Dimensionless well solutions: skin, radius of investigation and units

Quick refresher

There is one dimensionless solution for all wells. Pressure is in “units” of $qB\mu/(2\pi kh)$ , time is in units of $r_w^2/\eta$ (fractions of a second), and distances are in well radii. A few groups then describe the well and the reservoir: $S$ , $C_D$ , $r_{eD}$ , $L_D$ , $t_{DA}$ .
Dimensionless quantities do not depend on the system of units. $t_D$ and $C_D$ calculated in metric units and in oilfield units are the same, and this is how we check the constants and the conversions.
A line source (a well of zero radius) passes through the wall of the well only the share $\e^{-1/(4t_D)}$ of the rate: at $t_D = 1$ this is 78%, and at $t_D = 25$ it is 99%. A few seconds after start-up it already replaces the exact solution. In fact, the first minutes of the record are set by wellbore storage.
We can add the skin to the solution, or replace it by an effective wellbore radius, $r_w' = r_w\e^{-S}$ . At late times the two are the same. At the beginning, a negative skin is described correctly only by the “enlarged” well. The semilog straight line starts when $t_D\e^{2S}$ reaches about 100.
The radius of investigation, $r_i = \sqrt{4\eta t}$ , is a conventional boundary. There the rate of pressure decline is just passing through its maximum (the pressure falls fastest right at the well). 85% of the volume withdrawn from the reservoir comes from inside $r_i$ , and 95% comes from inside $1.33\,r_i$ . An accurate gauge in a neighbouring well notices a response about twice as far away.
The radius of drainage, $r_d = 0.75\,r_i$ , is the radius of steady-state flow with the same pressure drop. While it grows, the productivity index falls. After pseudo-steady state is reached, we calculate productivity with respect to the average reservoir pressure.

Key relationships

t_D = \dfrac{kt}{\phi\mu c_t r_w^2},\quad p_D = \dfrac{2\pi kh\,\Delta p}{qB\mu},\quad C_D = \dfrac{C}{2\pi\phi c_t h r_w^2}

dimensionless time, pressure and wellbore storage coefficient (SI)

q(r, t) = q\,\e^{-r^2/(4\eta t)}

flow rate through a circle of radius $r$ in the line-source solution

r_i = \sqrt{4\eta t},\qquad r_d = \sqrt{4\eta t/\gamma} = 0.75\,r_i

radius of investigation and radius of drainage; $\eta = k/(\phi\mu c_t)$ , $\gamma = 1.781$

\dfrac{V(r)}{V} = 1 - \e^{-X} + X\,E_1(X),\quad X = \dfrac{r^2}{4\eta t}

share of the withdrawn volume supplied by the reservoir inside radius $r$

J_t = \dfrac{2\pi kh}{B\mu\,[\ln(r_d/r_w) + S]}

transient productivity index (SI; in metric oilfield units $2\pi$ is replaced by 0.536)

Common pitfalls

Putting oilfield constants (2.637·10⁻⁴, 141.2, 0.8936) into metric data, or the other way round. The check: $t_D$ and $C_D$ in the two systems must be the same.
Adding a large negative skin to the finite-radius solution in the first moments. The result is a “rise” of pressure during production.
Taking the radius of investigation as a sharp boundary beyond which the pressure has not yet changed.
Mixing up the radius of drainage with the radius of investigation, or with the radius of the drainage area $r_e$ .
Taking the productivity from a short flow period as the stabilised one. It is too high while the radius of drainage is still growing.

After this lesson you can

list the dimensionless groups of well testing, and convert dimensionless solutions to metric and oilfield units with checked constants
explain how a finite-radius well differs from a line source (a well replaced by a line of zero radius), and use a numerical criterion to decide when a line source is good enough: at the wall of the well, away from it, and with a negative skin
put the skin into a dimensionless solution, either as an added term or as an effective wellbore radius, and estimate when a zone of changed permeability of finite size behaves like a thin skin
calculate the radius of investigation by several definitions, and explain why they give different numbers
tell the radius of drainage from the radius of investigation, and estimate the transient productivity index (the productivity index while the pressure is still changing)

Dimensionless groups of the well

The problem of a well in a homogeneous reservoir contains about ten dimensional quantities (quantities with units): permeability, thickness, porosity, viscosity, compressibility, well radius, rate, formation volume factor, wellbore volume and the distances to the boundaries. Dimensionless variables reduce them to a few groups (the dimensionless form of the equation). We measure distances in well radii, pressure in “units” of the radial-flow pressure drop, and time in units of $r_w^2/\eta$ . In that time the disturbance travels a distance of about one well radius. For an ordinary well this is a fraction of a second.

Each new feature of the well or the reservoir adds one group to the problem. The skin adds the number $S$ , the wellbore volume adds $C_D$ , and a boundary adds the dimensionless distance to it. The solution becomes a function of $t_D$ and these parameters, $p_D = f(t_D;\,S,\,C_D,\,r_{eD},\dots)$ . Two wells with the same groups have the same curve $p_D(t_D)$ , only on different pressure and time scales. Type curves (Type-curve matching) are built on this.

Dimensionless groups of well testing

Group: $r_D = r/r_w$
What it compares: the distance compared with the well radius
Typical values: 1 at the wall; $10^3$ – $10^5$ at the boundaries

Group: $t_D = \eta t/r_w^2$
What it compares: the time compared with the time the disturbance needs to cross the well radius
Typical values: from tens to millions per hour

Group: $p_D = \Delta p\cdot 2\pi kh/(qB\mu)$
What it compares: the pressure drop compared with the “unit” of radial flow
Typical values: 5–20 in radial flow

Group: $S$
What it compares: the extra pressure drop at the wellbore, in the same “units”
Typical values: from −5 to +20 and more

Group: $C_D = C/(2\pi\phi c_thr_w^2)$
What it compares: the storage capacity of the wellbore compared with that of the reservoir around the wellbore
Typical values: from single digits to $10^4$ and more

Group: $r_{eD} = r_e/r_w$ , $L_D = L/r_w$
What it compares: the size of the area and the distances to the boundaries
Typical values: $10^3$ – $10^5$

Group: $t_{DA} = \eta t/A$
What it compares: the time compared with the time needed to cover the area $A$
Typical values: 0.1: the start of pseudo-steady state in a circle

Group	What it compares	Typical values
$r_D = r/r_w$	the distance compared with the well radius	1 at the wall; $10^3$ – $10^5$ at the boundaries
$t_D = \eta t/r_w^2$	the time compared with the time the disturbance needs to cross the well radius	from tens to millions per hour
$p_D = \Delta p\cdot 2\pi kh/(qB\mu)$	the pressure drop compared with the “unit” of radial flow	5–20 in radial flow
$S$	the extra pressure drop at the wellbore, in the same “units”	from −5 to +20 and more
$C_D = C/(2\pi\phi c_thr_w^2)$	the storage capacity of the wellbore compared with that of the reservoir around the wellbore	from single digits to $10^4$ and more
$r_{eD} = r_e/r_w$ , $L_D = L/r_w$	the size of the area and the distances to the boundaries	$10^3$ – $10^5$
$t_{DA} = \eta t/A$	the time compared with the time needed to cover the area $A$	0.1: the start of pseudo-steady state in a circle

$\eta = k/(\phi\mu c_t)$ is the hydraulic diffusivity (how fast pressure spreads through the rock), and $\Delta p = p_i - p$ . For type curves we combine the groups: $t_D/C_D$ and $C_D\e^{2S}$ (Type-curve matching), and $t_D/L_D^2 = \eta t/L^2$ for boundaries (Semi-infinite systems: theory and test analysis).

