Semi-infinite systems: theory and test analysis

Quick refresher

A semi-infinite system is open in at least one direction. It has no pseudo-steady-state flow. After a long shut-in, the pressure returns to the initial pressure $p_i$ , not to the average pressure of the block as in a closed system.
Every straight boundary acts like a mirror and gives an image of the well. If one length $L$ sets the positions of all the images, their effect depends only on $\eta t/L^2$ . So one geometry needs only one type curve (a ready-made reference curve).
Product solution: in a rectangular system the derivative, divided by the plateau $m_e$ , equals the product of the factors for the two directions. An open direction gives 1. One boundary at distance $L$ gives $1 + \e^{-L^2/(\eta t)}$ . Two parallel boundaries give the strip factor, which grows like $\sqrt{4\pi\eta t}/W$ .
So the closed end of a channel acts on linear flow as a fault acts on radial flow: it doubles it. The slope of the line of $\Delta p$ against $\sqrt t$ becomes $2m_L$ . The most reliable way to find the distance to the closed end is one point of the derivative: $L_3 = \sqrt{\eta t\,\ln[1/(r - 1)]}$ .
A well right next to a boundary behaves like a well with a doubled rate in a system without that boundary. The skin refers to the real rate.
After a short producing time, the closed end of a channel leaves a trace on the build-up that is inversely proportional to $WL_3$ , not to $L^2$ as for a fault in a radial reservoir. So the closed end is easier to see, but it shows late: about $2L_3^2/\eta$ after shut-in.

Key relationships

\Delta p = m_e\Big[\ln t_D + 0.809 + 2S + \sum_j E_1\!\Big(\dfrac{d_j^2}{4\eta t}\Big)\Big]

pressure in a well with images after wellbore storage (the well acting as a small tank) has ended; $m_e = qB\mu/(4\pi kh)$

\dfrac{\Delta p'}{m_e} = F_x(t)\,F_y(t),\qquad F = 1 + \e^{-L^2/(\eta t)}

product solution; the factor of one boundary at distance $L$

L_3 = \sqrt{\eta t\,\ln\dfrac{1}{r - 1}},\qquad r = \dfrac{\Delta p'}{\tfrac12 m_L\sqrt t}

distance to the closed end of the channel from one point of the derivative

L_3 = \sqrt{\eta t_x/\pi} = 0.0336\sqrt{\dfrac{k\,t_x}{\phi\mu c_t}}

from the intersection of the lines of $\Delta p$ against $\sqrt t$ : m, mD, h, mPa·s, 1/MPa (needs a very long record)

\delta p_{\max} = \dfrac{qB\,t_p}{198\,\phi c_t h\,WL_3}

largest trace of the closed end on a pressure build-up after a short producing time $t_p$ : m³/d, h, 1/MPa, m, MPa

Common pitfalls

Calculating the channel width from the late $\sqrt t$ slope when the closed end already shows. The slope is doubled, so the width comes out half the true value.
Drawing the second line of $\Delta p$ against $\sqrt t$ through the last points and taking the distance to the closed end from the intersection. The late line forms slowly, so the distance comes out too small.
Mistaking a build-up that returns to $p_i$ more slowly than usual for depletion of the block, without checking for the open direction.
Doubling the skin together with the rate for a well that stands right next to a boundary.
Expecting a test of one well to show on which side the channel is closed. The record gives distances, not directions.

After this lesson you can

explain how a semi-infinite system differs from a closed one and from an infinite reservoir, and name its signs on the derivative and on a pressure build-up
write the pressure in a well with images in dimensionless form, using one characteristic length, and explain why the late part of the solution depends only on $\eta t/L^2$
build the derivative of a rectangular system of faults as the product of the factors for the two directions
recognise a channel closed at one end by the doubled slope of linear flow, and find the width of the channel and the distance to the closed end from the lines of $\Delta p$ against $\sqrt t$ and from one point of the derivative
estimate whether a test will see the closed end of the channel while the well produces and after shut-in, and say what the record cannot tell apart

Semi-infinite systems among the boundary models

The boundary models that we use to interpret well tests fall into three families. In an infinite reservoir, the pressure disturbance spreads without any obstacle. In a closed area, it meets sealing boundaries on all sides. After a transition period, the pressure then falls at the same rate everywhere. This is pseudo-steady-state flow (Reservoir limit test). Between the two lie semi-infinite systems. They have boundaries, but the reservoir is open in at least one direction. The disturbance goes off into the open side without limit, the volume that it covers grows without limit, and pseudo-steady-state flow never comes.

The simplest semi-infinite systems, and a closed block for comparison

System: One fault (half-plane)
Images: one, at $2L$
Late derivative: plateau $2m_e$
Where it is covered: A sealing fault near the well: the method of images

System: Right-angle corner
Images: three
Late derivative: plateau $4m_e$
Where it is covered: Several boundaries: parallel faults and wedge-shaped blocks and later in this lesson

System: Channel (two parallel boundaries)
Images: an endless row
Late derivative: slope ½
Where it is covered: Several boundaries: parallel faults and wedge-shaped blocks

System: Channel closed at one end
Images: two rows
Late derivative: slope ½, twice as high
Where it is covered: this lesson

System: Closed rectangle
Images: a grid
Late derivative: slope 1
Where it is covered: Reservoir limit test

System	Images	Late derivative	Where it is covered
One fault (half-plane)	one, at $2L$	plateau $2m_e$	A sealing fault near the well: the method of images
Right-angle corner	three	plateau $4m_e$	Several boundaries: parallel faults and wedge-shaped blocks and later in this lesson
Channel (two parallel boundaries)	an endless row	slope ½	Several boundaries: parallel faults and wedge-shaped blocks
Channel closed at one end	two rows	slope ½, twice as high	this lesson
Closed rectangle	a grid	slope 1	Reservoir limit test

$L$ is the distance to the fault, and $m_e = qB\mu/(4\pi kh)$ is the level of the derivative plateau in radial flow, the flow towards the well from all sides (Diagnosis with the pressure derivative). A wedge with any angle is also a semi-infinite system. Its late level $(360^\circ/\theta)\,m_e$ is covered in the lesson Several boundaries: parallel faults and wedge-shaped blocks.

What semi-infinite systems have in common is the open side. However long the well produces, somewhere in the reservoir the pressure stays at its initial value, and fluid keeps coming from parts that are farther and farther away. So after shut-in, the pressure in such a system builds up to the initial pressure $p_i$ , not to the average pressure of the block as in a closed area (Build-up in a bounded reservoir and average reservoir pressure). This is the main sign that tells a channel with one closed end from a long closed block. In a block, the late derivative sooner or later reaches slope 1, and the build-up reaches an average pressure below the initial one.

SchematicA channel closed at one end (a sketch, not to scale). The well is in the middle of the channel, at distance

L_3

from the closed end. The images in the walls form a vertical row, as for an open channel. The closed end reflects this whole row, so a second, mirror row appears at distance

2L_3

. There are no images on the side where the channel is open.

