Pressure build-up analysis

Quick refresher

Shutting in a well is the same as starting an imaginary injection well at the same point, at the same rate (the superposition principle: the effects of separate flows add up). From this comes the Horner equation: $p_{ws}$ is a straight-line function of $\lg[(t_p + \Delta t)/\Delta t]$ . The ratio $(t_p + \Delta t)/\Delta t$ is called the Horner ratio.
The slope $m$ of the Horner line per logarithmic cycle (per tenfold change of the ratio) gives the flow capacity: $kh = 2.149\,qB\mu/m$ (m³/d, mPa·s, MPa).
If we extend the line to a Horner ratio of 1, we get the pressure $p^*$ . In a new reservoir with little production, it equals the initial reservoir pressure.
The build-up pressures carry no skin effect. We find the skin from the difference between the pressure on the line at 1 h and the bottomhole pressure at the moment of shut-in.
The line does not appear at once. At the start, the inflow into the wellbore after shut-in gets in the way. At the end, the reservoir boundaries and the neighbouring wells do. We confirm the section of radial flow with a diagnostic plot (the pressure change against time on log–log axes).
The MDH method (named after Miller, Dyes and Hutchinson: $p_{ws}$ against $\lg\Delta t$ ) is simpler. But it is right only when the shut-in time is much shorter than the producing time, $\Delta t \ll t_p$ .

Key relationships

p_{ws} = p^* - m\,\lg\dfrac{t_p + \Delta t}{\Delta t}

the Horner equation ( $m$ is the change of pressure per logarithmic cycle)

kh = 2.149\,\dfrac{qB\mu}{m}

flow capacity, metric oilfield units (in oilfield units the constant is 162.6)

S = 1.151\left[\dfrac{p_{1\text{h}} - p_{wf}(\Delta t = 0)}{m} - \lg\dfrac{k}{\phi\mu c_t r_w^2} + 2.098\right]

skin factor, metric oilfield units (in oilfield units the constant is 3.23)

\Delta p_s = 0.869\,m\,S

skin pressure drop

Common pitfalls

Drawing the line through the early points, which wellbore storage has distorted. The slope comes out steeper and the permeability too low.
Taking $p_{1\text{h}}$ from the measurements and not from the line. One hour after shut-in, wellbore storage may not be over yet.
Mixing up the slope per decimal cycle and the slope per unit of natural logarithm. They differ by a factor of 2.303.
Taking $p^*$ for the average reservoir pressure in a reservoir that has been developed for a long time.
Using MDH when the shut-in time is comparable with the producing time of the well.

After this lesson you can

explain why the pressure in a shut-in well rises, and derive the Horner equation from the superposition principle (adding up the effects of separate flow rates)
draw a Horner plot, pick the section of radial flow on it (where fluid flows towards the well from all sides), and tell it from the distortions caused by wellbore storage (the well acting as a small tank) and by boundaries
find the flow capacity $kh$ , the permeability and the extrapolated pressure $p^*$ from the Horner line, in metric and oilfield units
calculate the skin factor from the pressure on the line at 1 h and the bottomhole pressure at shut-in
understand when the simplified MDH method is acceptable, and why it distorts the result when the producing time is short

The build-up test: what we measure

A well produces at a constant rate $q$ for the producing time $t_p$ . Then we shut it in and record the bottomhole pressure $p_{ws}$ as it changes with the shut-in time $\Delta t$ , the time since shut-in. This rise of pressure after shut-in is the pressure build-up curve.

The main advantage of a build-up test is that the rate during the record is known exactly: it is zero. In a drawdown test, where we record how the pressure falls after start-up (Pressure drawdown analysis), it is much harder to hold the rate strictly constant. The drawback is that the well does not produce while we record.

CalculatedCalculated bottomhole pressure during the test in the lesson's example: 60 h of production at 70 m³/d, then 24 h after shut-in. After shut-in the pressure first rises fast, then more and more slowly. During the record it does not return to the initial pressure.

For the analysis we need the rate history before shut-in, and the exact bottomhole pressure and time at the moment of shut-in, $p_{wf}(\Delta t = 0)$ (the last pressure of the flowing well). We also need the properties of the reservoir and the fluid: thickness $h$ , porosity $\phi$ , viscosity $\mu$ , formation volume factor $B$ (how much space a unit of fluid takes in the reservoir compared with the surface) and total compressibility $c_t$ . Last, we need the wellbore radius and the design of the well: the depth of the gauge, the pipe diameters and the position of the packer (the seal between the tubing and the casing).

Why the pressure builds up and how to describe it

While the well was producing, a drawdown cone formed around it (the region around the well where the pressure has dropped): the pressure at the wellbore is lower than the pressure far away. After shut-in the fluid still flows towards the well, because the pressure difference in the reservoir is still there. But the fluid cannot leave the reservoir now. It piles up at the wellbore and is compressed, so the pressure rises, and the cone slowly fills up. At the wall of the well the pressure gradient (the change of pressure with distance) falls to zero. As the pressures even out, the flows weaken. So the last stages of the build-up are slower and slower.

