Reservoir limit test

Quick refresher

In a closed area, after a transition period, the bottomhole pressure falls linearly: $p_{wf} = p_{int} - m^*t$ . The pressure decline rate $m^* = qB/(c_tV_p)$ depends only on the production rate, the compressibility and the pore volume. It does not depend on the permeability or the skin.
On the diagnostic plot (a log–log plot of the pressure change and of its derivative, the rate of change of the pressure with the logarithm of time), this stretch is a unit slope of the logarithmic derivative at the end of the record: a straight line of slope 1. The $\Delta p$ itself does not show a unit slope, because of the constant term.
Pore volume: $V_p = qB/(24\,c_tm^*)$ for $q$ in m³/d and $m^*$ in MPa/h. An error in $c_t$ passes into $V_p$ one to one.
The intercept $p_{int}$ of the line, together with the semilog line (the straight line of radial flow on a plot of pressure against the logarithm of time), gives the shape factor $C_A$ . But it is very sensitive: an error of one tenth of $m_e$ in $p_{int}$ changes $C_A$ by 10%.
The time to the line is set by the slowest process: the equalisation of the pressure along the long side of the area. This time grows as the square of the length of that side.
When the rate changes, we plot the rate-normalised pressure $(p_i - p_{wf})/q$ against the material balance time $Q/q$ . The slope of the line gives $B/(c_tV_p)$ .

Key relationships

m^* = \dfrac{qB}{c_tV_p}

pseudo-steady-state pressure decline rate; in metric units $m^*$ [MPa/h] $= qB/(24\,c_tV_p)$

p_{int} = p_i - \dfrac{qB\mu}{4\pi kh}\left(\ln\dfrac{4A}{\gamma C_Ar_w^2} + 2S\right)

intercept of the line on the pressure axis (SI, natural logarithm)

C_A = \dfrac{4\pi\,m_e}{m^*t_1}\,\exp\dfrac{p_{int} - p_1}{m_e}

shape factor from the two lines; $p_1$ is the pressure on the semilog line at $t_1$

\dfrac{p_i - p_{wf}}{q} = b + \dfrac{B}{c_tV_p}\,\dfrac{Q}{q}

rate-normalised pressure when the rate changes

Common pitfalls

Drawing the line through points of the transition period. The slope comes out steeper, and the volume too small.
Converting $V_p$ into hydrocarbons in place without checking that there is no inflow from outside. With edge water (water at the edge of the accumulation, which flows in as the pressure falls) there is no straight line, and the “volume” grows with time.
Using a total compressibility that leaves out the pore compressibility.
Dividing the slope in ordinary coordinates by the last rate, when the rate changed during the record.
Taking the shape factor from the intercept as exact, when the pressure error is comparable to a tenth of $m_e$ .

After this lesson you can

explain why in a closed area the bottomhole pressure in the end falls linearly with time, and link how fast it falls with the pore volume through the material balance
recognise the pseudo-steady-state stretch on a plot in ordinary (Cartesian) coordinates and on the diagnostic plot of the derivative
calculate the pore volume, the area and the hydrocarbons in place of the area from the slope of the line, and the shape factor from its intercept together with the results of the semilog analysis
estimate how long the test must last, and what we can say if the boundaries never showed
analyse a record with a changing rate, using the rate-normalised pressure and the material balance time
list what distorts the result: compressibility, water influx, neighbouring wells, gas, gauge drift

Why test a formation for its limits

An exploration well has found an accumulation, but seismic data leave questions. Does the formation continue behind the fault? Does it thin out after a few hundred metres? Do the neighbouring blocks communicate? We can measure the volume that the well “sees” directly. We put the well on production at a constant rate and wait until the pressure disturbance reaches all the boundaries. After that, the pressure in the whole area falls in the same way, as in a closed tank. From how fast the pressure falls we find the connected pore volume: the volume of pores from which the well really draws fluid.

Such a test is called a reservoir limit test. It is usually a part of an extended well test of an exploration or appraisal well. This is a test that lasts for weeks or months, while the well produces fluid. It answers questions that neither well logs nor core can answer: is the area closed, do the faults conduct fluid, and how much oil is in the connected volume.

A constant rate is the main and the hardest condition. If the rate changes, we use normalisation (the section below).
A downhole gauge with a high resolution and a small drift (a slow shift of its readings). In a day, the pressure in a large block may fall by only hundredths of a megapascal.
Time: from weeks to months, until the disturbance reaches the farthest boundary. The produced fluid must be taken away: flared (burnt), hauled away or sent to a gathering system. This often limits the duration.
The initial pressure, measured before start-up (with a formation tester, a tool that measures the pressure in the formation from inside the well, or from the first pressure build-up), and the properties of the formation and the fluid. Above all we need the total compressibility $c_t$ (how much the volume of the fluids and the pores changes when the pressure changes).

Stages of the record: from radial flow to depletion

A long record of the bottomhole pressure after start-up, a drawdown test, passes through several stages. First the pressure is set by wellbore storage (Wellbore storage), the well acting as a small tank. Then radial flow starts: flow towards the well from all sides. As in any test, it gives the flow capacity and the skin, a number that shows how much the zone around the well helps or hinders flow (Pressure drawdown analysis). When the disturbance reaches the nearer boundaries, the pressure derivative rises. In a long block between two long sides, there may be a stretch where it rises with a slope of about ½. This is linear flow along the block. At last the disturbance reaches the farthest boundary, and pseudo-steady-state flow starts. This is the stage in which the pressure falls at the same rate everywhere in the area (Pseudo-steady-state flow and the shape factor).

