Flow regimes · MunaiHub

Quick refresher

The regime is decided by the rate of pressure change. In steady state $\partial p/\partial t = 0$ at all points. In pseudo-steady state the rate is the same at all points. In a transient regime it depends both on time and on the distance from the well.
Without compressibility there would be no transient processes: a pressure drop at the bottom of the well would reach the boundary instantly. Compressibility makes the reservoir a “tank” that gives up its fluid gradually, starting with the zone near the well.
A pressure disturbance (the change of pressure that spreads out from the well) moves at a speed set by the hydraulic diffusivity $\eta = k/(\phi\mu c_t)$ . In time $t$ it clearly reaches a region of radius about $\sqrt{4\eta t}$ .
After a well starts, the regime is always transient at first. When the disturbance reaches the boundaries, the regime becomes pseudo-steady-state if the boundary is closed (no flow across it), or steady-state if as much fluid comes in through the boundary as is withdrawn.
In high-permeability reservoirs the transient period lasts hours and days. In tight reservoirs it lasts months and years. Wells in tight reservoirs therefore work for a long time with a falling rate, and the pseudo-steady-state inflow formulas must be used for them with care.

Key relationships

\frac{1}{r}\frac{\partial}{\partial r}\!\left(r\,\frac{\partial p}{\partial r}\right) = \frac{\phi\mu c_t}{k}\,\frac{\partial p}{\partial t}

diffusivity equation for radial flow

c_t = S_o c_o + S_w c_w + S_g c_g + c_f

total compressibility of the rocks and fluids

t_b \approx \dfrac{\phi\mu c_t\,r_e^2}{4k}

time for the disturbance to reach a boundary at a distance $r_e$ (SI)

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

rate of pressure decline in a closed area

Common pitfalls

Mixing up flow regimes, which describe how pressure depends on time, with the drive mechanisms of the accumulation (water drive, gas-cap drive and others).
Applying the pseudo-steady-state inflow formula to measurements made before this regime is reached. At that time the ratio of rate to drawdown (the pressure difference between the reservoir and the well) is higher, and the “permeability” calculated from it comes out too high.
Forgetting the pore compressibility. In an undersaturated reservoir (pressure above the bubble point, no free gas) it is often a third of the total compressibility or more.
Believing that a closed area can produce at a constant rate for ever without the pressure falling.

After this lesson you can

define the steady-state, pseudo-steady-state and transient regimes by how the pressure changes in time and in space, and name the boundary conditions that lead to them
explain the role of compressibility, and calculate the total compressibility and the hydraulic diffusivity of a reservoir
estimate how long the transient regime lasts and when the pseudo-steady-state or the steady-state regime begins
recognise the regime from the behaviour of the bottomhole pressure (the pressure at the bottom of the well), and choose the inflow formula that fits it
tell linear, radial and spherical flow geometry apart, and know where each of them occurs

Three regimes: what changes with time

Imagine a well that starts producing at a constant rate. Right after the start, fluid comes to the well only from the near zone. There the pressure has fallen, and the fluid and the rock have expanded a little and “given up” part of their volume. Farther out in the reservoir the pressure is still the initial pressure. With time, the zone of lowered pressure — the drawdown cone — grows wider. What happens next depends on what the cone meets at the outer boundary of the drainage area (drainage area).

Three flow regimes

Regime: Transient
Pressure: changes fastest at the well, still at its initial value far away
Boundary of the area: the disturbance has not reached it, so the reservoir behaves as if it were infinite
Where it occurs: after a start, a shut-in or a change of rate; well tests

Regime: Pseudo-steady state
Pressure: falls everywhere at the same rate, the shape of the profile does not change
Boundary of the area: closed to flow: a fault, a pinch-out, or the divide between the flows of neighbouring wells
Where it occurs: depletion of a closed block that gets no fluid from outside

Regime: Steady state
Pressure: does not change
Boundary of the area: as much fluid comes in through it as the well withdraws
Where it occurs: pressure maintenance by injection, an active aquifer (a water-bearing layer that feeds the reservoir)

Regime	Pressure	Boundary of the area	Where it occurs
Transient	changes fastest at the well, still at its initial value far away	the disturbance has not reached it, so the reservoir behaves as if it were infinite	after a start, a shut-in or a change of rate; well tests
Pseudo-steady state	falls everywhere at the same rate, the shape of the profile does not change	closed to flow: a fault, a pinch-out, or the divide between the flows of neighbouring wells	depletion of a closed block that gets no fluid from outside
Steady state	does not change	as much fluid comes in through it as the well withdraws	pressure maintenance by injection, an active aquifer (a water-bearing layer that feeds the reservoir)

In British literature the pseudo-steady-state regime is also called semi-steady state. In Russian-language literature the terms “pseudo-stationary” and “quasi-stationary” are also found.

