Inflow performance relationship (IPR)

Quick refresher

As long as the bottomhole pressure is above the bubble point pressure, oil or water flows in as a single-phase liquid (no free gas), and the IPR is a straight line: $q = J(\bar p_R - p_{wf})$ .
The productivity index $J$ is proportional to the transmissibility $kh/\mu$ (permeability times thickness, divided by viscosity). It is inversely proportional to the formation volume factor (the ratio of the volume of oil in the reservoir to its volume at the surface). It falls as the skin factor rises. The skin factor is a number that shows how much the zone right around the well helps or hinders flow.
Below the bubble point pressure, gas comes out of solution near the well. It becomes harder and harder for the oil to move. Each further drop in bottomhole pressure gives a smaller and smaller increase in rate, so the curve bends. Vogel's equation describes it.
If the reservoir pressure is above the bubble point pressure, the curve is composite. It is a straight line down to $p_b$ and a Vogel-type curve below it, joined without a kink.
With tests at several rates, Fetkovich's method finds the shape of the curve: $q = C(\bar p_R^2 - p_{wf}^2)^n$ . An exponent $n < 1$ points to extra losses at high velocities. In formulas with squares of pressure, the pressures are absolute (measured from zero pressure).

Key relationships

q = J\,(\bar p_R - p_{wf})

single-phase inflow, $p_{wf} \ge p_b$

\dfrac{q}{q_{max}} = 1 - 0.2\,\dfrac{p_{wf}}{\bar p_R} - 0.8\left(\dfrac{p_{wf}}{\bar p_R}\right)^2

Vogel's equation, $\bar p_R \le p_b$

q = J(\bar p_R - p_b) + \dfrac{J p_b}{1.8}\left[1 - 0.2\,\dfrac{p_{wf}}{p_b} - 0.8\left(\dfrac{p_{wf}}{p_b}\right)^2\right]

composite curve below the bubble point pressure, $p_{wf} < p_b < \bar p_R$

q = C\,(\bar p_R^2 - p_{wf}^2)^n

Fetkovich's method; pressures are absolute

Common pitfalls

Extending a straight-line IPR below the bubble point pressure. The rate will come out too high.
Comparing productivity indices measured at different drawdowns for a well with a curved IPR. Its $J$ depends on the bottomhole pressure.
Drawing the curve from measurements taken before the flow settled. Until the flow has stabilised, the point does not lie on the IPR.
Mixing up the axes of Fetkovich's log–log plot (both axes on logarithmic scales). The slope equals $n$ or $1/n$ , depending on what is plotted horizontally.

After this lesson you can

calculate the productivity index (the rate per unit of pressure drop) from a well test and from reservoir properties, and explain what it depends on
explain why the IPR bends below the bubble point pressure (the pressure at which gas starts to come out of the oil), and draw it with Vogel's equation
build a composite IPR (a straight line joined to a curve) when the reservoir pressure is above the bubble point pressure, including from a test made below the bubble point pressure
analyse a multi-rate test (a test at several flow rates) with Fetkovich's method: find the exponent $n$ , the coefficient $C$ and the absolute open flow (the rate at zero bottomhole pressure)
explain how the IPR changes when the reservoir is depleted, when the near-wellbore zone is damaged, and when several layers produce together

What the IPR shows

When a well operates in steady state (nothing changes with time), the reservoir delivers a rate $q$ to it at a bottomhole pressure $p_{wf}$ . Change the operating mode, for example with a different choke (a narrow opening that limits the flow), a different pump speed or a different wellhead pressure. After some time a different pair of rate and bottomhole pressure settles in. All such pairs at the same reservoir pressure lie on one curve: the IPR.

Two points on the curve are special. At $p_{wf} = \bar p_R$ there is no pressure difference, and the rate is zero. This is how engineers measure the reservoir pressure: they shut the well in and wait for the pressure to build up. At $p_{wf} = 0$ the rate is greatest. It is called the absolute open flow (AOF) and written $q_{max}$ . This mode can never be reached, because the weight of the fluid column always leaves some pressure at the bottomhole. But $q_{max}$ is a handy measure of the well's potential.

By convention, the rate is plotted along the horizontal axis and the bottomhole pressure along the vertical axis. Russian-language practice also uses an indicator diagram, which plots the rate against the drawdown $\Delta p = \bar p_R - p_{wf}$ (the pressure drop that pushes fluid into the well). It is the same information in other coordinates.

To get the curve, engineers test the well at several rates. They measure the bottomhole pressure with a downhole gauge, or calculate it from the wellhead pressure. Later they refine the curve during reservoir surveillance (regular monitoring of the field). Each measurement is made after the flow has stabilised. In the first hours and days after a change of rate the pressure is still changing, and the point does not fall on the curve (more in the lesson Well testing: purposes and types of test).

Single-phase inflow: the productivity index

If the bottomhole pressure is above the bubble point pressure, only liquid flows in the reservoir: oil without free gas, or water. By Darcy's law (the basic law of flow through porous rock), the inflow is proportional to the pressure difference. So the IPR is a straight line. The slope of the line defines the productivity index (PI): the rate per unit of drawdown.

Productivity index

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

where:

$J$ productivity index, m³/(d·MPa)
$q$ oil (liquid) rate at standard conditions, m³/d
$\bar p_R$ average reservoir pressure, MPa
$p_{wf}$ bottomhole pressure while the well is flowing, MPa

Units: metric oilfield units (m³/d, MPa)

Holds when

Stabilised flow.
Single-phase inflow: $p_{wf} \ge p_b$ .
The pressures are converted to the same depth. For a difference, absolute pressure (measured from zero) and gauge pressure (measured above atmospheric pressure) work equally well, as long as both are of the same kind.