Stages of the record of a reservoir limit test

Stage: Early
What sets the pressure: wellbore storage
Sign on the plots: unit slope of $\Delta p$ and of the derivative
What we get: wellbore storage coefficient

Stage: Radial flow
What sets the pressure: the formation around the well
Sign on the plots: derivative plateau; a straight line on the semilog plot
What we get: $kh$ , skin, $p_{1\text{h}}$

Stage: Transition
What sets the pressure: the nearer, then the farther boundaries
Sign on the plots: the derivative rises; in a long block, with a slope of about ½
What we get: distances to the boundaries, the shape

Stage: Pseudo-steady-state
What sets the pressure: the whole volume of the area
Sign on the plots: unit slope of the derivative; a straight line in ordinary coordinates
What we get: pore volume, $C_A$

Stage	What sets the pressure	Sign on the plots	What we get
Early	wellbore storage	unit slope of $\Delta p$ and of the derivative	wellbore storage coefficient
Radial flow	the formation around the well	derivative plateau; a straight line on the semilog plot	$kh$ , skin, $p_{1\text{h}}$
Transition	the nearer, then the farther boundaries	the derivative rises; in a long block, with a slope of about ½	distances to the boundaries, the shape
Pseudo-steady-state	the whole volume of the area	unit slope of the derivative; a straight line in ordinary coordinates	pore volume, $C_A$

Bottomhole pressure in an extended well test of an exploration well (assumed data)

$t$ , h: 0.1
$p_{wf}$ , MPa: 28.900
$t$ , h: 20
$p_{wf}$ , MPa: 28.442

$t$ , h: 0.2
$p_{wf}$ , MPa: 28.844
$t$ , h: 50
$p_{wf}$ , MPa: 28.266

$t$ , h: 0.5
$p_{wf}$ , MPa: 28.770
$t$ , h: 100
$p_{wf}$ , MPa: 28.023

$t$ , h: 1
$p_{wf}$ , MPa: 28.714
$t$ , h: 200
$p_{wf}$ , MPa: 27.601

$t$ , h: 2
$p_{wf}$ , MPa: 28.658
$t$ , h: 300
$p_{wf}$ , MPa: 27.211

$t$ , h: 5
$p_{wf}$ , MPa: 28.584
$t$ , h: 400
$p_{wf}$ , MPa: 26.830

$t$ , h: 10
$p_{wf}$ , MPa: 28.522
$t$ , h: 500
$p_{wf}$ , MPa: 26.452

$t$ , h
$p_{wf}$ , MPa
$t$ , h: 600
$p_{wf}$ , MPa: 26.074

$t$ , h	$p_{wf}$ , MPa	$t$ , h	$p_{wf}$ , MPa
0.1	28.900	20	28.442
0.2	28.844	50	28.266
0.5	28.770	100	28.023
1	28.714	200	27.601
2	28.658	300	27.211
5	28.584	400	26.830
10	28.522	500	26.452
		600	26.074

The rate is 200 m³/d of oil from start-up to the end of the record; the initial pressure is 30.0 MPa. The data are calculated from a model of a homogeneous closed block without wellbore storage, and rounded to 0.001 MPa.

CalculatedDiagnostic plot of the record in the table: the pressure change (points are readings, the line is the model) and its logarithmic derivative. Both scales are logarithmic. The plateau of the derivative is radial flow, up to 3 h. Then the derivative rises, and after 300 h it runs along a line of slope 1 (dashed): pseudo-steady-state flow.

A straight line in ordinary coordinates, and the pore volume

When the disturbance has reached all the boundaries, the fluid that the well produces is replaced only by the expansion of the rocks and fluids in the whole area. By the material balance, the average pressure falls at a rate $qB/(c_tV_p)$ . In pseudo-steady-state flow the bottomhole pressure falls at the same rate (closed area). Now plot the pressure against time in ordinary coordinates, that is, with both axes linear and not logarithmic (Cartesian coordinates). The late points lie on a straight line:

The pseudo-steady-state stretch of a drawdown test

p_{wf}(t) = p_{int} - m^*t,\qquad m^* = \frac{qB}{c_tV_p} = \frac{qB}{c_t\,\phi hA}

where:

$p_{int}$ intercept of the line on the pressure axis, that is, its continuation to $t = 0$ , Pa
$m^*$ pressure decline rate on the line (positive), Pa/s
$V_p = \phi hA$ pore volume connected to the well, m³
$q,\ B$ rate at standard conditions and formation volume factor; $qB$ is the rate at reservoir conditions, m³/s, m³/m³
$c_t$ total compressibility of the rocks and fluids, 1/Pa

Units: any consistent units (SI or, for example, m³/d, 1/MPa, m³ and MPa/d)

Holds when

a closed area with no inflow through the boundaries
a constant rate since start-up
a single-phase slightly compressible fluid, and $c_t$ is constant
pseudo-steady-state flow has started (we check it on the derivative)

The slope does not depend on the permeability and the skin. So we find the pore volume even if $kh$ is poorly known. The permeability, the skin and the shape of the area set the position of the line, that is, its intercept $p_{int}$ (next section).

In other unit systems (2)

Metric oilfield units

V_p\,[\text{m}^3] = \frac{q\,[\text{m}^3/\text{d}]\;B}{24\;c_t\,[\text{MPa}^{-1}]\;m^*\,[\text{MPa/h}]}

$q$ in m³/d at standard conditions, $B$ in m³/m³, $c_t$ in 1/MPa, $m^*$ in MPa/h; $V_p$ in m³.

24 is the number of hours in a day.

