Diagnosis with the pressure derivative

Quick refresher

The derivative $\Delta p' = \dd\Delta p/\dd\ln t$ is the slope of the semilog plot at each point. It is measured in units of pressure, and we take it with respect to the natural logarithm even if the axes are marked in decimal cycles.
If $\Delta p$ grows as $t^n$ , the derivative equals $n\,\Delta p$ . On a log–log plot both curves are then parallel straight lines with slope $n$ . This gives the unit slope of wellbore storage ( $n = 1$ , the curves coincide), the slope ½ of linear flow and the slope −½ of spherical flow. Radial flow ( $\Delta p$ grows as $\ln t$ ) gives a horizontal “plateau”.
The plateau level $m_e = qB\mu/(4\pi kh)$ gives the flow capacity without drawing a line. The ratio $\Delta p/m_e$ at any point of the plateau gives the skin. The dimensionless derivative on the plateau equals 0.5 for any well.
The hump between the wellbore-storage section and the plateau is higher the larger $C_De^{2S}$ is: the peak rises by about 0.44 units of $p_D'$ for every tenfold increase of the parameter. Both the skin and a large wellbore storage coefficient produce a hump.
The late shapes are these. A sealing fault gives a second plateau twice as high as the first. A closed area gives a rise with unit slope. A constant-pressure boundary gives a fall.
We calculate the derivative from measurements through points that are a differentiation interval $L$ apart (usually 0.1–0.3 of a logarithmic cycle). A narrow interval amplifies the noise. A wide interval cuts off the hump and shifts the transitions.

Key relationships

\Delta p' = \dfrac{\dd\,\Delta p}{\dd\ln t} = t\,\dfrac{\dd\,\Delta p}{\dd t}

logarithmic derivative; natural logarithm

kh = 0.933\,\dfrac{qB\mu}{m_e}

flow capacity from the plateau level $m_e$ : m³/d, mPa·s, MPa, mD·m (in oilfield units the constant is 70.6)

p_D' = \dfrac{kh\,\Delta p'}{1.866\,qB\mu}

dimensionless derivative; on the plateau it is 0.5 (oilfield: 141.2)

S = \dfrac12\left[\dfrac{\Delta p}{m_e} - \ln\dfrac{kt}{\phi\mu c_tr_w^2} + 4.831\right]

skin from the ratio of the pressure to the plateau: MPa, mD, h, mPa·s, 1/MPa, m

\Delta p' = \Delta p = \dfrac{qB}{24\,C}\,t

wellbore-storage section: m³/d, m³/MPa, h, MPa

Common pitfalls

Comparing a derivative taken with respect to the base-10 logarithm with the formulas for the natural one. The error is a factor of 2.303.
Differentiating between neighbouring readings. The noise is amplified, and the plateau cannot be seen.
Smoothing too much. The hump is cut off, the transitions shift, and the skin from the shape of the hump comes out too low.
Reading the plateau level from the last points of the record, where the differentiation interval is shortened or where the producing time before shut-in has an effect.
Taking a fall of the derivative at the end of a pressure build-up curve for a supply of fluid from outside (a constant-pressure boundary), without checking the producing time before shut-in and changes in the rates of neighbouring wells.

After this lesson you can

explain what the logarithmic derivative of pressure is, how it differs from the ordinary derivative with respect to time, and why we take it with respect to the natural logarithm
derive, and recognise on the diagnostic plot, the typical shapes: wellbore storage, the transition “hump”, radial flow, linear and spherical flow, a closed area, a constant-pressure boundary and a sealing fault, in real units and in dimensionless form
calculate the flow capacity (permeability times thickness) and the permeability from the level of the derivative “plateau”, and the skin from the ratio of the pressure to the plateau, in metric and oilfield units
explain what the height and the position of the hump in the transition section depend on
calculate the derivative from measurements with a differentiation interval, estimate the error from noise and the distortion from smoothing, and choose the interval
name the false signs on the derivative plot and ways to check them

The derivative is the slope of the semilog plot

The analysis of pressure drawdown and build-up curves rests on the semilog straight line. In radial flow the pressure changes by the same amount in every logarithmic cycle of time (Pressure drawdown analysis, Pressure build-up analysis). Everything that makes a real record differ from this line shows up first as a change of slope. This includes wellbore storage, a fracture, boundaries and heterogeneity. But on a plot of $\Delta p$ against $\lg t$ , a change of slope of 10–20% is almost invisible. The curve stays smooth, and it is easy to draw a “straight line” through its points in the wrong place.

So we plot the slope itself. The logarithmic derivative is the derivative of the pressure change with respect to the natural logarithm of time. We can also get it from the ordinary derivative with respect to time (the slope of the plot on ordinary axes) if we multiply that by the time:

The logarithmic derivative of pressure

\Delta p' = \frac{\dd\,\Delta p}{\dd\ln t} = t\,\frac{\dd\,\Delta p}{\dd t}

where:

$\Delta p'$ logarithmic derivative: the slope of the plot of $\Delta p$ against $\ln t$ at the given point, Pa (the same units as $\Delta p$ )
$\Delta p$ change of the bottomhole pressure since start-up or shut-in: $p_i - p_{wf}$ after start-up, $p_{ws} - p_{wf}(\Delta t = 0)$ after shut-in; always positive, Pa
$t$ time since start-up or shut-in (after shut-in, $\Delta t$ ), s
$\ln t$ natural logarithm of time; the derivative does not depend on the units of time, because changing them only shifts $\ln t$ by a constant, —

Units: any system of units: $\Delta p'$ is expressed in the same units as the pressure

Holds when

$\Delta p$ is counted from the pressure at the moment of start-up or shut-in, and $t$ from that moment.
The logarithm is natural. If the axis is built in base-10 logarithms, the slope per cycle is $\ln 10 = 2.303$ times larger than $\Delta p'$ .
For a build-up after a short producing time of the well, we take the derivative with respect to the equivalent time (Effect of producing time on build-up analysis).

$\Delta p'$ shows how much the pressure change grows when the time grows by a factor of $\e = 2.718$ . Per decimal cycle it grows by $2.303\,\Delta p'$ . The ordinary derivative $\dd\Delta p/\dd t$ in radial flow falls as $1/t$ , and it tells the eye little. Multiplying by $t$ turns it into a constant, and turns every departure from radial flow into a departure from this constant.

In other unit systems (1)

Through the base-10 logarithm

\Delta p' = \frac{1}{\ln 10}\,\frac{\dd\,\Delta p}{\dd\lg t} = 0.4343\,\frac{\dd\,\Delta p}{\dd\lg t}

the same units of pressure; $\lg t$ is the base-10 logarithm of time in any units.

$0.4343 = 1/\ln 10$ : the slope per decimal cycle is 2.303 times the derivative.

CalculatedThe build-up record from the lesson's examples on a semilog plot (points), and the tangents to it at three moments. The slope of a tangent per unit of the natural logarithm of time is the derivative: 0.16 MPa at 36 s, 0.075 MPa at 6 min and 0.030 MPa at 10 h.

Why we plot the derivative on log–log axes. Many flow regimes give a power law for the pressure against time (the pressure is proportional to a power of the time), and for such a law the derivative is proportional to the pressure itself:

The derivative of a power-law regime

\Delta p = A\,t^{\,n}\quad\Rightarrow\quad \Delta p' = n\,A\,t^{\,n} = n\,\Delta p,\qquad \lg\Delta p' = \lg\Delta p + \lg n

where:

$A$ constant of the regime, which depends on the properties of the reservoir and the well, Pa/sⁿ
$n$ exponent: 1 for wellbore storage or a closed area, ½ for linear flow, —
$\Delta p,\ \Delta p'$ change of pressure and its logarithmic derivative, Pa
$t$ time since start-up or shut-in, s

Units: any consistent system of units

Holds when

The regime lasts long enough for a straight section to form on the plot.
If a constant is added to the power term (for example, the skin pressure drop), the plot of $\Delta p$ is curved, but the derivative is not: the derivative of a constant is zero.