For a straight-line IPR, $J$ is constant. One measurement together with the reservoir pressure is then enough to draw the whole curve: $q = J(\bar p_R - p_{wf})$ . The line meets the rate axis at $J\bar p_R$ . This is the absolute open flow only for a liquid without dissolved gas, for example water. For oil, the curve bends earlier, below the bubble point pressure.

In other unit systems (1)

Oilfield units

J\,[\text{m}^3/(\text{d·MPa})] = 23.06\;J\,[\text{STB/(d·psi)}]

$q$ in barrels per day at standard conditions (STB/d), pressures in psi.

$23.06 = 0.158987/0.00689476$ : a barrel in cubic metres divided by a psi in megapascals.

To compare wells that penetrate different thicknesses of reservoir, use the specific productivity index $J/h$ , which is $J$ per metre of net thickness (the thickness of rock that can produce). It shows whether the wells differ in reservoir properties or only in thickness.

The radial inflow formula shows where the straight line comes from and what its slope depends on. (The lesson Rock permeability covers Darcy's law for radial flow, that is, flow towards a well from all sides.) Suppose the well has operated long enough for the effect of its start-up to reach the boundaries of the drainage area. The pressure then falls at the same rate everywhere in the area. This is pseudo-steady-state flow (Pseudo-steady-state flow and the shape factor). The rate can then be written neatly in terms of the average reservoir pressure.

Inflow to a well in pseudo-steady state

q = \frac{2\pi k h\,(\bar p_R - p_{wf})}{\mu B\left[\ln\dfrac{r_e}{r_w} - \dfrac{3}{4} + S\right]}

where:

$q$ oil rate at standard conditions, m³/s
$k$ permeability of the reservoir to oil, m²
$h$ net thickness of the reservoir, m
$\mu$ viscosity of the oil at reservoir conditions, Pa·s
$B$ formation volume factor of the oil, m³/m³
$r_e$ radius of the circular drainage area, m
$r_w$ wellbore radius, m
$S$ skin factor — the additional resistance of the near-wellbore zone (see Skin factor), —

Units: SI

Holds when

Pseudo-steady state in a circular drainage area with a closed (impermeable) outer boundary and the well at the centre. For an area of another shape, the term $\ln(r_e/r_w) - 3/4$ is replaced by an expression with a shape factor.
Single-phase inflow of a slightly compressible liquid, by Darcy's law.
A uniform reservoir (the same properties everywhere). Imperfect penetration (the well does not open the whole pay thickness) and damage to the near-wellbore zone are included in $S$ .

So $J = 2\pi k h/\left(\mu B[\ln(r_e/r_w) - 3/4 + S]\right)$ . The productivity index is proportional to the transmissibility $kh/\mu$ . It is inversely proportional to the sum of the “logarithmic” resistance of the reservoir and the skin factor. The term $-3/4$ appears because the formula uses the average pressure of the area, not the pressure at its boundary.

In other unit systems (2)

Metric oilfield units

J = \frac{0.536\,k h}{\mu B\left[\ln(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.

$0.536 = 2\pi\cdot 0.0853$ — the same constant as in the radial inflow formula. It converts mD → m², MPa → Pa, mPa·s → Pa·s and seconds to days.

Oilfield units

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

$J$ in STB/(d·psi), $k$ in mD, $h$ in ft, $\mu$ in cP, $B$ in rb/STB.

$7.08\cdot10^{-3} = 2\pi\cdot 1.127\cdot10^{-3}$ .

Worked exampleA straight-line IPR from a single test

Given (assumed data)

Average reservoir pressure $\bar p_R$: 18.0 MPa
Stabilised rate: $q_1 = 96$ m³/d at $p_{wf,1} = 14.0$ MPa
Bubble point pressure $p_b$: 9.0 MPa
Net thickness of the reservoir $h$: 12 m

Solution

Productivity index: $J = \dfrac{96}{18.0 - 14.0} = 24.0$ m³/(d·MPa), or $24.0/23.06 = 1.04$ STB/(d·psi).
Specific productivity index: $J/h = 2.0$ m³/(d·MPa·m).
Rate at $p_{wf} = 10.0$ MPa (still above $p_b$ ): $q = 24.0\cdot(18.0 - 10.0) = 192$ m³/d.
Below 9.0 MPa the formula does not apply: two-phase inflow (oil and gas together) begins, and the rate will be less than on the straight line. If we extend the line to zero pressure, it gives $J\bar p_R = 432$ m³/d. That is too high an estimate of the potential.

Answer:

J = 24.0

m³/(d·MPa). At a bottomhole pressure of 10 MPa the well will deliver about 192 m³/d.

What it means. One stabilised measurement and the reservoir pressure are enough to draw a straight line. But the line holds only while the inflow is single-phase. Before you use the line, check the bubble point pressure against the PVT data (laboratory data on the properties of the oil). The data in this example are assumed for illustration.

Worked exampleProductivity index from reservoir properties, and the flow efficiency of the well

Given (assumed data)

Permeability to oil $k$ ; thickness $h$: 80 mD; 12 m
Oil viscosity $\mu$ ; formation volume factor $B$: 1.5 mPa·s; 1.20 m³/m³
Drainage radius $r_e$ ; wellbore radius $r_w$: 300 m; 0.10 m
Actual productivity index (previous example): 24.0 m³/(d·MPa)

Solution

The logarithmic term: $\ln(r_e/r_w) - 0.75 = \ln 3000 - 0.75 = 7.256$ .
Productivity index of an undamaged well ( $S = 0$ ): $J_{\text{ideal}} = \dfrac{0.536\cdot 80\cdot 12}{1.5\cdot 1.20\cdot 7.256} = 39.4$ m³/(d·MPa).
The ratio of the actual productivity index to the ideal one is the flow efficiency (FE): $FE = 24.0/39.4 = 0.61$ .
This drop corresponds to a skin factor $S = 7.256\cdot(1/FE - 1) = 4.6$ .
Without damage, at the same bottomhole pressure of 14.0 MPa the rate would be $39.4\cdot 4.0 = 158$ m³/d — 64% more.

