Steady-state radial flow

Quick refresher

Steady state is possible only if the outer boundary replaces what the well takes out, by water injection or by an active aquifer. Without this recharge the pressure in the area falls, and the inflow becomes pseudo-steady-state.
The same flow rate passes through every circle around the well, but the area $2\pi rh$ gets smaller towards the wellbore. So the pressure gradient grows as $1/r$ , and the pressure varies with the logarithm of the distance. Each tenfold step in distance from the well takes the same share of the drawdown.
The Dupuit equation in consistent units (units that fit together, such as SI) is $q = 2\pi kh\,\Delta p/(\mu\ln(r_e/r_w))$ . Here $q$ is the flow rate at reservoir conditions, and the logarithm is the natural one.
The size of the area enters through a logarithm. Doubling $r_e$ changes the flow rate by only 7–9%. A skin of +4, with a logarithmic term of about 8, cuts the flow rate by a third.
The average pressure in the area is lower than the pressure at the outer boundary by half a “unit” of $q\mu/(2\pi kh)$ . In the equation written with the average pressure, $\ln(r_e/r_w) - 1/2$ replaces $\ln(r_e/r_w)$ .
For gas, the mass flow rate stays the same along the radius, not the volume flow rate. When the product of the viscosity and the gas deviation factor, $\mu Z$ , is constant, the inflow is proportional to the difference of the squares of the absolute pressures.

Key relationships

q = \dfrac{2\pi kh\,(p_e - p_w)}{\mu\ln(r_e/r_w)}

Dupuit equation, consistent units, flow rate at reservoir conditions

p(r) = p_w + \dfrac{q\mu}{2\pi kh}\ln\dfrac{r}{r_w}

pressure distribution around the well

\bar p - p_w = \dfrac{q\mu}{2\pi kh}\left(\ln\dfrac{r_e}{r_w} - \dfrac12\right)

in terms of the average pressure in the area

q_{sc} = \dfrac{\pi kh\,T_{sc}\,(p_e^2 - p_w^2)}{p_{sc}\,T\,\mu Z\ln(r_e/r_w)}

gas inflow, absolute pressures

Common pitfalls

Using the base-10 logarithm instead of the natural one. The flow rate comes out 2.3 times too large.
Forgetting the formation volume factor. In consistent units the equation gives the flow rate at reservoir conditions, not at standard conditions.
Using the average reservoir pressure but keeping $\ln(r_e/r_w)$ in the denominator. That term belongs to the pressure at the outer boundary.
Putting gauge pressures into the gas equation, or putting in the pressure difference instead of the difference of the squares.
Explaining a low flow rate by low permeability without checking the skin.

After this lesson you can

derive the Dupuit equation from Darcy's law, and write the pressure distribution around the well
calculate the flow rate and the productivity index in SI, metric oilfield and oilfield units, including cases with a skin factor, and explain where the constants come from
explain why the flow rate depends only a little on the radius of the drainage area but strongly on the near-wellbore zone, and estimate what share of the drawdown is used up near the well
tell apart the equations written with the pressure at the outer boundary and those written with the average reservoir pressure, and derive the difference between them
calculate steady-state gas inflow from the difference of the squares of the absolute pressures (pressures measured from zero, not above atmospheric)

When the inflow is steady-state

Inflow is steady-state when the flow rate and the pressure at every point of the reservoir do not change with time. Fluid does not build up anywhere and does not run out anywhere: as much enters any volume of the reservoir as leaves it. For a well, this means that inflow across the outer boundary of the drainage area replaces everything the well takes out. This happens under pressure maintenance by water injection, when injection wells surround a production well. It also happens in reservoirs with an active aquifer.

The simplest model of such a well is the basic inflow model:

The reservoir is homogeneous (the same everywhere). It has constant thickness $h$ and permeability $k$ , with impermeable rocks above and below.
A vertical well of radius $r_w$ cuts through the whole thickness of the reservoir. It stands at the centre of a circular area of radius $r_e$ .
A constant pressure $p_e$ is held at the outer boundary, which is the supply contour. In waterflooding (injecting water to push oil towards the wells), $r_e$ is roughly half the distance to the injection wells. Then $p_e$ is the pressure midway between the wells. A formation tester can measure it in an observation well (a well used only for monitoring).
Only one phase moves in the pores: undersaturated oil without free gas (oil above its bubble point pressure). The connate water does not move, so $k$ is the permeability to oil at irreducible water saturation (the water that stays in the pores).
The flow is laminar (smooth, without eddies) and obeys Darcy's law (Rock permeability).

SchematicSteady-state flow (schematic, not to scale). Below, a cross-section of the reservoir: fluid flows in towards the well, and across the outer boundary at distance

r_e

as much fluid enters as the well takes out. Above, the pressure distribution: a steep “funnel” at the wellbore, and a gentle rise to the pressure

p_e

at the boundary. The average pressure

\bar p

is only slightly below

p_e

. The reservoir has this pressure at a distance of about

0.61\,r_e

from the well.

If there is no recharge, the pressure in the whole area falls with time, and the inflow becomes pseudo-steady-state (the pressure falls at the same rate everywhere) (Pseudo-steady-state flow and the shape factor). Just after the well starts up, the inflow is transient (the pressure disturbance is still spreading) (Well testing: purposes and types of test). But near the well the pressure distribution is almost the same in all three regimes. So the conclusions of this lesson about the near-wellbore zone always hold.

