Pressure drawdown analysis

Quick refresher

We record a drawdown test after a well is started from rest, that is, from a shut-in state. We must know the reservoir pressure before start-up, either the initial pressure or the pressure after it has equalised, because we measure the pressure drop from it.
In radial flow, $p_{wf} = p_{1\text{h}} - m\,\lg t$ . The slope per decimal cycle (per tenfold change in time) gives the flow capacity: $kh = 2.149\,qB\mu/m$ (m³/d, mPa·s, MPa; in oilfield units the constant is 162.6).
We find the skin from the position of the line: for the same slope, the lower the pressure on the line at 1 h, the larger the skin. The slope does not depend on the skin.
The early points lie above the line. While the fluid in the wellbore is expanding (wellbore storage: the well acts as a small tank), the inflow from the reservoir lags behind the rate. In a closed area the late points lie below the line, because pseudo-steady-state depletion begins (the pressure falls everywhere at the same rate as the reservoir empties).
We choose the line between these sections and check it on the diagnostic plot. In radial flow, the logarithmic derivative of pressure (the slope of the pressure curve on a semilog plot) is horizontal.
The main difficulty of the method is holding the rate constant. That is why we analyse build-ups more often. We use drawdown tests for new wells and for long tests aimed at finding the boundaries.

Key relationships

p_{wf} = p_{1\text{h}} - m\,\lg t

semilog straight line; $m$ is the pressure drop per decimal cycle of time

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

flow capacity, metric oilfield units (162.6 in oilfield units)

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

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

\Delta p_s = 0.869\,m\,S

skin pressure drop

V_p = \dfrac{qB}{c_t\,|\dd p_{wf}/\dd t|}

pore volume of a closed area from the late straight section

Common pitfalls

Drawing the line through the early points, where wellbore storage is still acting. The slope comes out too steep, and the permeability too low.
Taking $p_{1\text{h}}$ from the measurements instead of from the line.
Mixing up the slope per decimal cycle with the slope per unit of natural logarithm. They differ by a factor of 2.303.
Extending the line beyond the radial-flow section, and mistaking the faster pressure decline near the boundaries for poorer reservoir properties.
Analysing a record as a drawdown test when the rate changed noticeably during the test.

After this lesson you can

explain why, at a constant rate, the bottomhole pressure falls in a straight line against the logarithm of time, and what sets the slope and the position of the line
draw a semilog plot, and tell the radial-flow section (where fluid flows towards the well from all sides) from the early points (distorted by wellbore storage) and the late points (distorted by the boundaries)
calculate the flow capacity, permeability and skin factor from the slope of the line and the pressure $p_{1\text{h}}$ (the pressure on the line at one hour), in metric and oilfield units, with base-10 and natural logarithms
estimate from the record the distance to the boundaries and the pore volume of the drainage area
name the conditions in which the method works poorly, and ways to get round these limits

The drawdown test

The test starts from rest. The well is shut in, and the reservoir pressure has equalised. It equals the initial pressure $p_i$ (in a virgin reservoir, one that has not been produced yet) or the current reservoir pressure. Then the well is opened at a constant rate $q$ , and we record how the bottomhole pressure $p_{wf}$ falls with the time $t$ after start-up. This record is called a drawdown test. In Russian-language practice it is also known as the “pressure stabilisation curve”.

CalculatedCalculated bottomhole pressure in the lesson's example test. The well is started at 120 m³/d from a reservoir at 27.5 MPa. The pressure drops sharply in the first minutes, then falls more and more slowly, and after 60–80 h it falls at a constant rate. The dashed straight line is drawn through the last two points.

The analysis needs the following: the rate, and confidence that it is constant; the reservoir pressure before start-up; the reservoir and fluid properties (thickness $h$ , porosity $\phi$ , viscosity $\mu$ , formation volume factor $B$ and total compressibility $c_t$ ); and the wellbore radius $r_w$ .

Drawdown and build-up: which to choose

Feature: What is done
Drawdown test: the well is started at a constant rate
Build-up test: the well is shut in after producing

Feature: Rate during the record
Drawdown test: must be held constant, which is difficult
Build-up test: zero, that is, known exactly

Feature: Production
Drawdown test: is not interrupted
Build-up test: is interrupted for the duration of the record

Feature: What must be known beforehand
Drawdown test: the reservoir pressure before start-up
Build-up test: the rate history and the pressure at shut-in

Feature: What is obtained
Drawdown test: $kh$ , skin; with a long record, the boundaries and the volume of the area
Build-up test: $kh$ , skin, reservoir pressure

Feature	Drawdown test	Build-up test
What is done	the well is started at a constant rate	the well is shut in after producing
Rate during the record	must be held constant, which is difficult	zero, that is, known exactly
Production	is not interrupted	is interrupted for the duration of the record
What must be known beforehand	the reservoir pressure before start-up	the rate history and the pressure at shut-in
What is obtained	$kh$ , skin; with a long record, the boundaries and the volume of the area	$kh$ , skin, reservoir pressure

Build-up analysis is in the lesson Pressure build-up analysis. The flow capacity and skin found from a drawdown test and from the build-up that follows it in the same well should agree. If they do not, check the rate measurements.