The pseudo-steady-state regime is a special case of the transient one. The pressure changes with time, but so evenly that the pressure differences between any two points stay constant. This is why the inflow formulas for it look like the steady-state ones, with only a different constant term (Pseudo-steady-state flow and the shape factor).

CalculatedCalculated profiles of the pressure decline in a closed circular area of 300 m radius (the data of the lesson's examples, rate 100 m³/d, skin 0). After 1 and 10 h (dashed lines) the drawdown cone is still growing: the regime is transient. After 100, 250 and 400 h (solid lines) the profile keeps its shape and shifts parallel to itself (the pressure falls at the same rate everywhere): the regime is pseudo-steady-state. The distance axis is logarithmic, starting from 1 m.

Analogy: a rod cooled at one end

The equation for the pressure in a reservoir is the same as the equation of heat conduction. Picture a long metal rod, with heat drawn out of one end continuously at a constant power. At first only the near part cools, and the cooled zone slowly moves along the rod. This is the transient regime. If the far end is insulated, then once the cooling reaches it, the whole rod cools at one rate and keeps the shape of its temperature distribution. This is the pseudo-steady-state regime. If the far end is dipped into a tank of hot water, the temperature distribution eventually stops changing. This is the steady-state regime. Pressure corresponds to temperature, mobility $k/\mu$ (how easily the fluid moves) corresponds to thermal conductivity, and the storage capacity $\phi c_t$ corresponds to heat capacity. The analogy is exact in mathematics and incomplete in physics: in a reservoir it is matter that moves, not heat.

Why compressibility is needed

If the fluid and the rock were perfectly incompressible, a pressure drop at the bottom of the well would pass to the very boundary at once. Not a single cubic metre could be withdrawn from a closed area. There would be no transient processes at all.

Real fluids and rock are compressible. When the pressure falls, every volume of the reservoir expands a little and gives up fluid. So the drawdown cone grows gradually. For the pressure to fall far from the well, the fluid closer to the well must first give up its excess, and viscosity (the internal friction of the fluid) slows the flow. The measure of this elastic store is called the total compressibility:

Total compressibility of the rocks and fluids

c_t = S_o c_o + S_w c_w + S_g c_g + c_f

where:

$c_t$ total compressibility, per unit of pore volume, 1/MPa
$S_o,\ S_w,\ S_g$ oil, water and gas saturation, fraction
$c_o,\ c_w,\ c_g$ compressibility of oil, water and gas, $c = -\frac{1}{V}\frac{\dd V}{\dd p}$ , 1/MPa
$c_f$ pore (rock) compressibility: the relative decrease of the pore volume when the pore pressure falls by one unit, 1/MPa

Units: any consistent units, for example 1/MPa or 1/psi ( $1\ \text{MPa}^{-1} = 6.895\cdot10^{-3}\ \text{psi}^{-1}$ )

Holds when

The compressibilities are measured at reservoir temperature and change little over the pressure range considered.
The saturations do not change noticeably over the time considered.
For oil below the bubble point pressure, $c_o$ also includes the release of gas. The compressibility then rises sharply, and such cases are described with multiphase models.

Each phase contributes in proportion to its share of the pores. The compressibility of gas is of the order of $1/p$ . At 10–20 MPa this is 0.05–0.1 MPa⁻¹, tens of times more than for oil, so even a few per cent of free gas increase $c_t$ several times. The product $\phi c_t$ is called the storativity of the reservoir. It shows how much fluid a unit volume of the reservoir gives up when the pressure falls by one unit.

Approximate compressibilities

Component: Undersaturated oil (above the bubble point)
Compressibility, MPa⁻¹: $(0.7\text{–}3)\cdot10^{-3}$
Comment: higher for light oils with a large amount of dissolved gas

Component: Formation water
Compressibility, MPa⁻¹: $(0.4\text{–}0.5)\cdot10^{-3}$
Comment: depends little on pressure

Component: Pores of cemented rocks
Compressibility, MPa⁻¹: $(0.3\text{–}1.5)\cdot10^{-3}$
Comment: several times higher for loose sands and high-porosity chalk

Component: Gas
Compressibility, MPa⁻¹: of the order of $1/p$
Comment: 0.05–0.1 at 10–20 MPa

Component	Compressibility, MPa⁻¹	Comment
Undersaturated oil (above the bubble point)	$(0.7\text{–}3)\cdot10^{-3}$	higher for light oils with a large amount of dissolved gas
Formation water	$(0.4\text{–}0.5)\cdot10^{-3}$	depends little on pressure
Pores of cemented rocks	$(0.3\text{–}1.5)\cdot10^{-3}$	several times higher for loose sands and high-porosity chalk
Gas	of the order of $1/p$	0.05–0.1 at 10–20 MPa

These are guide values for rough estimates. In a project, the oil compressibility is taken from PVT studies (laboratory studies of how the volume and the phase state of a fluid depend on pressure and temperature). The pore compressibility is taken from core experiments (Rock compressibility and the effect of stress on core properties).

