Effect of producing time on build-up analysis

Quick refresher

A build-up is the sum of two responses: the continuing pressure drop from the “producing” well, and the rise from an imaginary injection well. So we subtract from the drawdown response the further drop that would have occurred over the time $\Delta t$ if the well had kept producing.
The pressure rise after shut-in cannot exceed $p_i - p_{wf}(t_p)$ . In radial flow, the slope of the build-up with respect to $\ln\Delta t$ (the derivative $\Delta p'$ ) equals $m_e\,t_p/(t_p + \Delta t)$ . This is 9% below the plateau at $\Delta t = 0.1\,t_p$ , and half of the plateau at $\Delta t = t_p$ .
The Horner function allows for the producing time exactly. The MDH method is valid only for $\Delta t \ll t_p$ .
Equivalent time $\Delta t_e$ turns a build-up into a drawdown: the pressure rise a time $\Delta t$ after shut-in equals the pressure drop a time $\Delta t_e$ after the start of production. $\Delta t_e$ is always smaller than $t_p$ , so the build-up reproduces the drawdown only up to a time equal to the producing time.
The derivative with respect to $\ln\Delta t_e$ (the same as the derivative with respect to the Horner function) equals the plateau $m_e$ in radial flow, for any producing time. We plot it both against $\Delta t_e$ and against the elapsed time $\Delta t$ .
The time conversion requires the flow to have reached the radial regime before shut-in. With a varying rate, with boundaries, or in gas wells, we need superposition of the whole rate history, and a model.

Key relationships

\Delta t_e = \dfrac{t_p\,\Delta t}{t_p + \Delta t}

equivalent time; $t_p$ and $\Delta t$ in the same units

\Delta p_{BU}(\Delta t) = \Delta p_{DD}(\Delta t_e)

build-up as drawdown: radial flow in an infinite reservoir

\Delta p' = m_e\,\dfrac{t_p}{t_p + \Delta t}

the local slope of the build-up with respect to $\ln\Delta t$ . This is what the MDH method and the derivative with respect to $\Delta t$ see

\Delta p'_e = \dfrac{t_p + \Delta t}{t_p}\,\Delta p' = m_e

derivative with respect to equivalent time during radial flow

\Delta t = \dfrac{t_p\,T}{t_p - T}

how long to wait after shut-in for the equivalent time to reach $T$

Common pitfalls

Mistaking the fall of the derivative at the end of a build-up after a short producing time for a constant-pressure boundary or for pressure support (fluid flowing in from outside).
Using the MDH method on a build-up longer than a tenth of the producing time.
Matching a build-up to drawdown type curves against $\Delta t$ instead of $\Delta t_e$ .
Taking the length of a short last flow period after a long production history as $t_p$ . The time will be compressed too much.
Applying equivalent time when the flow had not reached the radial regime before shut-in.

After this lesson you can

use superposition (adding up the effects of separate rate changes) to explain why the pressure rise after a short producing time lags behind the pressure drop after the start of production and cannot exceed $p_i - p_{wf}(t_p)$
estimate how much the MDH method and the derivative with respect to $\ln\Delta t$ underestimate the slope for a given ratio $\Delta t/t_p$ , and choose between MDH, the Horner method and equivalent time
calculate the equivalent time and analyse a build-up as a drawdown: find the flow capacity, find the skin from the drawdown formula, and find the initial pressure on the straight line at $\Delta t_e = t_p$
calculate the derivative with respect to equivalent time from the readings, and find the flow capacity from its plateau (the flat part of the curve)
plan the flow and shut-in times so that the build-up reproduces the part of the drawdown that we need
state the conditions under which equivalent time does not apply, and what to do instead

Why a build-up “remembers” the producing time

The superposition principle gives a convenient picture of shut-in. The well appears to go on producing at the rate $q$ , while from the moment of shut-in an imaginary injection well with the same rate starts working at the same point (why the pressure builds up). So after shut-in two processes go on in the reservoir at once. At the well itself the pressure rises. But the drawdown cone created during production keeps spreading. Fluid flows towards the wellbore from ever more distant parts of the reservoir, and the pressure there goes on falling for some time. The shorter the producing time, the smaller the cone and the faster it fills in. There is almost nothing left to restore.

Let $\Delta p_{DD}(t) = p_i - p_{wf}(t)$ be the drop in bottomhole pressure a time $t$ after the start of production at a constant rate, from a reservoir with uniform pressure (the drawdown curve, DD). Let $\Delta p_{BU}(\Delta t)$ be the pressure rise a time $\Delta t$ after shut-in (the build-up, BU), that is, $p_{ws}(\Delta t) - p_{wf}(\Delta t = 0)$ . Superposition links them in a single line:

Build-up from the drawdown response

\Delta p_{BU}(\Delta t) = \Delta p_{DD}(\Delta t) - \Big[\Delta p_{DD}(t_p + \Delta t) - \Delta p_{DD}(t_p)\Big]

where:

$\Delta p_{BU}(\Delta t)$ rise in bottomhole pressure a time $\Delta t$ after shut-in: $p_{ws}(\Delta t) - p_{wf}(\Delta t = 0)$ , Pa
$\Delta p_{DD}(t)$ drop in bottomhole pressure a time $t$ after the start of production at a constant rate $q$ from a reservoir at pressure $p_i$ : $p_i - p_{wf}(t)$ , Pa
$t_p$ producing time at the rate $q$ before shut-in, s
$\Delta t$ time after shut-in, s

Units: any consistent system of units

Holds when

The problem is linear: the fluid is single-phase and slightly compressible, and the reservoir and fluid properties do not depend on pressure.
Before the start of production the pressure in the reservoir is uniform, and before shut-in the surface rate was constant.
$\Delta p_{DD}$ is the response of the same well, with the same skin and wellbore storage, shut in at the same point. Then the equality is exact, not approximate.

The square bracket is how much further the bottomhole pressure would have dropped over the time $\Delta t$ if the well had kept producing. While $\Delta t \ll t_p$ it is small: after a long producing time the pressure drop hardly grows over a short interval, and the build-up repeats the drawdown. When $\Delta t$ is comparable with $t_p$ , the bracket is comparable with $\Delta p_{DD}(\Delta t)$ itself, and the recovery visibly lags behind the pressure drop after the start of production.

The second consequence is a limit. In an infinite reservoir the pressure returns to its initial value after a very long shut-in. So the pressure rise cannot exceed $p_i - p_{wf}(t_p) = \Delta p_{DD}(t_p)$ , which is the drop accumulated during the producing time. The drawdown grows without limit, like the logarithm of time. The build-up, however, runs into a “ceiling”, and the shorter the producing time was, the lower the ceiling is. That is why a build-up after a short producing time flattens out earlier, and its logarithmic derivative falls.

