A sealing fault near the well: the method of images

Quick refresher

An image with the same rate, at a distance $2L$ , turns the line of the fault into a line of zero flow. The pressure in the well is the sum of its own response and the contribution of the image, $m_eE_1\big(\phi\mu c_tL^2/(kt)\big)$ , where $m_e = qB\mu/(4\pi kh)$ .
At first the well “knows nothing” about the fault. The first straight line and the first plateau of the derivative give $kh$ and the skin as usual. At the end, the contribution of the image itself grows as the logarithm of time, and the slope of the line and the level of the derivative double.
The derivative rises from $m_e$ to $2m_e$ according to $1 + \e^{-L^2/(\eta t)}$ , where $\eta = k/(\phi\mu c_t)$ . The transition shows up earlier than in the pressure itself, and it stretches over about 2.7 cycles (from 1% to 99% of the rise).
We find the distance from the time at which the lines intersect, $L = \sqrt{kt_x/(\gamma\phi\mu c_t)}$ , or from any point on the derivative step, $L = \sqrt{\eta t\,\ln[m_e/(\Delta p' - m_e)]}$ .
If we intersect the first line with a second line that has not yet fully formed, the distance comes out too small. It is more accurate to draw a line with the theoretical slope $2m$ through the last reading.
A fault a few metres from the well is hidden by wellbore storage. Only the doubled plateau is seen. If we do not know about the fault, the permeability comes out at half its true value, and the skin comes out falsely small or negative.

Key relationships

\Delta p = m_e\left[\ln t_D + 0.809 + 2S + E_1\!\left(\dfrac{L_D^2}{t_D}\right)\right]

pressure in a well near a fault after start-up; $L_D = L/r_w$

\Delta p' = m_e\left[1 + \e^{-L^2/(\eta t)}\right]

step in the logarithmic derivative; $\eta = k/(\phi\mu c_t)$

L = 0.0447\sqrt{\dfrac{k\,t_x}{\phi\mu c_t}}

from the time at which the lines intersect: m, mD, h, mPa·s, 1/MPa (in oilfield units the constant is 0.01217 and $L$ is in ft)

L = 0.0596\sqrt{\dfrac{k\,t}{\phi\mu c_t}\ln\dfrac{m_e}{\Delta p' - m_e}}

from a point on the derivative step (in oilfield units the constant is 0.01624)

L_{\max} = 0.0349\sqrt{\dfrac{qB\,t_p}{\phi c_t h\,\delta p}}

beyond this distance, a pressure build-up after $t_p$ hours of production will not show the fault at a detection threshold $\delta p$ (units: m³/d, h, 1/MPa, m, MPa)

Common pitfalls

Taking any doubling of the slope for a fault. A similar picture comes from a reservoir whose flow capacity far from the well is half as large, and from false “straight lines” in the wellbore-storage period.
Drawing the second line through points where the derivative has not yet reached the doubled level, and so getting too small a distance.
Processing the record of a well right next to a fault as ordinary radial flow.
Taking $p^*$ from the first straight line of a build-up near a fault as the initial reservoir pressure. It is too low by the contribution of the image.
Expecting a long build-up to always show a distant fault. The limit is set by the volume produced before shut-in and by the resolution of the gauge.

A fault as a reservoir boundary

A fault is a surface along which blocks of rock have moved relative to each other (Faults). For flow to a well, the displacement itself matters less than the transmissibility of the fault zone across the fault (how easily fluid can cross it). If this is negligible, the fault acts as a no-flow boundary: fluid flows along the fault, but not across it.

A fault can be sealing for different reasons. The simplest case is when the throw (the amount of displacement) is greater than the thickness of the reservoir. Then the rock opposite the reservoir, on the other side, is shale or tight rock. But even where permeable rocks are in contact, a fault may not pass fluid. The fault plane may be covered with clay smear (clay spread along the plane), the grains in the fault zone may be crushed, and the pores may be healed by secondary minerals (minerals that grew in the pores later). On the other hand, a fault may pass fluid partially if permeable layers on both sides of it are in contact (Partially communicating faults).

Why a fault may be sealing

Mechanism: Contact with impermeable rock
What happens: opposite the reservoir there is shale, salt or tight rock
When it is likely: the throw is greater than the thickness of the reservoir

Mechanism: Clay smear
What happens: clay smeared along the fault plane separates permeable rocks
When it is likely: the displaced sequence contains many shale beds

Mechanism: Cataclasis (crushing of grains)
What happens: the grains are crushed, the pores in the fault zone are small
When it is likely: clean sandstones, especially at great depth

Mechanism: Cementation
What happens: minerals that settled out of water solutions have sealed the pores of the fault zone
When it is likely: faults along which groundwater once moved

Mechanism	What happens	When it is likely
Contact with impermeable rock	opposite the reservoir there is shale, salt or tight rock	the throw is greater than the thickness of the reservoir
Clay smear	clay smeared along the fault plane separates permeable rocks	the displaced sequence contains many shale beds
Cataclasis (crushing of grains)	the grains are crushed, the pores in the fault zone are small	clean sandstones, especially at great depth
Cementation	minerals that settled out of water solutions have sealed the pores of the fault zone	faults along which groundwater once moved

The list is qualitative. We estimate the transmissibility of a particular fault from geological models, from the pressures in different blocks, and from well tests.

For well testing, it matters that faults are not the only sealing boundaries. A pinch-out of the reservoir (where it thins to nothing), an abrupt change of facies (sandstone passes into shale), and the edge of a channel sand body are also lines across which there is no flow. The method of images describes them all in the same way, as long as the boundary is straight enough on the scale of the distance to the well.

Why look for a fault with a well test if we have seismic surveys? Seismic sees large faults. But a fault with a small throw may be too small for its resolution, and seismic does not measure the transmissibility of a fault at all. A well test shows whether there is a boundary a few hundred metres away, and whether it passes fluid. The best result comes from combining the two: seismic suggests where to look, and the test confirms it and gives the distance.

The image well

The lesson on superposition covers the method of images. Beyond the fault, at the same distance, we place an imaginary well with the same rate history: the image well. The two identical wells are symmetrical about the fault line, so there is no pressure difference across it, and so no flow. Here we are interested in how this looks in the pressure record of the well itself.