The first start-up of a new well is the natural case for such a test, because the reservoir pressure has not yet been disturbed. The same analysis applies to the flow period of a drill stem test (Well testing: purposes and types of test) and to a long extended well test (a long trial production period). With the sign of the pressure change reversed, it also applies to the start of injection in an injection well.

Why the plot is a straight line

Until the disturbance reaches the boundaries, the bottomhole pressure follows the line-source solution (Line-source solution and the pressure function). We can read it as the Dupuit equation in which the radius of the outer boundary is replaced by the growing radius of the drawdown cone (the region where the pressure has dropped). The cone widens, and ever new rings of reservoir add to the resistance. In every tenfold interval of time, the radius of the cone grows by a factor of $\sqrt{10} \approx 3.2$ . The resistance of the ring between two circles depends only on the ratio of their radii. So every tenfold increase of time adds the same amount to the pressure drop, and the plot of pressure against the logarithm of time is a straight line.

Bottomhole pressure at a constant rate in an infinite reservoir

p_{wf} = p_i - \frac{qB\mu}{4\pi kh}\left[\ln\frac{kt}{\phi\mu c_t r_w^2} + 0.809 + 2S\right]

where:

$p_{wf}$ bottomhole pressure at time $t$ after start-up, Pa
$p_i$ reservoir pressure before start-up (initial reservoir pressure), Pa
$q,\ B,\ \mu$ rate at standard conditions, formation volume factor, viscosity, m³/s, m³/m³, Pa·s
$k,\ h$ permeability and net thickness of the reservoir, m², m
$\phi,\ c_t,\ r_w$ porosity, total compressibility, wellbore radius, —, 1/Pa, m
$S$ skin factor, —
$m,\ p_{1\text{h}}$ in the oilfield forms: the pressure drop per decimal cycle of time and the pressure on the line at $t$ = 1 h, MPa or psi

Units: SI; natural logarithm

Holds when

The rate is constant from the moment of start-up; before start-up the pressure throughout the reservoir equals $p_i$ .
The reservoir behaves as an infinite one: the disturbance has not reached the boundaries.
Wellbore storage is over. The condition $kt/(\phi\mu c_tr_w^2) > 25$ is met within seconds.
The reservoir is homogeneous, there is a single slightly compressible fluid, and the skin does not change with time.

Time enters only under the logarithm, so $p_{wf}$ plotted against $\ln t$ or $\lg t$ is a straight line. Its slope is set by the rate, the viscosity and the flow capacity $kh$ . Porosity, compressibility, wellbore radius and skin do not enter the slope. They only shift the line up or down.

In other unit systems (2)

Metric oilfield units (base-10 logarithm)

p_{wf} = p_{1\text{h}} - m\,\lg t,\quad m = 2.149\,\frac{qB\mu}{kh},\quad p_{1\text{h}} = p_i - m\left[\lg\frac{k}{\phi\mu c_t r_w^2} - 2.098 + 0.869\,S\right]

$q$ in m³/d at standard conditions, $\mu$ in mPa·s, $k$ in mD, $h$ and $r_w$ in m, $c_t$ in 1/MPa, $t$ in h; pressure and $m$ in MPa.

$2.149 = \ln 10/(4\pi\cdot0.0853)$ ; $2.098 = -\lg(3.553\cdot10^{-3}) - 0.809/\ln 10$ ; $0.869 = 2/\ln 10$ . The derivation is in the advanced section.

Oilfield units (base-10 logarithm)

p_{wf} = p_{1\text{h}} - m\,\lg t,\quad m = 162.6\,\frac{qB\mu}{kh},\quad p_{1\text{h}} = p_i - m\left[\lg\frac{k}{\phi\mu c_t r_w^2} - 3.23 + 0.869\,S\right]

$q$ in STB/d (stock-tank barrels of oil per day), $\mu$ in cP, $k$ in mD, $h$ and $r_w$ in ft, $c_t$ in 1/psi, $t$ in h; pressure and $m$ in psi.

$162.6 = 70.6\cdot\ln 10$ , where $70.6 = 1/(4\pi\cdot1.127\cdot10^{-3})$ ; $3.23 = -\lg(2.637\cdot10^{-4}) - 0.809/\ln 10$ .

The quantity $m$ is called the semilog slope. It is the pressure drop over one decimal cycle of time, for example from 2 to 20 h. Here, as in the lesson on pressure build-up, $m$ is positive. Some references take the slope with its sign. For a drawdown test it is then negative, and the signs in the formulas change.

Where the straight line is on the plot: its start and end

Below is the assumed record that we analyse in this lesson, and its semilog plot.