CalculatedCalculated logarithmic derivative of the build-up with respect to

\ln\Delta t

for the well of the lesson's examples, after production at the same rate for 0.5, 2, 6 and 24 h (solid lines; 6 h in orange), and the derivative of the drawdown (dashed). At the start all the curves are the same: the fall after wellbore storage, and then the plateau

m_e

. Then the derivative of each build-up turns downwards, and the shorter the producing time, the earlier it does so. The dot on each curve marks the moment

\Delta t = t_p

, where the derivative is half the plateau.

What changes if…

…the producing time is twice as long — the build-up follows the drawdown for longer: the same lag appears at twice the $\Delta t$ .
…the permeability is twice as high — the slope and the plateau of the derivative are half as large, but the moment the build-up starts to lag is the same: in radial flow the lag depends only on $\Delta t/t_p$ .
…the skin is larger — the build-up pressures $p_{ws}$ in radial flow are the same, but the bottomhole pressure at shut-in is lower. So the pressure rise measured from it is larger by the pressure drop across the skin. The lag of the slope, the factor $t_p/(t_p + \Delta t)$ , is unchanged.
…wellbore storage is larger — the early period is longer. If it covers the whole producing time, converting the time will no longer help (see the section on the limits of the method).

Local slope: the MDH method and the Horner plot

We can see how fast the build-up lags from the logarithmic approximation. As long as both the flow period and the shut-in take place in radial flow in an infinite reservoir, $\Delta p_{DD}(t) = m_e(\ln t_D + 0.809 + 2S)$ . Here $t_D$ is the dimensionless time, and $m_e = qB\mu/(4\pi kh)$ is the slope of the semilog straight line per unit of natural logarithm. It is also the level of the plateau of the logarithmic derivative (derivative plateau). We put this into the formula of the previous section and differentiate with respect to $\ln\Delta t$ :

Local slope of the build-up with respect to the logarithm of the shut-in time

\Delta p' = \frac{\dd\,\Delta p_{BU}}{\dd\ln\Delta t} = m_e\,\frac{t_p}{t_p + \Delta t}

where:

$\Delta p'$ logarithmic derivative of the build-up with respect to $\ln\Delta t$ — the slope of the MDH plot per unit of natural logarithm, Pa
$m_e$ level of the derivative plateau: $qB\mu/(4\pi kh)$ , Pa
$t_p,\ \Delta t$ producing time and time after shut-in, s

Units: any consistent system of units

Holds when

Radial flow in an infinite reservoir at the times $\Delta t$ and $t_p + \Delta t$ : wellbore storage is over and the boundaries have no effect.
The rate is constant before shut-in.

The factor $t_p/(t_p + \Delta t)$ depends neither on the reservoir nor on the well. It depends only on the ratio of the times. This is the producing-time effect on the slope. So on the MDH plot ( $p_{ws}$ against $\lg\Delta t$ ) the points bend downwards from the straight line, and the derivative with respect to $\ln\Delta t$ leaves the plateau, although the reservoir has no boundaries at all.

Local slope of the build-up with respect to

\ln\Delta t

as a fraction of the plateau in radial flow

$\Delta t/t_p$: 0.01
$t_p/(t_p + \Delta t)$: 0.990
What it means: the MDH method is practically exact

$\Delta t/t_p$: 0.05
$t_p/(t_p + \Delta t)$: 0.952
What it means: the slope is 5% lower

$\Delta t/t_p$: 0.1
$t_p/(t_p + \Delta t)$: 0.909
What it means: 9% lower

$\Delta t/t_p$: 0.5
$t_p/(t_p + \Delta t)$: 0.667
What it means: a third lower

$\Delta t/t_p$: 1
$t_p/(t_p + \Delta t)$: 0.500
What it means: half as large

$\Delta t/t_p$: 1.5
$t_p/(t_p + \Delta t)$: 0.400
What it means: the typical end of a build-up in a drill stem test

$\Delta t/t_p$: 3
$t_p/(t_p + \Delta t)$: 0.250
What it means: a quarter as large

$\Delta t/t_p$	$t_p/(t_p + \Delta t)$	What it means
0.01	0.990	the MDH method is practically exact
0.05	0.952	the slope is 5% lower
0.1	0.909	9% lower
0.5	0.667	a third lower
1	0.500	half as large
1.5	0.400	the typical end of a build-up in a drill stem test
3	0.250	a quarter as large

This is the instantaneous slope at the moment $\Delta t$ . The slope of a chord (a straight line drawn between two points) across a section of the build-up is the average of this factor over the section (in logarithms of time). So it is distorted less than the slope at the right-hand end of the section.

The Horner equation (Horner equation and plot) has no such distortion. The function $\lg[(t_p + \Delta t)/\Delta t]$ contains the producing time, and in radial flow the points fall on a straight line with the slope $m = 2.303\,m_e$ for any $t_p$ . The MDH method drops the term $\lg[(t_p + \Delta t)/t_p]$ , so it is valid only for $\Delta t \ll t_p$ . By the time $\Delta t$ , the points of the MDH plot lie below the continuation of the initial straight line by $m\lg[(t_p + \Delta t)/t_p]$ .

A false constant-pressure boundary

On the diagnostic plot of a build-up after a short producing time, the derivative with respect to $\ln\Delta t$ turns downwards at the end of the record. It does the same for a well near a constant-pressure boundary, such as a gas cap (free gas above the oil) or an active aquifer (a water-bearing layer that pushes water in; late-time shapes). Before you look for pressure support, compare the length of the record with the producing time. If the decline starts at about $\Delta t \approx 0.1\,t_p$ and by $\Delta t = t_p$ the derivative is half the plateau, this is the producing-time effect. The check is the derivative with respect to equivalent time (below): for a reservoir without boundaries it stays on the plateau.