Answer: The flow efficiency of the well is about 0.61. The near-wellbore zone is apparently damaged (skin factor about 4.6).

What it means. Comparing the actual and the calculated productivity index is a simple first check on whether the near-wellbore zone is worth treating. The conclusion is only as reliable as the permeability and the drainage radius. Well tests give the skin factor. Nodal analysis gives the real gain in rate after a treatment, because the lift system (the tubing and the surface equipment) may limit it. The data in this example are assumed for illustration.

Below the bubble point pressure: Vogel's equation

When the bottomhole pressure falls below the bubble point pressure $p_b$ , gas begins to come out of the oil already in the reservoir. This happens in the zone around the well where the pressure is below $p_b$ . Free gas occupies part of the pores and reduces the relative permeability to oil (how easily oil flows when gas is also present). Also, oil that has lost its gas becomes more viscous. The lower the bottomhole pressure, the wider this zone and the more gas there is in it. So each further drop in bottomhole pressure adds less and less oil, and the IPR bends: at low bottomhole pressures it runs more and more steeply downwards.

An exact calculation of such inflow needs a model of two-phase flow (oil and gas together). In 1968 J. Vogel used a numerical computer model to calculate how wells perform in solution gas drive reservoirs (where the energy comes from gas dissolved in the oil), for many combinations of rock and fluid properties. He found that if you divide the rate by $q_{max}$ and the bottomhole pressure by the reservoir pressure, the curves almost coincide. A simple formula describes them.

Vogel's equation

\frac{q}{q_{max}} = 1 - 0.2\,\frac{p_{wf}}{\bar p_R} - 0.8\left(\frac{p_{wf}}{\bar p_R}\right)^2

where:

$q$ oil rate at standard conditions at the bottomhole pressure $p_{wf}$ , m³/d
$q_{max}$ absolute open flow — the rate at $p_{wf} = 0$ , m³/d
$\bar p_R$ average reservoir pressure, MPa
$p_{wf}$ bottomhole pressure, MPa

Units: dimensionless form: the rates in any one unit, and the pressures likewise

Holds when

The reservoir pressure is not above the bubble point pressure ( $\bar p_R \le p_b$ ); solution gas drive.
Stabilised inflow. The equation is an empirical generalisation (a rule drawn from many calculated cases) of results for wells with an undamaged near-wellbore zone ( $S = 0$ ).
Low water cut (little water in the produced liquid). With a gas cap (a layer of free gas above the oil), a high water cut or high-viscosity oil, the accuracy is lower.
It is better to take absolute pressures. At reservoir pressures of tens of megapascals, the difference from gauge pressures is small.

The coefficients 0.2 and 0.8 add up to 1, and both terms enter with a minus sign. Only then is the rate zero at $p_{wf} = \bar p_R$ and $q_{max}$ at $p_{wf} = 0$ . The right-hand side can be written as the product $(1 - x)(1 + 0.8x)$ , where $x = p_{wf}/\bar p_R$ . The first factor is proportional to the drawdown. The second describes the bend of the curve.

To draw the curve, you need only the reservoir pressure and one stabilised measurement $(q_1, p_{wf,1})$ . From the equation, find $q_{max} = q_1/\left(1 - 0.2\,x_1 - 0.8\,x_1^2\right)$ , where $x_1 = p_{wf,1}/\bar p_R$ . Then calculate the rate at any bottomhole pressure.

A curved IPR has no single productivity index. The ratio $q/(\bar p_R - p_{wf})$ is the average index at a given drawdown, and it decreases as the drawdown grows. The instantaneous index is the slope of the curve, $-\dd q/\dd p_{wf} = (q_{max}/\bar p_R)(0.2 + 1.6\,p_{wf}/\bar p_R)$ . It falls even faster: from $1.8\,q_{max}/\bar p_R$ at a small drawdown to $0.2\,q_{max}/\bar p_R$ at zero bottomhole pressure, which is nine times less.

Worked exampleA Vogel curve from a single test

Given (assumed data)

Reservoir pressure $\bar p_R$ (equal to the bubble point pressure): 15.0 MPa
Stabilised rate: $q_1 = 55$ m³/d at $p_{wf,1} = 10.5$ MPa
To find: the rate at $p_{wf} = 6.0$ MPa

Solution

The relative bottomhole pressure of the test is $x_1 = 10.5/15.0 = 0.70$ . The Vogel factor is $1 - 0.2\cdot0.70 - 0.8\cdot0.70^2 = 0.468$ .
Absolute open flow: $q_{max} = 55/0.468 = 117.5$ m³/d.
At $p_{wf} = 6.0$ MPa: $x = 0.40$ , $q = 117.5\cdot(1 - 0.08 - 0.128) = 93$ m³/d.
For comparison, take a straight line through the same test: $J = 55/4.5 = 12.22$ m³/(d·MPa) and $q = 12.22\cdot 9.0 = 110$ m³/d. This is 18% more.
The instantaneous productivity index at the test point: $(117.5/15.0)(0.2 + 1.6\cdot 0.70) = 10.3$ m³/(d·MPa). This is less than the average 12.22.

Answer: At a bottomhole pressure of 6 MPa the well will deliver about 93 m³/d. The absolute open flow is about 117.5 m³/d.

What it means. A straight line drawn through a test at a small drawdown clearly overestimates the rate at large drawdowns. This is exactly how people overestimate the gain from putting a well on a pump that can lower the bottomhole pressure a lot. The data in this example are assumed for illustration.