Oilfield units

V_p\,[\text{bbl}] = \frac{q\,[\text{STB/d}]\;B}{24\;c_t\,[\text{psi}^{-1}]\;m^*\,[\text{psi/h}]}

$q$ in STB/d, $B$ in rb/STB, $c_t$ in 1/psi, $m^*$ in psi/h; $V_p$ in barrels (to get cubic feet, multiply by 5.615).

the same formula: the volume comes out in the same units in which the reservoir rate $qB$ is given.

Why do we recognise this stretch by the derivative and not by the pressure itself? The logarithmic derivative is $\Delta p' = t\,\dd\Delta p/\dd t = m^*t$ . On a log–log plot it is a line with slope 1 (late-time forms). The pressure drop itself is $\Delta p = (p_i - p_{int}) + m^*t$ . It contains a constant term, and it reaches unit slope only when the depletion $m^*t$ is much larger than the drawdown $p_i - p_{int}$ . On the other hand, on a plot in ordinary coordinates it is easy to “see” a line even in the transition period, because the curvature there is small. So we find the start of pseudo-steady-state flow from the derivative, and the slope and the intercept from the ordinary coordinates.

CalculatedThe same record in ordinary coordinates: the points are readings. The dashed line is drawn through the readings at 400 and 600 h and continued to

t = 0

, where the intercept

p_{int}

is read. The mark is the start of pseudo-steady-state flow (PSS) by the model, at about 350 h. Before it, the points lie above the line.

Worked exampleFlow capacity, skin and pore volume of the block

Given (assumed data)

Exploration well: extended well test of 600 h at a rate of 200 m³/d; $p_i$ = 30.0 MPa
$h$ ; $\phi$ ; $B$ ; $\mu$: 16 m; 0.22; 1.30 m³/m³; 0.8 mPa·s
$c_t$ ; $r_w$ ; initial water saturation $S_{wi}$: $1.7\cdot10^{-3}$ MPa⁻¹; 0.1 m; 0.25
Readings: the table above

Solution

Radial flow: the derivative is constant up to 3 h (figure). From the readings at 0.1 and 1 h, the slope is $m = 28.900 - 28.714 = 0.186$ MPa per cycle (a cycle is a tenfold range of time). Then $kh = 2.149\cdot\dfrac{200\cdot1.3\cdot0.8}{0.186} = 2403$ mD·m, and $k = 150.2$ mD.
Skin (Pressure drawdown analysis): $\lg\dfrac{k}{\phi\mu c_tr_w^2} = \lg\dfrac{150.2}{0.22\cdot0.8\cdot1.7\cdot10^{-3}\cdot0.01} = 7.701$ ; $\dfrac{p_i - p_{1\text{h}}}{m} = \dfrac{30.0 - 28.714}{0.186} = 6.914$ ; $S = 1.151\,(6.914 - 7.701 + 2.098) = 1.51$ .
The line in ordinary coordinates through the readings at 400 and 600 h: $m^* = \dfrac{26.830 - 26.074}{200} = 0.00378$ MPa/h, or 0.091 MPa/d.
Pore volume: $V_p = \dfrac{200\cdot1.3}{24\cdot1.7\cdot10^{-3}\cdot0.00378} = 1686$ thousand m³. The area is $V_p/(\phi h) = 47.9$ ha (hectares; 1 ha = 10,000 m²).
Oil in the connected volume: $V_p(1 - S_{wi})/B = 1686\cdot0.75/1.3 = 973$ thousand m³, about 827 thousand t at a density of 850 kg/m³.

Answer:

k \approx 150.2

mD, skin 1.51. The well drains a block with a pore volume of about 1686 thousand m³ (about 47.9 ha) and about 827 thousand t of oil.

What it means. The model that gave the data is a closed block of 1200 × 400 m (48 ha) with a pore volume of 1690 thousand m³. The analysis recovered it with an error of 0.2%. The exploration well stands in a small isolated block. For a development decision, this result matters more than any map, because the hydrocarbons in place come from the response of the formation itself. But it depends directly on $c_t$ (the last section). The data in this example are assumed for illustration.

The intercept and the shape factor

The position of the line carries information about the shape of the area. We continue the formula of pseudo-steady-state flow, written through the initial pressure (time to stabilisation), to $t = 0$ . This gives the intercept:

Intercept of the depletion line on the pressure axis

p_{int} = p_i - \frac{qB\mu}{4\pi kh}\left(\ln\frac{4A}{\gamma C_A r_w^2} + 2S\right)

where:

$p_{int}$ continuation of the line in ordinary coordinates to $t = 0$ , Pa
$C_A$ shape factor of the area with the given position of the well, —
$\gamma$ 1.781, the exponential of Euler's constant, —
$A,\ r_w,\ S$ area of the region, well radius, skin, m², m, —
$q,\ B,\ \mu,\ k,\ h$ as in the inflow formula, m³/s, m³/m³, Pa·s, m², m

Units: SI; $\ln$ is the natural logarithm

Holds when

the same as for the depletion line; a homogeneous formation, and the well penetrates the whole thickness.

The smaller $C_A$ is (the more elongated the area, or the closer the well to a boundary), the lower the line lies: the same production comes with a larger drawdown.