Equivalent time

The build-up lags the drawdown because the time after shut-in “weighs” less. So we can look for a time at which the two coincide. We put the logarithmic approximation into all three terms of the superposition. Of the three constants $0.809 + 2S$ , two cancel. The logarithms of the times combine into the logarithm of a single fraction. What remains is the drawdown formula, only with a different time:

Equivalent time

\Delta p_{BU}(\Delta t) = \Delta p_{DD}(\Delta t_e),\qquad \Delta t_e = \frac{t_p\,\Delta t}{t_p + \Delta t}

where:

$\Delta t_e$ equivalent time: the time after the start of production at which the pressure drop on the drawdown would equal the pressure rise on the build-up a time $\Delta t$ after shut-in, s
$t_p,\ \Delta t$ producing time and time after shut-in, s
$\Delta p_{BU},\ \Delta p_{DD}$ pressure rise after shut-in and pressure drop after the start of production, Pa

Units: any consistent system; $\Delta t_e$ is in the same units as $t_p$ and $\Delta t$

Holds when

Radial flow in an infinite reservoir at the times $t_p$ and $t_p + \Delta t$ : before shut-in the flow had reached the radial regime, and the disturbance had not reached the boundaries.
At the time $\Delta t$ the flow is also radial, or else $\Delta t \ll t_p$ . In that case $\Delta t_e \approx \Delta t$ , and the equality holds during wellbore storage as well.
The rate is constant before shut-in.

We can analyse a build-up plotted against $\Delta t_e$ with all the tools of drawdown analysis (Pressure drawdown analysis): the semilog straight line, the drawdown skin formula and drawdown type curves. R. Agarwal published the technique in 1980. The Horner equation has long contained the same function of time, and equivalent time merely writes it as a time.

In other unit systems (1)

Through the Horner ratio (decimal logarithms)

\lg\Delta t_e = \lg t_p - \lg\frac{t_p + \Delta t}{\Delta t}

$t_p$ and $\Delta t$ in the same units; $\Delta t_e$ in the same units.

There are no constants: a semilog plot against $\lg\Delta t_e$ is the Horner plot mirrored left to right and shifted by $\lg t_p$ . A Horner ratio of 1 corresponds to $\Delta t_e = t_p$ .

At the start of the build-up $\Delta t_e \approx \Delta t$ : at $\Delta t = 0.1\,t_p$ the equivalent time is 9% less than the elapsed time.
$\Delta t_e$ is always less than $t_p$ . It tends to $t_p$ as the shut-in becomes infinitely long. At $\Delta t = t_p$ it equals $t_p/2$ .
The point on the straight line at $\Delta t_e = t_p$ corresponds to an infinitely long shut-in. This is where we read $p^*$ , which in a virgin reservoir equals the initial pressure.
The longer the shut-in, the more crowded the points on the $\Delta t_e$ axis. The last hours of the record are squeezed into fractions of an hour of equivalent time. The Horner axis squeezes the points in the same way.

A build-up plotted against $\Delta t_e$ is a drawdown. So we calculate the skin from the drawdown formula (skin from a drawdown), using the pressure rise on the straight line at $\Delta t_e$ = 1 h. The correction for the producing time, which we add to the Horner skin formula, is already included here by the time conversion itself:

Skin from the straight line against equivalent time

S = 1.151\left[\frac{\Delta p_{1\text{h}}}{m} - \lg\frac{k}{\phi\,\mu\,c_t\,r_w^2} + 2.098\right]

where:

$\Delta p_{1\text{h}}$ pressure rise on the straight line (or its continuation) at $\Delta t_e$ = 1 h, measured from $p_{wf}(\Delta t = 0)$ , MPa
$m$ slope of the straight line against $\lg\Delta t_e$ per decimal cycle — the same as that of the Horner line, MPa
$k,\ \phi,\ \mu,\ c_t,\ r_w$ permeability (from the slope), porosity, viscosity, total compressibility, wellbore radius, mD, —, mPa·s, 1/MPa, m

Units: metric oilfield units: pressure in MPa, $k$ in mD, $\mu$ in mPa·s, $c_t$ in 1/MPa, $r_w$ in m, time in hours

Holds when

The conditions of equivalent time hold: before shut-in the flow had reached the radial regime.
We read $\Delta p_{1\text{h}}$ on the straight line, not from a measured point. If $t_p$ is less than an hour, the point $\Delta t_e$ = 1 h lies on the continuation of the line beyond $t_p$ .

The formula is the same as for a drawdown (Pressure drawdown analysis), with $\Delta p_{1\text{h}}$ in place of $p_i - p_{1\text{h}}$ . It coincides with the Horner formula if we add the term $\lg[(t_p + 1)/t_p]$ to the latter. Without that term, the Horner skin after a short producing time comes out too low.

In other unit systems (1)

Oilfield units

S = 1.151\left[\frac{\Delta p_{1\text{h}}}{m} - \lg\frac{k}{\phi\,\mu\,c_t\,r_w^2} + 3.23\right]

pressure and $m$ in psi, $k$ in mD, $\mu$ in cP, $c_t$ in 1/psi, $r_w$ in ft, time in hours.

$3.23 = -\lg(2.637\cdot10^{-4}) - 0.809/\ln 10$ , as in the drawdown skin formula; in metric units $2.098 = -\lg(3.553\cdot10^{-3}) - 0.351$ .

Final pressure build-up in a drill stem test (assumed data)

$\Delta t$ , h: 0.01
$\Delta t_e$ , h: 0.00998
$p_{ws}$ , MPa: 21.4875
$\Delta p$ , MPa: 2.6969

$\Delta t$ , h: 0.02
$\Delta t_e$ , h: 0.0199
$p_{ws}$ , MPa: 21.8868
$\Delta p$ , MPa: 3.0962

$\Delta t$ , h: 0.05
$\Delta t_e$ , h: 0.0496
$p_{ws}$ , MPa: 22.1990
$\Delta p$ , MPa: 3.4084

$\Delta t$ , h: 0.1
$\Delta t_e$ , h: 0.0984
$p_{ws}$ , MPa: 22.3875
$\Delta p$ , MPa: 3.5969

$\Delta t$ , h: 0.2
$\Delta t_e$ , h: 0.194
$p_{ws}$ , MPa: 22.5617
$\Delta p$ , MPa: 3.7711

$\Delta t$ , h: 0.5
$\Delta t_e$ , h: 0.462
$p_{ws}$ , MPa: 22.7777
$\Delta p$ , MPa: 3.9871

$\Delta t$ , h: 1
$\Delta t_e$ , h: 0.857
$p_{ws}$ , MPa: 22.9291
$\Delta p$ , MPa: 4.1385

$\Delta t$ , h: 1.5
$\Delta t_e$ , h: 1.20
$p_{ws}$ , MPa: 23.0110
$\Delta p$ , MPa: 4.2204

$\Delta t$ , h: 2
$\Delta t_e$ , h: 1.50
$p_{ws}$ , MPa: 23.0651
$\Delta p$ , MPa: 4.2745

$\Delta t$ , h: 3
$\Delta t_e$ , h: 2.00
$p_{ws}$ , MPa: 23.1348
$\Delta p$ , MPa: 4.3442