The diffusivity equation is linear, so we can add up its solutions for different rates. This is the superposition principle. We can picture a shut-in like this. The well keeps producing at the rate $q$ . From the time $t_p$ , an imaginary injection well at the same point starts to inject at the same rate. The two rates add up to zero. The change in pressure equals the difference of two “pressure drops”:

Pressure build-up as the superposition of two rates

p_i - p_{ws}(\Delta t) = \frac{qB\mu}{2\pi kh}\Big[p_D(t_p + \Delta t) - p_D(\Delta t)\Big]

where:

$p_i$ initial reservoir pressure, Pa
$p_{ws}(\Delta t)$ bottomhole pressure at the time $\Delta t$ after shut-in, Pa
$p_D(t)$ dimensionless pressure drop at the wall of a well that has produced at a constant rate for a time $t$ (a drop without units: the real drop divided by $qB\mu/(2\pi kh)$ ), with the skin included (Line-source solution and the pressure function), —
$t_p,\ \Delta t$ producing time before shut-in and time since shut-in, s
$q,\ B,\ \mu$ rate before shut-in at standard conditions, formation volume factor, viscosity, m³/s, m³/m³, Pa·s
$k,\ h$ permeability and thickness of the reservoir, m², m

Units: SI (any consistent system of units)

Holds when

The problem is linear: one slightly compressible fluid, and the properties of the reservoir and the fluid do not depend on the pressure.
The rate was constant before shut-in.
The inflow from the reservoir stops at once at shut-in (we ignore wellbore storage).

The first term is the pressure drop as if the well kept producing. The second is the “return” of pressure caused by the imaginary injection well. The skin enters both terms in the same way, so it cancels out. The pressures of an ideal build-up carry no skin effect.

The Horner equation and the Horner plot

Until the pressure disturbance reaches the reservoir boundaries, the reservoir behaves as an infinite one (“infinite-acting”). Then, within seconds of start-up, the dimensionless pressure drop at the wall of the well is described by the logarithm of time. We put this into the superposition equation and get the Horner equation. The skin and all the constants cancel, and only the logarithm of a ratio of times is left (the derivation is in the advanced sections).

The Horner equation

p_{ws} = p_i - \frac{qB\mu}{4\pi kh}\,\ln\frac{t_p + \Delta t}{\Delta t}

where:

$p_{ws}$ bottomhole pressure at the time $\Delta t$ after shut-in, Pa
$p_i$ initial reservoir pressure, Pa
$(t_p + \Delta t)/\Delta t$ the Horner time ratio; it has no units, —
$t_p,\ \Delta t$ producing time before shut-in and time since shut-in (in any equal units), s
$q,\ B,\ \mu,\ k,\ h$ rate before shut-in, formation volume factor, viscosity, permeability, thickness, m³/s, m³/m³, Pa·s, m², m

Units: SI (any consistent system of units)

Holds when

The reservoir behaves as an infinite one both while the well produces and during the shut-in: the disturbance has not reached the boundaries.
The rate before shut-in is constant. The reservoir is homogeneous, and there is one slightly compressible fluid.
Wellbore storage is over: the inflow from the reservoir into the wellbore has stopped after shut-in.

On the axes $p_{ws}$ against $\lg[(t_p + \Delta t)/\Delta t]$ the points lie on a straight line. Its slope depends on the flow capacity $kh$ . If we extend the line to a Horner ratio of 1 (an infinitely long shut-in), we get the initial pressure. The ratio of the times has no units, so we can take $t_p$ and $\Delta t$ in any equal units.

In other unit systems (2)

Metric oilfield units (base-10 logarithm)

p_{ws} = p^* - m\,\lg\frac{t_p + \Delta t}{\Delta t},\qquad m = 2.149\,\frac{qB\mu}{kh}

$q$ in m³/d at standard conditions, $B$ in m³/m³, $\mu$ in mPa·s, $k$ in mD, $h$ in m; $p$ and $m$ in MPa ( $m$ is the change of pressure per logarithmic cycle).

$2.149 = \ln 10/(4\pi\cdot 0.0853)$ : $\ln 10$ converts to the base-10 logarithm, and $0.0853$ is the constant of Darcy's law in these units. For the natural logarithm the constant is $0.933$ .

Oilfield units

p_{ws} = p^* - m\,\lg\frac{t_p + \Delta t}{\Delta t},\qquad m = 162.6\,\frac{qB\mu}{kh}

$q$ in STB/d (stock-tank barrels per day), $B$ in rb/STB (reservoir barrels per stock-tank barrel), $\mu$ in cP (centipoise), $k$ in mD, $h$ in ft; $p$ and $m$ in psi.

$162.6 = 70.6\cdot\ln 10$ , where $70.6 = 1/(4\pi\cdot 1.127\cdot10^{-3})$ is the constant for the natural logarithm.

In practice we use the base-10 logarithm and the slope $m$ : the change of pressure over one logarithmic cycle, that is, over a tenfold change of the Horner ratio (for example, from 100 down to 10). Here $m$ is positive. Some references take the slope as negative, and then the signs change in the formulas for $kh$ and for the skin. We usually draw the axis of the Horner ratio so that the ratio gets smaller to the right. Then the shut-in time grows from left to right, as on an ordinary plot.