The solution with images and one characteristic length

The pressure in a well near straight sealing boundaries is the sum of its own response and the contributions of the image wells (method of images). After wellbore storage (the well acting as a small tank) has ended, the well's own response is a logarithm of time. Each image at distance $d_j$ adds the line-source solution (the pressure response of an ideal, infinitely thin well):

Pressure in a well with images, in dimensionless form

p_{wD} = \tfrac12\left(\ln t_D + 0.809\right) + S + \tfrac12\sum_j E_1\!\left(\frac{\beta_j^2}{4\tau}\right),\qquad \tau = \frac{\eta t}{L^2} = \frac{t_D}{L_D^2},\quad \beta_j = \frac{d_j}{L}

where:

$p_{wD}$ dimensionless drop of the bottomhole pressure, $2\pi kh\,\Delta p/(qB\mu)$ , —
$t_D,\ S$ dimensionless time $kt/(\phi\mu c_tr_w^2)$ , and the skin (a number that shows how much the zone around the well helps or hinders flow), —
$L,\ L_D$ characteristic length of the system (for example, the distance to the nearest boundary), and the same length in well radii, m, —
$\tau$ dimensionless time based on $L^2$ ( $\eta$ is the hydraulic diffusivity: how fast a pressure change spreads through the rock), —
$d_j,\ \beta_j$ distance from the well to image number $j$ , and the same distance in units of $L$ , m, —
$E_1$ exponential integral (Line-source solution and the pressure function), —

Units: dimensionless

Holds when

the boundaries are straight and sealing, through the whole thickness of a homogeneous reservoir; the images are line sources ( $d_j \gg r_w$ )
the rate is constant, wellbore storage is over, $t_D > 25$
for images that go off to infinity (channels), we add terms until they become negligible

The well radius and the skin enter only the first terms. They shift the curve up or down. All the boundaries enter the sum, and the sum depends on one variable, $\tau$ , if all the distances are proportional to one length $L$ . So for each geometry we draw one type curve (a ready-made reference curve) in the coordinates of $\tau$ . Matching it in time gives $L$ , and matching it in pressure gives the flow capacity (permeability times thickness), as in ordinary type-curve matching (Type-curve matching). The derivative is $p_D' = \tfrac12\big[1 + \sum_j\e^{-\beta_j^2/(4\tau)}\big]$ (images of images).

If the number $n$ of images is finite, the late derivative settles on a plateau $(1 + n)\,m_e$ . The slope of the semilog straight line (the straight line on a plot of pressure against the logarithm of time) grows by the same factor: twice for one fault, four times in a right-angle corner. If there are infinitely many images, as in a channel, there is no new plateau. The number of images that the pressure “feels” grows like $\sqrt t$ , and the derivative runs with slope ½. The ratio of the late level to the early one, or the late slope, is the first sign that we use to choose the model.

The product solution for rectangular systems

If all the boundaries are parallel to one of two perpendicular directions, which we call $x$ and $y$ , the images stand at the points of a rectangular grid. The shift of each image is made up of a shift along $x$ and a shift along $y$ . The square of the distance is the sum of their squares, and the exponential of a sum is the product of the exponentials. So the sum over all the images splits into the product of two sums, one for each direction (product solution):

Product solution: the derivative of a rectangular system

\frac{\Delta p'}{m_e} = F_x(t)\,F_y(t),\qquad F_x = \sum_i \exp\!\left(-\frac{\xi_i^2}{4\eta t}\right),\quad F_y = \sum_l \exp\!\left(-\frac{\zeta_l^2}{4\eta t}\right)

where:

$\Delta p'$ logarithmic derivative of the pressure drop, $\dd\Delta p/\dd\ln t$ , Pa
$m_e$ derivative plateau in radial flow, $qB\mu/(4\pi kh)$ , Pa
$F_x,\ F_y$ direction factors: sums over the one-dimensional images along $x$ and along $y$ , including the well itself (shift 0), —
$\xi_i,\ \zeta_l$ shifts of the one-dimensional images along $x$ and along $y$ , m
$\eta$ hydraulic diffusivity $k/(\phi\mu c_t)$ (how fast a pressure change spreads through the rock), m²/s

Units: SI (any consistent system)

Holds when

all the boundaries are straight sealing planes, parallel to $x$ or $y$ , through the whole thickness of a homogeneous reservoir
the well is a line source, wellbore storage is over, and the rate is constant

Each direction brings in its own factor, independent of the other. An open direction gives 1 (one “well” with no images). One boundary at distance $L$ gives $1 + \e^{-L^2/(\eta t)}$ , an image at $2L$ . Two parallel boundaries give the strip factor. It is a sum over the row of images, and at late times it grows like $\sqrt{4\pi\eta t}/W$ (the derivation is in the advanced section).

Direction factors and the systems built from them

Along $x$: 1
Along $y$: $1 + \e^{-L^2/(\eta t)}$
System: one fault
Late derivative: $2m_e$

Along $x$: $1 + \e^{-L_A^2/(\eta t)}$
Along $y$: $1 + \e^{-L_B^2/(\eta t)}$
System: right-angle corner
Late derivative: $4m_e$

Along $x$: 1
Along $y$: strip of width $W$
System: channel
Late derivative: $m_e\sqrt{4\pi\eta t}/W$

Along $x$: $1 + \e^{-L_3^2/(\eta t)}$
Along $y$: strip of width $W$
System: channel closed at one end
Late derivative: $2m_e\sqrt{4\pi\eta t}/W$

Along $x$: strip of length $L$
Along $y$: strip of width $W$
System: closed rectangle
Late derivative: grows like $t$

Along $x$	Along $y$	System	Late derivative
1	$1 + \e^{-L^2/(\eta t)}$	one fault	$2m_e$
$1 + \e^{-L_A^2/(\eta t)}$	$1 + \e^{-L_B^2/(\eta t)}$	right-angle corner	$4m_e$
1	strip of width $W$	channel	$m_e\sqrt{4\pi\eta t}/W$
$1 + \e^{-L_3^2/(\eta t)}$	strip of width $W$	channel closed at one end	$2m_e\sqrt{4\pi\eta t}/W$
strip of length $L$	strip of width $W$	closed rectangle	grows like $t$

For a right-angle corner, the product solution gives $1 + \e^{-L_A^2/(\eta t)} + \e^{-L_B^2/(\eta t)} + \e^{-(L_A^2 + L_B^2)/(\eta t)}$ , exactly the three images of the corner. For a closed rectangle, the same strip factors describe the approach to pseudo-steady-state flow (Reservoir limit test).

CalculatedCalculated factors for a well in the middle of a channel 300 m wide that is closed 1200 m from it (

\eta

= 1480 m²/h, the data of the example below): the strip factor

F_y

, the end factor

F_x

, and their product, which is the derivative divided by the plateau

m_e

. Both scales are logarithmic.