The Dupuit equation

In steady-state flow the same flow rate $q$ passes through every circle of radius $r$ around the well. The area it passes through is the side surface of a cylinder, $2\pi rh$ , and it gets smaller towards the wellbore. So the Darcy velocity (the speed of the flow through the rock) $u = q/(2\pi rh)$ is inversely proportional to the distance from the well axis. It grows as we come closer to the well.

Darcy's law says that the velocity is proportional to the pressure gradient: $u = (k/\mu)\,\dd p/\dd r$ . (The pressure rises with distance from the well, while the flow is directed towards it.) We set the two expressions for the velocity equal and separate the variables: $\dd p = \dfrac{q\mu}{2\pi kh}\,\dfrac{\dd r}{r}$ . Integrating from the wall of the well ( $r_w$ , $p_w$ ) to the outer boundary ( $r_e$ , $p_e$ ) gives the pressure drop and, with it, the flow rate. This is the Dupuit equation:

Steady-state radial flow (Dupuit equation)

q = \frac{2\pi kh\,(p_e - p_w)}{\mu\ln\dfrac{r_e}{r_w}}

where:

$q$ flow rate at reservoir conditions; at standard conditions it equals $q/B$ , m³/s
$k$ permeability of the reservoir to the flowing fluid, m²
$h$ net reservoir thickness (only the layers that can flow), m
$p_e$ pressure at the outer boundary, Pa
$p_w$ pressure at the wall of the well (bottomhole), Pa
$\mu$ fluid viscosity at reservoir conditions, Pa·s
$r_e$ radius of the outer boundary, m
$r_w$ well radius, m
$B$ fluid formation volume factor (reservoir volume per unit of volume at the surface); it appears in the oilfield forms, where the flow rate is at standard conditions, m³/m³

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

Holds when

Steady state: a constant pressure $p_e$ is held at the outer boundary.
The reservoir is homogeneous and isotropic (its permeability is the same in all directions) and has constant thickness. The well stands at the centre of a circular area and penetrates the whole thickness.
The flow is single-phase. The fluid is incompressible or slightly compressible and obeys Darcy's law. The flow is horizontal.
The near-wellbore zone is unchanged, so the skin factor is zero (the lesson shows below how to account for it).

The drawdown is proportional to the flow rate and the viscosity, and inversely proportional to the flow capacity $kh$ (flow capacity of the reservoir). The ratio $kh/\mu$ is called the transmissibility. The geometry enters only through $\ln(r_e/r_w)$ . If a handbook equation uses the base-10 logarithm $\lg$ , it has the factor $\ln 10 = 2.303$ in front of it.

In other unit systems (2)

Metric oilfield units

q\,[\text{m}^3/\text{d}] = \frac{0.536\,k\,h\,(p_e - p_w)}{\mu\,B\,\ln(r_e/r_w)}

$k$ in mD, $h$ in m, $p$ in MPa, $\mu$ in mPa·s, $B$ in m³/m³; $q$ is the flow rate at standard conditions.

$0.536 = 2\pi\cdot 0.0853$ , where $0.0853$ is the constant of Darcy's law in these units. It comes from converting mD to m², MPa to Pa, mPa·s to Pa·s and seconds to days.

Oilfield units

q\,[\text{STB/d}] = \frac{7.08\cdot10^{-3}\,k\,h\,(p_e - p_w)}{\mu\,B\,\ln(r_e/r_w)} = \frac{k\,h\,(p_e - p_w)}{141.2\,\mu\,B\,\ln(r_e/r_w)}

$k$ in mD, $h$ in ft, $p$ in psi, $\mu$ in cP, $B$ in rb/STB (reservoir barrels per stock-tank barrel); $q$ is in barrels of oil at standard conditions per day.

$7.08\cdot10^{-3} = 2\pi\cdot 1.127\cdot10^{-3}$ , where $1.127\cdot10^{-3}$ is the constant of Darcy's law in oilfield units; $141.2 = 1/(7.08\cdot10^{-3})$ .

If we integrate to an arbitrary radius $r$ instead of to the outer boundary, we get the pressure distribution around the well:

Pressure distribution in steady-state radial flow

p(r) = p_w + \frac{q\mu}{2\pi kh}\ln\frac{r}{r_w}

where:

$p(r)$ reservoir pressure at distance $r$ from the well axis, Pa
$r$ distance from the well axis, $r_w \le r \le r_e$ , m
$q,\ \mu,\ k,\ h,\ p_w,\ r_w$ as in the Dupuit equation; $q$ is at reservoir conditions, m³/s, Pa·s, m², m, Pa, m

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

Holds when

The same as for the Dupuit equation.

It is handy to treat the quantity $q\mu/(2\pi kh)$ as the “unit” of pressure drop in radial flow. In these units the dimensionless pressure drop (a pressure drop with no units) $p_D = (p - p_w)/(q\mu/(2\pi kh))$ is simply $\ln(r/r_w)$ , and it depends only on geometry. For real wells $r_e/r_w$ ranges from a thousand to several tens of thousands. So the total dimensionless drop $\ln(r_e/r_w)$ is roughly 7 to 11.