A well near a sealing fault: dimensionless form

p_D = \underbrace{\tfrac12\left(\ln t_D + 0.809\right) + S}_{\text{well}} + \underbrace{\tfrac12\,E_1\!\left(\frac{L_D^2}{t_D}\right)}_{\text{image}},\qquad L_D = \frac{L}{r_w}

where:

$p_D$ dimensionless drop in bottomhole pressure $2\pi kh\,\Delta p/(qB\mu)$ , with skin — as in the lesson Line-source solution and the pressure function, —
$t_D$ dimensionless time $kt/(\phi\mu c_tr_w^2)$ , —
$L,\ L_D$ distance from the well to the fault, and the same distance in wellbore radii, m, —
$S$ skin factor of the real well; the image has no skin, —
$E_1$ exponential integral (the E₁ function), —

Units: no units; natural logarithm

Holds when

The fault is a straight sealing boundary through the whole thickness of a homogeneous reservoir, and the disturbance has not reached any other boundaries.
For the well itself the logarithmic approximation holds ( $t_D > 25$ ), and wellbore storage is over.
The image is a line source at a distance $2L$ . The argument of $E_1$ is $(2L_D)^2/(4t_D) = L_D^2/t_D$ .

We can write the argument of the image's contribution with the hydraulic diffusivity $\eta = k/(\phi\mu c_t)$ (how fast pressure spreads through the rock): $L_D^2/t_D = L^2/(\eta t)$ . So everything depends on a single number: time in units of $L^2/\eta$ . While $\eta t$ is small compared with $L^2$ , the contribution of the image is negligible. When $\eta t$ is much larger than $L^2$ , the function $E_1$ itself becomes a logarithm, and $p_D$ grows as $\ln t_D$ , which is twice as fast as for a single well.

In other unit systems (1)

Metric oilfield units

\Delta p = 0.933\,\frac{qB\mu}{kh}\left[\ln\frac{3.553\cdot10^{-3}\,kt}{\phi\mu c_tr_w^2} + 0.809 + 2S + E_1\!\left(\frac{L^2}{\eta t}\right)\right],\quad \eta = \frac{3.553\cdot10^{-3}\,k}{\phi\mu c_t}

$q$ in m³/d, $\mu$ in mPa·s, $k$ in mD, $h$ , $r_w$ and $L$ in m, $c_t$ in 1/MPa, $t$ in h; $\Delta p = p_i - p_{wf}$ in MPa; $\eta$ in m²/h.

$0.933 = 1/(4\pi\cdot0.0853)$ . The constant $3.553\cdot10^{-3}$ converts time in hours into dimensionless time (Pressure build-up analysis).

SchematicA well near a fault and its image, in plan view (schematic, not to scale). Circles 1 and 2 show the radius of investigation (how far the disturbance has travelled),

\sqrt{4\eta t}

, around the well. While it is smaller than

L

, the well behaves as in an infinite reservoir. When it reaches the fault, the derivative begins to rise. Arc 3 shows the same radius around the image. When this radius equals

2L

, the disturbance from the image has reached the well, and the contribution of the fault is noticeable in the pressure itself. After that the slope doubles. The lines are shown only on this side of the fault.

Stages of the record of a well near a fault

Stage: Infinite reservoir
$\eta t/L^2$: less than 0.2
Derivative: first plateau $m_e$
Semilog plot: first straight line, slope $m$

Stage: Start of the transition
$\eta t/L^2$: 0.2–0.45
Derivative: 1–10% above the plateau
Semilog plot: points slightly above the first line

Stage: Transition
$\eta t/L^2$: 0.45–10
Derivative: from $1.1\,m_e$ to $1.9\,m_e$
Semilog plot: a curve, no straight line

Stage: Doubled slope
$\eta t/L^2$: more than 10–25
Derivative: $2m_e$ to within 5% or better
Semilog plot: second straight line, slope $2m$

Stage	$\eta t/L^2$	Derivative	Semilog plot
Infinite reservoir	less than 0.2	first plateau $m_e$	first straight line, slope $m$
Start of the transition	0.2–0.45	1–10% above the plateau	points slightly above the first line
Transition	0.45–10	from $1.1\,m_e$ to $1.9\,m_e$	a curve, no straight line
Doubled slope	more than 10–25	$2m_e$ to within 5% or better	second straight line, slope $2m$

The boundaries of the stages come from the derivative step formula in the section below. At $\eta t/L^2$ = 0.217 the derivative is 1% above the plateau. At 0.434 it is 10% above. At 9.5 it has covered 90% of the step, and at 25 it has covered 96%. The time at which the two lines intersect corresponds to $\eta t/L^2 = 1.781$ .

Two straight lines on the semilog plot

When $\eta t$ is much larger than $L^2$ , we can replace the function $E_1$ by a logarithm, as for the well itself (logarithmic approximation). Two straight lines then describe the record:

The first and second semilog straight lines

\begin{aligned}\text{first:}\quad &\Delta p = m_e\left[\ln t_D + 0.809 + 2S\right]\\ \text{second:}\quad &\Delta p = 2m_e\left[\ln t_D + 0.809 + S - \ln\frac{2L}{r_w}\right]\end{aligned}

where:

$\Delta p$ drop in bottomhole pressure $p_i - p_{wf}$ , Pa
$m_e$ slope of the first line per unit of natural logarithm, $qB\mu/(4\pi kh)$ , Pa
$t_D,\ S$ dimensionless time and skin factor, —
$L,\ r_w$ distance to the fault and wellbore radius, m

Units: any consistent system of units; natural logarithm

Holds when

The first line holds while $\eta t/L^2 < 0.2$ , once wellbore storage is over.
The second line holds at $\eta t/L^2 \gg 1$ . Its slope differs from $2m$ by less than 5% only after $\eta t/L^2 \approx 10$ .
The skin enters the second line once, because only the real well has a pressure drop across the skin.

The slope of the second line is twice as large, $2m$ per decimal cycle, and its position depends on the distance to the fault. So, as usual, we find $kh$ and the skin from the first line (Pressure drawdown analysis). The second line serves as a check and gives the distance. If the same $kh$ and skin, put into the second line together with $L$ , reproduce its position, the picture is consistent.

If we equate the two lines, we get the moment at which they cross, the intersection time:

Distance to the fault from the intersection time of the lines

t_x = \frac{\gamma\,\phi\mu c_tL^2}{k}\quad\Rightarrow\quad L = \sqrt{\frac{k\,t_x}{\gamma\,\phi\mu c_t}},\qquad \gamma = 1.781

where:

$t_x$ intersection time of the extensions of the first and second lines, s
$L$ distance to the fault, m
$\gamma$ the exponential of Euler's constant, —
$k,\ \phi,\ \mu,\ c_t$ permeability (from the first line), porosity, viscosity, total compressibility, m², —, Pa·s, 1/Pa

Units: SI

Holds when

Both lines are identified reliably, and the second is drawn through points where the slope has really doubled.
$\phi$ and $c_t$ are known independently. Their error passes into $L$ as a square root.

At the moment of intersection, $\eta t_x/L^2 = \gamma$ . That is, the radius of investigation $\sqrt{4\eta t}$ equals $2.67\,L$ . By this time the derivative has already covered 57% of the way from the first plateau to the second. So the intersection is the middle of the transition, not its start.