Reservoir pressure above the bubble point pressure: the composite IPR

Often the reservoir is undersaturated: $\bar p_R > p_b$ , so all the gas is still dissolved in the oil. As long as the bottomhole pressure is above $p_b$ , the inflow is single-phase and the curve is a straight line. When $p_{wf}$ falls below $p_b$ , gas appears near the well and the curve bends. We build the composite IPR from two parts: a straight line down to the bubble point pressure and a Vogel-type curve below it. At the point $p_{wf} = p_b$ the two parts join without a kink: their slopes are the same.

Composite IPR

q = \begin{cases} J\,(\bar p_R - p_{wf}), & p_{wf} \ge p_b,\\[6pt] q_b + \dfrac{J p_b}{1.8}\left[1 - 0.2\,\dfrac{p_{wf}}{p_b} - 0.8\left(\dfrac{p_{wf}}{p_b}\right)^2\right], & p_{wf} < p_b, \end{cases}\qquad q_b = J\,(\bar p_R - p_b)

where:

$J$ productivity index of the single-phase part of the curve, m³/(d·MPa)
$p_b$ bubble point pressure, MPa
$q_b$ rate at $p_{wf} = p_b$ , m³/d
$\bar p_R,\ p_{wf}$ average reservoir pressure and bottomhole pressure, MPa

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

Holds when

$\bar p_R > p_b$ : above the bubble point pressure the inflow is single-phase.
Below the bubble point pressure the curve is of the Vogel type, but it is built from $p_b$ , not from the reservoir pressure.
Stabilised flow. The effect of the near-wellbore zone is included in $J$ .

The absolute open flow of the composite curve is $q_{max} = q_b + J p_b/1.8$ . The term $J p_b/1.8$ is the “Vogel” $q_{max}$ for the part of the curve below $p_b$ . It is chosen so that the slope of the curve at $p_b$ equals $J$ .

Where the 1.8 comes from: the slope of the Vogel curve at its top point is $1.8\,q_{max,V}/p_b$ (see the instantaneous productivity index above, with $p_b$ in place of the reservoir pressure). Set it equal to the slope of the straight line, $J$ . This gives $q_{max,V} = J p_b/1.8$ .

If the stabilised measurement $(q_1, p_{wf,1})$ was made below the bubble point pressure, put it into the second line of the formula to find the productivity index.

Productivity index from a test below the bubble point pressure

J = \frac{q_1}{(\bar p_R - p_b) + \dfrac{p_b}{1.8}\left[1 - 0.2\,\dfrac{p_{wf,1}}{p_b} - 0.8\left(\dfrac{p_{wf,1}}{p_b}\right)^2\right]}

where:

$q_1$ rate at the measured point, m³/d
$p_{wf,1}$ bottomhole pressure at this rate, $p_{wf,1} < p_b$ , MPa

Units: as for the composite curve

Holds when

The same as for the composite curve.
The bubble point pressure is known from a PVT study of an oil sample.

Then build the whole curve from the composite formula. If you treated the same measurement as a point on a straight line, the productivity index would come out too low, and the rate at low bottomhole pressures too high.

Worked exampleA composite IPR from a test below the bubble point pressure

Given (assumed data)

Reservoir pressure $\bar p_R$: 20.0 MPa
Bubble point pressure $p_b$ (from a PVT study): 14.0 MPa
Stabilised rate: $q_1 = 150$ m³/d at $p_{wf,1} = 11.0$ MPa

Solution

The test was made below $p_b$ , so we calculate the Vogel factor from the bubble point pressure: $x_1 = 11.0/14.0 = 0.786$ , $1 - 0.2x_1 - 0.8x_1^2 = 0.349$ .
Productivity index: $J = \dfrac{150}{(20.0 - 14.0) + \dfrac{14.0}{1.8}\cdot 0.349} = 17.2$ m³/(d·MPa).
The rate at the bubble point pressure is $q_b = 17.2\cdot 6.0 = 103$ m³/d. The Vogel part is $J p_b/1.8 = 134$ m³/d. So $q_{max} = 237$ m³/d.
Rate at $p_{wf} = 7.0$ MPa ( $x = 0.5$ ): $q = 103 + 134\cdot(1 - 0.1 - 0.2) = 197$ m³/d.
A straight line with the same $J$ , extended below $p_b$ , would give $17.2\cdot 13.0 = 224$ m³/d at 7 MPa (14% more). At zero bottomhole pressure it would give 344 m³/d instead of 237.

Answer:

J \approx 17.2

m³/(d·MPa). At a bottomhole pressure of 7 MPa the well will deliver about 197 m³/d, and the absolute open flow is about 237 m³/d.

What it means. The further the bottomhole pressure goes below the bubble point pressure, the bigger the error of the straight line. If a well is switched to a pump, choose the delivery rate and head (the pressure the pump adds) from the composite curve. Otherwise the pump will be too big for the reservoir. The data in this example are assumed for illustration.

CalculatedCalculated composite IPR for the data of the example: a straight line above the bubble point pressure (14 MPa) and a Vogel curve below it. The dashed line is the same straight line, wrongly extended below the bubble point pressure.

Multi-rate tests: Fetkovich's method

Vogel's equation needs one point, and its curve has a fixed shape. If a well has been tested at several stabilised rates, the data themselves can show the shape. In 1973 M. Fetkovich showed that the inflow of oil wells producing below the bubble point pressure is well described by the same relationship that had long been used for gas wells.

Fetkovich's inflow equation

q = C\,(\bar p_R^2 - p_{wf}^2)^n

where:

$q$ oil rate at standard conditions, m³/d
$C$ a coefficient reflecting the properties of the reservoir and the fluid; its dimension depends on $n$ , m³/(d·MPa²ⁿ)
$n$ exponent: 1 for purely viscous flow, down to 0.5 with strong inertial losses, —
$\bar p_R,\ p_{wf}$ average reservoir pressure and bottomhole pressure, absolute, MPa

Units: $q$ in m³/d, pressures in MPa (absolute)

Holds when

Stabilised rates (a flow-after-flow test: one stable rate after another) or isochronal rates (every rate held for the same time).
The pressures are absolute. The formula contains their squares, and a difference of squares of gauge pressures is not equal to a difference of squares of absolute ones.
The reservoir pressure is not above the bubble point pressure. If it is above, the formula describes only the part of the curve below $p_b$ .

