Testing in a developed reservoir: Slider's method

Quick refresher

Build-up = trend + response to the shut-in. The trend is the pressure that the well would have had if it had kept producing. The response is the pressure rise from an “injection well” with the rate $q$ , started at the same point at the moment of shut-in.
The response to the shut-in is the drawdown test of the well under test in the real formation, with its boundaries but without the neighbours. While the rates are constant, the work of the neighbours is all in the trend.
If the accumulation is in pseudo-steady-state flow and the rates are constant, the trend is a straight line: the pressure everywhere falls at the same speed, the decline rate $m^* = \sum qB/(24\,c_tV_p)$ MPa/h.
Slider's method: $\Delta p_{DD} = p_{ws} - p_{wf}(\Delta t = 0) + m^*\Delta t$ . We work with this quantity as with a drawdown test: the slope gives $kh$ , the point at 1 h gives the skin, and the derivative shows the flow regimes (radial flow, linear flow and so on) and the boundaries.
The correction $m^*\Delta t$ grows in proportion to time. It becomes equal to the level of the derivative plateau at $\Delta t_{DA_w} = 1/(4\pi) \approx 0.08$ , the time referred to the drainage area of the well. Without the correction, the derivative falls to zero by this time, and the pressure passes a maximum (later, if the response has already felt the boundaries).
The linear continuation of the trend is right only if the rates of the neighbours have not changed for a long time and do not change during the build-up. Otherwise we need superposition with the history of their rates, or a model.

Key relationships

p_{ws}(\Delta t) = p_{tr}(\Delta t) + \dfrac{qB\mu}{2\pi kh}\big[p_D(\Delta t_D) + S\big]

a build-up as the sum of the trend and the response to the shut-in

\Delta p_{DD} = p_{ws}(\Delta t) - p_{wf}(\Delta t = 0) + m^*\Delta t

Slider's method: a linear trend, $m^* > 0$ for a falling pressure

m^* = \dfrac{\sum q_jB}{24\,c_tV_p}

pressure decline rate of the block in pseudo-steady-state flow: MPa/h, m³/d, 1/MPa, m³

\dfrac{m^*\Delta t}{m_e} = 4\pi\,\Delta t_{DA_w}

the correction as a fraction of the plateau level; $A_w$ is the drainage area of the well

kh = 2.149\,\dfrac{qB\mu}{m},\qquad S = 1.151\left[\dfrac{\Delta p_{DD,1\text{h}}}{m} - \lg\dfrac{k}{\phi\mu c_tr_w^2} + 2.098\right]

flow capacity and skin from the corrected build-up (metric oilfield units)

Common pitfalls

Analysing the build-up of a well in a developed accumulation by the Horner method with the actual producing time, and taking $p^*$ or the maximum of the build-up for the reservoir pressure.
Taking a fall of the derivative at the end of a build-up for support from outside, without allowing for the pressure trend of the accumulation.
Estimating $m^*$ from the last hours before shut-in, when the rates of the neighbours changed recently.
Mixing up the sign: for a falling pressure the correction $m^*\Delta t$ is added to the pressure rise, and for a rising pressure (injection stronger than production) it is subtracted.
Thinking that the correction is needed only for very long build-ups: in the example of the lesson, after one day it is already a quarter of the plateau level.

After this lesson you can

explain why in a developed accumulation the pressure in a shut-in well, after rising, starts to fall, and how a pressure trend distorts the semilog plot and the derivative
write a build-up as the sum of the trend and the response to the shut-in, and get from it the drawdown test of the well under test
estimate the trend $m^*$ from the record before shut-in, from the material balance of the block or from reservoir pressure measurements, and calculate when the correction can be ignored
find the permeability, the skin and the signs of boundaries from the corrected build-up in metric and oilfield units
name the conditions in which the linear continuation of the trend does not work, and what to do then

A build-up in a developed accumulation: what changes

The Horner equation is derived for a well that started to produce in a formation with the same pressure everywhere, and that before shut-in “saw” only an infinite formation (Pressure build-up analysis). For an exploration well and the first production wells this is true. But when the accumulation is developed, we test wells that have produced for months and years. The disturbance from each of them reached the boundaries of its drainage area (the part of the formation from which the well draws fluid) long ago. The neighbouring wells keep taking oil and change the pressure around each other (well interference), and the pressure in the whole block falls. Shutting in one well does not stop this process.

So the build-up of such a well looks different. At first the pressure rises as usual, because the disturbance from the shut-in has not gone far. Then the rise slows down faster than it “should”, the pressure passes a maximum and starts to fall: the neighbours take oil from the area of the shut-in well too. On a Horner plot the points go down away from the line, and on the diagnostic plot (a log–log plot of the pressure change and its derivative) the derivative falls, as at a constant-pressure boundary (a boundary that holds the pressure constant). The late points of such a build-up no longer tell us the average reservoir pressure of the area: the neighbours change the pressure there too.

SchematicPlan of the tectonic block of the lesson's examples: four producing wells in a closed block of 2000 × 1000 m (faults on all sides). Well 1 is shut in for a build-up, and wells 2–4 keep producing. The numbers are the rates. The distances from well 1 to the south and west faults are 250 and 500 m.