On a log–log plot both curves are straight lines with slope $n$ , and the derivative lies below the pressure by $\lg(1/n)$ of a cycle. For $n = 1$ the curves coincide. For $n = 1/2$ the derivative is half as high. The derivative “does not see” constant terms: the skin pressure drop shifts $\Delta p$ but not $\Delta p'$ . So the shape of a regime shows up on the derivative earlier and more clearly than on the pressure itself.

Flow regimes on the diagnostic plot

Regime: Wellbore storage
How $\Delta p$ grows: $qBt/(24C)$
Derivative: slope 1, $\Delta p' = \Delta p$
What we find: $C$ (Wellbore storage)

Regime: Radial flow
How $\Delta p$ grows: as $\ln t$
Derivative: plateau $m_e$
What we find: $kh$ , skin

Regime: Linear flow
How $\Delta p$ grows: as $\sqrt t$
Derivative: slope ½, $\Delta p' = \Delta p/2$
What we find: fracture length or channel width

Regime: Spherical flow
How $\Delta p$ grows: towards a constant, the remainder as $1/\sqrt t$
Derivative: slope −½
What we find: spherical permeability

Regime: Closed area
How $\Delta p$ grows: at the end, as $t$
Derivative: slope 1
What we find: pore volume

Regime: Constant-pressure boundary
How $\Delta p$ grows: towards a constant
Derivative: falls to zero
What we find: distance to the boundary

Regime: Sealing fault
How $\Delta p$ grows: a second line with slope $2m$
Derivative: second plateau $2m_e$
What we find: distance to the fault

Regime	How $\Delta p$ grows	Derivative	What we find
Wellbore storage	$qBt/(24C)$	slope 1, $\Delta p' = \Delta p$	$C$ (Wellbore storage)
Radial flow	as $\ln t$	plateau $m_e$	$kh$ , skin
Linear flow	as $\sqrt t$	slope ½, $\Delta p' = \Delta p/2$	fracture length or channel width
Spherical flow	towards a constant, the remainder as $1/\sqrt t$	slope −½	spherical permeability
Closed area	at the end, as $t$	slope 1	pore volume
Constant-pressure boundary	towards a constant	falls to zero	distance to the boundary
Sealing fault	a second line with slope $2m$	second plateau $2m_e$	distance to the fault

$m_e = qB\mu/(4\pi kh)$ is the plateau level, and it is also the slope of the semilog straight line per unit of the natural logarithm; $m = 2.303\,m_e$ is the slope per decimal cycle. Linear flow (flow along parallel lines) comes from a hydraulic fracture (Inflow to a hydraulically fractured well) and from a narrow reservoir between two parallel faults (Several boundaries: parallel faults and wedge-shaped blocks). Spherical flow (flow that converges on a point from all directions) comes from partial penetration (Partial penetration and well deviation) and from the probe of a wireline formation tester (Supercharging and permeability from formation-tester pretests). The boundaries are in the section below.

The diagnostic plot

On a diagnostic plot with log–log axes, we draw two curves against the same time: the pressure change $\Delta p$ and its derivative $\Delta p'$ . Both are in the same units of pressure, so their positions relative to each other carry information too. In the wellbore-storage section the curves coincide. In radial flow the pressure lies above the derivative, and the larger the skin, the higher.

The scale is the same as for the type curves (Type-curve matching), and for the same reason: in logarithms, the change to dimensionless quantities is a shift. The dimensionless derivative is related to the dimensional one by the same factor as the pressures themselves:

The dimensionless derivative and its value in radial flow

p_D' = \frac{\dd p_D}{\dd\ln t_D} = \frac{2\pi kh}{qB\mu}\,\Delta p',\qquad p_D' = 0.5\ \ \text{(radial flow)}

where:

$p_D'$ dimensionless derivative, —
$p_D$ dimensionless change of pressure in the well itself, $2\pi kh\,\Delta p/(qB\mu)$ , with skin and wellbore storage, as in the lessons Line-source solution and the pressure function and Type-curve matching, —
$t_D$ dimensionless time $kt/(\phi\mu c_tr_w^2)$ ; the derivative with respect to $\ln t_D$ equals the derivative with respect to $\ln t$ , because $t_D$ and $t$ differ by a constant factor, —
$q,\ B,\ \mu$ change of the rate at standard conditions, formation volume factor (reservoir volume per unit of surface volume), viscosity, m³/s, m³/m³, Pa·s
$k,\ h$ permeability and thickness of the reservoir, m², m

Units: SI; the forms below are in oilfield units

Holds when

A constant rate after start-up (for a build-up: a constant rate before shut-in, and a shut-in time much shorter than the producing time).
The equality $p_D' = 0.5$ holds for radial flow in a homogeneous reservoir, where $p_D = \tfrac12(\ln t_D + 0.809) + S$ .

The dimensionless and dimensional derivatives differ by the same factor as the pressures do. So on the type-curve plot $p_D'$ shifts together with $p_D$ , and on the plot of the record $\Delta p'$ shifts together with $\Delta p$ . In radial flow $p_D$ grows by ½ per unit of $\ln t_D$ . This gives the level 0.5, the same for all wells and reservoirs. To match the plateau of the record to the level 0.5 is to find $kh$ without drawing a single line.

In other unit systems (2)

Metric oilfield units

p_D' = \frac{kh\,\Delta p'}{1.866\,qB\mu},\qquad m_e = 0.933\,\frac{qB\mu}{kh}

$q$ in m³/d at standard conditions, $\mu$ in mPa·s, $k$ in mD, $h$ in m, $\Delta p'$ and $m_e$ in MPa.

$1.866 = 1/(2\pi\cdot0.0853)$ and $0.933 = 1/(4\pi\cdot0.0853)$ , where 0.0853 is the constant of Darcy's law in these units (Rock permeability).

Oilfield units

p_D' = \frac{kh\,\Delta p'}{141.2\,qB\mu},\qquad m_e = 70.6\,\frac{qB\mu}{kh}

$q$ in STB/d (stock-tank barrels of oil per day), $\mu$ in cP, $k$ in mD, $h$ in ft, $\Delta p'$ and $m_e$ in psi.

$141.2 = 1/(2\pi\cdot1.127\cdot10^{-3})$ , $70.6 = 141.2/2$ .

Below is the assumed pressure build-up record that we analyse in this lesson. An oil well produced for 250 days at 150 m³/d, and then we shut it in at the wellhead. There is free gas in the tubing, so wellbore storage is noticeable. A downhole gauge recorded the pressure every second. For the analysis we took 20 readings per logarithmic cycle of time, which makes 105 readings in 50 h.

CalculatedDiagnostic plot of the lesson's record: the pressure rise after shut-in (points), and the derivative calculated from the measurements with an interval of 0.2 cycle (orange line). Both scales are logarithmic. The dashed lines are the results of the example below: the unit-slope line

\Delta p = qBt/(24C)

and the plateau level.

