Partially communicating faults

Quick refresher

The model of the fault is a thin vertical zone of width $l_f$ and permeability $k_f$ through the whole thickness of the formation, with no storativity of its own (it stores no fluid). The flow through it is proportional to the pressure jump across it.
The dimensionless transmissibility $\alpha = (k_f/l_f)/(k/L)$ compares the transmissibility of the fault zone with the transmissibility of the formation on the way from the well to the fault. $\alpha \to 0$ is a sealing fault, and $\alpha \to \infty$ means there is no fault.
What the fault reflects back is the same image as for a sealing fault, minus a “leak”. The leak is a line of imaginary injection wells behind the image, with a total rate $q$ , spread over a length of about $L/(2\alpha)$ .
The derivative rises as for a sealing fault and reaches a maximum below $2m_e$ (the lower, the larger $\alpha$ is). Then it returns to $m_e$ in a time of about $(L/\alpha)^2/\eta$ , where $\eta$ is the hydraulic diffusivity (how fast a pressure change spreads through the rock). The late semilog line (a straight line on a plot of pressure against the logarithm of time) is parallel to the first one and shifted by $2m_eS_f$ , where $S_f = \e^{4\alpha}E_1(4\alpha)$ .
From the height of the maximum we find $\alpha$ , and from the start of the rise the distance $L$ . Then $k_f/l_f = \alpha k/L$ . A test does not give the width and the permeability of the zone separately.
The resolution window (the range of $\alpha$ that a test can recognise) is limited. For $\alpha$ below about 0.01, a fault in a test of ordinary length is almost impossible to tell from a sealing one. For $\alpha$ above 2–3, its hump drowns in the noise of the derivative.

Key relationships

\alpha = \dfrac{k_f/l_f}{k/L} = \dfrac{k_fL}{k\,l_f}

dimensionless fault transmissibility (the same thickness on both sides)

\dfrac{\Delta p'}{m_e} = 1 + \e^{-1/t_{DL}}\Big[1 - 2\sqrt{\pi}\,\alpha\sqrt{t_{DL}}\;\mathrm{erfcx}\Big(2\alpha\sqrt{t_{DL}} + \dfrac{1}{\sqrt{t_{DL}}}\Big)\Big]

derivative after wellbore storage has ended; $t_{DL} = \eta t/L^2$

S_f = \e^{4\alpha}E_1(4\alpha)

apparent fault skin: the late line is shifted by $2m_eS_f$

\dfrac{k_f}{l_f} = \dfrac{\alpha\,k}{L}

transmissibility of the fault zone from the result of a test, mD/m

F = \dfrac{\alpha\,\Delta x}{\alpha\,\Delta x + L}

transmissibility multiplier of the fault for a grid with cells of size $\Delta x$ across the fault

Common pitfalls

Taking a maximum of the derivative below $2m_e$ as a sign that the fault is far away. The start of the rise gives the distance, and the leak sets the height of the maximum.
Thinking that the return of the derivative to the plateau means there is no fault. The late line is shifted by the apparent fault skin.
Expecting a test to give the permeability of the fault zone. It gives only $k_f/l_f$ .
Moving the fault transmissibility multiplier into a model with a different grid without recalculating it.
Mistaking a fall of the derivative at the end of a pressure build-up for a leak through the fault, without checking the effect of the well's producing time before shut-in.

After this lesson you can

describe the model of a thin semipermeable fault zone, and explain the meaning of the dimensionless transmissibility $\alpha$ as the ratio of two resistances
explain the response of the pressure and the derivative (the rise, the maximum below $2m_e$ , the return to the plateau, and the shift of the late straight line by the apparent fault skin) through an image with a “leak”
estimate $\alpha$ from the height of the derivative maximum and the distance to the fault from the start of the rise, and convert them into the transmissibility of the fault zone $k_f/l_f$ , the flow across the fault and the transmissibility multiplier of the model
find the range of $\alpha$ in which a test of a given length tells a partially communicating fault from a sealing one and from no fault
name the models that give a similar picture, and the data that help to tell them apart

A fault that lets fluid through

A fault is sealing when the rock opposite the reservoir is impermeable, or when the fault zone is sealed by clay smear (clay dragged along the fault), crushed grains or cement (the fault as a boundary). But often the throw (the displacement along the fault) is smaller than the thickness of the formation. Then the permeable layers on the two sides stay in contact, and the fault zone is only made worse: the grains are partly crushed, the clay is smeared into a thin film, and the pores are partly closed up. So fluid crosses the fault, but with extra resistance. The geology of faults is in the lesson Faults.

For development, this middle level of transmissibility is what matters. Across a partially communicating fault, injection in one block supports the pressure in the next block over time, but with a delay and with losses. The block behind the fault “gives” its oil to a well more slowly than it would without the fault. In a reservoir simulation model, we describe such a fault by a transmissibility multiplier between neighbouring grid cells, from 0 to 1 (transmissibility multiplier). A well test is one of the few ways to estimate it from measurements and not by analogy.

The simplest model is a semipermeable barrier: a vertical zone of width $l_f$ and permeability $k_f$ through the whole thickness of the formation. On both sides there is the same homogeneous formation. The zone is thin, and we ignore its own storativity (its ability to store fluid): all the fluid that enters it from one side leaves it on the other, and the pressure changes by a jump across it. The flow through a unit area of the zone follows Darcy's law for a layer of thickness $l_f$ :

Flow through the fault zone

u = \frac{k_f}{\mu}\cdot\frac{p_1 - p_2}{l_f}

where:

$u$ Darcy velocity through the fault: the flow rate at reservoir conditions per unit area of the zone, m/s
$k_f,\ l_f$ permeability and width of the fault zone (measured across the fault), m², m
$p_1,\ p_2$ pressure just on each side of the zone, Pa
$\mu$ fluid viscosity, Pa·s

Units: SI (any consistent system)

Holds when

the zone is thin compared with the distance to the well, and it stores no fluid
single-phase flow; the zone is the same along the whole length and thickness of the fault
on both sides there is a formation with the same properties and thickness

The zone works as a thin skin (a thin layer of extra resistance, like the one around a damaged well), only not around the well but on a plane in the formation. The extra pressure drop is proportional to the flow through it and to the ratio $l_f/k_f$ . So the width and the permeability of the zone enter the problem only through this ratio.