Darcy's law links the filtration velocity (the flow rate per unit area) to the pressure gradient. The mass balance says that the difference between the fluid flowing into an element of the reservoir and the fluid flowing out of it changes the mass of fluid in that element. We can express this change of mass through the change of pressure and $c_t$ . Together they give the diffusivity equation. For radial inflow to a vertical well in a uniform reservoir it reads:

Diffusivity equation for radial flow

\frac{1}{r}\frac{\partial}{\partial r}\!\left(r\,\frac{\partial p}{\partial r}\right) = \frac{1}{\eta}\,\frac{\partial p}{\partial t},\qquad \eta = \frac{k}{\phi\mu c_t}

where:

$p(r, t)$ pressure at the distance $r$ from the well axis at the time $t$ , Pa
$\eta$ hydraulic diffusivity, m²/s
$k$ permeability of the reservoir, m²
$\phi$ porosity, fraction
$\mu$ viscosity of the fluid, Pa·s
$c_t$ total compressibility, 1/Pa

Units: SI

Holds when

The reservoir is uniform and isotropic, of constant thickness, with horizontal radial flow.
There is one mobile fluid with a small, constant compressibility ( $c_t\,\Delta p \ll 1$ ), and the viscosity does not depend on pressure.
Darcy's law applies. Small terms with the square of the pressure gradient are neglected (the derivation is in the advanced section).

The left side is the “curvature” of the pressure profile, and the right side is the rate at which the pressure changes with time. Where the profile plotted against $\ln r$ is curved, the pressure changes. Where it is straight, the pressure stands still. The steady-state regime is the solution with a zero right side: the logarithmic profile of the Dupuit equation (the basic formula for steady-state radial flow; Steady-state radial flow). The pseudo-steady-state regime is the solution with a right side that is the same at all points.

In other unit systems (2)

Metric oilfield units (hydraulic diffusivity)

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

$k$ in mD, $\mu$ in mPa·s, $c_t$ in 1/MPa, $\eta$ in square metres per hour.

$3.553\cdot10^{-3} = 0.9869\cdot10^{-15}\cdot3600/(10^{-3}\cdot10^{-6})$ : it converts mD → m², hours → seconds, mPa·s → Pa·s and 1/MPa → 1/Pa. The same factor enters the dimensionless time of well tests (Well testing: purposes and types of test).

Oilfield units

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

$k$ in mD, $\mu$ in cP, $c_t$ in 1/psi, $\eta$ in square feet per hour.

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

The main consequence is that a pressure disturbance does not travel at a fixed speed, like sound. It spreads out, like heat. The distance it has clearly travelled in time $t$ is the radius of investigation $r_i = \sqrt{4\eta t}$ . It grows as the square root of time. To cover twice the distance, we need four times the time.

Worked exampleTotal compressibility, hydraulic diffusivity and radius of investigation

Given (assumed data)

Saturations: oil 0.75, irreducible water 0.25, no free gas
Compressibilities of oil, water and pores: $1.6\cdot10^{-3}$ ; $0.45\cdot10^{-3}$ ; $0.8\cdot10^{-3}$ MPa⁻¹
Permeability $k$ ; porosity $\phi$ ; oil viscosity $\mu$: 50 mD; 0.20; 1.2 mPa·s
Radius of the drainage area $r_e$: 300 m

Solution

Total compressibility in units of $10^{-3}$ MPa⁻¹: $c_t = 0.75\cdot1.6 + 0.25\cdot0.45 + 0.8 = 2.1125$ . Oil contributes 57%, water 5% and the pores 38%.
Hydraulic diffusivity: $\eta = 3.553\cdot10^{-3}\cdot\dfrac{50}{0.20\cdot1.2\cdot2.1125\cdot10^{-3}} = 350$ m²/h, or 0.097 m²/s.
Radius of investigation in an unbounded reservoir, $r_i = \sqrt{4\eta t}$ : after 1 h it is 37 m, after a day 183 m, and after 30 days 1005 m.
The disturbance reaches the boundary of the area in $t_b = r_e^2/(4\eta) = 300^2/(4\cdot350) = 64$ h, which is about 2.7 days.
If we forget the pore compressibility, we get $c_t = 1.3125\cdot10^{-3}$ MPa⁻¹, which is 1.61 times smaller. All the times of the transient processes would come out too short by the same factor.

Answer: The total compressibility is

2.1125\cdot10^{-3}

MPa⁻¹ and the hydraulic diffusivity is about 350 m²/h. The disturbance reaches the boundary of an area of 300 m radius in about 64 h.