In other unit systems (2)

Metric oilfield units

L = 0.0447\sqrt{\frac{k\,t_x}{\phi\mu c_t}}

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

$0.0447 = \sqrt{3.553\cdot10^{-3}/1.781}$ .

Oilfield units

L = 0.01217\sqrt{\frac{k\,t_x}{\phi\mu c_t}}

$L$ in ft, $k$ in mD, $t_x$ in h, $\mu$ in cP, $c_t$ in 1/psi.

$0.01217 = \sqrt{2.637\cdot10^{-4}/1.781}$ , where $2.637\cdot10^{-4}$ is the factor of dimensionless time in these units.

Let us work through an assumed record. An exploration well was put on an extended well test (a long trial production period) at a constant rate of 110 m³/d, and a downhole gauge recorded the pressure for about 400 h. Seismic data suggest a fault a few hundred metres from the well.

Pressure drop in an extended well test (assumed data, selected readings)

$t$: 36 s
$\Delta p$ , MPa: 0.5654
$\Delta p'$ , MPa: —

$t$: 6 min
$\Delta p$ , MPa: 0.7368
$\Delta p'$ , MPa: 0.0537

$t$: 30 min
$\Delta p$ , MPa: 0.8200
$\Delta p'$ , MPa: 0.0505

$t$: 1.00 h
$\Delta p$ , MPa: 0.8547
$\Delta p'$ , MPa: 0.0501

$t$: 2.00 h
$\Delta p$ , MPa: 0.8892
$\Delta p'$ , MPa: 0.0499

$t$: 5.01 h
$\Delta p$ , MPa: 0.9352
$\Delta p'$ , MPa: 0.0507

$t$: 10.0 h
$\Delta p$ , MPa: 0.9714
$\Delta p'$ , MPa: 0.0560

$t$: 20.0 h
$\Delta p$ , MPa: 1.0134
$\Delta p'$ , MPa: 0.0668

$t$: 39.8 h
$\Delta p$ , MPa: 1.0636
$\Delta p'$ , MPa: 0.0784

$t$: 100 h
$\Delta p$ , MPa: 1.1417
$\Delta p'$ , MPa: 0.0895

$t$: 200 h
$\Delta p$ , MPa: 1.2053
$\Delta p'$ , MPa: 0.0941

$t$: 398 h
$\Delta p$ , MPa: 1.2714
$\Delta p'$ , MPa: —

$t$	$\Delta p$ , MPa	$\Delta p'$ , MPa
36 s	0.5654	—
6 min	0.7368	0.0537
30 min	0.8200	0.0505
1.00 h	0.8547	0.0501
2.00 h	0.8892	0.0499
5.01 h	0.9352	0.0507
10.0 h	0.9714	0.0560
20.0 h	1.0134	0.0668
39.8 h	1.0636	0.0784
100 h	1.1417	0.0895
200 h	1.2053	0.0941
398 h	1.2714	—

$\Delta p = p_i - p_{wf}$ , and $p_i$ = 29 MPa. The data are calculated from a model of a homogeneous reservoir with wellbore storage, skin and a straight sealing fault, and rounded to 0.0001 MPa. The examples reveal the parameters of the model. We calculate the derivative from all 47 readings of the record (10 per cycle), with a differentiation interval of 0.2 cycle (Diagnosis with the pressure derivative). It is not given at the first and last points, because they have no neighbours at a distance of 0.2 cycle.

CalculatedSemilog plot of the record (dots). The dashed straight lines are the first line (through the readings at 0.5 h and 5 h) and the chord, the straight line through the last two readings from the example below. The solid grey line is the line with slope

2m

through the last reading. The points where these lines cross the first line are marked.

Worked examplePermeability, skin and distance to the fault from two straight lines

Given (assumed data)

Well and reservoir: $q$ = 110 m³/d; $B$ = 1.25; $\mu$ = 0.65 mPa·s; $h$ = 14 m; $\phi$ = 0.19; $c_t = 1.4\cdot10^{-3}$ MPa⁻¹; $r_w$ = 0.1 m
First line: readings at 0.501 h (0.8200 MPa) and 5.01 h (0.9352 MPa), exactly one cycle apart; the derivative between them is almost constant
Last readings: 199.5 h — 1.2053 MPa; 398.1 h — 1.2714 MPa (exactly 0.3 cycle apart)

Solution

The slope of the first line is $m = 0.9352 - 0.8200 = 0.1152$ MPa per cycle. Also $qB\mu = 110\cdot1.25\cdot0.65 = 89.375$ , so $kh = 2.149\cdot89.375/0.1152 = 1667$ mD·m and $k = 119.1$ mD.
The first line at 1 h gives $\Delta p_{1\text{h}} = 0.8200 + 0.3\cdot0.1152 = 0.8546$ MPa. For the skin, $\lg\dfrac{k}{\phi\mu c_tr_w^2} = 7.838$ , so $S = 1.151\,(0.8546/0.1152 - 7.838 + 2.098) = 1.93$ .
The slope from the last two readings is $(1.2714 - 1.2053)/0.3 = 0.2203$ MPa per cycle, which is 1.91 times the first. The doubling has almost taken place. This is a sign of a single sealing fault.
The intersection. The first line is $\Delta p = 0.8546 + 0.1152\lg t$ , and the chord is $\Delta p = 0.6986 + 0.2203\lg t$ . So $\lg t_x = (0.8546 - 0.6986)/(0.2203 - 0.1152) = 1.484$ and $t_x = 30.5$ h.
The distance. We have $\phi\mu c_t = 0.19\cdot0.65\cdot1.4\cdot10^{-3} = 1.729\cdot10^{-4}$ , so $L = 0.0447\sqrt{\dfrac{119.1\cdot30.5}{1.729\cdot10^{-4}}} = 205$ m.
The variant with the theoretical slope. We draw the line $\Delta p = 0.6724 + 0.2304\lg t$ , with slope $2m = 0.2304$ , through the last reading. The difference of the slopes equals $m$ , so $\lg t_x = (0.8546 - 0.6724)/0.1152 = 1.582$ , $t_x = 38.2$ h and $L = 229$ m.

Answer:

k \approx 119.1

mD, and the skin is about 1.93. The distance to the fault is 205 m from the chord of the last readings, and 229 m from the line with the doubled slope.

What it means. The record was calculated from a model with $k$ = 120 mD, $S$ = 2 and a fault at 230 m. The first line gave the permeability and skin with errors of 0.8% and 0.07, because we drew it before the fault had noticeably affected the pressure. The permeability is slightly low: at the ends of the interval, wellbore storage is still felt and the fault is just beginning to be felt. The distance from the chord is too small by 11%. By the end of the record the slope has not yet doubled, so the chord is flatter than the true second line and crosses the first line earlier. The line with the theoretical slope $2m$ through the last reading has a much smaller error. Here the error is almost zero, and on average for a record of this length it is about 3% (see the table below). The data in this example are assumed for illustration.