A simple rule follows from the logarithmic profile. Each tenfold step in distance from the well takes the same pressure drop: $\ln 10$ “units” of $q\mu/(2\pi kh)$ . For a well radius of 0.1 m, the path from its wall to 1 m “costs” as much pressure as the path from 10 to 100 m. But the volume of reservoir in these two rings is very different: the second ring holds ten thousand times more.

Share of the drawdown and share of the area within radius r (well radius 0.108 m, outer boundary 400 m)

Radius $r$: 0.5 m
Share of the drawdown: 19%
Share of the area: 0.00015%

Radius $r$: 1 m
Share of the drawdown: 27%
Share of the area: 0.00062%

Radius $r$: 3 m
Share of the drawdown: 40%
Share of the area: 0.0056%

Radius $r$: 10 m
Share of the drawdown: 55%
Share of the area: 0.062%

Radius $r$: 50 m
Share of the drawdown: 75%
Share of the area: 1.6%

Radius $r$: 100 m
Share of the drawdown: 83%
Share of the area: 6.2%

Radius $r$	Share of the drawdown	Share of the area
0.5 m	19%	0.00015%
1 m	27%	0.00062%
3 m	40%	0.0056%
10 m	55%	0.062%
50 m	75%	1.6%
100 m	83%	6.2%

The share of the drawdown is $\ln(r/r_w)/\ln(r_e/r_w)$ ; the share of the area is $(r^2 - r_w^2)/(r_e^2 - r_w^2)$ . Here each tenfold step in distance takes $\ln 10/\ln(r_e/r_w) \approx 28$ % of the drawdown.

Three practical consequences follow. First, the condition of the rock in the first few metres around the wellbore has a disproportionately strong effect on the flow rate. Drilling fluid (the mud used to drill the well) can damage this zone. A stimulation (a treatment that improves flow) can do the opposite. Either way, the flow rate changes more than it does for a change in permeability far out in the reservoir (Skin factor). Second, we need to know the radius of the outer boundary only roughly. Third, in high-rate wells the Darcy velocity is greatest at the wellbore, so Darcy's law fails there first (Inflow at high rates: non-Darcy flow).

Productivity index and the size of the area

If we divide the flow rate at standard conditions by the drawdown, we get the productivity index of steady-state inflow. It does not depend on the flow rate or on the drawdown. It depends only on the properties of the reservoir and the fluid, and on the geometry. So the inflow line is straight (Well inflow: the well performance diagram).

Productivity index in steady-state inflow

J = \frac{q_s}{p_e - p_w} = \frac{2\pi kh}{B\mu\ln\dfrac{r_e}{r_w}}

where:

$J$ productivity index with respect to the pressure at the outer boundary, m³/(s·Pa)
$q_s$ flow rate at standard conditions, m³/s
$B$ fluid formation volume factor, m³/m³
$k,\ h,\ \mu,\ r_e,\ r_w,\ p_e,\ p_w$ as in the Dupuit equation, m², m, Pa·s, m, m, Pa, Pa

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

Holds when

The conditions of the Dupuit equation hold, and the skin is zero.

Productivity is directly proportional to the flow capacity $kh$ , and inversely proportional to the viscosity and the formation volume factor. The radii appear under a logarithm, so they have little effect.

In other unit systems (2)

Metric oilfield units

J\,[\text{m}^3/(\text{d}\cdot\text{MPa})] = \frac{0.536\,k\,h}{\mu\,B\,\ln(r_e/r_w)}

$k$ in mD, $h$ in m, $\mu$ in mPa·s, $B$ in m³/m³.

$0.536 = 2\pi\cdot 0.0853$ , as in the Dupuit equation.

Oilfield units

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

$k$ in mD, $h$ in ft, $\mu$ in cP, $B$ in rb/STB. Conversion: 1 STB/(d·psi) = 23.06 m³/(d·MPa).

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

CalculatedCalculated flow rate of the well in the example below (drawdown 4.5 MPa) against the radius of the outer boundary, without skin and with a skin of +4. The radius scale is logarithmic. A tenfold increase in

r_e

reduces the flow rate less than a skin of +4 does: it adds

\ln 10 \approx 2.3

to the logarithmic term, while the skin adds 4.

The weak dependence on the radii has another side. Infill drilling (drilling extra wells between the existing ones) hardly increases the flow rate of each well. Its gain lies elsewhere: more wells, and better coverage of the reservoir. A larger bit (a wider hole) adds a few per cent. A hydraulic fracture, on the other hand, acts like a well with an enormous “effective” radius, and it increases the flow rate several times over (Skin factor).

Skin in steady-state inflow

Drilling, completion or stimulation almost always alters the real near-wellbore zone. The skin factor $S$ describes its effect. It is an extra resistance with no units, and it adds to the logarithmic term of the reservoir. The resistances of the reservoir and of the near-wellbore zone act one after the other (in series), like two sections of one pipe:

Steady-state inflow with a skin factor

q_s = \frac{2\pi kh\,(p_e - p_{wf})}{B\mu\left(\ln\dfrac{r_e}{r_w} + S\right)}

where:

$q_s$ flow rate at standard conditions, m³/s
$p_{wf}$ actual bottomhole flowing pressure of the producing well, Pa
$S$ skin factor: $S > 0$ means that inflow is hindered, $S < 0$ means that it is eased, —
$k,\ h,\ B,\ \mu,\ r_e,\ r_w,\ p_e$ as in the Dupuit equation, m², m, m³/m³, Pa·s, m, m, Pa

Units: SI; in metric and oilfield units the constants are the same as in the Dupuit equation

Holds when

The altered zone is small compared with the drainage area, and the flow in it can be treated as steady-state.
The skin does not depend on the flow rate (there are no inertial losses).