If $kh$ and the skin are known from radial flow, and the area from the slope, we can find $C_A$ from this equality. It is more convenient to eliminate the skin by taking the difference with the pressure $p_1$ on the semilog line at time $t_1$ (usually 1 h; the derivation is in the advanced sections):

Shape factor from two lines

C_A = \frac{4\pi\,m_e}{m^*\,t_1}\,\exp\frac{p_{int} - p_1}{m_e}

where:

$m_e$ slope of the semilog line per unit of natural logarithm, $m/2.303$ , MPa
$p_1$ pressure on the semilog line (or its continuation) at time $t_1$ , MPa
$m^*,\ p_{int}$ slope and intercept of the line in ordinary coordinates, MPa/h, MPa

Units: any consistent units: $m^*$ is in units of pressure per the unit of time in which $t_1$ is given

Holds when

both lines are drawn from one record with a constant rate
a homogeneous formation; radial flow and pseudo-steady-state flow are really visible in the record

The skin, the permeability and the area cancel. What remains is two slopes and the difference of two pressures. The catch is the exponential: an error $\delta$ in the difference $p_{int} - p_1$ changes $C_A$ by a factor of $\e^{\delta/m_e}$ . For $m_e$ = 0.08 MPa, an error of 0.01 MPa is enough to change $C_A$ by 13% (the table below). So we always check the result against the geology.

In other unit systems (1)

Base-10 logarithms

C_A = 5.458\,\frac{m}{m^*\,t_1}\cdot10^{(p_{int} - p_1)/m}

$m$ is the slope per decimal cycle; $m^*$ and $t_1$ are as above.

$5.458 = 4\pi/\ln 10$ .

How an error in the difference

p_{int} - p_1

changes

C_A

$\delta/m_e$: 0.05
$\delta$ at $m_e$ = 0.0808 MPa: 0.004 MPa
$C_A$ is multiplied by: 1.05
The shape pseudo-skin changes by: 0.025

$\delta/m_e$: 0.10
$\delta$ at $m_e$ = 0.0808 MPa: 0.008 MPa
$C_A$ is multiplied by: 1.11
The shape pseudo-skin changes by: 0.050

$\delta/m_e$: 0.20
$\delta$ at $m_e$ = 0.0808 MPa: 0.016 MPa
$C_A$ is multiplied by: 1.22
The shape pseudo-skin changes by: 0.100

$\delta/m_e$: 0.50
$\delta$ at $m_e$ = 0.0808 MPa: 0.040 MPa
$C_A$ is multiplied by: 1.65
The shape pseudo-skin changes by: 0.250

$\delta/m_e$	$\delta$ at $m_e$ = 0.0808 MPa	$C_A$ is multiplied by	The shape pseudo-skin changes by
0.05	0.004 MPa	1.05	0.025
0.10	0.008 MPa	1.11	0.050
0.20	0.016 MPa	1.22	0.100
0.50	0.040 MPa	1.65	0.250

The shape pseudo-skin $S_A = \tfrac12\ln(4\pi/(\gamma C_A)) + 3/4$ (shape factor), which writes the effect of the shape of the area as a skin, changes only by $\delta/(2m_e)$ . So an inexact $C_A$ has little effect on the forecast of productivity, but it is a poor guide to the shape.

Worked exampleShape factor from the intercept of the line

Given (assumed data)

Results of the previous example: $m$ = 0.186 MPa per cycle; $p_{1\text{h}}$ = 28.714 MPa; $m^*$ = 0.00378 MPa/h; $k$ = 150.2 mD; area 47.9 ha
Depletion line: through the readings at 400 h (26.830 MPa) and 600 h (26.074 MPa)

Solution

Intercept: $p_{int} = 26.074 + 0.00378\cdot600 = 28.342$ MPa.
Slope per unit of natural logarithm: $m_e = 0.186/2.303 = 0.0808$ MPa; $\dfrac{p_{int} - p_{1\text{h}}}{m_e} = \dfrac{28.342 - 28.714}{0.0808} = -4.604$ .
$C_A = \dfrac{4\pi\cdot0.0808}{0.00378\cdot1} \cdot \e^{-4.604} = 268.6\cdot0.01001 = 2.69$ .
Start of the line: the slope in ordinary coordinates between 200 and 300 h is 0.0039 MPa/h, and between 300 and 400 h it is 0.00381 MPa/h, which is still 0.8% more than later. Pseudo-steady-state flow was established at about 350 h. In dimensionless form, $t_{DA} = \dfrac{3.553\cdot10^{-3}\cdot150.2\cdot350}{0.22\cdot0.8\cdot1.7\cdot10^{-3}\cdot479{,}000} = 1.30$ .
Comparison with the table below: $C_A$ of about 2.5–2.7, with pseudo-steady-state flow reached at about $t_{DA}$ = 1.3, suits an elongated block of about 3:1 with the well nearer to one of the ends. A well at the centre of a 5:1 area also suits this $C_A$ , but it would reach pseudo-steady-state flow twice as early.

Answer:

C_A \approx 2.69

: the block is elongated, and the well is shifted towards its end.

What it means. In the model, $C_A$ = 2.54. The error of 6% came from the intercept being above the exact one by 0.005 MPa. This is only 0.06 $m_e$ (the readings are rounded to 0.001 MPa, and at 400 h the slope is still a little larger than the limit). The shape factor from the intercept is a hint, not a measurement. We check it against the map of faults, the time to reach pseudo-steady-state flow, and the shape of the transition stretch. The data in this example are assumed for illustration.