Wells of the block (assumed data)

Well: 1
$x$ ; $y$ , m: 500; 250
Rate, m³/d: 40
Note: well under test, $S$ = 1.5

Well: 2
$x$ ; $y$ , m: 1500; 250
Rate, m³/d: 50
Note: produces without breaks

Well: 3
$x$ ; $y$ , m: 500; 750
Rate, m³/d: 35
Note: produces without breaks

Well: 4
$x$ ; $y$ , m: 1500; 750
Rate, m³/d: 55
Note: produces without breaks

Well	$x$ ; $y$ , m	Rate, m³/d	Note
1	500; 250	40	well under test, $S$ = 1.5
2	1500; 250	50	produces without breaks
3	500; 750	35	produces without breaks
4	1500; 750	55	produces without breaks

The coordinates are counted from the west and south faults. Formation: $h$ = 10 m, $\phi$ = 0.20, $k$ = 40 mD, $\mu$ = 1.2 mPa·s, $c_t = 1.5\cdot10^{-3}$ MPa⁻¹, $B$ = 1.25, $r_w$ = 0.108 m; the initial pressure is 29.0 MPa. All the wells produce at constant rates for 5840 h (about eight months), and the block has reached pseudo-steady-state flow.

CalculatedCalculated bottomhole pressure of well 1 for two days before shut-in and after it (the points are the readings of the table below), and the continuation of the trend before shut-in (dashed). The build-up rises, passes a maximum near 110 h, and then falls together with the trend.

What the trend does to the ordinary processing of a build-up

Method: Horner or MDH (Miller–Dyes–Hutchinson) line
What we get: the slope depends on the chosen stretch; the permeability is overstated
Why: the continuing fall is mixed into the pressure rise

Method: The continuation of the line, $p^*$
What we get: neither the initial nor the average pressure
Why: the Horner equation knows nothing about boundaries or neighbours

Method: The derivative at the end of the record
What we get: falls, then goes into the negative
Why: $m^*\Delta t$ is subtracted from the derivative of the response

Method: The maximum of the build-up and the last reading
What we get: the last reading is below the maximum; neither is the average pressure
Why: the neighbours keep producing during the build-up

Method	What we get	Why
Horner or MDH (Miller–Dyes–Hutchinson) line	the slope depends on the chosen stretch; the permeability is overstated	the continuing fall is mixed into the pressure rise
The continuation of the line, $p^*$	neither the initial nor the average pressure	the Horner equation knows nothing about boundaries or neighbours
The derivative at the end of the record	falls, then goes into the negative	$m^*\Delta t$ is subtracted from the derivative of the response
The maximum of the build-up and the last reading	the last reading is below the maximum; neither is the average pressure	the neighbours keep producing during the build-up

The trend and the response to the shut-in

As with any build-up, we split the record by superposition (adding up the effects of separate events). A shut-in is a start-up of an injection well at the same point with the same rate. Everything that existed before it continues to act: the work of the well itself at the previous rate, the work of the neighbours, and the boundaries. The sum of everything except the imaginary injection well is the pressure that the well would have had if it had kept producing. It is called the trend, and is denoted $p_{tr}(\Delta t)$ .

A build-up as the sum of the trend and the response to the shut-in

p_{ws}(\Delta t) = p_{tr}(\Delta t) + \frac{qB\mu}{2\pi kh}\Big[p_D(\Delta t_D) + S\Big]

where:

$p_{ws}(\Delta t)$ pressure in the shut-in well a time $\Delta t$ after shut-in, Pa
$p_{tr}(\Delta t)$ the trend: the bottomhole pressure that the well would have had at the same moment if it had kept producing at the rate $q$ ; at the moment of shut-in it equals $p_{wf}(\Delta t = 0)$ , Pa
$p_D(\Delta t_D)$ dimensionless pressure drop at the wall of a well that was started at a constant rate in the same formation, with its boundaries but without the neighbours; without skin, —
$S$ skin of the well under test, —
$q,\ B,\ \mu,\ k,\ h$ rate before shut-in at standard conditions, formation volume factor, viscosity, permeability, thickness, m³/s, m³/m³, Pa·s, m², m

Units: SI

Holds when

the problem is linear: a single-phase slightly compressible fluid, and the properties of the formation do not depend on the pressure
the rates of the neighbours do not change during the build-up, and if they changed earlier, their history is in the trend
we leave out wellbore storage (the well acting as a small tank): the well is shut in at the bottom, or we do not use the early points

The difference $p_{ws} - p_{tr}$ is the reconstructed drawdown test of the well under test: it is as if this well alone had been started in the formation at the rate $q$ . The neighbours are not in it, because while the rates are constant their contribution is entirely in the trend. So we find from it the flow capacity, the skin and the real boundaries of the formation (faults, not the dividing lines of flow between wells). Subtracting from a build-up the continuation of what happened before shut-in is called desuperposition. For an exploration well the trend is the continuation of the logarithmic fall of pressure, and desuperposition is equivalent to the Horner method (Effect of producing time on build-up analysis, the advanced section).

What remains is to find the trend. In a developed accumulation it is given not by a formula but by the history of all the wells of the block. But there is an important special case in which it is simple.

Slider's method: a linear trend

If all the wells of the block have long produced at constant rates, the block is in pseudo-steady-state flow: the pressure at every point falls at the same speed, which the material balance of the whole block sets (Pseudo-steady-state flow and the shape factor). The trend is then a straight line, and the speed of the fall is the same for the bottomhole pressure of any well and for the average pressure:

The pressure decline rate of a block in pseudo-steady-state flow

m^* = -\frac{\dd p}{\dd t} = \frac{\sum_j q_jB}{c_tV_p}

where:

$m^*$ pressure decline rate in the block (the slope of the trend, taken with the opposite sign), Pa/s
$q_j$ rates of all the wells of the block at standard conditions; if the block has injection wells, their injection, converted to reservoir conditions, is subtracted from the sum $\sum q_jB$ , m³/s
$B$ oil formation volume factor, m³/m³
$c_t,\ V_p$ total compressibility and the pore volume of the whole connected block, 1/Pa, m³

Units: SI

Holds when

the block is closed, the fluid is single-phase, and the rates have been constant so long that the transient effects in the whole block have died out
there is no water influx (water flowing in from outside) and no gas cap (free gas above the oil)

The trend includes the production of all the wells of the block, not only the well under test. The more neighbours and the smaller the block, the steeper the trend. If injection exceeds production, $m^*$ is negative: the pressure rises.

