Well inflow: the well performance diagram

Quick refresher

The inflow model links three quantities: the flow rate $q$ , the bottomhole pressure $p_{wf}$ and the average reservoir pressure $\bar p$ . The drawdown $\bar p - p_{wf}$ (the pressure difference between the reservoir and the well) is the “driving force” of inflow.
The productivity index $J$ is the flow rate per unit of drawdown. When only one fluid flows, $J$ is constant, so the inflow line on the pressure–flow rate plot is a straight line with slope $-1/J$ .
Between the reservoir and the separator, the pressure is used up in two stages, one after the other: in the reservoir (inflow) and in the lift system (lift). The well works where the inflow line crosses the lift curve.
The flow squeezes in towards the wellbore, so a large part of the drawdown is used up in the first few metres around the well. This is why the near-wellbore zone and the skin factor (a number that shows how much the zone around the well helps or hinders flow) matter so much.
The lift system needs almost the same pressure at any flow rate. So when the reservoir pressure falls, the flow rate of a flowing well (one that flows with no pump) falls much faster than the pressure.
The number of production wells in a project is the planned production divided by the flow rate of one well. So well productivity shapes the whole project.

Key relationships

J = \dfrac{q}{\bar p - p_{wf}}

productivity index

p_{wf} = \bar p - \dfrac{q}{J}

inflow line for single-phase inflow

p_{wf} = p_{wh} + \rho g H + K q^2

lift curve for liquid without free gas

\textstyle\sum_j q_{w,j} B_w + q_{aq} = q_o B_o

injection balance under pressure maintenance (reservoir volumes)

Common pitfalls

Mixing up the initial pressure, the average reservoir pressure and the pressure at the outer boundary. Inflow equations written with different pressures have different constants.
Thinking that the reservoir alone sets the flow rate. The flow rate is set where inflow and lift cross.
Expecting a 15% fall in reservoir pressure to cut the flow rate of a flowing well by 15% too.
Comparing productivity indices that come from unstabilised measurements (taken before the flow settles) or from different “reservoir” pressures.

After this lesson you can

name the three quantities that the inflow model links, and tell apart the initial, average reservoir, bottomhole and outer-boundary pressures
explain why the flow near a well can be treated as radial (straight towards the well from all sides) and single-phase (one fluid only), and name the cases where this assumption fails
define the drawdown and the productivity index, and use them to draw the inflow line
find the operating point of a well on the performance diagram, where the inflow line crosses the lift curve, and explain how a fall in reservoir pressure moves it
estimate, from the flow rate of one well, how many production and injection wells the planned production needs

The inflow model: three quantities

A well produces because the pressure in its wellbore is lower than in the reservoir. Fluid moves from the region of high pressure to the region of low pressure, and it flows in towards the wellbore from all sides. To calculate how much will arrive, we link three quantities. This link is the well inflow model:

the flow rate $q$ — usually given at standard conditions (the fixed pressure and temperature at the surface that serve as a reference), in m³/d
the bottomhole flowing pressure $p_{wf}$ — the pressure in the working well opposite the reservoir, converted to the depth of the middle of the perforated interval (the section of casing with holes that let the fluid in)
the reservoir pressure — most often the average pressure $\bar p$ in the part of the reservoir that feeds the well, which is its drainage area

The difference $\bar p - p_{wf}$ is called the drawdown: it is the “driving force” of inflow. The flow rate divided by the drawdown is the productivity index (PI). It shows how many cubic metres per day each megapascal of drawdown gives.