Bottomhole pressure measurements after the well is started up (assumed data)

$t$ , h: 0.01
$p_{wf}$ , MPa: 26.782
$p_i - p_{wf}$ , MPa: 0.718

$t$ , h: 0.02
$p_{wf}$ , MPa: 26.418
$p_i - p_{wf}$ , MPa: 1.082

$t$ , h: 0.05
$p_{wf}$ , MPa: 26.020
$p_i - p_{wf}$ , MPa: 1.480

$t$ , h: 0.1
$p_{wf}$ , MPa: 25.863
$p_i - p_{wf}$ , MPa: 1.637

$t$ , h: 0.2
$p_{wf}$ , MPa: 25.763
$p_i - p_{wf}$ , MPa: 1.737

$t$ , h: 0.5
$p_{wf}$ , MPa: 25.654
$p_i - p_{wf}$ , MPa: 1.846

$t$ , h: 1
$p_{wf}$ , MPa: 25.576
$p_i - p_{wf}$ , MPa: 1.924

$t$ , h: 2
$p_{wf}$ , MPa: 25.501
$p_i - p_{wf}$ , MPa: 1.999

$t$ , h: 3
$p_{wf}$ , MPa: 25.458
$p_i - p_{wf}$ , MPa: 2.042

$t$ , h: 5
$p_{wf}$ , MPa: 25.403
$p_i - p_{wf}$ , MPa: 2.097

$t$ , h: 8
$p_{wf}$ , MPa: 25.353
$p_i - p_{wf}$ , MPa: 2.147

$t$ , h: 12
$p_{wf}$ , MPa: 25.310
$p_i - p_{wf}$ , MPa: 2.190

$t$ , h: 20
$p_{wf}$ , MPa: 25.256
$p_i - p_{wf}$ , MPa: 2.244

$t$ , h: 30
$p_{wf}$ , MPa: 25.212
$p_i - p_{wf}$ , MPa: 2.288

$t$ , h: 50
$p_{wf}$ , MPa: 25.154
$p_i - p_{wf}$ , MPa: 2.346

$t$ , h: 80
$p_{wf}$ , MPa: 25.085
$p_i - p_{wf}$ , MPa: 2.415

$t$ , h: 120
$p_{wf}$ , MPa: 24.998
$p_i - p_{wf}$ , MPa: 2.502

$t$ , h: 180
$p_{wf}$ , MPa: 24.868
$p_i - p_{wf}$ , MPa: 2.632

$t$ , h: 240
$p_{wf}$ , MPa: 24.737
$p_i - p_{wf}$ , MPa: 2.763

$t$ , h	$p_{wf}$ , MPa	$p_i - p_{wf}$ , MPa
0.01	26.782	0.718
0.02	26.418	1.082
0.05	26.020	1.480
0.1	25.863	1.637
0.2	25.763	1.737
0.5	25.654	1.846
1	25.576	1.924
2	25.501	1.999
3	25.458	2.042
5	25.403	2.097
8	25.353	2.147
12	25.310	2.190
20	25.256	2.244
30	25.212	2.288
50	25.154	2.346
80	25.085	2.415
120	24.998	2.502
180	24.868	2.632
240	24.737	2.763

Rate 120 m³/d, reservoir pressure before start-up 27.5 MPa. The data were calculated from a model of a well with wellbore storage and skin, at the centre of a closed circular area, and rounded to 0.001 MPa.

CalculatedSemilog plot for the data of the table. The time axis is logarithmic. The points of the first minutes lie above the line. This is wellbore storage. From 2 to 30 h the points fall on the radial-flow line. After 50 h they go below it, as the boundaries of the area make themselves felt. In the example below, the dashed line is drawn through the points at 2 and 20 h.

The start. Theory assumes that the inflow from the reservoir becomes equal to $q$ at once. In reality the well is opened at the surface, and for the first minutes it produces the fluid stored in the wellbore. The liquid and gas in the wellbore expand, and the inflow from the reservoir builds up gradually. This is wellbore storage. While it lasts, the reservoir is “under-loaded”, and the bottomhole pressure is higher than the formula gives: the early points lie above the line. The skin pressure drop also appears not at once but as the inflow grows.

The end. When the disturbance reaches the boundaries, the straight line ends. In a closed area the pressure begins to fall faster. The points go below the line, and on ordinary axes of pressure against time they fall on a new straight line. This is the pseudo-steady-state regime (Pseudo-steady-state flow and the shape factor). If fluid flows in across a boundary, the pressure does the opposite: it stabilises, and the points end up above the line. A single sealing fault (a fault that blocks flow) near the well doubles the slope (A sealing fault near the well: the method of images).

Sections of the drawdown test

Section: Early
What controls the pressure: wellbore volume and skin
Points on the semilog plot: above the line
What is obtained: wellbore storage coefficient (Wellbore storage)

Section: Middle
What controls the pressure: radial flow in an infinite reservoir
Points on the semilog plot: on the line
What is obtained: $kh$ , permeability, skin

Section: Late
What controls the pressure: boundaries of the drainage area
Points on the semilog plot: below the line in a closed area, above it where fluid is supplied
What is obtained: distance to the boundaries, pore volume (Reservoir limit test)

Section	What controls the pressure	Points on the semilog plot	What is obtained
Early	wellbore volume and skin	above the line	wellbore storage coefficient (Wellbore storage)
Middle	radial flow in an infinite reservoir	on the line	$kh$ , permeability, skin
Late	boundaries of the drainage area	below the line in a closed area, above it where fluid is supplied	distance to the boundaries, pore volume (Reservoir limit test)

The middle section is called infinite-acting radial flow (radial flow in a reservoir that behaves as if it had no boundaries). We see it most reliably on the diagnostic plot, where we plot the pressure drop $\Delta p = p_i - p_{wf}$ and its logarithmic derivative $t\,\dd\Delta p/\dd t$ on log–log axes (Diagnosis with the pressure derivative). On the radial-flow line the derivative is constant and equal to $m/2.303$ , the slope per unit of natural logarithm.