In dimensionless form, $q/q_{max} = [1 - (p_{wf}/\bar p_R)^2]^n$ , where $q_{max} = C\,\bar p_R^{2n}$ . At $n = 1$ the curve almost coincides with Vogel's curve. The difference between $1 - x^2$ and $1 - 0.2x - 0.8x^2$ is $0.2x(1 - x)$ , and it never exceeds 0.05. Values of $n < 1$ point to losses that grow faster than the rate. Most often these are inertial (turbulent) losses near the well.

Why squares of pressures? Below the bubble point pressure, the mobility of the oil, $k_{ro}/(\mu_o B_o)$ , falls roughly in line with the pressure. (Mobility shows how easily the oil moves: relative permeability divided by viscosity and formation volume factor.) In Darcy's law for radial inflow, integrate a mobility that is proportional to pressure from $p_{wf}$ to $\bar p_R$ . The result is a difference of squares of pressures. This is where the form of the equation with $n = 1$ comes from.

Analysing the test. For each rate, calculate $\Delta(p^2) = \bar p_R^2 - p_{wf}^2$ and plot the points on a log–log graph. Put $\Delta(p^2)$ on the horizontal axis and the rate on the vertical axis. The points then lie on a straight line with slope $n$ . If you swap the axes, the slope is $1/n$ , which is a frequent source of confusion. Find $C$ from any point of the line. Find $q_{max}$ by extending the line to $\Delta(p^2) = \bar p_R^2$ .

Worked exampleThe exponent n and the potential of a well from four rates

Given (assumed data)

Reservoir pressure $\bar p_R$ (absolute; the reservoir is saturated): 16.0 MPa
Rate 1: 40 m³/d at $p_{wf} = 14.0$ MPa
Rate 2: 65 m³/d at 12.0 MPa
Rate 3: 85 m³/d at 10.0 MPa
Rate 4: 101 m³/d at 8.0 MPa

Solution

Differences of squares of pressures ( $\bar p_R^2 = 256$ MPa²): 60, 112, 156 and 192 MPa².
The least-squares line (the straight line that fits the points best) in the coordinates $\log_{10}\Delta(p^2)$ and $\log_{10} q$ has a slope $n = 0.795$ . Check from the end points: $n = \log_{10}\!(101/40)/\log_{10}\!(192/60) = 0.796$ .
The coefficient from the same line: $C = 1.539$ m³/(d·MPa²ⁿ).
Absolute open flow: $q_{max} = C\,\bar p_R^{2n} = 1.539\cdot 256^{0.795} = 126$ m³/d. Rate at $p_{wf} = 4.0$ MPa: $q = 1.539\cdot 240^{0.795} = 120$ m³/d.
For comparison: Vogel's equation from rate 2 alone would give $q_{max} = 65/(1 - 0.15 - 0.45) = 163$ m³/d. This is 29% more.

Answer:

n \approx 0.795

q_{max} \approx 126

m³/d.

What it means. An exponent below one means that at high rates, part of the drawdown goes to inertial losses near the well, and the rate grows more slowly than $\Delta(p^2)$ . Vogel's equation does not allow for such losses, so from a single rate it would overestimate the potential of the well. In practice, engineers take at least three or four rates and check that the points really lie on a straight line. The data in this example are assumed for illustration.

CalculatedCalculated plot of the four-rate test for the data of the example, on log–log scales:

\Delta(p^2)

horizontally, the rate vertically. The slope of the line equals

n

. Extending the line to

\Delta(p^2) = \bar p_R^2 = 256

MPa² gives

q_{max}

Gas inflow

Gas is highly compressible: it shrinks a lot under pressure. Its viscosity and gas deviation factor (a correction for the way real gas differs from ideal gas) also depend on pressure. So a straight line does not suit a gas well, even in single-phase flow. Integrate Darcy's law for gas, using average values of $\mu Z$ . This leads to a difference of squares of pressures. The approximation works well at moderate pressures (roughly up to 14 MPa). In the general case, engineers use the real gas pseudo-pressure (a pressure-like quantity that allows for the changing properties of the gas) instead of squares of pressures. At high velocities, inertial losses are added to the viscous ones. In practice two kinds of equations are used.

Gas inflow equation with an exponent

q_g = C\,(\bar p_R^2 - p_{wf}^2)^n

where:

$q_g$ gas rate at standard conditions, thousand m³/d
$C$ a coefficient depending on the properties of the reservoir and the gas, thousand m³/(d·MPa²ⁿ)
$n$ exponent: 1 for purely viscous flow, down to 0.5 when inertial losses dominate, —
$\bar p_R,\ p_{wf}$ reservoir and bottomhole pressures, absolute, MPa

Units: $q_g$ in thousand m³/d, pressures in MPa (absolute)

Holds when

Stabilised or isochronal rates.
Absolute pressures.
An empirical relationship: extrapolation far beyond the tested drawdowns is unreliable.

This is historically the first equation of its kind (the back-pressure equation). It has been applied to gas wells since the 1930s. The absolute open flow of a gas well, $q_{max} = C\,\bar p_R^{2n}$ , is the standard measure of its potential.