Productivity index

J = \frac{q}{\bar p - p_{wf}}

where:

$J$ productivity index, m³/(d·MPa)
$q$ flow rate at standard conditions, m³/d
$\bar p$ average reservoir pressure in the drainage area, MPa
$p_{wf}$ bottomhole flowing pressure of the producing well, MPa

Units: metric oilfield units (m³/d, MPa); in oilfield units, stock-tank barrels per day per psi: 1 STB/(d·psi) = 23.06 m³/(d·MPa)

Holds when

Stabilised flow: the flow rate and the pressures do not change while they are measured.
Single-phase inflow: only one fluid flows. In an oil well the bottomhole flowing pressure is above the bubble point pressure (the pressure below which gas comes out of the oil).
Both pressures are converted to the same depth. For the difference, you can use absolute or gauge pressures (gauge pressure is measured above atmospheric pressure), as long as both are of the same kind.

When only one fluid flows, $J$ is constant: double the drawdown and the flow rate doubles. $J$ is proportional to the transmissibility of the reservoir $kh/\mu$ . Here $k$ is the permeability (how easily the rock lets fluid through), $h$ is the thickness and $\mu$ is the viscosity (how thick the fluid is). $J$ falls when the near-wellbore zone is damaged. The lesson Steady-state radial flow shows how to calculate $J$ from reservoir properties.

There is a trap here: several different pressures are called “reservoir pressure”. Inflow equations written with different pressures have different constants, so always make clear which pressure you mean.

Pressures that appear in inflow calculations

Pressure: Initial reservoir
Symbol: $p_i$
What it is: pressure in the reservoir before development begins
How it is found: wireline formation tester (a tool on a cable that measures the pressure in the rock), or the first pressure build-up (the pressure rise after shutting in) in a new well

Pressure: Average reservoir
Symbol: $\bar p$
What it is: pressure averaged over the volume of the drainage area
How it is found: from the pressure build-up curve with corrections, or from the material balance (a balance between the fluid produced and the fall in pressure)

Pressure: At the outer boundary
Symbol: $p_e$
What it is: pressure at the outer boundary of the drainage area
How it is found: mostly used in calculations; in waterflooding (injecting water to push the oil out) it is close to the pressure midway between the production well and the injection wells

Pressure: Bottomhole flowing
Symbol: $p_{wf}$
What it is: pressure in the working well opposite the reservoir
How it is found: a gauge lowered into the well, a sensor at the pump intake, or a conversion from the wellhead pressure

Pressure: Static
Symbol: $p_{ws}$
What it is: pressure in the shut-in well
How it is found: measured after the well is shut in; it rises with time and approaches the reservoir pressure

Pressure	Symbol	What it is	How it is found
Initial reservoir	$p_i$	pressure in the reservoir before development begins	wireline formation tester (a tool on a cable that measures the pressure in the rock), or the first pressure build-up (the pressure rise after shutting in) in a new well
Average reservoir	$\bar p$	pressure averaged over the volume of the drainage area	from the pressure build-up curve with corrections, or from the material balance (a balance between the fluid produced and the fall in pressure)
At the outer boundary	$p_e$	pressure at the outer boundary of the drainage area	mostly used in calculations; in waterflooding (injecting water to push the oil out) it is close to the pressure midway between the production well and the injection wells
Bottomhole flowing	$p_{wf}$	pressure in the working well opposite the reservoir	a gauge lowered into the well, a sensor at the pump intake, or a conversion from the wellhead pressure
Static	$p_{ws}$	pressure in the shut-in well	measured after the well is shut in; it rises with time and approaches the reservoir pressure

We convert all pressures to one depth: usually the middle of the perforated interval, or the reference depth chosen for the reservoir.

Why inflow to a well is treated as radial

Flow in a reservoir as a whole is complex. It is three-dimensional and often multiphase (several fluids move at once). It is influenced not only by pressure differences but also by gravity and by capillary forces (the forces that act where two fluids meet in tiny pores). Such problems are solved numerically, on reservoir simulation models (computer models of the reservoir). Near an individual well the picture often becomes as simple as it can be: only oil moves, along radii towards the axis of the wellbore, and only the pressure difference drives it.