What it means. In a reservoir of 50 mD, the transient regime at the scale of the drainage area lasts about three days. The disturbance reaches the boundary in 2.7 days, and the regime turns into pseudo-steady state a little later (the next example). For production this is a mere moment. For a well test it is plenty of time. The pores gave 38% of the total compressibility, so for undersaturated oil we cannot neglect them. The data in this example are assumed for illustration. The oil compressibility is the same as for the assumed oil of the lesson Black oil model: solution gas–oil ratio and formation volume factors.

How long each regime lasts

We estimate the duration of the transient regime from the same radius of investigation. The disturbance reaches the boundary at a distance $r_e$ in approximately this time:

Time for the disturbance to reach the boundary

t_b \approx \frac{\phi\mu c_t\,r_e^2}{4k}

where:

$t_b$ time from the start of the well to the arrival of the disturbance at the boundary, s
$r_e$ distance to the boundary; for a circular area, its radius, m
$\phi,\ \mu,\ c_t,\ k$ porosity, viscosity, total compressibility, permeability, —, Pa·s, 1/Pa, m²

Units: SI

Holds when

The reservoir is uniform, the well is at the centre of a circular area, and the rate is constant after the start.
For elongated areas and for wells shifted towards a boundary, the disturbance reaches the nearest boundary sooner, but the regime in the whole area sets in later (Pseudo-steady-state flow and the shape factor).

The time grows as the square of the size of the area and is inversely proportional to the mobility $k/\mu$ . A calculation of the full solution for a circular area shows that the bottomhole pressure starts to depart from the solution for an infinite reservoir at about $t \approx t_b$ . With a closed boundary, the pseudo-steady-state regime is practically established by $t \approx 1.3\,t_b$ . This is the condition $kt/(\phi\mu c_t A) \approx 0.1$ , where $A$ is the area of the region. With a constant pressure at the boundary, the bottomhole pressure comes within 1% of its steady-state value at about $t \approx 2\,t_b$ .

In other unit systems (2)

Metric oilfield units

t_b\,[\text{h}] = \frac{\phi\;\mu\,[\text{mPa·s}]\;c_t\,[\text{MPa}^{-1}]\;r_e^2\,[\text{m}^2]}{0.01421\,k\,[\text{mD}]}

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

$0.01421 = 4\cdot3.553\cdot10^{-3}$ .

Oilfield units

t_b\,[\text{h}] = \frac{\phi\;\mu\,[\text{cP}]\;c_t\,[\text{psi}^{-1}]\;r_e^2\,[\text{ft}^2]}{1.055\cdot10^{-3}\,k\,[\text{mD}]}

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

$1.055\cdot10^{-3} = 4\cdot2.637\cdot10^{-4}$ , where $2.637\cdot10^{-4}$ is the dimensionless-time factor in oilfield units.

Worked exampleWhen the pseudo-steady-state regime sets in: a permeable and a tight reservoir

Given (assumed data)

Reservoir and oil: as in the previous example: $\phi$ = 0.20, $\mu$ = 1.2 mPa·s, $c_t = 2.1125\cdot10^{-3}$ MPa⁻¹
Drainage area: a circle of 300 m radius, the well in the centre, a closed boundary
Permeability: 50 mD (reservoir A) or 0.5 mD (reservoir B)

Solution

The product $\phi\mu c_t = 0.20\cdot1.2\cdot2.1125\cdot10^{-3} = 5.07\cdot10^{-4}$ MPa⁻¹·mPa·s. The area of the circle is $A = \pi\cdot300^2 = 282.7$ thousand m².
Reservoir A: the disturbance reaches the boundary in $t_b = \dfrac{5.07\cdot10^{-4}\cdot300^2}{0.01421\cdot50} = 64$ h.
The pseudo-steady-state regime sets in at $kt/(\phi\mu c_t A) = 0.1$ : $t = \dfrac{0.1\cdot5.07\cdot10^{-4}\cdot282.7\cdot10^3}{3.553\cdot10^{-3}\cdot50} = 81$ h, or 3.4 days.
Reservoir B is 100 times less permeable, and all the times are 100 times longer: the disturbance reaches the boundary in 268 days, and the pseudo-steady-state regime sets in after 336 days.

Answer: In the 50 mD reservoir the transient regime ends after about 3.4 days. In the 0.5 mD reservoir it ends after about 336 days, which is almost a year.

What it means. A well in the permeable reservoir reaches the pseudo-steady-state regime almost at once, and after that its productivity is constant. A well in the tight reservoir works in the transient regime for almost a year. At a constant bottomhole pressure its rate falls all this time, and we cannot forecast it with the pseudo-steady-state inflow formula. We need solutions for transient inflow (Well testing: purposes and types of test). In ultra-low-permeability reservoirs the transient period stretches over years. The data in this example are assumed for illustration.

