Equivalent producing time and synthetic flowing time

Quick refresher

A build-up remembers the whole rate history. In radial flow (flow towards the well from all sides), $p_{ws}$ depends linearly on the superposition time function $X_{BU} = \sum w_j\lg[(\theta_j + \Delta t)/\Delta t]$ with weights $w_j = (q_j - q_{j-1})/q_n$ . The slope of the line gives $kh$ (with the last rate $q_n$ ), and its continuation to $X_{BU} = 0$ gives $p^*$ .
Each rate change affects the slope through the factor $\theta_j/(\theta_j + \Delta t)$ . At first the build-up answers to the last rate, and then to a more and more distant history. The recent changes are “forgotten” first.
The Horner equivalent time $t_p = N_p/q_n$ keeps the produced volume, and it is exact for very late readings. At the start of the build-up, a logarithmically weighted time would replace the history more accurately.
The ratio of slopes $R(\Delta t)$ shows the error of one producing time. For $\Delta t$ much shorter than the duration $\theta_n$ of the last flow period, it is about $(\Delta t/\theta_n)(q_{n-1}/q_n - 1)$ . A rate cut shortly before shut-in raises the derivative, like a fault.
After a long production in a closed area, the actual time gives a $p^*$ that exists nowhere in the formation. The synthetic time $t_{sDA} = e^{(p_e - \bar p)/m_e}/C_A$ , about half the time to reach pseudo-steady-state flow, turns $p^*$ into the pressure at the boundary of the area, $p_e$ .
Among working neighbours, a long build-up flattens out and starts to fall: the boundaries of the drainage areas move. From the early line we find $kh$ , the skin, $p^*$ and the average pressure. We treat the late trend separately.

Key relationships

X_{BU} = \sum_{j=1}^{n}\dfrac{q_j - q_{j-1}}{q_n}\,\lg\dfrac{\theta_j + \Delta t}{\Delta t}

superposition time function for a build-up; $\theta_j$ is the time from the start of flow period number $j$ to shut-in

p_{ws} = p^* - m\,X_{BU},\qquad m = 2.149\,\dfrac{q_nB\mu}{kh}

metric oilfield units (oilfield units: 162.6)

t_p = \dfrac{N_p}{q_n}

Horner equivalent producing time

R(\Delta t) = \dfrac{t_p + \Delta t}{t_p}\sum_{j}\dfrac{q_j - q_{j-1}}{q_n}\,\dfrac{\theta_j}{\theta_j + \Delta t}

how many times the slope with one time $t_p$ differs from the correct one

t_{sDA} = \dfrac{1}{C_A}\exp\dfrac{p_e - \bar p}{m_e}

synthetic time at which $p^* = p_e$

Common pitfalls

Putting into the Horner function the duration of the last flow period, when before it the well produced for a long time at another rate.
Not comparing the duration of the last flow period with the duration of the build-up.
Taking a rise of the derivative after a rate cut before shut-in for a fault.
Taking the $p^*$ found with the actual producing time in a developed accumulation (one with many producing wells) as the reservoir pressure.
Treating the late part of a long build-up among working neighbours as a build-up in a closed area.

After this lesson you can

write a build-up after several flow periods through the superposition time function in the Horner form, and find the flow capacity, $p^*$ and the skin from it
explain why the equivalent time $N_p/q_n$ usually works and when it is wrong, and estimate the error of the slope from the rate history
compare in an example the analysis with the superposition time function, with $N_p/q_n$ , with the duration of the last flow period and with the total producing time
recognise a false “fault” on the derivative after a rate cut before shut-in
calculate the synthetic flowing time of a well in a closed area, and explain what $p^*$ means with this choice
describe how a long build-up of a well among working neighbours behaves, and what we can get from it

A build-up after several flow periods: the superposition time function

If the rate changed before shut-in, each change is a separate step of the superposition, the adding up of the effects of separate events (shut-in after several flow periods). The disturbances from all the rate changes spread through the formation at the same time, and the pressure rise after shut-in is the sum of their continuing effects. In radial flow in an infinite formation, it is convenient to collect the logarithms of the times into one function. This is the multirate version of the Horner function:

A build-up after several flow periods in the Horner form

p_{ws}(\Delta t) = p^* - m\,X_{BU}(\Delta t),\qquad X_{BU} = \sum_{j=1}^{n}\frac{q_j - q_{j-1}}{q_n}\,\lg\frac{\theta_j + \Delta t}{\Delta t}

where:

$p_{ws}(\Delta t)$ bottomhole pressure a time $\Delta t$ after shut-in, MPa
$p^*$ the continuation of the line to $X_{BU} = 0$ , that is, to an infinitely long shut-in; in a new formation it equals the initial pressure, MPa
$m$ slope of the line per unit of $X_{BU}$ : $2.149\,q_nB\mu/(kh)$ , MPa
$X_{BU}$ superposition time function for a build-up: the sum of the decimal logarithms of the Horner ratios for each rate change, with weights $w_j = (q_j - q_{j-1})/q_n$ , —
$q_j$ rate of flow period number $j$ at standard conditions; $q_0 = 0$ , and $q_n$ is the last rate before shut-in, m³/d
$\theta_j$ time from the start of flow period number $j$ to shut-in: $t_n - t_{j-1}$ , where $t_n$ is the moment of shut-in, counted from the first start-up, h

Units: metric oilfield units: pressure and $m$ in MPa, $q$ in m³/d, $\mu$ in mPa·s, $k$ in mD, $h$ in m; time in any equal units

Holds when

radial flow in an infinite formation over the whole history and the build-up: the disturbance has not reached any boundaries
a linear problem (the effects of separate events add up): a single-phase slightly compressible fluid, a constant skin
wellbore storage (the well acting as a small tank) is over

Each term is the Horner ratio for its own rate change, with its own weight. The weights add up to 1. For one flow period the formula turns into the Horner equation (Horner equation). The points of the build-up against $X_{BU}$ lie on a line with a slope that corresponds to the last rate: $kh = 2.149\,q_nB\mu/m$ . This is the same superposition time function as in the lesson Superposition principle, where the shut-in enters the history as the last step with the rate $-q_n$ .