Worked exampleForecast of the derivative with the product solution

Given (assumed data)

Formation: $k$ = 100 mD; $\phi$ = 0.20; $\mu$ = 1.0 mPa·s; $c_t = 1.2\cdot10^{-3}$ MPa⁻¹
Channel from the geological model: width 300 m, the well in the middle; at 1200 m the channel seems to be closed by a cross-fault
Question: what the derivative will be after 2, 20 and 200 days of production at a constant rate

Solution

Hydraulic diffusivity: $\eta = 3.553\cdot10^{-3}\cdot100/(0.20\cdot1.0\cdot1.2\cdot10^{-3}) = 1480$ m²/h. The well is in the middle of the channel. So for $\eta t/W^2 > 0.2$ , the strip factor equals $2\sqrt{\pi\eta t/W^2}$ to better than 0.1%.
At 48 h: $\eta t = 71{,}040$ m², $\eta t/W^2 = 0.7893$ . The strip in linear flow gives $F_y = 2\sqrt{\pi\cdot0.7893} = 3.149$ . The closed end gives $L_3^2/(\eta t) = 20.270$ . The term $\e^{-20.270}$ is negligibly small, so $F_x = 1$ . The derivative is $3.149\,m_e$ : an open channel.
At 480 h: $\eta t/W^2 = 7.8933$ , $F_y = 9.959$ ; $L_3^2/(\eta t) = 2.027$ , $F_x = 1 + \e^{-2.027} = 1.1317$ . The derivative is $9.959\cdot1.1317 = 11.27\,m_e$ (the sum of all the images gives 11.27).
At 4800 h: $\eta t/W^2 = 78.9333$ , $F_y = 31.495$ ; $L_3^2/(\eta t) = 0.203$ , $F_x = 1.8163$ . The derivative is $57.2\,m_e$ .
The closed end raises the derivative by 10% when $\e^{-L_3^2/(\eta t)} = 0.1$ , that is, at $t = L_3^2/(\eta\ln10) = 423$ h. An almost complete doubling (90%) comes at $\eta t \approx 9.5\,L_3^2$ , after 385 days.

Answer: After 2 days the derivative is about 3.149

m_e

(an open channel). After 20 days it is 11.27

m_e

(the closed end has raised it by 13%). After 200 days it is 57.2

m_e

What it means. The strip factor and the end factor are calculated independently, and their product agrees with the sum over all the images to the last digit. A closed end 1200 m away shows after only 2–3 weeks of production. But the fully doubled line can be seen only after about a year. That is a job not for a well test but for permanent bottomhole gauges (pressure gauges that stay in the well) during production. The data in this example are assumed for illustration.

A channel closed at one end

A channel closed at one end is a long block between parallel faults that a cross-fault blocks off. Or it is a river-channel sand body that thins out, or that ends against a clay plug left by an abandoned river bed (Continental environments: aeolian and fluvial). In plan view, the boundaries form the letter U. That is why it has a second name, a U-shaped fault system. Take a well that stands at distances $L_1$ and $L_2$ from the walls ( $W = L_1 + L_2$ ) and at distance $L_3$ from the closed end.

By the product solution, the derivative of this system is the derivative of the open channel times the end factor $1 + \e^{-L_3^2/(\eta t)}$ . If the closed end is several times farther away than the width of the channel, the record passes through four stages. First comes the radial-flow plateau $m_e$ . Then the derivative rises while the walls make themselves felt. Then comes the linear flow of the open channel (flow that runs along the channel in parallel streamlines). The derivative is $\tfrac12m_L\sqrt t$ , with slope ½ (linear flow in a channel). Last, when the closed end has been “felt”, the derivative moves onto a parallel line twice as high, $m_L\sqrt t$ . The closed end does to linear flow what a fault does to radial flow. It doubles it, and by the same law.

A well in a channel closed at one end: late linear flow

\Delta p = 2m_L\sqrt t + \frac{qB\mu}{2\pi kh}\,(S + S_{ch}) - \frac{qB\mu}{khW}\,L_3,\qquad \Delta p' = m_L\sqrt t

where:

$\Delta p$ drop of the bottomhole pressure, $p_i - p_{wf}$ , Pa
$m_L$ slope of the line of $\Delta p$ against $\sqrt t$ for an open channel of the same width, $\dfrac{qB}{hW}\sqrt{\dfrac{\mu}{\pi k\phi c_t}}$ , Pa/√s
$S,\ S_{ch}$ skin of the well, and the geometric skin of the channel $\ln\frac{W}{2\pi r_w} - \ln\sin\frac{\pi L_1}{W}$ (channel geometric skin), —
$W,\ L_3$ width of the channel and distance from the well to the closed end, m
$q,\ B,\ \mu,\ k,\ h$ rate at standard conditions, formation volume factor, viscosity, permeability, thickness, m³/s, m³/m³, Pa·s, m², m

Units: SI

Holds when

the walls and the closed end are sealing planes through the whole thickness of a homogeneous reservoir; in the other direction the channel is unbounded
the closed end is farther away than the width of the channel ( $L_3 > W$ ), and $\eta t \gg L_3^2$ , so the closed end has long been “felt”; wellbore storage is over, and the rate is constant

The slope is doubled. Fluid reaches the well only from the open side, through one “pipe” of cross-section $Wh$ instead of two. The last term is the pressure drop that the whole rate would cause flowing along the channel over the length $L_3$ . The farther the closed end, the lower the line. The approach to this line is slow. The excess above it falls as $L_3^2/\sqrt{\pi\eta t}$ , not as $1/t$ . So the late points lie above the line for a long time.

CalculatedCalculated derivatives of a well in the middle of a channel of width

W

, without wellbore storage: an open channel, and channels closed at

L_3

W

3W

and

10W

from the well. The vertical axis is the ratio of the derivative to the plateau

m_e

. The dashed lines are the linear flow of the open channel and the line twice as high. Both scales are logarithmic.

We can find the distance to the closed end in three ways. The most reliable is to use one point of the derivative. The end factor $1 + \e^{-L_3^2/(\eta t)}$ is exact at any moment, and the ratio of the derivative to the continuation of the linear-flow line of the open channel gives $L_3$ at once. The second way is to intersect two lines on the plot of $\Delta p$ against $\sqrt t$ , like the intersection of two semilog lines for a fault (A sealing fault near the well: the method of images). But the late line takes a very long time to form, and without a correction the distance comes out too small. The third way is to fit a model to the whole record.