This holds as long as four conditions are met. The pressure everywhere is above the bubble point pressure, and there is no free gas in the reservoir. Water or gas has not yet reached the well. The connate water (the water that stays in the pores of the rock) does not move. The well penetrates the full thickness of the reservoir. Then the oil flows to the wellbore horizontally and from all sides, which is called radial flow.

Each production well collects fluid from its own area, called the drainage area. The neighbouring production and injection wells, the faults (breaks in the rock layers) and the edges of the reservoir set its size and shape. Flow divides (lines where the flow splits between wells) run between neighbouring production wells. Fluid does not cross them, so for each well they act as a no-flow boundary.

SchematicDrainage areas of production wells (schematic, not to scale). Flow divides run between neighbouring wells, and on the right a fault bounds the block. Near each well the flow converges radially. The circle of radius

r_e

is an imaginary round area of the same size, which inflow equations use instead of the real drainage area.

The main feature of radial inflow is that the flow converges. The whole flow rate passes through every circle around the well, but the area the fluid flows through, $2\pi rh$ , gets smaller towards the wellbore. So the Darcy velocity (the speed of the flow through the rock) and the pressure gradient (how fast the pressure changes with distance) are greatest at the wall of the well. A large part of the drawdown is therefore used up in the first few metres around the wellbore. For example, take a well of radius 0.1 m in a circular area of radius 300 m. Almost a third of the pressure difference between the outer boundary and the bottomhole is used up in the first metre around the wellbore, and more than half within 10 m. This is why the near-wellbore zone plays a special role (Skin factor). The lesson Steady-state radial flow analyses the pressure distribution in numbers.

The path of the fluid: where the pressure is spent

From the reservoir to the separator, the pressure falls from $\bar p$ to the separator pressure $p_{sep}$ . This drop is the whole pressure “budget” of a flowing well (a well that produces on its own, with no pump). The budget is used up along the way like this:

in the reservoir and the near-wellbore zone — the fluid needs pressure to flow through the rock to the wellbore. It also needs pressure to get through the damaged zone and the perforation tunnels (the channels shot through the casing into the rock). These losses are proportional to the flow rate, and they form the inflow line.
in the lift system — the fluid column in the tubing (the pipe inside the well that carries the fluid up) has a weight, and the pressure must hold it up. This weight hardly depends on the flow rate. Pressure is also lost to friction, which grows roughly as the square of the flow rate. Together these losses form the lift curve.
at the surface — at the choke (a valve with a small opening that limits the flow), in the flowline (the pipe from the well to the plant) and in the separation plant. When we calculate at the bottomhole, we include these losses through the wellhead pressure.

To calculate, it helps to split the system into two parts, one after the other: the reservoir, which delivers fluid to the well, and the lift system, which lifts it to the surface. The bottomhole pressure $p_{wf}$ is the dividing line between them. The lower $p_{wf}$ is, the more the reservoir delivers. The lift system works the other way: a higher flow rate needs a higher $p_{wf}$ . The well settles by itself at the bottomhole pressure where the reservoir delivers exactly as much as the lift system can lift.

The detailed calculation of every element of the system, and the choice of the point where it is split into parts, belong to nodal analysis. To see how a well works, two curves are enough.

The well performance diagram

The well performance diagram is two curves on one plot: flow rate on the horizontal axis, bottomhole pressure on the vertical axis. The first is the inflow performance relationship (the inflow line, IPR). It shows what flow rate the reservoir delivers at each bottomhole pressure. The second is the lift curve (outflow performance curve, VLP). It shows what bottomhole pressure is needed to lift each flow rate to the surface.

Inflow line (single-phase inflow)

p_{wf} = \bar p - \frac{q}{J}

where:

$p_{wf}$ bottomhole pressure at flow rate $q$ , MPa
$\bar p$ average reservoir pressure, MPa
$q$ flow rate at standard conditions, m³/d
$J$ productivity index, m³/(d·MPa)

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

Holds when

Single-phase inflow ( $p_{wf}$ above the bubble point pressure) and stabilised flow.
The productivity index does not depend on the flow rate: there are no inertial losses.