In other unit systems (2)

Metric oilfield units

m^*\,[\text{MPa/h}] = \frac{\sum_j q_jB}{24\,c_tV_p}

$q_j$ in m³/d, $c_t$ in 1/MPa, $V_p$ in m³.

24 is the number of hours in a day.

Oilfield units

m^*\,[\text{psi/h}] = \frac{\sum_j q_jB}{4.274\,c_tV_p}

$q_j$ in STB/d, $c_t$ in 1/psi, $V_p$ in ft³.

$4.274 = 24/5.615$ : hours in a day and cubic feet in a barrel.

With a linear trend $p_{tr}(\Delta t) = p_{wf}(\Delta t = 0) - m^*\Delta t$ , the reconstructed drawdown test comes from the record by one addition. This is Slider's method:

Slider's correction: the reconstructed drawdown test

\Delta p_{DD}(\Delta t) = p_{ws}(\Delta t) - p_{wf}(\Delta t = 0) + m^*\Delta t

where:

$\Delta p_{DD}(\Delta t)$ reconstructed pressure drop: the drawdown test of the well under test, counted from the moment of shut-in, Pa
$p_{ws}(\Delta t)$ the measured pressure after shut-in, Pa
$p_{wf}(\Delta t = 0)$ bottomhole pressure at the moment of shut-in, Pa
$m^*$ pressure decline rate before shut-in (the trend); positive if the pressure was falling, Pa/s
$\Delta t$ time after shut-in, s

Units: any consistent units: pressure in MPa, $m^*$ in MPa/h and $\Delta t$ in hours

Holds when

the trend is linear: the block is in pseudo-steady-state flow, and the rates of the neighbours did not change either before the build-up or during it
$m^*$ is found from the record before shut-in or from the material balance of the block (the section below)

The first two terms are the ordinary pressure rise $\Delta p_{BU}$ , and the third returns the fall that would have gone on without the shut-in. Then we analyse $\Delta p_{DD}$ as a drawdown test. On a semilog plot (pressure against $\lg\Delta t$ ) it is a line with the slope $m = 2.149\,qB\mu/(kh)$ , and the point at 1 h gives the skin. On the diagnostic plot the derivative has the plateau $m_e = m/2.303$ and the signs of boundaries (Pressure drawdown analysis, Diagnosis with the pressure derivative).

In other unit systems (2)

Metric oilfield units

kh = 2.149\,\frac{qB\mu}{m},\qquad S = 1.151\left[\frac{\Delta p_{DD,1\text{h}}}{m} - \lg\frac{k}{\phi\mu c_tr_w^2} + 2.098\right]

$m$ is the slope of the line $\Delta p_{DD}$ against $\lg\Delta t$ , MPa per cycle; $\Delta p_{DD,1\text{h}}$ is the value on the line at 1 h, MPa; $q$ in m³/d; $\mu$ in mPa·s; $k$ in mD; $h$ and $r_w$ in m; $c_t$ in 1/MPa.

$2.149 = \ln10/(4\pi\cdot0.0853)$ , $1.151 = \ln10/2$ , $2.098 = -\lg(3.553\cdot10^{-3}) - 0.809/\ln10$ are the constants of drawdown analysis.

Oilfield units

kh = 162.6\,\frac{qB\mu}{m},\qquad S = 1.151\left[\frac{\Delta p_{DD,1\text{h}}}{m} - \lg\frac{k}{\phi\mu c_tr_w^2} + 3.23\right]

$m$ and $\Delta p_{DD,1\text{h}}$ in psi; $q$ in STB/d; $\mu$ in cP; $k$ in mD; $h$ and $r_w$ in ft; $c_t$ in 1/psi.

$162.6 = 70.6\cdot\ln10$ ; $3.23 = -\lg(2.637\cdot10^{-4}) - 0.809/\ln10$ .

How large is the correction? It is convenient to compare it with the level of the derivative plateau $m_e$ , the slope per unit of natural logarithm. In pseudo-steady-state flow, each well drains a part of the block that is proportional to its rate. The ratio then reduces to the dimensionless time referred to this part:

The correction for the trend as a fraction of the plateau level

\frac{m^*\Delta t}{m_e} = 4\pi\,\Delta t_{DA_w},\qquad \Delta t_{DA_w} = \frac{k\,\Delta t}{\phi\mu c_tA_w},\qquad A_w = A\,\frac{q}{\sum_j q_j}

where:

$m_e$ level of the derivative plateau of the well under test, $qB\mu/(4\pi kh)$ , Pa
$\Delta t_{DA_w}$ time after shut-in referred to the area of the drainage region of the well, —
$A_w$ drainage area of the well under test: the share of the area of the block that is proportional to its rate, m²
$A,\ q,\ \sum_j q_j$ area of the block, rate of the well under test, and the sum of the rates of all the wells of the block, m², m³/s, m³/s

Units: dimensionless; in metric oilfield units $\Delta t_{DA_w} = 3.553\cdot10^{-3}\,k\,\Delta t/(\phi\mu c_tA_w)$ for $k$ in mD, $\Delta t$ in hours, $\mu$ in mPa·s, $c_t$ in 1/MPa, $A_w$ in m²

Holds when

the block is in pseudo-steady-state flow, and all the wells work at constant rates in one homogeneous formation.