$\Delta t$ , h: 4.5
$\Delta t_e$ , h: 2.57
$p_{ws}$ , MPa: 23.1956
$\Delta p$ , MPa: 4.4050

$\Delta t$ , h: 6
$\Delta t_e$ , h: 3.00
$p_{ws}$ , MPa: 23.2328
$\Delta p$ , MPa: 4.4422

$\Delta t$ , h: 9
$\Delta t_e$ , h: 3.60
$p_{ws}$ , MPa: 23.2768
$\Delta p$ , MPa: 4.4862

$\Delta t$ , h	$\Delta t_e$ , h	$p_{ws}$ , MPa	$\Delta p$ , MPa
0.01	0.00998	21.4875	2.6969
0.02	0.0199	21.8868	3.0962
0.05	0.0496	22.1990	3.4084
0.1	0.0984	22.3875	3.5969
0.2	0.194	22.5617	3.7711
0.5	0.462	22.7777	3.9871
1	0.857	22.9291	4.1385
1.5	1.20	23.0110	4.2204
2	1.50	23.0651	4.2745
3	2.00	23.1348	4.3442
4.5	2.57	23.1956	4.4050
6	3.00	23.2328	4.4422
9	3.60	23.2768	4.4862

The main flow period is $t_p$ = 6 h at an oil rate of 64 m³/d. The bottomhole pressure at the moment the valve is closed is $p_{wf}(\Delta t = 0)$ = 18.7906 MPa, and $\Delta p = p_{ws} - p_{wf}(\Delta t = 0)$ . The data are calculated from a model of a homogeneous infinite reservoir with skin and wellbore storage below the valve at the packer, and rounded to 0.0001 MPa, which is the resolution of a quartz gauge.

CalculatedThe build-up pressures from the table, plotted twice on a semilog plot: against the shut-in time

\Delta t

(open circles, the MDH plot) and against the equivalent time

\Delta t_e

(filled dots). Against

\Delta t

, the points bend downwards after 1–2 h. Against

\Delta t_e

, they lie on the radial-flow straight line (dashed), which the example below draws through the readings at 1.5 and 6 h. Its continuation to

\Delta t_e = t_p

= 6 h gives the initial pressure.

Worked exampleThree ways to analyse one build-up

Given (assumed data)

Main flow period: $q$ = 64 m³/d of oil, $t_p$ = 6 h; the final build-up lasts 9 h
Reservoir and fluid: $B$ = 1.21; $\mu$ = 1.35 mPa·s; $h$ = 9 m; $\phi$ = 0.18; $c_t = 1.5\cdot10^{-3}$ MPa⁻¹; $r_w$ = 0.108 m
Readings: the table above; we draw the line through the readings at 1.5 h (23.0110 MPa) and 6 h (23.2328 MPa), both after wellbore storage has ended

Solution

The pressure difference is $23.2328 - 23.0110 = 0.2218$ MPa, and $qB\mu = 64\cdot1.21\cdot1.35 = 104.544$ .
MDH method: $m = 0.2218/\lg(6/1.5) = 0.2218/0.60206 = 0.3684$ MPa per cycle, and $k = 2.149\cdot104.544/(0.3684\cdot9) = 67.8$ mD.
Horner method: the horizontal coordinates are $\lg(7.5/1.5) = 0.69897$ and $\lg(12/6) = 0.30103$ . Then $m = 0.2218/(0.69897 - 0.30103) = 0.5574$ MPa per cycle, and $k = 2.149\cdot104.544/(0.5574\cdot9) = 44.8$ mD.
Equivalent time: $\Delta t_e = 6\cdot1.5/7.5 = 1.2$ h and $6\cdot6/12 = 3$ h. Then $m = 0.2218/\lg(3/1.2) = 0.2218/0.39794 = 0.5574$ MPa. This is the same slope as the Horner one, because $\lg 2.5 = \lg 5 - \lg 2$ .
Skin from the drawdown formula. The pressure rise at the 1.5 h reading ( $\Delta t_e$ = 1.2 h) is $23.0110 - 18.7906 = 4.2204$ MPa. On the straight line at $\Delta t_e$ = 1 h it is $\Delta p_{1\text{h}} = 4.2204 - 0.5574\cdot\lg 1.2 = 4.2204 - 0.5574\cdot0.07918 = 4.1763$ MPa. Next, $\lg\dfrac{k}{\phi\mu c_tr_w^2} = \lg\dfrac{44.8}{0.18\cdot1.35\cdot1.5\cdot10^{-3}\cdot0.108^2} = 7.0227$ . So $S = 1.151\,(4.1763/0.5574 - 7.0227 + 2.098) = 2.96$ .
We read the initial pressure on the straight line at $\Delta t_e = t_p$ = 6 h. The pressure rise at the 6 h reading ( $\Delta t_e$ = 3 h) is $23.2328 - 18.7906 = 4.4422$ MPa. On the line at 6 h it is $4.4422 + 0.5574\cdot\lg(6/3) = 4.6100$ MPa. So $p_i = 18.7906 + 4.6100 = 23.4006$ MPa.

Answer: By the Horner method and by equivalent time,

k \approx 44.8

mD, the skin is 2.96 and

p_i \approx 23.40

MPa. The MDH method overestimated the permeability: it gave 67.8 mD, which is 51% too high.

What it means. The data were calculated from a model with a permeability of 45 mD, a skin of 3 and an initial pressure of 23.4 MPa. Horner and equivalent time recovered these values with an error of under 1% in permeability and 0.04 in skin. This small error comes from wellbore storage that had not quite finished by the 1.5 h reading, together with the rounding of the readings. MDH went wrong not because of the data. By 6 h the local slope with respect to $\lg\Delta t$ is only $6/(6 + 6) = 0.5$ of the full one, and the chord through 1.5 and 6 h is 1.5 times flatter than the straight line. The skin from the drawdown formula at $\Delta t_e$ = 1 h is the same as the skin from the Horner formula with the correction $\lg[(t_p + 1)/t_p]$ . Without the correction, the Horner formula would have given a skin lower by 0.08. The data in this example are assumed for illustration.