How to recognise the regime from measurements

In practice, we recognise the regime from how the bottomhole pressure of a well changes when it works at a constant rate (or from how the rate changes at a constant bottomhole pressure). As long as the disturbance has not reached the boundaries, the behaviour of the well is the same, whatever the boundary is like. The differences appear later.

CalculatedCalculated fall of the bottomhole pressure after the start of the well of the lesson's examples (rate 100 m³/d, skin 0) in a circular area of 300 m radius: with a closed boundary, with a constant pressure at the boundary (fluid is fed in through it), and in an infinite reservoir (grey dashed line). The time axis is logarithmic. For the first tens of hours the curves coincide. After that the pressure in the closed area falls faster and faster on this plot (it falls linearly with time), while with fluid fed in through the boundary it levels off.

Signs of the regimes in the bottomhole pressure at a constant rate

Regime: Transient radial
How $p_{wf}$ changes: by the same amount in each tenfold interval of time
Which plot straightens the record: $p_{wf}$ against $\log_{10} t$
What we find from it: flow capacity $kh$ and skin (Pressure build-up analysis)

Regime: Pseudo-steady state
How $p_{wf}$ changes: by the same amount in each hour
Which plot straightens the record: $p_{wf}$ against $t$ on an ordinary scale
What we find from it: pore volume of the area (Reservoir limit test)

Regime: Steady state
How $p_{wf}$ changes: does not change
Which plot straightens the record: —
What we find from it: productivity under pressure maintenance

Regime	How $p_{wf}$ changes	Which plot straightens the record	What we find from it
Transient radial	by the same amount in each tenfold interval of time	$p_{wf}$ against $\log_{10} t$	flow capacity $kh$ and skin (Pressure build-up analysis)
Pseudo-steady state	by the same amount in each hour	$p_{wf}$ against $t$ on an ordinary scale	pore volume of the area (Reservoir limit test)
Steady state	does not change	—	productivity under pressure maintenance

The regime is seen most reliably on the diagnostic plot of the pressure derivative (the rate of pressure change per unit of the logarithm of time) on a double-logarithmic scale (Diagnosis with the pressure derivative). For transient radial flow the derivative is horizontal. For pseudo-steady state it rises with a slope of 1. With fluid fed in through the boundary it falls.

Rate of pressure decline in a closed area

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

where:

$\bar p$ pore-volume-averaged pressure in the drainage area, MPa
$q$ rate at standard conditions, m³/d
$B$ formation volume factor of the fluid, m³/m³
$c_t$ total compressibility, 1/MPa
$V_p$ pore volume of the drainage area, m³

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

Holds when

The area is closed, with no inflow across the boundaries, and the rate is constant.
The fluid is single-phase with a constant compressibility.

This is the material balance of a “closed tank”: the withdrawal is made up only by the expansion of the rocks and fluids. For the average pressure the equality holds at any moment. In the pseudo-steady-state regime the pressure at every point, including the bottom of the well, falls at the same rate, so the pressure record becomes a straight line on ordinary axes (Well testing: purposes and types of test).

Flow geometry

We also tell flows apart by geometry, that is, by how the streamlines (the paths of the fluid) converge. There are four idealised schemes: linear, radial, hemispherical and spherical. The geometry decides how the pressure drop grows in the transient regime, and so which plot straightens the record.

Flow geometries

Geometry: Linear
Where it occurs: inflow of water from an aquifer to a long contact; narrow, elongated reservoirs; flow to a hydraulic fracture (a crack opened in the rock by pumping fluid under high pressure)
Pressure drop at a constant rate: grows as $\sqrt t$

Geometry: Radial
Where it occurs: a vertical well that penetrates the whole thickness of the reservoir
Pressure drop at a constant rate: grows as $\ln t$

Geometry: Spherical and hemispherical
Where it occurs: a well that penetrates only a small part of the thickness; the probe of a wireline formation tester (a tool on a cable that measures the pressure in the rock)
Pressure drop at a constant rate: quickly approaches a constant value; the difference from it falls as $1/\sqrt t$

Geometry	Where it occurs	Pressure drop at a constant rate
Linear	inflow of water from an aquifer to a long contact; narrow, elongated reservoirs; flow to a hydraulic fracture (a crack opened in the rock by pumping fluid under high pressure)	grows as $\sqrt t$
Radial	a vertical well that penetrates the whole thickness of the reservoir	grows as $\ln t$
Spherical and hemispherical	a well that penetrates only a small part of the thickness; the probe of a wireline formation tester (a tool on a cable that measures the pressure in the rock)	quickly approaches a constant value; the difference from it falls as $1/\sqrt t$

Real inflow often passes through several geometries in turn. For a well with a hydraulic fracture, the flow is first linear towards the fracture and then radial. We recognise the geometry from the slope of the pressure derivative on a double-logarithmic plot: 1/2 means linear flow, 0 radial, and −1/2 spherical.