The line starts at the point $q = 0$ , $p_{wf} = \bar p$ . It falls by $1/J$ MPa for each cubic metre per day. The more productive the well, the flatter the line. Below the bubble point pressure the line bends; see Inflow performance relationship (IPR).

Lift curve for liquid without free gas

p_{wf} = p_{wh} + \rho g H + K q^2

where:

$p_{wh}$ wellhead (tubing head) pressure, MPa
$\rho$ density of the liquid in the tubing, kg/m³
$g$ acceleration due to gravity, 9.81, m/s²
$H$ true vertical depth of the middle of the perforated interval, m
$K$ friction loss coefficient in the tubing, MPa/(m³/d)²
$q$ liquid flow rate, m³/d

Units: pressures in MPa; the product $\rho g H$ comes out in pascals and is divided by $10^6$

Holds when

Vertical tubing down to the perforated interval; single-phase liquid without free gas, constant density.
The wellhead pressure is held constant — by the choke or by the pressure in the gathering system (the network of pipes that collects the production).
$K$ depends on the tubing diameter and on the density and viscosity of the liquid. The lesson Nodal analysis of the production system shows how to calculate it.

At zero flow rate the bottomhole pressure only has to hold up the liquid column: $p_{wh} + \rho g H$ . As the flow rate grows, friction adds to it and the curve bends upwards. In gassy oil the column is lighter, so we calculate the curve with multiphase flow correlations, which are formulas fitted to measurements of gas–liquid flow in pipes (Multiphase flow in tubing).

The two curves cross at the operating point of the well. This is the only pair of flow rate and bottomhole pressure at which the reservoir delivers as much as the lift system lifts. The drawdown at this point is the one that the lift system “allows”.

CalculatedCalculated well performance diagram for the data of the example below: inflow lines at average reservoir pressures of 24, 20 and 15 MPa (productivity index 40 m³/(d·MPa)) and the lift curve. A and B are operating points. At 15 MPa the inflow line lies below the lift curve, so the well cannot flow naturally.

Worked exampleOperating point of a flowing well as the reservoir pressure declines

Given (assumed data)

Productivity index $J$ (single-phase inflow): 40 m³/(d·MPa)
Average reservoir pressure $\bar p$: 24 MPa now, 20 MPa in a few years
Wellhead pressure $p_{wh}$: 1.2 MPa
Density of the liquid in the tubing $\rho$ (no free gas): 840 kg/m³
True vertical depth of the middle of the perforated interval $H$: 1800 m
Friction coefficient in the tubing $K$: $1.2\cdot10^{-5}$ MPa/(m³/d)²

Solution

Weight of the liquid column: $\rho g H = 840\cdot 9.81\cdot 1800$ Pa $= 14.83$ MPa. Even at zero flow rate the bottomhole needs a pressure of $p_{wh} + \rho g H = 16.03$ MPa.
Operating point at $\bar p$ = 24 MPa: set $24 - q/40$ equal to $16.03 + Kq^2$ . We solve the equation $Kq^2 + q/40 - 7.97 = 0$ and take the positive solution: $q_{\text{A}} = 281$ m³/d, $p_{wf} = 16.98$ MPa.
Where the 24 MPa goes: drawdown 7.02 MPa (29%), weight of the column 14.83 MPa (62%), friction 0.95 MPa (4%), wellhead pressure 1.2 MPa (5%).
At $\bar p$ = 20 MPa the same equation gives $q_{\text{B}} = 148$ m³/d, $p_{wf} = 16.30$ MPa. The reservoir pressure has fallen by 17%, while the flow rate has fallen by 47%.
Natural flow stops when the reservoir pressure falls to $p_{wh} + \rho g H = 16.03$ MPa. Below this value the reservoir cannot hold up the liquid column, even at zero flow rate.