Two-term gas inflow equation

\bar p_R^2 - p_{wf}^2 = a\,q_g + b\,q_g^2

where:

$a$ viscous flow resistance coefficient; includes the effect of the skin factor, MPa²/(thousand m³/d)
$b$ inertial resistance coefficient, MPa²/(thousand m³/d)²
$q_g$ gas rate at standard conditions, thousand m³/d

Units: pressures in MPa (absolute), $q_g$ in thousand m³/d

Holds when

Stabilised rates.
$a$ and $b$ are found from a test at several rates: the dependence of $(\bar p_R^2 - p_{wf}^2)/q_g$ on $q_g$ is a straight line with slope $b$ and intercept $a$ on the ordinate axis.

The two-term formula is the main form of the gas inflow equation in Russian-language practice. It follows from a law of flow through porous rock that has an inertial term, and both of its coefficients have a clear physical meaning. More in the lesson Inflow at high rates: non-Darcy flow.

The same technique works for oil wells with high rates. Using several rates, plot $(\bar p_R - p_{wf})/q$ against $q$ : the points lie on a straight line. The intercept of the line on the ordinate axis shows the viscous resistance of the reservoir together with the skin factor. The slope shows the inertial losses. A horizontal line means that there are no inertial losses and the IPR is linear.

How the IPR changes

Depletion of the reservoir. When the reservoir pressure falls below the bubble point pressure, the curve “shrinks”: both $\bar p_R$ and $q_{max}$ decrease. The cause is not only the lower pressure. The gas saturation of the reservoir (the share of the pore space filled with gas) grows, and it becomes harder and harder for the oil to move. In the dimensionless coordinates $q/q_{max}$ against $p_{wf}/\bar p_R$ , the curves of different years are close. This is what made it possible to generalise them with Vogel's equation. The advanced section shows how to forecast future curves.
Oil viscosity. A more viscous oil gives a smaller rate at the same drawdown. But the dimensionless curves of two oils are almost identical: viscosity changes the scale of the curve, not its shape.
Condition of the near-wellbore zone. The skin factor or the flow efficiency (FE) describes damage to the near-wellbore zone or its improvement. FE is the ratio of the actual productivity index to that of the same well without damage. $FE < 1$ means the zone is damaged. $FE > 1$ means it has been improved by an acid treatment or hydraulic fracturing.
Several layers in one well. Suppose layers with different permeabilities and pressures produce together. The curve of the well is then the sum of the curves of the layers at the same bottomhole pressure. At a small drawdown, a layer with a lower pressure may take in fluid from the others (crossflow between layers). That is why such wells are also tested layer by layer.
Rising water cut. As the share of water grows, the relative permeabilities and the density and viscosity of the mixture change. For a well with a high water cut, the curve is usually drawn for the liquid rate, and the share of water is handled separately.

Flow efficiency

FE = \frac{\bar p_R - p_{wf} - \Delta p_S}{\bar p_R - p_{wf}} = \frac{J_{\text{actual}}}{J_{\text{ideal}}}

where:

$\Delta p_S$ additional pressure loss in the near-wellbore zone at the given rate (from well tests), MPa
$J_{\text{actual}}$ actual productivity index, m³/(d·MPa)
$J_{\text{ideal}}$ productivity index of the same well without damage ( $S = 0$ ), m³/(d·MPa)

Units: dimensionless; the pressures in the same units

Holds when

Single-phase inflow, stabilised flow.
The additional pressure loss is proportional to the rate (the skin factor does not depend on the rate).

For pseudo-steady-state inflow, $FE = \dfrac{\ln(r_e/r_w) - 0.75}{\ln(r_e/r_w) - 0.75 + S}$ . A skin factor of $+5$ with a logarithmic term of about 7.5 leaves the well with 60% of the ideal productivity.

Terms and symbols

Inflow performance relationship (IPR): How the rate depends on the bottomhole pressure at a given reservoir pressure.
Productivity index $J$: The rate per unit of drawdown. It is constant for a straight-line IPR. Unit: m³/(d·MPa).
Specific productivity index: The productivity index per metre of net thickness, $J/h$ .
Absolute open flow $q_{max}$ (AOF): The rate at zero bottomhole pressure. It is a conventional measure of the potential of a well.
Bubble point pressure $p_b$: The pressure below which gas comes out of the oil. It marks the boundary between the straight and the curved parts of the IPR.
Composite IPR: A straight line above the bubble point pressure and a Vogel curve below it, joined without a kink.
Flow efficiency (FE): The ratio of the actual productivity index to that of the same well without damage to the near-wellbore zone.
Exponent $n$: The exponent in equations of the form $q = C\,\Delta(p^2)^n$ . Values below one point to inertial losses.

Common mistakes and misconceptions

Wrong

“At a drawdown of 5 MPa the well gives 50 m³/d, so at 10 MPa it will give 100.”

In fact

Only if both bottomhole pressures are above the bubble point pressure. Below it the curve bends and the rate grows more slowly than the drawdown. Use the composite curve or Vogel's equation for the calculation.

Wrong

“ $q_{max}$ is the rate the well can really deliver.”

In fact

It is a conventional rate at zero bottomhole pressure. The real maximum is always lower. The pressure of the liquid column remains at the bottomhole, and sand production, water and gas coning, and the strength of the casing limit the drawdown that is allowed.

Wrong

“The well's productivity index has fallen, so the near-wellbore zone is damaged.”

In fact

Not necessarily. The average productivity index falls by itself if the bottomhole pressure has dropped below the bubble point pressure. It also falls as the reservoir pressure drops and the gas saturation grows. Well tests confirm damage: they give the skin factor.

Wrong

“On a Fetkovich plot the slope of the line is always $n$ .”

In fact

Only if the difference of the squares of the pressures is on the horizontal axis and the rate on the vertical axis. With the axes the other way round, the slope is $1/n$ . Always check what is plotted on which axis.

Wrong

“Gauge pressures can be substituted into Fetkovich's equation.”