The share of the drawdown lost in the near-wellbore zone is $S/(\ln(r_e/r_w) + S)$ . With a logarithmic term of about 8, a skin of +4 takes a third of the drawdown, and a skin of +8 takes a half. The lesson Skin factor explains where the skin comes from and how to calculate it from the size and permeability of the altered zone.

Worked exampleFlow rate from the Dupuit equation: without skin, with skin and in oilfield units

Given (assumed data)

Permeability $k$ ; thickness $h$: 60 mD; 18 m
Oil viscosity $\mu$ ; formation volume factor $B$: 1.8 mPa·s; 1.25 m³/m³
Well radius $r_w$ (216 mm bit); radius of the outer boundary $r_e$: 0.108 m; 400 m
Pressure at the outer boundary $p_e$ ; bottomhole pressure $p_{wf}$: 22.0 MPa; 17.5 MPa
Skin factor (from a well test): +4

Solution

Logarithmic term (natural logarithm): $\ln(400/0.108) = 8.217$ .
Without skin: $q = \dfrac{0.536\cdot 60\cdot 18\cdot 4.5}{1.8\cdot 1.25\cdot 8.217} = 141$ m³/d; productivity index $141/4.5 = 31.3$ m³/(d·MPa).
With a skin of +4, the denominator contains $8.217 + 4 = 12.217$ . Then $q = 141\cdot 8.217/12.217 = 95$ m³/d, which is 67% of the ideal value (the value without skin).
Pressure drop across the skin: $\Delta p_s = 4.5\cdot 4/12.217 = 1.47$ MPa.
Check in oilfield units: 18 m = 59.06 ft, 4.5 MPa = 652.7 psi; $q = \dfrac{7.08\cdot10^{-3}\cdot 60\cdot 59.06\cdot 652.7}{1.8\cdot 1.25\cdot 8.217} = 886$ STB/d.
Converting barrels to cubic metres: $886\cdot 0.159 = 141$ m³/d. This is the same result as in metric units.

Answer: Without skin the well would deliver about 141 m³/d. With a skin of +4 it delivers about 95 m³/d. Of the 4.5 MPa drawdown, 1.47 MPa is lost in the near-wellbore zone.

What it means. A moderate skin of +4 took away a third of the flow rate. The logarithmic term is only about 8, so four units of skin add half as much again to the resistance of the reservoir. The two systems of units give the same flow rate if you convert all the quantities carefully. The constants 0.536 and $7.08\cdot10^{-3}$ differ only because the units differ. The data in this example are assumed for illustration.

Average pressure in the drainage area

It is hard to measure the pressure at the outer boundary directly. The average pressure in the drainage area, $\bar p$ , is found from the pressure build-up after the well is shut in (the pressure rises once the flow stops) and from the material balance (a balance between the fluid produced and the fall in pressure). So it is more convenient to write the inflow equation with $\bar p$ . We average over the pore volume: each ring of the reservoir counts with the weight of its volume, $2\pi rh\phi\,\dd r$ . We put the logarithmic profile into this average and integrate from $r_w$ to $r_e$ . The result is:

Average pressure in steady-state radial flow

\bar p - p_w = \frac{q_s B\mu}{2\pi kh}\left(\ln\frac{r_e}{r_w} - \frac12\right),\qquad p_e - \bar p = \frac12\cdot\frac{q_s B\mu}{2\pi kh}

where:

$\bar p$ pressure averaged over the pore volume of the drainage area, Pa
$q_s$ flow rate at standard conditions, m³/s
$p_w,\ p_e$ pressure at the wall of the well and at the outer boundary, Pa
$B,\ \mu,\ k,\ h,\ r_e,\ r_w$ as in the Dupuit equation, m³/m³, Pa·s, m², m, m, m
$J$ productivity index based on the average pressure (metric form), m³/(d·MPa)
$S$ skin factor, if the near-wellbore zone is altered, —

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

Holds when

Steady-state inflow in a circular area; porosity and thickness are constant.
$r_w \ll r_e$ : terms of the order of $(r_w/r_e)^2$ are left out.

Most of the pore volume lies far from the well, where the pressure is close to $p_e$ . So the average pressure is only half a “unit” of radial pressure drop below the pressure at the outer boundary. The logarithmic profile reaches the value $\bar p$ at a distance of $r_e\,\e^{-1/2} \approx 0.61\,r_e$ .

In other unit systems (1)

Metric oilfield units (productivity index)

J = \frac{q_s}{\bar p - p_w} = \frac{0.536\,k\,h}{\mu\,B\left(\ln\dfrac{r_e}{r_w} - \dfrac12 + S\right)}

$J$ in m³/(d·MPa), $k$ in mD, $h$ in m, $\mu$ in mPa·s, $B$ in m³/m³; the skin is added in the same way as in the equation written with $p_e$ .