Pressure rise after shut-in and its derivative (assumed data, a selection)

$\Delta t$: 1.8 s
$\Delta p$ , MPa: 0.0134
$\Delta p'$ , MPa: 0.0130

$\Delta t$: 3.6 s
$\Delta p$ , MPa: 0.0256
$\Delta p'$ , MPa: 0.0262

$\Delta t$: 11 s
$\Delta p$ , MPa: 0.0770
$\Delta p'$ , MPa: 0.0723

$\Delta t$: 36 s
$\Delta p$ , MPa: 0.2080
$\Delta p'$ , MPa: 0.1605

$\Delta t$: 1.9 min
$\Delta p$ , MPa: 0.4323
$\Delta p'$ , MPa: 0.1942

$\Delta t$: 6.0 min
$\Delta p$ , MPa: 0.5941
$\Delta p'$ , MPa: 0.0789

$\Delta t$: 19 min
$\Delta p$ , MPa: 0.6495
$\Delta p'$ , MPa: 0.0369

$\Delta t$: 1.00 h
$\Delta p$ , MPa: 0.6880
$\Delta p'$ , MPa: 0.0315

$\Delta t$: 3.16 h
$\Delta p$ , MPa: 0.7224
$\Delta p'$ , MPa: 0.0301

$\Delta t$: 10.0 h
$\Delta p$ , MPa: 0.7569
$\Delta p'$ , MPa: 0.0306

$\Delta t$: 31.6 h
$\Delta p$ , MPa: 0.7920
$\Delta p'$ , MPa: 0.0299

$\Delta t$	$\Delta p$ , MPa	$\Delta p'$ , MPa
1.8 s	0.0134	0.0130
3.6 s	0.0256	0.0262
11 s	0.0770	0.0723
36 s	0.2080	0.1605
1.9 min	0.4323	0.1942
6.0 min	0.5941	0.0789
19 min	0.6495	0.0369
1.00 h	0.6880	0.0315
3.16 h	0.7224	0.0301
10.0 h	0.7569	0.0306
31.6 h	0.7920	0.0299

$\Delta p = p_{ws} - p_{wf}(\Delta t = 0)$ . The data were calculated from a model of a homogeneous infinite reservoir with wellbore storage and skin. Random gauge noise with a standard deviation of 0.0004 MPa was added, and then the values were rounded to 0.0001 MPa. The parameters of the model are given in the examples. The derivative is calculated from all 105 readings with an interval of 0.2 cycle (the method is in the section on calculation from measurements).

How to read this plot. In the first seconds both curves follow one unit-slope line: only the wellbore works. Then the pressure flattens, but the derivative keeps rising, and about a minute and a half after shut-in it passes a hump (a peak) about 0.21 MPa high, seven times higher than the future plateau. After 1–2 h the derivative is almost horizontal at about 0.030 MPa. This is radial flow, and these are the points from which we calculate the flow capacity and the skin. There are no signs of boundaries up to the end of the record: in the last cycle the derivative neither rises nor falls by more than the noise allows.

The plateau: flow capacity and skin without a line

In radial flow the derivative is constant and equal to the slope of the semilog straight line per unit of the natural logarithm. This horizontal section is the derivative plateau. We find the flow capacity directly from its level, without choosing points for a line and without the risk of drawing the line through the transition section:

Flow capacity from the level of the derivative plateau

kh = \frac{qB\mu}{4\pi\,m_e}

where:

$kh$ flow capacity, m³
$m_e$ level of the derivative plateau; it is also the slope of the semilog straight line per unit of the natural logarithm, $m_e = m/\ln 10$ , Pa
$q,\ B,\ \mu$ rate at standard conditions (after start-up or before shut-in), formation volume factor, viscosity, m³/s, m³/m³, Pa·s

Units: SI

Holds when

The derivative is horizontal for at least half a logarithmic cycle. We average its level over the points of this section.
A constant rate. For a build-up, the shut-in time is much shorter than the producing time of the well.
The plateau refers to radial flow through the whole thickness of the reservoir. With partial penetration, an early plateau can refer to only part of the thickness (Partial penetration and well deviation).

This is the same formula as for the flow capacity from the slope of a line (Pressure drawdown analysis), but the slope is read as the level of a horizontal section. We see a plateau at once, and a departure from it by a few per cent is visible too. So the derivative makes it easy to check whether all the points chosen for a line really belong to radial flow.

In other unit systems (2)

Metric oilfield units

kh = 0.933\,\frac{qB\mu}{m_e}

$q$ in m³/d, $\mu$ in mPa·s, $m_e$ in MPa; $kh$ in mD·m.

$0.933 = 1/(4\pi\cdot0.0853)$ ; with the slope per decimal cycle $m = 2.303\,m_e$ , the same formula gives the familiar $kh = 2.149\,qB\mu/m$ .

Oilfield units

kh = 70.6\,\frac{qB\mu}{m_e}

$q$ in STB/d, $\mu$ in cP, $m_e$ in psi; $kh$ in mD·ft.

$70.6 = 1/(4\pi\cdot1.127\cdot10^{-3})$ ; for the slope per decimal cycle it is 162.6.

We find the skin from the position of the pressure curve. In radial flow $\Delta p = m_e(\ln t_D + 0.809 + 2S)$ , so the ratio of the pressure to the plateau level at any point of radial flow gives the skin:

Skin factor from the ratio of the pressure to the derivative plateau

S = \frac12\left[\frac{\Delta p(t)}{m_e} - \ln\frac{kt}{\phi\mu c_tr_w^2} - 0.809\right]

where:

$\Delta p(t)$ change of pressure at the time $t$ , taken on the plateau, Pa
$m_e$ level of the derivative plateau, Pa
$k,\ \phi,\ \mu,\ c_t,\ r_w$ permeability (from the plateau), porosity, viscosity, total compressibility, wellbore radius, m², —, Pa·s, 1/Pa, m
$t$ time since start-up or shut-in, s

Units: SI; natural logarithm

Holds when

The point $t$ lies on the plateau: wellbore storage is over, and the boundaries have not yet had an effect.
For a build-up, $\Delta p = p_{ws} - p_{wf}(\Delta t = 0)$ , and the shut-in time is much shorter than the producing time.
The skin refers to the wellbore radius $r_w$ that we put into the formula.

This is the skin formula from the line of a drawdown test, written with the natural logarithm (Pressure drawdown analysis). $\Delta p/m_e$ is how many “units” of pressure drop have built up by the time $t$ . Of these, $\ln t_D + 0.809$ would be given by a reservoir without skin, and half of the remainder is the skin. On the diagnostic plot this is the vertical distance between the pressure curve and the derivative curve: the larger the skin, the higher the pressure above the derivative.

In other unit systems (1)

Metric oilfield units

S = \frac12\left[\frac{\Delta p(t)}{m_e} - \ln\frac{k\,t}{\phi\mu c_tr_w^2} + 4.831\right]

$t$ in h, $k$ in mD, $\mu$ in mPa·s, $c_t$ in 1/MPa, $r_w$ in m; $\Delta p$ and $m_e$ in MPa.

$-\ln(3.553\cdot10^{-3}) = 5.640$ and $5.640 - 0.809 = 4.831$ ; the factor $3.553\cdot10^{-3}$ converts time in hours into dimensionless time, as in the lesson Pressure drawdown analysis.

Worked exampleWellbore storage, flow capacity and skin from the diagnostic plot

Given (assumed data)

Well and reservoir: $q$ = 150 m³/d; $B$ = 1.28; $\mu$ = 0.6 mPa·s; $h$ = 20 m; $\phi$ = 0.24; $c_t = 2.1\cdot10^{-3}$ MPa⁻¹; $r_w$ = 0.1 m
Producing time before shut-in: 6000 h, much longer than the record
From the plot: a reading on the unit slope: 1.8 s ( $5.01\cdot10^{-4}$ h), $\Delta p$ = 0.0134 MPa; the plateau is the mean value of the derivative over all the points from 3.2 to 32 h, $m_e$ = 0.02996 MPa; $\Delta p$ at 10 h is 0.7569 MPa