Distance to the closed end of the channel: from a point of the derivative and from the intersection of the lines of Δp against √t

L_3 = \sqrt{\eta t\,\ln\frac{1}{r - 1}},\quad r = \frac{\Delta p'(t)}{\tfrac12 m_L\sqrt t};\qquad L_3 = \sqrt{\frac{\eta\,t_x}{\pi}}

where:

$r$ ratio of the derivative at time $t$ to the continuation of the linear-flow line of the open channel, $1 < r < 2$ , —
$\Delta p'(t)$ logarithmic derivative of the pressure at time $t$ , Pa
$t_x$ time at which the line of the open channel (slope $m_L$ ) and the late line (slope $2m_L$ ) intersect on the plot of $\Delta p$ against $\sqrt t$ , s
$\eta$ hydraulic diffusivity $k/(\phi\mu c_t)$ , m²/s

Units: SI

Holds when

the record contains the linear flow of the open channel. We draw the line $\tfrac12m_L\sqrt t$ along it on the derivative, and the line with slope $m_L$ on the plot of $\Delta p$ against $\sqrt t$
the closed end is farther away than the width of the channel; the permeability is known from the radial-flow plateau
the second formula holds if the late line has really been reached. If we draw a line with slope $2m_L$ through the last reading at time $t_{end}$ , it gives $L_x = L_3 - L_3^2/\sqrt{\pi\eta t_{end}}$ (the correction is in the example)

The first formula is the same as for the derivative step at a fault. Only here we take the ratio to the line with slope ½, not to the plateau. The second formula comes from equating the two lines. The difference of their intercepts equals $qB\mu L_3/(khW)$ , and $m_L = \tfrac{qB\mu}{khW}\sqrt{\eta/\pi}$ .

In other unit systems (2)

Metric oilfield units

L_3 = 0.0596\sqrt{\frac{k\,t}{\phi\mu c_t}\ln\frac{1}{r - 1}},\qquad L_3 = 0.0336\sqrt{\frac{k\,t_x}{\phi\mu c_t}}

$L_3$ in m, $k$ in mD, $t$ and $t_x$ in h, $\mu$ in mPa·s, $c_t$ in 1/MPa.

$0.0596 = \sqrt{3.553\cdot10^{-3}}$ , as in the formula for the step at a fault (A sealing fault near the well: the method of images); $0.0336 = \sqrt{3.553\cdot10^{-3}/\pi}$ .

Oilfield units

L_3 = 0.01624\sqrt{\frac{k\,t}{\phi\mu c_t}\ln\frac{1}{r - 1}},\qquad L_3 = 0.00916\sqrt{\frac{k\,t_x}{\phi\mu c_t}}

$L_3$ in ft, $k$ in mD, $t$ and $t_x$ in h, $\mu$ in cP, $c_t$ in 1/psi.

$0.01624 = \sqrt{2.637\cdot10^{-4}}$ ; $0.00916 = \sqrt{2.637\cdot10^{-4}/\pi}$ .

An extended well test in a river-channel body

Here is an assumed record of an extended well test (a long test in which the well produces for days or weeks). An exploration well has penetrated the sandstone of a river-channel body. Seismic data and the neighbouring wells show that the body is narrow and long, but we do not know whether it is open at both ends. The well was started by opening a downhole valve, at a constant rate of 150 m³/d. The pressure was recorded for about 26 days.

Pressure drop in a well of a river-channel body (assumed data, a selection of the readings)

$t$: 36 s
$\sqrt t$ , √h: 0.10
$\Delta p$ , MPa: 0.4148
$\Delta p'$ , MPa: 0.0398

$t$: 6 min
$\sqrt t$ , √h: 0.32
$\Delta p$ , MPa: 0.5031
$\Delta p'$ , MPa: 0.0377

$t$: 1.00 h
$\sqrt t$ , √h: 1.00
$\Delta p$ , MPa: 0.5970
$\Delta p'$ , MPa: 0.0553

$t$: 2.00 h
$\sqrt t$ , √h: 1.41
$\Delta p$ , MPa: 0.6419
$\Delta p'$ , MPa: 0.0775

$t$: 6.31 h
$\sqrt t$ , √h: 2.51
$\Delta p$ , MPa: 0.7615
$\Delta p'$ , MPa: 0.1380

$t$: 10.0 h
$\sqrt t$ , √h: 3.16
$\Delta p$ , MPa: 0.8324
$\Delta p'$ , MPa: 0.1753

$t$: 20.0 h
$\sqrt t$ , √h: 4.47
$\Delta p$ , MPa: 0.9785
$\Delta p'$ , MPa: 0.2663

$t$: 50.1 h
$\sqrt t$ , √h: 7.08
$\Delta p$ , MPa: 1.3207
$\Delta p'$ , MPa: 0.5280

$t$: 100 h
$\sqrt t$ , √h: 10.00
$\Delta p$ , MPa: 1.7898
$\Delta p'$ , MPa: 0.8742

$t$: 200 h
$\sqrt t$ , √h: 14.13
$\Delta p$ , MPa: 2.5474
$\Delta p'$ , MPa: 1.3703

$t$: 398 h
$\sqrt t$ , √h: 19.95
$\Delta p$ , MPa: 3.7065
$\Delta p'$ , MPa: 2.0535

$t$: 631 h
$\sqrt t$ , √h: 25.12
$\Delta p$ , MPa: 4.7740
$\Delta p'$ , MPa: —

$t$	$\sqrt t$ , √h	$\Delta p$ , MPa	$\Delta p'$ , MPa
36 s	0.10	0.4148	0.0398
6 min	0.32	0.5031	0.0377
1.00 h	1.00	0.5970	0.0553
2.00 h	1.41	0.6419	0.0775
6.31 h	2.51	0.7615	0.1380
10.0 h	3.16	0.8324	0.1753
20.0 h	4.47	0.9785	0.2663
50.1 h	7.08	1.3207	0.5280
100 h	10.00	1.7898	0.8742
200 h	14.13	2.5474	1.3703
398 h	19.95	3.7065	2.0535
631 h	25.12	4.7740	—

$\Delta p = p_i - p_{wf}$ , $p_i$ = 24 MPa. The data are calculated from a model of a homogeneous reservoir with wellbore storage, skin and three sealing boundaries (two walls and a closed end), and rounded to 0.0001 MPa. The parameters of the model are given in the example. The derivative is calculated from all 59 readings (10 per cycle, where a cycle is a tenfold range of time), with a differentiation interval of 0.2 cycle (each value comes from the readings 0.2 cycle before and after it). The last two readings have no derivative.

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

m_e

, the linear-flow line of the open channel

\tfrac12m_L\sqrt t

, and the line twice as high,

m_L\sqrt t

CalculatedThe same record in the coordinates of

\Delta p

against

\sqrt t

(points). The first line is drawn through the readings at 2 and 6.3 h (the linear flow of the open channel). The second line, with twice the slope, is drawn through the last reading. Their intersection is marked. The light dashed line is the asymptote of the model: the straight line that the model curve approaches at very late times. The record gets close to it only after very long production. The late points still lie above it, so the line through the last reading intersects the first one earlier, at

\sqrt{t_x}

= 10.8 instead of 12.9.