CalculatedDiagnostic plot of the same test. The calculated curves are the pressure drop (upper) and its logarithmic derivative (lower). The dots are the measurements from the table. Both axes are logarithmic. In the first seconds both curves follow the dashed unit-slope line, because only the wellbore is working. From 2 to 30 h the derivative lies on the plateau at 0.106 MPa: this is radial flow. After that it rises again with a slope of 1: this is pseudo-steady-state flow.

Flow capacity, permeability and skin

Flow capacity from the slope of the semilog straight line

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

where:

$kh$ flow capacity: the product of permeability and the thickness of the producing part of the reservoir, m³
$m$ pressure drop over one decimal cycle of time on the line, Pa
$q,\ B,\ \mu$ rate 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 infinite-acting radial flow.
The rate is constant. If the thickness of the producing part of the reservoir is not known, we report the result as $kh$ .

The formula is the same as for the Horner line, because in both cases the slope is set by radial flow in an infinite reservoir. An error in the rate passes directly into an error in the flow capacity. Permeability is $kh$ divided by the net thickness. It is an average over the volume of the reservoir within the radius of investigation (how far the pressure disturbance has travelled).

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 decimal cycle; $kh$ in mD·m.

$2.149 = \ln 10/(4\pi\cdot0.0853)$ . For a slope per unit of natural logarithm, the constant is 0.933.

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, $\mu$ in cP, $m$ in psi per decimal cycle; $kh$ in mD·ft.

$162.6 = 70.6\cdot\ln 10$ ; for the natural logarithm, 70.6.

We find the skin from the position of the line. It is convenient to use the pressure $p_{1\text{h}}$ , the value on the line, or on its continuation, at $t$ = 1 h (one-hour pressure). We compare the difference $p_i - p_{1\text{h}}$ with what a homogeneous reservoir without skin would have given by that time:

Skin factor from a drawdown test

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

where:

$p_i$ reservoir pressure before the well is started up, MPa
$p_{1\text{h}}$ pressure on the line (or its continuation) at $t$ = 1 h, MPa
$m$ pressure drop per decimal cycle of time on the line, 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, 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

We read $p_{1\text{h}}$ on the line and do not take it from the measurements, because wellbore storage may not be over an hour after start-up.
The well stood idle long enough before start-up for the pressure in the reservoir to equalise. Otherwise we need superposition with the previous history (adding up the effects of earlier rate changes) in place of $p_i$ .
$m$ and $k$ are found from the same straight section.

The fraction shows how many “cycles” of pressure drop had built up by the first hour of production. We subtract what a homogeneous reservoir without skin would have given. The remainder, multiplied by $1.151 = \ln 10/2$ , is the skin. An error in $p_i$ equal to $m$ changes the skin by 1.15, so we must know the pressure before start-up accurately.

In other unit systems (2)

Oilfield units

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

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

$3.23 = -\lg(2.637\cdot10^{-4}) - 0.809/\ln 10$ .

SI, natural logarithm, from any point of the line

S = \frac12\left[\frac{p_i - p_{wf}(t)}{m_e} - \ln\frac{k\,t}{\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, $t$ 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 share of the drawdown that is lost in the near-wellbore zone, as in the lesson on skin factor: $\Delta p_s = 0.869\,mS$ .

Worked examplePermeability and skin from a drawdown test

Given (assumed data)

Oil rate $q$ ; pressure before start-up $p_i$: 120 m³/d; 27.5 MPa
Formation volume factor $B$ ; viscosity $\mu$ ; thickness $h$: 1.25 m³/m³; 0.9 mPa·s; 14 m
Porosity $\phi$ ; total compressibility $c_t$ ; wellbore radius $r_w$: 0.17; $1.9\cdot10^{-3}$ MPa⁻¹; 0.108 m
Measurements: the table and plots above

Solution

First choose the line. The derivative is horizontal from 2 to 30 h, and on the semilog plot these points lie on a straight line. Take two points exactly one cycle apart: 2 h (25.501 MPa) and 20 h (25.256 MPa).
The slope per cycle is $m = 25.501 - 25.256 = 0.245$ MPa.
The flow capacity: $qB\mu = 120\cdot1.25\cdot0.9 = 135$ , so $kh = 2.149\cdot135/0.245 = 1184$ mD·m. The permeability is $k = 1184/14 = 84.6$ mD.
The pressure on the line at 1 h is $p_{1\text{h}} = 25.501 + 0.245\cdot\lg 2 = 25.5748$ MPa. The measurement at 1 h (25.576 MPa) almost coincides with it, because by that time wellbore storage is practically over.
The terms of the skin formula are $\dfrac{p_i - p_{1\text{h}}}{m} = \dfrac{27.5 - 25.5748}{0.245} = 7.858$ and $\lg\dfrac{k}{\phi\mu c_tr_w^2} = 7.397$ .
The skin is $S = 1.151\,(7.858 - 7.397 + 2.098) = 2.95$ .
The skin pressure drop is $\Delta p_s = 0.869\cdot0.245\cdot2.95 = 0.63$ MPa. This is 28% of the pressure drop at 20 h (2.24 MPa).

Answer:

kh \approx 1184

mD·m,

k \approx 84.6

mD, and the skin is about 2.95. The near-wellbore zone accounts for 0.63 MPa.