In field-development calculations two models are used most often: the radial one for inflow to a well, and the linear or radial one for the inflow of water from an aquifer to the accumulation (Aquifer models: water influx). A complicated reservoir shape and the heterogeneity of the reservoir are handled numerically, on reservoir simulation models (What a reservoir simulation model is and why it is needed).

Which model to choose for calculations

Regime and calculation model

Task: Interpretation of pressure drawdown and build-up tests (the pressure falls while the well produces, and rises after it is shut in)
Regime: transient
Model: solution for an infinite reservoir (Well testing: purposes and types of test)

Task: Productivity of a well for a production forecast
Regime: pseudo-steady state
Model: formula with the term $\ln(r_e/r_w) - 3/4$ , or with a shape factor (Pseudo-steady-state flow and the shape factor)

Task: A well under pressure maintenance by injection
Regime: close to steady state
Model: Dupuit equation (Steady-state radial flow)

Task: Depletion of an accumulation, material balance
Regime: pseudo-steady state for the accumulation as a whole
Model: one average pressure for the whole volume (General material balance equation for oil reservoirs)

Task: Water influx from the aquifer
Regime: usually transient
Model: aquifer models (Aquifer models: water influx)

Task: Wells in tight reservoirs
Regime: transient for a long time
Model: solutions for transient inflow, analysis of the rate decline

Task	Regime	Model
Interpretation of pressure drawdown and build-up tests (the pressure falls while the well produces, and rises after it is shut in)	transient	solution for an infinite reservoir (Well testing: purposes and types of test)
Productivity of a well for a production forecast	pseudo-steady state	formula with the term $\ln(r_e/r_w) - 3/4$ , or with a shape factor (Pseudo-steady-state flow and the shape factor)
A well under pressure maintenance by injection	close to steady state	Dupuit equation (Steady-state radial flow)
Depletion of an accumulation, material balance	pseudo-steady state for the accumulation as a whole	one average pressure for the whole volume (General material balance equation for oil reservoirs)
Water influx from the aquifer	usually transient	aquifer models (Aquifer models: water influx)
Wells in tight reservoirs	transient for a long time	solutions for transient inflow, analysis of the rate decline

Terms and symbols

Transient regime: The pressure changes with time in a different way at different points. The disturbance has not yet reached the boundaries, and the reservoir behaves as an infinite one (infinite-acting).
Pseudo-steady-state regime: The pressure in a closed area falls everywhere at the same rate, and the pressure differences are constant.
Steady-state regime: The pressure and the flow rate at every point do not change: fluid entering through the boundary replaces what the well withdraws.
Drawdown cone: The zone of lowered pressure around a working well.
Total compressibility $c_t$: The sum of the fluid compressibilities, weighted by the saturations, and the pore compressibility.
Hydraulic diffusivity $\eta$: $k/(\phi\mu c_t)$ . It sets the speed at which pressure disturbances spread, in m²/s.
Radius of investigation $r_i$: The distance to which the pressure disturbance has clearly advanced in time $t$ , $\sqrt{4\eta t}$ .

Common mistakes and misconceptions

Wrong

“In a transient regime Darcy's law does not work.”

In fact

Darcy's law holds at every point and at every moment: the diffusivity equation is built on it. What does not apply in a transient regime is not the law itself but the Dupuit equation, which was derived on the assumption that the same rate passes through any circle around the well.

Wrong

“Pseudo-steady state is almost steady state: the pressure has stabilised.”

In fact

The pressure keeps falling, and at the same rate at all points. Only the pressure differences are constant. A well in a closed area cannot hold its rate for ever without the pressure falling.

Wrong

“The rock compressibility is small, so we can neglect it.”

In fact

In an undersaturated reservoir the pore compressibility is comparable to the oil compressibility: in the lesson's example it gave 38% of the total compressibility. Without it, the times of the transient processes and the store of elastic energy come out too low.

Wrong

“The test lasted a day, so it covered the whole reservoir.”

In fact

The radius of investigation grows as the square root of time and of the mobility $k/\mu$ . In one day it is about 180 m in a 50 mD reservoir and about 18 m in a 0.5 mD reservoir.

In depth: advanced detail

Derivation of the diffusivity equation

Take linear flow along the $x$ axis through a reservoir with a cross-section area $A$ . The mass of fluid in an element of length $\dd x$ is $\rho\phi A\,\dd x$ . Its change per unit time equals the difference between the mass flow rates entering and leaving the element:

Mass balance in an element of the reservoir

-\frac{\partial (\rho u)}{\partial x} = \frac{\partial (\rho\phi)}{\partial t},\qquad u = -\frac{k}{\mu}\,\frac{\partial p}{\partial x}

where:

$\rho$ density of the fluid, kg/m³
$u$ filtration velocity (the flow rate divided by the cross-section area), m/s
$\phi$ porosity, which depends on pressure through the pore compressibility, —
$k,\ \mu$ permeability and viscosity, m², Pa·s

Units: SI

Holds when

The flow is only along $x$ , with one fluid, and Darcy's law applies without gravity.