The correction as a fraction of the plateau does not depend on the size of the rates or on how fast the pressure falls in megapascals. It is set by the hydraulic diffusivity of the formation and the area per well, that is, by how much of “its” area the disturbance has covered since the shut-in. At $\Delta t_{DA_w} = 1/(4\pi) \approx 0.08$ it equals the plateau level. The derivative of the raw build-up (the build-up before the correction) falls to zero, and the pressure passes a maximum (if the boundaries have not yet raised the derivative of the response). We can ignore the correction as long as $4\pi\Delta t_{DA_w}$ is only a few per cent.

Worked exampleWhen the trend becomes noticeable

Given (assumed data)

Block and wells: the table at the beginning of the lesson: $A$ = 200 ha (1 ha = 10,000 m²), rates 40, 50, 35 and 55 m³/d; the well under test is well 1
Formation: $k$ = 40 mD, $\phi\mu c_t = 0.20\cdot1.2\cdot1.5\cdot10^{-3} = 3.6\cdot10^{-4}$ (mPa·s/MPa); $h$ = 10 m, $B$ = 1.25

Solution

Drainage area of well 1: $A_w = 2\cdot10^6\cdot40/180 = 444{,}444$ m², about 44.4 ha.
After one day: $\Delta t_{DA_w} = 3.553\cdot10^{-3}\cdot40\cdot24/(3.6\cdot10^{-4}\cdot444{,}444) = 0.02132$ ; $4\pi\cdot0.02132 = 0.268$ . The correction is 27% of the plateau level.
A direct check: $m^* = 180\cdot1.25/(24\cdot1.5\cdot10^{-3}\cdot4\cdot10^6) = 0.0015625$ MPa/h, $m_e = 0.933\cdot40\cdot1.25\cdot1.2/(40\cdot10) = 0.1400$ MPa; $0.0015625\cdot24/0.1400 = 0.268$ .
The correction equals the plateau level at $\Delta t_{DA_w} = 1/(4\pi)$ : $\Delta t = 0.0796\cdot3.6\cdot10^{-4}\cdot444{,}444/(3.553\cdot10^{-3}\cdot40) = 89.6$ h.

Answer: After one day the correction is 27% of the plateau level. At about 90 h it becomes equal to the plateau level.

What it means. For a build-up that lasts a few hours, the trend hardly matters. For a build-up of a day or more, it matters. Note that the correction depends on the area per well, and not on how many wells there are in the block or how fast it is depleted in megapascals per year. The closer the wells stand to each other, the earlier the trend is noticeable on a build-up. The data in this example are assumed for illustration.

Where to get the trend

A record before shut-in. This is the best source. We lower a gauge into the producing well a day or two before shut-in, and record the fall of the bottomhole pressure at a constant rate. The slope of this line is $m^*$ . At the same time we see whether the pressure falls along a straight line, and whether the neighbours changed their rates.
The material balance of the block. $m^* = \sum qB/(24\,c_tV_p)$ , if we know the rates of all the wells, the pore volume and the compressibility. The uncertainty of $V_p$ , and especially of $c_t$ , passes into $m^*$ one to one (Reservoir limit test).
Reservoir pressure measurements. Repeated measurements in the block show how fast the pressure falls with time. They come from build-ups of neighbouring wells, or from a formation tester (a tool that measures the formation pressure from inside the well) in new wells (Pressure measurements in producing reservoirs: barriers and accuracy). In pseudo-steady-state flow this decline rate is the same as for the bottomhole pressure.
A reservoir simulation model. A model of the accumulation (a computer model), matched to the production history (History matching), calculates the trend at the point of the well, allowing for all the wells, even when the trend is not a straight line. This is the general case of the method.
We can ignore the trend if $4\pi\Delta t_{DA_w}$ over the whole build-up is less than a few per cent. This happens, for example, for a short build-up with wide well spacing, or under pressure maintenance (water or gas is injected to hold the pressure), when production and injection are balanced.

An example: a build-up of a well in a block with four wells

Well 1 from the table at the beginning of the lesson was shut in by a downhole valve after 5840 h of production. The neighbours keep producing at the same rates. The gauge was in the well for two days before shut-in. Below are the readings and the columns that we need for the correction.

Build-up of well 1 and the correction for the trend (assumed data)

$\Delta t$ , h: 0.5
$p_{ws}$ , MPa: 19.465
$m^*\Delta t$ , MPa: 0.001
$\Delta p_{DD}$ , MPa: 1.897

$\Delta t$ , h: 1
$p_{ws}$ , MPa: 19.561
$m^*\Delta t$ , MPa: 0.002
$\Delta p_{DD}$ , MPa: 1.994

$\Delta t$ , h: 2
$p_{ws}$ , MPa: 19.656
$m^*\Delta t$ , MPa: 0.003
$\Delta p_{DD}$ , MPa: 2.090

$\Delta t$ , h: 5
$p_{ws}$ , MPa: 19.780
$m^*\Delta t$ , MPa: 0.008
$\Delta p_{DD}$ , MPa: 2.219

$\Delta t$ , h: 10
$p_{ws}$ , MPa: 19.869
$m^*\Delta t$ , MPa: 0.016
$\Delta p_{DD}$ , MPa: 2.316

$\Delta t$ , h: 20
$p_{ws}$ , MPa: 19.951
$m^*\Delta t$ , MPa: 0.031
$\Delta p_{DD}$ , MPa: 2.413

$\Delta t$ , h: 30
$p_{ws}$ , MPa: 19.992
$m^*\Delta t$ , MPa: 0.047
$\Delta p_{DD}$ , MPa: 2.470

$\Delta t$ , h: 50
$p_{ws}$ , MPa: 20.033
$m^*\Delta t$ , MPa: 0.078
$\Delta p_{DD}$ , MPa: 2.542