Error of the distance from the intersection time versus the length of the record (calculation without wellbore storage)

End of record, $\eta t/L^2$: 5
Slope ratio: 1.75
$L$ from the chord / true: 0.74
$L$ from slope $2m$ / true: 0.91

End of record, $\eta t/L^2$: 10
Slope ratio: 1.87
$L$ from the chord / true: 0.83
$L$ from slope $2m$ / true: 0.95

End of record, $\eta t/L^2$: 25
Slope ratio: 1.94
$L$ from the chord / true: 0.91
$L$ from slope $2m$ / true: 0.98

End of record, $\eta t/L^2$: 50
Slope ratio: 1.97
$L$ from the chord / true: 0.94
$L$ from slope $2m$ / true: 0.99

End of record, $\eta t/L^2$: 100
Slope ratio: 1.99
$L$ from the chord / true: 0.97
$L$ from slope $2m$ / true: 1.00

End of record, $\eta t/L^2$	Slope ratio	$L$ from the chord / true	$L$ from slope $2m$ / true
5	1.75	0.74	0.91
10	1.87	0.83	0.95
25	1.94	0.91	0.98
50	1.97	0.94	0.99
100	1.99	0.97	1.00

We draw the chord through readings at the end of the record that are a factor of two apart in time. We draw the line with slope $2m$ through the last reading. The first line is exact. For the record of the lesson, $\eta t/L^2 \approx 18.6$ , and the same calculation gives an error of about 12% for the chord and about 3% for the line with slope $2m$ .

The derivative step

A fault shows up earlier and more clearly on the derivative. The derivative of the function $E_1(a/t)$ with respect to $\ln t$ is $\e^{-a/t}$ (derivative near a straight boundary). So after wellbore storage is over, a simple formula describes the derivative, and the distance follows from it directly:

The derivative step at a sealing fault, and the distance from a single point on it

\Delta p' = m_e\left[1 + \exp\left(-\frac{L^2}{\eta t}\right)\right]\quad\Rightarrow\quad L = \sqrt{\eta\,t\,\ln\frac{m_e}{\Delta p' - m_e}}

where:

$\Delta p'$ logarithmic derivative at time $t$ on the step, $m_e < \Delta p' < 2m_e$ , Pa
$m_e$ level of the first plateau of the derivative, $qB\mu/(4\pi kh)$ , Pa
$\eta$ hydraulic diffusivity $k/(\phi\mu c_t)$ , m²/s
$L,\ t$ distance to the fault and time since start-up, m, s

Units: SI

Holds when

The reservoir is homogeneous and there is one straight sealing fault.
Wellbore storage is over, and $m_e$ and $k$ are known from the first plateau.
We take points in the middle of the step, where $\Delta p'/m_e$ is between 1.2 and 1.7. At the start and at the end of the step, small errors in the derivative change $L$ greatly.

The step is one and the same curve for any fault, if time is measured in units of $L^2/\eta$ . So every point on it gives the distance, and we do not need to wait for the second line. The middle of the step ( $\Delta p' = 1.5\,m_e$ ) comes at $\eta t = L^2/\ln 2$ .

In other unit systems (2)

Metric oilfield units

L = 0.0596\sqrt{\frac{k\,t}{\phi\mu c_t}\,\ln\frac{m_e}{\Delta p' - m_e}}

$L$ in m, $k$ in mD, $t$ in h, $\mu$ in mPa·s, $c_t$ in 1/MPa; $\Delta p'$ and $m_e$ in the same units.

$0.0596 = \sqrt{3.553\cdot10^{-3}}$ .

Oilfield units

L = 0.01624\sqrt{\frac{k\,t}{\phi\mu c_t}\,\ln\frac{m_e}{\Delta p' - m_e}}

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

$0.01624 = \sqrt{2.637\cdot10^{-4}}$ .

CalculatedThe derivative of the lesson's record divided by the level of the first plateau (dots), and the step from the formula above for a fault at 230 m (line). The time scale is logarithmic. The marks show the 10% rise, the middle of the step, the time at which the two lines intersect, and 90% of the step.

Worked exampleDistance to the fault from a single point on the step

Given (assumed data)

First plateau of the derivative: mean from 1 to 3.2 h — $m_e$ = 0.04996 MPa
Point on the step: at 25.12 h the derivative is 0.07068 MPa
Hydraulic diffusivity from the first line: $\eta = 3.553\cdot10^{-3}\cdot119.1/(0.19\cdot0.65\cdot1.4\cdot10^{-3}) = 2447$ m²/h

Solution

The ratio to the plateau is $\Delta p'/m_e = 0.07068/0.04996 = 1.4147$ . This is the middle of the step, a convenient point.
Then $\ln\dfrac{m_e}{\Delta p' - m_e} = \ln\dfrac{1}{1.4147 - 1} = 0.880$ .
So $L = \sqrt{\eta t\cdot0.880} = \sqrt{2447\cdot25.12\cdot0.880} = 233$ m.
Sensitivity: if we read the derivative 2% of the plateau higher, the ratio is 1.4347 and $L$ = 226 m, which is an error of 2.7%.
The same formula at 6.31 h, where the rise has only just begun ( $\Delta p'/m_e = 1.0366$ ), gives $L$ = 226 m. But an error in the derivative of the same 2% of the plateau would already give 211 m, which is 7% less.

Answer: From a point in the middle of the step, the distance to the fault is about 233 m. The model has 230 m.

What it means. The derivative step contains the distance at every point, so we can find a fault without waiting for the second line. Here 40 h of record would have been enough instead of 400 (the derivative at the 25 h point is calculated from readings up to 40 h). The price is dependence on the model. The formula is valid for a homogeneous reservoir with one straight boundary, and we must know the level of the first plateau accurately. In practice, we do the same by fitting a model with an image to the whole record. The software compares the calculated step with the measured one at all points at once (Type-curve matching). The data in this example are assumed for illustration.

Worked exampleHow long to record to see a fault

Given (assumed data)

Reservoir: as in the examples above; $\eta$ ≈ 2447 m²/h
Fault from seismic: 230 m or 460 m from the well

Solution

The time scale $L^2/\eta$ : 21.6 h for 230 m and 86.5 h for 460 m.
A 10% rise of the derivative ( $\eta t/L^2$ = 0.434) comes after 9.4 and 38 h.
The middle of the step ( $\eta t/L^2 = 1/\ln 2 = 1.443$ ) comes after 31 and 125 h.
The slope has doubled to within 5% ( $\eta t/L^2$ ≈ 10) after 216 and 865 h (that is, 36 days for the fault at 460 m).

Answer: To see a fault at 230 m with confidence, 31 h of record is enough (the middle of the step). The second line needs about 220 h. For a fault at 460 m all the times are four times longer.