What it means. The data in this example are assumed for illustration. They were calculated from a model with a permeability of 85 mD and a skin of 3.0. The analysis returned these values with an error of 0.5% in permeability and 0.05 in skin. This error comes from the wellbore storage that had not quite died out by 2 h, and from rounding the measurements to 0.001 MPa. A moderate positive skin takes more than a quarter of the pressure drop. A treatment of the near-wellbore zone would noticeably raise productivity, unless the lift equipment (the equipment that raises fluid to the surface) limits the rate (Skin factor).

What the late section gives

The end of the line is also a result. The time when the points leave the line shows how far away the boundary is. By that moment the radius of investigation is comparable with the distance to the boundary. This is only an estimate of the order of magnitude. We find the distance more accurately by fitting a model with a boundary (A sealing fault near the well: the method of images). If the well works long enough and the area is closed, the pressure begins to fall linearly in time. The rate of decline then gives the pore volume of the area, as in a reservoir limit test (rate of pressure decline).

Worked exampleThe start and end of the line, the distance to the boundary and the volume of the area

Given (assumed data)

Results of the previous example: $k$ = 84.6 mD; the line runs from 2 to 30 h
Reservoir: $\phi$ = 0.17; $\mu$ = 0.9 mPa·s; $c_t = 1.9\cdot10^{-3}$ MPa⁻¹; $h$ = 14 m; $r_w$ = 0.108 m
Late measurements: 180 h — 24.868 MPa; 240 h — 24.737 MPa

Solution

The start of the line. At 2 h the dimensionless time $t_D = 3.553\cdot10^{-3}\,kt/(\phi\mu c_tr_w^2) = 177{,}298$ is thousands of times larger than 25. The logarithmic formula itself is valid from the first seconds. Wellbore storage sets the start of the line.
The end of the line. The radius of investigation $r_i = 0.119\sqrt{kt/(\phi\mu c_t)}$ is 352 m at 30 h and 454 m at 50 h. The 30 h point is still on the line, and the 50 h point is already slightly below it. So the boundary is at a distance of about 350–450 m.
The late section. Between 180 and 240 h the pressure falls by $24.868 - 24.737 = 0.131$ MPa in 60 h, that is, by 0.0524 MPa/d.
The pore volume is $V_p = \dfrac{qB}{c_t\,|\dd p/\dd t|} = \dfrac{150}{1.9\cdot10^{-3}\cdot0.0524} = 1507$ thousand m³.
The radius of a circle with this volume is $r_e = \sqrt{V_p/(\pi h\phi)} = 449$ m. This agrees with the estimate from the radius of investigation.

Answer: Radial flow occupies the interval from 2 to 30 h. The boundary of the closed area is about 449 m from the well, and the pore volume of the area is about 1507 thousand m³.

What it means. In the model from which the data were calculated, the well stands at the centre of a closed circle of radius 450 m with a pore volume of 1514 thousand m³. Both values were recovered with an error of less than 1%. Note that the line ended before the pseudo-steady-state regime set in. The slope began to grow at $kt/(\phi\mu c_tA) \approx 0.05$ , and the linear decline was established after 0.1 (time to reach the regime). In a real record, the shape of this transition tells us about the shape of the area and the position of the well in it. The data in this example are assumed for illustration.

The same formulas in oilfield units

The reports of service companies and many interpretation programs use oilfield units: barrels per day, feet and psi (pounds per square inch). The formulas are the same, and only the constants change. Let us check this on the same record.

Worked examplePermeability and skin in oilfield units

Given (assumed data)

Unit conversions: 1 m³ = 6.2898 bbl; 1 m = 3.2808 ft; 1 MPa = 145.04 psi
Test data: the same as in the example with metric units
Line: $m$ = 0.245 MPa per cycle; $p_{1\text{h}}$ = 25.575 MPa; $p_i$ = 27.5 MPa

Solution

Convert the data: $q = 754.8$ STB/d, $h = 45.93$ ft, $r_w = 0.3543$ ft, $c_t = 1.310\cdot10^{-5}$ psi⁻¹.
Convert the pressures: $m = 35.53$ psi per cycle and $p_i - p_{1\text{h}} = 279.2$ psi.
The flow capacity is $kh = 162.6\cdot\dfrac{754.8\cdot1.25\cdot0.9}{35.53} = 3885$ mD·ft, and $k = 3885/45.93 = 84.6$ mD.
The terms of the skin formula are $\dfrac{p_i - p_{1\text{h}}}{m} = 7.858$ , which is the same number as in metric units, and $\lg\dfrac{k}{\phi\mu c_tr_w^2} = 8.527$ .
The skin is $S = 1.151\,(7.858 - 8.527 + 3.23) = 2.95$ .

Answer: In oilfield units we get the same

k \approx 84.6

mD and

S \approx 2.95

as in metric units (84.6 mD and 2.95).

What it means. Permeability is in millidarcies in both systems, and skin has no units, so the results must agree. They do, to within the rounding of the constants. The logarithm in the skin formula is different in the two systems (8.527 against 7.397), because it contains a quantity that has units. The other constant makes up for the difference in units: 3.23 instead of 2.098. The data in this example are assumed for illustration.