In other unit systems (1)

Oilfield units

p_{ws} = p^* - m\,X_{BU},\qquad m = 162.6\,\frac{q_nB\mu}{kh}

$q_n$ in STB/d, $B$ in rb/STB, $\mu$ in cP, $k$ in mD, $h$ in ft; pressure and $m$ in psi.

$162.6$ is the same constant as in the Horner equation (Pressure build-up analysis); $X_{BU}$ is dimensionless.

With a changing rate, we take the derivative of a build-up not against the logarithm of time, but against the superposition time function. This is the superposition derivative. Its link with the ordinary logarithmic derivative shows how the formation “remembers” the history:

The superposition derivative

\Delta p'_X = -\frac{\dd\,\Delta p}{\dd\,(X_{BU}\ln 10)} = \frac{\Delta p'}{\displaystyle\sum_{j=1}^{n}\frac{q_j - q_{j-1}}{q_n}\,\frac{\theta_j}{\theta_j + \Delta t}}

where:

$\Delta p'_X$ derivative of the pressure rise against the superposition time function, reduced to the natural logarithm, MPa
$\Delta p'$ ordinary logarithmic derivative of the build-up against $\ln\Delta t$ , MPa
$\Delta p$ pressure rise after shut-in: $p_{ws}(\Delta t) - p_{wf}(\Delta t = 0)$ , MPa

Units: any consistent system; derivatives in units of pressure

Holds when

the conditions of the previous equation; the minus sign is there because $X_{BU}$ decreases as $\Delta t$ grows.

In radial flow, $\Delta p'_X = m_e = q_nB\mu/(4\pi kh)$ for any rate history: a plateau. The denominator is the “memory” of the rate changes. Each term is close to its weight while $\Delta t \ll \theta_j$ , and goes to zero when $\Delta t \gg \theta_j$ . So the ordinary derivative of a build-up first answers to the last rate, and later to a more and more distant history: recent rate changes are forgotten first. For one flow period, the denominator equals $t_p/(t_p + \Delta t)$ , and $\Delta p'_X$ is the derivative against the equivalent time (Effect of producing time on build-up analysis).

CalculatedThe rate history of the well of the lesson's examples (solid): an extended well test on three chokes, 30, 55 and 80 m³/d, 28 days in all, and then a shut-in for a build-up. The dashed line is the Horner replacement: production at the last rate of 80 m³/d for

t_p = N_p/q_n

= 402 h. The areas under both lines, the produced volume of 1340 m³, are equal.

The Horner equivalent time

Horner proposed to replace the history by one flow period at the last rate $q_n$ , lasting as long as it takes to produce the same volume. This is the equivalent producing time:

The Horner equivalent producing time

t_p = \frac{N_p}{q_n}

where:

$t_p$ equivalent producing time that we put into the Horner function, d
$N_p$ cumulative production by the moment of shut-in, at standard conditions, m³
$q_n$ the last steady rate before shut-in, m³/d

Units: $N_p$ in m³, $q_n$ in m³/d, $t_p$ in days; for the Horner ratio in hours we multiply by 24

Holds when

the last rate was held long enough compared with the duration of the build-up (an estimate below)
the history went on in an infinitely acting formation; after a long production in a bounded area we use the synthetic time (the sections below)

The replacement keeps the produced volume, and this volume decides by how much the pressure in the formation has fallen on the whole. So for very late readings it is exact. When the time after shut-in is much longer than the whole history, each Horner ratio is $\lg[(\theta_j + \Delta t)/\Delta t] \approx \theta_j/(\Delta t\ln 10)$ , and the sum depends only on $\sum w_j\theta_j = N_p/q_n$ . At the start of the build-up, what matters is not the sum of the times but the sum of their logarithms. There, the history would be replaced more accurately by a time $T$ with $\lg T = \sum w_j\lg\theta_j$ . For the history of the example this is 257 h instead of 402 h.

No single producing time replaces the history over the whole build-up. How wrong a given $t_p$ is ( $N_p/q_n$ , the duration of the last flow period, or the total time since the first start-up) is shown by the ratio of slopes: the local slope of the Horner plot with this $t_p$ (it is also the derivative against $\ln\Delta t_e$ ), divided by the correct one:

The error of the slope with one producing time

R(\Delta t) = \frac{t_p + \Delta t}{t_p}\,\sum_{j=1}^{n}\frac{q_j - q_{j-1}}{q_n}\,\frac{\theta_j}{\theta_j + \Delta t}

where:

$R(\Delta t)$ ratio of the local slope of the Horner plot with the time $t_p$ to the correct slope $m$ , —
$t_p$ producing time put into the Horner function, h
$\theta_j,\ q_j$ age of rate change number $j$ at the moment of shut-in, and the rates of the flow periods, h, m³/d

Units: dimensionless; times in the same units

Holds when

radial flow in an infinite formation; wellbore storage is over.

With $t_p = N_p/q_n$ , the ratio equals 1 both at $\Delta t \to 0$ and for $\Delta t$ much longer than the whole history, but not in between. While $\Delta t$ is much shorter than the duration $\theta_n$ of the last flow period, and $\theta_n$ is much shorter than the earlier flow periods, the error is about $(\Delta t/\theta_n)(q_{n-1}/q_n - 1)$ . If the last rate is higher than the previous one, the slope is too low (the permeability is overstated) and the derivative goes down. If it is lower, the slope is too high and the derivative goes up. With the duration of the last flow period instead of $N_p/q_n$ , the error is about $(q_{n-1}/q_n)\,\Delta t/\theta_n$ : the derivative rises, whichever way the rate was changed.