What it means. The time grows as the square of the distance. If we only need to confirm a fault at 460 m, not measure it accurately, a few days will do. The derivative step itself will show the beginning of the doubling. But for this, wellbore storage must end well before, and the rate must stay constant for all these days. That is why we get long records at a constant rate in extended well tests, and in other cases we rely on a pressure build-up (see the section below). The data in this example are assumed for illustration.

A fault very close: hemiradial flow

The closer the fault, the earlier the transition begins. If $L$ is only a few metres or tens of metres, the first plateau may fit into a few minutes, and wellbore storage may hide it entirely. The record then shows only one plateau, the doubled one, and one straight line with slope $2m$ . Fluid comes to the well from one side only, as to a well at the edge of a half-plane. This regime is called hemiradial flow (half-circular flow).

Hemiradial flow: flow capacity and skin for a known distance to the fault

kh = \frac{qB\mu}{2\pi\,\Delta p_2'},\qquad S = \frac{\Delta p(t)}{\Delta p_2'} - \ln\frac{kt}{\phi\mu c_tr_w^2} - 0.809 + \ln\frac{2L}{r_w}

where:

$\Delta p_2'$ level of the doubled plateau of the derivative, $2m_e = qB\mu/(2\pi kh)$ , Pa
$\Delta p(t)$ pressure drop at time $t$ on the doubled plateau, Pa
$L$ distance to the fault, known from geology or seismic, m
$k,\ \phi,\ \mu,\ c_t,\ r_w$ permeability (from the doubled plateau), porosity, viscosity, total compressibility, wellbore radius, m², —, Pa·s, 1/Pa, m

Units: SI; natural logarithm

Holds when

The first plateau is hidden by wellbore storage, and the doubled plateau has formed.
The distance $L$ is known independently, because we cannot find it from such a record.
The reservoir is homogeneous, and the fault is straight and sealing.

This is the second straight line of the previous section, but without the first. If we do not know about the fault and process the record as ordinary radial flow, we take the slope $2m$ for $m$ . The flow capacity then comes out at half its true value, and the skin comes out as $S_{\text{app}} = \tfrac12\left[S + \ln(r_w/L)\right]$ . This is smaller than the true skin and often negative (the derivation is in the advanced section).

In other unit systems (1)

Metric oilfield units

kh = 1.866\,\frac{qB\mu}{\Delta p_2'},\qquad S = 2.303\left[\frac{\Delta p_{1\text{h}}}{m_2} - \lg\frac{k}{\phi\mu c_tr_w^2} + 2.098 + \lg\frac{2L}{r_w}\right]

$q$ in m³/d, $\mu$ in mPa·s, $k$ in mD, $c_t$ in 1/MPa, $r_w$ and $L$ in m; $m_2$ is the slope of the line per decimal cycle, $\Delta p_{1\text{h}}$ is the pressure drop on it at 1 h, in MPa.

$1.866 = 2\cdot0.933$ . The factor $2.303$ replaces the usual $1.151$ because the slope is doubled, $m_2 = 2.303\cdot\Delta p_2'$ . The constant $2.098$ is the same as in the skin formula for a drawdown test (Pressure drawdown analysis).

CalculatedDiagnostic plot of a well 10 m from a fault. It shows the pressure drop (dots) and the derivative calculated from the readings (orange line). The grey line is the derivative of the same well without a fault. After the wellbore-storage hump, the derivative goes straight to the doubled level

2m_e

. The first plateau

m_e

is absent from the record.

Worked exampleA well next to a fault, mistaken for an ordinary well

Given (assumed data)

Well: the same design and the same reservoir as in the examples above, but 10 m from the fault according to seismic data
Record: a straight line from 2 to 20 h: 1.0547 and 1.2829 MPa (exactly one cycle apart); the derivative on the plateau is about 0.0990 MPa

Solution

Processing without the fault: $m = 1.2829 - 1.0547 = 0.2282$ MPa per cycle and $k = 2.149\cdot89.375/(0.2282\cdot14) = 60.1$ mD.
Then $\Delta p_{1\text{h}} = 1.0547 - 0.3\cdot0.2282 = 0.9862$ MPa and $\lg\dfrac{k}{\phi\mu c_tr_w^2} = 7.541$ , so $S = 1.151\,(0.9862/0.2282 - 7.541 + 2.098) = -1.29$ . The well looks stimulated.
With the fault taken into account, the slope is $2m$ , so $k = 2\cdot60.1 = 120.2$ mD. For the skin, $\lg\dfrac{k}{\phi\mu c_tr_w^2} = 7.842$ and $\lg(2L/r_w) = \lg 200 = 2.301$ , so $S = 2.303\,(0.9862/0.2282 - 7.842 + 2.098 + 2.301) = 2.02$ .
The link between the two answers: $\tfrac12[S + \ln(r_w/L)] = \tfrac12\,(2.02 + \ln 0.01) = -1.29$ . This is exactly the apparent skin from the second step.

Answer: Without knowing about the fault, we would get

k \approx 60.1

mD and a skin of −1.29. With the fault taken into account,

k \approx 120.2

mD and the skin is 2.02.

What it means. The record looks plausible: a wellbore-storage hump, a small dip after it, and a flat plateau. We can easily mistake the dip for the end of the hump. Nothing in the record points to the fault. Only a comparison with other data gives it away: a permeability half of that from core and neighbouring wells, and a negative skin in a well that was never stimulated. That is why we interpret wells near known faults with the geology in mind. If we suspect a fault, we reduce wellbore storage, by shutting the well in downhole or by measuring the inflow with a downhole flowmeter (Wellbore storage), so that the first plateau has time to appear. The data in this example are assumed for illustration.

Pressure build-up near a fault

We more often look for faults on a pressure build-up (the pressure rise after the well is shut in). The rate during the record is known exactly: it is zero. The calculation now needs two superpositions (two additions of solutions): one in time (start-up and shut-in) and one in space (the well and its image). After producing for a time $t_p$ and then shutting in for a time $\Delta t$ , the pressure is:

Build-up of a well near a sealing fault

p_i - p_{ws} = m_e\left[\ln\frac{t_p + \Delta t}{\Delta t} + E_1\!\left(\frac{L^2}{\eta\,(t_p + \Delta t)}\right) - E_1\!\left(\frac{L^2}{\eta\,\Delta t}\right)\right]

where:

$p_{ws}$ bottomhole pressure a time $\Delta t$ after shut-in, Pa
$p_i$ pressure in the reservoir before start-up, Pa
$t_p,\ \Delta t$ producing time before shut-in and time after shut-in, s
$m_e,\ \eta,\ L$ level of the derivative plateau, hydraulic diffusivity, distance to the fault, Pa, m²/s, m

Units: SI; natural logarithm

Holds when

Before start-up the pressure in the reservoir was $p_i$ . The rate stays constant during $t_p$ .
For the well itself the logarithmic approximation holds, and wellbore storage is over. There are no other boundaries.