In Russian-language literature the same relation is often written with the transmissibility $\varepsilon = kh/\mu$ (the flow capacity divided by the viscosity). The slope of the line per decimal logarithmic cycle is then $i = 0.183\,Q/\varepsilon$ . Here $Q = qB$ is the rate at reservoir conditions, and $0.183 = \ln 10/(4\pi)$ (in consistent units).

Below is the assumed build-up that we analyse in this lesson, and its Horner plot.

Pressure measurements after shut-in (assumed data)

$\Delta t$ , h: 0.05
Horner ratio $(t_p + \Delta t)/\Delta t$: 1201
$p_{ws}$ , MPa: 20.209

$\Delta t$ , h: 0.1
Horner ratio $(t_p + \Delta t)/\Delta t$: 601
$p_{ws}$ , MPa: 21.385

$\Delta t$ , h: 0.2
Horner ratio $(t_p + \Delta t)/\Delta t$: 301
$p_{ws}$ , MPa: 22.024

$\Delta t$ , h: 0.5
Horner ratio $(t_p + \Delta t)/\Delta t$: 121
$p_{ws}$ , MPa: 22.470

$\Delta t$ , h: 1
Horner ratio $(t_p + \Delta t)/\Delta t$: 61
$p_{ws}$ , MPa: 22.739

$\Delta t$ , h: 2
Horner ratio $(t_p + \Delta t)/\Delta t$: 31
$p_{ws}$ , MPa: 22.989

$\Delta t$ , h: 3
Horner ratio $(t_p + \Delta t)/\Delta t$: 21
$p_{ws}$ , MPa: 23.130

$\Delta t$ , h: 4
Horner ratio $(t_p + \Delta t)/\Delta t$: 16
$p_{ws}$ , MPa: 23.227

$\Delta t$ , h: 6
Horner ratio $(t_p + \Delta t)/\Delta t$: 11
$p_{ws}$ , MPa: 23.360

$\Delta t$ , h: 8
Horner ratio $(t_p + \Delta t)/\Delta t$: 8.5
$p_{ws}$ , MPa: 23.451

$\Delta t$ , h: 12
Horner ratio $(t_p + \Delta t)/\Delta t$: 6
$p_{ws}$ , MPa: 23.574

$\Delta t$ , h: 16
Horner ratio $(t_p + \Delta t)/\Delta t$: 4.75
$p_{ws}$ , MPa: 23.656

$\Delta t$ , h: 24
Horner ratio $(t_p + \Delta t)/\Delta t$: 3.5
$p_{ws}$ , MPa: 23.763

$\Delta t$ , h	Horner ratio $(t_p + \Delta t)/\Delta t$	$p_{ws}$ , MPa
0.05	1201	20.209
0.1	601	21.385
0.2	301	22.024
0.5	121	22.470
1	61	22.739
2	31	22.989
3	21	23.130
4	16	23.227
6	11	23.360
8	8.5	23.451
12	6	23.574
16	4.75	23.656
24	3.5	23.763

Producing time before shut-in $t_p$ = 60 h, rate 70 m³/d, bottomhole pressure at shut-in 15.816 MPa. The data were calculated from a model of a homogeneous infinite reservoir with wellbore storage, and rounded to 0.001 MPa.

CalculatedHorner plot for the data of the table. The axis of the Horner ratio is logarithmic, and the shut-in time grows to the right. The points of the first 1–2 h lie below the line: this is wellbore storage. After that they lie on the radial-flow line. The dashed line is the line drawn in the example below through the points at 4 and 24 h, extended to the ratio 1, where we read

p^*

Where the straight line is on the plot: early, middle and late sections

The theory assumes that the inflow from the reservoir stops at the moment of shut-in. In reality we usually close the well at the wellhead (the top of the well). The reservoir keeps feeding fluid into the wellbore for some time, until it has compressed the liquid or gas there. This is the afterflow, and it is a sign of wellbore storage (the well acting as a small tank). While it lasts, the pressure rises more slowly than in an ideal build-up, and the skin pressure drop does not disappear at once. So the early points lie below the line. In the plot above, these are the first 1–2 h.

The larger the volume of the wellbore, the more compressible its contents (gas especially), and the lower the transmissibility of the reservoir, the longer this section lasts. A shut-in valve at the bottom of the well, as in drill stem testers (tools that test a formation on the drill pipe), cuts it down to minutes.

The middle section is the radial-flow line in an infinite reservoir. Only this section gives $kh$ , $p^*$ and the skin. When the pressure disturbance reaches the boundaries, the line ends. A sealing fault (a fault that blocks flow) makes the slope steeper, up to twice as steep (A sealing fault near the well: the method of images). In a closed area the pressure levels off at the average reservoir pressure (Build-up in a bounded reservoir and average reservoir pressure). The work of neighbouring wells also distorts the late points.