In fact

In formulas with squares of pressures, use absolute pressures: $(a + 0.1)^2 - (b + 0.1)^2 \ne a^2 - b^2$ . For a straight-line IPR, where only the pressure difference matters, it makes no difference.

In depth: advanced detail

Forecasting the IPR as the reservoir pressure falls

As the reservoir is depleted, the IPR changes. A pump or gas lift chosen for today's operating mode may be unsuitable in a year or two. M. Standing proposed a convenient forecasting method. He introduced the productivity index at zero drawdown, $J^*$ , which is the slope of the Vogel curve at its top point. At a vanishingly small drawdown, the pressure everywhere in the drainage area equals $\bar p_R$ . So $J^*$ is proportional to the oil mobility at reservoir pressure.

Productivity index at zero drawdown

J^* = \frac{1.8\,q_{max}}{\bar p_R} = \frac{1.8\,J}{1 + 0.8\,p_{wf}/\bar p_R}

where:

$J^*$ productivity index at zero drawdown, m³/(d·MPa)
$J$ average productivity index from a test, $q/(\bar p_R - p_{wf})$ , m³/(d·MPa)
$q_{max},\ \bar p_R,\ p_{wf}$ absolute open flow, and the reservoir and bottomhole pressures of the test, m³/d, MPa

Units: any consistent units

Holds when

The shape of the curve is of the Vogel type ( $\bar p_R \le p_b$ ).

The second equality follows from writing $1 - 0.2x - 0.8x^2$ as $(1 - x)(1 + 0.8x)$ . Then the average index $J$ equals $(q_{max}/\bar p_R)(1 + 0.8x)$ , where $x = p_{wf}/\bar p_R$ .

Conversion to a future reservoir pressure

J^*_f = J^*_p\,\frac{\left(k_{ro}/\mu_o B_o\right)_f}{\left(k_{ro}/\mu_o B_o\right)_p},\qquad q_{max,f} = \frac{J^*_f\,\bar p_{R,f}}{1.8}

where:

$p,\ f$ subscripts for the present and the future state, —
$k_{ro}$ relative permeability to oil at the average saturation of the reservoir, —
$\mu_o$ viscosity of the oil at reservoir pressure, mPa·s
$B_o$ formation volume factor of the oil, m³/m³
$\bar p_{R,f}$ future reservoir pressure, MPa

Units: the ratio is dimensionless; $J^*$ and $q_{max}$ are in the units of the current curve

Holds when

The shape of the curve will also be of the Vogel type in the future.
$k_{ro}$ comes from the saturation forecast (material balance, an accounting of the fluids in the reservoir) and the relative permeability curves. $\mu_o$ and $B_o$ come from PVT data at the future pressure.
The properties of the near-wellbore zone do not change.

The productivity index at zero drawdown is proportional to the oil mobility $k_{ro}/(\mu_o B_o)$ . The future curve is $q = q_{max,f}\,(1 - 0.2x - 0.8x^2)$ with $x = p_{wf}/\bar p_{R,f}$ .

Fetkovich's method gives a rougher estimate, if you take the coefficient $C$ as proportional to the reservoir pressure: $q_{max,f} = q_{max,p}\,(\bar p_{R,f}/\bar p_{R,p})^{1+2n}$ . At $n = 1$ this is the cube of the pressure ratio. It needs no relative permeability data, so it suits a quick check. The lesson Performance prediction for solution gas drive reservoirs covers how to forecast the pressure and saturation of a solution gas drive reservoir.

Worked exampleThe IPR in a few years' time

Given (assumed data)

Now (the well of the Vogel example): $\bar p_R = 15.0$ MPa, $q_{max} = 117.5$ m³/d
Now: $k_{ro}$ ; $\mu_o$ ; $B_o$: 0.75; 1.6 mPa·s; 1.20
In the future: $\bar p_R$: 12.0 MPa
In the future (forecast): $k_{ro}$ ; $\mu_o$ ; $B_o$: 0.55; 1.9 mPa·s; 1.15

Solution

Now: $J^*_p = 1.8\cdot 117.5/15.0 = 14.10$ m³/(d·MPa).
Oil mobility: now $0.75/(1.6\cdot1.20) = 0.391$ , in the future $0.55/(1.9\cdot1.15) = 0.252$ (mPa·s)⁻¹; the ratio is 0.644.
In the future: $J^*_f = 14.10\cdot0.644 = 9.09$ m³/(d·MPa) and $q_{max,f} = 9.09\cdot 12.0/1.8 = 60.6$ m³/d.
Rate at $p_{wf} = 6.0$ MPa: now 93 m³/d. In the future ( $x = 0.5$ ) it is $60.6\cdot(1 - 0.1 - 0.2) = 42$ m³/d.
Check by Fetkovich's method at $n = 1$ : $117.5\cdot(12.0/15.0)^3 = 60.2$ m³/d, which is close.

Answer: The potential of the well will fall by about half:

q_{max}

will go from 117.5 to 60.6 m³/d. At a bottomhole pressure of 6 MPa the rate will fall from 93 to 42 m³/d.

What it means. The reservoir pressure falls by 20%, but that is not the only reason for the drop. The oil has also lost mobility: the gas saturation of the reservoir has grown, and the oil viscosity has increased. You need such a forecast to choose a pump with a safety margin, and to plan when to switch the well to another lift method. The data in this example are assumed for illustration.

Vogel's curve for a well with a damaged or treated near-wellbore zone

Vogel's equation was found for undamaged wells. For a well with a flow efficiency $FE \ne 1$ , M. Standing proposed putting into it an “ideal” bottomhole pressure. This is the pressure that the well would have at the same rate without damage: $p'_{wf} = \bar p_R - FE\,(\bar p_R - p_{wf})$ . This substitution gives the following formula.