Constants of semilog analysis

Constant: In the slope: $m\,kh/(qB\mu)$
Metric, $\lg$: 2.149
Oilfield, $\lg$: 162.6
Metric, $\ln$: 0.933
Oilfield, $\ln$: 70.6

Constant: In the skin formula
Metric, $\lg$: 2.098
Oilfield, $\lg$: 3.23
Metric, $\ln$: 4.831
Oilfield, $\ln$: 7.432

Constant: Factor before the bracket in the skin formula
Metric, $\lg$: 1.151
Oilfield, $\lg$: 1.151
Metric, $\ln$: 0.5
Oilfield, $\ln$: 0.5

Constant	Metric, $\lg$	Oilfield, $\lg$	Metric, $\ln$	Oilfield, $\ln$
In the slope: $m\,kh/(qB\mu)$	2.149	162.6	0.933	70.6
In the skin formula	2.098	3.23	4.831	7.432
Factor before the bracket in the skin formula	1.151	1.151	0.5	0.5

Metric oilfield units: m³/d, mPa·s, mD, m, MPa, 1/MPa, hours. Oilfield units: STB/d, cP, mD, ft, psi, 1/psi, hours. The “ $\ln$ ” columns are for a plot of pressure against the natural logarithm of time: the slope is per unit of $\ln t$ , and the pressure is the value on the line at $t$ = 1 h.

When the method works poorly

The rate is not constant. After start-up the rate usually declines smoothly along with the bottomhole pressure, and the choke (a narrow opening that limits the flow) and the separator (the vessel that splits the oil, gas and water) add fluctuations. A change in rate of a few per cent shifts the pressure by an amount comparable with the slope of the line. If the rate changes smoothly, we plot the ratio $(p_i - p_{wf})/q$ instead of the pressure. For stepwise changes we use superposition (Superposition principle, Equivalent producing time and synthetic flowing time).
Skin changes. In its first hours and days a new well cleans itself of filtrate and particles of the drilling fluid. The skin decreases, and the pressure falls more slowly than the formula says, or even rises. A record with a changing skin does not fall on a straight line (Nonideal wellbore storage, well clean-up and changing skin).
The pressure before start-up is unknown or has not equalised. The slope of the line, and so the permeability, does not depend on $p_i$ , but without $p_i$ we cannot find the skin. In a reservoir that has produced for a long time, the pressure around the well before start-up is seldom uniform. Build-up analysis is then more reliable.
Small pressure changes. In a high-permeability reservoir the slope can be hundredths of a megapascal per cycle, at pressures of tens of megapascals. This needs high-resolution gauges and a stable well regime.
Bottomhole pressure below the bubble point pressure. Below this pressure (the pressure at which gas starts to come out of the oil), gas comes out of solution near the wellbore. The flow becomes two-phase, and the formulas for a single slightly compressible fluid stop being valid (assumptions of the equation).

Because of the first difficulty, we analyse producing wells more often on build-up, where the rate is zero. The drawdown test remains the main method where a shut-in is impossible or undesirable, at the first start-up of a well, and in long tests whose aim is to find the boundaries and the volume of the area.

Terms and symbols

Drawdown test: A record of bottomhole pressure after a well is started from rest at a constant rate.
Semilog plot: A plot of bottomhole pressure against the logarithm of time. In radial flow it is a straight line.
Slope $m$: The pressure drop on the line over one decimal cycle of time. It is inversely proportional to the flow capacity $kh$ .
Pressure $p_{1\text{h}}$: The pressure on the line, or on its continuation, 1 h after start-up. We need it to calculate the skin.
Infinite-acting radial flow: The middle section of the record, where wellbore storage is over and the disturbance has not reached the boundaries. The pressure derivative is horizontal.
Diagnostic plot: A plot of the pressure drop and its logarithmic derivative on log–log axes. We use it to choose the section for the straight line.
Late linear section: In a closed area, the bottomhole pressure falls at a constant rate $qB/(c_tV_p)$ . From this rate we find the pore volume.

Common mistakes and misconceptions

Wrong

“The line is drawn through all the points of the record.”

In fact

Only the radial-flow section obeys the formula: the part after wellbore storage has ended and before the boundaries make themselves felt. In the lesson's example these are the points from 2 to 30 h of a record that is 240 h long.

Wrong

“The effect of wellbore storage on a drawdown test looks the same as on a build-up: the early points are below the line.”

In fact

On a drawdown the early points lie above the line, because the pressure falls more slowly than the ideal. On a build-up they lie below the line, because the pressure rises more slowly than the ideal. In both cases the change in pressure lags behind the calculated one.

Wrong

“A large skin makes the line steeper.”

In fact

Skin shifts the line down by $0.869\,mS$ without changing the slope. The line becomes steeper at a lower flow capacity $kh$ , a higher viscosity or a higher rate.

Wrong

“The pressure began to fall faster — so the reservoir properties are worse away from the well.”

In fact

First check the simpler explanation: the disturbance has reached a sealing boundary. A sign of a closed area is a straight line on ordinary axes of pressure against time, together with a derivative that has a slope of 1.

Wrong

“ $p_{1\text{h}}$ is the pressure measured an hour after start-up.”