$\Delta t$ , h: 70
$p_{ws}$ , MPa: 20.053
$m^*\Delta t$ , MPa: 0.109
$\Delta p_{DD}$ , MPa: 2.593

$\Delta t$ , h: 100
$p_{ws}$ , MPa: 20.063
$m^*\Delta t$ , MPa: 0.156
$\Delta p_{DD}$ , MPa: 2.650

$\Delta t$ , h: 130
$p_{ws}$ , MPa: 20.063
$m^*\Delta t$ , MPa: 0.203
$\Delta p_{DD}$ , MPa: 2.697

$\Delta t$ , h: 160
$p_{ws}$ , MPa: 20.055
$m^*\Delta t$ , MPa: 0.250
$\Delta p_{DD}$ , MPa: 2.736

$\Delta t$ , h: 200
$p_{ws}$ , MPa: 20.038
$m^*\Delta t$ , MPa: 0.313
$\Delta p_{DD}$ , MPa: 2.782

$\Delta t$ , h: 250
$p_{ws}$ , MPa: 20.009
$m^*\Delta t$ , MPa: 0.391
$\Delta p_{DD}$ , MPa: 2.831

$\Delta t$ , h	$p_{ws}$ , MPa	$m^*\Delta t$ , MPa	$\Delta p_{DD}$ , MPa
0.5	19.465	0.001	1.897
1	19.561	0.002	1.994
2	19.656	0.003	2.090
5	19.780	0.008	2.219
10	19.869	0.016	2.316
20	19.951	0.031	2.413
30	19.992	0.047	2.470
50	20.033	0.078	2.542
70	20.053	0.109	2.593
100	20.063	0.156	2.650
130	20.063	0.203	2.697
160	20.055	0.250	2.736
200	20.038	0.313	2.782
250	20.009	0.391	2.831

The bottomhole pressure before shut-in: 48 h before, 17.644 MPa; 24 h before, 17.607 MPa; at the moment of shut-in, 17.569 MPa. $m^*$ is from these readings (the example below), and $\Delta p_{DD} = p_{ws} - 17.569 + m^*\Delta t$ . The data are calculated from a model of a closed homogeneous block with four wells and rounded to 0.001 MPa; there is no wellbore storage.

Worked examplePermeability, skin and boundaries from a build-up among working neighbours

Given (assumed data)

Well 1: $q$ = 40 m³/d, $B$ = 1.25, $\mu$ = 1.2 mPa·s, $h$ = 10 m; $\phi\mu c_tr_w^2 = 0.20\cdot1.2\cdot1.5\cdot10^{-3}\cdot0.108^2 = 4.199\cdot10^{-6}$
Record: the table above; the producing time of 5840 h is much longer than the build-up, so the Horner plot practically coincides with the MDH plot
From the diagnostic plot: the derivative of the raw build-up at 50 h — 0.068 MPa; the derivative after the correction at 100 h — 0.169 MPa

Solution

The trend from the record before shut-in: $m^* = (17.644 - 17.569)/48 = 1.5625\cdot10^{-3}$ MPa/h. By the material balance of the block: $180\cdot1.25/(24\cdot1.5\cdot10^{-3}\cdot4\cdot10^6) = 1.5625\cdot10^{-3}$ MPa/h. It is the same: the block is in pseudo-steady-state flow.
Without the correction, a line through 2 and 20 h: the slope is $19.951 - 19.656 = 0.295$ MPa per cycle, and $k = 2.149\cdot40\cdot1.25\cdot1.2/(0.295\cdot10) = 43.7$ mD. Through 5 and 50 h: $20.033 - 19.780 = 0.253$ , and $k$ = 51.0 mD. The slope depends on the chosen stretch.
With the correction: $\Delta p_{DD}$ at 2, 20, 5 and 50 h is 2.090, 2.413, 2.219 and 2.542 MPa (the table). The slopes $2.413 - 2.090 = 0.323$ and $2.542 - 2.219 = 0.323$ MPa per cycle are the same: the points lie on one line. $k = 128.94/(0.323\cdot10) = 39.9$ mD.
Skin: $\Delta p_{DD}$ at 1 h is 1.994 MPa (a point on the line); $\lg(39.9/4.199\cdot10^{-6}) = 6.978$ ; $S = 1.151\,(1.994/0.323 - 6.978 + 2.098) = 1.49$ . Without the correction we would have got 2.13.
Boundaries. The plateau level is $m_e = 0.323/2.303 = 0.140$ MPa; $kt/(\phi\mu c_t) = 5.542\cdot10^6$ at 50 h, and $1.108\cdot10^7$ at 100 h. The raw derivative at 50 h is half the plateau. If we take this for a constant-pressure boundary, then $1 - \e^{-L^2/(\eta t)} = 0.068/0.140$ , so $L^2/(\eta t) = 0.665$ and $L = 0.0596\sqrt{5.542\cdot10^6\cdot0.665} = 114$ m. After the correction the derivative rises: $\ln[0.140/(0.169 - 0.140)] = 1.574$ , and $L = 0.0596\sqrt{1.108\cdot10^7\cdot1.574} = 249$ m (distance to a fault from the derivative).

Answer: With the correction:

k \approx 39.9

mD, skin about 1.49, and a sealing fault about 249 m away. Without it: a permeability of 43.7–51.0 mD, depending on the stretch, and a false constant-pressure boundary at 114 m.

What it means. The model that gave the record has $k$ = 40 mD, $S$ = 1.5 and a south fault 250 m from the well, and the correction recovered all this. Without it, the error of the permeability grows with the length of the stretch. The “constant-pressure boundary” at 114 m is inside a closed block, where there is no support from outside and there cannot be. The maximum of the build-up, 20.064 MPa at 110 h, is not the reservoir pressure. It equals neither the average pressure of the block (19.875 MPa by the material balance) nor the average pressure of the area of the well. We find the average pressure of the area by the MBH method, with an area proportional to the rate (Build-up in a bounded reservoir and average reservoir pressure). The data in this example are assumed for illustration.

