Several boundaries: parallel faults and wedge-shaped blocks

Quick refresher

The images are reflected in each other until the picture is symmetric about all the boundaries. A channel gives an endless row with a step of $2W$ . A wedge with an angle $180^\circ/n$ gives $2n$ wells on a circle round the apex of the angle (the point where the faults cross), the real well included.
When wellbore storage (the well acting as a small tank) is over, the derivative equals $m_e\big[1 + \sum_j \e^{-d_j^2/(4\eta t)}\big]$ . Each image that the disturbance has “reached” adds one more $m_e$ to the level.
In a channel the number of images “felt” grows as $\sqrt t$ . So $\Delta p$ grows as $\sqrt t$ , and the derivative runs with a slope of ½. The slope of the line $\Delta p$ against $\sqrt t$ gives the width of the channel, if we know the permeability from radial flow.
The intercept of the same line contains the skin of the well (a number that shows how much the zone right around the well helps or hinders flow) and the geometric skin of the channel (an extra pressure drop that comes only from the shape of the flow), $\ln\tfrac{W}{2\pi r_w} - \ln\sin\tfrac{\pi L_1}{W}$ . We use it to check the position of the well in the channel.
In a wedge with an angle $\theta$ (any angle, not only $180^\circ/n$ ), the derivative reaches the level $(360^\circ/\theta)\,m_e$ . The approach to it is slow: the shortfall is about $d^2/(2\eta t)$ , where $d$ is the distance from the well to the apex of the angle. 95% of the level comes at about $\eta t \approx 10\,d^2$ .
The steps from different boundaries merge if the distances to them differ by less than a factor of 6–7. The test gives distances, but not directions. Where the faults run is shown by seismic surveys and geology.

Key relationships

\Delta p' = m_e\Big[1 + \sum_j \exp\Big(-\dfrac{d_j^2}{4\eta t}\Big)\Big]

the derivative for any system of images; $d_j$ are the distances to the images, $\eta = k/(\phi\mu c_t)$

\Delta p = m_L\sqrt t + 1.866\,\dfrac{qB\mu}{kh}\,(S + S_{ch})

linear flow in a channel; $\Delta p' = \tfrac12 m_L\sqrt t$

W = 0.394\,\dfrac{qB}{h\,m_L}\sqrt{\dfrac{\mu}{k\phi c_t}}

width of the channel: m³/d, m, MPa/√h, mPa·s, mD, 1/MPa (oilfield: 8.13; ft, psi/√h, cP, 1/psi)

S_{ch} = \ln\dfrac{W}{2\pi r_w} - \ln\sin\dfrac{\pi L_1}{W}

geometric skin of the channel

\theta = 360^\circ\,\dfrac{m_e}{\Delta p'_{\infty}}

the angle between the faults from the late level of the derivative $\Delta p'_\infty$

Common pitfalls

Taking the late slope ½ for a hydraulic fracture, or, the other way round, calculating the channel width from an early slope ½ that a fracture gave.
Calculating the angle between the faults from the last point of a derivative that is still rising: the angle comes out too large.
Calculating the channel width when radial flow is not visible in the record: the $\sqrt t$ slope gives only the product $W\sqrt k$ .
Looking for an intermediate plateau $2m_e$ where the distances to two boundaries differ by only a factor of 2–3: the steps merge.
Orienting the boundaries from a test of a single well: it gives distances, not directions.

After this lesson you can

build the system of images for a channel and for a wedge with an angle $180^\circ/n$ , and explain why the late pressure derivative “counts” the images
recognise a channel by the slope ½ of the derivative; find the width of the channel from the slope of the line $\Delta p$ against $\sqrt t$ , the distance to the nearer boundary from the first step of the derivative, and the distance to the farther one from the difference; and use the intercept of the line to check the position of the well through the geometric skin
find the angle between crossing faults from the late level of the derivative, and estimate it from a rise that has not yet finished
estimate when linear flow and the late level of a wedge appear in a record, and calculate the duration of the test that we need
list what a test of a single well cannot tell apart: the directions of the faults, the side of the nearer boundary, the width of a channel without the permeability, and a wedge versus a reservoir whose flow capacity (permeability times thickness) gets worse farther away

Images of images

One straight sealing boundary is replaced by one image well (A sealing fault near the well: the method of images): two identical wells that are symmetric about the line of the fault, so no fluid crosses it. If there are two boundaries, one image is not enough. The image behind fault A breaks the condition at fault B: relative to B, nothing balances it. So we reflect it in B too. The new image we reflect in A again, and so on, until the whole picture is symmetric about both lines (the method of images).

For two cases, this chain is simple. Parallel faults at a distance $W$ from each other give an endless row of images along the perpendicular to the faults. The picture repeats with a step of $2W$ , as between two parallel mirrors. Crossing faults with an angle $\theta = 180^\circ/n$ ( $n$ is a whole number) give a finite number of images: $2n - 1$ , all on a circle round the apex of the angle. The radius of the circle is the distance $d$ from the well to the apex. For other angles the images fall inside the reservoir itself, and we find the solution in another way, as a sum of terms called the eigenfunctions of the wedge (the basic solutions that fit the shape of the wedge). The late level of the derivative is the same (section below).

SchematicSystems of images (a sketch, not to scale). Left: a channel between two parallel faults. The real channel is highlighted, the dashed lines are its mirror copies, and each copy has an image of the well. The row goes on in both directions. Right: a wedge with an angle of 60° between faults A and B. The well and its five images lie on a circle round the apex of the angle.

Each image is one more well with the same rate. Until its disturbance has reached the real well, it has no effect. After that its contribution to the pressure grows like one more line-source solution (the solution for a well treated as an infinitely thin line). The derivative of the contribution of one image well at a distance $d$ , with respect to the logarithm of time, is $m_e\e^{-d^2/(4\eta t)}$ . This is the same step of the derivative as at a single fault. So the general formula is:

The pressure derivative of a well with several sealing boundaries

\Delta p' = m_e\left[1 + \sum_{j} \exp\left(-\frac{d_j^2}{4\eta t}\right)\right],\qquad m_e = \frac{qB\mu}{4\pi kh},\quad \eta = \frac{k}{\phi\mu c_t}

where:

$\Delta p'$ logarithmic derivative of the pressure drop, $\dd\Delta p/\dd\ln t$ , Pa
$m_e$ level of the derivative plateau in radial flow, Pa
$d_j$ distance from the well to the $j$ -th image, m
$\eta$ hydraulic diffusivity (how fast a pressure change spreads through the rock), m²/s
$t$ time since start-up, s

Units: SI (any consistent system)

Holds when

The boundaries are straight sealing planes through the whole thickness of a homogeneous reservoir. Images describe exactly a channel and a wedge with an angle $180^\circ/n$ .
Wellbore storage is over. The images are line sources ( $d_j$ is much larger than the wellbore radius).
The rate is constant. For a build-up we add up the same terms by superposition, and we allow for the producing time (Pressure build-up analysis).