Vogel's equation with the flow efficiency of the well

\frac{q}{q_{max}^{FE=1}} = 1.8\,FE\left(1 - \frac{p_{wf}}{\bar p_R}\right) - 0.8\,FE^2\left(1 - \frac{p_{wf}}{\bar p_R}\right)^2

where:

$q_{max}^{FE=1}$ absolute open flow of the same well without damage, m³/d
$FE$ flow efficiency, —
$q,\ \bar p_R,\ p_{wf}$ rate, reservoir pressure and bottomhole pressure, m³/d, MPa

Units: dimensionless form

Holds when

A Vogel-type inflow for an undamaged well.
The additional pressure loss in the near-wellbore zone is proportional to the rate.
At $FE > 1$ the formula applies as long as $p'_{wf} \ge 0$ , that is, at $p_{wf} \ge \bar p_R\,(1 - 1/FE)$ . Below that it gives a physically meaningless result.

At $FE = 1$ the formula coincides with Vogel's equation. Draw curves for the actual FE and for the FE after a treatment, and you get the expected gain in inflow. Then check the real gain in rate with nodal analysis.

Connections to other topics

Nodal analysis of the production system — how the IPR, together with the outflow curve, sets the operating mode of a well.
Skin factor — what the skin factor is and how to find it from well tests.
Pseudo-steady-state flow and the shape factor — derivation of the inflow formula with the average reservoir pressure, and the shape factors of the drainage area.
Relative permeability — why free gas reduces the inflow of oil.
Inflow at high rates: non-Darcy flow — inertial losses and the two-term gas inflow formula.
Performance prediction for solution gas drive reservoirs — how the pressure and saturation of a reservoir are forecast for future curves.
Rock permeability — Darcy's law and radial inflow — the basis of the productivity index formula.

Check yourself

1
The reservoir pressure is 20 MPa and the bubble point pressure is 12 MPa. A well delivered 60 m³/d at a bottomhole pressure of 16 MPa. What rate should be expected at a bottomhole pressure of 14 MPa?
1. A.120 m³/d
2. B.90 m³/d
3. C.It cannot be estimated without Vogel's equation
4. D.About 80 m³/d: the curve is already bending
Show the answer
Correct answer: B. 90 m³/d
Both bottomhole pressures are above the bubble point pressure, so the inflow is single-phase and the curve is a straight line. Then $J = 60/(20 - 16) = 15$ m³/(d·MPa) and $q = 15\cdot(20 - 14) = 90$ m³/d.
2
A well in a reservoir with $\bar p_R = p_b = 16$ MPa delivered 40 m³/d at $p_{wf} = 12$ MPa. Estimate $q_{max}$ from Vogel's equation and compare it with a straight line through the same measurement.
Show the answer
$x = 12/16 = 0.75$ . The Vogel factor is $1 - 0.2\cdot0.75 - 0.8\cdot0.75^2 = 1 - 0.15 - 0.45 = 0.40$ . So $q_{max} = 40/0.40 = 100$ m³/d. A straight line would give $J = 40/4 = 10$ m³/(d·MPa) and $q_{max} = 160$ m³/d, which is 60% more.
3
Why does the composite IPR below the bubble point pressure contain the factor $J p_b/1.8$ ?
1. A.It is the absolute open flow of the whole well
2. B.It is an empirical correction for water cut
3. C.It makes the slopes of the line and of the curve coincide at the point $p_{wf} = p_b$
4. D.It accounts for the skin factor
Show the answer
Correct answer: C. It makes the slopes of the line and of the curve coincide at the point $p_{wf} = p_b$
The slope of the Vogel curve at its top point is $1.8\,q_{max,V}/p_b$ . For the curve to continue the straight line without a kink, set this slope equal to the productivity index $J$ . This gives $q_{max,V} = J p_b/1.8$ . The absolute open flow of the whole well is larger: $q_b + J p_b/1.8$ .
4
On the plot of a multi-rate test, $\log_{10} q$ is plotted horizontally and $\log_{10}(\bar p_R^2 - p_{wf}^2)$ vertically, and the slope of the line is 1.25. What is the exponent $n$ , and what does it say about the inflow?
Show the answer
With the axes this way round, the slope equals $1/n$ , so $n = 1/1.25 = 0.8$ . The exponent is below one. At high rates, part of the drawdown goes to inertial losses near the well, and the rate grows more slowly than $\Delta(p^2)$ .
5
How will the IPR change after a successful treatment of the near-wellbore zone, if the reservoir pressure has not changed?
1. A.It will rotate about the point $q = 0$ , $p_{wf} = \bar p_R$ : at the same bottomhole pressure the rate will be greater
2. B.It will shift up: the reservoir pressure will rise
3. C.It will not change: the treatment affects only the lift system
4. D.It will shift down parallel to itself
Show the answer
Correct answer: A. It will rotate about the point $q = 0$ , $p_{wf} = \bar p_R$ : at the same bottomhole pressure the rate will be greater
The treatment lowers the extra resistance near the well and raises the productivity index. The reservoir pressure, which is the point of the curve at zero rate, stays the same. So the curve rotates about this point and becomes flatter.
6
A well has penetrated two layers, both above the bubble point pressure: the first has $\bar p_R = 18$ MPa and $J = 10$ , the second $\bar p_R = 15$ MPa and $J = 20$ m³/(d·MPa). What is the rate of the well at a bottomhole pressure of 14 MPa and at 16 MPa?
Show the answer
Add the curves of the layers at the same bottomhole pressure. At 14 MPa: $10\cdot 4 + 20\cdot 1 = 60$ m³/d. At 16 MPa the second layer no longer delivers. It takes in fluid: $20\cdot(15 - 16) = -20$ m³/d. The first layer delivers $10\cdot 2 = 20$ m³/d. Nothing reaches the surface, because all the production of the first layer goes into the second. This is crossflow between layers.

Updated: 19 October 2026