Solution

Wellbore storage coefficient from the unit slope: $qB = 150\cdot1.28 = 192$ ; $C = \dfrac{qB\,\Delta t}{24\,\Delta p} = \dfrac{192\cdot5.01\cdot10^{-4}}{24\cdot0.0134} = 0.299$ m³/MPa. The derivative from the measurements at this point is 0.0130 MPa, almost the same as $\Delta p$ : on the unit slope the curves merge.
Flow capacity from the plateau: $qB\mu = 192\cdot0.6 = 115.2$ ; $kh = 0.933\cdot115.2/0.02996 = 3588$ mD·m; $k = 3588/20 = 179.4$ mD.
The same in oilfield units (1 m³ = 6.2898 bbl, 1 MPa = 145.04 psi, 1 m = 3.2808 ft): $q$ = 943.5 STB/d, $m_e$ = 4.345 psi; $kh = 70.6\cdot943.5\cdot1.28\cdot0.6/4.345 = 11{,}774$ mD·ft, that is, 3589 mD·m.
Skin from the point at 10 h: $\Delta p/m_e = 0.7569/0.02996 = 25.26$ ; $\ln\dfrac{kt}{\phi\mu c_tr_w^2} = \ln\dfrac{179.4\cdot10}{0.24\cdot0.6\cdot2.1\cdot10^{-3}\cdot0.01} = 20.20$ ; $S = \tfrac12\,(25.26 - 20.20 + 4.831) = 4.95$ .
Dimensionless wellbore storage coefficient: $2\pi\phi c_thr_w^2 = 2\pi\cdot0.24\cdot2.1\cdot10^{-3}\cdot20\cdot0.1^2 = 6.333\cdot10^{-4}$ m³/MPa; $C_D = 0.299/(6.333\cdot10^{-4}) = 472$ ; the parameter of the type curve is $C_De^{2S} = 9.4\cdot10^{6}$ .

Answer:

C \approx 0.299

m³/MPa,

kh \approx 3588

mD·m (

k \approx 179.4

mD), and the skin is about 4.95.

What it means. We calculated the record from a model with $C$ = 0.3 m³/MPa, $k$ = 180 mD and $S$ = 5. The three numbers came back with errors of 0.3%, 0.3% and 0.05 in skin. We did not have to draw a single line: the plateau level is averaged over a whole logarithmic cycle of values of the derivative, and the skin comes from the ratio of the pressure to the plateau at one point. If the skin were zero, the pressure at 10 h would be smaller by $2m_eS$ = 0.30 MPa, and on the diagnostic plot the pressure curve would drop noticeably towards the derivative. The data in this example are assumed for illustration.

Wellbore storage and the transition hump

While only the wellbore works, the pressure grows in proportion to time (Wellbore storage). The derivative of a linear function with respect to the logarithm of time equals the function itself:

The derivative in the wellbore-storage section

\Delta p' = t\,\frac{\dd\,\Delta p}{\dd t} = \Delta p = \frac{qB}{C}\,t,\qquad p_D' = p_D = \frac{t_D}{C_D}

where:

$q,\ B$ change of the rate at the wellhead at standard conditions, and formation volume factor, m³/s, m³/m³
$C$ wellbore storage coefficient, m³/Pa
$t$ time since start-up or shut-in, s
$t_D/C_D$ dimensionless time divided by the dimensionless wellbore storage coefficient, —

Units: SI; in metric oilfield units $\Delta p = qBt/(24C)$ : m³/d, m³/MPa, h, MPa

Holds when

The inflow from the reservoir has not changed yet: this is the section of pure wellbore storage.
The coefficient $C$ is constant.

On a log–log plot the pressure and the derivative lie on one unit-slope line. As soon as the inflow from the reservoir starts to change, the curves part. The pressure flattens, and the derivative keeps rising for some time.

Where the hump comes from. By the end of the transition, the pressure must reach the line of radial flow. In the variables of the type curves (Type-curve matching), on this line $p_D$ equals $\tfrac12[\ln(t_D/C_D) + 0.809]$ plus $\tfrac12\ln(C_De^{2S})$ . So the line lies higher the larger $C_De^{2S}$ is, that is, the larger the skin and the wellbore storage coefficient. While the wellbore gives out or takes in fluid, the pressure change builds up slowly. Then the pressure “catches up” with the line of radial flow, and on the scale of the logarithm of time this is a fast rise. A fast rise on a semilog plot is exactly the hump of the derivative, and its peak is the inflection point of the semilog plot, where the plot is steepest. For large $C_De^{2S}$ the peak comes about when the unit-slope line, extended upwards, would cross the line of radial flow.

CalculatedCalculated derivatives of the type curves of a well with wellbore storage and skin:

p_D'

against

t_D/C_D

for

C_De^{2S}

10^2

10^4

10^6

10^8

and

10^{10}

(from bottom to top). Both scales are logarithmic. All the curves start on the unit-slope line and end on the plateau 0.5. Between them is a hump, which is higher and later the larger the parameter.

The derivative hump for different

C_De^{2S}

(calculated from the lesson's model)

$C_De^{2S}$: $10^{2}$
Peak of $p_D'$: 1.25
Peak / 0.5: 2.51
$t_D/C_D$ of the peak: 5.1
Entry into ±5% of 0.5: 210

$C_De^{2S}$: $10^{4}$
Peak of $p_D'$: 2.15
Peak / 0.5: 4.30
$t_D/C_D$ of the peak: 7.1
Entry into ±5% of 0.5: 320

$C_De^{2S}$: $10^{6}$
Peak of $p_D'$: 3.04
Peak / 0.5: 6.08
$t_D/C_D$ of the peak: 9.4
Entry into ±5% of 0.5: 420

$C_De^{2S}$: $10^{8}$
Peak of $p_D'$: 3.93
Peak / 0.5: 7.85
$t_D/C_D$ of the peak: 11.7
Entry into ±5% of 0.5: 530

$C_De^{2S}$: $10^{10}$
Peak of $p_D'$: 4.80
Peak / 0.5: 9.61
$t_D/C_D$ of the peak: 14.1
Entry into ±5% of 0.5: 630

$C_De^{2S}$	Peak of $p_D'$	Peak / 0.5	$t_D/C_D$ of the peak	Entry into ±5% of 0.5
$10^{2}$	1.25	2.51	5.1	210
$10^{4}$	2.15	4.30	7.1	320
$10^{6}$	3.04	6.08	9.4	420
$10^{8}$	3.93	7.85	11.7	530
$10^{10}$	4.80	9.61	14.1	630

$t_D/C_D = 0.02232\,kh\,t/(\mu C)$ in metric oilfield units (Wellbore storage). The last column is the value of $t_D/C_D$ after which the derivative stays within 5% of the plateau level. This is 3 to 8 times later than the classic estimate of the start of the semilog straight line, $t_D \approx (60 + 3.5S)\,C_D$ (how long wellbore storage lasts). For $C_De^{2S}$ of about 1 and less, there is practically no hump.

What changes if…

…the skin is larger by one: the parameter $C_De^{2S}$ is 7.4 times larger, the peak of the hump is higher by about 0.38 units of $p_D'$ (three quarters of the plateau level), and the hump comes a little later.
…the wellbore storage coefficient is twice as large: the peak is higher by only 0.13, but the whole early part of the record shifts to the right. In real time the hump comes about twice as late.
…the flow capacity is twice as large: both curves shift down by a factor of 2 and to the left by a factor of 2, and the shape does not change. The permeability is not in $C_De^{2S}$ .
…the skin is negative and the wellbore storage is small ( $C_De^{2S}$ of the order of one): there is no hump. The derivative comes up to the plateau from below, or reaches it almost at once.

Boundaries and other flow geometries

When the pressure disturbance reaches the reservoir boundaries, the derivative leaves the plateau, and the direction in which it leaves at once tells what kind of boundary it is. At a sealing boundary, nothing replaces the fluid that the well takes out, so the pressure falls faster and the derivative rises. Across a constant-pressure boundary, the fluid that the well takes out is replaced, so the pressure stabilises and the derivative falls. The shape of the transition at a straight boundary comes from the method of images (image wells), where an imaginary mirror-image well stands in for the boundary. The contribution of the image is $\pm m_eE_1\big(\phi\mu c_tL^2/(kt)\big)$ , and the derivative of the function $E_1(a/t)$ with respect to $\ln t$ is a simple exponential, $\e^{-a/t}$ .