Worked examplePermeability, channel width and distance to the closed end

Given (assumed data)

Well and formation: $q$ = 150 m³/d; $B$ = 1.2; $\mu$ = 0.8 mPa·s; $h$ = 12 m; $\phi$ = 0.22; $c_t = 1.4\cdot10^{-3}$ MPa⁻¹; $r_w$ = 0.1 m
Derivative plateau: average from 0.1 to 0.2 h: $m_e$ = 0.03759 MPa; reading at 0.1 h: $\Delta p$ = 0.5031 MPa
Linear flow of the open channel: readings 1.995 h — 0.6419 MPa, 6.310 h — 0.7615 MPa; on this stretch the derivative divided by $\sqrt t$ averages 0.05489 MPa/√h
Late data: derivative at 100 h — 0.8742 MPa; last reading at 631 h — 4.7740 MPa

Solution

Flow capacity: $qB\mu = 150\cdot1.2\cdot0.8 = 144$ ; $kh = 0.933\cdot144/0.03759 = 3574$ mD·m, and $k = 297.8$ mD. Skin: $\ln\dfrac{kt}{\phi\mu c_tr_w^2} = 16.308$ at $t$ = 0.1 h; $S = \tfrac12\,(0.5031/0.03759 - 16.308 + 4.831) = 0.95$ (Diagnosis with the pressure derivative).
Hydraulic diffusivity: $\eta = 3.553\cdot10^{-3}\cdot297.8/(0.22\cdot0.8\cdot1.4\cdot10^{-3}) = 4294$ m²/h.
First line of $\Delta p$ against $\sqrt t$ : $m_L = (0.7615 - 0.6419)/(2.5119 - 1.4125) = 0.10879$ MPa/√h. Width of the channel: $\sqrt{\mu/(k\phi c_t)} = 2.9533$ ; $W = 0.394\cdot180/(12\cdot0.10879)\cdot2.9533 = 160$ m (channel width).
The closed end from a point of the derivative: we continue the line of the open channel, drawn along the derivative itself (this way the error of numerical differentiation cancels). At 100 h it is $0.05489\cdot\sqrt{100}$ ; $r = 0.8742/(0.05489\cdot10) = 1.593$ ; $\ln[1/(1.593 - 1)] = 0.523$ ; $L_3 = \sqrt{4294\cdot100\cdot0.523} = 474$ m.
From the intersection of the lines: the intercept of the first line is $b_1 = 0.6419 - 0.10879\cdot1.4125 = 0.4882$ MPa. The line with slope $2m_L$ through the last reading has $b_2 = 4.7740 - 2\cdot0.10879\cdot25.119 = -0.6914$ MPa. The intersection is at $\sqrt{t_x} = (b_1 - b_2)/m_L = 10.843$ , $t_x = 117.6$ h; $L_x = 0.0336\sqrt{297.8\cdot117.6/(0.22\cdot0.8\cdot1.4\cdot10^{-3})} = 401$ m.
Correction for the slow approach of the late line: $X = \sqrt{\pi\eta t_{end}} = \sqrt{\pi\cdot4294\cdot631} = 2918$ m; $L_3 = \tfrac{X}{2}\left(1 - \sqrt{1 - 4L_x/X}\right) = 480$ m.

Answer:

k \approx 297.8

mD, skin 0.95, and a channel about 160 m wide. The closed end is 474 m away by the point of the derivative, and 480 m by the intersection of the lines with the correction (401 m without the correction).

What it means. The record was calculated from a model with $k$ = 300 mD, $S$ = 1, a channel 160 m wide and a closed end at 480 m. The permeability comes out about 1% too low, because the plateau is still raised slightly by the tail of wellbore storage. The point of the derivative gave the distance to the closed end with an error of about 1% already at 100 h. The intersection of the lines without the correction made it too small by 16%, because in 26 days the late points have not yet settled on the asymptote. If we had taken the channel for an open one and used the slope of the last points, the width would have come out about 80 m, almost half. The record does not say on which side of the well the closed end lies. Seismic data and the geological model give that. The data in this example are assumed for illustration.

A well right next to a boundary: a doubled rate

If a well stands a few metres from a fault, its image almost coincides with the well itself. As soon as the disturbance has covered the distance to the boundary (a time of about $\delta^2/\eta$ , seconds or minutes), the two work as one well with a doubled rate. Fluid comes from one side only, and the resistance to flow is twice as large (hemiradial flow: flow to the well through half a circle). This lets us reduce systems with the well at a wall to simpler ones, without a new type curve. A well at one wall of a channel of width $W$ is a well with a doubled rate in the middle of an open channel of width $2W$ . A well at the closed end of a channel is a well with a doubled rate in an open channel of the same width. A well in a corner is a well with four times the rate in an infinite reservoir.

A well at a sealing boundary as a well with a doubled rate

p_{wD} = 2\,p_D^{*}(t_D) - \ln\frac{2\delta}{r_w} + S

where:

$p_{wD}$ dimensionless drop of the bottomhole pressure of the well at the boundary, —
$p_D^{*}(t_D)$ dimensionless solution without skin for the same system without the nearest boundary (all the other boundaries and their images stay), —
$\delta$ distance from the axis of the well to the nearest boundary, m
$S$ skin of the well: it enters once, because it refers to the real rate, —

Units: dimensionless

Holds when

the boundary is straight and sealing; $\delta$ is small compared with the other dimensions of the system
the time is longer than a few $\delta^2/\eta$ : the image at distance $2\delta$ has already merged with the well (the derivation is in the advanced section)

The doubled rate doubles the whole pressure drop in the formation, but not the skin. The term $-\ln(2\delta/r_w)$ corrects for the fact that the image is not in the well itself but at $2\delta$ from it. The equivalent well with the doubled rate has a radius $\sqrt{2\delta r_w}$ , the geometric mean (the square root of the product) of the well radius and the distance to the image. This is the same correction as in the formula for hemiradial flow.

Here is a trap. A well at the closed end of a channel first gives a plateau $2m_e$ , and then linear flow with a doubled slope $2m_L$ . If we take it for an ordinary well in an open channel, the plateau gives a flow capacity that is half the true one, and the channel width from the slope comes out $\sqrt2$ times too small. The slope is twice as large, but $\sqrt{1/k}$ in the width formula is also $\sqrt2$ times larger. Meanwhile the record looks perfect. Only a disagreement with the core, the neighbouring wells and the geology gives it away.