Sections of a build-up on the Horner plot

Section: Early
What controls the pressure: afterflow into the wellbore after shut-in, skin, reservoir properties near the wellbore
What it looks like: points below the line; on the log–log plot of $\Delta p$ , slope 1
What we get from it: wellbore storage coefficient; not used for $kh$ and the skin

Section: Middle
What controls the pressure: radial flow in an infinite reservoir
What it looks like: a straight line
What we get from it: $kh$ , permeability, $p^*$ , skin

Section: Late
What controls the pressure: reservoir boundaries, neighbouring wells, depletion
What it looks like: a bend: the slope grows near a fault, and the pressure levels off in a closed area
What we get from it: distance to the boundaries, average reservoir pressure

Section	What controls the pressure	What it looks like	What we get from it
Early	afterflow into the wellbore after shut-in, skin, reservoir properties near the wellbore	points below the line; on the log–log plot of $\Delta p$ , slope 1	wellbore storage coefficient; not used for $kh$ and the skin
Middle	radial flow in an infinite reservoir	a straight line	$kh$ , permeability, $p^*$ , skin
Late	reservoir boundaries, neighbouring wells, depletion	a bend: the slope grows near a fault, and the pressure levels off in a closed area	distance to the boundaries, average reservoir pressure

The condition of an infinite reservoir has two parts. Neither the disturbance from the producing well nor the disturbance from the shut-in may reach the boundaries. Suppose the well produced for so long that it “felt” the boundaries of its drainage area (the part of the reservoir from which it draws fluid). We can still pick out a Horner line, but its extension $p^*$ is then equal neither to the initial reservoir pressure nor to the average reservoir pressure.

Flow capacity, p* and skin from the Horner line

Flow capacity from the slope of the Horner line

kh = \frac{qB\mu\,\ln 10}{4\pi m}

where:

$kh$ flow capacity: the product of the permeability and the thickness of the producing part of the reservoir, m³
$m$ slope of the line per decimal logarithmic cycle, Pa
$q,\ B,\ \mu$ rate before shut-in at standard conditions, formation volume factor, viscosity, m³/s, m³/m³, Pa·s

Units: SI

Holds when

The slope is measured on the section of radial flow.
$h$ is the thickness of the producing part of the reservoir. If it is not known, the result stays $kh$ .

The test gives exactly the product $kh$ , the flow capacity of the whole producing thickness. The permeability is $kh$ divided by the net thickness (the part of the thickness that can produce). It is averaged over a large volume of reservoir and refers to the mobile fluid. So it can differ noticeably from the permeability of a core (a rock sample taken from the well) (Rock permeability).

In other unit systems (2)

Metric oilfield units

kh\,[\text{mD·m}] = 2.149\,\frac{q\,[\text{m}^3/\text{d}]\;B\;\mu\,[\text{mPa·s}]}{m\,[\text{MPa per cycle}]}

$q$ in m³/d at standard conditions, $\mu$ in mPa·s, $m$ in MPa per logarithmic cycle; $kh$ in mD·m.

$2.149 = \ln 10/(4\pi\cdot 0.0853)$ , as in the Horner equation.

Oilfield units

kh\,[\text{mD·ft}] = 162.6\,\frac{q\,[\text{STB/d}]\;B\;\mu\,[\text{cP}]}{m\,[\text{psi per cycle}]}

$q$ in STB/d, $B$ in rb/STB, $\mu$ in cP, $m$ in psi per cycle; $kh$ in mD·ft.

$162.6 = 70.6\cdot\ln 10$ .

If we extend the line to a Horner ratio of 1, we get the extrapolated pressure $p^*$ . In a new reservoir from which little fluid has been taken, $p^* \approx p_i \approx \bar p$ , where $\bar p$ is the average reservoir pressure. This is how the very first build-up gives the initial reservoir pressure. After long production, $p^*$ equals neither the initial pressure nor the average pressure in the drainage area. We find the average pressure with corrections for a bounded reservoir (Build-up in a bounded reservoir and average reservoir pressure).

The build-up pressures carry no skin effect. So we find the skin by comparing the line with the bottomhole pressure at the moment of shut-in, $p_{wf}(\Delta t = 0)$ . Before shut-in the pressure followed the drawdown equation with a skin. If we replace $p_i$ in it with $p^*$ , and $qB\mu/(4\pi kh)$ with the slope of the line, we get the skin formula. It is convenient to use the pressure $p_{1\text{h}}$ . This is the value on the line, or on its extension, 1 h after shut-in, that is, at the Horner ratio $t_p + 1$ ( $t_p$ in hours):

Skin factor from a build-up

S = 1.151\left[\frac{p_{1\text{h}} - p_{wf}(\Delta t = 0)}{m} - \lg\frac{k}{\phi\,\mu\,c_t\,r_w^2} + 2.098\right]

where:

$p_{1\text{h}}$ pressure on the Horner line (or its extension) at $\Delta t$ = 1 h, MPa
$p_{wf}(\Delta t = 0)$ bottomhole pressure at the moment of shut-in, MPa
$m$ slope of the line per logarithmic cycle, MPa
$k$ permeability found from the slope, mD
$\phi$ porosity, fraction
$\mu$ viscosity of the fluid in the reservoir, mPa·s
$c_t$ total compressibility of the reservoir and the fluids, 1/MPa
$r_w$ wellbore radius, 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 producing time $t_p$ is much longer than 1 h. In the general case we add the term $\lg[(t_p + 1)/t_p]$ to the brackets.
Both the production and the shut-in take place as radial flow in an infinite reservoir.
We read $p_{1\text{h}}$ on the line, even if the measurement itself at 1 h is still distorted by wellbore storage.