Shape factor and characteristic times of rectangular areas (calculated by the model)

Shape and position of the well: Square, well at the centre
$C_A$: 30.88
Pseudo-steady-state flow from $t_{DA}$: 0.15
Radial flow up to $t_{DA}$: 0.042

Shape and position of the well: Rectangle 2:1, well at the centre
$C_A$: 21.84
Pseudo-steady-state flow from $t_{DA}$: 0.27
Radial flow up to $t_{DA}$: 0.024

Shape and position of the well: Rectangle 2:1, well on the axis at 1/4 of the length from the short side
$C_A$: 4.52
Pseudo-steady-state flow from $t_{DA}$: 0.93
Radial flow up to $t_{DA}$: 0.022

Shape and position of the well: Rectangle 2:1, well on the axis at 1/8 of the length from the short side
$C_A$: 0.58
Pseudo-steady-state flow from $t_{DA}$: 1.04
Radial flow up to $t_{DA}$: 0.007

Shape and position of the well: Rectangle 3:1, well on the axis at 0.3 of the length from the short side
$C_A$: 2.54
Pseudo-steady-state flow from $t_{DA}$: 1.29
Radial flow up to $t_{DA}$: 0.016

Shape and position of the well: Rectangle 4:1, well at the centre
$C_A$: 5.38
Pseudo-steady-state flow from $t_{DA}$: 0.54
Radial flow up to $t_{DA}$: 0.012

Shape and position of the well: Rectangle 5:1, well at the centre
$C_A$: 2.36
Pseudo-steady-state flow from $t_{DA}$: 0.67
Radial flow up to $t_{DA}$: 0.009

Shape and position of the well	$C_A$	Pseudo-steady-state flow from $t_{DA}$	Radial flow up to $t_{DA}$
Square, well at the centre	30.88	0.15	0.042
Rectangle 2:1, well at the centre	21.84	0.27	0.024
Rectangle 2:1, well on the axis at 1/4 of the length from the short side	4.52	0.93	0.022
Rectangle 2:1, well on the axis at 1/8 of the length from the short side	0.58	1.04	0.007
Rectangle 3:1, well on the axis at 0.3 of the length from the short side	2.54	1.29	0.016
Rectangle 4:1, well at the centre	5.38	0.54	0.012
Rectangle 5:1, well at the centre	2.36	0.67	0.009

The well is a line source in a homogeneous closed rectangle. The times follow the criteria of this lesson. Pseudo-steady-state flow starts when the slope in ordinary coordinates differs from the limit by less than 1%. Radial flow lasts while the derivative differs from the plateau by less than 1%. The classical tabulated $C_A$ agree with these within 0.4%. The times in the reference tables (shape factor) are defined by the error of the pressure $p_D$ itself and with other tolerances. So they differ from these, larger for some shapes and smaller for others.

The transition period: why we must wait so long

Between radial flow and the depletion line lies the transition period. The disturbance has reached the nearer boundaries, but the pressure along the area has not yet equalised. In a rectangular area, the slope in ordinary coordinates approaches the limit as a sum of decaying exponentials. Each of them describes the equalisation of the pressure in its own “mode” (a pattern of unevenness in the pressure, like one tone of a vibrating string): along the length, along the width, and their combinations. The slowest to decay is the equalisation along the long side:

How the slope in ordinary coordinates approaches its limit

\frac{|\dd p_{wf}/\dd t|}{m^*} - 1 \approx 2\cos^2\frac{\pi x_w}{L}\;\exp\left(-\frac{\pi^2\eta\,t}{L^2}\right)

where:

$L$ length of the area (the long side), m
$x_w$ distance from the well to the nearer short side, m
$\eta = k/(\phi\mu c_t)$ hydraulic diffusivity of the formation (how fast a pressure change spreads through the rock), m²/s
$m^*$ limiting rate of the pressure decline, $qB/(c_tV_p)$ , Pa/s

Units: SI; in metric units $\eta = 3.553\cdot10^{-3}\,k/(\phi\mu c_t)$ m²/h for $t$ in hours

Holds when

a closed rectangular area, a homogeneous formation
a late transition period: the faster “modes” have already decayed
the well is not at the middle of the length: for a well at the middle of the length, on the long axis, this mode does not appear, and the equalisation goes four times faster

The time to the line grows as the square of the length of the area, and it hardly depends on the width. A block that is twice as long reaches pseudo-steady-state flow four times later, although its area is only twice as large. We set the right-hand side equal to 0.01 (the slope differs from the limit by 1%). This gives an estimate of the start of the line: the formula below.

Estimate of the time to reach pseudo-steady-state flow

t_{pss} \approx \frac{L^2}{\pi^2\eta}\,\ln\left(200\cos^2\frac{\pi x_w}{L}\right)

where:

$t_{pss}$ time from which the slope in ordinary coordinates differs from the limit by less than 1%, h
$L,\ x_w,\ \eta$ as in the previous formula, m, m, m²/h

Units: metric: $L$ and $x_w$ in m, $\eta$ in m²/h, $t_{pss}$ in hours

Holds when

a closed rectangular area; the well is clearly shifted from the middle of the length ( $\cos^2(\pi x_w/L)$ is not small)
for a well at the middle of the length, we use the table of $t_{DA}$ of the previous section.

For the block of the examples ( $L$ = 1200 m, $x_w$ = 360 m, $\eta$ = 1781 m²/h) we get $81.9\cdot\ln 69.1 \approx 347$ h. The exact calculation of the model gives the same. The factor $L^2/(\pi^2\eta)$ is the time in which the unevenness of the pressure along the length of the area falls by a factor of $\e$ .

CalculatedCalculated deviation of the slope in ordinary coordinates from the limit for several rectangular areas (the model of the lesson); both scales are logarithmic. The dotted line is 1%: from this moment we can draw the line. The dashed line next to the curve of the block of the examples is its slowest “mode” by the formula above.

On the diagnostic plot, the same transition shows as a rise of the derivative from the plateau to a line of slope 1. In a long block, between them there may be a stretch where the derivative grows with a slope of about ½. The flow goes along the block, as in a channel between two parallel faults (Several boundaries: parallel faults and wedge-shaped blocks). From it we estimate the width of the block, and from the moment when the derivative leaves the plateau, the distance to the nearest boundary (A sealing fault near the well: the method of images). The shape of the transition is the main hint about the geometry. We compare the record with models, and not with one number $C_A$ .