A comparison in an example: the superposition time function and one producing time

A build-up after an extended well test with three flow periods (assumed data)

$\Delta t$ , h: 0.05
$p_{ws}$ , MPa: 23.0377
$X_{BU}$: 3.7104

$\Delta t$ , h: 0.1
$p_{ws}$ , MPa: 23.4632
$X_{BU}$: 3.4096

$\Delta t$ , h: 0.2
$p_{ws}$ , MPa: 23.7848
$X_{BU}$: 3.1089

$\Delta t$ , h: 0.5
$p_{ws}$ , MPa: 24.1588
$X_{BU}$: 2.7120

$\Delta t$ , h: 1
$p_{ws}$ , MPa: 24.4261
$X_{BU}$: 2.4126

$\Delta t$ , h: 2
$p_{ws}$ , MPa: 24.6864
$X_{BU}$: 2.1149

$\Delta t$ , h: 4
$p_{ws}$ , MPa: 24.9412
$X_{BU}$: 1.8203

$\Delta t$ , h: 8
$p_{ws}$ , MPa: 25.1895
$X_{BU}$: 1.5315

$\Delta t$ , h: 16
$p_{ws}$ , MPa: 25.4284
$X_{BU}$: 1.2529

$\Delta t$ , h: 24
$p_{ws}$ , MPa: 25.5619
$X_{BU}$: 1.0971

$\Delta t$ , h: 48
$p_{ws}$ , MPa: 25.7754
$X_{BU}$: 0.8476

$\Delta t$ , h: 72
$p_{ws}$ , MPa: 25.8898
$X_{BU}$: 0.7138

$\Delta t$ , h	$p_{ws}$ , MPa	$X_{BU}$
0.05	23.0377	3.7104
0.1	23.4632	3.4096
0.2	23.7848	3.1089
0.5	24.1588	2.7120
1	24.4261	2.4126
2	24.6864	2.1149
4	24.9412	1.8203
8	25.1895	1.5315
16	25.4284	1.2529
24	25.5619	1.0971
48	25.7754	0.8476
72	25.8898	0.7138

The history: 30 m³/d for 10 days, 55 m³/d for 16 days, 80 m³/d for 2 days; $t_n$ = 672 h, $q_n$ = 80 m³/d; the bottomhole pressure at the moment of shut-in is 19.2270 MPa. The data are calculated from a model of a homogeneous infinite formation with skin and wellbore storage (shut-in at the wellhead), and rounded to 0.0001 MPa.

Worked exampleAn extended well test: four ways to allow for the history

Given (assumed data)

Well and formation: $B$ = 1.16; $\mu$ = 1.8 mPa·s; $h$ = 12 m; we may take the formation as infinite: during the production the disturbance did not reach the boundaries (an assumption of the example)
History: 30 m³/d from 0 to 240 h, 55 m³/d up to 624 h, 80 m³/d until shut-in at 672 h
Readings for the line: 4 h — 24.9412 MPa, 16 h — 25.4284 MPa (the table above)

Solution

Weights and the ages of the rate changes: $w_1 = 30/80 = 0.375$ , $w_2 = (55 - 30)/80 = 0.3125$ , $w_3 = (80 - 55)/80 = 0.3125$ ; $\theta_1 = 672$ , $\theta_2 = 432$ , $\theta_3 = 48$ h. Cumulative production: $N_p = 30\cdot10 + 55\cdot16 + 80\cdot2 = 1340$ m³, and $t_p = N_p/q_n = 1340/80 = 16.75$ days, or 402 h.
Superposition time function: $X_{BU}(4) = 0.375\lg\dfrac{676}{4} + 0.3125\lg\dfrac{436}{4} + 0.3125\lg\dfrac{52}{4} = 1.8203$ ; $X_{BU}(16) = 0.375\lg 43 + 0.3125\lg 28 + 0.3125\lg 4 = 1.2529$ .
The line against $X_{BU}$ : $m = (25.4284 - 24.9412)/(1.8203 - 1.2529) = 0.4872/0.5674 = 0.8587$ MPa; $q_nB\mu = 80\cdot1.16\cdot1.8 = 167.04$ ; $k = 2.149\cdot167.04/(0.8587\cdot12) = 34.8$ mD; $p^* = 25.4284 + 0.8587\cdot1.2529 = 26.504$ MPa.
Horner with $t_p$ = 402 h: abscissas $\lg(406/4) = 2.00647$ and $\lg(418/16) = 1.41706$ ; $m = 0.4872/(2.00647 - 1.41706) = 0.8266$ MPa; $k = 36.2$ mD; $p^* = 25.4284 + 0.8266\cdot1.41706 = 26.600$ MPa.
The same with the duration of the last flow period (48 h): $\lg 13 = 1.11394$ , $\lg 4 = 0.60206$ , $m = 0.9518$ MPa, $k = 31.4$ mD, $p^* = 26.001$ MPa. And with the total time (672 h): $\lg 169 = 2.22789$ , $\lg 43 = 1.63347$ , $m = 0.8196$ MPa, $k = 36.5$ mD, $p^* = 26.767$ MPa.

Answer: The superposition time function gives

k \approx 34.8

mD and

p^* = 26.504

MPa, the initial pressure. The Horner equivalent time gives 36.2 mD and 26.600 MPa. The last flow period gives 31.4 mD and 26.001 MPa. The total time gives 36.5 mD and 26.767 MPa.

What it means. The data were calculated from a model with a permeability of 35 mD and an initial pressure of 26.5 MPa. The superposition time function recovered them almost exactly (what is left is the tail of wellbore storage in the reading at 4 h). The Horner equivalent time was wrong by 4% in permeability and by 0.10 MPa in $p^*$ . For this history that is tolerable, although $p^*$ is above the initial pressure, which cannot happen in a new formation. The duration of the last flow period is the worst choice: two days at 80 m³/d do not cancel 26 days of production at smaller rates, and the slope came out 11% steeper. The data in this example are assumed for illustration.