CalculatedThe diagnostic plot of the build-up of well 1: the derivative of the raw record (orange) and after the correction for the trend (green), the plateau level and the line

m^*\Delta t

(dashed). The derivative of the raw record is the derivative after the correction minus

m^*\Delta t

. It falls where the line

m^*\Delta t

approaches the plateau, and by 110 h it becomes negative. On the corrected derivative, a rise is visible after 30 h: the effect of the south fault.

The derivative after the correction: boundaries instead of support

Why the raw derivative misleads is clear from a simple relation. The derivative of the trend against $\ln\Delta t$ is $-m^*\Delta t$ , so

The derivative of a raw build-up in a developed accumulation

\Delta p_{BU}' = \Delta p_{DD}' - m^*\Delta t

where:

$\Delta p_{BU}'$ logarithmic derivative of the pressure rise after shut-in, $\dd\Delta p_{BU}/\dd\ln\Delta t$ , Pa
$\Delta p_{DD}'$ logarithmic derivative of the reconstructed drawdown test, Pa
$m^*\Delta t$ derivative of the correction for the trend against $\ln\Delta t$ , Pa

Units: any consistent units

Holds when

a linear trend.

On a log–log plot, $m^*\Delta t$ is a line of unit slope. While it is far below the plateau, the raw and the corrected derivatives coincide. When it approaches the plateau, the raw derivative falls and goes into the negative. The picture is that of a constant-pressure boundary. And when the pressure in the accumulation rises ( $m^* < 0$ ), it is the other way round: a rise, as at a fault or in a closed area.

After the correction, the derivative shows what is around the well. First there is radial flow. Then come the real boundaries of the formation: here the south fault, 250 m away, raises the derivative to the doubled plateau level, and later the other sides of the block make themselves felt. There are no dividing lines of flow between the wells on the corrected derivative: the neighbours are allowed for in the trend. If we wait very long, the corrected derivative reaches a unit slope that corresponds to the pore volume of the whole block: the response to the shut-in is the start-up of one well in a closed block (Reservoir limit test). But such times are almost unreachable in a developed accumulation: the neighbours will change their rates first.

Why not the Horner time and not the equivalent time

The Horner function and the equivalent time both allow for the rate history of the well itself, assuming that before shut-in there was an infinite formation around it (Effect of producing time on build-up analysis). In a developed accumulation, the boundaries of the area and the neighbours had long been acting before shut-in, and these time functions do not remove the trend. Another way is to put into the Horner function the synthetic flowing time, at which $p^*$ becomes the pressure at the boundary of the drainage area (Equivalent producing time and synthetic flowing time). So we find the pressure at the boundary and, with the MBH correction, the average pressure of the area. But the synthetic time does not correct the late part of the record and its derivative.

When a linear trend is not right

The method rests on one assumption: if the well had not been shut in, its pressure would have kept falling as it did before shut-in. This is true only if nothing changes in the block. A change of the rate of a neighbour, the start or end of injection (pumping water into the formation), or the arrival of edge water makes the trend nonlinear, and the line $m^*\Delta t$ is then a wrong correction.

CalculatedCalculated derivative of the build-up of well 1 after the correction for a linear trend found from two days of record before shut-in. The neighbours did not change their rates (green). The rate of well 3, 500 m away, was raised from 35 to 70 m³/d 48 h before well 1 was shut in (orange). Well 3 was shut in 24 h after the start of the build-up (dashed).

Worked exampleWhat a rate change of a neighbour does to the correction

Given (assumed data)

Case 1: the neighbours did not change their rates
Case 2: the rate of well 3 (500 m from well 1) was raised from 35 to 70 m³/d 48 h before well 1 was shut in
Case 3: well 3 was shut in 24 h after the start of the build-up of well 1
Correction: a linear trend from the readings 48 h before and at the moment of shut-in; $m_e$ = 0.140 MPa; the formation as in the previous example

Solution

The trend measured before shut-in: 1.5625, 1.5833 and 1.5625 $\cdot10^{-3}$ MPa/h. In the second case it is a little steeper: the influence of the neighbour has already come, but is still growing. In the third case the record before shut-in knows nothing of the future shut-in of the neighbour.
The error of the reconstructed drawdown test at 200 h, $\Delta p_{DD}$ minus the exact value: 0.000, −0.053 and 0.034 MPa.
The derivative after the correction at 100 h: 0.169, 0.140 and 0.189 MPa. The distance to the fault from it: 249 m in the first case and 203 m in the third. In the second case the derivative hardly differs from the plateau, and the fault is not seen.

Answer: A change in the rate of a neighbour by tens of cubic metres a day, two days before the build-up, hides the fault. Its shut-in during the build-up makes the fault look 46 m closer. The formation is the same in both cases.

What it means. The disturbance from a neighbour 500 m away does not reach well 1 at once. For a hydraulic diffusivity of about 400 m²/h, its influence grows over tens and hundreds of hours ( $r^2/(4\eta) \approx 160$ h). So changes a few days before the test are dangerous, and so are changes during it. This gives practical rules. For the duration of the build-up and for several days before it, keep the rates of the neighbours constant and record them. If this is impossible, do not continue the trend by a straight line. Subtract instead a response calculated by superposition with the history of the neighbours' rates (Superposition principle), or by a model of the accumulation. The data in this example are assumed for illustration.