How long the test must last

We plan the duration in advance from the expected size of the block. The time to reach pseudo-steady-state flow comes from the table of $t_{DA}$ (the formula for hours) or from the estimate through the length of the area. We add about the same time again to measure the slope reliably. Then we decide whether logistics and the budget can stand it: where to put the produced fluid, how much it costs to keep the drilling rig or the equipment waiting, and when the answer is needed.

If during the whole test the boundaries never showed, this is also a result. The connected pore volume is not smaller than the pore volume inside a circle whose radius equals the radius of investigation (how far the disturbance has travelled) at the end of the record:

Lower estimate of the connected pore volume

V_p \gtrsim \pi r_i^2 h\phi,\qquad r_i = 0.119\sqrt{\frac{k\,t}{\phi\mu c_t}}

where:

$r_i$ radius of investigation at the end of the record, m
$t$ duration of production at a constant rate, h
$k,\ \phi,\ \mu,\ c_t,\ h$ permeability, porosity, viscosity, total compressibility, thickness, mD, —, mPa·s, 1/MPa, m

Units: metric oilfield units: $k$ in mD, $t$ in h, $\mu$ in mPa·s, $c_t$ in 1/MPa; $r_i$ in m

Holds when

there are no signs of boundaries in the whole record: the derivative is on the plateau
an estimate of the order of magnitude: different definitions of the radius of investigation differ by tens of per cent

This is a lower limit, not an estimate of the volume: the accumulation may be much larger. The constant $0.119 = 2\sqrt{3.553\cdot10^{-3}}$ is the same as in the lesson Pressure drawdown analysis.

Worked exampleHow long to wait if the block is longer

Given (assumed data)

Formation and well: as in the examples above: $k$ = 150 mD, $\phi\mu c_t = 2.992\cdot10^{-4}$ MPa⁻¹·mPa·s, $h$ = 16 m, $\phi$ = 0.22
Hypothesis from seismic: a block twice as long, 2400 × 400 m; the well on the axis, 720 m from the short side
Second question: what we can say if no boundary showed in 600 h

Solution

Hydraulic diffusivity: $\eta = 3.553\cdot10^{-3}\cdot150/(2.992\cdot10^{-4}) = 1781$ m²/h.
The slowest mode: $L^2/(\pi^2\eta) = 2400^2/(9.87\cdot1781) = 328$ h; $\cos^2(\pi\cdot720/2400) = 0.345$ , $\ln(200\cdot0.345) = 4.23$ .
Start of the line: $t_{pss} \approx 328\cdot4.23 = 1387$ h, or 58 days. We need about the same time again to measure the slope reliably. In all, that is about 116 days against 29 days for the block of 1200 m.
If the boundaries are not visible for 600 h: $r_i = 0.119\sqrt{150\cdot600/(2.992\cdot10^{-4})} = 2064$ m. The connected pore volume is not smaller than $\pi\cdot2064^2\cdot16\cdot0.22 = 47$ million m³.

Answer: For a block of 2400 × 400 m, the depletion line will appear after about 58 days, and the test will stretch over almost four months. A record of 25 days with no signs of boundaries would mean a connected pore volume larger than 47 million m³.

What it means. Doubling the length at the same width made the time four times longer, and the area only twice as large: the length decides. So long blocks are more often studied not by one extended well test, but by a combination of build-ups, interference tests between neighbouring wells and the production history. The data in this example are assumed for illustration.

If the rate changed: rate-normalised pressure

It is hard to keep the rate constant for weeks. The operator changes the choke (the valve that sets the flow rate), the water cut grows (the share of water in the produced fluid), or the separator limits the flow. If the rate was reduced during the record, the slope in ordinary coordinates misleads. After the reduction, the bottomhole pressure for a time rises, or falls more slowly, and a line through the points of different periods comes out flatter than the true one. For pseudo-steady-state flow there is a simple way out. We assume that at each new rate pseudo-steady-state flow is established at once (the quasi-pseudo-steady-state approximation):

Rate-normalised pressure and material balance time

\frac{p_i - p_{wf}(t)}{q(t)} = b + \frac{B}{c_tV_p}\,t_{mb},\qquad t_{mb} = \frac{Q(t)}{q(t)}

where:

$q(t)$ rate at the moment of the reading, at standard conditions, m³/d
$Q(t)$ cumulative production up to the moment of the reading, m³
$t_{mb}$ material balance time, d
$b$ drawdown per unit rate: $0.933\,\frac{B\mu}{kh}\left(\ln\frac{4A}{\gamma C_Ar_w^2} + 2S\right)$ for $k$ in mD, $\mu$ in mPa·s, $h$ in m; $0.933 = 1/(4\pi\cdot0.0853)$ is the constant of the slope per unit of natural logarithm (where the constants come from), MPa/(m³/d)
$B,\ c_t,\ V_p$ formation volume factor, total compressibility, pore volume, m³/m³, 1/MPa, m³

Units: metric: for $q$ in m³/d, $Q$ in m³ and $c_t$ in 1/MPa, the time $t_{mb}$ is in days, and the slope is in MPa/(m³/d) per day

Holds when

a closed area, pseudo-steady-state flow
the rate changes slowly compared with the time of pressure equalisation; we do not use the points right after a sharp change of the rate (tens of hours)
a single-phase fluid, and $B$ and $c_t$ are constant

The first term is the drawdown at the current rate. The second is the depletion by the material balance: $p_i - \bar p = BQ/(c_tV_p)$ . In the coordinates “rate-normalised pressure against material balance time”, the points of all the periods lie on one line with the slope $B/(c_tV_p)$ . And the intercept $b$ , together with $kh$ and the skin, again gives $C_A$ . The same method is the basis of the modern analysis of production from daily readings of rate and pressure.