The derivative “counts” the images. Each one adds one $m_e$ to the level when $\eta t$ becomes comparable with $d_j^2$ (the middle of its step is at $\eta t = d_j^2/(4\ln 2)$ ). A finite number of images gives a new plateau, and a growing number gives a rise: with a slope of ½ in a channel and with a slope of 1 in a closed area.

In other unit systems (1)

Metric oilfield units

\Delta p' = m_e\left[1 + \sum_{j} \exp\left(-\frac{\phi\mu c_t\,d_j^2}{0.01421\,k\,t}\right)\right],\qquad m_e = 0.933\,\frac{qB\mu}{kh}

$d_j$ in m, $k$ in mD, $t$ in h, $\mu$ in mPa·s, $c_t$ in 1/MPa, $q$ in m³/d, $h$ in m; $\Delta p'$ and $m_e$ in MPa.

$0.01421 = 4\cdot3.553\cdot10^{-3}$ ; $0.933 = 1/(4\pi\cdot0.0853)$ (Diagnosis with the pressure derivative).

Systems of boundaries and the late derivative

Boundaries: A single fault
Images: one, at $2L$
Late derivative: $2m_e$
What we find: the distance $L$

Boundaries: Two parallel faults (a channel)
Images: an endless row, step $2W$
Late derivative: slope ½
What we find: $W$ , $L_1$ , $L_2$

Boundaries: Two at an angle $\theta = 180^\circ/n$
Images: $2n - 1$ on a circle
Late derivative: $(360^\circ/\theta)\,m_e$
What we find: the angle $\theta$

Boundaries: A channel closed at one end
Images: two rows
Late derivative: slope ½, twice as high as in an open channel
What we find: Semi-infinite systems: theory and test analysis

Boundaries: A closed rectangle
Images: a two-dimensional lattice
Late derivative: slope 1
What we find: pore volume (Reservoir limit test)

Boundaries	Images	Late derivative	What we find
A single fault	one, at $2L$	$2m_e$	the distance $L$
Two parallel faults (a channel)	an endless row, step $2W$	slope ½	$W$ , $L_1$ , $L_2$
Two at an angle $\theta = 180^\circ/n$	$2n - 1$ on a circle	$(360^\circ/\theta)\,m_e$	the angle $\theta$
A channel closed at one end	two rows	slope ½, twice as high as in an open channel	Semi-infinite systems: theory and test analysis
A closed rectangle	a two-dimensional lattice	slope 1	pore volume (Reservoir limit test)

$L$ is the distance to the fault, $W$ is the width of the channel, and $L_1$ and $L_2$ are the distances to its walls. For a wedge with an angle that is not $180^\circ/n$ , the late level is the same, $(360^\circ/\theta)\,m_e$ , but images do not give the exact solution.

A channel: linear flow

A channel can be a graben (a block dropped between two parallel faults), a narrow block between parallel sealing faults, or a river-channel sand body surrounded by the clays of floodplain deposits (mud laid down beside the river) (Continental environments: aeolian and fluvial). The width of such bodies is from tens of metres to a few kilometres, and their length is many times larger.

Until the pressure disturbance has reached the walls, the well works as in an infinite reservoir, and the derivative has the plateau $m_e$ . If the well is closer to one wall, that wall acts first: the derivative starts to double, as at a single fault. When the disturbance reaches both walls, the streamlines straighten out along the channel. The fluid comes to the well from both sides along two “pipes” of cross-section $Wh$ . The disturbed length of each pipe grows as $\sqrt{\eta t}$ . To push a constant rate through an ever longer pipe, we need a proportionally larger pressure difference. So in linear flow (linear flow) the pressure drop grows as the square root of time:

A well in a channel: linear flow

p_D = 2\sqrt{\pi t_{DW}} + S + S_{ch},\qquad t_{DW} = \frac{kt}{\phi\mu c_tW^2}\quad\Leftrightarrow\quad \Delta p = m_L\sqrt t + \frac{qB\mu}{2\pi kh}\,(S + S_{ch})

where:

$p_D$ dimensionless pressure drop at the bottom, $2\pi kh\,\Delta p/(qB\mu)$ , —
$t_{DW}$ dimensionless time based on the square of the channel width, —
$W$ channel width: the distance between the walls, $W = L_1 + L_2$ , m
$m_L$ slope of the line $\Delta p$ against $\sqrt t$ , $\dfrac{qB}{hW}\sqrt{\dfrac{\mu}{\pi k\phi c_t}}$ , Pa/√s
$S,\ S_{ch}$ skin of the well and geometric skin of the channel (formula below), —
$q,\ B,\ \mu,\ k,\ h,\ \phi,\ c_t$ rate, formation volume factor (reservoir volume per unit of surface volume), viscosity, permeability, thickness, porosity, total compressibility, m³/s, m³/m³, Pa·s, m², m, —, 1/Pa

Units: SI

Holds when

The walls are parallel sealing boundaries through the whole thickness of a homogeneous reservoir; along the channel the reservoir is unbounded.
The disturbance has reached both walls: the derivative lies on a line with slope ½ (by the table below, at $\eta t/W^2$ from 0.1 to 0.4, depending on the position of the well).
A constant rate; wellbore storage is over.

The slope $m_L$ depends on the width of the channel, but not on the position of the well in it. The position and the skin enter only the constant term. The logarithmic derivative $\Delta p' = \tfrac12 m_L\sqrt t$ is a line with slope ½ on a log–log plot (power-law regimes). The pressure drop $\Delta p$ itself is curved on this plot, because it contains the constant term.

In other unit systems (2)

Metric oilfield units

\Delta p = 0.394\,\frac{qB}{hW}\sqrt{\frac{\mu\,t}{k\phi c_t}} + 1.866\,\frac{qB\mu}{kh}\,(S + S_{ch})

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

$0.394 = 4\cdot0.933\cdot\sqrt{\pi\cdot3.553\cdot10^{-3}}$ ; $1.866 = 2\cdot0.933$ .

Oilfield units

\Delta p = 8.13\,\frac{qB}{hW}\sqrt{\frac{\mu\,t}{k\phi c_t}} + 141.2\,\frac{qB\mu}{kh}\,(S + S_{ch})

$q$ in STB/d, $h$ and $W$ in ft, $\mu$ in cP, $k$ in mD, $c_t$ in 1/psi, $t$ in h; $\Delta p$ in psi.

$8.13 = 4\cdot70.6\cdot\sqrt{\pi\cdot2.637\cdot10^{-4}}$ ; $141.2 = 2\cdot70.6$ .

Geometric skin of a channel

S_{ch} = \ln\frac{W}{2\pi r_w} - \ln\sin\frac{\pi L_1}{W}

where:

$S_{ch}$ the extra pressure drop (in units of $qB\mu/(2\pi kh)$ ) for the change from linear flow along the channel to radial flow at the well, —
$r_w$ wellbore radius, m
$L_1$ distance from the well to one of the walls (the formula is symmetric: $L_1$ and $W - L_1$ give the same), m

Units: any equal units of length

Holds when

Steady linear flow; the well is a line source in a homogeneous channel (a pseudo-skin of the geometry, and not damage of the reservoir).