$0.536$ is the same constant; in oilfield units it is $7.08\cdot10^{-3}$ .

The logarithmic term in radial inflow equations

Regime: Steady state
Pressure in the equation: at the outer boundary $p_e$
Logarithmic term: $\ln(r_e/r_w) + S$

Regime: Steady state
Pressure in the equation: average $\bar p$
Logarithmic term: $\ln(r_e/r_w) - 1/2 + S$

Regime: Pseudo-steady state
Pressure in the equation: at the outer boundary $p_e$
Logarithmic term: $\ln(r_e/r_w) - 1/2 + S$

Regime: Pseudo-steady state
Pressure in the equation: average $\bar p$
Logarithmic term: $\ln(r_e/r_w) - 3/4 + S$

Regime	Pressure in the equation	Logarithmic term
Steady state	at the outer boundary $p_e$	$\ln(r_e/r_w) + S$
Steady state	average $\bar p$	$\ln(r_e/r_w) - 1/2 + S$
Pseudo-steady state	at the outer boundary $p_e$	$\ln(r_e/r_w) - 1/2 + S$
Pseudo-steady state	average $\bar p$	$\ln(r_e/r_w) - 3/4 + S$

The lesson Pseudo-steady-state flow and the shape factor derives the pseudo-steady-state inflow equations. The difference between the constants is small compared with the logarithm, but you must allow for it when you compare productivities.

Worked exampleWhy permeability cannot be found from a steady-state rate

Given (assumed data)

Steady-state operation under pressure maintenance: $q$ = 120 m³/d at $p_{wf}$ = 17.0 MPa
From a pressure build-up test in the same well: $\bar p$ = 21.3 MPa; flow capacity $kh$ = 1500 mD·m
$B$ ; $\mu$ ; thickness $h$: 1.20; 2.0 mPa·s; 12 m
$r_e$ ; $r_w$: 350 m; 0.1 m

Solution

Logarithmic term for the average pressure: $\ln(350/0.1) - 0.5 = 7.661$ .
If we assume zero skin, the steady-state rate gives the flow capacity $kh = qB\mu(\ln(r_e/r_w) - 1/2)/(0.536\,\Delta p)$ .
Putting in the numbers: $kh = \dfrac{120\cdot 1.2\cdot 2.0\cdot 7.661}{0.536\cdot 4.3} = 957$ mD·m. So the “apparent” permeability is $957/12 = 79.8$ mD.
The build-up test gives a flow capacity of 1500 mD·m, which means $k$ = 125 mD. The skin explains the difference: $S = \dfrac{0.536\cdot 1500\cdot 4.3}{120\cdot 1.2\cdot 2.0} - 7.661 = 4.34$ .

Answer: The steady-state rate gives an “apparent” permeability of 79.8 mD. This is 36% lower than the value from the build-up test. A skin of about 4.34 accounts for the difference.

What it means. A steady-state rate gives a single number, the productivity, and permeability and skin cannot be told apart in it. The “apparent” permeability calculated with zero skin is too low for a damaged well and too high for a stimulated one. Recording the pressure in the transient regime separates them (Well testing: purposes and types of test). The data in this example are assumed for illustration.

Steady-state gas inflow

Gas is highly compressible (its volume changes a lot with pressure), and this changes the derivation. The same quantity of gas still passes through every circle around the well. So the mass flow rate is constant, and so is the flow rate at standard conditions $q_{sc}$ . The volume flow rate in the reservoir, however, increases towards the well, because the pressure falls and the gas expands. At reservoir conditions the volume of gas is $q_{sc}B_g$ . Here the formation volume factor $B_g = p_{sc}ZT/(pT_{sc})$ is inversely proportional to pressure.

We put this volume into Darcy's law. Under the integral we then get $p\,\dd p/(\mu Z)$ instead of $\dd p$ . Suppose that the product $\mu Z$ of the viscosity and the gas deviation factor (which shows how far a real gas is from an ideal gas) changes little over the pressure range. Then we can take it out of the integral, at its value at the average pressure. The inflow turns out to be proportional to the difference of the squares of the pressures:

Steady-state radial gas inflow

q_{sc} = \frac{\pi kh\,T_{sc}\,(p_e^2 - p_w^2)}{p_{sc}\,T\,\bar\mu\bar Z\ln\dfrac{r_e}{r_w}}

where:

$q_{sc}$ gas flow rate at standard conditions, m³/s
$p_e,\ p_w$ absolute pressures at the outer boundary and at the bottomhole, Pa
$T$ reservoir temperature, K
$T_{sc},\ p_{sc}$ temperature and pressure at standard conditions, K, Pa
$\bar\mu,\ \bar Z$ gas viscosity and gas deviation factor at the average pressure $(p_e + p_w)/2$ , Pa·s, —
$k,\ h,\ r_e,\ r_w$ as in the Dupuit equation, m², m, m, m

Units: SI; absolute pressures only; $\ln$ is the natural logarithm

Holds when

Steady-state inflow at constant temperature (isothermal). The reservoir is homogeneous, and the well penetrates its whole thickness.
$\mu Z$ changes little between $p_w$ and $p_e$ , which holds roughly for pressures up to 14 MPa. At higher pressures we use the pseudo-pressure (advanced section).
Darcy's law holds. In gas wells at high flow rates, inertial losses are added (Inflow at high rates: non-Darcy flow). The skin is added to the logarithm in the same way as for oil.