Answer: Now the well delivers about 281 m³/d. After the reservoir pressure falls to 20 MPa, it delivers about 148 m³/d. Below 16.03 MPa it cannot flow naturally.

What it means. The lift system “takes” about 16.03 MPa at any flow rate. Only the difference $\bar p - 16.03$ is left for the drawdown: about 8.0 MPa at 24 MPa, and about 4.0 MPa at 20 MPa, which is half as much. This is why the flow rate of a flowing well falls much faster than the reservoir pressure. Injecting water to maintain the pressure helps natural flow last longer. When this is not enough, the well is switched to artificial lift (a pump or gas lift), which lowers the lift curve. The data in this example are assumed for illustration.

Any change in either curve moves the operating point. As the reservoir is depleted, the inflow line moves down parallel to itself, if $J$ does not change. Damage to the near-wellbore zone turns the line about the point $\bar p$ and makes it steeper. Stimulation (treating the zone to improve flow) or hydraulic fracturing (breaking the rock with high-pressure fluid) turns it the other way and makes it flatter. A lower wellhead pressure, a pump or gas lift lowers the lift curve. A rising water cut (the share of water in the produced liquid) raises it, because the column becomes heavier. The lesson Nodal analysis of the production system gives the full list, with calculations.

From the flow rate of a well to a development plan

The productivity of a well decides not only how much that well produces, but also the shape of the whole project. A large field, especially an offshore one, is usually brought to a plateau (a period of constant production) as fast as possible. It is then held there until water reaches the wells. The plateau level is often set at about 10% of the recoverable reserves (the oil that can actually be produced) per year. This is a guideline, not a rule: each project chooses its level on economic grounds.

We find the number of production wells by dividing the planned production by the flow rate of one well. So an error of a factor of two in productivity means twice as many wells, or half as many. Offshore, it also means a different number and size of platforms. When wells are expensive, a project makes sense only if one well produces more than some minimum profitable flow rate. This minimum depends on the volume of reserves and the oil price, on the water depth, on the distance to existing fields and pipelines, and on the cost of transport. So it is calculated separately for each project.

To hold the plateau, the reservoir pressure is usually maintained by injecting water. The injection wells must replace the volume that the production wells take out. We make these balances at reservoir conditions, that is, in the volumes that the fluids occupy in the reservoir.

Injection balance under reservoir pressure maintenance

\sum_{j} q_{w,j} B_w + q_{aq} = q_o B_o

where:

$q_{w,j}$ water injection rate in the $j$ -th injection well at standard conditions, m³/d
$B_w$ formation volume factor of the injected water (its volume in the reservoir for each m³ at the surface), close to 1, m³/m³
$q_{aq}$ water influx from the aquifer beyond the oil–water contour (the water-bearing rock outside the oil zone), at reservoir conditions, m³/d
$q_o$ total oil production rate at standard conditions, m³/d
$B_o$ oil formation volume factor, m³/m³

Units: any identical flow-rate units, for example m³/d

Holds when

We consider the period before water breaks through to the production wells. The pressure is above the bubble point pressure, so only oil with dissolved gas is produced.
The reservoir pressure is held constant: the withdrawal is fully replaced.

The injection wells and the edge water (water from the aquifer around the oil zone) together replace the reservoir volume of the oil produced. We cannot find the number of injection wells without estimating the water influx, and this is one of the most uncertain quantities in field development. The ratio of the volume injected to the volume withdrawn, both at reservoir conditions, is called voidage replacement (voidage replacement ratio, VRR). Russian-language practice gives it as a percentage.