The first term is the ordinary Horner line (Pressure build-up analysis). The two terms with $E_1$ come from the image well. It started working a time $t_p + \Delta t$ ago and “stopped” a time $\Delta t$ ago. Which of the two terms matters more depends on whether the disturbance had time to reach the fault during production. That is set by the ratio $L^2/(\eta t_p)$ .

Two cases of a build-up near a fault

: Start of the build-up
The fault was felt during production: $L^2 < \eta t_p$: Horner line with the correct slope $m$ , but $p^*$ is below the initial pressure by $m_eE_1\big(L^2/(\eta t_p)\big)$
Not felt: $L^2 > \eta t_p$: Horner line with $p^* \approx p_i$

: End of the build-up
The fault was felt during production: $L^2 < \eta t_p$: the points leave the line upwards; at $\eta\Delta t \gg L^2$ the slope is $2m$
Not felt: $L^2 > \eta t_p$: a small dip below the line and a return to it

: What it gives
The fault was felt during production: $L^2 < \eta t_p$: the distance from the intersection of the Horner lines
Not felt: $L^2 > \eta t_p$: at best a sign of a boundary and a rough estimate of the distance

	The fault was felt during production: $L^2 < \eta t_p$	Not felt: $L^2 > \eta t_p$
Start of the build-up	Horner line with the correct slope $m$ , but $p^*$ is below the initial pressure by $m_eE_1\big(L^2/(\eta t_p)\big)$	Horner line with $p^* \approx p_i$
End of the build-up	the points leave the line upwards; at $\eta\Delta t \gg L^2$ the slope is $2m$	a small dip below the line and a return to it
What it gives	the distance from the intersection of the Horner lines	at best a sign of a boundary and a rough estimate of the distance

Distance to the fault from the intersection of the Horner lines

L = \sqrt{\frac{\eta\,\Delta t_{ex}}{\gamma}},\qquad \Delta t_{ex} = \frac{t_p\,\Delta t_x}{t_p + \Delta t_x}

where:

$\Delta t_x$ shut-in time at which the extensions of the first and second Horner lines meet, s
$\Delta t_{ex}$ the matching equivalent time, which corrects the shut-in time for the length of production (Effect of producing time on build-up analysis), s
$\eta,\ \gamma$ hydraulic diffusivity and $\gamma = 1.781$ , m²/s, —

Units: SI

Holds when

The fault was felt long before shut-in: $L^2 \ll \eta t_p$ . Then $E_1\big(L^2/(\eta t_p)\big)$ can be replaced by a logarithm.
Both Horner lines have fully formed. A second Horner line that has not fully formed gives too small a value of $L$ , as in a drawdown test.

Call $H = (t_p + \Delta t)/\Delta t$ the Horner ratio. The first line is $p_i - p_{ws} = m_e(\ln H + \ln H_x)$ , where $H_x = \eta t_p/(\gamma L^2)$ . The second is $p_i - p_{ws} = 2m_e\ln H$ . The lines meet at $H = H_x$ . Solving this for $L$ and writing the result through $\Delta t_x$ gives the formula. If $\Delta t_x \ll t_p$ , the equivalent time is almost equal to $\Delta t_x$ , and the formula becomes the same as the one for a drawdown test.

CalculatedHorner plot of the build-up after 398 h of an extended well test (dots). The dashed straight line passes through the readings at 1 h and 5 h. We extend it to a Horner ratio of 1 and read

p^*

there. The horizontal line at the top is the initial pressure, 29 MPa. The late points leave the line upwards.

Worked exampleBuild-up after an extended well test:

p^*

below the initial pressure

Given (assumed data)

Well: the well from the examples above; after 398.1 h of production, we shut it in at the wellhead
Build-up readings: 1 h — 28.5831 MPa; 5.012 h — 28.6626 MPa
Initial pressure: 29.0 MPa (measured before start-up)

Solution

Horner ratios: $(398.1 + 1)/1 = 399.1$ and $(398.1 + 5.012)/5.012 = 80.43$ . The difference of their decimal logarithms is 0.6957.
Slope: $m = (28.6626 - 28.5831)/0.6957 = 0.1143$ MPa per cycle. This is almost the same as for the first line of the drawdown test, so the permeability from it is right.
Extending the line to a ratio of 1: $p^* = 28.6626 + 0.1143\cdot\lg 80.43 = 28.880$ MPa. This is 0.120 MPa below the initial pressure.
The underestimate we expect: $m_e = 0.1143/2.303 = 0.0496$ MPa. With $L$ = 230 m and $\eta$ = 2447 m²/h, we get $L^2/(\eta t_p) = 0.0543$ , $E_1(0.0543) = 2.390$ and $m_eE_1 = 0.119$ MPa.

Answer:

p^*

of the first line is 28.880 MPa. That is 0.120 MPa below the initial pressure of 29.0 MPa. It comes from the image well, not from depletion of the reservoir (a fall in pressure caused by production).

What it means. In 400 h we produce too little from the reservoir to lower the pressure in the accumulation noticeably. Yet $p^*$ is still below the initial pressure. The fault keeps the drawdown cone (the area of lowered pressure around the well) on the well's side, and the Horner line through the early points “does not know” about it. If we did not know the initial pressure, we might take this underestimate for depletion. We might also take it for a sign of communication with a part of the accumulation that is already being developed. The late points of this build-up leave the line upwards and, after hundreds of hours, approach the slope $2m$ . We find the distance from them and from the intersection of the Horner lines, with the same reservations about a second line that has not fully formed. The data in this example are assumed for illustration.

If the disturbance did not reach the fault during production, the build-up shows the fault only as a small deviation from the Horner line. The deviation grows, reaches a maximum about $\Delta t \approx L^2/\eta$ after shut-in, and then falls again. For $\eta t_p \ll L^2$ its largest value has a simple formula:

Largest deviation of the build-up from the line, and the maximum distance at which a fault can be detected

\delta p_{\max} = \frac{qB\,t_p}{4\pi\e\,\phi c_th\,L^2}\quad\Rightarrow\quad L_{\max} = \sqrt{\frac{qB\,t_p}{4\pi\e\,\phi c_th\,\delta p}}

where:

$\delta p_{\max}$ largest deviation of the build-up pressure from the Horner line of an infinite reservoir (one with no boundaries), Pa
$\delta p$ smallest deviation that we can tell apart reliably (it depends on the resolution and the noise of the gauge), Pa
$q,\ B,\ t_p$ rate before shut-in, formation volume factor (reservoir volume per unit of surface volume), producing time, m³/s, m³/m³, s
$\phi,\ c_t,\ h$ porosity, total compressibility, reservoir thickness, —, 1/Pa, m
$\e$ base of natural logarithms, 2.718, —

Units: SI

Holds when

The fault was not felt during production: $\eta t_p \ll L^2$ . In practice the formula already works for $L^2 > 3\eta t_p$ .
The record lasts at least $L^2/\eta$ after shut-in.
The reservoir is homogeneous. There are no other disturbances, such as neighbouring wells in operation or gauge drift (a slow creep of the gauge reading).