Injection and edge water. The response of an accumulation to a change in injection, or to the arrival of water, develops over months and is never linear. We take the trend from a model or from the history of pressure measurements.
Transient flow. If a well or a block has not yet reached pseudo-steady-state flow (a new accumulation, wells put on production recently), the pressure is still changing from place to place and the trend is curved. We can use a straight line only on a short stretch, where the curvature is small.
Gas, and a pressure below the bubble point pressure. Below this pressure, gas comes out of the oil. The compressibility and the mobility (permeability divided by viscosity) change with the pressure, the problem is nonlinear, and the sum “trend + response” is an approximation. For gas, we work in pseudo-pressures (quantities that allow for how the gas properties change with pressure).
Wellbore storage. The correction does not fix an early stretch with a changing wellbore storage coefficient. For a build-up in a developed accumulation, a shut-in at the bottom (with a downhole valve) is especially useful (Nonideal wellbore storage, well clean-up and changing skin).
The accuracy of $m^*$ . An error $\delta m^*$ gives an error $\delta m^*\Delta t$ , which grows with time. For a build-up several days long, we need $m^*$ to within a few per cent, and a gauge with a small drift (a slow shift of its readings).

Terms and symbols

Developed accumulation: An accumulation in which many wells work. The pressure in each depends on the production of its neighbours and falls together with the pressure in the block.
Pressure trend $p_{tr}$: The pressure that the well would have had after shut-in if it had kept producing. In pseudo-steady-state flow it is a line with the slope $-m^*$ .
Slider's method: Analysis of a build-up with a correction for a linear trend: $\Delta p_{DD} = p_{ws} - p_{wf}(\Delta t = 0) + m^*\Delta t$ is analysed as a drawdown test.
Reconstructed drawdown test $\Delta p_{DD}$: The response of the formation to the shut-in, expressed as the pressure drop on start-up. It contains no contribution of the neighbours.
$m^*$: The pressure decline rate in the block, $\sum qB/(c_tV_p)$ . In pseudo-steady-state flow it is the same for the bottomhole and the average pressure.

Common mistakes and misconceptions

Wrong

“The derivative at the end of the build-up falls, so there is edge water nearby.”

In fact

In a developed accumulation, this is how the continuing depletion looks: $m^*\Delta t$ is subtracted from the derivative of the response. First we correct for the trend, and only then look for boundaries.

Wrong

“The maximum of the build-up is the reservoir pressure in the area of the well.”

In fact

The maximum comes when the rise from the shut-in equals the fall from the neighbours. It equals neither the average pressure of the block nor the average pressure of the area of the well.

Wrong

“We can take the trend from the last two hours before shut-in.”

In fact

The trend must be steady. If the neighbours changed their rates in the last day or two, their influence is still growing, and a short record gives a wrong $m^*$ .

Wrong

“The correction is needed only for very long build-ups.”

In fact

Its size as a fraction of the plateau is $4\pi\Delta t_{DA_w}$ : with closely spaced wells it is noticeable after a day.

Wrong

“After the correction for the trend, the derivative will show the dividing lines between the wells.”

In fact

No: the neighbours are in the trend. The corrected derivative shows the response of one well in a formation with its real boundaries: the faults and the pinch-outs (places where a bed thins out to nothing).

In depth: advanced detail

The Horner method as a special case of the correction for a trend

Take a well that produced for a time $t_p$ in an infinite formation before shut-in, with no neighbours. The trend is the continuation of its own drawdown test: $p_{tr}(\Delta t) = p_i - m\left[\lg(t_p + \Delta t) + b\right]$ . Here $b$ collects the constants and the skin: $b = \lg[k/(\phi\mu c_tr_w^2)] - 2.098 + 0.869S$ in metric oilfield units. The response to the shut-in is the drawdown test of the same well started at the moment of shut-in, with the same $b$ and skin: $p_{ws} - p_{tr} = m\left[\lg\Delta t + b\right]$ . Adding the two, we get $p_{ws} = p_i - m\lg\left[(t_p + \Delta t)/\Delta t\right]$ . The constants and the skin cancel, and this is the Horner equation.

So a Horner plot is a correction for the trend in which the trend is a logarithm. Slider's method is the same correction when the trend is a straight line. The general recipe is to build the trend from everything that we know about the history of the well and its neighbours, to subtract it, and to analyse the rest as a drawdown test. The more exact the trend, the longer the remainder stays a fingerprint of the formation.

Where

4\pi\Delta t_{DA_w}

comes from

In metric oilfield units, $m^* = \sum q_jB/(24\,c_tV_p)$ , $m_e = qB\mu/(4\pi\cdot0.0853\,kh)$ , and $V_p = \phi hA$ . The ratio is $m^*\Delta t/m_e = (\sum q_j/q)\cdot 4\pi\cdot0.0853\,k\Delta t/(24\,\phi\mu c_tA)$ . The factor $0.0853/24$ , or more exactly $0.08527/24 = 3.553\cdot10^{-3}$ , is the same one that turns hours into dimensionless time. So $m^*\Delta t/m_e = 4\pi\cdot3.553\cdot10^{-3}\,k\Delta t/[\phi\mu c_tA\,(q/\sum q_j)] = 4\pi\,\Delta t_{DA_w}$ .

The area $A_w = A\,q/\sum q_j$ is the drainage area of the well in pseudo-steady-state flow. Each well takes from its area as much as the area gives out at the pressure decline rate that is common to the block. The same factor $1/(4\pi) \approx 0.08$ appears in the lesson on the average pressure. There it is the producing time at which a Horner line after a long production stays straight for longest (producing time in the Horner function). Behind it is the same link between depletion and radial flow.