Worked exampleHow many wells the plateau needs

Given (assumed data)

Recoverable oil reserves; stock-tank oil density: 20 million t; 850 kg/m³
Plateau level: 10% of recoverable reserves per year
Well productivity index $J$: 30 m³/(d·MPa)
Average reservoir pressure $\bar p$ (maintained by injection): 25 MPa
Bottomhole pressure required by the lift system at the working flow rate: 18 MPa
Oil formation volume factor $B_o$: 1.25 m³/m³
Water influx from the aquifer (estimate); injectivity of an injection well: 1500 m³/d; 700 m³/d

Solution

Reserves as a volume: 20 million t is $20\cdot10^{9}$ kg, and $20\cdot10^{9}/850 = 23.53\cdot10^{6}$ m³. The plateau is 10% a year: 2.35 million m³ a year, or 6446 m³/d.
Flow rate of one well: $q = J(\bar p - p_{wf}) = 30\cdot(25 - 18) = 210$ m³/d.
Number of production wells: $6446/210 = 30.7$ , rounded up to 31.
Suppose that opening up the reservoir with less damage raises $J$ to 45 m³/(d·MPa). Then the flow rate of a well is 315 m³/d, and the number of wells falls to 21.
Underground withdrawal (the produced volume at reservoir conditions): $q_o B_o = 6446\cdot 1.25 = 8058$ m³/d. We subtract the water influx, so the injection we need is $8058 - 1500 = 6558$ m³/d (we take the water formation volume factor as 1).
Number of injection wells: $6558/700 = 9.4$ , rounded up to 10.

Answer: For a plateau of about 6446 m³/d, the number of production wells is 31 (21 at

J

= 45) and the number of injection wells is 10.

What it means. A productivity 1.5 times higher saves about a third of the production wells. This is the main argument for opening the reservoir with less damage and for good-quality well completion (the equipment and work that make a drilled well ready to produce). The number of injection wells does not change: the underground withdrawal and the uncertain estimate of the edge water decide it. The calculation is simplified. In practice, engineers also allow for these things: well downtime; the higher friction in the tubing at a higher flow rate; the difference between wells in the centre and at the edge of the reservoir; and the fall in flow rate after water breakthrough. The data in this example are assumed for illustration.

The three inflow regimes

The relationship between flow rate and pressures depends on two things: what happens at the outer boundary of the drainage area, and how much time has passed since the last change in the well's operating conditions. We distinguish three flow regimes.

Inflow regimes

Regime: Transient
What happens at the boundary of the area: the disturbance has not yet reached the boundary: the reservoir behaves as if it were infinite
How the pressure changes: at the well it falls linearly with the logarithm of time
Where it is used: well testing: Well testing: purposes and types of test

Regime: Steady state
What happens at the boundary of the area: as much fluid enters across the boundary as the well takes out: injection, an aquifer
How the pressure changes: does not change with time
Where it is used: inflow under pressure maintenance: Steady-state radial flow

Regime: Pseudo-steady state
What happens at the boundary of the area: the boundary is no-flow, and there is no recharge
How the pressure changes: falls at the same rate everywhere
Where it is used: depletion of a closed area: Pseudo-steady-state flow and the shape factor

Regime	What happens at the boundary of the area	How the pressure changes	Where it is used
Transient	the disturbance has not yet reached the boundary: the reservoir behaves as if it were infinite	at the well it falls linearly with the logarithm of time	well testing: Well testing: purposes and types of test
Steady state	as much fluid enters across the boundary as the well takes out: injection, an aquifer	does not change with time	inflow under pressure maintenance: Steady-state radial flow
Pseudo-steady state	the boundary is no-flow, and there is no recharge	falls at the same rate everywhere	depletion of a closed area: Pseudo-steady-state flow and the shape factor

After a well is started up or its flow rate is changed, the inflow is at first transient. When the pressure disturbance reaches the boundaries, the inflow becomes steady-state or pseudo-steady-state, depending on whether there is recharge (a supply of fluid or pressure from outside). In all three regimes the pressure near the well varies in the same way, with the logarithm of the distance. So the productivity index and the skin factor work in each regime. Only the logarithmic term of the inflow equation changes, and in the transient regime this term also grows with time.