In other unit systems (1)

Metric oilfield units

q_r = 0.0853\,\frac{k_f}{l_f}\cdot\frac{A\,(p_1 - p_2)}{\mu}

$q_r$ in m³/d at reservoir conditions, $k_f$ in mD, $l_f$ in m, $A$ is the area of the zone (the length of a stretch of the fault times the thickness of the formation), m²; pressures in MPa; $\mu$ in mPa·s.

$0.0853$ is the constant of Darcy's law in these units (Rock permeability).

Dimensionless fault transmissibility

How visible a fault is to a well depends not on its transmissibility alone, but on a comparison with the transmissibility of the formation. The resistance of the fault zone per unit area is proportional to $l_f/k_f$ . The resistance of the formation on the way from the well to the fault is proportional to $L/k$ . Their ratio is the main parameter of the problem:

Dimensionless fault transmissibility

\alpha = \frac{k_f/l_f}{k/L} = \frac{k_f\,L}{k\,l_f}

where:

$\alpha$ dimensionless fault transmissibility, —
$k_f,\ l_f$ permeability and width of the fault zone, mD, m
$k$ permeability of the formation, mD
$L$ distance from the well to the fault, m

Units: dimensionless; $k_f$ and $k$ , and $l_f$ and $L$ , in the same units

Holds when

the thickness of the formation is the same on both sides of the fault (otherwise the numerator and the denominator contain their own $kh$ )
the fluid has the same properties on both sides

$\alpha = 0$ is a sealing fault, and $\alpha \to \infty$ means there is no fault for the flow. The parameter grows with distance. The same fault looks more sealing to a near well and more “transparent” to a distant one. The reason is that the resistance of the formation up to the fault grows, and the resistance of the zone does not.

How a fault looks at different

\alpha

$\alpha$: below 0.01
Maximum of $\Delta p'/m_e$: above 1.84
What the test shows: almost like a sealing fault; the return to the plateau takes tens of thousands of $L^2/\eta$

$\alpha$: 0.01–0.3
Maximum of $\Delta p'/m_e$: 1.84–1.32
What the test shows: a clear hump and a return to the plateau; $\alpha$ can be found most reliably

$\alpha$: 0.3–2
Maximum of $\Delta p'/m_e$: 1.32–1.08
What the test shows: a small hump, visible with a good gauge

$\alpha$: above 2–3
Maximum of $\Delta p'/m_e$: below 1.08–1.06
What the test shows: the fault is lost in the noise of the derivative

$\alpha$	Maximum of $\Delta p'/m_e$	What the test shows
below 0.01	above 1.84	almost like a sealing fault; the return to the plateau takes tens of thousands of $L^2/\eta$
0.01–0.3	1.84–1.32	a clear hump and a return to the plateau; $\alpha$ can be found most reliably
0.3–2	1.32–1.08	a small hump, visible with a good gauge
above 2–3	below 1.08–1.06	the fault is lost in the noise of the derivative

The values of the maximum are from the table of the derivative in the section below.

Worked exampleTwo hypotheses about the fault zone

Given (assumed data)

Formation: $k$ = 80 mD, $\phi$ = 0.18, $\mu$ = 1.2 mPa·s, $c_t = 1.5\cdot10^{-3}$ MPa⁻¹; by the seismic data, the fault is 200 m from the well
Hypothesis 1: crushed zone: width $l_f$ = 3 m, $k_f$ = 0.6 mD
Hypothesis 2: clay smear: width $l_f$ = 0.05 m, $k_f$ = 0.002 mD

Solution

Hypothesis 1: $\alpha = 0.6\cdot200/(80\cdot3) = 0.5$ . Hypothesis 2: $\alpha = 0.002\cdot200/(80\cdot0.05) = 0.1$ .
Transmissibility of the zone $k_f/l_f$ : 0.20 and 0.04 mD/m. The second zone lets fluid through 5 times worse, although it is 60 times thinner.
Time scale: $\eta = 3.553\cdot10^{-3}\cdot80/(0.18\cdot1.2\cdot1.5\cdot10^{-3}) = 877$ m²/h, and $L^2/\eta = 45.6$ h.
From the table in the section on the derivative: at $\alpha$ = 0.5 the maximum is 1.238 $m_e$ at $t_{DL}$ = 1.85, that is, after 84 h. At $\alpha$ = 0.1 it is 1.527 $m_e$ at $t_{DL}$ = 3.94, after 180 h.

Answer: The crushed zone would give a small hump (1.238

m_e

after about 3.5 days). The clay smear would give a clear one (1.527

m_e

after about 7.5 days). A test that lasts a week tells these hypotheses apart. By the fourth day, the first has already passed its maximum, and the second has risen above 1.4

m_e

and keeps growing.

What it means. A thin film with a tiny permeability can resist more strongly than a wide crushed zone. Only the ratio $k_f/l_f$ matters. So estimates “from core samples of the fault zone”, made without knowing its width and continuity, say little about transmissibility. A well test measures it directly, though as one number. The data in this example are assumed for illustration.

An image with a leak

At a sealing fault, one image (an imaginary mirror well) with the same rate describes the pressure on the well side (A sealing fault near the well: the method of images). A partially communicating fault does not reflect the disturbance completely. Part of it passes into the next block. The exact solution (the derivation is in the advanced section) turns out to be easy to picture. What the fault reflects back is the same image as for a sealing fault, minus a “leak”. Behind the image, on the continuation of the line from the well to the image, stands a line of imaginary injection wells. Their total rate equals $q$ . Their density falls off exponentially with distance from the image, over a length scale of $L/(2\alpha)$ .

SchematicA partially communicating fault (a sketch, not to scale). Top: the formation in plan view: the well, the thin fault zone of width

l_f

and permeability

k_f

, and the flow across it. Bottom: the equivalent picture for the well side. It shows an image with the same rate as for a sealing fault, and behind it a line of imaginary injection wells (the “leak”) with a total rate

q

. The smaller the circle, the smaller the share of the rate.

This picture explains the whole response. Until the disturbance reaches the fault, there is neither an image nor a leak. Then the image starts to act. The injection wells of the leak stand farther away than the image, so their influence comes later. At first the fault looks sealing. Then the disturbance reaches the line of the leak, and the leak feeds the well. In the end, the image and the leak, seen from afar as almost one point, almost cancel each other. The derivative returns to the plateau, and a constant extra pressure drop stays in the pressure.