Derivative with respect to equivalent time

For the diagnostic plot of a build-up after a short producing time, we take the derivative not with respect to $\ln\Delta t$ but with respect to $\ln\Delta t_e$ . Since $\ln\Delta t_e = \ln t_p - \ln[(t_p + \Delta t)/\Delta t]$ , this is the same derivative as the one with respect to the Horner function, that is, with respect to the superposition time function for a single rate. We get it from the ordinary derivative by multiplication:

Derivative with respect to equivalent time

\Delta p'_e = \frac{\dd\,\Delta p_{BU}}{\dd\ln\Delta t_e} = \frac{t_p + \Delta t}{t_p}\cdot\frac{\dd\,\Delta p_{BU}}{\dd\ln\Delta t}

where:

$\Delta p'_e$ logarithmic derivative of the build-up with respect to equivalent time, Pa
$\dd\,\Delta p_{BU}/\dd\ln\Delta t$ ordinary logarithmic derivative of the build-up with respect to the time after shut-in, $\Delta p'$ , Pa
$t_p,\ \Delta t$ producing time and time after shut-in, s

Units: any consistent system; the derivatives are in units of pressure

Holds when

$t_p$ is the producing time at a constant rate before shut-in. With a varying rate, we take the derivative with respect to the superposition function of the whole history (Equivalent producing time and synthetic flowing time).
The logarithm is natural. The relation follows from $\dd\ln\Delta t_e/\dd\ln\Delta t = t_p/(t_p + \Delta t)$ .

The factor $(t_p + \Delta t)/t_p$ exactly cancels the factor $t_p/(t_p + \Delta t)$ of the local slope. So in radial flow, $\Delta p'_e = m_e$ for any producing time. While $\Delta t \ll t_p$ , the two derivatives coincide. At the end of a long build-up, the derivative with respect to equivalent time is larger than the ordinary one, by the same factor by which $t_p + \Delta t$ is larger than $t_p$ . For the same neighbouring readings, the noise is amplified by the same factor.

We can plot it in two ways. We can plot both $\Delta p_{BU}$ and $\Delta p'_e$ against $\Delta t_e$ , as Agarwal proposed for the pressure. The build-up can then be overlaid on drawdown type curves. Interpretation software usually plots them against the elapsed time $\Delta t$ . Then the late part of the record is not squeezed towards $t_p$ , and radial flow is still seen as a plateau. The values of the derivative are the same in both cases. Only the time axis differs.

We calculate it from the readings with the same three-point formula as the ordinary derivative (how the derivative is calculated from readings), with $\ln\Delta t_e$ in place of $\ln\Delta t$ . We also measure the differentiation interval in $\ln\Delta t_e$ . At the end of the record the readings crowd together on this axis, so for the same interval the neighbouring points are ever further apart in elapsed time.

CalculatedDiagnostic plot of the build-up in the table. It shows the pressure rise (dots) and two derivatives, calculated from the lesson's model and plotted against the elapsed time

\Delta t

: the derivative with respect to

\ln\Delta t

(grey) and the derivative with respect to

\ln\Delta t_e

(orange). The dashed line is the plateau

m_e

. After 0.5 h the grey derivative turns downwards, while the orange one stays on the plateau to the end of the record.

Worked exampleDerivative from readings: with respect to Δt and to Δt_e

Given (assumed data)

Readings: from the table: pressure rise 4.2204 MPa at 1.5 h, 4.3442 MPa at 3 h, 4.4422 MPa at 6 h, 4.4862 MPa at 9 h
Test: $t_p$ = 6 h; $qB\mu$ = 104.544; $h$ = 9 m

Solution

First, with respect to $\ln\Delta t$ at the 3 h point. The neighbours at 1.5 and 6 h are each $\ln 2 = 0.6931$ away from it, so the derivative is a central difference (the difference between the two neighbours, divided by the distance between them): $\Delta p' = (4.4422 - 4.2204)/(2\cdot0.6931) = 0.1600$ MPa.
Now with respect to $\ln\Delta t_e$ . The equivalent times of the same readings are 1.2, 2 and 3 h. The distances are $\ln(2/1.2) = 0.5108$ and $\ln(3/2) = 0.4055$ . The chord slopes are $s_l = (4.3442 - 4.2204)/0.5108 = 0.2424$ MPa and $s_r = (4.4422 - 4.3442)/0.4055 = 0.2417$ MPa.
The weighted average of the chord slopes is $\Delta p'_e = (0.2424\cdot0.4055 + 0.2417\cdot0.5108)/(0.5108 + 0.4055) = 0.2420$ MPa. As a check with the factor, $0.1600\cdot(6 + 3)/6 = 0.2400$ MPa.
At the 6 h point (neighbours at 3 and 9 h) the same formula gives $\Delta p' = 0.1206$ MPa with respect to $\ln\Delta t$ and $\Delta p'_e = 0.2414$ MPa with respect to $\ln\Delta t_e$ .
The flow capacity from the derivative with respect to equivalent time is $kh = 0.933\cdot104.544/0.2420 = 403$ mD·m, so $k = 44.8$ mD. The derivative with respect to $\ln\Delta t$ at the same point would have given 610 mD·m, which is 1.5 times higher.

Answer: The derivative with respect to

\ln\Delta t

falls: 0.1600 MPa at 3 h and 0.1206 MPa at 6 h. The derivative with respect to

\ln\Delta t_e

sits on the plateau: 0.2420 and 0.2414 MPa. From it,

k \approx 44.8

mD.

What it means. The fall of the derivative with respect to $\ln\Delta t$ is exactly the factor $t_p/(t_p + \Delta t)$ : 6/9 = 0.67 at 3 h and 6/12 = 0.5 at 6 h. There is no boundary in the reservoir, and the derivative with respect to equivalent time shows this. It is slightly above the plateau of the model (0.2409 MPa), because of the tail of wellbore storage and the rounding of the readings. If the fall had remained on the derivative with respect to $\Delta t_e$ as well, we would need to look for its cause in the reservoir. The data in this example are assumed for illustration.

Type curves, time compression and planning the flow period

Type curves for a well with wellbore storage and skin (Type-curve matching) are calculated for the start of production at a constant rate. For a build-up we would have to construct a separate family of curves for each dimensionless producing time $t_{pD} = kt_p/(\phi\mu c_tr_w^2)$ . We would also need to know the permeability before matching, to choose the right family. Equivalent time removes this problem. We plot the pressure rise and the derivative $\Delta p'_e$ against $\Delta t_e$ and match them to the same curves as a drawdown. The pressure match gives $kh$ , and the time match gives the wellbore storage coefficient $C$ (we use $\Delta t_e$ in the formula instead of $\Delta t$ ). The curve parameter gives the skin.