The constants: $1.151 = \ln 10/2$ ; $2.098 = -\lg(3.553\cdot10^{-3}) - 0.351$ , where $3.553\cdot10^{-3}$ converts the time in hours into dimensionless time in these units, and $0.351 = 0.809/\ln 10$ comes from the logarithmic approximation of the solution. The more the difference $p_{1\text{h}} - p_{wf}$ exceeds what a homogeneous reservoir would give, the larger the skin.

In other unit systems (2)

Oilfield units

S = 1.151\left[\frac{p_{1\text{h}} - p_{wf}(\Delta t = 0)}{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.

$3.23 = -\lg(2.637\cdot10^{-4}) - 0.351$ ; $2.637\cdot10^{-4}$ is the factor of dimensionless time in oilfield units.

SI, natural logarithm, through

p^*

S = \frac12\left[\frac{p^* - p_{wf}(\Delta t = 0)}{m_e} - \ln\frac{k\,t_p}{\phi\,\mu\,c_t\,r_w^2} - 0.809\right],\qquad m_e = \frac{qB\mu}{4\pi kh}

pressure in Pa, $k$ in m², $t_p$ in s, $\mu$ in Pa·s, $c_t$ in 1/Pa, $r_w$ in m; $m_e$ is the slope per unit of natural logarithm.

$0.809 = \ln(4/\gamma)$ , where $\gamma = 1.781$ is the exponential of Euler's constant.

We calculate the skin pressure drop and the flow efficiency (how much of the ideal flow the well really gets, compared with a well without skin) as in the lesson on the skin factor: $\Delta p_s = 0.869\,m\,S$ and $\text{FE} = (p^* - p_{wf} - \Delta p_s)/(p^* - p_{wf})$ . For a new well, $p^*$ replaces the average reservoir pressure.

Worked examplePermeability, p* and skin from a build-up

Given (assumed data)

Oil rate before shut-in $q$ and producing time $t_p$: 70 m³/d; 60 h
Formation volume factor $B$ , viscosity $\mu$ , thickness $h$: 1.20 m³/m³; 1.6 mPa·s; 9 m
Porosity $\phi$ , total compressibility $c_t$ , wellbore radius $r_w$: 0.21; $1.4\cdot10^{-3}$ MPa⁻¹; 0.108 m
Bottomhole pressure at shut-in $p_{wf}(\Delta t = 0)$: 15.816 MPa
Measurements after shut-in: the table and the plot above

Solution

Choosing the line: on the plot, the points after 2 h lie on a line, and the early ones lie below it (wellbore storage). We take two points on the line: $\Delta t$ = 4 h (Horner ratio 16; 23.227 MPa) and 24 h (3.5; 23.763 MPa).
Slope per cycle: $m = \dfrac{23.763 - 23.227}{\lg 16 - \lg 3.5} = \dfrac{0.536}{0.6601} = 0.812$ MPa.
Flow capacity and permeability: $kh = 2.149\cdot\dfrac{70\cdot 1.2\cdot 1.6}{0.812} = 356$ mD·m, $k = 356/9 = 39.5$ mD.
Extrapolated pressure: $p^* = 23.763 + 0.812\cdot\lg 3.5 = 24.205$ MPa.
Pressure on the line at 1 h (Horner ratio 61): $p_{1\text{h}} = 24.205 - 0.812\cdot\lg 61 = 22.755$ MPa. The measurement at 1 h (22.739 MPa) is lower, because wellbore storage is still acting there.
Skin: $\lg\dfrac{k}{\phi\mu c_t r_w^2} = \lg\dfrac{39.5}{0.21\cdot 1.6\cdot 1.4\cdot10^{-3}\cdot 0.108^2} = 6.857$ , $S = 1.151\left[\dfrac{22.755 - 15.816}{0.812} - 6.857 + 2.098\right] = 4.36$ .
Skin pressure drop and flow efficiency: $\Delta p_s = 0.869\cdot 0.812\cdot 4.36 = 3.1$ MPa, $\text{FE} = \dfrac{24.205 - 15.816 - 3.1}{24.205 - 15.816} = 0.63$ .

Answer:

kh \approx 356

mD·m,

k \approx 39.5

mD. The initial reservoir pressure is about 24.20 MPa. The skin is about 4.4, so the near-wellbore zone takes 3.1 MPa of the 8.39 MPa of drawdown.

What it means. The data in this example are assumed for illustration. We calculated them from a model of a homogeneous infinite reservoir with a permeability of 40 mD and a skin of 4.5, with wellbore storage. The analysis returned these values with an error of about 1% in permeability and 3% in skin. The error comes from wellbore storage, which had not quite finished by 4 h, and from the rounding of the measurements. In practice we draw the line through all the points of the radial section by the method of least squares (a fit that makes the total of the squared gaps as small as possible). More than a third of the drawdown goes to the near-wellbore zone. If we removed the damage, the rate would grow by a factor of about 1.6, unless the lift system (the pump and tubing that bring the liquid up) limits it.