A well near a partially communicating fault: dimensionless pressure

p_D = \tfrac12\left(\ln t_D + 0.809\right) + S + \underbrace{\tfrac12 E_1\!\left(\frac{1}{t_{DL}}\right)}_{\text{image}} - \underbrace{\alpha\int_0^\infty \e^{-2\alpha\xi}\,E_1\!\left(\frac{(2 + \xi)^2}{4t_{DL}}\right)\dd\xi}_{\text{leak}}

where:

$p_D$ dimensionless drop of the bottomhole pressure $2\pi kh\,\Delta p/(qB\mu)$ , with skin, —
$t_D,\ t_{DL}$ dimensionless time $kt/(\phi\mu c_tr_w^2)$ , and the same time based on $L^2$ : $t_{DL} = kt/(\phi\mu c_tL^2) = \eta t/L^2$ , —
$\xi$ distance behind the image in units of $L$ (the variable of integration), —
$\alpha$ dimensionless fault transmissibility, —
$E_1$ exponential integral (function E₁), —

Units: dimensionless; natural logarithm

Holds when

the fault is a straight semipermeable barrier through the whole thickness of a homogeneous formation that is the same on both sides; there are no other boundaries
for the well itself the logarithmic approximation holds, and wellbore storage is over
the image and the leak are line sources ( $L$ is much larger than the well radius)

At $\alpha = 0$ the integral vanishes, and the formula of a sealing fault remains. For large $\alpha$ , the leak almost completely cancels the image. We can also write the last term as an integral over time with a complementary error function. It is the same quantity in another form (advanced section).

Stages of the response of a well near a partially communicating fault

$t_{DL} = \eta t/L^2$: below 0.2
What happens: the disturbance has not reached the fault
Derivative: plateau $m_e$

$t_{DL} = \eta t/L^2$: 0.2 to the time of the maximum
What happens: the image acts; the leak is still small
Derivative: a rise almost as for a sealing fault

$t_{DL} = \eta t/L^2$: near the time of the maximum
What happens: the leak gains strength
Derivative: a maximum below $2m_e$

$t_{DL} = \eta t/L^2$: much larger than $1/\alpha^2$
What happens: the image and the leak almost cancel each other
Derivative: return to $m_e$ ; the pressure is shifted by $2m_eS_f$

$t_{DL} = \eta t/L^2$	What happens	Derivative
below 0.2	the disturbance has not reached the fault	plateau $m_e$
0.2 to the time of the maximum	the image acts; the leak is still small	a rise almost as for a sealing fault
near the time of the maximum	the leak gains strength	a maximum below $2m_e$
much larger than $1/\alpha^2$	the image and the leak almost cancel each other	return to $m_e$ ; the pressure is shifted by $2m_eS_f$

The time of the maximum is from 1.2 to 16 units of $L^2/\eta$ for $\alpha$ from 2 to 0.01 (the table in the next section), and about 70 for $\alpha$ = 0.001.

Pressure and derivative

The derivative of the contribution of the image against $\ln t$ is the familiar step $\tfrac12\e^{-1/t_{DL}}$ . The derivative of the leak is expressed through $\mathrm{erfcx}(x) = \e^{x^2}\mathrm{erfc}(x)$ , a scaled form of the complementary error function (a standard mathematical function). Together:

Pressure derivative near a partially communicating fault

\frac{\Delta p'}{m_e} = 1 + \e^{-1/t_{DL}}\left[1 - 2\sqrt{\pi}\,\alpha\sqrt{t_{DL}}\;\mathrm{erfcx}\left(2\alpha\sqrt{t_{DL}} + \frac{1}{\sqrt{t_{DL}}}\right)\right]

where:

$\Delta p'$ logarithmic derivative of the pressure drop, Pa
$m_e$ level of the radial-flow plateau, $qB\mu/(4\pi kh)$ , Pa
$t_{DL}$ dimensionless time $\eta t/L^2$ , —
$\mathrm{erfcx}$ scaled complementary error function, $\e^{x^2}\mathrm{erfc}(x)$ ; for large $x$ it is close to $1/(x\sqrt\pi)$ , —

Units: dimensionless

Holds when

the same as for the pressure; wellbore storage is over.

At $\alpha = 0$ the square bracket equals 1, and we get the step of a sealing fault, $1 + \e^{-1/t_{DL}}$ . For any $\alpha > 0$ , the bracket is first close to 1, and then falls to zero. So the derivative passes a maximum and returns to the plateau.

In other unit systems (1)

Metric oilfield units

t_{DL} = \frac{3.553\cdot10^{-3}\,k\,t}{\phi\mu c_tL^2},\qquad m_e = 0.933\,\frac{qB\mu}{kh}

$k$ in mD, $t$ in h, $\mu$ in mPa·s, $c_t$ in 1/MPa, $L$ and $h$ in m, $q$ in m³/d; $m_e$ in MPa.

As in the lesson Diagnosis with the pressure derivative.

CalculatedCalculated derivative of a well near a partially communicating fault, divided by the plateau, without wellbore storage, for different

\alpha

. The time scale is logarithmic. The top curve is a sealing fault (

\alpha = 0

The maximum of the derivative and the return to the plateau

$\alpha$: 0.01
Maximum of $\Delta p'/m_e$: 1.844
$t_{DL}$ of the maximum: 15.57
$t_{DL}$ of the return to 1.1: 9797
$S_f$: 2.79

$\alpha$: 0.02
Maximum of $\Delta p'/m_e$: 1.774
$t_{DL}$ of the maximum: 10.10
$t_{DL}$ of the return to 1.1: 2568
$S_f$: 2.20

$\alpha$: 0.05
Maximum of $\Delta p'/m_e$: 1.646
$t_{DL}$ of the maximum: 5.83
$t_{DL}$ of the return to 1.1: 467
$S_f$: 1.49

$\alpha$: 0.1
Maximum of $\Delta p'/m_e$: 1.527
$t_{DL}$ of the maximum: 3.94
$t_{DL}$ of the return to 1.1: 139
$S_f$: 1.05

$\alpha$: 0.2
Maximum of $\Delta p'/m_e$: 1.397
$t_{DL}$ of the maximum: 2.76
$t_{DL}$ of the return to 1.1: 45.5
$S_f$: 0.69

$\alpha$: 0.5
Maximum of $\Delta p'/m_e$: 1.238
$t_{DL}$ of the maximum: 1.85
$t_{DL}$ of the return to 1.1: 11.7
$S_f$: 0.36

$\alpha$: 1
Maximum of $\Delta p'/m_e$: 1.144
$t_{DL}$ of the maximum: 1.46
$t_{DL}$ of the return to 1.1: 4.2
$S_f$: 0.21

$\alpha$: 2
Maximum of $\Delta p'/m_e$: 1.081
$t_{DL}$ of the maximum: 1.24
$t_{DL}$ of the return to 1.1: —
$S_f$: 0.11

$\alpha$	Maximum of $\Delta p'/m_e$	$t_{DL}$ of the maximum	$t_{DL}$ of the return to 1.1	$S_f$
0.01	1.844	15.57	9797	2.79
0.02	1.774	10.10	2568	2.20
0.05	1.646	5.83	467	1.49
0.1	1.527	3.94	139	1.05
0.2	1.397	2.76	45.5	0.69
0.5	1.238	1.85	11.7	0.36
1	1.144	1.46	4.2	0.21
2	1.081	1.24	—	0.11

Calculated from the formula of the derivative. “—” means that the maximum is below 1.1. The time is in units of $L^2/\eta$ . $S_f$ is the apparent fault skin (the formula is below).