The price is time compression. The $\Delta t_e$ axis ends at $t_p$ . However long the shut-in lasts, the build-up reproduces the drawdown only up to a time equal to the producing time, and the late hours of the record are squeezed towards this limit. In the lesson's example, 9 h of build-up gave an equivalent time of only 3.6 h. This gives a simple planning rule: for the build-up to show an equivalent time $T$ , the shut-in must last

Shut-in duration for a given equivalent time

\Delta t = \frac{t_p\,T}{t_p - T},\qquad T < t_p

where:

$T$ equivalent time up to which the build-up must reproduce the drawdown, h
$t_p$ producing time before shut-in, h
$\Delta t$ required shut-in duration, h

Units: any identical units of time

Holds when

The reservoir is infinite and the flow is radial. The formula is the inverse of $\Delta t_e(\Delta t)$ .
We choose $T$ according to what we need to see: for example, one cycle of the derivative plateau after wellbore storage has ended.

The closer $T$ is to $t_p$ , the longer the shut-in. At $T = t_p/2$ it equals $t_p$ , and at $T = 0.8\,t_p$ it is already $4t_p$ . The whole test, $t_p + \Delta t$ , is shortest when the flow lasts $2T$ and the shut-in also lasts $2T$ . The derivative of the sum with respect to $t_p$ is $1 - T^2/(t_p - T)^2$ , and it vanishes at $t_p = 2T$ .

Worked exampleWhich flow period to choose

Given (assumed data)

Goal: to see the derivative plateau on the build-up up to an equivalent time $T$ = 5 h (wellbore storage ends within half an hour)
Options: a main flow period of 6, 10 or 15 h

Solution

$t_p$ = 6 h: $\Delta t = 6\cdot5/(6 - 5) = 30$ h, so the whole test takes 36 h.
$t_p$ = 10 h ( $= 2T$ ): $\Delta t = 10\cdot5/(10 - 5) = 10$ h, so the total is 20 h.
$t_p$ = 15 h: $\Delta t = 15\cdot5/(15 - 5) = 7.5$ h, so the total is 22.5 h.

Answer: The shortest test has a flow period of 10 h and a shut-in of 10 h. It takes 20 h in total, instead of 36 h with a six-hour flow period.

What it means. A longer flow period pays off as long as the shut-in gets shorter faster than the flow period grows. But an hour of flow costs differently from an hour of shut-in. The production in a test is usually burned off in a flare, the risk of sand production grows, and the test programme limits the volume produced. The calculation shows what each extra hour of equivalent time costs. The final decision is a compromise. The data in this example are assumed for illustration.

When equivalent time does not help

Equivalent time is derived from the logarithmic approximation of radial flow in an infinite reservoir. Wherever this approximation fails for even one of the three terms of the superposition, the time conversion stops being exact.

The flow has not reached the radial regime. Suppose the well was shut in during the transition period of wellbore storage. Then the bottomhole pressure at shut-in has not yet settled on the semilog straight line. The build-up against $\Delta t_e$ does not coincide with the drawdown curve. The late points are squeezed towards $\Delta t_e \approx t_p$ , where the type curve is still on its hump, and the skin calculated from the pressure at shut-in comes out too low. The derivative with respect to $\ln\Delta t_e$ at the end of a long build-up does still reach the plateau (we plot it against $\Delta t$ ). We describe the record as a whole with a model that has the whole rate history (example below).
Boundaries. If the disturbance from the start of production or from the shut-in has reached the boundaries, the logarithmic approximation is wrong for $\Delta p_{DD}(t_p + \Delta t)$ . In a closed region the build-up levels out at the average pressure (Build-up in a bounded reservoir and average reservoir pressure), whereas the drawdown goes to a unit-slope line (Reservoir limit test). No choice of time will turn one into the other. Near a sealing fault, the derivative with respect to $\ln\Delta t_e$ rises with time to twice the plateau level. But against $\Delta t_e$ this rise is squeezed towards $t_p$ , so we look for boundaries on the plot against $\Delta t$ .
Varying rate. We derived the formula for a single rate. With several rates, we take the superposition function of the whole history, or we choose a single notional producing time (Equivalent producing time and synthetic flowing time). A dangerous mistake is to take only the last flow period as $t_p$ after a long production history and a short intermediate shut-in. The time will then be compressed too much, and the derivative will be distorted.
Gas wells. The gas equation is nonlinear. So we apply superposition and equivalent time to the pseudo-pressure (real gas pseudo-pressure) and, for large pressure changes, to pseudo-time as well. Pseudo-time allows for the change of $\mu c_t$ with pressure.
Changing skin and wellbore storage. Clean-up of the near-wellbore region during the flow period, or the redistribution of gas and liquid in the wellbore after shut-in, breaks the linearity of the problem (Nonideal wellbore storage, well clean-up and changing skin).

Worked exampleA flow period shorter than wellbore storage

Given (assumed data)

Well: the same reservoir and the same skin, but after completion: a 1 h clean-up flow with shut-in at the surface, free gas in the tubing, $C$ = 0.15 m³/MPa
Readings: $p_{wf}(\Delta t = 0)$ = 19.3643 MPa; 10 h after shut-in — 23.3761 MPa, 40 h after shut-in — 23.3940 MPa

Solution

The start of the semilog straight line, from the estimate in the wellbore storage lesson (duration of wellbore storage), is $t \approx 44.8\,(60 + 3.5\cdot3)\,\dfrac{1.35\cdot0.15}{45\cdot9} = 1.58$ h. This is longer than the whole flow period.
The Horner line through 10 and 40 h: the horizontal coordinates are $\lg(11/10) = 0.04139$ and $\lg(41/40) = 0.01072$ . Then $m = (23.3940 - 23.3761)/(0.04139 - 0.01072) = 0.5836$ MPa per cycle, and $k = 2.149\cdot104.544/(0.5836\cdot9) = 42.8$ mD.
Continuing the line gives $p^* = 23.3940 + 0.5836\cdot0.01072 = 23.4003$ MPa.
Skin from the Horner formula with the correction for the producing time, $\lg[(1 + 1)/1] = 0.30103$ . We have $p_{1\text{h}} = 23.4003 - 0.5836\cdot0.30103 = 23.2246$ MPa and $\lg\dfrac{k}{\phi\mu c_tr_w^2} = 7.0029$ . So $S = 1.151\,[(23.2246 - 19.3643)/0.5836 - 7.0029 + 2.098 + 0.30103] = 2.31$ .
The equivalent time of the readings is $1\cdot10/11 = 0.909$ h and $1\cdot40/41 = 0.976$ h. Both are right at the limit $t_p$ = 1 h.

Answer: The permeability of 42.8 mD and

p^*

= 23.40 MPa are close to the correct values (45 mD and 23.4 MPa), but the skin is 2.31 instead of 3.