The MDH method and the equivalent producing time

Suppose the shut-in time is much shorter than the producing time, $\Delta t \ll t_p$ . Then $(t_p + \Delta t)/\Delta t \approx t_p/\Delta t$ , and the Horner equation turns into a relation between the pressure and the logarithm of the shut-in time alone. The MDH method (the Miller–Dyes–Hutchinson method) is built on this: we plot $p_{ws}$ against $\lg\Delta t$ .

The MDH line

p_{ws} = p_{1\text{h}} + m\,\lg\Delta t

where:

$p_{ws}$ bottomhole pressure at the time $\Delta t$ after shut-in, MPa
$\Delta t$ time since shut-in, h
$p_{1\text{h}}$ pressure on the MDH line at $\Delta t$ = 1 h, MPa
$m$ slope per logarithmic cycle, the same as for the Horner line, MPa

Units: metric oilfield units (pressure in MPa, time in hours)

Holds when

$\Delta t \ll t_p$ over the whole section we use. If the shut-in time is no more than about a tenth of the producing time, the slope is distorted by no more than a few per cent.
The other conditions are the same as for the Horner method.

We calculate the flow capacity and the skin by the same formulas as for the Horner line. We do not need to know the producing time, and this is the convenience of the method. In Russian-language practice this way of processing a build-up is often called the “tangent method”, because we draw a tangent (a straight line that touches the curve) to its late part.

Worked exampleMDH on the same data

Given (assumed data)

Measurements: the same as in the previous example; producing time 60 h
Points for the line: $\Delta t$ = 4 h (23.227 MPa) and 24 h (23.763 MPa)

Solution

Slope on the plot of $p_{ws}$ against $\lg\Delta t$ : $m_{\text{MDH}} = \dfrac{23.763 - 23.227}{\lg 24 - \lg 4} = \dfrac{0.536}{0.7782} = 0.689$ MPa.
Permeability: $k = 2.149\cdot\dfrac{70\cdot 1.2\cdot 1.6}{0.689\cdot 9} = 46.6$ mD. This is 18% more than by Horner (39.5 mD).
Skin with $p_{1\text{h}}$ on the MDH line ( $23.227 - 0.689\cdot\lg 4 = 22.812$ MPa): $S = 6.13$ instead of 4.36 by Horner.
The reason: MDH drops the term $m\,\lg[(t_p + \Delta t)/t_p]$ . By 24 h it equals 0.12 MPa. So the late points on the MDH plot lie lower than the extension of the early ones, and the line through them comes out flatter than the true one.

Answer: By MDH the permeability is 46.6 mD and the skin is 6.13. Both are too high: with

t_p

= 60 h and

\Delta t

up to 24 h, the condition

\Delta t \ll t_p

is not met.

What it means. MDH is handy when the well produced for a long time, hundreds or thousands of hours, and the exact producing time is not known, or the rate changed long before the shut-in. Then the term that MDH drops is small. After a short producing period, as in exploration wells, we use the Horner method. The data in this example are assumed for illustration.

If the rate changed before shut-in, we put the equivalent producing time $t_p = N_p/q$ into the Horner formula. Here we divide the cumulative production $N_p$ by the last stable rate $q$ . This shortcut does not work when the rate changes sharply. One example is a well that overflows at the wellhead (liquid spills out) while the liquid level in the tubing is still rising. Then we take the whole rate history into account by superposition (Equivalent producing time and synthetic flowing time).

How to record a build-up that can be analysed

Before shut-in, the well produces at a constant, stabilised rate, and the rate history is recorded.
The pressure and the time at the moment of shut-in are recorded exactly. We measure $\Delta t$ from that moment, and we use that pressure to calculate the skin.
The readings are frequent at the start (every few seconds) and less and less frequent later, so that the points are spread evenly on a logarithmic time scale.
The gauge has enough resolution (it can detect small changes of pressure). In high-permeability reservoirs the whole build-up can be only a fraction of a megapascal, and we need quartz gauges with a resolution of the order of 0.0001 MPa.
If possible, the well is shut in at the bottom, with a valve near the reservoir. This reduces wellbore storage.
The test is long enough to record at least one logarithmic cycle of radial flow after wellbore storage ends. The calculations made when the test is planned give a guide.
The design of the well is known: the depth of the gauge, the pipe diameters, the position of the packer and the perforated interval (the part of the casing with holes through which the fluid enters).

Terms and symbols

Producing time $t_p$: How long the well produced at a constant rate before shut-in. If the rate changed, it is the equivalent time $N_p/q$ .
Horner ratio: $(t_p + \Delta t)/\Delta t$ . It has no units, and it falls from very large values towards 1 as the shut-in time grows.
Extrapolated pressure $p^*$: The pressure on the extension of the Horner line at a Horner ratio of 1. In a new reservoir it is close to the initial reservoir pressure.
$p_{1\text{h}}$: The pressure on the line (or on its extension) 1 h after shut-in. We need it to calculate the skin.
Afterflow: The inflow from the reservoir into the wellbore that continues after the well is closed at the wellhead. It is a sign of wellbore storage, and it distorts the early points of the build-up.
Radial flow (infinite reservoir): The regime in which fluid flows towards the well from all sides and the pressure disturbance has not yet reached any boundary. On the Horner plot it appears as a straight line.