Neither permeability nor viscosity appears in the formula. The “signal” of the fault depends only on the volume produced $qBt_p$ , the storativity of the reservoir $\phi c_th$ (how much fluid the reservoir gives up per unit of pressure drop) and the distance. Permeability only decides when to wait for the maximum. So the radius at which a build-up can still notice a boundary depends less on the properties of the reservoir than on the volume produced before shut-in and the resolution of the gauge.

In other unit systems (2)

Metric oilfield units

\delta p_{\max} = \frac{qB\,t_p}{820\,\phi c_th\,L^2},\qquad L_{\max} = 0.0349\sqrt{\frac{qB\,t_p}{\phi c_th\,\delta p}}

$q$ in m³/d, $t_p$ in h, $c_t$ in 1/MPa, $h$ and $L$ in m; pressures in MPa.

$820 = 24\cdot4\pi\e$ (24 is the number of hours in a day); $0.0349 = 1/\sqrt{820}$ .

Oilfield units

L_{\max} = 0.0827\sqrt{\frac{qB\,t_p}{\phi c_th\,\delta p}}

$q$ in STB/d, $t_p$ in h, $c_t$ in 1/psi, $h$ and $L$ in ft, $\delta p$ in psi.

$0.0827 = \sqrt{5.615/820}$ , where 5.615 is the number of cubic feet in a barrel.

Worked exampleWill a short test see the fault?

Given (assumed data)

Test: 6 h of flow at a rate of 110 m³/d, then a build-up; the same reservoir: $B$ = 1.25; $\phi$ = 0.19; $c_t = 1.4\cdot10^{-3}$ MPa⁻¹; $h$ = 14 m; $\eta$ ≈ 2447 m²/h
Gauge: reliably detects deviations of 0.5 kPa (0.0005 MPa) or more
Fault: 230 m from the well

Solution

$\eta t_p = 14{,}682$ m², while $L^2 = 52{,}900$ m². So the fault was not felt during the flow period.
Largest deviation: $qBt_p = 110\cdot1.25\cdot6 = 825$ and $\phi c_th = 0.19\cdot1.4\cdot10^{-3}\cdot14 = 3.724\cdot10^{-3}$ m/MPa. Then $\delta p_{\max} = \dfrac{825}{820\cdot3.724\cdot10^{-3}\cdot52{,}900} = 0.0051$ MPa, that is, 5.1 kPa. It occurs about $L^2/\eta = 22$ h after shut-in.
The full build-up formula for a well near a fault gives 5.1 kPa at 19 h. So the simple formula is right.
Maximum distance at a threshold of 0.5 kPa: $L_{\max} = 0.0349\sqrt{\dfrac{825}{3.724\cdot10^{-3}\cdot0.0005}} = 735$ m. To see a fault at this distance, the build-up must last about $L_{\max}^2/\eta = 221$ h.

Answer: A fault at 230 m gives a deviation of about 5.1 kPa almost a day after shut-in. We can notice it if we extend the build-up. This test will not show a fault farther than 735 m, however long the record lasts.

What it means. A deviation of a few kilopascals is a fraction of a per cent of the pressure rise in the build-up. On the Horner plot we see it only with a good gauge and a quiet record. To “reach” farther, we must produce more. The maximum distance grows as the square root of the volume produced, so four times the volume doubles the radius. The data in this example are assumed for illustration.

We plot the derivative of a build-up near a fault in the same way as the derivative of a drawdown test, but against the equivalent time or the superposition function. Otherwise, after a short producing period, the derivative falls at the end of the record by itself. This fall can hide the rise caused by the fault (Effect of producing time on build-up analysis).

Limits of the method

Both lines are needed. For the intersection method the record must reach $\eta t/L^2 \approx 10$ –25. That is tens of times longer than the time at which the fault began to have an effect. Otherwise the second line is flatter than the true one, and the distance comes out too small (see the table above). The derivative step gives the distance sooner, but it needs confidence in the model.
Wellbore storage can hide a close fault. If the first plateau has no time to appear, we cannot tell a well near a fault from a well in a reservoir with half the flow capacity. Only independent information about the fault, or reducing wellbore storage, can help.
A distant fault is not seen. In a drawdown test, we do not see it until the radius of investigation approaches the fault. In a build-up after a short producing period, we do not see it beyond the maximum detection distance, however long we wait.
Doubling is not proof. Other models also give a doubled step. One example is a reservoir whose flow capacity is half as large beyond a certain radius (the transition is similar but not identical). Another is false “straight lines” in the wellbore-storage period. To check, use geology, seismic data, neighbouring wells, and the agreement of $kh$ and skin from both lines.
A fault may let fluid through. Then the derivative first rises and then returns to the first plateau, because flow across the fault begins (Partially communicating faults).
There may be more than one boundary. If the derivative rises above the doubled level, or keeps rising with a slope of ½ or 1, other boundaries have joined the first fault (Several boundaries: parallel faults and wedge-shaped blocks).
The reservoir beyond the fault and along it. The method assumes a homogeneous reservoir and a straight boundary through the whole thickness. Several things change both the time and the shape of the transition: a change of thickness or permeability along the boundary, a tilt of the fault plane, or different permeability in different directions. So the distance from the simple model is an estimate, not a measurement.

Terms and symbols

Sealing fault: A fault across which no fluid flows. For flow to a well, it acts as a straight no-flow boundary.
Image well: An imaginary well with the same rate history. We place it as a mirror image beyond the fault, at a distance $2L$ from the real well.
Slope doubling: The change from the first semilog straight line, with slope $m$ , to the second, with slope $2m$ . On the derivative, it is the change from the plateau $m_e$ to the plateau $2m_e$ .
Intersection time $t_x$: The moment at which the extensions of the two straight lines meet. The distance to the fault follows from it: $L = \sqrt{kt_x/(\gamma\phi\mu c_t)}$ .
Derivative step: The rise of the derivative from $m_e$ to $2m_e$ according to $1 + \e^{-L^2/(\eta t)}$ . It takes about 2.7 logarithmic cycles (2.7 tenfold changes of time).
Hemiradial flow: Half-circular flow to a well right next to a sealing boundary. The slope of the line is doubled from the start.
Maximum detection distance: The largest distance to a boundary at which its effect on a build-up still exceeds the resolution of the gauge. It grows as the square root of the volume produced.