What it means. The late readings already lie in radial flow, so the slope and $p^*$ are nearly right. The Horner method does not need the flow to have reached the straight line before shut-in. The skin does need it. The formula assumes that at shut-in the bottomhole pressure lay on the semilog straight line of the drawdown. In fact it was still above the line, because the reservoir had not had time to “feel” the full drawdown. On type curves against $\Delta t_e$ , these readings would fall at $\Delta t_e \approx 1$ h. There the drawdown curve of the well is still on its hump (its derivative is 1.9 times higher than the plateau), while the build-up derivative with respect to $\ln\Delta t_e$ is already almost on the plateau. The match would be meaningless. The right way is a model with wellbore storage, skin and the rate history, fitted to the whole record. The data in this example are assumed for illustration.

Terms and symbols

Producing time $t_p$: The length of production at a constant rate before shut-in. With a varying rate, it is a notional time that replaces the history (Equivalent producing time and synthetic flowing time).
Equivalent time $\Delta t_e$: The time after the start of production at which the drawdown would show the same pressure change as the build-up a time $\Delta t$ after shut-in, $t_p\Delta t/(t_p + \Delta t)$ . It is always less than $t_p$ .
Pressure rise after shut-in $\Delta p_{BU}$: The rise in pressure after shut-in, $p_{ws}(\Delta t) - p_{wf}(\Delta t = 0)$ . It cannot exceed the pressure drop over the producing time, $p_i - p_{wf}(t_p)$ .
Derivative with respect to equivalent time $\Delta p'_e$: The derivative $\dd\Delta p_{BU}/\dd\ln\Delta t_e$ . It is equal to the derivative with respect to the Horner function, and in radial flow it is equal to the plateau $m_e$ for any $t_p$ .
Time compression: A property of the Horner function and of equivalent time: the late hours of a build-up fit into a narrow interval near $\Delta t_e = t_p$ .
Desuperposition: Recovering the drawdown from the build-up by adding the continuation of the pressure drop that the producing well would have given (see the advanced section).

Common mistakes and misconceptions

Wrong

“The derivative at the end of the build-up has halved — the reservoir has a constant-pressure boundary.”

In fact

First compare the length of the record with the producing time. After a short producing time, the derivative with respect to $\ln\Delta t$ falls by the factor $(t_p + \Delta t)/t_p$ even without any boundaries. The derivative with respect to equivalent time shows whether there is a real decline.

Wrong

“The MDH method is simpler, and the result is the same.”

In fact

Only when $\Delta t \ll t_p$ . In the lesson's example, after a six-hour flow period, MDH applied to the readings at 1.5 and 6 h overestimated the permeability by a factor of 1.5.

Wrong

“The longer the build-up, the deeper into the reservoir the test sees, however long the flow period was.”

In fact

Equivalent time does not exceed $t_p$ . The build-up reproduces the drawdown only up to a time equal to the producing time. The late information is squeezed towards $t_p$ and needs very accurate readings. To see further, lengthen the flow period.

Wrong

“Equivalent time is suitable for any build-up.”

In fact

Only if the flow had reached the radial regime before shut-in, and the boundaries had no effect during the test. After a one-hour clean-up flow with gas in the tubing, the skin from the build-up in the lesson's example came out at 2.3 instead of 3.

Wrong

“Equivalent time and equivalent producing time are the same thing.”

In fact

$\Delta t_e$ converts the time after shut-in, while $N_p/q$ replaces the producing time when the rate varies. We use the first in place of $\Delta t$ , and the second in place of $t_p$ .

In depth: advanced detail

Derivation of equivalent time and its error

In radial flow in an infinite reservoir, $\Delta p_{DD}(t) = m_e\,[\ln(at) + 0.809 + 2S]$ , where $a = k/(\phi\mu c_tr_w^2)$ converts time into dimensionless time. We put this into $\Delta p_{BU} = \Delta p_{DD}(\Delta t) + \Delta p_{DD}(t_p) - \Delta p_{DD}(t_p + \Delta t)$ . Of the three terms $0.809 + 2S$ , only one remains, and the logarithms combine into $\ln a + \ln[t_p\Delta t/(t_p + \Delta t)]$ . The result is $\Delta p_{BU} = m_e[\ln(a\Delta t_e) + 0.809 + 2S] = \Delta p_{DD}(\Delta t_e)$ .

Where the approximation is not exact, it is useful to know by how much. Let $\varepsilon(t)$ be the deviation of the drawdown from its semilog straight line. It is negative during wellbore storage and zero in radial flow. From the same superposition we get:

Error of equivalent time

\Delta p_{BU}(\Delta t) - \Delta p_{DD}(\Delta t_e) = \big[\varepsilon(\Delta t) - \varepsilon(\Delta t_e)\big] + \big[\varepsilon(t_p) - \varepsilon(t_p + \Delta t)\big]

where:

$\varepsilon(t)$ deviation of the drawdown from its semilog straight line: $\Delta p_{DD}(t) - m_e[\ln(at) + 0.809 + 2S]$ , Pa
$\Delta t,\ \Delta t_e,\ t_p$ time after shut-in, equivalent time, producing time, s

Units: any consistent system

Holds when

The reservoir is infinite, wellbore storage and skin are constant, and the problem is linear.

The first bracket vanishes when $\Delta t \ll t_p$ (then $\Delta t_e \approx \Delta t$ ) or when both times are already in radial flow. The second vanishes when the flow period itself has reached the radial regime. So the main condition is a flow period several times longer than wellbore storage. In the lesson's test ( $t_p$ = 6 h, wellbore storage lasting tenths of an hour), the difference over the whole record is no more than 0.00085 MPa. After a one-hour clean-up flow with gas in the tubing (the example in the section on limits), both brackets are large, because both the producing time and the equivalent time fall in the transition from wellbore storage to radial flow. At 10 h after shut-in the first bracket is 0.154 MPa and the second is −0.132 MPa. The build-up lies above the drawdown curve by their sum, 0.022 MPa.

Desuperposition: stretching the pressure instead of compressing the time

We can make the same correction in the pressure instead of the time. We rewrite the superposition as $\Delta p_{DD}(\Delta t) = \Delta p_{BU}(\Delta t) + [\Delta p_{DD}(t_p + \Delta t) - \Delta p_{DD}(t_p)]$ . To the pressure rise after shut-in, we add the amount by which the pressure in a producing well would have dropped over the time $\Delta t$ . The result is a true drawdown, stretched to the full length of the shut-in. This technique is called desuperposition.