How the data of the lesson are calculated

Each well is a line source in a closed homogeneous rectangle. The pressure at the point of well 1 is the sum of the responses of the rectangle to the work of all four wells. The Green's function of the rectangle (its response to a point source) uses reflected sources for small times, and a series of cosines for large times, as in the lesson Build-up in a bounded reservoir and average reservoir pressure. The shut-in is the response to a rate $-q$ from the moment of shut-in. The calculation was checked by an independent summation of the exponential integrals of the reflected wells: the pressures agree to $10^{-4}$ MPa, and the average pressure over the block equals the material balance. In pseudo-steady-state flow, the trend comes out linear to hundredths of a per cent. So here the correction by a straight line and the exact subtraction of the trend coincide.

Connections to other topics

Build-up in a bounded reservoir and average reservoir pressure — a build-up in a closed area, the MBH method and the average pressure of the area of a well.
Reservoir limit test — pseudo-steady-state flow, the pressure decline rate and the pore volume.
Superposition principle — a build-up as a sum of responses, the influence of neighbouring wells and the delay of disturbances.
Effect of producing time on build-up analysis — desuperposition for an exploration well, and the equivalent time.
Equivalent producing time and synthetic flowing time — another way to allow for the producing history: choosing the time in the Horner function.
Diagnosis with the pressure derivative — the pressure derivative and false signs of boundaries.
A sealing fault near the well: the method of images — the distance to a fault from the derivative step.
Pseudo-steady-state flow and the shape factor — why in a closed block the pressure falls at the same speed everywhere.
Pressure drawdown analysis — the analysis of a drawdown test, which we apply to the reconstructed curve.

Check yourself

1
A build-up of a well in a block under development: the pressure rose for 4 days and then started to fall slowly, and the derivative has been falling since the middle of the record. What explanation should we check first?
1. A.The depletion trend of the block: the neighbours keep producing
2. B.Dual porosity
3. C.A constant-pressure boundary near the well
4. D.A faulty gauge
Show the answer
Correct answer: A. The depletion trend of the block: the neighbours keep producing
In a developed accumulation, shutting in one well does not stop the depletion of the block. $m^*\Delta t$ is subtracted from the derivative of the response, the derivative falls, and the pressure passes a maximum. We first correct for the trend. Only if the derivative still falls after that do we look for support from outside.
2
Before shut-in, the bottomhole pressure fell in 36 h from 21.430 to 21.358 MPa at constant rates of all the wells. 50 h after shut-in the pressure is 23.508 MPa. Find $m^*$ and the reconstructed drawdown test $\Delta p_{DD}$ at 50 h.
Show the answer
$m^* = (21.430 - 21.358)/36 = 0.072/36 = 0.0020$ MPa/h. $\Delta p_{DD} = 23.508 - 21.358 + 0.0020\cdot50 = 2.150 + 0.100 = 2.250$ MPa. Without the correction, the pressure rise (2.150 MPa) is too low by 0.100 MPa.
3
The hydraulic diffusivity of the formation is 300 m²/h, and each well has 40 ha. What fraction of the plateau level will the correction for the trend be after 24 h, and when will the derivative of the raw build-up fall to zero (if the boundaries do not show earlier)?
Show the answer
$\Delta t_{DA_w} = \eta\Delta t/A_w = 300\cdot24/400{,}000 = 0.018$ ; $4\pi\cdot0.018 = 0.226$ , about 23% of the plateau level. The derivative of the raw build-up becomes zero when $4\pi\Delta t_{DA_w} = 1$ : $\Delta t = A_w/(4\pi\eta) = 400{,}000/(12.57\cdot300) \approx 106$ h.
4
In a block with pressure maintenance (water or gas is injected to hold the pressure), injection exceeds production, and the reservoir pressure rises. How will the derivative of a build-up of a producing well look at the end of the record, without the correction for the trend?
1. A.It will become negative
2. B.It will rise, as at a sealing fault or in a closed area
3. C.It will not change
4. D.It will fall, as at a constant-pressure boundary
Show the answer
Correct answer: B. It will rise, as at a sealing fault or in a closed area
For a rising pressure, $m^* < 0$ , and $\Delta p_{BU}' = \Delta p_{DD}' - m^*\Delta t = \Delta p_{DD}' + |m^*|\Delta t$ : a rising line is added to the derivative of the response. The result is a rise like the effect of a sealing boundary. In this case we subtract the correction.
5
A neighbouring well 400 m away was started after a repair a day before the build-up of the well under test. Why is Slider's method unreliable here, and what should we do?
Show the answer
The disturbance from the start-up of the neighbour arrives with a delay of about $r^2/(4\eta)$ , and keeps growing for a long time. A trend measured in the last hours does not reflect the future fall, and the line $m^*\Delta t$ understates it. The corrected derivative is then distorted: a fault may vanish, or a false one may appear. It is better to postpone the build-up until the neighbour produces at a steady rate. If that is impossible, we subtract the response calculated by superposition with the rate history of the neighbour, or we use a model of the accumulation.
6
After the correction for the trend, the plateau level is 0.12 MPa, and at 80 h the derivative is 0.15 MPa; $k$ = 50 mD, $\phi\mu c_t = 3.0\cdot10^{-4}$ (mPa·s/MPa). How far away is the fault?
Show the answer
$L = 0.0596\sqrt{\dfrac{k\,t}{\phi\mu c_t}\ln\dfrac{m_e}{\Delta p' - m_e}} = 0.0596\sqrt{\dfrac{50\cdot80}{3.0\cdot10^{-4}}\ln\dfrac{0.12}{0.03}}$ . The expression under the root is $1.333\cdot10^7\cdot1.386 = 1.848\cdot10^7$ , the root is 4299, and $L \approx 256$ m. From the raw derivative, without the correction, the rise would be smaller or not visible at all.

Updated: 19 October 2026