In fact

It is the value on the radial-flow line, or on its continuation, at $t$ = 1 h. If wellbore storage lasts longer than an hour, the measurement and $p_{1\text{h}}$ differ markedly, and putting the measurement into the formula distorts the skin.

In depth: advanced detail

Where the constants 2.149, 2.098, 162.6 and 3.23 come from

In the bottomhole pressure formula, we change from the natural logarithm to the base-10 logarithm: $\ln x = \ln 10\cdot\lg x$ . The factor in front of the bracket becomes $m = \ln 10\cdot qB\mu/(4\pi kh)$ , and the terms in the brackets are divided by $\ln 10 = 2.303$ :

Bottomhole pressure through the base-10 logarithm

p_i - p_{wf} = m\left[\lg t + \lg\frac{k}{\phi\mu c_t r_w^2} + \lg a + 0.351 + 0.869\,S\right],\qquad m = \frac{qB\mu\,\ln 10}{4\pi kh}

where:

$a$ a factor that converts $kt/(\phi\mu c_tr_w^2)$ into dimensionless time in the chosen units; in SI $a = 1$ , —
$0.351,\ 0.869$ $0.809/\ln 10$ and $2/\ln 10$ , —
$m$ pressure drop over one decimal cycle of time, pressure units

Units: any system of units; only the factor $a$ and the constant in $m$ depend on it

Holds when

The same as for the bottomhole pressure formula in the main section.

In metric oilfield units, $a = 3.553\cdot10^{-3}$ and $\lg a + 0.351 = -2.449 + 0.351 = -2.098$ . In oilfield units, $a = 2.637\cdot10^{-4}$ and $\lg a + 0.351 = -3.579 + 0.351 = -3.228 \approx -3.23$ . If we put $t$ = 1 h into the formula, we get an expression for $p_{1\text{h}}$ . If we solve it for $S$ , we get the skin formula with the factor $1/0.869 = 1.151$ .

We get the constant in the slope by converting the units in $\ln 10/(4\pi)$ . In metric oilfield units we divide the rate by 86,400 (days to seconds), multiply the viscosity by $10^{-3}$ and the permeability by $0.9869\cdot10^{-15}$ , and divide the pressure by $10^{6}$ . Together this is $1/0.0853$ , so $m = 2.303\,qB\mu/(4\pi\cdot0.0853\,kh) = 2.149\,qB\mu/(kh)$ . In oilfield units, $1.127\cdot10^{-3}$ stands in place of 0.0853, and the constant is 162.6.

Analysis with the natural logarithm

The natural logarithm is the natural choice for theory. It is also convenient when we fit the line by least squares in a spreadsheet, with $\ln t$ on the horizontal axis. The slope of the line is then $m_e = qB\mu/(4\pi kh)$ , and the working formulas in metric oilfield units become:

Flow capacity and skin from a plot of pressure against the natural logarithm of time

kh = 0.933\,\frac{qB\mu}{m_e},\qquad S = \frac12\left[\frac{p_i - p_{1\text{h}}}{m_e} - \ln\frac{k}{\phi\mu c_t r_w^2} + 4.831\right]

where:

$m_e$ pressure drop on the line when $\ln t$ increases by one, that is, when time grows by a factor of 2.718, MPa
$p_{1\text{h}}$ pressure on the line at $t$ = 1 h ( $\ln t = 0$ ), MPa
$p_i,\ q,\ B,\ \mu$ pressure before start-up, rate, formation volume factor, viscosity, MPa, m³/d, m³/m³, mPa·s
$k,\ h,\ \phi,\ c_t,\ r_w$ permeability, thickness, porosity, total compressibility, wellbore radius, mD, m, —, 1/MPa, m

Units: metric oilfield units; time in hours

Holds when

The same as for the formulas with the base-10 logarithm.

The constants are $0.933 = 1/(4\pi\cdot0.0853)$ and $4.831 = -\ln(3.553\cdot10^{-3}) - 0.809$ . In oilfield units they are 70.6 and 7.432. For the lesson's record, $m_e = 0.245/2.303 = 0.1064$ MPa, and both formulas give the same flow capacity of 1184 mD·m and the same skin of 2.95.

A least-squares line

Two points are convenient for a hand calculation, but each carries a measurement error. In practice we fit the line through all the points of the radial-flow section by least squares (the method that finds the line with the smallest total squared distance from the points), after first inspecting the plot and excluding outliers (points that are clearly wrong).

Worked exampleA line through seven points

Given (assumed data)

Points of the radial-flow section: 2, 3, 5, 8, 12, 20 and 30 h from the table in the main section
Coordinates: $x = \lg t$ , $y = p_{wf}$

Solution

The number of points is $n = 7$ . Linear regression $y = a - m\,x$ gives the slope $m = 0.2455$ MPa per cycle and the intercept $a = p_{1\text{h}} = 25.5749$ MPa.
The flow capacity is $kh = 2.149\cdot135/0.2455 = 1182$ mD·m, and $k = 84.4$ mD.
For the skin, $\dfrac{27.5 - 25.5749}{0.2455} = 7.842$ and $S = 1.151\,(7.842 - 7.396 + 2.098) = 2.93$ .

Answer: From seven points

k \approx 84.4

mD and

S \approx 2.93

. This is practically the same as from two points (84.6 mD and 2.95).