The late part of the record also carries information about the fault. When $t_{DL}$ is much larger than $1/\alpha^2$ , the derivative approaches the plateau slowly, in inverse proportion to time. The pressure moves onto a line that is parallel to the first one but shifted upward: the fault acts as an extra skin.

The late stretch: the rest of the derivative and the apparent fault skin

\frac{\Delta p'}{m_e} - 1 \approx \frac{1 + 4\alpha}{8\alpha^2\,t_{DL}},\qquad \Delta p = m_e\left[\ln t_D + 0.809 + 2S + 2S_f\right],\quad S_f = \e^{4\alpha}E_1(4\alpha)

where:

$S_f$ apparent fault skin: the extra pressure drop on the late line, in units of $2m_e$ , —
$E_1$ exponential integral, —
$t_D,\ t_{DL},\ S$ dimensionless time based on the well radius and based on the distance to the fault, and the skin of the well, —

Units: the remainder and $S_f$ are dimensionless; $\Delta p$ and $m_e$ are in the same pressure units; natural logarithm

Holds when

$t_{DL} \gg 1/\alpha^2$ and $t_{DL} \gg 1$ : the remainder of the derivative is small; the formula for the remainder is approximate (it overstates it by a few per cent)
the disturbance has not reached any other boundaries

$S_f$ falls with $\alpha$ : about 2.8 at $\alpha$ = 0.01, about 1 at 0.1, and 0.2 at 1. If the record shows both the first and the late parallel line, the vertical distance between them, $2m_eS_f$ , gives $\alpha$ without reading the maximum. But we must wait for the late line until the derivative almost returns to the plateau. That takes tens of units of $L^2/\eta$ at $\alpha$ near 0.3, and hundreds at 0.1 and below. This works only for faults that conduct rather well.

CalculatedCalculated contribution of the fault to the dimensionless pressure of the well

p_D

, for different

\alpha

. The time scale is logarithmic. For a sealing fault the contribution grows without end (the slope doubles). For a partially communicating fault it settles at a constant

S_f

(the dashed lines).

What changes if…

…the transmissibility of the zone $k_f/l_f$ is twice as large. Then $\alpha$ is twice as large: the hump is lower, the maximum comes earlier, and the return to the plateau is 3–4 times earlier (in units of $L^2/\eta$ ).
…the well is twice as far from the same fault. The rise starts four times later, and $\alpha$ is twice as large: the hump is lower, and the fault looks more “transparent”.
…the permeability of the formation is twice as high. The rise starts twice as early, but $\alpha$ is twice as small: the same fault looks more sealing against a good formation.
…the rate is twice as large. The times stay the same, and only the pressure differences are twice as large. This makes it easier to deal with noise, but it does not change the shape of the hump.

Fault transmissibility from a test

Order of interpretation

1
Radial flow
From the derivative plateau we get $kh$ and $k$ , and from the ratio of the pressure to the plateau the skin (Diagnosis with the pressure derivative).
2
Distance
From the very start of the rise (the derivative 10–20% above the plateau) we find the distance $L$ by the formula for one fault (derivative step). The leak acts there too, and $L$ comes out too large. If we use the point where the derivative is 10% above the plateau, $L$ is too large by about 2% at $\alpha$ = 0.1, by 5% at 0.3 and by 9% at 0.5. Later points overstate it more. We refine the final $L$ from the time of the maximum and by fitting a model.
3
Fault transmissibility
From the height of the derivative maximum we find $\alpha$ in the table of the previous section. The time of the maximum is a check of $L$ .
4
Fault zone
$k_f/l_f = \alpha k/L$ . If we estimate the width of the zone from core samples or from similar faults, we get $k_f$ .
5
Model fit
We fit the model of a fault with finite transmissibility to the whole record at once, starting from the values we found (Type-curve matching). We also check whether other models give the same picture (the last section).

Here is an assumed record. An exploration well has penetrated the formation 150–200 m from a fault that the seismic survey shows. We need to find out whether the well's block communicates with the neighbouring block. The well was started by opening a downhole valve, at a rate of 150 m³/d, and the pressure was recorded for 13 days.

Pressure drop near a fault (assumed data, a selection of the readings)

$t$: 36 s
$\Delta p$ , MPa: 0.5245
$\Delta p'$ , MPa: 0.07730

$t$: 6 min
$\Delta p$ , MPa: 0.6711
$\Delta p'$ , MPa: 0.05841

$t$: 30 min
$\Delta p$ , MPa: 0.7638
$\Delta p'$ , MPa: 0.05700

$t$: 1.00 h
$\Delta p$ , MPa: 0.8031
$\Delta p'$ , MPa: 0.05678

$t$: 3.16 h
$\Delta p$ , MPa: 0.8686
$\Delta p'$ , MPa: 0.05776

$t$: 6.31 h
$\Delta p$ , MPa: 0.9096
$\Delta p'$ , MPa: 0.06286

$t$: 10.0 h
$\Delta p$ , MPa: 0.9398
$\Delta p'$ , MPa: 0.06894

$t$: 20.0 h
$\Delta p$ , MPa: 0.9908
$\Delta p'$ , MPa: 0.07774

$t$: 31.6 h
$\Delta p$ , MPa: 1.0277
$\Delta p'$ , MPa: 0.08110

$t$: 50.1 h
$\Delta p$ , MPa: 1.0655
$\Delta p'$ , MPa: 0.08176

$t$: 100 h
$\Delta p$ , MPa: 1.1213
$\Delta p'$ , MPa: 0.07872

$t$: 200 h
$\Delta p$ , MPa: 1.1738
$\Delta p'$ , MPa: 0.07329

$t$: 316 h
$\Delta p$ , MPa: 1.2067
$\Delta p'$ , MPa: —

$t$	$\Delta p$ , MPa	$\Delta p'$ , MPa
36 s	0.5245	0.07730
6 min	0.6711	0.05841
30 min	0.7638	0.05700
1.00 h	0.8031	0.05678
3.16 h	0.8686	0.05776
6.31 h	0.9096	0.06286
10.0 h	0.9398	0.06894
20.0 h	0.9908	0.07774
31.6 h	1.0277	0.08110
50.1 h	1.0655	0.08176
100 h	1.1213	0.07872
200 h	1.1738	0.07329
316 h	1.2067	—

