_e/r_w) - 1/2$ , and the term $r^2/(2r_e^2)$ gives a further $-1/4$ . This is how $-3/4$ arises. The skin is added to the bracket as an additional pressure drop in a thin zone at the wellbore.

Shape factor of a rectangle: an approximate equation

For a rectangular area of length $L$ and width $W$ , we can split the logarithmic term into two parts. The first is linear flow along the long axis: fluid from the whole area drains towards the cross-section where the well stands. The second is the convergence of this flow onto the well itself, in a strip of width $W$ :

Approximate shape factor of a rectangular area

\frac12\ln\frac{4A}{\gamma C_A r_w^2} \approx \frac{2\pi L}{W}\left(\frac13 - x_D + x_D^2\right) + \ln\frac{W}{2\pi r_w\sin(\pi y_D)}

where:

$L,\ W$ length and width of the rectangle, $L \ge W$ ; area $A = LW$ , m
$x_D$ distance from the well to one of the short sides, divided by $L$ , —
$y_D$ distance from the well to one of the long sides, divided by $W$ , —
$C_A,\ \gamma,\ r_w$ shape factor, the constant 1.781, well radius, —, —, m

Units: dimensionless; $L$ , $W$ and $r_w$ in the same units; $\ln$ is the natural logarithm

Holds when

Pseudo-steady-state regime, homogeneous reservoir, the well penetrates the whole thickness.
The well is at least the width $W$ away from both short sides. Then the error of the logarithmic term is below 0.01. For a well on the long axis ( $y_D = 1/2$ ), half the width is enough.
Closer to a short side the equation underestimates the resistance. For an off-axis well at a distance of $W/2$ the error reaches 0.1–0.2. For a well in a corner of the area the equation is unsuitable, and tabulated values are needed.

The first term is the resistance of linear inflow. It grows in proportion to $L/W$ and is smallest when the well is in the middle ( $x_D = 1/2$ ). The second term is the resistance of the convergence to the well. It grows as the well approaches a long side. For a well at the centre of 2:1, 4:1 and 5:1 rectangles, the equation reproduces the tabulated $C_A$ to within about 0.1%.

Worked exampleCheck against the table: a well in a 4:1 rectangle

Given (assumed data)

4:1 rectangle, case 1: well at the centre: $x_D = y_D = 1/2$
4:1 rectangle, case 2: well on the long axis at a quarter of the length from a short side: $x_D = 3/4$ , $y_D = 1/2$

Solution

For $A = 4W^2$ the left-hand side is $\tfrac12\ln\dfrac{16}{\gamma C_A} + \ln\dfrac{W}{r_w}$ . The term $\ln(W/r_w)$ cancels with the second term on the right. For a well at the centre, what remains is $\tfrac12\ln\dfrac{16}{\gamma C_A} = 2\pi\cdot4\cdot\tfrac{1}{12} - \ln 2\pi = 0.2565$ .
So $\ln C_A = \ln(16/1.781) - 2\cdot0.2565 = 1.6824$ and $C_A = 5.378$ . The table gives 5.38.
Well at a quarter of the length: $\tfrac12\ln\dfrac{16}{\gamma C_A} = 2\pi\cdot4\cdot\left(\tfrac13 - \tfrac34 + \tfrac{9}{16}\right) - \ln 2\pi = 1.8273$ .
$\ln C_A = \ln(16/1.781) - 2\cdot1.8273 = -1.4592$ , and $C_A = 0.232$ (the tabulated value is 0.232). The shape pseudo-skin is $S_A = 2.46$ .

Answer: The approximate equation gives

C_A

= 5.378 and 0.232, practically the tabulated values.

What it means. Shifting the well by a quarter of the length towards the short side in a 4:1 area makes $C_A$ more than twenty times smaller, and the shape pseudo-skin rises from 0.89 to 2.46. In elongated blocks (narrow blocks between faults, sandstones laid down in old river channels) it pays to place the well in the middle. It also pays to position the neighbouring wells so that their areas are closer to squares. The geometry of this example is assumed for illustration.