We have $p_e^2 - p_w^2 = 2\bar p_m(p_e - p_w)$ with $\bar p_m = (p_e + p_w)/2$ . So the equation is the Dupuit equation for the volume flow rate at the average pressure, divided by $B_g$ at that pressure. This gives a rule: for gas, we evaluate everything that depends on pressure at the average pressure.

In other unit systems (2)

Metric oilfield units

q_{sc}\,[\text{m}^3/\text{d}] = \frac{0.2679\,k\,h\,T_{sc}\,(p_e^2 - p_w^2)}{p_{sc}\,T\,\bar\mu\bar Z\ln(r_e/r_w)}

$k$ in mD, $h$ in m, $p$ in MPa (absolute), $T$ in K, $\mu$ in mPa·s. For standard conditions of 20 °C and 0.101325 MPa, the factor is $0.2679\,T_{sc}/p_{sc} = 775$ .

$0.2679 = \pi\cdot 0.08527$ is half the constant 0.536, because $\pi = 2\pi/2$ .

Oilfield units

q_{sc}\,[\text{Mscf/d}] = \frac{k\,h\,(p_e^2 - p_w^2)}{1422\,T\,\bar\mu\bar Z\ln(r_e/r_w)}

$k$ in mD, $h$ in ft, $p$ in psia (absolute), $T$ in degrees Rankine (the absolute temperature scale that goes with °F), $\mu$ in cP; the flow rate is in thousand standard cubic feet per day.

$1422$ comes from the same constant of Darcy's law at standard conditions of 14.7 psia and 60 °F (520 °R).

Worked exampleFlow rate of a gas well in steady-state inflow

Given (assumed data)

Permeability $k$ ; thickness $h$: 8 mD; 12 m
Pressures at the outer boundary and at the bottomhole (absolute): 12.0 MPa and 9.0 MPa
Reservoir temperature $T$: 358 K (85 °C)
At the average pressure of 10.5 MPa: gas viscosity; gas deviation factor $Z$: 0.016 mPa·s; 0.88
$r_e$ ; $r_w$: 500 m; 0.1 m
Standard conditions: 293.15 K and 0.101325 MPa

Solution

Logarithmic term: $\ln(500/0.1) = 8.517$ . Difference of the squares of the pressures: $12^2 - 9^2 = 63$ MPa².
Factor for standard conditions: $0.2679\cdot 293.15/0.101325 = 775$ .
Flow rate: $q_{sc} = 775\cdot\dfrac{8\cdot 12\cdot 63}{358\cdot 0.016\cdot 0.88\cdot 8.517} = 109$ thousand m³/d.
Check using the formation volume factor at the average pressure: $B_g = \dfrac{0.101325\cdot 0.88\cdot 358}{10.5\cdot 293.15} = 0.01037$ .
The Dupuit equation for the volume in the reservoir gives $q = \dfrac{0.536\cdot 8\cdot 12\cdot 3.0}{0.016\cdot 8.517} = 1133$ m³/d. Dividing by $B_g$ : $1133/0.01037 = 109$ thousand m³/d.

Answer: The well delivers about 109 thousand m³ of gas per day at standard conditions.

What it means. In the reservoir the same gas takes up about 1133 m³ per day, which is a hundred times less than at the surface. Both ways give the same result, because the equation with squared pressures is the Dupuit equation with the gas volume converted to the average pressure. At the real flow rates of gas wells, inertial losses at the wellbore add to this viscous resistance. So engineers find the potential of gas wells from tests at several flow rates (Inflow performance relationship (IPR)). The data in this example are assumed for illustration.

Terms and symbols

Steady-state flow: A flow regime in which the pressure and the speed at every point of the reservoir do not change with time. The inflow across the outer boundary replaces the withdrawal.
Outer boundary, or supply contour ( $r_e$ , $p_e$ ): The outer boundary of the drainage area, where injection or an aquifer holds the pressure at $p_e$ .
Dupuit equation: The equation for the flow rate of steady-state radial inflow: $q = 2\pi kh\,\Delta p/(\mu\ln(r_e/r_w))$ .
Flow capacity $kh$ and transmissibility $kh/\mu$: The flow capacity is the product of permeability and reservoir thickness. The transmissibility is that product divided by the viscosity. Both determine the flow rate at a given drawdown.
Dimensionless pressure $p_D$: The pressure drop measured in “units” of $q\mu/(2\pi kh)$ . In steady-state flow $p_D = \ln(r/r_w)$ .
Average reservoir pressure $\bar p$: The pressure averaged over the pore volume of the drainage area. In steady-state flow it is below $p_e$ by half a “unit” of radial pressure drop.
Absolute pressure: Pressure measured from vacuum (zero). Gauge pressure is measured from atmospheric pressure. Only absolute pressures go into equations with squared pressures.

Common mistakes and misconceptions

Wrong

“If we double the radius of the drainage area, the flow rate will halve.”