$\Delta p = p_i - p_{wf}$ , $p_i$ = 24 MPa. The data are calculated from a model of a homogeneous formation with wellbore storage, skin and a straight fault of finite transmissibility, and rounded to 0.0001 MPa. The parameters of the model are given in the example. The derivative is calculated from all 56 readings (10 per cycle, where a cycle is a tenfold range of time), with a differentiation interval of 0.2 cycle (each value comes from the readings 0.2 cycle before and after it). The last two readings have no derivative.

CalculatedDiagnostic plot of the record: the pressure drop (points) and the derivative from the readings (orange line). The grey dashed line is the derivative of the same well near a sealing fault at the same distance. The horizontal line is the plateau

m_e

Worked exampleDimensionless fault transmissibility and the transmissibility of its zone

Given (assumed data)

Well and formation: $q$ = 150 m³/d; $B$ = 1.3; $\mu$ = 0.7 mPa·s; $h$ = 15 m; $\phi$ = 0.2; $c_t = 1.6\cdot10^{-3}$ MPa⁻¹; $r_w$ = 0.1 m
Derivative plateau: average from 0.32 to 1.6 h: $m_e$ = 0.05695 MPa; reading at 1 h: $\Delta p$ = 0.8031 MPa
Start of the rise: at 6.31 h the derivative is 0.06286 MPa
Maximum of the derivative: 0.08176 MPa at 50.1 h

Solution

Flow capacity: $qB\mu = 150\cdot1.3\cdot0.7 = 136.5$ ; $kh = 0.933\cdot136.5/0.05695 = 2236$ mD·m, and $k = 149.1$ mD. Skin: $\ln\dfrac{kt}{\phi\mu c_tr_w^2} = 18.014$ at 1 h; $S = \tfrac12\,(0.8031/0.05695 - 18.014 + 4.831) = 0.46$ .
Distance from the start of the rise: $\eta = 3.553\cdot10^{-3}\cdot149.1/(0.2\cdot0.7\cdot1.6\cdot10^{-3}) = 2365$ m²/h; $\Delta p'/m_e = 1.104$ , $\ln[1/(1.104 - 1)] = 2.263$ ; $L = \sqrt{2365\cdot6.31\cdot2.263} = 184$ m.
The maximum: $0.08176/0.05695 = 1.436$ , clearly below 2. A sealing fault would have given almost $2m_e$ by this time. So the fault lets fluid through.
In the table, the maximum 1.436 lies between the rows $\alpha$ = 0.1 (1.527) and $\alpha$ = 0.2 (1.397). The share of the interval is $(1.436 - 1.527)/(1.397 - 1.527) = 0.700$ . Interpolating on $\lg\alpha$ (the base-10 logarithm): $\lg\alpha = -1 + 0.700\cdot0.301 = -0.789$ , so $\alpha \approx 0.163$ .
A check by the time of the maximum: in the same way, $\lg t_{DL} = 0.5955 + 0.700\cdot(0.4409 - 0.5955) = 0.4873$ , $t_{DL}$ = 3.07; $L = \sqrt{\eta t/t_{DL}} = \sqrt{2365\cdot50.1/3.07} = 196$ m. This is within the accuracy of reading a flat maximum.
Transmissibility of the zone: $k_f/l_f = \alpha k/L = 0.163\cdot149.1/184 = 0.132$ mD/m. If the width of the zone is about 2 m, $k_f \approx$ 0.26 mD. If it is a film 0.2 m wide, $k_f$ is about 0.026 mD.

Answer:

k \approx 149.1

mD, skin 0.46. The fault, 184 m from the well, is partially communicating:

\alpha \approx 0.163

k_f/l_f \approx 0.132

mD/m.

What it means. The record was calculated from a model with $k$ = 150 mD, $S$ = 0.5, a fault at 180 m and $\alpha$ = 0.15 ( $k_f/l_f$ = 0.125 mD/m). The distance is found with an error of about 2%, and $\alpha$ with an error of about 9%. The error of $\alpha$ comes from the height of the maximum. The ratio from the readings is about 1% below the exact maximum of 1.451: the derivative from the readings smooths the flat top, and the plateau is slightly raised by the tail of wellbore storage. If we work back from this ratio with the exact formula, we get 0.163, so the interpolation in the table is not to blame. The sensitivity is large: 1% in the ratio is about 8% in $\alpha$ . The time of the maximum is read roughly, because the top is flat and there are only 10 readings per cycle. So the distance from it (196 m) is worse than the one from the start of the rise. The record ended at $t_{DL} \approx 22$ . The return to the plateau is still ahead, but the difference from a sealing fault is already beyond doubt. The data in this example are assumed for illustration.

From dimensionless transmissibility to a fault in the model

A test gives $\alpha$ , and together with $k$ and $L$ it gives the ratio $k_f/l_f$ . This is all we can get. The pressure in the well does not tell the width of the zone and its permeability apart. We estimate the width of fault zones from the core of wells that crossed the fault, from outcrops (places where the rock shows at the surface) and from seismic attributes (properties calculated from the seismic data). It ranges from centimetres (a film of clay smear) to metres and tens of metres (a crushed zone). So the $k_f$ “from the test” always rests on an assumed width, and it changes if the width changes.