Terms and symbols

Inflow model: The inflow model links the flow rate, the bottomhole pressure and the reservoir pressure of a given well.
Drawdown: The difference between the average reservoir pressure and the bottomhole pressure, $\bar p - p_{wf}$ . It is the “driving force” of inflow.
Productivity index $J$: The flow rate per unit of drawdown, in m³/(d·MPa). It is constant in single-phase inflow.
Drainage area: The part of the reservoir from which fluid flows to a given well. In equations it is replaced by a circle of radius $r_e$ .
Lift curve: The bottomhole pressure needed to lift a given flow rate to the surface at a given wellhead pressure, drawn as a curve.
Operating point: The point where the inflow line crosses the lift curve. It gives the actual flow rate and bottomhole pressure of the well.
Production plateau: The period of constant production of a field after it reaches the design level.
Voidage replacement: The ratio of the volume of injected water to the volume of fluids withdrawn, at reservoir conditions.

Common mistakes and misconceptions

Wrong

“The reservoir decides the flow rate of a well: the higher the productivity, the higher the flow rate.”

In fact

The flow rate is set where inflow and lift cross. If the lift system uses up most of the pressure, higher productivity gives a smaller gain in flow rate than you expect; see Nodal analysis of the production system.

Wrong

“The reservoir pressure has fallen by 17%, so the flow rate will fall by 17%.”

In fact

In a flowing well, the lift system takes almost the same pressure at any flow rate. The pressure left for the drawdown therefore shrinks by a much larger share than the reservoir pressure does. In this lesson's example the flow rate fell by almost half.

Wrong

“The reservoir pressure is one definite number.”

In fact

There are the initial pressure, the average over the drainage area, the pressure at the outer boundary and the static pressure. Inflow equations written with different pressures have different constants, so always say which pressure you mean.

Wrong

“A stabilised measurement of flow rate and pressure gives the permeability of the reservoir.”

In fact

Such a measurement gives only the productivity index, in which permeability and skin cannot be told apart. Well testing separates them by studying the transient response, which is how the pressure changes with time (Well testing: purposes and types of test).

In depth: advanced detail

Underground withdrawal in the general case

The balance in the main section holds only while nothing but oil with dissolved gas is produced. After water breakthrough, and at pressures below the bubble point pressure, the underground withdrawal (the produced volume at reservoir conditions) also includes produced water and free gas. Free gas is gas in excess of what was dissolved in the oil.

Underground withdrawal of oil, water and free gas

V = q_o B_o + q_w B_w + q_o\,(R_p - R_s)\,B_g

where:

$V$ underground withdrawal (withdrawal at reservoir conditions), m³/d
$q_o,\ q_w$ oil and water flow rates at standard conditions, m³/d
$B_o,\ B_w,\ B_g$ formation volume factors of oil, water and gas at reservoir pressure, m³/m³
$R_p$ producing gas–oil ratio: the gas produced per cubic metre of oil, at standard conditions, m³/m³
$R_s$ solution gas–oil ratio (the gas dissolved in each cubic metre of oil) at the current reservoir pressure, m³/m³

Units: m³/d; the producing and solution gas–oil ratios are in m³ of gas per m³ of oil at standard conditions

Holds when

The flow rates are at standard conditions; the formation volume factors and the solution gas–oil ratio are at the current reservoir pressure.
Free gas is the gas in excess of the dissolved gas, $(R_p - R_s)\,q_o$ at standard conditions.

The free-gas term grows quickly when the pressure falls below the bubble point pressure, because gas occupies a large volume at reservoir pressure. So it is usually more profitable to keep the pressure above the bubble point pressure than to replace, later, the huge reservoir volume of the gas that has come out of the oil. The lesson Black oil model: solution gas–oil ratio and formation volume factors covers formation volume factors and solution gas–oil ratio.