Results of the example from the readings at 4 and 16 h

Time function: superposition time function
$k$ , mD: 34.8
$p^*$ , MPa: 26.504

Time function: $N_p/q_n$ = 402 h
$k$ , mD: 36.2
$p^*$ , MPa: 26.600

Time function: last flow period, 48 h
$k$ , mD: 31.4
$p^*$ , MPa: 26.001

Time function: total time, 672 h
$k$ , mD: 36.5
$p^*$ , MPa: 26.767

Time function	$k$ , mD	$p^*$ , MPa
superposition time function	34.8	26.504
$N_p/q_n$ = 402 h	36.2	26.600
last flow period, 48 h	31.4	26.001
total time, 672 h	36.5	26.767

The model: $k$ = 35 mD, initial pressure 26.5 MPa. The slope and $p^*$ depend on the choice of the time. The pressures of the build-up do not.

CalculatedThe derivative of the build-up of the example, taken against different time functions, as a fraction of the radial-flow plateau

m_e

(calculated by the model of the lesson). Against the superposition time function it stands on the plateau (the small rise in the first hour is the tail of wellbore storage). Against the Horner function with

N_p/q_n

and with the total time, it falls by 14–19% towards the end of the record, as it would for an improvement of the formation far from the well, or for support from outside. With the duration of the last flow period, it rises, as at a fault.

A false fault: the rate was cut before shut-in

The most dangerous history is a long production at a high rate, followed by a rate cut shortly before shut-in. The well is moved to a smaller choke, or restricted because of sand or a high gas–oil ratio (a lot of gas with each unit of oil). The weight of the old, large rate, $w_1 = q_1/q_n$ , is then larger than one, and the weight of the last change is negative. After several hours of shut-in, the slope of the build-up starts to answer to the old rate. The derivative taken with one producing time rises. When the rate is cut by half, it rises almost to the doubled level, like at a sealing fault (A sealing fault near the well: the method of images).

Worked exampleA day on a smaller choke

Given (assumed data)

Well and formation: the same as in the previous example
History: 60 days at 70 m³/d, then one day at 35 m³/d; shut-in at $t_n$ = 1464 h
Readings: 4 h after shut-in — 24.8965 MPa, 16 h — 25.1780 MPa

Solution

Weights: $w_1 = 70/35 = 2$ ( $\theta_1$ = 1464 h), $w_2 = (35 - 70)/35 = -1$ ( $\theta_2$ = 24 h). $N_p = 70\cdot60 + 35\cdot1 = 4235$ m³, and $t_p = N_p/q_n = 4235/35 = 121$ days $= 2904$ h.
The ratio of slopes with this $t_p$ at 8 h: $R = \left(\dfrac{2\cdot1464}{1472} - \dfrac{24}{32}\right)\dfrac{2912}{2904} = (1.98913 - 0.75000)\cdot1.00275 = 1.243$ ; at 24 h: $(1.96774 - 0.50000)\cdot1.00826 = 1.480$ .
The first-order estimate at $\Delta t$ = 2 h: $(2/24)\cdot(70/35 - 1) = 0.083$ , against the exact $R - 1 = 0.075$ .
The Horner line with $t_p$ = 2904 h through the readings at 4 and 16 h: $\lg(2908/4) = 2.86153$ , $\lg(2920/16) = 2.26126$ ; $m = 0.2815/(2.86153 - 2.26126) = 0.4690$ MPa; $q_nB\mu = 35\cdot1.16\cdot1.8 = 73.08$ ; $k = 2.149\cdot73.08/(0.4690\cdot12) = 27.9$ mD.
The superposition time function: $X_{BU}(4) = 2\lg(1468/4) - \lg(28/4) = 4.2842$ , $X_{BU}(16) = 2\lg(1480/16) - \lg(40/16) = 3.5343$ ; $m = 0.2815/(4.2842 - 3.5343) = 0.3754$ MPa; $k = 34.9$ mD.

Answer: With the Horner equivalent time, the slope is too high, and the permeability is 27.9 mD instead of 34.9 mD by the superposition time function. By 24 h the derivative is 1.48 times as high as the plateau, and it keeps growing.

What it means. A rise of the derivative by half (to 1.5 times its level) over one or two decades of time (a decade is a tenfold range of time) is easy to take for a sealing fault not far from the well. The cause is the history: two months at 70 m³/d left a deep drawdown cone, and ten or so hours after shut-in its filling counts for more than the day at 35 m³/d. The equivalent time keeps the volume, but it cannot keep the order of events. By the first-order estimate, for the error of the slope up to 16 h not to exceed 5%, the last rate had to be held for at least $16\cdot1/0.05 = 320$ h, about 13 days. The data in this example are assumed for illustration.

CalculatedThe derivative of the build-up after a day at half the rate, as a fraction of the plateau

m_e

(calculated by the model of the lesson): against the Horner function with

N_p/q_n

(orange) and against the superposition time function (black). The first one rises to 1.6 times the plateau by the end of the two-day record, and keeps growing. The dashed line is the level that the derivative reaches at a sealing fault.

Record the rate history, and not only the cumulative production: the rates and the moments of the rate changes over a time comparable with several durations of the build-up are a necessary part of the test data.
Check the duration of the last flow period. If, by the first-order estimate, the last flow period would have to last longer than it did to keep the error within 5%, we analyse with the superposition time function.
Do not trust the late part of a derivative taken with one producing time, if the rate changed shortly before shut-in: a rise and a fall can be traces of the history, not of boundaries.
An inexact history gives an inexact answer. Rates that are allocated among wells by calculation, and not measured, carry their error into the flow capacity and into the shape of the derivative.

Synthetic time in a closed area

This is a different problem: a well that has produced for months in a bounded area, and has long ago reached pseudo-steady-state flow. If we put into the Horner function the actual producing time (or $N_p/q_n$ , which is the same for a constant rate), the line comes out short, and $p^*$ goes above any pressure in the formation. Then we find the average pressure with the MBH correction (which producing time to use). But we can do otherwise, and give $p^*$ a meaning.