Pressure build-up in a semi-infinite system

A pressure build-up (the rise of pressure after the well is shut in) is a superposition of the start-up and the shut-in: the effects of the two events add up. Each image is also “shut in” together with the well (Pressure build-up analysis). Here is the formula for one fault (build-up near a fault) generalised to any set of images:

Build-up of a well in a system with images

p_i - p_{ws} = m_e\left[\ln\frac{t_p + \Delta t}{\Delta t} + \sum_j\left(E_1\!\left(\frac{d_j^2}{4\eta\,(t_p + \Delta t)}\right) - E_1\!\left(\frac{d_j^2}{4\eta\,\Delta t}\right)\right)\right]

where:

$p_{ws}$ bottomhole pressure $\Delta t$ after shut-in, Pa
$t_p,\ \Delta t$ producing time at constant rate, and time after shut-in, s
$d_j$ distances to the images, m
$m_e,\ \eta$ level of the derivative plateau, and hydraulic diffusivity, Pa, m²/s

Units: SI; natural logarithm

Holds when

before start-up the pressure in the formation was $p_i$ ; the rate is constant during $t_p$ ; for the well itself the logarithmic approximation holds, and wellbore storage is over
for channels the sum is infinite but converges: distant images give equal terms at $t_p + \Delta t$ and at $\Delta t$ , and the difference vanishes

The result now depends on two dimensionless times, $\eta t_p/L^2$ and $\eta\Delta t/L^2$ , not on one as for a pressure drawdown test (the record while the well produces). A boundary shows on a build-up only if the disturbance has had time to reach it during the producing time, or has time to reach it during the recording. As $\Delta t \to \infty$ , all the terms go to zero, and $p_{ws} \to p_i$ . So a semi-infinite system builds up to the initial pressure. The continuation of the early Horner line (the build-up plotted against $\lg[(t_p + \Delta t)/\Delta t]$ ) gives an extrapolated pressure $p^*$ that is below $p_i$ by $m_e\sum_j E_1\big(d_j^2/(4\eta t_p)\big)$ .

For a channel, the late points of the build-up lie on a straight line in the coordinates $p_{ws}$ against $(\sqrt{t_p + \Delta t} - \sqrt{\Delta t})$ (build-up in a channel). For a channel closed at one end, the slope is twice as large, $2m_L$ , once both $t_p$ and $\Delta t$ are already in the stage of doubled linear flow. We calculate the derivative of such a build-up with the producing time allowed for (Effect of producing time on build-up analysis). More often, however, we see the closed end on a build-up after a short producing time, as in a drill stem test (a test made through the drill pipe). Then the disturbance has hardly reached the closed end during the flow period. The trace of the closed end is a small lag of the pressure behind the build-up of an open channel, and we can estimate its largest size in advance:

Largest trace of the closed end of a channel on a build-up after a short producing time

\delta p_{\max} = \frac{\e^{-1/2}}{2\sqrt{2\pi}}\cdot\frac{qB\,t_p}{\phi c_t h\,WL_3} = 0.121\,\frac{qB\,t_p}{\phi c_t h\,WL_3},\qquad \Delta t_{\max} \approx \frac{2L_3^2}{\eta}

where:

$\delta p_{\max}$ largest lag of the build-up pressure behind the build-up of the same channel without the closed end, Pa
$qB\,t_p$ volume produced during the producing time, at reservoir conditions, m³
$\phi,\ c_t,\ h$ porosity, total compressibility, thickness, —, 1/Pa, m
$W,\ L_3$ width of the channel and distance to the closed end, m
$\Delta t_{\max}$ time after shut-in at which the lag is largest, s

Units: SI

Holds when

during the producing time the closed end has hardly been felt: $\eta t_p \ll L_3^2$ (in practice it is enough that $L_3^2 > 3\eta t_p$ )
the closed end is farther away than the width of the channel, and by the time $\Delta t_{\max}$ the channel has long been in linear flow
a homogeneous formation; no gauge drift and no interference from neighbouring wells

Compare this with a single fault in a radial reservoir. There the trace falls as $1/L^2$ , and here as $1/(WL_3)$ . In a channel the disturbance does not spread out over a plane, so the closed end acts on the pressure much more strongly than a fault at the same distance. But the signal comes just as late, because the time grows as $L_3^2$ (the derivation is in the advanced section).

In other unit systems (1)

Metric oilfield units

\delta p_{\max} = \frac{qB\,t_p}{198\,\phi c_t h\,WL_3}

$q$ in m³/d, $t_p$ in h, $c_t$ in 1/MPa, $h$ , $W$ and $L_3$ in m; $\delta p_{\max}$ in MPa.

$198 = 24/0.121$ (24 is the number of hours in a day).

CalculatedCalculated lag of the build-up pressure behind the build-up of an open channel, after a flow period of 6, 12 and 24 h in the river-channel body of the example above (width 160 m, closed end at 480 m). The points are the estimate

\delta p_{\max}

from the formula at

\Delta t = 2L_3^2/\eta

. The time scale is logarithmic.

Worked exampleWill a drill stem test see the closed end of the river-channel body?

Given (assumed data)

Test: flow of 12 h at 150 m³/d ( $B$ = 1.2), then a build-up; the gauge resolves 0.5 kPa
Formation: as in the example above: $k$ = 300 mD, $\phi$ = 0.22, $\mu$ = 0.8 mPa·s, $c_t = 1.4\cdot10^{-3}$ MPa⁻¹, $h$ = 12 m
Channel: width 160 m, the closed end 480 m from the well

Solution

Hydraulic diffusivity $\eta = 4326$ m²/h; $\eta t_p = 51{,}912$ m² against $L_3^2$ = 230,400 m². During the flow period the closed end has hardly been felt ( $\eta t_p/L_3^2 = 0.23$ ).
Storage capacity (capacitance) of the part of the channel up to the closed end, that is, the volume of fluid that it gives out when the pressure falls by 1 MPa: $\phi c_thWL_3 = 0.22\cdot1.4\cdot10^{-3}\cdot12\cdot160\cdot480 = 283.85$ m³/MPa; $qBt_p = 150\cdot1.2\cdot12 = 2160$ (m³/d·h).
Largest trace of the closed end: $\delta p_{\max} = 2160/(198\cdot283.85) = 0.0384$ MPa, that is 38.4 kPa, at about $2L_3^2/\eta = 107$ h after shut-in.
A calculation with the full sum of images gives 38.4 kPa at 101 h. The formula is right.
For comparison: a fault at the same 480 m in a radial reservoir would leave a trace $qBt_p/(820\,\phi c_thL^2) = 3.1$ kPa (detection limit of a fault). That is about 12 times smaller.

Answer: A closed end at 480 m will bend the build-up by about 38.4 kPa. That is tens of times more than the smallest change that the gauge can detect. But the maximum comes only 107 h after shut-in.

What it means. The signal of the closed end is large because the whole production comes from a narrow strip. The difficulty is something else: time. To see the maximum, the build-up must last four or five days, almost ten times as long as the flow period. A shorter build-up shows only the start of the lag, and we must tell it from the producing-time effect and from noise. It helps to extend the flow period, or to repeat the test during an extended well test. The data in this example are assumed for illustration.