The first term is large: for a channel 400 m wide and a well of radius 0.1 m it is about 6.5. The flow must “gather” from the width of the channel to the wellbore. The second term is small until the well is right at a wall. In the middle of the channel $\sin = 1$ , at a quarter of the width the addition is 0.35, and at a twentieth it is 1.86.

CalculatedCalculated derivatives of a well without wellbore storage in a channel of width

W

: in the middle of the channel and at

0.1W

from one wall. For comparison, the same well at the centre of a closed square with side

W

. The vertical axis is the ratio of the derivative to the plateau level

m_e

, and both scales are logarithmic.

When linear flow begins in a channel

$L_1/W$: 0.50
Derivative within 5% of the line with slope ½, $\eta t/W^2$: 0.09
Within 1%: 0.13
$-\ln\sin(\pi L_1/W)$: 0.00

$L_1/W$: 0.25
Derivative within 5% of the line with slope ½, $\eta t/W^2$: 0.30
Within 1%: 0.47
$-\ln\sin(\pi L_1/W)$: 0.35

$L_1/W$: 0.10
Derivative within 5% of the line with slope ½, $\eta t/W^2$: 0.36
Within 1%: 0.53
$-\ln\sin(\pi L_1/W)$: 1.17

$L_1/W$: 0.05
Derivative within 5% of the line with slope ½, $\eta t/W^2$: 0.37
Within 1%: 0.53
$-\ln\sin(\pi L_1/W)$: 1.86

$L_1/W$	Derivative within 5% of the line with slope ½, $\eta t/W^2$	Within 1%	$-\ln\sin(\pi L_1/W)$
0.50	0.09	0.13	0.00
0.25	0.30	0.47	0.35
0.10	0.36	0.53	1.17
0.05	0.37	0.53	1.86

Calculated from the derivative formula with images (a well without wellbore storage). After this time the difference only decreases. A well in the middle of the channel reaches linear flow earliest of all: its images stand symmetrically, and the slowest part of the transition drops out.

The channel width from a record

Here is an assumed record. A new well has penetrated a sandstone in a narrow block. By seismic data, the block is bounded by two parallel faults, but the distances to them are known only roughly. We started the well by opening a downhole valve at a constant rate of 60 m³/d, and recorded the pressure for about 500 h.

Pressure drop in a well in a channel (assumed data, a selection)

$t$: 36 s
$\sqrt t$ , √h: 0.10
$\Delta p$ , MPa: 0.9480
$\Delta p'$ , MPa: 0.2697

$t$: 6 min
$\sqrt t$ , √h: 0.32
$\Delta p$ , MPa: 1.3101
$\Delta p'$ , MPa: 0.1126

$t$: 30 min
$\sqrt t$ , √h: 0.71
$\Delta p$ , MPa: 1.4834
$\Delta p'$ , MPa: 0.1049

$t$: 1.00 h
$\sqrt t$ , √h: 1.00
$\Delta p$ , MPa: 1.5554
$\Delta p'$ , MPa: 0.1038

$t$: 2.00 h
$\sqrt t$ , √h: 1.41
$\Delta p$ , MPa: 1.6269
$\Delta p'$ , MPa: 0.1038

$t$: 5.01 h
$\sqrt t$ , √h: 2.24
$\Delta p$ , MPa: 1.7249
$\Delta p'$ , MPa: 0.1136

$t$: 10.0 h
$\sqrt t$ , √h: 3.16
$\Delta p$ , MPa: 1.8094
$\Delta p'$ , MPa: 0.1343

$t$: 20.0 h
$\sqrt t$ , √h: 4.47
$\Delta p$ , MPa: 1.9109
$\Delta p'$ , MPa: 0.1598

$t$: 50.1 h
$\sqrt t$ , √h: 7.08
$\Delta p$ , MPa: 2.0750
$\Delta p'$ , MPa: 0.2031

$t$: 100 h
$\sqrt t$ , √h: 10.00
$\Delta p$ , MPa: 2.2334
$\Delta p'$ , MPa: 0.2683

$t$: 200 h
$\sqrt t$ , √h: 14.13
$\Delta p$ , MPa: 2.4510
$\Delta p'$ , MPa: 0.3754

$t$: 316 h
$\sqrt t$ , √h: 17.78
$\Delta p$ , MPa: 2.6437
$\Delta p'$ , MPa: 0.4727

$t$: 501 h
$\sqrt t$ , √h: 22.39
$\Delta p$ , MPa: 2.8864
$\Delta p'$ , MPa: —

$t$	$\sqrt t$ , √h	$\Delta p$ , MPa	$\Delta p'$ , MPa
36 s	0.10	0.9480	0.2697
6 min	0.32	1.3101	0.1126
30 min	0.71	1.4834	0.1049
1.00 h	1.00	1.5554	0.1038
2.00 h	1.41	1.6269	0.1038
5.01 h	2.24	1.7249	0.1136
10.0 h	3.16	1.8094	0.1343
20.0 h	4.47	1.9109	0.1598
50.1 h	7.08	2.0750	0.2031
100 h	10.00	2.2334	0.2683
200 h	14.13	2.4510	0.3754
316 h	17.78	2.6437	0.4727
501 h	22.39	2.8864	—

$\Delta p = p_i - p_{wf}$ , $p_i$ = 26 MPa. The data were calculated from a model of a homogeneous reservoir with wellbore storage, skin and two parallel sealing boundaries, and rounded to 0.0001 MPa. The parameters of the model are given in the example. The derivative comes from all 58 readings of the record (10 per cycle) with an interval of 0.2 cycle. It is missing for the last two readings.

CalculatedDiagnostic plot of the record: the pressure drop (points) and the derivative from the measurements (line). The dashed lines are the plateau

m_e

from the early section, and the line with slope ½ drawn through the late derivative.

The channel width from the slope of the line Δp against √t

W = \frac{qB}{h\,m_L}\sqrt{\frac{\mu}{\pi k\phi c_t}}

where:

$m_L$ slope of the line of the pressure drop in the coordinates $\Delta p$ against $\sqrt t$ in the section of linear flow, Pa/√s
$k$ permeability from the derivative plateau (radial flow), m²
$q,\ B,\ h,\ \mu,\ \phi,\ c_t$ as in the formula of linear flow, m³/s, m³/m³, m, Pa·s, —, 1/Pa

Units: SI

Holds when

The line is drawn through the points where the derivative already has a slope of ½.
The permeability was found independently: from the radial flow of the same record or from other data.
$\phi$ , $c_t$ and $h$ are known: their error enters $W$ as $1/\sqrt{\phi c_t}$ and $1/h$ .

We find the width from the slope, and we do not wait for the points to “straighten out” in the pressure itself: the section with slope ½ on the derivative shows which points to draw the line through. The distance to the far wall is $L_2 = W - L_1$ , where $L_1$ is the distance to the nearer wall from the first step of the derivative.

In other unit systems (2)

Metric oilfield units

W = 0.394\,\frac{qB}{h\,m_L}\sqrt{\frac{\mu}{k\phi c_t}}

$W$ and $h$ in m, $q$ in m³/d, $m_L$ in MPa/√h, $\mu$ in mPa·s, $k$ in mD, $c_t$ in 1/MPa.