In pseudo-steady-state flow the shape of the drawdown cone does not change. The pressure in the whole area falls at the same rate, and the pressure distribution at the moment of shut-in is the average pressure plus an unchanging profile. The build-up depends only on this distribution, and not on how many months it took to form. Almost the same distribution would be created by a well that started at the same rate in a formation whose pressure equals the present pressure at the boundary of the area, $p_e$ , and produced for a short time, until the disturbance covered the area. The duration of this imaginary flow period is the synthetic flowing time $t_s$ . For the imaginary formation, the Horner line with the producing time $t_s$ continues to its initial pressure $p_e$ . The pressure distributions are almost the same, so the same is true for the real build-up: with $t_s$ in the Horner function, $p^*$ equals the pressure at the boundary of the area at the moment of shut-in. The exact condition is given by the formula below.

CalculatedPressure along the diagonal of the drainage area of the well in the example below (a square of 800 × 800 m, the well at the centre) at the moment of shut-in after a year of production (solid), and after a synthetic flow period (dashed). The synthetic flow period is a start-up in a formation with a pressure equal to the pressure at the corner of the area, followed by 289 h of production at the same rate. Near the well the curves coincide to 0.003 MPa, and at the corner they differ by 0.04 MPa: this part of the area did not have enough time in the synthetic flow period. Calculated by the model of the lesson.

The time to give to the synthetic flow period follows from the formula for the average pressure after production in pseudo-steady-state flow (average pressure from the continuation of the line): $p^* - \bar p = m_e\ln(C_At_{pDA})$ for any producing time that we put in. We want $p^* = p_e$ :

Synthetic flowing time

t_{sDA} = \frac{k\,t_s}{\phi\mu c_tA} = \frac{1}{C_A}\,\exp\frac{p_e - \bar p}{m_e}

where:

$t_s$ synthetic flowing time, the time that we put into the Horner function, s
$t_{sDA}$ the same, referred to the area of the region, —
$C_A$ shape factor of the area with the given position of the well, —
$p_e - \bar p$ excess of the pressure at the chosen point of the boundary over the average pressure in pseudo-steady-state flow, Pa
$m_e$ slope of radial flow per unit of natural logarithm, $qB\mu/(4\pi kh)$ , Pa
$A,\ k,\ \phi,\ \mu,\ c_t$ area of the region, permeability, porosity, viscosity, total compressibility, m², m², —, Pa·s, 1/Pa

Units: SI; in metric oilfield units $t_{sDA} = 3.553\cdot10^{-3}\,kt_s/(\phi\mu c_tA)$ : mD, h, mPa·s, 1/MPa, m²

Holds when

before shut-in the well had reached pseudo-steady-state flow
we analyse the build-up by the radial-flow line, while the shut-in has not yet been felt at the boundaries
the ratio $(p_e - \bar p)/m_e$ depends only on the shape of the area, the position of the well and the chosen point of the boundary. We calculate it with a model of the area (the table below)

For a well at the centre of a circle, $p_e - \bar p = m_e/2$ , and $t_{sDA} = e^{0.5}/31.62 = 0.052$ . In a non-circular area the pressure along the boundary is not the same, so the synthetic time depends on which point of the boundary $p^*$ should match. Usually we take the farthest one, where the pressure in the area is highest. The synthetic time comes out at about half the time to reach pseudo-steady-state flow. With it we also calculate the average pressure: $\bar p = p^* - m\lg(C_At_{sDA})$ .

Synthetic time for a well at the centre of the area (calculated by the model of the lesson)

Area: circle, well at the centre
$C_A$: 31.62
Point of the boundary: $(p_e - \bar p)/m_e$: boundary: 0.500
$t_{sDA}$: 0.052

Area: square, at the centre
$C_A$: 30.88
Point of the boundary: $(p_e - \bar p)/m_e$: corner: 0.693
$t_{sDA}$: 0.065

Area: square, at the centre
$C_A$: 30.88
Point of the boundary: $(p_e - \bar p)/m_e$: middle of a side: 0.347
$t_{sDA}$: 0.046

Area: rectangle 2:1, at the centre
$C_A$: 21.84
Point of the boundary: $(p_e - \bar p)/m_e$: corner: 1.055
$t_{sDA}$: 0.132

Area: rectangle 2:1, at the centre
$C_A$: 21.84
Point of the boundary: $(p_e - \bar p)/m_e$: middle of the long side: −0.708
$t_{sDA}$: 0.023

Area	$C_A$	Point of the boundary: $(p_e - \bar p)/m_e$	$t_{sDA}$
circle, well at the centre	31.62	boundary: 0.500	0.052
square, at the centre	30.88	corner: 0.693	0.065
square, at the centre	30.88	middle of a side: 0.347	0.046
rectangle 2:1, at the centre	21.84	corner: 1.055	0.132
rectangle 2:1, at the centre	21.84	middle of the long side: −0.708	0.023

At the middle of the long side of the rectangle, the pressure is lower than the average: that point is closer to the well than the far parts of the area. For a circle the values are exact. For the square and the rectangle they come from the model of a closed rectangle. The $C_A$ of the square and of the 2:1 rectangle agree with the known values of the Dietz shape factor.