Common mistakes and misconceptions

Wrong

“The Horner line is any straight stretch of the plot.”

In fact

A section that wellbore storage has distorted can look straight too. Its slope is steeper, and the permeability comes out too low. We confirm radial flow with the diagnostic plot and the pressure derivative.

Wrong

“ $p^*$ is the average reservoir pressure.”

In fact

This is true only in a new reservoir with little production. After long development, $p^*$ is higher than the average pressure in the drainage area. We find the average pressure with corrections for a bounded reservoir.

Wrong

“We can find the skin from a single build-up, without data on how the well produced.”

In fact

The build-up pressures carry no skin effect. We need the exact bottomhole pressure at the moment of shut-in and the rate before it.

Wrong

“The slope per natural logarithm and the slope per decimal cycle are the same thing.”

In fact

They differ by a factor of 2.303. The constants 2.149 and 162.6 belong to the decimal cycle, and 0.933 and 70.6 belong to the natural logarithm.

Wrong

“MDH and Horner always give the same permeability.”

In fact

They do only when $\Delta t \ll t_p$ . In the lesson's example, with 60 h of production and a record up to 24 h, MDH overestimates the permeability by almost 20%.

In depth: advanced detail

Derivation of the Horner equation and the skin formula

Take a well that produces at a constant rate in an infinite reservoir. Within seconds of start-up, the dimensionless pressure at the wall is $p_D(t) = \tfrac12\left(\ln t_D + 0.809\right) + S$ , where $t_D = \dfrac{kt}{\phi\mu c_t r_w^2}$ (SI). We put this into the superposition: $p_D(t_p + \Delta t) - p_D(\Delta t) = \tfrac12\ln\dfrac{t_p + \Delta t}{\Delta t}$ . The constant 0.809, the factor $k/(\phi\mu c_t r_w^2)$ and the skin cancel. We multiply by $qB\mu/(2\pi kh)$ and get the Horner equation.

At the moment of shut-in, $p_{wf} = p_i - \dfrac{qB\mu}{4\pi kh}\left(\ln\dfrac{k t_p}{\phi\mu c_t r_w^2} + 0.809 + 2S\right)$ . We replace $p_i$ with $p^*$ and $qB\mu/(4\pi kh)$ with the slope $m_e$ , and solve for $S$ . This is the form through $p^*$ . Switching to base-10 logarithms and to $p_{1\text{h}} = p^* - m\lg(t_p + 1)$ gives the working formula with the constant 2.098.

The logarithmic approximation does not hold in the first moments after shut-in, while the dimensionless time is below about 25. In theory the very earliest points lie above the line. In practice these are seconds, and wellbore storage hides them completely. Wellbore storage makes the early points lie below the line instead.

A build-up as a probe: which pressure in the reservoir a reading corresponds to

We can read a build-up as a snapshot of the reservoir at the moment of shut-in. We equate the build-up pressure with the pressure in the reservoir while the well was producing (both in the logarithmic approximation). The result is this: a reading taken a time $\Delta t$ after shut-in equals the pressure that, at the moment of shut-in, existed at a distance $r_p$ from the well.

Radius of the “probe” in a build-up

r_p = \sqrt{\frac{4k\,\Delta t_e}{\gamma\,\phi\mu c_t}},\qquad \Delta t_e = \frac{t_p\,\Delta t}{t_p + \Delta t}

where:

$r_p$ distance from the well at which the pressure at the moment of shut-in was equal to the reading $p_{ws}(\Delta t)$ , m
$\Delta t_e$ equivalent time (Effect of producing time on build-up analysis); for $\Delta t \ll t_p$ it is almost equal to $\Delta t$ , s
$t_p,\ \Delta t$ producing time before shut-in and time since shut-in, s
$\gamma$ the exponential of Euler's constant, 1.781, —
$k,\ \phi,\ \mu,\ c_t$ permeability, porosity, viscosity, total compressibility, m², —, Pa·s, 1/Pa

Units: SI

Holds when

The reservoir is infinite. At the point $r_p$ the logarithmic approximation holds: the disturbance passed it long ago.
Wellbore storage is over. The skin changes the pressure only in the well itself, and it does not enter the formula.

The longer the build-up lasts, the farther from the well lies the point whose pressure the reading “shows”. For the lesson's example, the readings at 1, 4 and 24 h correspond to distances of about 26, 50 and 108 m. We use this, for example, when we compare the pressures in the cells of a reservoir simulation model with the results of a build-up (Well model in a grid block).

In other unit systems (1)

Metric oilfield units

r_p\,[\text{m}] = 0.0893\sqrt{\frac{k\,[\text{mD}]\;\Delta t_e\,[\text{h}]}{\phi\;\mu\,[\text{mPa·s}]\;c_t\,[\text{MPa}^{-1}]}}

$k$ in mD, $\Delta t_e$ in h, $\mu$ in mPa·s, $c_t$ in 1/MPa, $r_p$ in m.

$0.0893 = \sqrt{4\cdot 3.553\cdot10^{-3}/1.781}$ , where $3.553\cdot10^{-3}$ is the factor of dimensionless time in these units.