The derivative at a straight boundary

\Delta p' = m_e\left[1 \pm \exp\left(-\frac{\phi\mu c_tL^2}{kt}\right)\right],\qquad p_D' = \frac12\left[1 \pm \e^{-L_D^2/t_D}\right]

where:

$m_e$ plateau level, $qB\mu/(4\pi kh)$ , Pa
$L$ distance from the well to the boundary; the image is at the distance $2L$ , m
$L_D$ dimensionless distance $L/r_w$ , —
$\pm$ plus: a sealing boundary (a fault); minus: a constant-pressure boundary, —
$\phi,\ \mu,\ c_t,\ k,\ t$ porosity, viscosity, total compressibility, permeability, time, —, Pa·s, 1/Pa, m², s

Units: SI

Holds when

A straight boundary through the whole thickness of a homogeneous reservoir; there are no other boundaries.
Wellbore storage is over, and the logarithmic approximation holds for the well itself.
The image is a line source: $L$ is much larger than the wellbore radius.

At a fault the derivative rises from $m_e$ to $2m_e$ . At a constant-pressure boundary it falls from $m_e$ towards zero, and at the end with a slope of −1. One and the same exponential describes the transition. Its middle falls at $kt/(\phi\mu c_tL^2) = 1/\ln 2 = 1.44$ . From 10 to 90% of the transition takes 1.3 logarithmic cycles, and from 1 to 99% takes 2.7 cycles. The lesson A sealing fault near the well: the method of images shows how we find the distance to the fault from this transition.

In other unit systems (1)

Metric oilfield units

\Delta p' = m_e\left[1 \pm \exp\left(-\frac{\phi\mu c_tL^2}{3.553\cdot10^{-3}\,k\,t}\right)\right]

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

$3.553\cdot10^{-3}$ converts time in hours into dimensionless time (Pressure build-up analysis).

In a closed area, when the disturbance has reached all the boundaries, pseudo-steady-state flow begins. The pressure falls everywhere at one rate, $qB/(c_tV_p)$ (Pseudo-steady-state flow and the shape factor). The derivative then grows in proportion to time:

The derivative in a closed area (pseudo-steady-state flow)

\Delta p' = t\,\frac{\dd\,\Delta p}{\dd t} = \frac{qB}{c_tV_p}\,t,\qquad p_D' = 2\pi\,t_{DA}

where:

$V_p$ pore volume of the closed area, m³
$c_t$ total compressibility, 1/Pa
$q,\ B$ rate at standard conditions and formation volume factor, m³/s, m³/m³
$t_{DA}$ dimensionless time related to the area of the region $A$ : $kt/(\phi\mu c_tA)$ , —

Units: SI; in metric oilfield units $\Delta p' = qBt/(24\,c_tV_p)$ : m³/d, h, 1/MPa, m³, MPa

Holds when

A closed area with no inflow across the boundaries; the disturbance has reached all the boundaries.
A constant rate; a single-phase fluid with a constant compressibility.

A unit slope again, as for wellbore storage, but at the end of the record. The “tank” that now discharges is the whole pore volume of the area. From the position of this line we find the pore volume, and from the moment it appears and the shape of the transition we find the shape of the area (Reservoir limit test). For a build-up in the same area, it is the other way round: after shut-in the pressure levels off at the average value, and the derivative falls (Build-up in a bounded reservoir and average reservoir pressure).

CalculatedCalculated dimensionless derivatives of a well without wellbore storage and skin: an infinite reservoir (dashed), a sealing fault and a constant-pressure boundary at a distance of

1000\,r_w

, and a closed circle of radius

1000\,r_w

with the well at the centre. Both scales are logarithmic. Up to

t_D \approx 10^5

the curves coincide (the plateau 0.5). Then they part: at the fault towards a second plateau at 1.0, at the circle onto a unit-slope line, and at the constant-pressure boundary towards zero.

Other flow geometries give power-law regimes (flow geometry). In linear flow, towards a hydraulic fracture or along a narrow reservoir between two parallel faults, the pressure drop grows as $\sqrt t$ . In spherical flow towards a short opened interval, the pressure quickly comes close to a constant value, and the remainder falls as $1/\sqrt t$ :

The derivative in linear and spherical flow

\Delta p = A_l\sqrt{t} + \Delta p_s\ \Rightarrow\ \Delta p' = \tfrac12\,A_l\sqrt{t};\qquad \Delta p = \Delta p_\infty - \frac{A_s}{\sqrt t}\ \Rightarrow\ \Delta p' = \frac{A_s}{2\sqrt t}

where:

$A_l,\ A_s$ constants of the linear and the spherical regimes: they contain the permeability, the storativity of the reservoir (its ability to store fluid by elastic expansion) and the size of the fracture or of the opened interval, Pa/s^½, Pa·s^½
$\Delta p_s$ a constant pressure drop, for example across the skin; the derivative does not depend on it, Pa
$\Delta p_\infty$ the value to which the pressure drop tends in spherical flow, Pa

Units: any consistent system of units

Holds when

Linear flow: the flow is along parallel lines towards the plane of a fracture or along a channel, and the disturbance has not reached the ends of the fracture or other boundaries.
Spherical flow: a homogeneous reservoir around a small inflow interval, and the disturbance has not yet reached the top and the bottom of the reservoir.

The slopes ½ and −½ on the derivative are signs of the flow geometry. In linear flow the derivative runs half as high as the pressure drop without $\Delta p_s$ , and parallel to it. The pressure drop itself with a skin is curved on a log–log plot, so it is more reliable to recognise the regime from the derivative. The calculation from the constants $A_l$ and $A_s$ is in the lessons on hydraulic fractures, partial penetration and formation testers.

How we calculate the derivative from measurements

A gauge gives not a function but a series of readings with noise and a finite resolution. If we take the slope between neighbouring readings, the noise is amplified many times: the difference of two close pressures is comparable with the noise itself. So we calculate the derivative at each point from the readings that are a given differentiation interval $L$ away from it. We measure the interval in fractions of a logarithmic cycle of time.

The derivative from three readings

\Delta p_i' = \frac{s_l\,\Delta X_r + s_r\,\Delta X_l}{\Delta X_l + \Delta X_r},\qquad s_l = \frac{\Delta p_i - \Delta p_l}{\Delta X_l},\quad s_r = \frac{\Delta p_r - \Delta p_i}{\Delta X_r}

where:

$i$ the reading at which we calculate the derivative, —
$l,\ r$ the nearest readings to the left and to the right that are at least $L$ decimal cycles from $i$ : $\lg(t_i/t_l) \ge L$ , $\lg(t_r/t_i) \ge L$ , —
$\Delta X_l,\ \Delta X_r$ distances in the natural logarithm of time: $\ln(t_i/t_l)$ and $\ln(t_r/t_i)$ , —
$s_l,\ s_r$ slopes of the chords (the straight lines between neighbouring points) to the left and to the right, Pa
$L$ differentiation interval, in fractions of a decimal logarithmic cycle, —

Units: $\Delta p'$ and the slopes are in units of pressure; $\Delta X$ has no units

Holds when

The readings are ordered in time, and the pressure is counted from the moment of start-up or shut-in.
The interval is the same for the whole record. In the first and the last $L$ cycles of the record there is no neighbour on one side. The derivative is not calculated there, or it is calculated with a shortened interval and is less reliable.

Each chord is the slope on its own segment. The weighted mean of the two chords is the slope of the parabola through the three readings, taken at the middle point. When $\Delta X_l$ and $\Delta X_r$ are equal, the formula turns into the central difference $(\Delta p_r - \Delta p_l)/(\Delta X_l + \Delta X_r)$ . The wider the interval, the larger the difference of pressure between the outer readings compared with the noise, and the more the real bends of the curve are smoothed.