The same 0.394 as in the formula of linear flow.

Oilfield units

W = 8.13\,\frac{qB}{h\,m_L}\sqrt{\frac{\mu}{k\phi c_t}}

$W$ and $h$ in ft, $q$ in STB/d, $m_L$ in psi/√h, $\mu$ in cP, $k$ in mD, $c_t$ in 1/psi.

CalculatedThe same record in the coordinates

\Delta p

against

\sqrt t

(points). The line is drawn through the readings at 199.5 and 501.2 h and extended to

\sqrt t = 0

, where we read the intercept

b

. The early points lie below the line: there is no linear flow there yet.

Worked examplePermeability, skin, channel width and the position of the well

Given (assumed data)

Well and reservoir: $q$ = 60 m³/d; $B$ = 1.18; $\mu$ = 1.4 mPa·s; $h$ = 10 m; $\phi$ = 0.21; $c_t = 1.3\cdot10^{-3}$ MPa⁻¹; $r_w$ = 0.1 m
Derivative plateau: the mean from 0.5 to 2 h: $m_e$ = 0.10403 MPa; the reading at 1 h: $\Delta p$ = 1.5554 MPa
First step: at 10 h the derivative is 0.13431 MPa
Late readings: 199.5 h — 2.4510 MPa; 501.2 h — 2.8864 MPa

Solution

Flow capacity: $qB\mu = 60\cdot1.18\cdot1.4 = 99.12$ ; $kh = 0.933\cdot99.12/0.10403 = 889$ mD·m, $k = 88.9$ mD.
The skin from the ratio of the pressure to the plateau (Diagnosis with the pressure derivative): $\ln\dfrac{kt}{\phi\mu c_tr_w^2} = 16.962$ at $t$ = 1 h; $S = \tfrac12\,(1.5554/0.10403 - 16.962 + 4.831) = 1.41$ .
The nearer wall from the point of the first step (the derivative step): $\eta = 3.553\cdot10^{-3}\cdot88.9/(0.21\cdot1.4\cdot1.3\cdot10^{-3}) = 826$ m²/h; $\Delta p'/m_e = 1.291$ , $\ln[1/(1.291 - 1)] = 1.234$ ; $L_1 = \sqrt{826\cdot10\cdot1.234} = 101$ m. The far wall has no effect at this moment.
Linear flow: between 100 and 316 h (half a cycle) the derivative grew from 0.2683 to 0.4727 MPa. The slope on the log–log plot is $\lg(0.4727/0.2683)/0.5 = 0.49$ , that is, ½. So the late readings lie on the line $\Delta p$ against $\sqrt t$ .
The slope of the line: $m_L = (2.8864 - 2.4510)/(22.387 - 14.125) = 0.05270$ MPa/√h.
The width: $qB = 70.8$ ; $\sqrt{\mu/(k\phi c_t)} = \sqrt{1.4/(88.9\cdot0.21\cdot1.3\cdot10^{-3})} = 7.595$ ; $W = 0.394\cdot70.8/(10\cdot0.05270)\cdot7.595 = 402$ m. The far wall: $L_2 = 402 - 101 = 301$ m.
Check by the intercept: $b = 2.8864 - 0.05270\cdot22.387 = 1.7066$ MPa; $S + S_{ch} = b/(2m_e) = 8.20$ , hence $S_{ch} = 6.79$ . By the formula of the geometric skin for $W$ = 402 m and $L_1$ = 101 m: $6.461 + 0.343 = 6.80$ . The position of the well agrees.

Answer:

k \approx 88.9

mD, skin 1.41; a channel about 402 m wide, with the well 101 m from one wall and 301 m from the other.

What it means. We calculated the record from a model with $k$ = 90 mD, $S$ = 1.5, and walls at 100 and 300 m (a width of 400 m). The permeability is underestimated by about 1%: the plateau is still raised a little by the tail of wellbore storage. The width was found to about 0.5%, because the late readings lie deep in linear flow ( $\eta t/W^2$ of about 1 and 2.6). The intercept of the line is a good check, but a poor way to look for $L_1$ : the sine near the middle of the channel is flat, and an error of 0.1 in the skin shifts $L_1$ by 10–15 m. A record cannot tell which side the nearer wall is on, and in which direction the channel extends. Seismic surveys, the geological model and the neighbouring wells give that. The data in this example are assumed for illustration.

A channel is visible on the pressure build-up curve too. Suppose the flow in the channel is already linear both at the end of production and at the time $\Delta t$ after shut-in. Then the superposition of the two solutions gives a simple relation: the constant terms with the skin cancel.

A build-up in a channel in the section of linear flow

p_i - p_{ws} = m_L\left(\sqrt{t_p + \Delta t} - \sqrt{\Delta t}\right)

where:

$p_{ws}$ bottomhole pressure at the time $\Delta t$ after shut-in, Pa
$t_p,\ \Delta t$ producing time before shut-in and time since shut-in, s
$m_L$ slope of linear flow, the same as while the well produces, Pa/√s

Units: SI (metric oilfield units: MPa, h, $m_L$ in MPa/√h)

Holds when

A constant rate during $t_p$ ; before start-up the pressure in the reservoir was $p_i$ .
Both $t_p$ and $\Delta t$ are large enough for both terms of the superposition to be in the linear-flow section ( $\eta\Delta t/W^2$ not less than 0.1–0.4 by the table above).

If we plot $p_{ws}$ against $\sqrt{t_p + \Delta t} - \sqrt{\Delta t}$ , the late points lie on a line with slope $-m_L$ . If we extend the line to the zero of this function (an infinitely long shut-in), it gives the initial pressure $p_i$ . Along the axis the channel is unbounded, and the pressure in it recovers in time to the initial one. We calculate the derivative of such a build-up so that it allows for the producing time: against the superposition time function (a time scale that adds up all the earlier rate changes) or against the equivalent time (Effect of producing time on build-up analysis).

Crossing faults: a wedge

A block bounded by two crossing faults is a common case in fault zones. Two faults of different directions come together, and the well is inside the angle between them. If the angle is $\theta = 180^\circ/n$ , the images close up: the well and its $2n - 1$ images stand on a circle of radius $d$ round the apex of the angle, where $d$ is the distance from the well to the apex. For example, a right angle gives three images, and an angle of 60° gives five. A single straight fault is also a “wedge”, with an angle of 180°: one image.

When the disturbance has covered the whole neighbourhood of the apex, all the images are “felt”, and the derivative reaches the level $2n\,m_e = (360^\circ/\theta)\,m_e$ . This result does not depend on whether the angle fits the series $180^\circ/n$ . Far from the apex the flow is radial, but it goes not through the whole circle, only through the share $\theta/360^\circ$ of it. The same rate passes through a cross-section that is that many times smaller, and the pressure drop for each logarithmic cycle is that many times larger (the derivation is in the advanced section).