To calculate the flow between blocks, we do not need the width at all. The flow through a section of the fault of area $A$ , for a given pressure jump across it, is set by the same ratio $k_f/l_f$ :

Transmissibility of a section of the fault

T_f = \frac{k_f}{l_f}\,A,\qquad q_r = \frac{T_f}{\mu}\,(p_1 - p_2)

where:

$T_f$ transmissibility of the section of the fault: the geometric part, without viscosity, like the transmissibility between cells $T_{ij}$ in the lesson Pressure equation for single-phase flow, m³
$A$ area of the section: its length along the fault (the strike) times the thickness of the formation that is in contact across the fault, m²
$q_r$ flow across the section at reservoir conditions, m³/s
$p_1 - p_2$ pressure jump across the fault zone itself, Pa
$\mu$ fluid viscosity, Pa·s

Units: SI

Holds when

single-phase flow; the zone is the same over the area $A$
the jump across the zone is smaller than the difference of the average pressures in the blocks: on the way to the fault and away from it, some pressure drop is also lost in the blocks themselves

The ratio $T_f/\mu$ is a “productivity index” of the fault between blocks: the flow per unit of pressure jump. In a model we describe a fault by a multiplier on the transmissibility between the cells on the two sides of the fault.

In other unit systems (1)

Metric oilfield units

q_r = 0.0853\,\frac{T_f}{\mu}\,(p_1 - p_2),\qquad T_f = \frac{k_f}{l_f}\,A

$T_f$ in mD·m ( $k_f/l_f$ in mD/m, $A$ in m²), $\mu$ in mPa·s, pressures in MPa; $q_r$ in m³/d at reservoir conditions.

$0.0853$ is the constant of Darcy's law in these units. $0.0853\,T_f/\mu$ is the flow per 1 MPa of the jump, in m³/(d·MPa), like the coupling coefficient $a_{ij}$ of the cells of a model.

In a grid model, we calculate the transmissibility between two neighbouring cells from their sizes and permeabilities, by adding the resistances of the two half-cells (Interblock flow and transmissibility). The fault zone adds its own resistance to them. The ratio of the transmissibilities with and without the fault is the multiplier:

Fault transmissibility multiplier

F = \frac{\Delta x/k}{\Delta x/k + l_f/k_f} = \frac{\alpha\,\Delta x}{\alpha\,\Delta x + L}

where:

$F$ multiplier of the transmissibility between the cells on opposite sides of the fault, from 0 to 1, —
$\Delta x$ size of the cells across the fault (equal cells), m
$k$ permeability of the cells (the same as for the formation in the test), mD
$\alpha,\ L$ dimensionless fault transmissibility, and the distance to which it refers, —, m

Units: dimensionless; $\Delta x$ and $L$ in the same units

Holds when

neighbouring cells have the same size and permeability; for different cells, we add the half-resistances $\Delta x_i/(2k_i)$ separately
the width of the zone is small compared with the cell

The multiplier depends on the grid. When the cells get smaller, the resistance of the formation between the cell centres falls, and the resistance of the zone does not. So the same fault needs a smaller multiplier. We cannot carry a multiplier from one model to another without recalculating it. We can carry $k_f/l_f$ .

Worked exampleTransmissibility multiplier and the flow between blocks

Given (assumed data)

Test result: $\alpha$ ≈ 0.163 at the distance $L$ = 184 m; $k$ = 149.1 mD; $k_f/l_f$ ≈ 0.132 mD/m
Model: cells across the fault 100 m or 50 m
Section of the fault: length 1500 m; the layers in contact across the fault are 15 m thick; oil viscosity 0.7 mPa·s

Solution

Cells of 100 m: $F = 0.163\cdot100/(0.163\cdot100 + 184) = 0.081$ . Cells of 50 m: $F = 0.163\cdot50/(0.163\cdot50 + 184) = 0.042$ .
Area of the section: $A = 1500\cdot15 = 22{,}500$ m². Transmissibility: $T_f = 0.132\cdot22{,}500 = 2970$ mD·m. Flow per 1 MPa of jump: $0.0853\cdot2970/0.7 = 362$ m³/(d·MPa).
For a pressure jump across the fault of 0.5 MPa, the flow is $q_r = 362\cdot0.5 = 181$ m³/d at reservoir conditions. That is about 139 m³/d of oil at standard conditions for $B$ = 1.3.

Answer: The multiplier is 0.081 for cells of 100 m and 0.042 for cells of 50 m. The fault passes about 362 m³/(d·MPa) through the section of 1500 m.

What it means. A fault that clearly “holds” the pressure in a two-week test passes, on the scale of a field, hundreds of cubic metres per day for each megapascal of jump. For development this means that over the years the pressure on both sides will equalise, and injection on one side will support the other, but with a delay and with losses on the jump. The practical conclusion is that a test answers the question “how much”, not “yes or no”. We pass the result to the model as $k_f/l_f$ , and recalculate the multiplier for each grid. The data in this example are assumed for illustration.

What a test cannot tell apart

A partially communicating fault shows only if the record is long enough for the derivative to fall clearly behind the sealing case, and if the hump is higher than the noise. This sets the resolution window: the range of $\alpha$ that a given test can estimate.

Worked exampleWhich faults will a test show

Given (assumed data)

Formation and fault: from the example above: $\eta$ ≈ 2365 m²/h, $L$ ≈ 184 m
Duration: 13 days (316 h) or 3 days (72 h)
Criteria: at the end of the record the derivative is below the sealing case by at least $0.1\,m_e$ ; the hump is at least 5% above the plateau

Solution

The end of the record in units of $L^2/\eta$ : $t_{DL} = 2365\cdot316/184^2 = 22.1$ and $2365\cdot72/184^2 = 5.03$ .
The smallest $\alpha$ at which the derivative at the end of the record falls behind the sealing case by $0.1\,m_e$ (calculated from the formula of the derivative): 0.008 for 13 days and 0.026 for 3 days.
The largest $\alpha$ at which the maximum is still 5% above the plateau (from the formula of the derivative): about 3.4. It does not depend on the duration, if the record reached the maximum.

Answer: A 13-day test can recognise a partially communicating fault if

\alpha

is from about 0.008 to 3.4. A 3-day test can do so if

\alpha

is from 0.026 to 3.4.

What it means. The lower limit is set by the duration. The shorter the record, the more faults look sealing, and for them we can give only an upper limit of $\alpha$ . The upper limit is set by the noise of the derivative. If we can see a hump of 3%, it rises to about 6. If we need a hump of 10%, it falls to 1.6. The criteria themselves are a choice: in each test we set them from the actual noise. The data in this example are assumed for illustration.