Worked exampleThe middle well of three in a closed block

Given (assumed data)

Block: closed, 2400 × 800 m; three wells on its axis, 400, 1200 and 2000 m from the end, produced for a year at 40 m³/d each; the middle one was shut in for a build-up, and the neighbours keep producing
Formation and fluid: $k$ ≈ 20 mD, from the derivative plateau of the same build-up (it does not depend on the choice of producing time); $h$ = 20 m; $\phi$ = 0.22; $\mu$ = 1.5 mPa·s; $c_t = 1.5\cdot10^{-3}$ MPa⁻¹; $B$ = 1.2
Readings: 4 h — 21.2479 MPa, 16 h — 21.4751 MPa

Solution

Drainage area: for equal rates, the dividing lines of flow (the lines between the drainage areas of neighbouring wells) split the block into three squares of 800 × 800 m, and the middle well is at the centre of its square. $A$ = 640,000 m², $C_A$ = 30.88. $\phi\mu c_tA = 0.22\cdot1.5\cdot1.5\cdot10^{-3}\cdot640{,}000 = 316.8$ . In an hour, $t_{DA}$ grows by $3.553\cdot10^{-3}\cdot20/316.8 = 2.2431\cdot10^{-4}$ .
Synthetic time for the corner of the square ( $(p_e - \bar p)/m_e$ = 0.693 from the table): $t_{sDA} = e^{0.693}/30.88 = 2.000/30.88 = 0.0648$ ; $t_s = 0.0648/(2.2431\cdot10^{-4}) = 288.9$ h.
The Horner line with $t_s$ : $\lg\dfrac{288.9 + 4}{4} = 1.86466$ , $\lg\dfrac{288.9 + 16}{16} = 1.28004$ ; $m = (21.4751 - 21.2479)/(1.86466 - 1.28004) = 0.3886$ MPa; $k = 2.149\cdot40\cdot1.2\cdot1.5/(0.3886\cdot20) = 19.9$ mD; $p^* = 21.4751 + 0.3886\cdot1.28004 = 21.9725$ MPa.
With the actual producing time of 8760 h: $\lg(8764/4) = 3.34064$ , $\lg(8776/16) = 2.73918$ ; $m = 0.3777$ MPa, $k = 20.5$ mD, $p^* = 22.5097$ MPa.
Average pressure of the area: with $t_s$ , $\bar p = 21.9725 - 0.3886\cdot\lg(30.88\cdot0.0648) = 21.9725 - 0.3886\cdot0.30125 = 21.8554$ MPa. With the actual time, $t_{pDA} = 8760\cdot2.2431\cdot10^{-4} = 1.9650$ , $\lg(30.88\cdot1.9650) = 1.78304$ , and $\bar p = 22.5097 - 0.3777\cdot1.78304 = 21.8362$ MPa.

Answer: With the synthetic time,

p^*

= 21.97 MPa, the pressure at the corner of the drainage area at the moment of shut-in (by the model, 21.969 MPa). With the actual time it is 22.51 MPa, above the pressure at any point of the block. The average pressure of the area is about 21.86 MPa.

What it means. By the material balance of the block, the average pressure at the moment of shut-in is 21.852 MPa, and at the middle of the side of the square it is 21.910 MPa. The synthetic time gave not only a meaningful $p^*$ but also a longer straight line. With the actual time, the slope from the readings at 4–16 h is already distorted by the continuing depletion (the slope after production in pseudo-steady-state flow). The permeability came out 3% too high, and the average pressure 0.016 MPa below the material balance. The pressure of 21.97 MPa can be compared with a formation tester reading (a pressure measurement in the formation, taken from inside the well) in a new well drilled near the corner of the area, if the reading is made near the moment of shut-in. Later the pressure there will keep changing. The data in this example are assumed for illustration.

A well among working neighbours

In a developed accumulation, the drainage area of each well is limited not only by faults but also by the dividing lines of flow between neighbours. These lines behave like no-flow boundaries (boundaries that let no fluid through). In pseudo-steady-state flow, the pore volume that a well drains is proportional to its rate (several wells). While all the wells work, the area of a well behaves as a closed one. When one well is shut in, the dividing lines start to move: the neighbours gradually take over its area. The first hours and days of the build-up do not notice it, and everything said above holds: the synthetic time, $p^*$ at the boundary, and the MBH correction from the early line. Then the pressure rise slows down, and the shut-in well turns into an observation well (a well that only records the pressure): its pressure starts to fall together with the depletion that the neighbours continue.

CalculatedThe 60-day build-up of the middle well of the example (calculated by the model of the lesson) on one Horner plot, twice: with the actual producing time of 8760 h (circles) and with the synthetic time of 289 h (dots). The dashed lines are the lines through the readings at 4 and 16 h. The pressure grows up to 21.84 MPa by about the 17th day, and then falls: by the 60th day it is 21.67 MPa. On the plot with the actual time, this fall starts while the Horner ratios are still large.

What we get from such a build-up. The early line gives the flow capacity and the skin. With the synthetic time, $p^*$ equals the pressure at the boundary of the drainage area at the moment of shut-in. With the MBH correction for the shape of the area at the moment of shut-in, we get the average pressure of this area (Build-up in a bounded reservoir and average reservoir pressure).
What we do not do. We do not treat the late part of the record as a build-up in a closed area: the levelling off and the fall come from the work of the neighbours, not from the shape of the area. We deal separately with the pressure trend (the slow background change of the pressure) on which the build-up runs (Testing in a developed reservoir: Slider's method).
How we choose the area. We find the shape and the size of the drainage area at the moment of shut-in from the layout of the wells, their rates and the faults. Without them we can neither calculate the synthetic time nor correct $p^*$ to the average pressure.

Terms and symbols

Superposition time function for a build-up $X_{BU}$: $\sum w_j\lg[(\theta_j + \Delta t)/\Delta t]$ with weights $w_j = (q_j - q_{j-1})/q_n$ . For one flow period, it is the logarithm of the Horner ratio.
Superposition derivative $\Delta p'_X$: The derivative of the pressure rise against $X_{BU}\ln 10$ , with a minus sign. In radial flow it equals the plateau $m_e$ for any rate history.
Equivalent producing time $t_p = N_p/q_n$: The length of a flow period at the last rate that produces the same cumulative production. It is the classical replacement of the rate history in the Horner function.
Synthetic flowing time $t_s$: The producing time that we put into the Horner function after a long production in a bounded area, so that $p^*$ is the pressure at the boundary of the drainage area.
Ratio of slopes $R(\Delta t)$: How many times the local slope of the Horner plot with one producing time differs from the correct one.
Dividing line of flow: The boundary between the drainage areas of neighbouring wells. In pseudo-steady-state flow it lets no fluid through, and after one of the wells is shut in it moves.