Connections to other topics

Nodal analysis of the production system — detailed calculation of all elements of the system from the reservoir to the separator.
Inflow performance relationship (IPR) — how to draw the inflow line when it is not straight.
Steady-state radial flow — the inflow equation written with reservoir properties, and the pressure distribution around the well.
Skin factor — why the near-wellbore zone has such a strong effect on productivity.
Well testing: purposes and types of test — how to split the productivity index into permeability and skin.
Pseudo-steady-state flow and the shape factor — inflow during the depletion of a closed area.
Black oil model: solution gas–oil ratio and formation volume factors — formation volume factors for converting flow rates to reservoir conditions.

Check yourself

1
At an average reservoir pressure of 22 MPa, a well delivers 90 m³/d at a bottomhole pressure of 18 MPa. The inflow is single-phase. What flow rate will it deliver at a bottomhole pressure of 16 MPa?
1. A.160 m³/d
2. B.It cannot be found without the permeability of the reservoir
3. C.120 m³/d
4. D.135 m³/d
Show the answer
Correct answer: D. 135 m³/d
$J = 90/(22 - 18) = 22.5$ m³/(d·MPa). At a drawdown of 6 MPa the flow rate is $22.5\cdot 6 = 135$ m³/d. In single-phase inflow the inflow line is straight, so we do not need to know the permeability.
2
What on the well performance diagram sets the actual flow rate of the well?
1. A.The intersection of the inflow line and the lift curve
2. B.The slope of the lift curve
3. C.The reservoir pressure alone
4. D.The point where the inflow line crosses the flow rate axis
Show the answer
Correct answer: A. The intersection of the inflow line and the lift curve
Where the inflow line crosses the flow rate axis, the flow rate is the one at zero bottomhole pressure, and a real well can never reach it. The well really works where the reservoir delivers as much as the lift system lifts, at the same bottomhole pressure.
3
In the example of this lesson, the reservoir pressure falls from 24 to 20 MPa. Why does the flow rate of the flowing well fall by almost half, although its productivity index has not changed?
Show the answer
The lift system needs at least $p_{wh} + \rho g H \approx 16$ MPa at any flow rate. What is left for the drawdown is the reservoir pressure minus this value: about 8 MPa at 24 MPa and about 4 MPa at 20 MPa. The flow rate is proportional to the drawdown (with a small correction for friction), so it falls by almost half.
4
The planned production of a field is 5000 m³/d of oil, and one well delivers 180 m³/d. How many production wells are needed? How will the answer change if the flow rate of a well can be raised to 250 m³/d?
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$5000/180 \approx 27.8$ , so 28 wells are needed. With 250 m³/d, $5000/250 = 20$ wells. A rise in flow rate of almost 40% saves 8 wells. This is why the economics of the whole project depend on well productivity.
5
Which assumption of the single-phase radial inflow model fails first if the bottomhole pressure is lowered below the bubble point pressure?
1. A.The flow being horizontal
2. B.Immobile connate water
3. C.Single-phase flow: free gas comes out of the oil near the well
4. D.Constant reservoir thickness
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Correct answer: C. Single-phase flow: free gas comes out of the oil near the well
Gas starts to come out of the oil where the pressure is lowest, which is right at the well. Free gas lowers the permeability to oil, and the inflow line bends: the productivity index stops being constant.
6
A reservoir is developed with full pressure maintenance. Oil production is 4000 m³/d, with $B_o$ = 1.3. Water has not broken through, and no edge water comes in. How much water must be injected?
1. A.3080 m³/d
2. B.5200 m³/d
3. C.4000 m³/d
4. D.As much as the liquid produced at the surface
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Correct answer: B. 5200 m³/d
We balance the volumes at reservoir conditions: $q_o B_o = 4000\cdot 1.3 = 5200$ m³/d. The water formation volume factor is about 1, so we must inject the same amount of water at standard conditions.

Updated: 30 September 2026