The late level of the derivative in a wedge, and the angle from it

\Delta p'_{\infty} = \frac{360^\circ}{\theta}\,m_e\quad\Rightarrow\quad \theta = 360^\circ\,\frac{m_e}{\Delta p'_{\infty}}

where:

$\Delta p'_{\infty}$ the late level of the derivative, when the disturbance has covered the neighbourhood of the apex of the angle, Pa
$m_e$ plateau level of radial flow, $qB\mu/(4\pi kh)$ , Pa
$\theta$ angle between the faults on the side of the well, degrees

Units: any equal units of pressure

Holds when

Two straight sealing faults through the whole thickness of a homogeneous reservoir meet at one point (in plan view). The disturbance has reached no other boundaries.
The level has really been reached: on the way to it, the derivative is still lower, and the angle from it comes out too large.
The angle is less than 180°: the well is inside an “acute” (less than 90°) or “obtuse” (more than 90°) angle, not beyond a bend of a fault.

The ratio of the late level to the plateau is the “number of wells” including the images: 2 for a single fault, 4 for a right angle, 6 for an angle of 60°. For any angle it is simply $360^\circ/\theta$ : a level 2.5 times as high as the plateau means an angle of 144°.

Wedges with an angle

180^\circ/n

: the well on the bisector (the line that halves the angle), the distance to each fault is

L

Angle: $180^\circ$ (single fault)
Images: 1
Late level, $m_e$: 2
95% of the level at $\eta t/L^2$: 9.5
The same in units of $d^2$: —

Angle: $90^\circ$
Images: 3
Late level, $m_e$: 4
95% of the level at $\eta t/L^2$: 19
The same in units of $d^2$: 9.6

Angle: $60^\circ$
Images: 5
Late level, $m_e$: 6
95% of the level at $\eta t/L^2$: 38
The same in units of $d^2$: 9.6

Angle: $45^\circ$
Images: 7
Late level, $m_e$: 8
95% of the level at $\eta t/L^2$: 66
The same in units of $d^2$: 9.6

Angle: $30^\circ$
Images: 11
Late level, $m_e$: 12
95% of the level at $\eta t/L^2$: 144
The same in units of $d^2$: 9.6

Angle	Images	Late level, $m_e$	95% of the level at $\eta t/L^2$	The same in units of $d^2$
$180^\circ$ (single fault)	1	2	9.5	—
$90^\circ$	3	4	19	9.6
$60^\circ$	5	6	38	9.6
$45^\circ$	7	8	66	9.6
$30^\circ$	11	12	144	9.6

Calculated from the derivative formula with images. $d = L/\sin(\theta/2)$ is the distance from the well to the apex of the angle. The time to reach the level is set by the distance to the apex and not to the faults: in units of $d^2$ it is the same for all angles.

CalculatedCalculated derivatives of a well on the bisector of a wedge, the distance to each fault

L

, without wellbore storage: 180° (a single fault), 90°, 60° and 45°. The vertical axis is the ratio of the derivative to the plateau level. The dashed lines are the late levels 2, 4, 6 and 8.

Why does a narrow wedge reach its level so slowly? For late times in the derivative formula we can replace each term $\e^{-x}$ by $1 - x$ , and the sum of the squares of the distances from the well to all its images in the wedge is $4nd^2$ . We get a simple law of approach, which holds also for angles that are not $180^\circ/n$ :

The approach of the derivative to the late level of a wedge, and the angle from two late points

\frac{\Delta p'}{m_e} \approx R_\infty\left(1 - \frac{a}{t}\right),\quad a = \frac{d^2}{2\eta};\qquad a = \frac{r_1 - r_2}{r_1/t_2 - r_2/t_1},\quad R_\infty = \frac{r_2}{1 - a/t_2}

where:

$R_\infty$ the late level in units of $m_e$ , $360^\circ/\theta$ , —
$r_1,\ r_2$ the ratios $\Delta p'/m_e$ at the times $t_1 < t_2$ on the way to the level, —
$d$ distance from the well to the apex of the angle, m
$a$ approach constant: at the time $t$ the shortfall to the late level is the share $a/t$ , s
$\eta$ hydraulic diffusivity, m²/s

Units: SI; $t$ and $a$ in the same units of time

Holds when

The shortfall is small, no more than 15–20%. At a larger one, the next terms of the expansion matter, and the estimate of $R_\infty$ comes out too low and the angle too large.
The derivative from measurements has no noticeable noise: the difference $r_2 - r_1$ is small, and its error strongly affects the result.
The disturbance has reached no other boundaries than the two faults.

Two late points give two unknowns: the level and the approach constant. The level is the angle between the faults, and the constant is the distance to the apex, $d = \sqrt{2\eta a}$ (the estimate is usually too low by 5–15%: the next terms of the expansion reduce the apparent shortfall). For 95% of the level we need $\eta t \approx 10\,d^2$ .

The angle between faults from a record

A well in a high-permeability reservoir has penetrated a block near the crossing of two faults. By seismic data, the nearer fault A runs about a hundred metres away, and the second fault B several hundred metres away. The angle between them on the map is from 50 to 80°. We started the well by opening a downhole valve at a constant rate of 400 m³/d, and recorded the pressure for about 500 h.

An extended well test near the crossing of faults (assumed data, a selection)

$t$: 36 s
$\Delta p$ , MPa: 0.6265
$\Delta p'$ , MPa: 0.05754
$\Delta p'/m_e$: 1.025

$t$: 6 min
$\Delta p$ , MPa: 0.7569
$\Delta p'$ , MPa: 0.05613
$\Delta p'/m_e$: 1.000

$t$: 12 min
$\Delta p$ , MPa: 0.7956
$\Delta p'$ , MPa: 0.05613
$\Delta p'/m_e$: 1.000

$t$: 1.00 h
$\Delta p$ , MPa: 0.8896
$\Delta p'$ , MPa: 0.06569
$\Delta p'/m_e$: 1.170

$t$: 1.58 h
$\Delta p$ , MPa: 0.9215
$\Delta p'$ , MPa: 0.07405
$\Delta p'/m_e$: 1.319

$t$: 3.16 h
$\Delta p$ , MPa: 0.9774
$\Delta p'$ , MPa: 0.08751
$\Delta p'/m_e$: 1.559

$t$: 10.0 h
$\Delta p$ , MPa: 1.0909
$\Delta p'$ , MPa: 0.11487
$\Delta p'/m_e$: 2.046

$t$: 31.6 h
$\Delta p$ , MPa: 1.2605
$\Delta p'$ , MPa: 0.19076
$\Delta p'/m_e$: 3.399

$t$: 100 h
$\Delta p$ , MPa: 1.5314
$\Delta p'$ , MPa: 0.27219
$\Delta p'/m_e$: 4.849

$t$: 316 h
$\Delta p$ , MPa: 1.8731
$\Delta p'$ , MPa: 0.31323
$\Delta p'/m_e$: 5.580

$t$	$\Delta p$ , MPa	$\Delta p'$ , MPa	$\Delta p'/m_e$
36 s	0.6265	0.05754	1.025
6 min	0.7569	0.05613	1.000
12 min	0.7956	0.05613	1.000
1.00 h	0.8896	0.06569	1.170
1.58 h	0.9215	0.07405	1.319
3.16 h	0.9774	0.08751	1.559
10.0 h	1.0909	0.11487	2.046
31.6 h	1.2605	0.19076	3.399
100 h	1.5314	0.27219	4.849
316 h	1.8731	0.31323	5.580

$\Delta p = p_i - p_{wf}$ , $p_i$ = 31 MPa. The data were calculated from a model of a homogeneous reservoir with wellbore storage, skin and two sealing faults that meet at an angle, and rounded to 0.0001 MPa. The parameters of the model are given in the example. The derivative is from all the readings (10 per cycle) with an interval of 0.2 cycle. The ratio is to the plateau level from the example.