CalculatedThe record of the same well when the rate was cut twice (200 → 180 → 160 m³/d): the rate-normalised pressure against the material balance time. The points of the three periods lie on one line (dashed, through the readings at 300 and 600 h) if we drop the first tens of hours after a rate change. The first two readings (100 and 140 h) lie a little below it: the transition period has not ended yet.

Worked exampleA variable rate: the volume from the rate-normalised pressure

Given (assumed data)

The same well, the rate was reduced: 200 m³/d up to 150 h, then 180 m³/d up to 350 h, then 160 m³/d
Reading at 300 h: 27.430 MPa; produced 2375 m³
Reading at 600 h: 26.670 MPa; produced 4416.7 m³
$B$ ; $c_t$ ; $p_i$: 1.3; $1.7\cdot10^{-3}$ MPa⁻¹; 30.0 MPa

Solution

At 300 h: the rate-normalised pressure is $(30.0 - 27.430)/180 = 0.0142778$ MPa/(m³/d), and $t_{mb} = 2375/180 = 13.194$ days.
At 600 h: $(30.0 - 26.670)/160 = 0.0208125$ MPa/(m³/d), and $t_{mb} = 4416.7/160 = 27.604$ days.
Slope: $\dfrac{0.0208125 - 0.0142778}{27.604 - 13.194} = 0.0004535$ MPa/(m³/d) per day.
Pore volume: $V_p = \dfrac{1.3}{1.7\cdot10^{-3}\cdot0.0004535} = 1686$ thousand m³.
For comparison, the slope in ordinary coordinates between the same readings: $(27.430 - 26.670)/300 = 0.002533$ MPa/h. With the last rate, 160 m³/d, it gives $V_p = \dfrac{160\cdot1.3}{24\cdot1.7\cdot10^{-3}\cdot0.002533} = 2013$ thousand m³.

Answer: The rate-normalised pressure gives 1686 thousand m³ (in the model it is 1690 thousand m³). The slope in ordinary coordinates gives 2013 thousand m³, which is 19% more.

What it means. Between 300 and 600 h, the rate was cut, the bottomhole pressure for a time almost stopped falling, and the average slope came out smaller than the true depletion rate. That is why the volume is too large. Normalisation allows for the rate change both through the drawdown and through the cumulative production. The points in the first tens of hours after a rate change move away from the line: at 400 h, 50 h after the second cut, the rate-normalised pressure is still 0.4% above the line. The data in this example are assumed for illustration.

What distorts the result

The most vulnerable point is the compressibility. $V_p$ is inversely proportional to $c_t$ , and $c_t$ is the sum of the compressibilities of oil, water and pores: $c_t = S_oc_o + S_wc_w + c_f$ . The pore compressibility from core is known, at best, to within tens of per cent (Rock compressibility and the effect of stress on core properties). For example, take $S_o$ = 0.75, $c_o = 1.6\cdot10^{-3}$ and $c_w = 0.45\cdot10^{-3}$ MPa⁻¹. If we replace $c_f = 0.4\cdot10^{-3}$ by $0.8\cdot10^{-3}$ MPa⁻¹, $c_t$ rises from $1.71\cdot10^{-3}$ to $2.11\cdot10^{-3}$ MPa⁻¹, and the pore volume from the same record becomes 19% smaller.

Signs of a non-ideal record

Sign: The slope in ordinary coordinates decreases with time; the derivative grows more slowly than with slope 1
Possible cause: support from an aquifer (a water-bearing formation) or a gas cap (free gas above the oil); a neighbouring block behind a conducting fault
What to do: do not calculate the volume from one pair of points; use a model with inflow or with two blocks (Aquifer models: water influx, Partially communicating faults)

Sign: The slope changed sharply
Possible cause: a neighbouring well started or stopped; the rate changed
What to do: check the history of the neighbours and the rates; superposition (Superposition principle)

Sign: The volume changes noticeably with the choice of the pair of points
Possible cause: the transition period has not ended; gauge drift; fluctuations of the rate
What to do: check the derivative; normalise by the rate; check the gauge

Sign: The pressure fell below the bubble point pressure (the pressure at which gas starts to come out of the oil)
Possible cause: there is free gas in the formation: the compressibility has risen sharply
What to do: on such a stretch the single-phase formula overstates the volume; we limit the analysis to the stretch above the bubble point pressure

Sign	Possible cause	What to do
The slope in ordinary coordinates decreases with time; the derivative grows more slowly than with slope 1	support from an aquifer (a water-bearing formation) or a gas cap (free gas above the oil); a neighbouring block behind a conducting fault	do not calculate the volume from one pair of points; use a model with inflow or with two blocks (Aquifer models: water influx, Partially communicating faults)
The slope changed sharply	a neighbouring well started or stopped; the rate changed	check the history of the neighbours and the rates; superposition (Superposition principle)
The volume changes noticeably with the choice of the pair of points	the transition period has not ended; gauge drift; fluctuations of the rate	check the derivative; normalise by the rate; check the gauge
The pressure fell below the bubble point pressure (the pressure at which gas starts to come out of the oil)	there is free gas in the formation: the compressibility has risen sharply	on such a stretch the single-phase formula overstates the volume; we limit the analysis to the stretch above the bubble point pressure

A gas well. The compressibility of gas is close to $1/p$ and changes noticeably during the test. So there is no straight line on a plot of pressure against time. We work in pseudo-pressures and pseudo-time (quantities that allow for how the gas properties change with pressure), or directly with the material balance $p/Z$ (pressure divided by the gas deviation factor) against cumulative production (Material balance for gas reservoirs).
Gauge drift and temperature. Over weeks, the drift of a sensor can give a “pressure decline” comparable to the depletion of a large block. For such tests we use downhole quartz gauges, and check them before they go into the well and after they come out.
Intermediate build-ups. If the rate cannot be held, we interrupt the production with short build-ups, and plot average pressures against cumulative production (repeated tests).
The volume is pore space, not hydrocarbons in place. To go to the hydrocarbons in place, we need the saturation and the formation volume factor. To put the area on a map, we need the thickness and the porosity over the area (Volumetric estimation of hydrocarbons in place).