Two errors pull in different directions. Noise with a standard deviation $\sigma$ (its typical size) gives an error of about $0.307\,\sigma/L$ in the central difference: the wider the interval, the smaller the error. The distortion of the shape, on the other hand, grows as the square of the interval: a chord differs from a tangent more, the more sharply the curve bends. Where the curve is almost straight, on the plateau, we can take a wide interval. Where it bends sharply, at the hump, a wide interval cuts off the peak and shifts the transitions (advanced section).

Worked exampleThe derivative at one point with three intervals

Given (assumed data)

Readings of the lesson's record: 2.51 h — 0.7162 MPa; 5.01 h — 0.7364; 7.08 h — 0.7472; 7.94 h — 0.7501; 8.91 h — 0.7545; 12.59 h — 0.7643; 25.1 h — 0.7846
Recording step: 20 readings per cycle: neighbouring readings are 0.05 cycle apart
Gauge noise: $\sigma$ = 0.0004 MPa

Solution

$L$ = 0.05 cycle, neighbouring readings 7.08 and 8.91 h: $\Delta X_l + \Delta X_r = 0.1\cdot\ln 10 = 0.2303$ ; $\Delta p' = (0.7545 - 0.7472)/0.2303 = 0.03170$ MPa.
$L$ = 0.2 cycle, readings 5.01 and 12.59 h: $0.4\cdot\ln 10 = 0.9210$ ; $\Delta p' = (0.7643 - 0.7364)/0.9210 = 0.03029$ MPa.
$L$ = 0.5 cycle, readings 2.51 and 25.1 h: $\ln 10 = 2.3026$ ; $\Delta p' = (0.7846 - 0.7162)/2.3026 = 0.02971$ MPa.
The model from which the record was calculated gives 0.03003 MPa at this point. The errors are 5.6, 0.9 and −1.1%.
Over the whole plateau, from 3.2 to 32 h, the root-mean-square deviation of the derivative from the model is 6.5% for $L$ = 0.05, 4.1% for 0.1, 1.6% for 0.2 and 0.7% for 0.5. The estimate $0.307\,\sigma/L$ as a share of the plateau is 8.2, 4.1, 2.1 and 0.8%.

Answer: From neighbouring readings, the derivative “jumps” by several per cent (root-mean-square 6.5%). With an interval of 0.2–0.5 cycle, the scatter is less than two per cent.

What it means. The noise of 0.4 kPa is negligible compared with the pressure rise itself (about 0.75 MPa), but not compared with the pressure change between neighbouring readings: on the plateau the pressure grows by only 3.4 kPa in 0.05 cycle. The estimate of the scatter by the simple formula is close to the actual one. A single point can by chance be closer to the model than with a wide interval. So we judge the quality of the derivative over a whole section, not at one point. The data in this example are assumed for illustration.

CalculatedThe derivative of the lesson's record for different differentiation intervals. Top: the plateau on an enlarged scale. At

L

= 0.05 cycle (the grey zigzag line) the scatter is a few per cent, and at

L

= 0.5 (orange) it is less than one per cent; the dashed line is the model. Bottom: the hump. At

L

= 0.2 (dark line) the peak is cut by 3%, and at

L

= 0.5 (orange) by 15% and shifted to an earlier time.

Worked exampleWhat smoothing does to the hump

Given (assumed data)

Hump of the model: peak 0.2086 MPa at 85 s after shut-in
From the plateau example: $m_e$ = 0.02996 MPa; $C_D$ = 472; the skin from the ratio of the pressure to the plateau 4.95
Table of humps: peak / 0.5: 4.30 for $C_De^{2S} = 10^4$ , 6.08 for $10^6$ , 7.85 for $10^8$

Solution

Peak of the derivative from the measurements: 0.2069 MPa at $L$ = 0.1; 0.2022 MPa at 0.2; 0.1765 MPa at 0.5. These are below the model by 0.8, 3.1 and 15%.
$L$ = 0.1: the height of the hump as a share of the plateau is 0.2069/0.02996 = 6.91, between $10^6$ and $10^8$ . Interpolation in the base-10 logarithm of the parameter: $\lg(C_De^{2S}) = 6 + 2\,(6.91 - 6.08)/(7.85 - 6.08) = 6.94$ ; $S = \tfrac12\ln(10^{6.94}/472) = 4.91$ .
$L$ = 0.2: the ratio is 6.75, $\lg(C_De^{2S}) = 6.76$ , $S = 4.70$ .
$L$ = 0.5: the ratio is 5.89, already between $10^4$ and $10^6$ : $\lg(C_De^{2S}) = 4 + 2\,(5.89 - 4.30)/(6.08 - 4.30) = 5.79$ , $S = 3.6$ .

Answer: From the hump, the skin is 4.91, 4.70 and 3.6 at intervals of 0.1, 0.2 and 0.5 cycle. From the ratio of the pressure to the plateau, it is 4.95.

What it means. The hump carries the same information about the skin as the position of the pressure curve: both depend on $C_De^{2S}$ . With a narrow interval the two estimates agree, and this supports the model of a homogeneous reservoir with a constant wellbore storage coefficient. But the hump is the sharpest bend of the record, and a wide interval cuts off its peak. At 0.5 cycle the skin came out smaller by 1.4. Matching to type curves gives the same error if we compare it with an over-smoothed derivative. If the hump is distorted by a redistribution of gas and liquid in the wellbore (Nonideal wellbore storage, well clean-up and changing skin), we do not estimate the skin from it at all. The data in this example are assumed for illustration.

Recalculate with several intervals. We start with 0.1–0.2 cycle and compare with 0.05 and with 0.3–0.5. The shapes that stay belong to the record, and the shapes that change belong to the processing.
Do not draw conclusions from the edges of the record. In the last $L$ cycles the derivative is either not calculated or calculated with a shortened interval. This is where imaginary boundaries most often “appear”.
Set the moment of start-up or shut-in, and the pressure at that moment, exactly. An error in them distorts the early part of the derivative more than the noise does (Wellbore storage).
Remember the resolution of the gauge (the smallest change it can show). On the plateau the pressure changes by $0.23\,m_e$ in 0.1 cycle. If this is comparable with the resolution and the noise, the derivative needs a wide base, and in high-permeability reservoirs a quartz gauge.
Allow for the rate history. If the rate before shut-in or during the record changed, we take the derivative not with respect to $\ln t$ but with respect to the superposition time function (Superposition principle), a time scale that adds up the effects of all the earlier rate changes. For a build-up after a short producing time, we use the equivalent time (Effect of producing time on build-up analysis).

False signs and how to check them

The derivative is sensitive to everything that changes the slope of the semilog plot, not only to the reservoir. Before we ascribe a shape to the reservoir, we check the simpler causes.