The first equality is conservation of mass, and the second is Darcy's law. Together they link the change of pressure in space with the change of the fluid mass in time.

We express the right side through the compressibilities: $\partial\rho/\partial p = \rho c$ for the fluid and $\partial\phi/\partial p = \phi c_f$ for the pores. So the rate of change of $\rho\phi$ equals $\rho\phi\,(c + c_f)$ times $\partial p/\partial t$ . The sum $c + c_f$ is the total compressibility of a single-phase system. If the pores also hold immobile phases, we put $c_t$ in its place. On the left side, the derivative of the product gives a term with $\partial^2 p/\partial x^2$ and a term with the square of the gradient, $c\,(\partial p/\partial x)^2$ . The second one is small when the compressibility is small: the density hardly changes over the whole pressure range, $c\,\Delta p \ll 1$ . Dropping it, and taking $k$ and $\mu$ as constant, we get a linear equation:

Diffusivity equation for linear flow

\frac{\partial^2 p}{\partial x^2} = \frac{\phi\mu c_t}{k}\,\frac{\partial p}{\partial t}

where:

$p(x, t)$ pressure, Pa
$\phi,\ \mu,\ c_t,\ k$ porosity, viscosity, total compressibility, permeability, —, Pa·s, 1/Pa, m²

Units: SI

Holds when

The reservoir is uniform, and $k$ and $\mu$ do not depend on pressure.
The compressibility is small and constant. This holds for liquids, but not for gas.

This equation describes the inflow of water from a linear aquifer, and core experiments in which a pump is started. For radial flow, the same derivation for a ring of radius $r$ and area $2\pi rh$ gives the equation of the main section.

To solve it we need an initial condition (the pressure everywhere equals the initial pressure $p_i$ ) and two boundary conditions. At the outer boundary we set either zero flow (a closed boundary, $\partial p/\partial r = 0$ ) or a constant pressure (fluid is fed in through the boundary). At the well we set a constant rate or a constant bottomhole pressure. The first outer condition leads, after the transient period, to the pseudo-steady-state regime, and the second to the steady-state regime.

Dimensionless variables

It is convenient to write the solutions of the diffusivity equation in dimensionless variables (variables without units). Then one solution fits any reservoir:

Dimensionless radius, time and pressure

r_D = \frac{r}{r_w},\qquad t_D = \frac{k\,t}{\phi\mu c_t r_w^2},\qquad p_D = \frac{2\pi kh\,(p_i - p)}{qB\mu}

where:

$r_D,\ t_D,\ p_D$ dimensionless distance, time and pressure drop, —
$r_w$ radius of the well, m
$p_i$ initial reservoir pressure, Pa
$q,\ B$ rate at standard conditions and formation volume factor, m³/s, m³/m³

Units: SI; in metric oilfield units $t_D = 3.553\cdot10^{-3}\,kt/(\phi\mu c_t r_w^2)$ with $t$ in hours

Holds when

The conditions of the diffusivity equation; a constant rate.

For a well at the centre of a circle of radius $r_e$ , the transient regime lasts up to about $t_D/r_{eD}^2 \approx 1/4$ , where $r_{eD} = r_e/r_w$ . In a closed circle the pseudo-steady-state regime sets in at $t_D r_w^2/A \approx 0.1$ .

Diffusivity equation in dimensionless form

\frac{1}{r_D}\frac{\partial}{\partial r_D}\!\left(r_D\,\frac{\partial p_D}{\partial r_D}\right) = \frac{\partial p_D}{\partial t_D}

where:

$r_D,\ t_D,\ p_D$ dimensionless distance, time and pressure drop, defined above, —

Units: dimensionless

Holds when

The same as for the dimensional equation. At the well, $r_D\,\partial p_D/\partial r_D = -1$ at a constant rate.

The equation contains no reservoir parameter at all. A solution found once is converted to any reservoir through the definitions of the dimensionless variables.

The plots of this lesson were calculated in exactly this way. The dimensionless equation was solved numerically for a circular area with the well at the centre and two kinds of outer boundary. The curve of the infinite reservoir was built from the logarithmic approximation $p_D = \tfrac12(\ln t_D + 0.809)$ . The result was converted to hours and megapascals using the data of the examples. The solutions are analysed in detail in the lessons Diffusivity equation and Line-source solution and the pressure function.