Constants in three systems of units

To go from a dimensionless solution back to quantities with units, we do two things. We multiply the dimensionless time by the characteristic time $\phi\mu c_tr_w^2/k$ , and we multiply the dimensionless pressure by the pressure “unit”. In metric oilfield units and in oilfield units, these scales contain constants: conversion factors collected into a single number (formula constants).

Conversion of a dimensionless solution to pressure drop and time

\Delta p = \frac{qB\mu}{2\pi kh}\,p_D,\qquad t = \frac{\phi\mu c_t r_w^2}{k}\,t_D

where:

$\Delta p$ pressure decline $p_i - p$ at the point considered, Pa
$t$ time since start-up of the well, s
$p_D,\ t_D$ dimensionless pressure and time read from the solution or the type curve, —
$q,\ B$ rate at standard conditions and formation volume factor; $qB$ is the rate at reservoir conditions, m³/s, m³/m³
$k,\ h,\ \mu$ permeability, reservoir thickness, viscosity, m², m, Pa·s
$\phi,\ c_t,\ r_w$ porosity, total compressibility, well radius, —, 1/Pa, m

Units: SI

Holds when

The conditions of the diffusivity equation hold: the fluid is single-phase and slightly compressible, and the properties of the reservoir and the fluid are constant (Diffusivity equation).
$\Delta p$ is positive for production. For an injection well the sign is opposite.

Both formulas are the definitions of $p_D$ and $t_D$ , solved for the quantities with units. Everything that makes one well differ from another with the same dimensionless parameters sits in two scales: the pressure “unit” $qB\mu/(2\pi kh)$ and the characteristic time $\phi\mu c_tr_w^2/k$ .

In other unit systems (2)

Metric oilfield units

\Delta p = 1.866\,\frac{qB\mu}{kh}\,p_D,\qquad t = \frac{\phi\mu c_t r_w^2}{3.553\cdot10^{-3}\,k}\,t_D

$q$ in m³/d at standard conditions, $\mu$ in mPa·s, $k$ in mD, $h$ and $r_w$ in m, $c_t$ in 1/MPa; $\Delta p$ in MPa, $t$ in h.

$1.866 = 1/(2\pi\cdot0.0853)$ , where $0.0853$ is the constant of Darcy's law in these units, and $3.553\cdot10^{-3} = 0.9869\cdot10^{-15}\cdot3600/(10^{-3}\cdot10^{-6})$ .

Oilfield units

\Delta p = 141.2\,\frac{qB\mu}{kh}\,p_D,\qquad t = \frac{\phi\mu c_t r_w^2}{2.637\cdot10^{-4}\,k}\,t_D

$q$ in STB/d (stock-tank barrels per day), $\mu$ in cP (centipoise), $k$ in mD, $h$ and $r_w$ in ft, $c_t$ in 1/psi; $\Delta p$ in psi, $t$ in h.

$141.2 = 1/(2\pi\cdot1.127\cdot10^{-3})$ , where $1.127\cdot10^{-3}$ is the constant of Darcy's law in oilfield units. $2.637\cdot10^{-4}$ is the same kind of conversion, for millidarcy, centipoise, psi, feet and hours.

The constant

a

of the dimensionless groups and test formulas in three systems of units

Quantity: $t_D$
Formula: $a\,kt/(\phi\mu c_tr_w^2)$
SI: 1
Metric oilfield: $3.553\cdot10^{-3}$
Oilfield: $2.637\cdot10^{-4}$

Quantity: $p_D$
Formula: $a\,kh\,\Delta p/(qB\mu)$
SI: $2\pi$
Metric oilfield: $1/1.866$
Oilfield: $1/141.2$

Quantity: $C_D$
Formula: $a\,C/(\phi c_thr_w^2)$
SI: $1/(2\pi)$
Metric oilfield: $1/(2\pi)$
Oilfield: $0.8936$

Quantity: $t_D/C_D$
Formula: $a\,kh\,t/(\mu C)$
SI: $2\pi$
Metric oilfield: $1/44.8$
Oilfield: $2.951\cdot10^{-4}$

Quantity: $r_i$
Formula: $a\sqrt{kt/(\phi\mu c_t)}$
SI: 2
Metric oilfield: $0.119$
Oilfield: $0.0325$

Quantity: $r_d$
Formula: $a\sqrt{kt/(\phi\mu c_t)}$
SI: $1.499$
Metric oilfield: $0.0893$
Oilfield: $0.0243$

Quantity: $m$
Formula: $a\,qB\mu/(kh)$
SI: $0.1832$
Metric oilfield: $2.149$
Oilfield: $162.6$

Quantity: $m_e$
Formula: $a\,qB\mu/(kh)$
SI: $1/(4\pi)$
Metric oilfield: $0.933$
Oilfield: $70.6$

Quantity	Formula	SI	Metric oilfield	Oilfield
$t_D$	$a\,kt/(\phi\mu c_tr_w^2)$	1	$3.553\cdot10^{-3}$	$2.637\cdot10^{-4}$
$p_D$	$a\,kh\,\Delta p/(qB\mu)$	$2\pi$	$1/1.866$	$1/141.2$
$C_D$	$a\,C/(\phi c_thr_w^2)$	$1/(2\pi)$	$1/(2\pi)$	$0.8936$
$t_D/C_D$	$a\,kh\,t/(\mu C)$	$2\pi$	$1/44.8$	$2.951\cdot10^{-4}$
$r_i$	$a\sqrt{kt/(\phi\mu c_t)}$	2	$0.119$	$0.0325$
$r_d$	$a\sqrt{kt/(\phi\mu c_t)}$	$1.499$	$0.0893$	$0.0243$
$m$	$a\,qB\mu/(kh)$	$0.1832$	$2.149$	$162.6$
$m_e$	$a\,qB\mu/(kh)$	$1/(4\pi)$	$0.933$	$70.6$

Metric oilfield units: $k$ in mD, $h$ and $r$ in m, $t$ in h, $\mu$ in mPa·s, $c_t$ in 1/MPa, pressure in MPa, $q$ in m³/d at standard conditions, $C$ in m³/MPa. Oilfield units: $k$ in mD, $h$ and $r$ in ft, $t$ in h, $\mu$ in cP, $c_t$ in 1/psi, pressure in psi, $q$ in STB/d, $C$ in bbl/psi. Here $m$ is the slope of the semilog straight line per decimal cycle (Pressure drawdown analysis), $m_e$ is the level of the derivative plateau (the flat part of the pressure derivative curve; the derivative is the slope of the pressure against the logarithm of time; Diagnosis with the pressure derivative), and $r_d$ is the radius of drainage (see the section below). $t_{DA}$ has the same constant as $t_D$ , with the area $A$ in place of $r_w^2$ (in oilfield units the area is in square feet; 1 acre = 43,560 ft²). All constants are recalculated from the exact definitions: 1 ft = 0.3048 m, 1 bbl = 0.158987 m³, 1 psi = 6894.757 Pa, 1 mD = $0.9869233\cdot10^{-15}$ m², 1 cP = 1 mPa·s.