What can imitate a boundary or heterogeneity

What we see: A fall of the derivative at the end of a build-up
Causes other than the reservoir: a short producing time before shut-in; pressure equalising in a closed area; the work of a neighbouring producing well, or the shut-in of an injection well
How to check: the derivative with respect to the equivalent time (Effect of producing time on build-up analysis); the rate history of the neighbouring wells

What we see: A rise of the derivative at the end of the record
Causes other than the reservoir: in a drawdown test: the start-up of a neighbouring producing well, or a rise of its rate; in a build-up: the shut-in of a neighbouring producing well, or the start of an injection well; gauge drift (a slow wandering of the gauge reading)
How to check: the rate history; a comparison of the drawdown test and the build-up of the same well

What we see: A hump at a small skin
Causes other than the reservoir: a large wellbore storage coefficient; a redistribution of gas and liquid in the wellbore
How to check: $C$ from the unit slope; the shape of the early section (Nonideal wellbore storage, well clean-up and changing skin)

What we see: A dip between two sections of the plateau
Causes other than the reservoir: a changing wellbore storage coefficient; dual porosity (a fractured reservoir); a layered reservoir with crossflow (flow between the layers)
How to check: the design and the contents of the wellbore; core and geology

What we see: Teeth and oscillations
Causes other than the reservoir: noise, a narrow interval, rate fluctuations, tidal fluctuations of the pressure in the reservoir
How to check: recalculation with another interval; the rate record

What we see: An early section with a slope other than 1
Causes other than the reservoir: an inexact moment of start-up or shut-in; a changing $C$
How to check: refine the moment from the record of the gauge (Wellbore storage)

What we see	Causes other than the reservoir	How to check
A fall of the derivative at the end of a build-up	a short producing time before shut-in; pressure equalising in a closed area; the work of a neighbouring producing well, or the shut-in of an injection well	the derivative with respect to the equivalent time (Effect of producing time on build-up analysis); the rate history of the neighbouring wells
A rise of the derivative at the end of the record	in a drawdown test: the start-up of a neighbouring producing well, or a rise of its rate; in a build-up: the shut-in of a neighbouring producing well, or the start of an injection well; gauge drift (a slow wandering of the gauge reading)	the rate history; a comparison of the drawdown test and the build-up of the same well
A hump at a small skin	a large wellbore storage coefficient; a redistribution of gas and liquid in the wellbore	$C$ from the unit slope; the shape of the early section (Nonideal wellbore storage, well clean-up and changing skin)
A dip between two sections of the plateau	a changing wellbore storage coefficient; dual porosity (a fractured reservoir); a layered reservoir with crossflow (flow between the layers)	the design and the contents of the wellbore; core and geology
Teeth and oscillations	noise, a narrow interval, rate fluctuations, tidal fluctuations of the pressure in the reservoir	recalculation with another interval; the rate record
An early section with a slope other than 1	an inexact moment of start-up or shut-in; a changing $C$	refine the moment from the record of the gauge (Wellbore storage)

One and the same shape of the derivative can have different causes. We choose the model from all the signs of the record, and we check it with geology, seismic data and the data of neighbouring wells (what can go wrong).

Terms and symbols

Logarithmic derivative $\Delta p'$: The slope of the plot of $\Delta p$ against $\ln t$ at a given point, $t\,\dd\Delta p/\dd t$ . It is in units of pressure.
Diagnostic plot: A plot of $\Delta p$ and $\Delta p'$ against time on log–log axes. We recognise the flow regimes by the shape of the curves.
Derivative plateau $m_e$: The horizontal section of the derivative at the level $qB\mu/(4\pi kh)$ . It is the sign of radial flow. The dimensionless level is 0.5.
Derivative hump: The maximum of the derivative between the wellbore-storage section and the plateau. Its height grows with $C_De^{2S}$ .
Dimensionless derivative $p_D'$: $\dd p_D/\dd\ln t_D = 2\pi kh\,\Delta p'/(qB\mu)$ . On the type-curve plot it shifts together with $p_D$ .
Differentiation interval $L$: The distance, in fractions of a logarithmic cycle, to the readings from which we calculate the derivative. It sets the compromise between noise and smoothing.

Common mistakes and misconceptions

Wrong

“The derivative plateau is at 0.030 MPa, so the slope of the semilog line is 0.030 MPa per cycle.”

In fact

The derivative is the slope per unit of the natural logarithm. The slope per decimal cycle is 2.303 times larger: 0.069 MPa. If we mix them up, the flow capacity comes out wrong by a factor of 2.3.

Wrong

“A hump of the derivative means a positive skin.”

In fact

The height of the hump is set by $C_De^{2S}$ . With a large wellbore storage coefficient there is a hump even at zero skin. We find the skin from the position of the pressure curve relative to the plateau, or by matching to a model when we know $C_D$ .

Wrong

“The smoother the derivative, the better it was calculated.”

In fact

Smoothness is bought by smoothing: a wide interval cuts off the hump and stretches the transitions. In the lesson's example, with an interval of 0.5 cycle the peak of the hump is 15% lower, and the skin estimated from the hump is smaller by about 1.4.

Wrong

“The derivative went down at the end of the build-up, so the reservoir has support from outside.”

In fact

A short producing time before shut-in, and the equalising of pressure in a closed area, look the same. We first plot the derivative against the equivalent time and check the rate history, and only then think about a constant-pressure boundary.

Wrong

“On a log–log plot the late deviations look small, so we can neglect them.”

In fact

On the pressure curve, yes. But the derivative amplifies them: a fault doubles its level, and this is clearly seen even on a logarithmic scale. This is why we look for boundaries on the derivative.

In depth: advanced detail

How the curves of the lesson were calculated

The curves of the lesson were calculated from solutions in Laplace space (a mathematical method that turns the equations of time into simpler equations of a variable $s$ ), as in the lesson on wellbore storage. We do not differentiate numerically. If $\bar p_{wD}(s)$ is the Laplace transform of the pressure, then the transform of its derivative with respect to time is $s\,\bar p_{wD}(s)$ , because $p_{wD} = 0$ at the initial moment.

The logarithmic derivative through the Laplace transform

p_D' = t_D\,\frac{\dd p_{wD}}{\dd t_D} = t_D\,\mathcal{L}^{-1}\left\{s\,\bar p_{wD}(s)\right\}

where:

$\bar p_{wD}(s)$ Laplace transform of the dimensionless pressure in the well, —
$s$ variable of the Laplace transform, corresponding to $t_D$ , —
$\mathcal{L}^{-1}$ the inverse transform; numerically, the Stehfest algorithm with sixteen terms (Stehfest algorithm), —

Units: dimensionless

Holds when

$p_{wD}(0) = 0$ : before start-up the pressure in the reservoir is equalised.
The numerical inversion gives the derivative with three or four correct digits.

For a well near a straight boundary, we add to the transform of the well in an infinite reservoir the transform of the image, $\pm K_0(2L_D\sqrt s)/s$ . It goes in the same place as the solution without wellbore storage and skin (the transform with wellbore storage and skin). For a closed circle we take the solution with a sealing outer boundary (the Laplace transform).

The model of a well of finite radius shows a detail that the pressure curve does not show. A line source (a well treated as an infinitely thin line) reaches the plateau of 0.5 within 1% already at $t_D \approx 25$ , but a well of finite radius without wellbore storage reaches it only at $t_D \approx 330$ (within 2% at $t_D \approx 140$ ). At the very start, for $t_D$ below about 0.01, the flow to the wall of the well is almost parallel, and the derivative runs with slope ½ at half the pressure, as for a fracture. In a real well this is fractions of a second, hidden by wellbore storage. Such a picture becomes noticeable at a large effective radius, that is, for a well with a strongly negative skin. This is why a well like that and a well with a hydraulic fracture are described by similar models (Inflow to a hydraulically fractured well).

The error of the numerical derivative and the choice of the interval

We estimate both errors of the central difference, which uses equal steps $\delta$ to the left and to the right in the natural logarithm of time, $\delta = L\ln 10$ . Expanding the function in a Taylor series (a sum of powers of the step) gives a correction proportional to the second derivative of $\Delta p'$ . The noise of two independent readings gives a random error:

The two errors of a numerical derivative

\varepsilon_{\text{noise}} \approx \frac{\sigma}{\sqrt2\,\delta},\qquad \varepsilon_{\text{smooth}} \approx \frac{\delta^2}{6}\,\frac{\dd^2\Delta p'}{\dd X^2},\qquad \delta = L\ln 10,\quad X = \ln t

where:

$\varepsilon_{\text{noise}}$ standard deviation of the error of the derivative caused by the noise of the readings, Pa
$\varepsilon_{\text{smooth}}$ systematic error of smoothing (a chord instead of a tangent); it has the sign of the second derivative, Pa
$\sigma$ standard deviation of the noise of one reading, Pa
$\delta$ step of the central difference to each side, in the natural logarithm of time, —
$\dd^2\Delta p'/\dd X^2$ curvature of the derivative plot in the coordinates $\ln t$ , Pa

Units: any consistent system of units

Holds when

A central difference with equal steps to each side; the noise is independent from reading to reading.
The curvature on the segment is almost constant. For a very wide interval the estimate of the smoothing becomes rough.