Gas: when the compressibility is not small

The compressibility of gas is of the order of $1/p$ and depends strongly on pressure, and the gas viscosity changes with pressure too. The condition $c\,\Delta p \ll 1$ is not met, so the diffusivity equation for gas is nonlinear. We bring it to an almost linear form by replacing the pressure with the real gas pseudo-pressure (pseudo-pressure). This is a quantity calculated from the pressure that allows for the way the gas properties change with pressure. For large pressure changes we also replace time with pseudo-time, which accounts for the change of $\mu c_t$ . After such a replacement, the gas solutions look the same as the solutions for a liquid, and the idea of the three regimes holds (Steady-state radial flow, Gas formation volume factor, compressibility and viscosity).

Connections to other topics

Rock permeability — Darcy's law is one of the two foundations of the diffusivity equation.
Steady-state radial flow — steady-state inflow: the Dupuit equation and the pressure distribution.
Pseudo-steady-state flow and the shape factor — inflow in a closed area, the average pressure and shape factors.
Well testing: purposes and types of test — how the transient regime is used to test wells.
Well inflow: the well performance diagram — the three inflow regimes in a well model.
Reservoir drive mechanisms — the sources of reservoir energy: an idea that must not be confused with flow regimes.
Black oil model: solution gas–oil ratio and formation volume factors — the compressibility of oil above the bubble point pressure.
Diffusivity equation — the full derivation of the equation and its dimensionless form.

Check yourself

1
A well works at a constant rate. Its bottomhole pressure falls by 0.05 MPa per day, and in an observation well of the same block the pressure also falls by 0.05 MPa per day. Which regime is this?
1. A.Transient: the disturbance has not yet reached the boundaries
2. B.Solution gas drive
3. C.Steady state
4. D.Pseudo-steady state
Show the answer
Correct answer: D. Pseudo-steady state
The pressure is falling, so the regime is not steady state. The rate of fall is the same at different points, which is the sign of the pseudo-steady-state regime of a closed area. “Solution gas drive” is a source of energy of the accumulation, not a flow regime.
2
How does the time for the pressure disturbance to reach the boundary of the area change if the radius of the area is doubled and the permeability is halved?
1. A.It doubles
2. B.It increases 8 times
3. C.It increases 4 times
4. D.It does not change
Show the answer
Correct answer: B. It increases 8 times
$t_b$ is proportional to $r_e^2/k$ . A radius twice as large gives a factor of 4, and a permeability twice as small gives another 2, so 8 times in all.
3
Calculate the total compressibility for $S_o$ = 0.70, $S_w$ = 0.30, $c_o = 1.2\cdot10^{-3}$ , $c_w = 0.45\cdot10^{-3}$ , $c_f = 0.6\cdot10^{-3}$ MPa⁻¹. How does it change if, at a pressure of 15 MPa, 3% of free gas appears in the pores and the oil saturation falls to 0.67?
Show the answer
Without gas: $0.70\cdot1.2 + 0.30\cdot0.45 + 0.6 = 0.84 + 0.135 + 0.6 = 1.575$ , that is $1.575\cdot10^{-3}$ MPa⁻¹. With gas: oil $0.67\cdot1.2 = 0.804$ ; gas $0.03\cdot(1/15) \approx 0.0020$ MPa⁻¹, that is $2.0\cdot10^{-3}$ ; in all about $3.54\cdot10^{-3}$ MPa⁻¹, which is 2.2 times more. Three per cent of gas doubled the elastic store of the reservoir. Below the bubble point the oil also releases gas, and the effective compressibility is higher still.
4
Which plot straightens the record of the bottomhole pressure in the pseudo-steady-state regime?
1. A. $\log_{10} p_{wf}$ against $\log_{10} t$
2. B. $p_{wf}$ against $\sqrt t$
3. C. $p_{wf}$ against $t$ on an ordinary scale
4. D. $p_{wf}$ against $\log_{10} t$
Show the answer
Correct answer: C. $p_{wf}$ against $t$ on an ordinary scale
In the pseudo-steady-state regime the pressure falls at a constant rate $qB/(c_tV_p)$ , that is, linearly with time. A straight line on $p_{wf}$ against $\log_{10} t$ is typical of transient radial inflow, and on $p_{wf}$ against $\sqrt t$ of linear inflow.
5
Why can the inflow to a producing well be taken as roughly steady-state when water is injected into the surrounding injection wells, but not in a closed block without injection?
Show the answer
The steady-state regime requires that as much fluid enters through the boundary of the area as the well withdraws. Then the pressure at every point is constant. Injection around the edge of the area makes up the withdrawal. In a closed block that gets no fluid from outside, the withdrawal is covered only by the expansion of the rocks and fluids, and for that the pressure must keep falling. This is the pseudo-steady-state regime.

Updated: 1 October 2026