CalculatedThe derivative of the record divided by the plateau level (points), and the calculation by the model (line). The dashed lines are the levels 1, 2 and 6. The readings at 100 and 316 h, from which the example finds the angle, are marked.

Worked exampleThe angle between the faults and the distance to the apex

Given (assumed data)

Well and reservoir: $q$ = 400 m³/d; $B$ = 1.25; $\mu$ = 0.6 mPa·s; $h$ = 20 m; $\phi$ = 0.22; $c_t = 1.5\cdot10^{-3}$ MPa⁻¹
Derivative plateau: the mean from 0.1 to 0.2 h: $m_e$ = 0.05613 MPa
Rise from the nearer fault: at 1.58 h the derivative is 0.07405 MPa
Late points: at 100 h, 0.27219 MPa; at 316.2 h, 0.31323 MPa

Solution

Flow capacity: $qB\mu = 400\cdot1.25\cdot0.6 = 300$ ; $kh = 0.933\cdot300/0.05613 = 4987$ mD·m, $k = 249.4$ mD; $\eta = 3.553\cdot10^{-3}\cdot249.4/(0.22\cdot0.6\cdot1.5\cdot10^{-3}) = 4475$ m²/h.
The nearer fault: $\Delta p'/m_e = 1.319$ , $\ln[1/(1.319 - 1)] = 1.143$ ; $L_A = \sqrt{4475\cdot1.58\cdot1.143} = 90$ m.
The late ratios: $r_1 = 0.27219/0.05613 = 4.849$ , $r_2 = 0.31323/0.05613 = 5.580$ . The derivative is still rising. From the last point the angle would be $360^\circ/5.580 = 64.5^\circ$ , an upper limit.
The approach to the level: $a = \dfrac{4.849 - 5.580}{4.849/316.2 - 5.580/100} = 18.1$ h; $R_\infty = 5.580/(1 - 18.1/316.2) = 5.919$ .
The angle: $\theta = 360^\circ/5.919 = 60.8^\circ$ .
The distance to the apex: $d = \sqrt{2\cdot4475\cdot18.1} = 402$ m, a lower estimate.

Answer: The angle between the faults is about 60.8° (from the last point, 64.5°). The nearer fault is at 90 m, and the apex of the angle is no nearer than 402 m.

What it means. The model of the record has an angle of 60°, faults 90 and 330 m from the well, and the apex at 442 m. After 316 h the derivative has covered only 93% of the way to the level 6, and the angle from the last point is too high by 4–5°. Extrapolating from two points almost removed the error. The distance to the apex is underestimated by about 10%, as the approach formula warns. The angle and the distance to the nearer fault agree with the seismic data. We can pass the conclusion “an angle of about 60°, the apex at 400–450 m” to the geological model. To read the level directly to within 5%, the derivative would have to reach about $10\,d^2/\eta$ , that is, about 440 h, and the record would have to go another 0.2 cycle further, almost to 700 h. The data in this example are assumed for illustration.

How boundaries add up

Each boundary comes into the record at about the time when $\eta t$ becomes comparable with the square of the distance to it. The derivative rises by 1% at $\eta t \approx 0.22\,L^2$ , and it passes half of the step at $\eta t \approx 1.44\,L^2$ (A sealing fault near the well: the method of images). The steps from different boundaries separate only if their times differ by more than 1.6–1.7 logarithmic cycles, that is, if the distances differ by a factor of 6–7 or more.

A right angle: the length of the intermediate plateau

2m_e

when the well is near one of the faults

$L_B/L_A$: 3
Plateau within ±5%, logarithmic cycles: 0.13

$L_B/L_A$: 5
Plateau within ±5%, logarithmic cycles: 0.22

$L_B/L_A$: 7
Plateau within ±5%, logarithmic cycles: 0.34

$L_B/L_A$: 10
Plateau within ±5%, logarithmic cycles: 0.59

$L_B/L_A$	Plateau within ±5%, logarithmic cycles
3	0.13
5	0.22
7	0.34
10	0.59

Calculated from the derivative formula with images for an angle of 90°. $L_A$ is the distance to the nearer fault, and $L_B$ to the farther one. At $L_B/L_A$ = 3 the derivative passes the level $2m_e$ almost without a pause.

So a record with two steps, “1 → 2 → 4”, is rare. Usually we see a smooth staircase, and we fit it with a model as a whole. From this comes a practical rule. From the first, earliest part of the rise we find the distance to the nearer boundary: while the far ones do not yet have an effect, the formula of a single fault is exact. From the late behaviour (slope ½, a new level or slope 1) we find the type of the system and its second dimension.

A channel closed at one end (a U-shaped system of faults): the images in the end wall double the late linear flow. The derivative runs with a slope of ½, but twice as high as in an open channel of the same width. We cover such semi-infinite systems separately (Semi-infinite systems: theory and test analysis).
A closed rectangle: the images fill the plane with a lattice, and their number grows in proportion to $t$ . The derivative reaches a slope of 1, and from the rate of pressure decline we find the pore volume (Reservoir limit test).
A fault and a constant-pressure boundary: the image from the edge of an active edge-water aquifer (water around the reservoir that keeps the pressure almost constant) is taken with the opposite sign. After the rise from the fault, the derivative falls.
A fault that lets fluid through: the derivative rises, does not reach $2m_e$ , and returns to the plateau (Partially communicating faults).

Worked exampleHow long to record to see the channel and the angle

Given (assumed data)

Channel: from the first example: $W$ ≈ 402 m, $L_1/W$ ≈ 0.25, $\eta$ ≈ 826 m²/h
Wedge: from the second example: $d$ ≈ 402 m, $\eta$ ≈ 4475 m²/h

Solution

Channel: by the table, the derivative lies on the line with slope ½ (within 5%) at $\eta t/W^2$ = 0.30, that is, after $0.30\cdot402^2/826 = 59$ h.
To draw the line $\Delta p$ against $\sqrt t$ through a section of at least one logarithmic cycle, we extend the record tenfold: 590 h, about 25 days.
Wedge: 95% of the late level is reached at $\eta t \approx 9.6\,d^2$ , after $9.6\cdot402^2/4475 = 347$ h, about 14 days.

Answer: Linear flow in this channel starts after about 59 h. For a confident width we need about 25 days of record. The level of the wedge can be read directly after 14 days. This is why in the example we found the angle by extrapolation.