Connections to other topics

Steady-state radial flow — steady-state inflow, and the logarithmic terms of the inflow equations collected in one table.
Flow regimes — how the transient regime gives way to the pseudo-steady-state one.
Skin factor — skin and flow efficiency in the pseudo-steady-state inflow equation.
Well inflow: the well performance diagram — the productivity index on the well performance diagram.
Inflow performance relationship (IPR) — the inflow performance curve when the inflow stops being single-phase.
Well testing: purposes and types of test — the reservoir limit test and the rate of pressure decline.
Reservoir limit test — the pore volume and shape of the area from a long pressure record.
Build-up in a bounded reservoir and average reservoir pressure — how the average pressure of the area is found from the pressure build-up curve.
Reservoir drive mechanisms — the compressibility drive of a reservoir as pseudo-steady-state depletion.

Check yourself

1
We calculate the productivity index from the average reservoir pressure for a well at the centre of a circular closed area. Which term stands in the denominator of the equation (without skin)?
1. A. $\ln(r_e/r_w) - 3/4$
2. B. $\lg(r_e/r_w) - 3/4$
3. C. $\ln(r_e/r_w)$
4. D. $\ln(r_e/r_w) - 1/2$
Show the answer
Correct answer: A. $\ln(r_e/r_w) - 3/4$
With the average pressure, the term in the pseudo-steady-state regime is $-3/4$ . The term $-1/2$ belongs to the pressure at the outer boundary. $\ln(r_e/r_w)$ without a correction belongs to steady-state inflow with a constant pressure at the outer boundary. The base-10 logarithm without the factor 2.303 is wrong here.
2
A single well produces from a closed block with a pore volume of 2.0 million m³. Its flow rate is 150 m³/d, $B$ = 1.25 and $c_t = 2.0\cdot10^{-3}$ MPa⁻¹. By how much will the reservoir pressure fall in a year?
Show the answer
The rate of decline is $150\cdot1.25/(2.0\cdot10^{-3}\cdot2.0\cdot10^6) = 187.5/4000 \approx 0.047$ MPa/d, which is about 17 MPa over 365 days. Without recharge, such a withdrawal quickly uses up the pressure margin: if the bubble point pressure is 10 MPa below the initial pressure, the reservoir reaches it after about 213 days.
3
A well stands in a 2:1 rectangular area on the axis at 1/8 of the length from the short side ( $C_A$ = 0.58). What shape pseudo-skin does this give?
1. A.About 0
2. B.About 0.2
3. C.About 20
4. D.About 2
Show the answer
Correct answer: D. About 2
$S_A = \tfrac12\ln\dfrac{4\pi}{1.781\cdot0.58} + 0.75 = \tfrac12\ln 12.2 + 0.75 \approx 1.25 + 0.75 = 2.0$ . With a logarithmic term of about 7.3 this reduces the productivity by about a fifth.
4
In a block with several wells in the pseudo-steady-state regime, why do we calculate the average pressure as the average of the pressures of the drainage areas, weighted by flow rate?
Show the answer
The pressure in all areas falls at the same rate $q_iB/(c_tV_i)$ , so the volume that each well drains is proportional to its flow rate. The pore-volume average is then the same as the average weighted by flow rate. Flow rates, unlike volumes, are measured all the time. The rule applies within one hydrodynamically connected block (a block where fluid can flow between all its parts).
5
How long does a well at the centre of a square area of 0.5 km² need to reach the pseudo-steady-state regime, if $k$ = 20 mD and $\phi\mu c_t = 5\cdot10^{-4}$ MPa⁻¹·mPa·s?
1. A.About 350 h
2. B.About 3500 h
3. C.About 4 h
4. D.About 35 h
Show the answer
Correct answer: A. About 350 h
For a square with the well at the centre $t_{DA}$ = 0.1: $t = 0.1\cdot5\cdot10^{-4}\cdot5\cdot10^5/(3.553\cdot10^{-3}\cdot20) = 25/0.0711 \approx 352$ h, about two weeks.

Updated: 1 October 2026