What else gives a rise of the derivative and a return to the plateau

Model: Sealing fault of finite length
What is similar: the flow goes round the end of the fault, so the derivative rises and returns
What is different: the shape depends on the position of the end; the length can be seen on seismic

Model: Sealing fault with a “window”
What is similar: flow through the window gives the same picture
What is different: almost impossible to tell apart with one well; seismic helps

Model: A formation that is worse beyond a line, with no barrier
What is similar: a rise of the derivative
What is different: it stays at an intermediate level, with no return

Model: A short producing time before a build-up
What is similar: the derivative of the build-up falls at the end
What is different: it disappears if we use the equivalent time, a time scale that allows for the producing time (Effect of producing time on build-up analysis)

Model: Rate changes of neighbouring wells
What is similar: humps and dips in the late part
What is different: they coincide in time with the rate changes of the neighbours

Model	What is similar	What is different
Sealing fault of finite length	the flow goes round the end of the fault, so the derivative rises and returns	the shape depends on the position of the end; the length can be seen on seismic
Sealing fault with a “window”	flow through the window gives the same picture	almost impossible to tell apart with one well; seismic helps
A formation that is worse beyond a line, with no barrier	a rise of the derivative	it stays at an intermediate level, with no return
A short producing time before a build-up	the derivative of the build-up falls at the end	it disappears if we use the equivalent time, a time scale that allows for the producing time (Effect of producing time on build-up analysis)
Rate changes of neighbouring wells	humps and dips in the late part	they coincide in time with the rate changes of the neighbours

We often cannot tell these models apart from one record. Geology, seismic and the data of other wells help.

Only a ratio. A test gives $\alpha$ , that is, $k_f/l_f$ for known $k$ and $L$ . It does not give the width and the permeability of the zone separately.
Link with the distance. The error of $L$ passes straight into $k_f/l_f$ . The error of $\phi c_t$ passes through $L$ as a square root. If the start of the rise cannot be read (wellbore storage hides it), $L$ and $\alpha$ can be found only together, by fitting, and the answer is not unique.
Direct measurements of communication. An interference test across the fault shows the flow itself. It uses an active well (a well whose rate is changed) on one side and a well that responds on the other. Pressure readings taken again and again, layer by layer, in wells of the neighbouring blocks (Pressure measurements in producing reservoirs: barriers and accuracy), and the production history in the model (History matching), check the result on the scale of the field.

Terms and symbols

Partially communicating fault: A fault across which fluid flows with extra resistance. In the model it is a thin semipermeable zone.
Dimensionless fault transmissibility $\alpha$: $k_fL/(kl_f)$ : the ratio of the transmissibility of the fault zone to the transmissibility of the formation on the way from the well to the fault.
Transmissibility of the fault zone $k_f/l_f$: The only property of the zone that a test gives. Its unit is mD/m.
Leak: In the method of images, the line of imaginary injection wells behind the image, with a total rate $q$ . It describes the flow across the fault.
Apparent fault skin $S_f$: $\e^{4\alpha}E_1(4\alpha)$ : the constant extra pressure drop on the late line after the derivative returns to the plateau, in units of $2m_e$ .
Fault transmissibility multiplier $F$: The ratio of the transmissibility between the cells of a model with the fault and without it. It depends on the size of the cells.
Resolution window: The range of $\alpha$ in which a test of a given length and quality can tell a partially communicating fault from both a sealing fault and no fault.

Common mistakes and misconceptions

Wrong

“The derivative did not reach $2m_e$ , so the fault is far away, and the record is just short.”

In fact

The start of the rise shows the distance. If the rise started on time, but the maximum is passed below $2m_e$ and the derivative goes down, the fault lets fluid through. The height of the maximum gives $\alpha$ .

Wrong

“The derivative returned to the plateau, so there is no fault.”

In fact

The late line is parallel to the first, but shifted by $2m_eS_f$ . The fault stays an extra resistance, like a skin. No fault at all would mean no hump at all.

Wrong

“The test gave the permeability of the fault zone.”

In fact

It gave $k_f/l_f$ . We get the permeability of the zone only by assuming its width, and it changes with this assumption.

Wrong

“The fault transmissibility multiplier is a property of the fault, and we can take it from another model.”

In fact

The multiplier depends on the size of the cells: $F = \alpha\Delta x/(\alpha\Delta x + L)$ . We should carry $k_f/l_f$ between models and recalculate the multiplier for each grid.

Wrong

“A fall of the derivative at the end of a build-up is a sign of a leak through the fault.”

In fact

A short producing time of the well before shut-in gives the same fall. First we build the derivative with the equivalent time or the superposition time function (a time scale that adds up the effects of all the earlier rate changes). We discuss a leak only if the fall stays.

In depth: advanced detail

Derivations: the image with a leak, the derivative and the late asymptotes

Reflection coefficient. We split the pressure disturbance along the fault into harmonics $\e^{\mathrm i\omega y}$ . In the Laplace domain (after the Laplace transform of the time variable), each of them decays across the fault as $\e^{-\beta|x|}$ , $\beta = \sqrt{s + \omega^2}$ (lengths in units of $L$ ). On the well side, a reflected wave with a coefficient $R$ is added to the incident wave. Behind the fault goes the transmitted wave with a coefficient $T$ . Continuity of the flow across the fault gives $T = 1 - R$ . The condition on the zone, that the flow is proportional to the pressure jump, $\partial p/\partial x = \alpha\,(p_+ - p_-)$ , gives $\beta(1 - R) = 2\alpha R$ , that is, $R = \beta/(\beta + 2\alpha)$ . At $\alpha = 0$ we get $R = 1$ , full reflection (a sealing fault), and at $\alpha \to \infty$ we get $R = 0$ .

The leak as a line of wells. We write $R = 1 - 2\alpha/(\beta + 2\alpha)$ and represent $1/(\beta + 2\alpha)$ as the integral $\int_0^\infty \e^{-(\beta + 2\alpha)\xi}\dd\xi$ . The factor $\e^{-\beta\xi}$ is a shift of the source by a further $\xi$ . So the reflected field is the image minus sources that are distributed behind it with the density $2\alpha\e^{-2\alpha\xi}$ , and their total rate is $q$ . In the well, the contribution of each source at distance $2 + \xi$ is $\tfrac12E_1\big((2 + \xi)^2/(4t_{DL})\big)$ . This gives the pressure formula of the lesson. In the Laplace domain it is the same thing: $s\bar p = \int_0^\infty \e^{-2B}/(B + 2\alpha)\,\dd u$ , $B = \sqrt{s + u^2}$ . At $\alpha = 0$ this is $K_0(2\sqrt s)$ , the Laplace transform of one image.