What it means. Both times grow as the square of the size of the system: a channel twice as wide needs a record four times longer. For the wedge we took $d$ from the second example. It is a lower estimate, so the time is a lower limit too: with the real 442 m it is about 420 h (17–18 days). The time that the disturbance needs to reach the apex of a wedge is set not by the angle but by the distance $d$ . For a well far from the apex, we cannot read the level from a record of reasonable length, and what is left is extrapolation or fitting a model. We get long records at a constant rate in an extended well test. For a build-up, the possibilities are limited by the volume produced before shut-in (the build-up near a fault). The data in this example are assumed for illustration.

What a record cannot tell apart

The late part of a record is the most informative and the most ambiguous. One and the same shape of the derivative can come from different reservoirs, and the same numbers can come from different geometries. Here is what a test of a single well gives and does not give:

Late signs and their possible causes

What we see: Slope ½ at the end of the record
Possible causes: a channel; an elongated closed block before it reaches slope 1
How to check: geology, seismic data, length of the record

What we see: Slope ½ at the start of the record
Possible causes: a hydraulic fracture or a natural fracture
How to check: the history of the well; the duration of the section

What we see: A new level $N\,m_e$
Possible causes: a wedge with an angle $360^\circ/N$ ; a flow capacity that gets $N$ times worse farther away
How to check: the map of faults; the shape of the transition; neighbouring wells

What we see: A rise, then a fall back to the plateau
Possible causes: a partially communicating fault; a fault of finite length
How to check: Partially communicating faults

What we see: A doubling with no plateau $m_e$ before it
Possible causes: a well right at a fault, with the first plateau hidden by wellbore storage
How to check: a fault close to the well

What we see	Possible causes	How to check
Slope ½ at the end of the record	a channel; an elongated closed block before it reaches slope 1	geology, seismic data, length of the record
Slope ½ at the start of the record	a hydraulic fracture or a natural fracture	the history of the well; the duration of the section
A new level $N\,m_e$	a wedge with an angle $360^\circ/N$ ; a flow capacity that gets $N$ times worse farther away	the map of faults; the shape of the transition; neighbouring wells
A rise, then a fall back to the plateau	a partially communicating fault; a fault of finite length	Partially communicating faults
A doubling with no plateau $m_e$ before it	a well right at a fault, with the first plateau hidden by wellbore storage	a fault close to the well

The list is not complete: heterogeneities of the reservoir, and changes of the rates of neighbouring wells during the record, give similar shapes too.

Directions. The record depends only on the distances to the images. A well 100 m from the wall of a channel that runs north, and the same well in a channel that runs east, give the same pressure. The test cannot say which side the nearer wall is on either.
Width without permeability. Without radial flow, the $\sqrt t$ slope gives only $W\sqrt k$ (the callout above). An error of 20% in $\phi c_t$ changes the width by 10%.
An unfinished transition. The angle from a derivative that is still rising is too large, and a second line drawn too early underestimates the distances. Extrapolation and fitting a model help, but they do not replace a long record.
A wedge or a heterogeneity. A level of $4m_e$ comes from a right angle between faults, and also from a reservoir whose flow capacity beyond some radius is four times lower. The shapes of the transitions differ a little, but with gauge noise they are easy to confuse.
Straight boundaries are a model. Real faults are curved, they fade out along their length, and they displace the reservoir not through its whole thickness. The distances and the angle from a simple model are estimates that we check against the map.

Terms and symbols

System of images: A set of image wells for which the pressure picture is symmetric about all the sealing boundaries. For a channel it is an endless row, and for a wedge with an angle $180^\circ/n$ it is $2n - 1$ images.
Channel: A reservoir between two parallel sealing boundaries: a graben, a narrow block, or a river-channel sand body.
Linear flow in a channel: The late regime in which the flow goes to the well along the channel: $\Delta p$ grows as $\sqrt t$ , and the derivative has a slope of ½.
Geometric skin of a channel $S_{ch}$: The constant $\ln\tfrac{W}{2\pi r_w} - \ln\sin\tfrac{\pi L_1}{W}$ in the intercept of the line $\Delta p$ against $\sqrt t$ . It grows when the well comes closer to a wall.
Wedge: A block between two crossing sealing faults. The late level of the derivative is $(360^\circ/\theta)\,m_e$ .
Distance to the apex $d$: The distance from the well to the point where the faults cross, in plan view. It sets the time at which the derivative reaches the late level, $\eta t \approx 10\,d^2$ .
Intermediate plateau: A section of the derivative at the level $2m_e$ between two steps. It is visible if the second boundary is 6–7 times farther than the first.

Common mistakes and misconceptions

Wrong

“A slope ½ on the derivative means a well with a hydraulic fracture.”

In fact

A fracture gives a slope ½ at the start of the record, before radial flow. A slope ½ at the end of the record, after the plateau, is a sign of a channel or an elongated block. From it we calculate a width, and not a fracture length.

Wrong

“The angle between the faults is 360° divided by the last ratio of the derivative to the plateau.”

In fact

Only if the derivative has already reached the level. While it is rising, such an angle is an upper limit: in the lesson's example it is 64.5° instead of 60°. We estimate the level from two late points or by fitting a model.

Wrong

“We can find the channel width from the $\sqrt t$ slope alone.”

In fact

The slope gives $W\sqrt k$ . We need the permeability, from the radial-flow plateau of the same record or from other data. Without it the width cannot be found.

Wrong

“A test shows where the faults run.”

In fact

It shows how far from the well the images are. The directions of the faults and the side of the nearer wall are given by seismic surveys, the geological model and the neighbouring wells.

Wrong

“If the well in a channel is not in the centre, the derivative must have an intermediate plateau $2m_e$ .”

In fact

Only if the far wall is 6–7 times farther than the near one. In the lesson's example the walls are at 100 and 300 m, and the derivative passes the level $2m_e$ without a pause.

In depth: advanced detail

Derivations: linear flow in a channel, the level and the approach in a wedge

Channel. The Green's function of a strip $0 < y < W$ with sealing edges (the response of the strip to a point source) is a series in the cosines $\cos(n\pi y/W)$ with the factors $\e^{-n^2\pi^2\eta t/W^2}$ . Along the channel it is the one-dimensional solution $1/\sqrt{4\pi\eta t}$ . The logarithmic derivative at the well equals $2\pi\eta t$ times their product at the point of the well:

The derivative in a channel: a series in the eigenfunctions of the strip

p_D' = \sqrt{\pi t_{DW}}\left[1 + 2\sum_{n=1}^{\infty} \e^{-n^2\pi^2t_{DW}}\cos^2\frac{n\pi L_1}{W}\right]

where:

$p_D'$ dimensionless derivative, $\Delta p'/(2m_e)$ , —
$t_{DW}$ dimensionless time $\eta t/W^2$ , —
$L_1,\ W$ distance to the wall and width of the channel, m

Units: dimensionless

Holds when

The well is a line source; wellbore storage is over.

For large times the sum vanishes, and what is left is $\sqrt{\pi t_{DW}}$ , the slope ½. The slowest correction is the one with $n = 1$ . It decreases as $\e^{-\pi^2 t_{DW}}$ , and in the middle of the channel its factor $\cos^2(\pi/2)$ is zero. This series and the sum of images are two forms of one solution. Numerically they agree to the eighth digit.