Worked exampleA report in oilfield units: checking the dimensionless groups

Given (assumed data)

Well and reservoir (from a service company's report): $q$ = 503 STB/d; $B$ = 1.25; $\mu$ = 1.1 cP; $k$ = 40 mD; $h$ = 32.8 ft; $\phi$ = 0.19
Compressibility, radius, wellbore, skin: $c_t = 1.10\cdot10^{-5}$ psi⁻¹; $r_w$ = 0.328 ft; $C = 5.2\cdot10^{-4}$ bbl/psi; $S$ = 2
Conversions: 1 ft = 0.3048 m; 1/psi = 145.04 1/MPa; 1 bbl/psi = 23.06 m³/MPa

Solution

In oilfield units, the dimensionless time per hour is found like this: $\phi\mu c_tr_w^2 = 0.19\cdot1.1\cdot1.10\cdot10^{-5}\cdot0.328^2 = 2.4734\cdot10^{-7}$ , and $t_D = 2.637\cdot10^{-4}\cdot40/(2.4734\cdot10^{-7}) = 42{,}646$ .
The dimensionless wellbore storage coefficient is $C_D = 0.8936\cdot5.2\cdot10^{-4}/(0.19\cdot1.10\cdot10^{-5}\cdot32.8\cdot0.328^2) = 63.0$ .
Now the same data in metric units: $c_t = 1.5954\cdot10^{-3}$ MPa⁻¹, $r_w = 0.09997$ m, $h = 9.997$ m, and $C = 5.2\cdot10^{-4}\cdot23.06 = 0.01199$ m³/MPa.
The metric calculation gives $t_D = 3.553\cdot10^{-3}\cdot40/(0.19\cdot1.1\cdot1.5954\cdot10^{-3}\cdot0.09997^2) = 42{,}648$ per hour and $C_D = 0.01199/(2\pi\cdot0.19\cdot1.5954\cdot10^{-3}\cdot9.997\cdot0.09997^2) = 63.0$ .
By the rule $t_D \approx (60 + 3.5S)\,C_D$ (Wellbore storage), wellbore storage ends at $t_D = 67\cdot63.0 = 4221$ . That is $4221/42{,}648 = 0.0990$ h, or about 5.9 min after start-up.

Answer: In both systems

t_D \approx 42{,}648

per hour and

C_D \approx 63.0

. The semilog straight line can start about 5.9 min after start-up.

What it means. The check is that the dimensionless groups agree to the third or fourth significant digit. A lost or extra conversion factor would give a difference of several times. The small difference that remains comes from rounding the report data (0.328 ft is 0.09997 m). The conversion of bbl/psi to m³/MPa (the factor 23.06) is a frequent source of errors when the wellbore storage coefficient is carried over from foreign reports. The data in this example are assumed for illustration.

The finite-radius well and the line source

The solutions represent a well in two ways. The line source is a line of zero radius through which the rate is withdrawn. Its pressure is written with the exponential integral (a special mathematical function; Line-source solution and the pressure function). The finite-radius well (finite wellbore radius solution, cylindrical source) is a cylinder of radius $r_w$ with the same rate specified on its wall. Its solution is written with Bessel functions (another family of special functions). The lesson on the line source compares the two solutions at the wall of the well. Here we look at why they differ, and where exactly.

The reason can be seen in the flow rate. In the line-source solution, only a part of the well's rate passes through a circle of radius $r$ , not the whole rate:

Flow rate through a circle in the line-source solution

\frac{q(r, t)}{q} = \exp\!\left(-\frac{r^2}{4\eta t}\right) = \exp\!\left(-\frac{r_D^2}{4t_D}\right)

where:

$q(r, t)$ flow rate of fluid through a cylindrical surface of radius $r$ at time $t$ , reduced to standard conditions, m³/s
$q$ well rate at standard conditions, m³/s
$\eta$ hydraulic diffusivity $k/(\phi\mu c_t)$ , m²/s
$r_D,\ t_D$ dimensionless distance and time, —

Units: no units (a ratio of rates)

Holds when

This is the line-source solution: the reservoir is infinite and homogeneous, and the rate is constant from start-up (the derivation is in the advanced section).

The rest of the rate is supplied by the fluid and rock expanding inside the circle. For a circle the size of the well radius, this means that the line-source model takes part of the rate “from the wellbore”, that is, from rock that is not there. At $t_D = 1$ only 78% of the rate passes through the wall. At $t_D = 5$ it is 95%, at $t_D = 25$ it is 99%, and at $t_D = 100$ it is 99.75%. This gives the condition $t_D > 25$ . After that point, the two formulations differ by less than one per cent of the rate.

The exact solution for a finite-radius well in Laplace space (the result of a mathematical transform of the time variable) is written for any point of the reservoir:

Finite-radius well: pressure in the reservoir in Laplace space

\bar p_D(r_D, s) = \frac{K_0\!\left(r_D\sqrt{s}\right)}{s^{3/2}\,K_1\!\left(\sqrt{s}\right)},\qquad r_D \ge 1

where:

$\bar p_D(r_D, s)$ Laplace transform of the dimensionless pressure decline at the distance $r_D$ , —
$s$ variable of the Laplace transform with respect to $t_D$ , —
$K_0,\ K_1$ modified Bessel functions of the second kind of order zero and one, —

Units: no units

Holds when

The reservoir is infinite, the rate at the wall $r_D = 1$ is constant, and there is no skin and no wellbore storage. We return to the function of time numerically (Stehfest algorithm).

At $r_D = 1$ this is the solution at the wall given in the lesson on the line source. For the line source in the same variables, $\bar p_D = K_0(r_D\sqrt s)/s$ . The factor $1/(\sqrt s\,K_1(\sqrt s))$ tends to one at small $s$ , that is, at large times. Then the two solutions coincide.

CalculatedCalculated dimensionless pressure decline near the well at

t_D

= 1, 10 and 100: the finite-radius well (solid lines) and the line source (dashed). The distance scale is logarithmic, from the wall of the well (

r_D = 1

) to 30 well radii. The difference is large only in the first moments and right at the well.

Dimensionless pressure near the well: finite radius and line source

$r_D$: 1
$t_D = 10$ : exact: 1.651
line: 1.568
$t_D = 100$ : exact: 2.723
line: 2.708

$r_D$: 2
$t_D = 10$ : exact: 0.975
line: 0.911
$t_D = 100$ : exact: 2.032
line: 2.019

$r_D$: 5
$t_D = 10$ : exact: 0.244
line: 0.216
$t_D = 100$ : exact: 1.138
line: 1.128

$r_D$: 10
$t_D = 10$ : exact: 0.016
line: 0.012
$t_D = 100$ : exact: 0.529
line: 0.522

$r_D$	$t_D = 10$ : exact	line	$t_D = 100$ : exact	line
1	1.651	1.568	2.723	2.708
2	0.975	0.911	2.032	2.019
5	0.244	0.216	1.138	1.128
10	0.016	0.012	0.529	0.522

“Exact” is the finite-radius solution, and “line” is the line source, $\tfrac12E_1(r_D^2/(4t_D))$ . At $t_D = 100$ the difference over the whole near zone is about 0.01: from 0.015 at the wall to 0.007 at ten radii. It decreases more slowly than $1/t_D$ : at the wall, from $t_D$ = 10 to 100, it falls by almost a factor of six.