Where the metric constants come from

In SI, the slope per unit of natural logarithm is $qB\mu/(4\pi kh)$ . We put $q$ in m³/d (dividing by 86,400), $\mu$ in mPa·s (multiplying by $10^{-3}$ ) and $k$ in mD (multiplying by $0.9869\cdot10^{-15}$ m²), and we express the pressure in MPa (dividing by $10^6$ ). The factor is $\dfrac{10^{-3}}{86{,}400\cdot 4\pi\cdot 0.9869\cdot10^{-15}\cdot10^{6}} = 0.933$ . Multiplying by $\ln 10$ gives 2.149 for the decimal cycle.

In SI, the dimensionless time is $t_D = \dfrac{kt}{\phi\mu c_t r_w^2}$ . We take $t$ in hours (multiplying by 3600), $k$ in mD, $\mu$ in mPa·s and $c_t$ in 1/MPa (multiplying by $10^{-6}$ ). The factor is then $\dfrac{0.9869\cdot10^{-15}\cdot 3600}{10^{-3}\cdot10^{-6}} = 3.553\cdot10^{-3}$ . From it comes the constant of the skin formula: $-\lg(3.553\cdot10^{-3}) - 0.809/\ln 10 = 2.449 - 0.351 = 2.098$ .

Connections to other topics

Skin factor — what the skin is, and how to turn it into a pressure drop and a loss of rate.
Rock permeability — what the permeability found from a build-up means, and how it differs from the permeability of a core.
Superposition principle — the superposition principle, on which the method is built.
Pressure drawdown analysis — the same semilog analysis for the period when the well produces.
Wellbore storage — why the early points of a build-up lie below the line.
Diagnosis with the pressure derivative — how to find the section of radial flow reliably.
Build-up in a bounded reservoir and average reservoir pressure — the average reservoir pressure from a build-up in a bounded reservoir.
Effect of producing time on build-up analysis — how a short producing time distorts a build-up.
A sealing fault near the well: the method of images — how the slope changes when a fault affects the test.

Check yourself

1
Why can we put the producing time and the shut-in time into the Horner ratio in any equal units?
1. A.We cannot: only in hours
2. B.Because the slope does not depend on the rate
3. C.The ratio has no units
4. D.Because the base-10 logarithm is used
Show the answer
Correct answer: C. The ratio has no units
$(t_p + \Delta t)/\Delta t$ is a ratio of two times, and the units cancel. We need hours only where time appears on its own: in $p_{1\text{h}}$ and in the constant of the skin formula.
2
Through which points do we draw the Horner line?
1. A.Through the last point and the initial pressure
2. B.Through all the points
3. C.Through the earliest ones: they are the most accurate
4. D.Through the points of the middle section, after wellbore storage has ended and before the boundaries have an effect
Show the answer
Correct answer: D. Through the points of the middle section, after wellbore storage has ended and before the boundaries have an effect
The early points are distorted by the inflow into the wellbore after shut-in, and the late points by the boundaries and the neighbouring wells. Only the section of radial flow obeys the Horner equation. We confirm it on the diagnostic plot and with the pressure derivative.
3
The slope of the Horner line is 0.30 MPa per cycle, the rate before shut-in is 50 m³/d, $B$ = 1.25, $\mu$ = 0.9 mPa·s and $h$ = 6 m. Find the flow capacity and the permeability.
Show the answer
$kh = 2.149\cdot 50\cdot 1.25\cdot 0.9/0.30 \approx 403$ mD·m; $k \approx 403/6 \approx 67$ mD.
4
Why can't we find the skin from a build-up alone, without the bottomhole pressure at the moment of shut-in?
Show the answer
At shut-in the pressure drop across the skin disappears, and in the superposition the skin enters both terms in the same way. So the build-up pressures carry no trace of it. The skin shows itself only in the bottomhole pressure of the producing well. We find it from the difference between the line (through $p_{1\text{h}}$ or $p^*$ ) and $p_{wf}(\Delta t = 0)$ .
5
A well produced at a constant rate for 2000 h, and the build-up was recorded for 48 h. Can we process it by the MDH method?
1. A.Only if the skin is zero
2. B.No, MDH applies only to exploration wells
3. C.Yes: $\Delta t \ll t_p$ , so the error is small
Show the answer
Correct answer: C. Yes: $\Delta t \ll t_p$ , so the error is small
By 48 h the term that MDH drops, $m\lg[(t_p + \Delta t)/t_p]$ , is only $\lg(2048/2000) \approx 0.01$ of a cycle. The slope by MDH differs from the Horner slope by less than 1%.
6
In a reservoir that has been developed for many years, a build-up gave the extrapolated pressure $p^*$ . How does it compare with the average pressure in the drainage area of the well?
1. A.Always lower than the bottomhole pressure
2. B.Usually higher than it
3. C.Equal to it
Show the answer
Correct answer: B. Usually higher than it
The Horner equation is built for an infinite reservoir, where after an infinitely long shut-in the pressure would return to the initial pressure. In a bounded, partly depleted area the pressure levels off at the average value, and the extension of the line goes higher. We find the average pressure with corrections for a bounded reservoir.

Updated: 19 October 2026