In fact

The radii enter through a logarithm: for $r_w$ = 0.1 m, increasing $r_e$ from 250 to 500 m reduces the flow rate by only 8%.

Wrong

“The Dupuit equation gives the flow rate at the surface directly.”

In fact

In consistent units $q$ is the flow rate at reservoir conditions. At standard conditions it equals $q/B$ . The metric and oilfield forms show the formation volume factor explicitly.

Wrong

“The pressure at the outer boundary and the average reservoir pressure are the same thing.”

In fact

In steady-state flow $p_e - \bar p$ equals half a “unit” of $q\mu/(2\pi kh)$ . In the equation written with $\bar p$ , the term $\ln(r_e/r_w) - 1/2$ replaces $\ln(r_e/r_w)$ .

Wrong

“Gas flows to the well in the same way as oil: it is enough to put in its viscosity.”

In fact

The volume flow rate of gas increases towards the well. The mass flow rate is the one that stays constant. That is why the equation contains the difference of the squares of the absolute pressures (or of the pseudo-pressures), and why the viscosity and $Z$ are taken at the average pressure.

Wrong

“The well delivers less than the Dupuit equation predicts, so the permeability must be lower.”

In fact

More often the near-wellbore zone is to blame: damage, partial penetration and perforation all add skin. Well testing separates permeability from skin.

In depth: advanced detail

The logarithm from the continuity equation

The continuity equation (the rule that fluid is neither created nor lost) gives the same result. In steady-state flow of an incompressible fluid, nothing builds up in any volume. For radially symmetric flow (the same in every direction around the well) this is written as $\dfrac{1}{r}\dfrac{\dd}{\dd r}\!\left(r\,\dfrac{\dd p}{\dd r}\right) = 0$ . So the product $r\,\dd p/\dd r$ is constant (it is proportional to the flow rate), and the general solution is $p = a\ln r + b$ . We find the constants $a$ and $b$ from two boundary conditions: the pressures at the wall of the well and at the outer boundary, or the flow rate at the wall.

The same mathematics describes heat conduction through a pipe wall. The thermal resistance of a cylindrical layer is proportional to $\ln(r_2/r_1)$ , and the resistances of concentric layers add up. It is the same in the reservoir: the resistances of successive rings add up. The skin equation for a zone of altered permeability is based on this (Skin factor).

A steady-state profile within transient inflow

The steady-state inflow equations are useful even when the reservoir as a whole is far from steady state. After the well starts up, the pressure disturbance moves outwards into the reservoir. In the region it reached long ago, the flow rate through the circles hardly varies with radius, because the fluid there hardly expands. So near the well the pressure is distributed in the same way as in steady-state inflow. Only its level falls with time:

Quasi-steady-state pressure profile near the well

p(r, t) = p_{wf}(t) + \frac{q_sB\mu}{2\pi kh}\left(\ln\frac{r}{r_w} + S\right)

where:

$p(r,t)$ pressure at distance $r$ from the well at time $t$ (outside the altered zone), Pa
$p_{wf}(t)$ bottomhole pressure at the same time, Pa
$q_s,\ B,\ \mu,\ k,\ h,\ S$ flow rate at standard conditions, formation volume factor, viscosity, permeability, thickness, skin, m³/s, m³/m³, Pa·s, m², m, —

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

Holds when

Constant flow rate after start-up; a homogeneous reservoir.
$r$ is much smaller than the radius of investigation $\sqrt{4kt/(\phi\mu c_t)}$ (how far into the reservoir the disturbance has travelled): the disturbance reached this point long ago.

We can write the bottomhole pressure in transient inflow in the steady-state form if we replace $r_e$ by the radius of drainage $r_d = \sqrt{4kt/(\gamma\phi\mu c_t)} \approx 0.75\,r_i$ , where $r_i$ is the radius of investigation and $\gamma = 1.781$ . This radius grows with time. This is why the skin, defined through the steady-state pressure drop in a thin zone at the wellbore, enters the well-testing equations in the same way (Pressure build-up analysis).

Gas at high pressures: pseudo-pressure

If $\mu Z$ changes a lot over the pressure range, we cannot take it out of the integral. Then we use the real gas pseudo-pressure, a function that takes this dependence into account:

Gas inflow in terms of pseudo-pressure

m(p) = 2\int_{p_0}^{p}\frac{p'\,\dd p'}{\mu(p')\,Z(p')},\qquad q_{sc} = \frac{\pi kh\,T_{sc}\,\big[m(p_e) - m(p_w)\big]}{p_{sc}\,T\,\ln\dfrac{r_e}{r_w}}

where:

$m(p)$ real gas pseudo-pressure, Pa²/(Pa·s)
$p_0$ any reference pressure: only the difference of the pseudo-pressures enters the equation, Pa
$\mu(p),\ Z(p)$ gas viscosity and gas deviation factor as functions of pressure at reservoir temperature, Pa·s, —

Units: SI; absolute pressures

Holds when

Steady-state inflow at constant temperature, obeying Darcy's law. The gas properties are known from PVT data (laboratory measurements of how pressure, volume and temperature are linked).