Skin in the dimensionless solution: a term or an effective radius

Skin is an additional pressure drop in a thin zone at the wellbore (definition of skin). The dimensionless solution can account for it in two ways. We can add it to the pressure at the wall. Or we can replace the well by another one, without skin but with an effective wellbore radius $r_w' = r_w\e^{-S}$ (effective radius).

Two ways of accounting for skin in the solution for a well

p_{wD}(t_D) = p_D(t_D) + S\qquad\text{or}\qquad p_{wD}(t_D) = p_D\!\left(t_D\,\e^{2S}\right)

where:

$p_{wD}$ dimensionless decline of the bottomhole pressure of a well with skin, —
$p_D(t_D)$ solution for a finite-radius well without skin, at the wall, —
$S$ skin factor, —
$\e^{2S}$ the ratio $(r_w/r_w')^2$ : the dimensionless time of a well with the effective radius is $t_D\e^{2S}$ , —

Units: no units

Holds when

The first form is the skin of an infinitely thin zone that follows the rate instantly.
The second form is a well without skin, of radius $r_w' = r_w\e^{-S}$ , in the same reservoir. The pressure is still referred to the $k$ and $h$ of the reservoir.
Wellbore storage is not included. It can be added to either form (Wellbore storage).

In the logarithmic approximation the two forms are the same: $\tfrac12[\ln(t_D\e^{2S}) + 0.809] = \tfrac12(\ln t_D + 0.809) + S$ . They differ at the beginning. An added negative skin gives a negative $p_{wD}$ at small $t_D$ , which is a “rise” of pressure during production. A well with an enlarged radius stays physical (it behaves like a real well): its pressure comes to the same straight line when $t_D\e^{2S}$ becomes larger than 25–100.

A practical consequence follows for treated wells (wells whose near-wellbore zone has been stimulated). The semilog straight line starts not at $t_D \approx 100$ but at $t_D \approx 100\,\e^{-2S}$ : at about 5500 for $S = -2$ , and at about 40,000 for $S = -3$ . For a damaged well ( $S > 0$ ) the effective radius is tiny, so the condition is met at once. But a real damaged zone has a size and its own transient (see the advanced section).

Worked exampleWhen the semilog straight line starts in a treated well

Given (assumed data)

The lesson's well after an acid treatment: $S$ = −2; $r_w$ = 0.1 m; $\eta$ = 425 m²/h; per hour $t_D$ = 42,500
Wellbore: $C$ = 0.012 m³/MPa, $C_D$ = 62.8

Solution

The effective radius is $r_w' = 0.1\cdot\e^{2} = 0.739$ m.
The condition $t_D\e^{2S} > 100$ is the same as $r_i > 20\,r_w'$ . So $t > 100\,r_w'^2/\eta = 100\cdot0.739^2/425 = 0.1285$ h, which is about 7.7 min.
By the rule $(60 + 3.5S)\,C_D$ , wellbore storage lasts until $t_D = 53\cdot62.8 = 3328$ , that is, 4.7 min. This rule is a guide for $C_D\e^{2S} > 10^3$ (Wellbore storage). Here, however, $C_D\e^{2S} = 1.15$ . The wellbore is small compared with the storage of the reservoir around the “enlarged” well, so the rule has been applied outside its range.
Check with the solution for a well of radius $r_w'$ . After 2 min ( $t_D$ = 1417, $t_D\e^{-4}$ = 25.9), $p_{wD} = 2.0793$ , while the semilog formula $\tfrac12(\ln t_D + 0.809) + S$ gives 2.0326, which is less by 2.2%. After 10 min the two values are 2.8500 and 2.8373, a difference of 0.45%.
Now include both effects: the same well of radius $r_w'$ , with wellbore storage, $C_D\e^{2S} = 1.15$ . Then $p_{wD}$ is lower than the semilog value by 0.9% after 4.7 min and by 0.56% after 7.7 min. Wellbore storage and the size of the well act in opposite directions here, on the same scale. But the conclusion is the same: before 7.7 min the straight line is no more accurate than half a per cent.

Answer: The semilog straight line of this well starts after about 7.7 min. This is later than wellbore storage ends by the usual rule (about 4.7 min).

What it means. In a treated well, the start of the straight line can be set not by the wellbore but by the “size” of the well. The more negative the skin, the longer the pressure remembers the near zone. The effective radius is a simplification (the treated zone is treated as if it were infinitely permeable). A zone of finite permeability comes to the same straight line in a similar time (see the advanced section). In a well with a hydraulic fracture, the early flow is not radial at all. It is linear towards the fracture, and the effective radius describes only the late radial flow (Inflow to a hydraulically fractured well). The data in this example are assumed for illustration.

The radius of investigation: how far a test “sees”

The disturbance has no sharp front. According to the diffusivity equation, the pressure starts to change at once throughout the reservoir. Far away it changes by a vanishingly small amount. So the question “how far does a test see?” has several answers, and each answers a different question. The radius of investigation $r_i = \sqrt{4\eta t}$ , used in practice (planning of tests), has a simple physical meaning. At a point at this distance, the rate of pressure decline is passing through its maximum at the given moment: before that it was growing, and afterwards it will fall.

Rate of pressure decline at a point in the reservoir

\frac{\partial\,\Delta p}{\partial t} = \frac{qB\mu}{4\pi kh}\cdot\frac1t\,\exp\!\left(-\frac{r^2}{4\eta t}\right)

where:

$\Delta p$ pressure decline $p_i - p$ at the distance $r$ from the well, Pa
$t$ time since start-up, s
$q,\ B,\ \mu,\ k,\ h$ rate at standard conditions, formation volume factor, viscosity, permeability, thickness, m³/s, m³/m³, Pa·s, m², m
$\eta$ hydraulic diffusivity $k/(\phi\mu c_t)$ , m²/s

Units: SI

Holds when

This is the line-source solution: the reservoir is infinite, and the rate is constant.

At a point at the distance $r$ , the rate of pressure decline first grows and then falls. Its maximum is exactly at $t = r^2/(4\eta)$ , that is, when $r = r_i$ . So $r_i$ is the place where the drawdown cone (the region where the pressure has dropped) is deepening, right now, at the highest rate for that point. It is a moving “front” of the disturbance (overall, the pressure falls fastest right at the well). The pressure decline there already equals $0.11$ of a “unit” $qB\mu/(2\pi kh)$ . After that it continues, only more slowly.

The second question is where the withdrawn fluid came from. It is supplied by the fluid and rock expanding throughout the drawdown cone. The share supplied inside radius $r$ is written with the same function $E_1$ :

The third question is where a gauge will notice the response. The answer depends on the rate and on the gauge resolution. A pressure decline $\delta p$ is reached at the distance where $m_eE_1(X) = \delta p$ , with $m_e = qB\mu/(4\pi kh)$ the level of the derivative plateau (Diagnosis with the pressure derivative). The lesson on the line source shows how this is used in interference testing. The fourth question is at what distance a boundary already changes the pressure in the well itself. We answer it with the pressure derivative: by how many per cent it rises above the plateau (the derivative step at a fault).