The derivative. We take the derivative of the contribution of the leak against $\ln t$ , using the fact that $t\,\dd E_1(a/t)/\dd t = \e^{-a/t}$ . The substitution $w = (2 + \xi)/(2\sqrt{t_{DL}})$ and completing the square in the exponent give an expression with the complementary error function $\mathrm{erfc}$ . The transition to $\mathrm{erfcx}$ turns it into the formula of the derivative in the lesson. When we integrate the derivative over time, we get the contribution of the leak to the pressure in the form in which the exact solution is usually written:

Contribution of the leak to the pressure: the form with the complementary error function

\alpha\int_0^\infty \e^{-2\alpha\xi}E_1\!\left(\frac{(2+\xi)^2}{4t_{DL}}\right)\dd\xi = \sqrt{\pi}\,\alpha\int_0^{t_{DL}} \e^{4\alpha^2u + 4\alpha}\,\mathrm{erfc}\left(2\alpha\sqrt u + \frac{1}{\sqrt u}\right)\frac{\dd u}{\sqrt u}

where:

$u$ dimensionless time $t_{DL}$ under the integral sign, —
$\mathrm{erfc}$ complementary error function, —
$\alpha,\ \xi,\ t_{DL}$ as in the pressure formula of the lesson, —

Units: dimensionless

Holds when

the same as for the pressure formula.

The left side is the leak as a line of sources, and the right side is the same quantity as an integral over time. The expression under the integral on the right is the rate at which the contribution of the leak grows in time.

Late asymptotes. For large arguments, $\mathrm{erfcx}(x) \approx \tfrac{1}{x\sqrt\pi}\big(1 - \tfrac{1}{2x^2}\big)$ . Substituted into the square bracket, it gives the rest of the derivative, $(1 + 4\alpha)/(8\alpha^2t_{DL})$ . The limit of the contribution of the fault to the pressure comes from the final value theorem: $\lim s\bar p$ as $s \to 0$ , that is, the integral $\int_0^\infty \e^{-2u}\dd u/(u + 2\alpha)$ , which equals $\e^{4\alpha}E_1(4\alpha)$ . This is $S_f$ . All three forms of the solution (the integral over the leak, the integral with erfc, and the inverse Laplace transform) agree numerically to the sixth digit or better.

Connections to other topics

A sealing fault near the well: the method of images — the sealing fault: the image, the derivative step, and the distance from the start of the rise.
Several boundaries: parallel faults and wedge-shaped blocks — several boundaries: channels and wedges; a fault that lets fluid through, among the other late signs.
Superposition principle — the method of images, on which the model with a leak is built.
Diagnosis with the pressure derivative — the plateau, calculating the derivative from the readings, smoothing and noise.
Skin factor — the extra resistance of a thin zone: the same idea as the skin around a well.
History matching — fault transmissibility multipliers when a model is matched to the production history.
Interblock flow and transmissibility — the transmissibility between the cells of a model, to which the fault multiplier is applied.
Faults — the geology of faults: displacements, which layers face each other across the fault, crushed zones and smears.
Effect of producing time on build-up analysis — why the derivative of a build-up falls at the end, and how to tell this from a leak.
Pressure measurements in producing reservoirs: barriers and accuracy — pressure measurements by block and by layer to check communication.

Check yourself

1
The derivative plateau is 0.050 MPa, the maximum after it is 0.076 MPa, and then the derivative goes down. Estimate the dimensionless fault transmissibility from the table of the lesson.
Show the answer
The ratio $0.076/0.050 = 1.52$ is almost the row $\alpha$ = 0.1 (1.527). So $\alpha \approx 0.1$ : interpolation between the rows 0.1 and 0.2 gives about 0.10. We take the distance from the start of the rise, and then $k_f/l_f = \alpha k/L$ .
2
A fault zone 0.5 m wide has a permeability of 0.05 mD, the formation has 40 mD, and the well is 300 m from the fault. Find $\alpha$ and estimate whether a test will notice this fault.
Show the answer
$\alpha = 0.05\cdot300/(40\cdot0.5) = 0.75$ . In the table, the maximum of the derivative is between 1.24 ( $\alpha$ = 0.5) and 1.14 ( $\alpha$ = 1), that is, about 1.18 of the plateau. A hump of 15–20% is visible with a good gauge and a clean record, but $\alpha$ from it will be inexact: in this region the maximum changes slowly.
3
The same fault: well A stands 100 m from it, and well B 400 m. In which well will a test show the higher hump of the derivative?
1. A.In A
2. B.In B
3. C.The same: it is one fault
4. D.Neither will have a hump
Show the answer
Correct answer: A. In A
$\alpha = k_fL/(kl_f)$ is proportional to the distance. For A it is four times smaller, and the fault is closer to sealing for it, so the hump is higher. For B the rise starts 16 times later ( $L^2$ is 16 times larger), and the hump is lower. The resistance of the formation on the way to the fault grows with distance, and the resistance of the zone does not.
4
A test gave $\alpha$ = 0.2 at a distance of 250 m. What multiplier should we set for cells of 200 m across the fault, and what for cells of 50 m?
Show the answer
$F = \alpha\Delta x/(\alpha\Delta x + L)$ . For 200 m: $40/(40 + 250) \approx 0.14$ . For 50 m: $10/(10 + 250) \approx 0.038$ . Smaller cells need a smaller multiplier: the resistance of the formation between their centres is smaller, and the resistance of the fault zone is the same.
5
A build-up after a two-day flow period of the well: the derivative rose above the plateau and went down at the end of the record. Can we at once conclude that the fault is partially communicating?
1. A.Yes, if the rise was more than 50%
2. B.Yes, a fall after a rise is a sign of a leak
3. C.No: we must first rule out the effect of the short producing time before shut-in
4. D.No: a fall of the derivative means a channel
Show the answer
Correct answer: C. No: we must first rule out the effect of the short producing time before shut-in
After a short producing time, the derivative of a build-up goes down at the end by itself. We build it with the equivalent time or the superposition time function (Effect of producing time on build-up analysis). If the fall disappears, the fault may even be sealing. We discuss a leak only if the fall stays.
6
Why does a well test give only the ratio $k_f/l_f$ , and not the permeability of the fault zone?
Show the answer
In the model of a thin zone, the flow through it is proportional to the pressure jump times $k_f/l_f$ . The width and the permeability enter only in this combination, and any combination with the same ratio gives the same pressure in the well. We get the permeability of the zone by assuming its width from core, outcrops or analogues.

Updated: 19 October 2026