What the record cannot tell apart

Directions. The record depends only on the distances to the images. It does not say on which side the closed end lies, or in which direction the channel stretches. Seismic data, the geological model and the neighbouring wells give that.
A closed end or a worsening formation. If on one side of the well the channel narrows further, gets thinner or loses permeability, the $\sqrt t$ slope also grows. At late times it grows by a factor of $2/(1 + \sqrt{k_2/k})$ , if the permeability on that side is $k_2$ and the porosity and the cross-section are the same (one-dimensional flow to the well from two sides). With half the permeability, the slope is about 17% larger. With 10 times lower permeability it is 1.5 times larger, and with 100 times lower it is 1.8 times larger. It is easy to take such a record for a doubling from a closed end that is not yet complete. The length of the record, the geology and the neighbouring wells help to tell them apart.
A closed end that lets fluid through. If the cross-fault lets some fluid through, the derivative rises towards the doubled line and then comes back to the line of the open channel (Partially communicating faults).
A closed block. A long closed block also gives slope ½, before it reaches slope 1. The second closed end shows as a late rise with slope 1, and as a build-up that goes not to $p_i$ but to an average pressure (Reservoir limit test).
An unfinished transition. On the derivative, the closed end follows the same law as a fault. But the late line of $\Delta p$ against $\sqrt t$ approaches its asymptote slowly. So without a correction, the distance from the second line comes out too small, by a fraction of about $L_3/\sqrt{\pi\eta t}$ . To keep the error below 5%, $\eta t$ must exceed $L_3^2$ by more than a hundred times. A point of the derivative and a model fit are more reliable.
Straight boundaries are a model. River-channel bodies bend, and they change in width and thickness. The distances from the simple model are estimates, and we check them against the map.

Terms and symbols

Semi-infinite system: A reservoir with sealing boundaries on some sides but not on all: a half-plane, a corner, a channel, or a channel closed at one end. It has no pseudo-steady-state flow.
Closed-end channel: A reservoir between two parallel boundaries, with a third boundary across one end (a U-shaped fault system). Its late linear flow is twice as strong as in an open channel: the slope is doubled.
Characteristic length $L$: A size to which all the distances to the images are proportional. The late part of the solution depends only on $\eta t/L^2$ .
Product solution: In a rectangular system, the derivative divided by the plateau equals the product of the factors for the two directions.
Direction factor: The sum of $\exp(-\xi^2/(4\eta t))$ over the one-dimensional images along one direction. It is 1 if the direction is open, $1 + \e^{-L^2/(\eta t)}$ for one boundary, and the strip factor for two boundaries.
Closed end of the channel, $L_3$: The distance from the well to the closed end of the channel. We find it from a point of the derivative or from the intersection of the lines of $\Delta p$ against $\sqrt t$ .
Doubled rate: A technique for a well right next to a boundary. Its image merges with it, so it works as a well with a doubled rate in a system without that boundary. The skin is not doubled.

Common mistakes and misconceptions

Wrong

“If the derivative runs with slope ½ after the plateau, it is an open channel, and we can calculate the width from the last points.”

In fact

First check whether the derivative has moved onto a parallel line twice as high. In the example of this lesson, the width from the late points would come out at about 80 m instead of 160.

Wrong

“The second line on the plot of $\Delta p$ against $\sqrt t$ is the last points of the record.”

In fact

The late line at a closed end forms slowly: the excess above it falls as $1/\sqrt t$ . In the example of the lesson, the line with slope $2m_L$ through the last reading gave the closed end at 401 m instead of 480. A correction, or a point of the derivative, fixes this.

Wrong

“The pressure after shut-in did not reach the initial pressure in a week, so the block is closed and is being depleted.”

In fact

In a channel closed at one end, the pressure builds up to $p_i$ , but slowly: the closed end delays it by a time of about $2L_3^2/\eta$ . Only a build-up that goes to an average pressure below $p_i$ , and slope 1 during production, prove that the block is closed.

Wrong

“For a well next to a fault, both the rate and the skin are doubled.”

In fact

The image doubles only the formation part of the pressure drop. The skin is tied to the real rate and enters once. If the model doubles the skin too, the skin fitted to the record comes out half the true one, and the damage is underestimated.

Wrong

“The closed end is far away, so a short test will not see it, just as it will not see a distant fault.”

In fact

In a channel the trace of the closed end falls as $1/(WL_3)$ , not as $1/L^2$ . In the example of the lesson, a 12-hour flow would give a trace of about 38 kPa on the build-up. That is 12 times more than a fault at the same distance in a radial reservoir. The limit is time: the maximum comes after $2L_3^2/\eta$ .

In depth: advanced detail

Derivations: the product solution, the late line, the doubled rate and the trace of the closed end

Product solution. We get an image in a rectangular system by reflecting the well several times in the boundaries along $x$ and along $y$ . Its shift is a pair $(\xi_i, \zeta_l)$ , where $\xi_i$ runs over the one-dimensional images along $x$ (including 0), and $\zeta_l$ over those along $y$ . Every pair occurs exactly once, and $\sum_{i,l}\exp\big(-(\xi_i^2 + \zeta_l^2)/(4\eta t)\big) = \sum_i\exp(-\xi_i^2/(4\eta t))\cdot\sum_l\exp(-\zeta_l^2/(4\eta t))$ . The derivative against $\ln t$ of the line-source solution with an image at distance $d$ equals $m_e\e^{-d^2/(4\eta t)}$ . This gives the rule. For a strip of width $W$ , the series over $\zeta_l$ (step $2W$ ) turns, by the Poisson summation formula, into a series in cosines. Its main term is $\sqrt{4\pi\eta t}/W$ . So the strip factor grows like the square root of time.

The late line and the intersection. When $\eta t \gg W^2$ , the row of images in the walls acts as a plane source. The one-dimensional solution for a row at distance $2L_3$ then gives a contribution $\tfrac{qB\mu}{khW}\big[\sqrt{\eta t/\pi}\,\e^{-a} - L_3\,\mathrm{erfc}\sqrt a\big]$ , with $a = L_3^2/(\eta t)$ . For small $a$ , this is $m_L\sqrt t - \tfrac{qB\mu}{khW}\,L_3 + \tfrac{qB\mu}{khW}\cdot\tfrac{L_3^2}{\sqrt{\pi\eta t}} + \dots$ — the asymptote of the main section, and the slowly falling excess above it. We equate the lines $m_L\sqrt t + b$ and $2m_L\sqrt t + b - \tfrac{qB\mu}{khW}L_3$ , and use $m_L = \tfrac{qB\mu}{khW}\sqrt{\eta/\pi}$ . This gives $\sqrt{t_x} = L_3\sqrt{\pi/\eta}$ . The line with slope $2m_L$ through a reading at time $t_{end}$ is raised by the excess, and the intersection gives $L_x = L_3 - L_3^2/X$ , with $X = \sqrt{\pi\eta t_{end}}$ . The solution of this quadratic equation is the correction in the example.