Common mistakes and misconceptions

Wrong

“If the rate changed, it is enough to put in $N_p/q$ : this is always right.”

In fact

The equivalent time is exact for late readings, and tolerable if the last rate was held for a long time. If the rate changed noticeably shortly before shut-in, the slope and the derivative are distorted. The superposition time function is more reliable.

Wrong

“The producing time is the duration of the last flow period.”

In fact

The last flow period does not cancel the previous ones. In the example of the lesson, two days at 80 m³/d after 26 days at smaller rates gave a permeability 10% lower, if we take only them for $t_p$ .

Wrong

“The derivative of the build-up rose by half, so there is a fault nearby.”

In fact

First check the rate history. A rate cut a day before shut-in gives the same rise of a derivative taken with one producing time. Against the superposition time function it disappears.

Wrong

“The $p^*$ from the Horner plot with the actual producing time is the formation pressure at the boundary of the area.”

In fact

After a long production in a closed area, such a $p^*$ is above the pressure at any point of the formation. It equals the pressure at the boundary only with the synthetic time. We find the average pressure by the MBH correction.

Wrong

“The pressure at the end of the build-up fell, so the gauge has failed.”

In fact

In a developed accumulation, a shut-in well becomes, with time, an observation well, and it “sees” the depletion from its neighbours. This is information about the formation, not a defect of the record.

In depth: advanced detail

A short shut-in in the middle of production: how long it is remembered

A common case: the well produced at a constant rate, was shut in briefly (for a measurement or to run in instruments), produced again, and only then was shut in for the build-up. Let the intermediate shut-in last $\tau$ and end a time $T$ before the main one. If we put the total producing time into the Horner function, the derivative of the build-up deviates from the plateau by an amount given by the same memory formula:

The distortion of the derivative by a short intermediate shut-in

\frac{\delta\,\Delta p'}{m_e} = -\frac{\tau\,\Delta t}{(T + \tau + \Delta t)(T + \Delta t)},\qquad \max\left|\frac{\delta\,\Delta p'}{m_e}\right| \approx \frac{\tau}{4T}

where:

$\delta\,\Delta p'$ deviation of the derivative of the build-up from the plateau because of the intermediate shut-in, MPa
$\tau$ duration of the intermediate shut-in, h
$T$ time from the end of the intermediate shut-in to the main one, h
$\Delta t$ time after the main shut-in, h

Units: dimensionless; times in the same units

Holds when

radial flow; before the intermediate shut-in the well produced much longer than $T$ and $\Delta t$
the estimate of the maximum is for $\tau \ll T$ ; it is reached near $\Delta t = \sqrt{T(T + \tau)}$ .

The derivative dips less, the shorter the intermediate shut-in and the earlier it was. For example, a three-hour shut-in that ended 30 h before the main one distorts the derivative by no more than 2.4%, at about 31 h after shut-in. For the distortion to be below 2%, the last flow period must last longer than $12.5\,\tau$ . Otherwise we include the intermediate shut-in in the superposition time function.

Skin after several flow periods

We find the skin from a build-up after several flow periods in the same way as after one (skin from a build-up): from the difference between the line and the bottomhole pressure at the moment of shut-in. Only the correction for the producing time changes. It is built from all the rate changes:

Skin from the line against the superposition time function

S = 1.151\left[\frac{p_{1\text{h}} - p_{wf}(\Delta t = 0)}{m} - \lg\frac{k}{\phi\mu c_tr_w^2} + 2.098 + \sum_{j=1}^{n}w_j\lg\frac{\theta_j + 1}{\theta_j}\right]

where:

$p_{1\text{h}}$ pressure on the line against $X_{BU}$ at $\Delta t$ = 1 h: $p^* - m\,X_{BU}(1\ \text{h})$ , MPa
$w_j,\ \theta_j$ weights and ages of the rate changes; $\theta_j$ in hours, —, h
$m,\ k,\ \phi,\ \mu,\ c_t,\ r_w$ slope of the line, permeability, porosity, viscosity, total compressibility, well radius, MPa, mD, —, mPa·s, 1/MPa, m

Units: metric oilfield units; in oilfield units, instead of 2.098 we use 3.23

Holds when

radial flow over the whole history; the skin has not changed
for one flow period the correction equals $\lg[(t_p + 1)/t_p]$ , as in the Horner formula.

The correction is noticeable only if some flow period lasted just a few hours. Usually it is thousandths, and the formula coincides with the formula for one flow period.

Worked exampleSkin of the well of the example with three flow periods

Given (assumed data)

From the example: $m$ = 0.8587 MPa, $p^*$ = 26.504 MPa, $k$ = 34.8 mD; $p_{wf}(\Delta t = 0)$ = 19.2270 MPa
Formation: $\phi$ = 0.19; $\mu$ = 1.8 mPa·s; $c_t = 1.3\cdot10^{-3}$ MPa⁻¹; $r_w$ = 0.1 m

Solution

$X_{BU}(1) = 0.375\lg 673 + 0.3125\lg 433 + 0.3125\lg 49 = 2.4126$ ; $p_{1\text{h}} = 26.504 - 0.8587\cdot2.4126 = 24.4323$ MPa.
$\lg\dfrac{k}{\phi\mu c_tr_w^2} = \lg\dfrac{34.8}{0.19\cdot1.8\cdot1.3\cdot10^{-3}\cdot0.01} = 6.8936$ ; the correction is $0.375\lg\dfrac{673}{672} + 0.3125\lg\dfrac{433}{432} + 0.3125\lg\dfrac{49}{48} = 0.0034$ .
$S = 1.151\,[(24.4323 - 19.2270)/0.8587 - 6.8936 + 2.098 + 0.0034] = 1.46$ .