Geometric skin. If we integrate the same solution over time, we get $2\sqrt{\pi t_{DW}}$ plus a sum over $n$ with the factors $\tfrac2n\cos^2(n\pi L_1/W)\,\e^{-n\pi r_w/W}$ and decaying terms. The sum is a series that has a known formula, with the terms $\cos(nx)\,\e^{-nu}/n$ . For $u = \pi r_w/W \to 0$ it gives $\ln\tfrac{W}{\pi r_w} - \ln\big(2\sin\tfrac{\pi L_1}{W}\big)$ , that is, $S_{ch}$ .

The level of a wedge for any angle. Far from the apex (at distances $r \gg d$ ) the flow converges on the apex through a sector with an arc $\theta r$ : $q = (kh\,\theta r/\mu)\,\partial p/\partial r$ . From this, $p = \tfrac{q\mu}{kh\theta}\ln r + \text{const}$ . In an infinite reservoir $2\pi$ stands in place of $\theta$ . So the pressure drop per unit of $\ln r$ is larger by a factor of $2\pi/\theta$ . It is the same per unit of $\ln t$ , since the radius of the disturbance grows as $\sqrt t$ . In the well, where the pressure changes following the far region, the derivative reaches the same level, $(2\pi/\theta)\,m_e$ .

The approach to the level. For angles $180^\circ/n$ , the well and the images stand at points of a circle of radius $d$ with polar angles $\pm\varphi_0 + 2j\theta$ . The square of the distance to a point with an angular shift $\psi$ is $2d^2(1 - \cos\psi)$ , and the sum of the cosines over all $2n$ points is zero (each group of $n$ points is spaced evenly). So $\sum_j d_j^2 = 4nd^2$ , and for small $d_j^2/(4\eta t)$ the derivative is $2n\,m_e\,[1 - d^2/(2\eta t)]$ . For an arbitrary angle, the exact solution of the wedge is a series in Bessel functions of the orders $k\pi/\theta$ . Its main term gives the same correction $1 - d^2/(2\eta t)$ , and the others, for $\theta < 180^\circ$ , decay faster.

Connections to other topics

A sealing fault near the well: the method of images — one sealing boundary: the image, the derivative step, and the distance from one point of the step.
Superposition principle — the method of images and the adding up of solutions on which the lesson is built.
Diagnosis with the pressure derivative — the diagnostic plot, the plateau, the slopes ½ and 1, and the calculation of the derivative from measurements.
Partially communicating faults — a fault that lets fluid through: a rise of the derivative and a return to the plateau.
Semi-infinite systems: theory and test analysis — channels closed at one end, and other semi-infinite systems of faults.
Reservoir limit test — closed blocks: slope 1, pore volume and shape factor.
Inflow to a hydraulically fractured well — linear flow to a fracture: the same slope ½, but at the start of the record.
Flow regimes — the geometry of flow: radial, linear and pseudo-steady-state regimes.
Faults — the geology of faults: types, crossings, sealing.
Well testing: purposes and types of test — kinds of tests and planning their duration.

Check yourself

1
After the plateau of radial flow, the derivative rose and at the end of the record it runs as a straight line with slope ½ on a log–log plot. What is it most likely?
1. A.A channel between two parallel sealing boundaries
2. B.A wedge between crossing faults
3. C.A constant-pressure boundary
4. D.A single sealing fault
Show the answer
Correct answer: A. A channel between two parallel sealing boundaries
In a channel the number of “felt” images grows as $\sqrt t$ , the pressure does too, and the derivative has a slope of ½. A single fault and a wedge give a new horizontal plateau ( $2m_e$ and $(360^\circ/\theta)m_e$ ), and a constant-pressure boundary lowers the derivative. A similar late slope also occurs in an elongated closed block before it reaches slope 1. We check this with the length of the record and with geology.
2
The plateau level of radial flow is 0.045 MPa. The late derivative has risen to 0.108 MPa and has stayed at this level for the third logarithmic cycle. What angle between the faults does the block form?
Show the answer
The ratio of the levels is $0.108/0.045 = 2.4$ . The angle is $\theta = 360^\circ/2.4 = 150^\circ$ , an obtuse angle, for example a bend of a fault or two faults that meet at a wide angle. The images for such an angle do not give the exact solution, but the late level $(360^\circ/\theta)\,m_e$ is right for any angle. Since the derivative has stayed at the level for a long time, the approach to it is over, and the angle is not too large.
3
A well in a channel: $q$ = 90 m³/d, $B$ = 1.2, $h$ = 8 m, $\mu$ = 2 mPa·s, $\phi$ = 0.18, $c_t = 1.5\cdot10^{-3}$ MPa⁻¹; from the derivative plateau $k$ = 50 mD. The late points lie on a line $\Delta p$ against $\sqrt t$ with a slope of 0.20 MPa/√h. Find the width of the channel.
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Under the root: $\mu/(k\phi c_t) = 2/(50\cdot0.18\cdot1.5\cdot10^{-3}) = 148.1$ , and the root is 12.17. The factor $qB/(h\,m_L) = 90\cdot1.2/(8\cdot0.20) = 67.5$ . The width is $W = 0.394\cdot67.5\cdot12.17 \approx 324$ m. If the permeability were twice as small, with the same slope the width would come out $\sqrt2$ times larger, about 458 m.
4
From a record we found distances of 100 and 300 m to the two parallel walls of a channel. Can the same record tell on which side of the well the nearer wall is, and in which direction the channel extends?
1. A.Yes, from the intercept of the line $\Delta p$ against $\sqrt t$
2. B.No: the record depends only on the distances to the images
3. C.Yes, from the shape of the derivative step
4. D.Yes, if the record is longer than 1000 h
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Correct answer: B. No: the record depends only on the distances to the images
The pressure at the well is set by the distances to the images, and not by their directions: turning the whole system of faults does not change the record. Even the intercept of the line is symmetric: the geometric skin is the same for $L_1$ and $W - L_1$ . The directions are given by seismic surveys, the geological model and the neighbouring wells.
5
Why does a 30° wedge need a much longer record than a right angle, if the distance from the well on the bisector to each fault is the same?
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The time to reach the late level is set by the distance to the apex of the angle: 95% of the level at $\eta t \approx 10\,d^2$ . On the bisector $d = L/\sin(\theta/2)$ : for 90° this is $1.41L$ , and for 30° it is $3.86L$ . The ratio of the squares is $(3.86/1.41)^2 \approx 7.5$ . So the 30° wedge reaches its level (12 $m_e$ ) about 7.5 times later than a right angle reaches its level of 4 $m_e$ .
6
At 200 h the ratio of the derivative to the plateau is 3.4 and still rising. The estimate of the angle $360^\circ/3.4 = 106^\circ$ : is it an underestimate or an overestimate of the true angle?
1. A.An exact estimate
2. B.An overestimate
3. C.An underestimate
4. D.We cannot say
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Correct answer: B. An overestimate
While the derivative is rising, it is below the late level $360^\circ/\theta$ , and the angle is inversely proportional to the level. So 106° is an upper limit, and the true angle is smaller. We estimate the level from two late points with the law $R_\infty(1 - a/t)$ , or we extend the record.

Updated: 19 October 2026