The doubled rate. The image of the well in the nearest boundary stands at distance $2\delta$ . For $t_D \gg \delta_D^2$ it gives $\tfrac12E_1(\delta_D^2/t_D) \approx \tfrac12\big[\ln(t_D/\delta_D^2) - 0.5772\big]$ . Together with the well, this is $\ln t_D + 0.809 - \ln(2\delta_D)$ , that is, $2\cdot\tfrac12\big[\ln\big(t_D/(2\delta_D)\big) + 0.809\big]$ : twice the solution for a well of radius $\sqrt{2\delta r_w}$ . All the other images also stand in pairs at almost coinciding points, and they are doubled.

The trace of the closed end on a build-up. On a build-up, the contribution of the mirror row is the difference $g(t_p + \Delta t) - g(\Delta t)$ of the one-dimensional solution from the second paragraph. For $t_p \ll \Delta t$ it is close to $t_p\,\dd g/\dd t$ , and $\dd g/\dd t = \tfrac{qB\mu}{khW}\cdot\tfrac12\sqrt{\eta/(\pi t)}\,\e^{-L_3^2/(\eta t)}$ . The function $t^{-1/2}\e^{-L_3^2/(\eta t)}$ has its maximum at $t = 2L_3^2/\eta$ . Substitution gives $\delta p_{\max} = \tfrac{\e^{-1/2}}{2\sqrt{2\pi}}\cdot\tfrac{qB\,t_p}{\phi c_thWL_3}$ , because $\tfrac{qB\mu\eta}{kh} = \tfrac{qB}{\phi c_th}$ .

Connections to other topics

A sealing fault near the well: the method of images — one fault: the image, the doubling of the slope, the derivative step, the detection limit on a build-up.
Several boundaries: parallel faults and wedge-shaped blocks — the channel and the wedge: linear flow, the width of the channel, the geometric skin, the level $360^\circ/\theta$ .
Partially communicating faults — a closed end or a wall that lets some fluid through.
Dimensionless well solutions: skin, radius of investigation and units — dimensionless solutions, the radius of investigation, and when a boundary is visible on the derivative.
Superposition principle — the method of images, and adding solutions in time and in space.
Diagnosis with the pressure derivative — the diagnostic plot, the plateau and the slopes of the derivative.
Reservoir limit test — closed blocks: slope 1 and the pore volume.
Build-up in a bounded reservoir and average reservoir pressure — why a closed block builds up to the average pressure and not to the initial one.
Continental environments: aeolian and fluvial — river-channel bodies: width, length, thinning out.
Faults — the geology of the faults that form channels and closed ends.

Check yourself

1
After the radial-flow plateau, the derivative runs with slope ½ for about a cycle, then rises and settles on a parallel line about twice as high. Which system is the most likely?
1. A.A constant-pressure boundary
2. B.A channel closed at one end
3. C.A right-angle corner between faults
4. D.A single sealing fault
Show the answer
Correct answer: B. A channel closed at one end
Slope ½ is linear flow in a channel. The move onto a line twice as high, with the same slope, is the end factor $1 + \e^{-L_3^2/(\eta t)}$ growing to 2. A fault and a corner give horizontal plateaus, and a constant-pressure boundary pulls the derivative down.
2
In the record of a well in a channel, at 200 h the derivative is 1.42 times higher than the continuation of the linear-flow line of the open channel; $\eta$ = 2000 m²/h. How far away is the closed end?
Show the answer
$\ln[1/(1.42 - 1)] = \ln 2.381 = 0.868$ . $L_3 = \sqrt{2000\cdot200\cdot0.868} = \sqrt{347{,}200} \approx 589$ m. The ratio is already clearly greater than one, so the estimate holds up against noise of a few per cent in the derivative.
3
In the late part of a record, where the derivative is already on the doubled line, the $\sqrt t$ slope is 0.30 MPa/√h. The interpreter calculated the channel width, taking the channel for an open one, and got 110 m. What is the real width?
1. A.55 m
2. B.About 155 m
3. C.110 m
4. D.About 220 m
Show the answer
Correct answer: D. About 220 m
On the doubled line the slope equals $2m_L$ , and the width is inversely proportional to the slope. The true $m_L$ is 0.15 MPa/√h, and the width is twice as large, about 220 m (if the permeability from the plateau is right).
4
A well was drilled 8 m from the closed end of a channel, and the first plateau is hidden by wellbore storage. The interpreter processed the record as a well in an open channel. By what factors will the permeability and the channel width be wrong?
Show the answer
A well at a closed end works like a well with a doubled rate in an open channel. The visible plateau is $2m_e$ , so the permeability comes out half of the true one. The $\sqrt t$ slope is also doubled, and the width formula contains $\sqrt{1/k}$ , which is $\sqrt2$ times larger: $W_{\text{apparent}} = W\cdot\tfrac12\cdot\sqrt2 \approx 0.71\,W$ . The skin will be distorted too. Core data, the neighbouring wells and the geology help to reveal the error.
5
After a long producing time, a well in a channel closed at one end was shut in. To what pressure does the build-up tend?
1. A.To the initial reservoir pressure $p_i$
2. B.To the pressure $p^*$ of the first Horner line
3. C.To the average pressure of the part of the channel up to the closed end
4. D.To the bottomhole pressure at the moment of shut-in
Show the answer
Correct answer: A. To the initial reservoir pressure $p_i$
The channel is open on one side, the volume covered by the disturbance grows without limit, and as $\Delta t \to \infty$ all the terms of the superposition vanish: $p_{ws} \to p_i$ . It happens more slowly than in an open channel, because the closed end delays the equalisation. The $p^*$ of the first line is below $p_i$ by the contribution of the images.
6
A test: 8 h of flow at 100 m³/d ( $B$ = 1.3), $\phi$ = 0.18, $c_t = 1.6\cdot10^{-3}$ MPa⁻¹, $h$ = 10 m; a channel 200 m wide is closed 600 m from the well; $\eta$ = 1500 m²/h. What is the largest trace of the closed end on the build-up, and when do we expect it?
Show the answer
$qBt_p = 100\cdot1.3\cdot8 = 1040$ ; $\phi c_thWL_3 = 0.18\cdot1.6\cdot10^{-3}\cdot10\cdot200\cdot600 = 345.6$ m³/MPa. $\delta p_{\max} = 1040/(198\cdot345.6) = 0.0152$ MPa, about 15 kPa. Time: $2L_3^2/\eta = 2\cdot360{,}000/1500 = 480$ h, which is 20 days. The condition $\eta t_p = 12{,}000 \ll L_3^2 = 360{,}000$ holds. The signal is noticeable, but a build-up of such a length after an 8-hour flow is rare. It is more likely that the closed end will be seen during an extended well test.

Updated: 19 October 2026