Answer: The skin is about 1.46. The correction for the rate history is 0.0034, so it changes almost nothing.

What it means. The data were calculated with a skin of 1.5. The largest share of the correction comes from the last flow period: two days is only 48 h. The data in this example are assumed for illustration.

Connections to other topics

Superposition principle — adding in time, and the superposition time function, on which the lesson is built.
Pressure build-up analysis — the Horner equation, $p^*$ and the skin from a build-up after one flow period.
Effect of producing time on build-up analysis — a short production before shut-in, and the equivalent time after it.
Build-up in a bounded reservoir and average reservoir pressure — the average pressure by the MBH correction, and the choice of the time for a long line.
Pseudo-steady-state flow and the shape factor — the drainage areas of several wells, and the shape factor.
Diagnosis with the pressure derivative — the diagnostic plot, on which the rate history draws false boundaries.
A sealing fault near the well: the method of images — a real fault: what not to confuse with a rise of the derivative after a rate cut.
Testing in a developed reservoir: Slider's method — a build-up in a developed accumulation against a pressure trend.
Reservoir limit test — the material balance time $Q/q$ : the same quantity in the analysis of a long production.

Check yourself

1
A well produced for 20 days at 50 m³/d, then for 5 days at 100 m³/d, and then was shut in. Find the equivalent producing time, and estimate how much the slope of the Horner plot with it differs from the correct one 24 h after shut-in.
Show the answer
$N_p = 50\cdot20 + 100\cdot5 = 1500$ m³, and $t_p = 1500/100 = 15$ days = 360 h (in fact the well produced for 600 h). The weights are $w_1 = 0.5$ ( $\theta_1$ = 600 h) and $w_2 = 0.5$ ( $\theta_2$ = 120 h): $R(24) = \dfrac{384}{360}\left(0.5\cdot\dfrac{600}{624} + 0.5\cdot\dfrac{120}{144}\right) = 1.0667\cdot0.8974 \approx 0.957$ . The slope is 4% smaller than the correct one, and the permeability from it is 4% larger.
2
A well produced for a month at 90 m³/d. A day before the build-up the rate was cut to 45 m³/d. The derivative calculated from the Horner function with $N_p/q$ started to rise after a few hours. What is the most likely cause?
1. A.The effect of the rate history
2. B.A sealing fault
3. C.A constant-pressure boundary
4. D.The end of wellbore storage
Show the answer
Correct answer: A. The effect of the rate history
After the rate cut, the weight of the old rate is $w_1 = 90/45 = 2$ , and that of the last change is $-1$ . After a few hours, the recent change is “forgotten”, and the slope starts to answer to the old rate, which is twice as large: the derivative grows towards the doubled level, like at a fault. The check is the derivative against the superposition time function of the whole history: if the rise disappears, there is no fault.
3
Before shut-in, a well produced for 36 h at the rate $q_n$ , and before that for a long time at a rate 1.5 times larger. The build-up was analysed with $t_p = N_p/q_n$ from readings up to 6 h. Estimate the error of the slope, and how long the last flow period had to be to keep it within 3%.
Show the answer
The first-order estimate: $R - 1 \approx (\Delta t/\theta_n)(q_{n-1}/q_n - 1) = (6/36)\cdot0.5 \approx 0.083$ . The slope is overstated by about 8%, and the permeability is understated. For an error of 3% we need $\theta_n \ge 6\cdot0.5/0.03 = 100$ h, that is, more than four days.
4
After a year of production in a closed block, the build-up was processed by Horner with the actual producing time. What is the $p^*$ that we got?
1. A.A quantity with no direct physical meaning, above the pressure at any point of the area
2. B.The initial reservoir pressure
3. C.The pressure at the boundary of the area at the moment of shut-in
4. D.The average pressure in the drainage area
Show the answer
Correct answer: A. A quantity with no direct physical meaning, above the pressure at any point of the area
After a long production in pseudo-steady-state flow, the continuation of the line with the actual time goes above any pressure in the formation (in the example of the lesson, 22.51 MPa when the highest pressure in the block is 21.97 MPa). We get the average pressure with the MBH correction, and the pressure at the boundary by continuing the line with the synthetic time.
5
A well at the centre of a circular drainage area of 0.6 km² has long been producing in pseudo-steady-state flow: $k$ = 25 mD, $\phi$ = 0.18, $\mu$ = 1.2 mPa·s, $c_t = 1.6\cdot10^{-3}$ MPa⁻¹. What synthetic time must we put into the Horner function so that $p^*$ is the pressure at the boundary of the area?
Show the answer
For a circle, $t_{sDA} = e^{0.5}/31.62 \approx 0.052$ . $\phi\mu c_tA = 0.18\cdot1.2\cdot1.6\cdot10^{-3}\cdot600{,}000 = 207.4$ , and $t_s = 0.052\cdot207.4/(3.553\cdot10^{-3}\cdot25) = 10.78/0.0888 \approx 121$ h, about 5 days.
6
In a long build-up of a well among working neighbours, the pressure stopped rising after three weeks and started to fall. Why?
1. A.The shut-in well has become an observation well: the dividing lines of flow have moved, and it sees the depletion from its neighbours
2. B.The gauge is faulty
3. C.Wellbore storage has ended
4. D.This is a sign of a sealing fault
Show the answer
Correct answer: A. The shut-in well has become an observation well: the dividing lines of flow have moved, and it sees the depletion from its neighbours
While all the wells were producing, the area of the shut-in well was closed. After shut-in, the neighbours gradually take over its area, and the pressure in it follows the general depletion of the block. From the early line of the build-up we can still find $kh$ , the skin, $p^*$ and the average pressure. We deal with the late fall separately.

Updated: 19 October 2026