For constant $\mu Z$ the difference of the pseudo-pressures equals $(p_e^2 - p_w^2)/(\mu Z)$ , so we get back the equation with squared pressures. At high pressures the ratio $p/(\mu Z)$ is almost constant, and the inflow becomes proportional simply to the pressure difference, as for a liquid. The pseudo-pressure works over the whole range, so it is the standard choice in software that interprets gas well tests.

Connections to other topics

Rock permeability — Darcy's law, on which the derivation is built.
Skin factor — the additional resistance of the near-wellbore zone and the effective wellbore radius.
Well inflow: the well performance diagram — how the inflow equation and the lift curve together set the flow rate.
Pseudo-steady-state flow and the shape factor — inflow without recharge: the constant −3/4 and the shape factors of the area.
Inflow performance relationship (IPR) — the inflow performance curve when the inflow stops being single-phase.
Well testing: purposes and types of test — how permeability and skin are separated.
Inflow at high rates: non-Darcy flow — inertial losses in gas wells and high-rate wells.
Gas formation volume factor, compressibility and viscosity — the formation volume factor, viscosity and compressibility of gas.

Check yourself

1
A well of radius 0.1 m operates in steady state. How will its flow rate change at the same drawdown if the radius of the outer boundary increases from 250 to 500 m?
1. A.It will increase by about 8%
2. B.It will decrease by about 8%
3. C.It will not change
4. D.It will halve
Show the answer
Correct answer: B. It will decrease by about 8%
The flow rate is inversely proportional to $\ln(r_e/r_w)$ : $\ln 2500 = 7.82$ , $\ln 5000 = 8.52$ , and the ratio is $7.82/8.52 = 0.92$ . The size of the area enters through a logarithm, so it has little effect.
2
A well of radius 0.1 m, outer boundary at 300 m. What share of the drawdown is used up within 1 m of the well axis?
1. A.About 29%
2. B.About 58%
3. C.About 0.3%
4. D.About 3%
Show the answer
Correct answer: A. About 29%
The share is $\ln 10/\ln 3000 = 2.30/8.01 \approx 0.29$ . Almost a third of the drawdown is lost in a ring less than a metre thick. That ring holds only thousandths of a per cent of the volume of the area.
3
Why does the steady-state gas inflow equation contain the difference of the squares of the pressures, and why must the pressures be absolute?
Show the answer
It is the mass flow rate, not the volume flow rate, that stays constant: the volume of gas in the reservoir is inversely proportional to pressure. When we put the volume $q_{sc}B_g$ into Darcy's law, we get $p\,\dd p$ under the integral, and its integral is the difference of squares. The volume of gas is inversely proportional to the absolute pressure, not to the gauge pressure. Also, atmospheric pressure does not cancel in a difference of squares. So gauge pressures give an error, which is especially noticeable at low pressures.
4
We calculate the productivity index in steady-state operation, using the average reservoir pressure from a pressure build-up test. Which logarithmic term should stand in the denominator (without skin)?
1. A. $\lg(r_e/r_w)$
2. B. $\ln(r_e/r_w) - 3/4$
3. C. $\ln(r_e/r_w)$
4. D. $\ln(r_e/r_w) - 1/2$
Show the answer
Correct answer: D. $\ln(r_e/r_w) - 1/2$
In steady-state inflow the average pressure is below the pressure at the outer boundary by half a “unit” of $q\mu/(2\pi kh)$ . So with $\bar p$ the term is $\ln(r_e/r_w) - 1/2$ . The term $-3/4$ belongs to the pseudo-steady-state regime. The base-10 logarithm without the factor 2.303 is wrong here in any case.
5
Calculate the flow rate in oilfield units: $kh$ = 1000 mD·ft, drawdown 500 psi, $\mu$ = 1.0 cP, $B$ = 1.2, $\ln(r_e/r_w)$ = 8.0, zero skin. How many cubic metres per day is that?
Show the answer
The denominator is $141.2\cdot 1.0\cdot 1.2\cdot 8.0 = 1355.5$ , and $q = 5\cdot10^{5}/1355.5 \approx 369$ STB/d, or $369\cdot 0.159 \approx 59$ m³/d at standard conditions. With the constant $7.08\cdot10^{-3}$ the result is the same: $3540/9.6 \approx 369$ .
6
A well in steady-state operation delivers 40% less than the Dupuit equation predicts with the core permeability. Name the possible causes, and say how to check them.
Show the answer
Most likely the cause is a positive skin: damage to the near-wellbore zone, partial penetration of the reservoir, too few perforation tunnels, or tunnels that are too short. Other causes are possible. The permeability measured on core (rock samples from the well) may differ from the effective permeability of the reservoir to oil. Gas may come out of the oil near the bottomhole (pressure below the bubble point pressure). The pressure at the outer boundary, or the thickness of the part of the reservoir that contributes to flow, may have been estimated wrongly. To check, run a well test, which gives the flow capacity $kh$ and the skin separately, and run production logging (flow measurements along the well) of the flowing intervals.

Updated: 26 October 2026

Quick refresher

After this lesson you can

When the inflow is steady-state

The Dupuit equation

Where the drawdown is spent

Productivity index and the size of the area

Skin in steady-state inflow

Average pressure in the drainage area

Steady-state gas inflow

Terms and symbols

Common mistakes and misconceptions

In depth: advanced detail

Connections to other topics

Check yourself