Radii that answer different questions, in fractions of the radius of investigation

Criterion: The rate of pressure decline passes its maximum
Radius: $r_i$
Remark: the pressure decline there is $0.11$ of a “unit” $qB\mu/(2\pi kh)$

Criterion: Steady-state flow with the same pressure drop
Radius: $0.75\,r_i$
Remark: the radius of drainage (the section below)

Criterion: Half of the withdrawn volume comes from within
Radius: $0.52\,r_i$
Remark: almost the whole pressure drop is inside this circle

Criterion: 85% of the volume comes from within
Radius: $r_i$
Remark: the share inside $r_i$ is 85.2%

Criterion: 95% of the volume comes from within
Radius: $1.33\,r_i$
Remark: 99% is inside $1.75\,r_i$

Criterion: The pressure decline is 1% of $m_e$
Radius: $1.79\,r_i$
Remark: a gauge in a neighbouring well; depends on the rate

Criterion: A fault raises the derivative by 1%
Radius: $1.07\,r_i$
Remark: by 10% — a fault not farther than $0.76\,r_i$

Criterion: A closed circle raises the derivative by 1%
Radius: $1.32\,r_i$
Remark: the radius of the circle; by 10% — $1.06\,r_i$

Criterion	Radius	Remark
The rate of pressure decline passes its maximum	$r_i$	the pressure decline there is $0.11$ of a “unit” $qB\mu/(2\pi kh)$
Steady-state flow with the same pressure drop	$0.75\,r_i$	the radius of drainage (the section below)
Half of the withdrawn volume comes from within	$0.52\,r_i$	almost the whole pressure drop is inside this circle
85% of the volume comes from within	$r_i$	the share inside $r_i$ is 85.2%
95% of the volume comes from within	$1.33\,r_i$	99% is inside $1.75\,r_i$
The pressure decline is 1% of $m_e$	$1.79\,r_i$	a gauge in a neighbouring well; depends on the rate
A fault raises the derivative by 1%	$1.07\,r_i$	by 10% — a fault not farther than $0.76\,r_i$
A closed circle raises the derivative by 1%	$1.32\,r_i$	the radius of the circle; by 10% — $1.06\,r_i$

Calculated from the line-source solution; for a closed circle with the well at the centre, from the exact solution. The ratios do not depend on the properties of the reservoir, because everything is expressed through $r_i = \sqrt{4\eta t}$ . A boundary begins to show on the derivative at about the time when the radius of investigation reaches it. The influence must reach the boundary and return, and a fault at the distance $r_i$ raises the derivative by only about 2%.

CalculatedCalculation for the lesson's well one day after start-up (

r_i

= 202 m). The orange curve is the share of the pressure drop: the pressure decline at the distance

r

divided by the decline at the wall of the well (without skin). The teal curve is the share of the withdrawn volume supplied by the reservoir inside radius

r

. The points mark the radius of drainage, the radius of investigation, and the radius inside which 95% of the volume is supplied.

Worked exampleHow far a one-day flow period “sees”: four answers

Given (assumed data)

The lesson's well: $q$ = 80 m³/d; $B$ = 1.25; $\mu$ = 1.1 mPa·s; $k$ = 40 mD; $h$ = 10 m; $\phi$ = 0.19; $c_t = 1.6\cdot10^{-3}$ MPa⁻¹; $\eta$ = 425 m²/h
Time since start-up: 24 h; there are no boundaries nearby, so the reservoir behaves as an infinite one
Gauge in a neighbouring well: reliably detects a pressure change of 0.5 kPa (0.0005 MPa)

Solution

The radius of investigation is $r_i = \sqrt{4\cdot425\cdot24} = 202$ m. The radius of drainage is $0.749\cdot202 = 151$ m.
Where the fluid comes from: half of the withdrawn volume is supplied inside $0.517\,r_i = 104$ m, and 95% inside $1.335\,r_i = 270$ m.
The neighbouring well: $m_e = 0.933\cdot80\cdot1.25\cdot1.1/(40\cdot10) = 0.2566$ MPa. We need $E_1(X) = 0.0005/0.2566 = 0.00195$ , so $X = 4.55$ and $r = \sqrt{X}\,r_i = 431$ m.
The boundary: by the end of the day a fault will raise the derivative by 10% if it is closer than $0.759\,r_i = 153$ m, and by at least 1% if it is closer than $1.073\,r_i = 217$ m.

Answer: In a day the test “reached” anywhere from 151 to 431 m, depending on what we count. The radius of drainage is 151 m and the radius of investigation is 202 m. 95% of the volume comes from within 270 m. The response in the neighbouring well extends to 431 m. A fault is noticeable on the derivative up to 217 m.

What it means. For productivity and inflow formulas we need the radius of drainage. For the question “what volume did the oil come from?” we need the radii of the volume shares. For interference testing we need the radius at which the response exceeds the gauge resolution. For finding boundaries we need the derivative criterion. The last one is slightly wider than $r_i$ at the 1% level and narrower at the 10% level. So we can recognise a fault confidently on a record only when it is noticeably closer than the radius of investigation. The data in this example are assumed for illustration.

The radius of drainage and the transient productivity

Near the well, where the disturbance arrived long ago, the pressure profile is almost steady-state. The flow rate through the circles hardly changes with the radius, and the pressure grows with distance as the logarithm, as in the Dupuit equation (quasi-steady-state profile). Let us continue this logarithmic straight line out to the initial pressure $p_i$ . The radius at which the line reaches $p_i$ is called the radius of drainage. The Dupuit equation with $r_e = r_d$ gives the same bottomhole pressure as transient flow. This is not the radius of the drainage area $r_e$ , which in Russian-language practice is sometimes also called the radius of drainage or the radius of the supply contour. $r_e$ is set by the boundaries, while $r_d$ grows with time.

The radius of drainage in transient flow

r_d = \sqrt{\frac{4\eta t}{\gamma}} = 0.749\,r_i,\qquad p_i - p_{wf} = \frac{qB\mu}{2\pi kh}\left(\ln\frac{r_d}{r_w} + S\right)

where:

$r_d$ radius of drainage at time $t$ , m
$\gamma$ $\e^{0.5772} = 1.781$ , —
$r_i$ radius of investigation $\sqrt{4\eta t}$ , m
$p_i,\ p_{wf}$ initial reservoir pressure and bottomhole flowing pressure, Pa
$q,\ B,\ \mu,\ k,\ h,\ S$ rate at standard conditions, formation volume factor, viscosity, permeability, thickness, skin, m³/s, m³/m³, Pa·s, m², m, —

Units: SI; the logarithm is natural

Holds when

The rate is constant, and the reservoir is infinite and homogeneous.
$t_D > 25$ , so the logarithmic approximation is valid at the wall of the well.

The second formula is the same pressure function, $\tfrac12(\ln t_D + 0.809) + S$ , rewritten as the Dupuit equation. The radius of drainage grows as the square root of time, and it is always three quarters of the radius of investigation. Beyond it, the pressure already differs from the logarithmic straight line. That line “rests” on $p_i$ at $r_d$ , while the real profile approaches $p_i$ smoothly farther out.