Common mistakes and misconceptions

Wrong

“A fault shows up as soon as the radius of investigation reaches it.”

In fact

On the derivative, yes: when the radius of investigation equals $L$ , the derivative is already 2% above the plateau. In the pressure itself, the effect of the image becomes noticeable later, when the radius of investigation approaches $2L$ . The doubling of the slope is established only at $\eta t/L^2 \approx 10$ , which is 40 times later.

Wrong

“The second line is simply the last points of the record.”

In fact

Only if the slope has really doubled by the end of the record. In the example of this lesson, the chord (the straight line through the last two readings) is only 1.91 times steeper than the first line. The distance from it comes out 11% too small.

Wrong

“A negative skin from a test means the well has been cleaned up thoroughly.”

In fact

If we take a well next to a fault for an ordinary well, the apparent skin is $\tfrac12[S - \ln(L/r_w)]$ , and the permeability comes out at half its true value. A negative skin in a well that was not stimulated is a reason to check the geology.

Wrong

“The longer the build-up, the farther it ‘sees’.”

In fact

Only up to a limit. The largest deviation from the Horner line depends on the volume produced before shut-in and on the distance. If it is smaller than the resolution of the gauge, a build-up of any length will not show the fault.

Wrong

“ $p^*$ from the Horner line in a new accumulation (one not yet produced) equals the initial pressure.”

In fact

If the fault was felt during production, $p^*$ of the first line is below the initial pressure by $m_eE_1\big(L^2/(\eta t_p)\big)$ . In the example of this lesson, that is 0.12 MPa.

Connections to other topics

Superposition principle — the method of images and the adding of solutions in time and in space, which this lesson is built on.
Diagnosis with the pressure derivative — the diagnostic plot, the plateau and the step of the derivative, and how to calculate the derivative from readings.
Pressure drawdown analysis — the first straight line: flow capacity and skin from a drawdown test.
Pressure build-up analysis — the Horner plot and $p^*$ , which behave differently near a fault.
Several boundaries: parallel faults and wedge-shaped blocks — several faults: channels, corners and closed blocks.
Partially communicating faults — faults that let some fluid through, and their signs on the derivative.
Faults — the geology of faults: types, displacements, sealing.
Wellbore storage — why a close fault may be hidden under the early part of the record and how to reduce wellbore storage.
Effect of producing time on build-up analysis — equivalent time for the derivative of a build-up.
Well testing: purposes and types of test — types of tests and how to plan their duration.

Check yourself

1
How do the slope of the semilog straight line and the level of the logarithmic derivative change when the pressure disturbance has completely “felt” a single straight sealing fault?
1. A.The slope increases fourfold
2. B.Both double
3. C.The slope doubles, the derivative does not change
4. D.Both fall to zero
Show the answer
Correct answer: B. Both double
At $\eta t \gg L^2$ the image's own contribution grows as $m_e\ln t$ . That is, it adds another $m_e$ to the slope per unit of natural logarithm. The derivative is the same slope, so it too goes to $2m_e$ . A fall to zero is a sign of a constant-pressure boundary.
2
The first straight line gives $k$ = 60 mD. Also $\phi$ = 0.21, $\mu$ = 1.1 mPa·s and $c_t = 1.6\cdot10^{-3}$ MPa⁻¹. The extensions of the first and second lines meet 100 h after start-up. Find the distance to the fault.
Show the answer
$\phi\mu c_t = 0.21\cdot1.1\cdot1.6\cdot10^{-3} = 3.70\cdot10^{-4}$ . $L = 0.0447\sqrt{60\cdot100/(3.70\cdot10^{-4})} = 0.0447\sqrt{1.62\cdot10^7} \approx 0.0447\cdot4030 \approx 180$ m. If we drew the second line through a slope that had not yet doubled, the true distance is larger.
3
The first plateau of the derivative is at 0.040 MPa. At 30 h after start-up the derivative is 0.062 MPa, and $\eta$ = 1500 m²/h. Find the distance to the fault. Then estimate when the derivative will reach 0.076 MPa (90% of the step).
Show the answer
$\Delta p'/m_e = 1.55$ , and $\ln[1/(1.55 - 1)] = \ln 1.818 = 0.598$ . So $L = \sqrt{1500\cdot30\cdot0.598} \approx \sqrt{26{,}900} \approx 164$ m. A ratio of 1.9 occurs at $\eta t/L^2 = 1/\ln(1/0.9) = 9.5$ , that is, after $9.5\cdot26{,}900/1500 \approx 170$ h.
4
A well of radius 0.1 m is 8 m from a fault, and wellbore storage lasts about 2 h. We processed a drawdown test as ordinary radial flow and got $k$ = 40 mD and a skin of −1.8. What is most likely the real situation?
1. A.The record cannot be processed
2. B.Permeability is about 80 mD, skin about +0.8
3. C.Permeability is about 20 mD, skin about −3.6
4. D.The well really is stimulated, $k$ = 40 mD
Show the answer
Correct answer: B. Permeability is about 80 mD, skin about +0.8
Wellbore storage hides the first plateau, and we see only the doubled one. We took the slope $2m$ for $m$ , so the true permeability is twice as large. The apparent skin is $S' = \tfrac12[S + \ln(r_w/L)]$ . From this, $S = 2S' - \ln(r_w/L) = -3.6 + \ln 80 \approx -3.6 + 4.38 \approx 0.8$ .
5
By what factor must we increase the volume produced before a build-up to double the maximum detection distance of a fault? Does the time at which we expect the largest deviation from the Horner line change for a fault at this new distance?
Show the answer
$L_{\max}$ is proportional to the square root of $qBt_p$ , so we need four times the volume. The maximum of the deviation occurs about $L^2/\eta$ after shut-in. For a fault twice as far away, that is four times later. For the same fault, the time of the maximum does not depend on the volume produced. Only the size of the deviation grows.
6
The derivative rose above the first plateau, but after one and a half cycles it returned to its former level. What is this most likely to be?
1. A.A constant-pressure boundary
2. B.A closed area
3. C.A partially communicating fault
4. D.A sealing fault
Show the answer
Correct answer: C. A partially communicating fault
A sealing fault raises the derivative to $2m_e$ and leaves it there. A closed area keeps raising it with a slope of 1. A constant-pressure boundary lowers it below the plateau. A return to the first plateau means that, with time, flow across the boundary has begun. This is how a partially communicating fault behaves.

A sealing fault near the well: the method of images

Quick refresher

After this lesson you can

A fault as a reservoir boundary

The image well

Two straight lines on the semilog plot

The derivative step

A fault very close: hemiradial flow

Pressure build-up near a fault

Limits of the method

Terms and symbols

Common mistakes and misconceptions

In depth: advanced detail

Connections to other topics

Check yourself