The first error falls with the interval, and the second grows as its square. Their sum is smallest at $\delta^3 \approx 3\sigma/(\sqrt2\,|\dd^2\Delta p'/\dd X^2|)$ . On the plateau the curvature is close to zero, and we can widen the interval almost without loss. At the peak of the hump the curvature is negative and large, so a wide interval underestimates the peak. So in practice we use an interval of 0.1–0.3 cycle, and we check that the shapes stay the same when the interval changes.

Worked exampleThe error budget for the lesson's record

Given (assumed data)

Noise: $\sigma$ = 0.0004 MPa
Curvature of the derivative plot by the model: at the peak of the hump −0.178 MPa; on the plateau near 10 h about 0.00011 MPa

Solution

$L$ = 0.2 ( $\delta$ = 0.461): the noise gives $0.0004/(\sqrt2\cdot0.461) = 0.00061$ MPa, and the smoothing at the peak gives $0.461^2/6\cdot(-0.178) = -0.0063$ MPa, that is, 3.0% of the peak.
$L$ = 0.5 ( $\delta$ = 1.151): the noise is 0.00025 MPa, and the smoothing at the peak by the formula is −0.039 MPa (19%). The exact central difference by the model underestimates the peak by 16%: with so wide an interval, the formula shows only the order of magnitude.
On the plateau at $L$ = 0.5, the smoothing gives $1.151^2/6\cdot0.00011 = 0.00002$ MPa, which is 0.1% of the plateau level, and the noise of 0.00025 MPa is 0.8%.
The best interval at the peak: $\delta^3 \approx 3\cdot0.0004/(\sqrt2\cdot0.178)$ , $\delta \approx 0.17$ , that is, $L \approx 0.07$ cycle.

Answer: At the peak of the hump, a narrow interval is better (about 0.07 cycle). On the plateau, a wide one is better: there the smoothing distorts almost nothing, and the noise is suppressed.

What it means. So one interval for the whole record is always a compromise. Interpretation programs let us change it and compare the results. Only the shapes of the derivative that stay when we change the interval within reason can be trusted. The data in this example are assumed for illustration.

Connections to other topics

Type-curve matching — matching the pressure and the derivative to type curves: the derivative makes the match less ambiguous.
Wellbore storage — the unit slope, and how long wellbore storage lasts before the plateau begins.
Pressure drawdown analysis — the semilog straight line, whose slope is the level of the derivative plateau.
Pressure build-up analysis — the Horner plot, and the choice of the section of the line by the derivative.
A sealing fault near the well: the method of images — the step of the derivative at a fault and the distance to it.
Several boundaries: parallel faults and wedge-shaped blocks — channels, angles between faults and their slopes on the derivative.
Reservoir limit test — a closed area: the unit slope of the derivative and the pore volume.
Effect of producing time on build-up analysis — the derivative of a build-up after a short producing time, and the equivalent time.
Nonideal wellbore storage, well clean-up and changing skin — the humps and dips that a changing wellbore storage coefficient gives.
Flow regimes — the flow regimes and geometries that we recognise from the derivative.
Superposition principle — images at boundaries, and the derivative with respect to the superposition time function when the rate changes.

Check yourself

1
On a diagnostic plot the derivative is horizontal at the level 0.18 MPa from 2 to 40 h. What does this mean, and what can we find from this section?
1. A.Radial flow: from the level we find the flow capacity $kh$
2. B.Pure wellbore storage: we find $C$
3. C.Linear flow towards a fracture
4. D.Pseudo-steady-state flow of a closed area
Show the answer
Correct answer: A. Radial flow: from the level we find the flow capacity $kh$
A constant derivative is a constant slope of the semilog plot, that is, radial flow in an infinite reservoir. The plateau level equals $qB\mu/(4\pi kh)$ . From it we find $kh$ , and from the ratio of the pressure to the plateau we find the skin. Wellbore storage and a closed area correspond to a slope of 1, and linear flow to ½.
2
The plateau level is 0.18 MPa; the rate is 60 m³/d, $B$ = 1.15, $\mu$ = 2.2 mPa·s, and the thickness of the reservoir is 9 m. Find the flow capacity, the permeability and the slope of the semilog straight line per decimal cycle.
Show the answer
$qB\mu = 60\cdot1.15\cdot2.2 = 151.8$ ; $kh = 0.933\cdot151.8/0.18 \approx 787$ mD·m; $k \approx 87$ mD. The slope per cycle is $m = 2.303\cdot0.18 \approx 0.415$ MPa. Check with the formula for a line: $2.149\cdot151.8/0.415 \approx 786$ mD·m.
3
A day after start-up the derivative started to rise, and after about three logarithmic cycles it reached a new horizontal level, twice as high as the first. What is it most likely?
1. A.A constant-pressure boundary
2. B.A closed area
3. C.The end of wellbore storage
4. D.A single straight sealing fault
Show the answer
Correct answer: D. A single straight sealing fault
The image, with the same rate, in the end adds one more $m_e$ to the slope, and the derivative reaches $2m_e$ . The transition from 1 to 99% takes about 2.7 cycles. In a closed area the derivative would rise with slope 1 and have no new horizontal level. At a constant-pressure boundary it would fall. Wellbore storage ends at the start of the record.
4
Why does a wide differentiation interval distort the hump of the derivative more than the plateau?
Show the answer
The smoothing error is proportional to the curvature of the derivative plot and to the square of the interval. On the plateau the curvature is close to zero, and a wide interval changes almost nothing, but it suppresses the noise. At the peak of the hump the curvature is large and negative, and the chord is clearly below the tangent. In the lesson's example, at an interval of 0.5 cycle the peak is 15% lower, and the scatter on the plateau is less than one per cent.
5
On the plateau at 10 h after start-up, $\Delta p$ = 0.92 MPa and the plateau level is 0.040 MPa; $k$ = 25 mD, $\phi$ = 0.16, $\mu$ = 1.4 mPa·s, $c_t = 1.5\cdot10^{-3}$ MPa⁻¹ and $r_w$ = 0.1 m. Find the skin.
Show the answer
$\Delta p/m_e = 0.92/0.040 = 23.0$ . $\dfrac{kt}{\phi\mu c_tr_w^2} = \dfrac{25\cdot10}{0.16\cdot1.4\cdot1.5\cdot10^{-3}\cdot0.01} = 7.44\cdot10^7$ , and its natural logarithm is 18.12. $S = \tfrac12(23.0 - 18.12 + 4.831) \approx 4.9$ .
6
In linear flow towards a fracture (with the skin pressure drop subtracted), how do the pressure and the derivative relate on a log–log plot?
1. A.They coincide on a line with slope 1
2. B.The derivative is 2.303 times above the pressure
3. C.Parallel straight lines with slope ½, and the derivative is half as high
4. D.The derivative is horizontal, and the pressure grows
Show the answer
Correct answer: C. Parallel straight lines with slope ½, and the derivative is half as high
For $\Delta p = A\sqrt t$ the derivative is $\dd\Delta p/\dd\ln t = \tfrac12 A\sqrt t = \Delta p/2$ . Both curves are straight lines with slope ½, and the derivative is lower by $\lg 2 \approx 0.3$ of a cycle. A coincidence on a unit slope is a sign of wellbore storage or a closed area.

Updated: 7 October 2026