In other unit systems (2)

Metric oilfield units

r_d = 0.0893\sqrt{\frac{k\,t}{\phi\mu c_t}}

$r_d$ in m, $k$ in mD, $t$ in h, $\mu$ in mPa·s, $c_t$ in 1/MPa.

$0.0893 = \sqrt{4\cdot3.553\cdot10^{-3}/1.781}$ .

Oilfield units

r_d = 0.0243\sqrt{\frac{k\,t}{\phi\mu c_t}}

$r_d$ in ft, $k$ in mD, $t$ in h, $\mu$ in cP, $c_t$ in 1/psi.

$0.0243 = \sqrt{4\cdot2.637\cdot10^{-4}/1.781}$ .

In transient flow, the ratio of the rate to the drawdown is called the transient productivity index. It is the same productivity formula as for steady-state flow, only with a growing radius of drainage:

Transient productivity index

J_t = \frac{q}{p_i - p_{wf}(t)} = \frac{2\pi kh}{B\mu\left[\ln(r_d/r_w) + S\right]}

where:

$J_t$ productivity index at time $t$ , with respect to the initial pressure, m³/(s·Pa)
$p_{wf}(t)$ bottomhole flowing pressure at time $t$ at a constant rate, Pa
$r_d$ radius of drainage at time $t$ , m
$k,\ h,\ B,\ \mu,\ r_w,\ S$ permeability, thickness, formation volume factor, viscosity, well radius, skin, m², m, m³/m³, Pa·s, m, —

Units: SI

Holds when

The flow is transient: the disturbance has not reached the boundaries. We measure the drawdown from the initial pressure $p_i$ , because the average pressure in an infinite reservoir is not defined.
If the drawdown is held constant, the rate falls by the same formula, with a small correction (the rate at constant pressure).

While the drawdown cone expands, the same rate needs an ever larger pressure drop, so the productivity falls slowly. For every tenfold increase in time, the logarithm in the denominator grows by 1.151. In a closed area, the fall in productivity referred to $p_i$ does not stop, because the area is being depleted. The productivity referred to the average pressure reaches the constant value of pseudo-steady-state flow, where $\ln(r_e/r_w) - \tfrac34$ stands in place of $\ln(r_d/r_w)$ (Pseudo-steady-state flow and the shape factor). It is as if the radius of drainage stops at $r_e\e^{-3/4} = 0.47\,r_e$ .

In other unit systems (2)

Metric oilfield units

J_t = \frac{0.536\,kh}{B\mu\left[\ln(r_d/r_w) + S\right]}

$J_t$ in m³/(d·MPa) at standard conditions, $k$ in mD, $h$ in m, $\mu$ in mPa·s.

$0.536 = 2\pi\cdot0.0853$ .

Oilfield units

J_t = \frac{7.08\cdot10^{-3}\,kh}{B\mu\left[\ln(r_d/r_w) + S\right]}

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

$7.08\cdot10^{-3} = 2\pi\cdot1.127\cdot10^{-3} = 1/141.2$ .

CalculatedCalculated productivity index of the lesson's well (

S

= 2) at the centre of a closed circular area of radius 600 m. It is shown with respect to the initial pressure (orange line) and to the average pressure of the area (teal). The dashed horizontal line is the pseudo-steady-state value. The time scale is logarithmic. Up to 100 h, the productivity with respect to the initial pressure is the same as in an infinite reservoir. With respect to the average pressure it is slightly higher from the first hours (by 0.6% after a day, and by 2.4% by 100 h), because the average pressure of a closed area declines from the very start-up.

Worked exampleProductivity after an hour, after a day and after stabilisation

Given (assumed data)

The lesson's well: $kh$ = 400 mD·m; $B\mu$ = 1.375 mPa·s; $S$ = 2; $r_w$ = 0.1 m; $\phi\mu c_t = 3.344\cdot10^{-4}$ MPa⁻¹·mPa·s
Drainage area: a closed circle of radius 600 m, the well at the centre

Solution

The factor is $0.536\,kh/(B\mu) = 0.536\cdot400/1.375 = 155.9$ m³/(d·MPa).
After 1 h: $r_d = 0.0893\sqrt{40/(3.344\cdot10^{-4})} = 30.9$ m, $\ln(r_d/r_w) = 5.733$ , and $J_t = 155.9/(5.733 + 2) = 20.16$ m³/(d·MPa).
After 24 h: $r_d = 151.3$ m, $\ln(r_d/r_w) = 7.322$ , and $J_t = 155.9/(7.322 + 2) = 16.72$ .
After pseudo-steady state is reached: $\ln(600/0.1) - 0.75 = 7.950$ , and $J = 155.9/(7.950 + 2) = 15.67$ .
The transient productivity becomes equal to the stabilised one when $r_d = 600\,\e^{-0.75} = 283.4$ m, that is, at $t = 1.781\cdot283.4^2/(4\cdot425) = 84$ h.

Answer: After an hour the productivity is 20.16, after a day 16.72, and after stabilisation 15.67 m³/(d·MPa). A short flow period overestimates it by 29% and 6.7%.

What it means. The reservoir and the skin have not changed in this time. Only the radius of drainage has. So we cannot carry the productivity measured in the first hours of a test over to production. We recalculate it from permeability and skin, for the size of the future drainage area. In a closed area, the productivity referred to the initial pressure keeps falling after 84 h, while the productivity referred to the average pressure stops at 15.67 (see the figure above). The data in this example are assumed for illustration.

When a line source is enough

A summary for practice. The line-source solution and its logarithmic approximation are the working tools of almost all well tests. We have to replace them in only a few cases, but these cases are important.

Where the line-source solution is valid and where it is not

Case: Bottomhole pressure at $t_D > 25$ (or 100)
Line source: an error of about 2% (0.5%)
What to do otherwise: —

Case: The first seconds after start-up
Line source: underestimates the pressure drop
What to do otherwise: the finite-radius solution; but on a record the wellbore hides everything

Case: Negative skin
Line source: valid at $t_D\e^{2S} > 100$
What to do otherwise: the effective radius or a radial composite model

Case: A neighbouring well, reflections at boundaries
Line source: valid: $r \gg r_w$
What to do otherwise: allow for the wellbore storage of the active well

Case: A hydraulic fracture, a horizontal wellbore
Line source: not valid at the beginning
What to do otherwise: Inflow to a hydraulically fractured well, Horizontal wells

Case: An accumulation as a “well” in an aquifer
Line source: not valid
What to do otherwise: solutions with a finite radius (Unsteady-state water influx: the van Everdingen–Hurst method)

Case: The disturbance has reached the boundaries
Line source: only together with reflections
What to do otherwise: Superposition principle, Pseudo-steady-state flow and the shape factor

Case	Line source	What to do otherwise
Bottomhole pressure at $t_D > 25$ (or 100)	an error of about 2% (0.5%)	—
The first seconds after start-up	underestimates the pressure drop	the finite-radius solution; but on a record the wellbore hides everything
Negative skin	valid at $t_D\e^{2S} > 100$	the effective radius or a radial composite model
A neighbouring well, reflections at boundaries	valid: $r \gg r_w$	allow for the wellbore storage of the active well
A hydraulic fracture, a horizontal wellbore	not valid at the beginning	Inflow to a hydraulically fractured well, Horizontal wells
An accumulation as a “well” in an aquifer	not valid	solutions with a finite radius (Unsteady-state water influx: the van Everdingen–Hurst method)
The disturbance has reached the boundaries	only together with reflections	Superposition principle, Pseudo-steady-state flow and the shape factor

The criteria come from comparing with the exact solutions calculated in this lesson and in the lesson Line-source solution and the pressure function.