What it means. For these smooth calculated data, the regression improves nothing. The points at 2–5 h are still slightly above the true line, and the point at 30 h is already slightly below it. Both errors make the fitted line a little steeper. On a real record, with gauge noise and rate fluctuations, a regression over many points is more reliable than two arbitrarily chosen points. The data in this example are assumed for illustration.

Connections to other topics

Line-source solution and the pressure function — the solution from which the logarithmic dependence of pressure on time follows.
Pressure build-up analysis — the same analysis for a record taken after a well is shut in: the Horner and Miller–Dyes–Hutchinson (MDH) plots.
Skin factor — what the skin that was found means and how much rate it “costs”.
Well testing: purposes and types of test — types of tests, sections of the record and planning of the duration.
Wellbore storage — why the early points lie above the line and how to estimate the duration of this section.
Diagnosis with the pressure derivative — the diagnostic plot and the choice of the radial-flow section.
Reservoir limit test — pore volume and the shape of the area from the late section of a long record.
Pseudo-steady-state flow and the shape factor — the regime that replaces radial flow in a closed area.
Superposition principle — how to allow for a changing rate and the previous history of the well.

Check yourself

1
On the semilog plot of a drawdown test the first points lie clearly above the line drawn through the later ones. What is the most likely cause?
1. A.The permeability of the reservoir near the well is higher than farther away
2. B.The disturbance has reached a sealing boundary
3. C.Wellbore storage: the inflow from the reservoir has not yet become equal to the rate at the surface
4. D.Skin decreases with time
Show the answer
Correct answer: C. Wellbore storage: the inflow from the reservoir has not yet become equal to the rate at the surface
For the first minutes the well produces the fluid stored in the wellbore, and the reservoir is “under-loaded”. The bottomhole pressure is higher than calculated, so the points lie above the line. A boundary, on the contrary, shows itself at the end of the record. A heterogeneity of the reservoir changes the slope, and it does not lift only the early points.
2
The slope of the line is 0.30 MPa per decimal cycle. The rate is 45 m³/d, $B$ = 1.2, $\mu$ = 3.0 mPa·s, and the thickness of the reservoir is 10 m. Find the flow capacity and the permeability.
Show the answer
$qB\mu = 45\cdot1.2\cdot3.0 = 162$ ; $kh = 2.149\cdot162/0.30 \approx 1160$ mD·m; $k \approx 1160/10 = 116$ mD.
3
For the same well, $p_i$ = 20.0 MPa, the pressure on the line at 1 h is 17.6 MPa, $\phi$ = 0.20, $c_t = 1.5\cdot10^{-3}$ MPa⁻¹ and $r_w$ = 0.1 m. Find the skin factor.
Show the answer
The first fraction is $(20.0 - 17.6)/0.30 = 8.0$ . Under the logarithm, $\phi\mu c_tr_w^2 = 0.20\cdot3.0\cdot1.5\cdot10^{-3}\cdot0.01 = 9.0\cdot10^{-6}$ , and $\lg\dfrac{116}{9.0\cdot10^{-6}} = \lg(1.29\cdot10^{7}) = 7.11$ . The skin is $S = 1.151\,(8.0 - 7.11 + 2.098) \approx 3.4$ . The skin pressure drop is $0.869\cdot0.30\cdot3.4 \approx 0.9$ MPa.
4
The rate in the test was doubled. How will the slope of the line and the permeability and skin calculated from it change (single-phase flow, Darcy's law holds)?
1. A.The slope doubles; permeability and skin do not change
2. B.The slope does not change; permeability doubles
3. C.The slope doubles; the skin doubles
4. D.Nothing changes
Show the answer
Correct answer: A. The slope doubles; permeability and skin do not change
All pressure drops are proportional to the rate, so the slope doubles. In the flow capacity formula the doubled rate is divided by the doubled slope. In the skin formula the ratio $(p_i - p_{1\text{h}})/m$ stays the same. So the results are the same. If the skin grew with the rate, that would point to inertial losses near the wellbore.
5
After 60 h of production the points on the semilog plot go below the line, and on ordinary axes of pressure against time they fall on a straight line. What does this mean and what can be determined?
1. A.The gauge is drifting; the record must be rejected
2. B.The disturbance has reached the boundaries of a closed area; the pore volume is found from the rate of pressure decline
3. C.Fluid flows in across the boundary; the pressure at the boundary is found
4. D.The well is cleaning up; a new skin is found
Show the answer
Correct answer: B. The disturbance has reached the boundaries of a closed area; the pore volume is found from the rate of pressure decline
A pressure decline that is linear in time is a sign of the pseudo-steady-state regime in a closed area: $\dd p_{wf}/\dd t = -qB/(c_tV_p)$ . If we know the rate, the formation volume factor and the compressibility, we can find the pore volume. With support from outside, the pressure would instead have stabilised.
6
Why can permeability be found from a drawdown test without knowing the reservoir pressure before start-up, while skin cannot?
Show the answer
The slope of the line sets the permeability, that is, how much the pressure changes from one moment to another. A constant shift of the whole record does not affect it. The position of the line relative to the pressure before start-up sets the skin, because the formula contains the difference $p_i - p_{1\text{h}}$ . An error in $p_i$ passes entirely into this difference, and so into the skin.

Updated: 19 October 2026