Desuperposition in radial flow

\Delta p_{DD}(\Delta t) \approx \Delta p_{BU}(\Delta t) + m\,\lg\frac{t_p + \Delta t}{t_p}

where:

$\Delta p_{DD}(\Delta t)$ reconstructed pressure drop, as after the start of production, MPa
$m$ slope of the semilog straight line per decimal cycle, MPa
$\Delta p_{BU},\ t_p,\ \Delta t$ pressure rise after shut-in, producing time and time after shut-in, MPa, h, h

Units: pressure in any units (MPa), time in any identical units

Holds when

From the moment of shut-in onwards, the well would have “gone on” producing in the radial-flow regime. Otherwise we take the correction from a recorded drawdown or from a model.

We do not know the slope $m$ in advance, so the calculation is iterative (we repeat it until the answer stops changing). The first pass is without the correction (this is the MDH method). Then we apply the correction with the slope we found, and so on. For a pair of points, the slope error at each pass is multiplied by the share that the difference of the corrections makes up of the difference of the logarithms of time. In the lesson's example (readings at 1.5 and 6 h) this share is 0.34, and the iterations converge to the Horner slope. Equivalent time does the same at once, without iterations, which is why it is used more often. Desuperposition is convenient when we take the continuation of the drawdown not from a formula but from the record itself, for example from the pressure trend in a developed reservoir (Testing in a developed reservoir: Slider's method).

Connections to other topics

Pressure build-up analysis — the Horner equation, $p^*$ and the skin from a build-up — the foundation this lesson builds on.
Diagnosis with the pressure derivative — the logarithmic derivative, the plateau and false signatures on the diagnostic plot.
Type-curve matching — the type curves to which a build-up is matched against equivalent time.
Superposition principle — the addition of responses from which both the Horner equation and equivalent time are derived.
Wellbore storage — how long wellbore storage lasts and when the flow reaches the straight line.
Pressure drawdown analysis — the drawdown formulas that are applied to a build-up after the time conversion.
Equivalent producing time and synthetic flowing time — which producing time to use if the rate varied or the well produced for a long time in a bounded area.
Well testing: purposes and types of test — the drill stem test: flow periods and pressure build-ups.
Build-up in a bounded reservoir and average reservoir pressure — a build-up in a closed region, where the time conversion does not help.
Nonideal wellbore storage, well clean-up and changing skin — when skin and wellbore storage change.

Check yourself

1
A well produced for 10 h and the build-up was recorded for 15 h; there are no boundaries in the reservoir. By what factor is the derivative with respect to $\ln\Delta t$ at the end of the record below the radial-flow plateau?
1. A.Not below: in radial flow the derivative is constant
2. B.15 times
3. C.2.5 times
4. D.1.5 times
Show the answer
Correct answer: C. 2.5 times
The local slope with respect to $\ln\Delta t$ is $m_e\,t_p/(t_p + \Delta t) = m_e\cdot10/25 = 0.4\,m_e$ , which is 2.5 times below the plateau. The derivative with respect to equivalent time would have stayed on the plateau at this point.
2
A well produced for 8 h. Calculate the equivalent time of the readings at 2, 8 and 40 h after shut-in. What does this mean for the analysis of the late readings?
Show the answer
$\Delta t_e = 8\cdot2/10 = 1.6$ h, $8\cdot8/16 = 4$ h, and $8\cdot40/48 \approx 6.67$ h. The last 32 h of the record (from 8 to 40 h) fit into the equivalent-time interval from 4 to 6.67 h, which is less than a quarter of a logarithmic cycle. The late readings are squeezed towards $t_p$ = 8 h. Through them the build-up reproduces the drawdown only up to 6.67 h, and for the derivative plot it is better to plot them against $\Delta t$ .
3
Which method of analysing a build-up after a short producing time gives the correct radial-flow slope?
1. A.Any of them: the slope does not depend on the producing time
2. B.The MDH method on the points after wellbore storage has ended
3. C.Matching to drawdown type curves against $\Delta t$
4. D.The Horner plot or a plot against $\lg\Delta t_e$
Show the answer
Correct answer: D. The Horner plot or a plot against $\lg\Delta t_e$
The Horner function and equivalent time contain the producing time, so in radial flow the points fall on a straight line with the correct slope. MDH and matching against $\Delta t$ do not allow for it: the slope with respect to $\lg\Delta t$ is underestimated by the factor $(t_p + \Delta t)/t_p$ , and the permeability is overestimated.
4
After a 4 h flow period, a straight line was drawn on the build-up against $\lg\Delta t_e$ . The slope is 0.42 MPa per cycle, the pressure rise on the line at $\Delta t_e$ = 1 h is 3.70 MPa, and the bottomhole pressure at shut-in is 17.30 MPa. The reservoir is virgin (not yet produced), and no boundaries are seen. Find the initial pressure, and explain why we read it at $\Delta t_e$ = 4 h.
Show the answer
On the straight line at $\Delta t_e = t_p$ = 4 h, the pressure rise is $3.70 + 0.42\lg 4 = 3.70 + 0.253 \approx 3.95$ MPa, so $p_i \approx 17.30 + 3.95 = 21.25$ MPa. The value $\Delta t_e = t_p$ corresponds to an infinitely long shut-in (a Horner ratio of 1). In an infinite reservoir the pressure would return to its initial value, and the pressure rise would reach $p_i - p_{wf}(t_p)$ .
5
Why does interpretation software usually plot the derivative with respect to equivalent time against the elapsed time $\Delta t$ , rather than against $\Delta t_e$ ?
Show the answer
The values of the derivative do not depend on the choice of axis. Against $\Delta t_e$ , the late part of a long build-up is squeezed near $t_p$ . Signs of boundaries that appear at $\Delta t$ greater than $t_p$ fit into a narrow strip and are hard to see. Against $\Delta t$ , the late readings stay where they are, and radial flow is still seen as a plateau.
6
The clean-up flow of a new well lasted 40 minutes, with shut-in at the surface. The wellbore is partly filled with gas, and an estimate shows that the semilog straight line on the drawdown would have started only after 2 h. What can we reliably get from the following 30 h build-up?
1. A.The flow capacity and $p^*$ from the late points; the skin calculated from the pressure at shut-in will come out too low
2. B.The skin exactly, but not the flow capacity
3. C.Everything, if the record is analysed against $\Delta t_e$
4. D.Nothing
Show the answer
Correct answer: A. The flow capacity and $p^*$ from the late points; the skin calculated from the pressure at shut-in will come out too low
The late points of the build-up lie in radial flow, so we can find the slope and $p^*$ from them. The skin requires the bottomhole pressure at shut-in to lie on the semilog straight line of the drawdown, but the flow had not reached it. In the lesson's example, after a one-hour clean-up flow, the skin came out at 2.3 instead of 3. Converting the time to $\Delta t_e$ does not help here. We need a model with the whole rate history.

Updated: 19 October 2026