Terms and symbols

Dimensionless solution: A solution of the diffusivity equation in the variables $r_D$ , $t_D$ , $p_D$ and the dimensionless parameters of the well and the reservoir. It is the same for all wells with the same parameters.
Finite-radius well: A formulation in which the rate is specified at the wall of a cylinder of radius $r_w$ . Unlike the line source, it does not take fluid “from inside the wellbore”.
Effective wellbore radius $r_w'$: The radius of a well without skin that has the same bottomhole pressure, $r_w\e^{-S}$ . It lets us describe a negative skin at the start of flow.
Radius of investigation $r_i$: The distance, $\sqrt{4\eta t}$ , at which the rate of pressure decline passes through its maximum at the given moment. It is a conventional boundary of the zone covered by the test.
Radius of drainage $r_d$: The radius of steady-state flow with the same bottomhole pressure, $\sqrt{4\eta t/\gamma} = 0.75\,r_i$ .
Transient productivity index $J_t$: The ratio of the rate to the difference between the initial and bottomhole pressures in transient flow. It falls while the radius of drainage grows.
Radial composite model: A reservoir with a zone of different permeability around the well. When the zone is large, it replaces the skin and gives two levels on the derivative.

Common mistakes and misconceptions

Wrong

“The radius of investigation is a boundary beyond which the pressure has not yet changed.”

In fact

The pressure changes throughout the reservoir immediately after start-up. $r_i$ is the place where the rate of its decline is just passing through the maximum. In the example of the lesson, $r_i$ = 202 m after a day, and an accurate gauge in a neighbouring well would notice a response even at 430 m.

Wrong

“A negative skin can simply be subtracted from the solution from the very start-up.”

In fact

Only at late times. At the beginning, an added negative skin gives a “rise” of pressure during production. The correct approach is a well with an effective radius, whose semilog straight line starts at $t_D\e^{2S} \approx 100$ . For $S = -2$ that is after a few minutes.

Wrong

“The productivity from the first hours of operation is the one at which the well will work.”

In fact

While the radius of drainage grows, the productivity falls. In the example of the lesson, after an hour it is 29% higher than the stabilised one, and after a day it is 7% higher.

Wrong

“If pressure is converted to psi, the constant 141.2 can be used.”

In fact

Only if all the quantities have been converted to oilfield units: rate, thickness, viscosity, compressibility and radius. The check: the dimensionless $t_D$ and $C_D$ calculated in both systems must be the same.

Wrong

“Skin always acts instantly.”

In fact

A zone with a permeability 3–5 times lower than that of the reservoir behaves like a thin skin when the radius of investigation is about 10 times its radius. A half-metre zone does so after a minute, but a fifteen-metre zone only after half a day (see the advanced section).

In depth: advanced detail

Derivations: flow through a circle, the volume share and the maximum rate

Flow through a circle. In the line-source solution, $\Delta p = \tfrac{qB\mu}{4\pi kh}E_1(x)$ , where $x = r^2/(4\eta t)$ . We know that $\dd E_1/\dd x = -\e^{-x}/x$ and $\dd x/\dd r = 2x/r$ . So the pressure gradient is $\partial p/\partial r = -\partial\Delta p/\partial r = \tfrac{qB\mu}{2\pi kh}\cdot\tfrac{\e^{-x}}{r}$ . By Darcy's law, the flow through a cylinder of radius $r$ is $\tfrac{2\pi rhk}{\mu}\,\tfrac{\partial p}{\partial r} = qB\,\e^{-x}$ at reservoir conditions. That is $q\,\e^{-x}$ at standard conditions.

Volume share. The elastic storage released inside radius $R$ (the fluid and rock expanding as the pressure falls) is $\phi c_th\int_0^R 2\pi r\,\Delta p\,\dd r$ . We substitute $u = r^2/(4\eta t)$ , which gives $2\pi r\,\dd r = 4\pi\eta t\,\dd u$ . The integral becomes $qBt\int_0^X E_1(u)\,\dd u$ . Integrating by parts gives $\int_0^X E_1(u)\,\dd u = XE_1(X) + 1 - \e^{-X}$ . As $X \to \infty$ this is 1, which is the whole withdrawn volume $qBt$ . So the material balance is satisfied.

The maximum rate of pressure decline. The derivative of $\Delta p$ with respect to time is $\tfrac{qB\mu}{4\pi kh}\cdot\tfrac{\e^{-a/t}}{t}$ , where $a = r^2/(4\eta)$ . Its derivative with respect to $t$ is proportional to $\e^{-a/t}(a/t - 1)/t^2$ . This is zero at $t = a$ , that is, when $r^2 = 4\eta t$ .

A skin zone of finite size: when the skin is “instantaneous”

Skin treated as an infinitely thin zone assumes two things: the changed permeability is concentrated right at the wall, and fluid does not accumulate in this zone. A real zone of radius $r_s$ and permeability $k_s$ has its own transient. As long as the disturbance has not gone far beyond the zone, the well “sees” the permeability of the zone. The exact solution is the radial composite model of the reservoir: inside $r_s$ the permeability is $k_s$ , outside it is $k$ , and at the boundary between the zones the pressure and the flow rate are continuous (they do not jump).

A well with a zone of different permeability: the Laplace transform

\bar p_{wD}(s) = \frac{K_0(\sigma) + \rho\,I_0(\sigma)}{s\,\kappa\sigma\left[K_1(\sigma) - \rho\,I_1(\sigma)\right]},\qquad \sigma = \sqrt{s/\kappa}

where:

$\kappa$ the ratio $k_s/k$ of the permeability of the zone to that of the reservoir, —
$\rho$ a coefficient found from the continuity of pressure and flow rate at the boundary of the zone $R = r_s/r_w$ : $\rho = -\dfrac{u\lambda K_0(\sigma R) - \kappa\sigma K_1(\sigma R)}{u\lambda I_0(\sigma R) + \kappa\sigma I_1(\sigma R)}$ , $u = \sqrt s$ , $\lambda = K_1(uR)/K_0(uR)$ , —
$I_0,\ I_1,\ K_0,\ K_1$ modified Bessel functions, —

Units: no units; $p_D$ is referred to the permeability $k$ of the outer zone

Holds when

The flow is single-phase, porosity, compressibility and viscosity are the same in both zones, and there is no wellbore storage.

At $\kappa = 1$ we get the finite-radius solution. At early times, if the zone is large enough, the derivative goes to the level $0.5/\kappa$ (the flow goes through the zone). At late times it goes to 0.5, and the pressure goes to $p_D + S$ with the Hawkins skin $S = (1/\kappa - 1)\ln R$ (the Hawkins formula).

CalculatedCalculated pressure derivative of the lesson's well with a zone of reduced permeability, divided by the radial-flow plateau

m_e

, without wellbore storage. One zone has a radius of 0.5 m and permeability

k/5

. The other has a radius of 15 m and permeability

k/3

. The grey dashed line is the well without a zone, and the light line is the plateau level. Both scales are logarithmic.

When a zone of altered permeability begins to behave as a thin skin

Zone: damage, 0.5 m
Permeability: $k/5$
Skin: 6.4
Derivative within 5% of the plateau: 71 s
$r_i/r_s$ at that moment: 11.6

Zone: zone, 15 m
Permeability: $k/3$
Skin: 10.0
Derivative within 5% of the plateau: 11.7 h
$r_i/r_s$ at that moment: 9.4

Zone	Permeability	Skin	Derivative within 5% of the plateau	$r_i/r_s$ at that moment
damage, 0.5 m	$k/5$	6.4	71 s	11.6
zone, 15 m	$k/3$	10.0	11.7 h	9.4

The lesson's well: $\eta$ = 425 m²/h, $r_w$ = 0.1 m. The skin is found by the Hawkins formula. The values are calculated with the composite model, without wellbore storage.

Here is the rule that follows from the calculation. A zone with a permeability 3–5 times lower than that of the reservoir acts as a thin skin when the radius of investigation is about ten times its radius (the derivative is within 5% of the plateau). The stronger the contrast, the longer it takes: at $k/10$ we need about $20\,r_s$ , while a zone of improved permeability “disappears” already by $5$ – $6\,r_s$ (calculated with the same model). A half-metre zone of mud damage “disappears” in a minute, long before wellbore storage ends, so the formula with a skin is accurate for such zones. A zone of tens of metres gives two levels on the derivative: first $0.5/\kappa$ , then 0.5. Such a zone may be water near a former injection well, condensate that has dropped out near a gas-condensate well (Retrograde condensation, wax, asphaltenes and high-pressure fluids), or gas released below the bubble point pressure (the pressure below which gas comes out of the oil). This is what a radial composite reservoir looks like. We cannot process such a record as radial flow with a skin, because the first plateau would give the permeability of the zone instead of that of the reservoir.

Connections to other topics

Line-source solution and the pressure function — the line-source solution, the logarithmic approximation and the comparison at the wall of the well.
Diffusivity equation — where the dimensionless variables come from.
Systems of units: SI, oilfield and metric oilfield — the exact definitions of units and the derivation of formula constants.
Skin factor — skin, the Hawkins formula and the effective radius.
Wellbore storage — wellbore storage, which hides the first minutes of the record.
Type-curve matching — the dimensionless groups $t_D/C_D$ and $C_D\e^{2S}$ on type curves.
Well testing: purposes and types of test — the radius of investigation in test planning.
Steady-state radial flow — the Dupuit equation and the quasi-steady-state profile at the well.
Pseudo-steady-state flow and the shape factor — productivity after pseudo-steady state is reached.
A sealing fault near the well: the method of images — how a boundary raises the derivative and when it can be seen.
Semi-infinite systems: theory and test analysis — the dimensionless time $\eta t/L^2$ for systems of faults.

Check yourself

1
A well of radius 0.1 m, a reservoir with a hydraulic diffusivity of 360 m²/h. How long after start-up does 99% of the rate pass through the wall of the well in the line-source solution?
1. A.About 2.5 s
2. B.Only after pseudo-steady state is reached
3. C.About 2.5 min
4. D.About 2.5 h
Show the answer
Correct answer: A. About 2.5 s
The share of the rate, $\e^{-1/(4t_D)}$ , equals 0.99 at $t_D = 24.9$ . The time is $t = t_D r_w^2/\eta = 24.9\cdot0.01/360 = 6.9\cdot10^{-4}$ h, that is, 2.5 s. The difference between the line source and the finite-radius well lasts only for the first seconds.
2
A report gives a wellbore storage coefficient of 0.02 bbl/psi. Find $C$ in m³/MPa and $C_D$ for $\phi$ = 0.2, $c_t = 1.5\cdot10^{-3}$ MPa⁻¹, $h$ = 15 m, $r_w$ = 0.108 m.
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$C = 0.02\cdot23.06 = 0.461$ m³/MPa. The denominator is $2\pi\phi c_thr_w^2 = 2\pi\cdot0.2\cdot1.5\cdot10^{-3}\cdot15\cdot0.108^2 = 3.30\cdot10^{-4}$ m³/MPa. $C_D = 0.461/(3.30\cdot10^{-4}) \approx 1400$ . In oilfield units the same number comes out, and this is the check of the conversion.
3
After a treatment the skin of a well is $S = -3$ . From what dimensionless time (based on the wellbore radius) can the semilog formula be used with an error of about 0.5%?
1. A. $t_D \approx 25$
2. B. $t_D \approx 100$
3. C. $t_D \approx 40{,}000$
4. D. $t_D \approx 4\cdot10^8$
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Correct answer: C. $t_D \approx 40{,}000$
The well behaves like a well of radius $r_w' = r_w\e^{3}$ . For this well, the condition $t_D' > 100$ means $t_D\e^{2S} > 100$ , that is, $t_D > 100\,\e^{6} \approx 40{,}000$ . For an ordinary well this is minutes, not seconds.
4
The flow period lasted 12 h in a reservoir with $\eta$ = 800 m²/h. Find the radius of investigation, the radius of drainage and the radius inside which 95% of the withdrawn volume is supplied. At what distance must a fault be for the derivative to rise by 10% by the end of the flow period?
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$r_i = \sqrt{4\cdot800\cdot12} = 196$ m, $r_d = 0.749\cdot196 = 147$ m, and 95% of the volume is supplied from within $1.335\cdot196 = 262$ m. The derivative step is $\e^{-L^2/(\eta t)} = 0.1$ at $L = \sqrt{\eta t\ln10} = 0.759\,r_i = 149$ m. So the fault must be no farther than about 150 m.
5
Why is the productivity index of a new well over the first hour of work higher than after a month, if the reservoir and the skin have not changed?
1. A.The oil viscosity falls
2. B.The radius of drainage grows, so the same rate needs a larger pressure drop
3. C.The skin grows
4. D.The permeability of the reservoir falls
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Correct answer: B. The radius of drainage grows, so the same rate needs a larger pressure drop
The denominator of the transient productivity index contains $\ln(r_d/r_w) + S$ , and the radius of drainage $r_d$ grows as $\sqrt t$ . In the example of the lesson, the productivity after an hour is 29% higher than the stabilised one. So we recalculate the productivity from a short test for the future drainage area.
6
Around a well of radius 0.1 m there is a zone of radius 3 m with a permeability four times lower than that of the reservoir. The hydraulic diffusivity of the reservoir is 200 m²/h. After how long can the zone be regarded as a thin skin?
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By the rule of the lesson, the zone acts as a thin skin when the radius of investigation is about ten times the radius of the zone, that is, about 30 m: $t \approx r_i^2/(4\eta) = 900/800 \approx 1.1$ h. Before that, the derivative is above the radial-flow plateau, and the permeability found from it would be too low. The skin of such a zone by the Hawkins formula is $3\ln 30 \approx 10.2$ .

Updated: 19 October 2026