Terms and symbols

Reservoir limit test: A long production of a well at a constant rate, until pseudo-steady-state flow starts. From how fast the pressure falls, we find the connected pore volume.
Connected pore volume $V_p$: The volume of pores from which the well draws fluid. It may be smaller than the volume of the accumulation on the map.
Depletion rate $m^*$: The speed at which the pressure falls along the line in ordinary coordinates: $qB/(c_tV_p)$ .
Intercept $p_{int}$: The continuation of the depletion line to $t = 0$ . Together with $kh$ and the skin it gives the shape factor.
Material balance time $t_{mb}$: The ratio of the cumulative production to the current rate, $Q/q$ . It lets us analyse a record with a changing rate.
Rate-normalised pressure: $(p_i - p_{wf})/q$ , the change of pressure per unit of the current rate.
Extended well test: A long production of an exploration or appraisal well, with the fluid taken out, to study the formation.

Common mistakes and misconceptions

Wrong

“The points in ordinary coordinates lay on a straight line, so pseudo-steady-state flow is established.”

In fact

In the transition period the curvature is small, and a “line” is seen too early. Its slope is steeper, and the volume smaller. We check the start of pseudo-steady-state flow on the derivative: slope 1 at the end of the record.

Wrong

“The pore volume that we found is the volume of the whole accumulation.”

In fact

It is the volume connected to the well during the test. Behind a sealing fault there may be more oil, and with water support the “volume” is not constant.

Wrong

“The shape factor from the intercept gives the shape of the area unambiguously.”

In fact

An error of a tenth of $m_e$ in the pressure difference changes $C_A$ by 10%, and different shapes give close values of $C_A$ . We find the shape together with the map, the time to reach pseudo-steady-state flow and the transition stretch.

Wrong

“The larger the area of the block, the longer the test.”

In fact

The length decides: the time to reach pseudo-steady-state flow grows as the square of the long side. A block twice as long with the same width needs four times as much time.

Wrong

“If the rate changed, it is enough to take the last rate and the slope in ordinary coordinates.”

In fact

A slope through the points of different periods is distorted by the change of the drawdown. In the example of the lesson, such a calculation overstated the volume by 19%. The rate-normalised pressure against the material balance time gave an error of less than 1%.

In depth: advanced detail

Derivation of the depletion line and of the shape factor formula

In pseudo-steady-state flow, the difference between the average and the bottomhole pressure is constant: $\bar p - p_{wf} = \frac{qB\mu}{2\pi kh}\big(\tfrac12\ln\frac{4A}{\gamma C_Ar_w^2} + S\big)$ (shape factor). The average pressure from the material balance is $\bar p = p_i - qBt/(c_tV_p)$ . Adding the two, we get $p_{wf} = p_{int} - m^*t$ with $m^* = qB/(c_tV_p)$ and the intercept from the main section.

On the semilog line of the same record, $p_1 = p_i - m_e\big(\ln\frac{kt_1}{\phi\mu c_tr_w^2} + 0.809 + 2S\big)$ , where $m_e = qB\mu/(4\pi kh)$ . We subtract: $p_{int} - p_1 = m_e\big(\ln\frac{kt_1}{\phi\mu c_tr_w^2} + \ln\frac4\gamma - \ln\frac{4A}{\gamma C_Ar_w^2}\big) = m_e\ln\big(C_At_{1DA}\big)$ . The skin and the well radius cancel. The dimensionless time $t_{1DA} = kt_1/(\phi\mu c_tA)$ is expressed through the slopes: $m^*/(4\pi m_e) = k/(\phi\mu c_tA)$ , so $t_{1DA} = m^*t_1/(4\pi m_e)$ . From this, $C_A = \frac{4\pi m_e}{m^*t_1}\exp\frac{p_{int} - p_1}{m_e}$ .

This formula is also found with a constant of 5.456 or 5.458 before the ratio of the slopes. It is right only if $m$ is the slope per decimal cycle, and the exponent is written through the decimal logarithm ( $10^{(p_{int} - p_1)/m}$ or $\e^{2.303(p_{int} - p_1)/m}$ ). A mixture, the slope per natural logarithm with the constant for the decimal one, makes $C_A$ 2.3 times too small.

Exact solutions for a closed area

For a well in the centre of a closed circle, the diffusivity equation can be solved exactly. In the Laplace domain (after the Laplace transform of the time variable), the solution is written through modified Bessel functions. In time it is a series of decaying exponentials, whose exponents are the roots of an equation with Bessel functions of the first and second kind (boundaries of the formation). While the disturbance has not reached the boundary, the sum coincides with the solution for an infinite formation. Soon after that the series decays, and the pseudo-steady-state formula remains. We need the exact solution only in the short transition period. For a circle it is short, and for elongated areas it is long.

For a rectangle, the solution is built from two “strips” with sealing edges (this is how the curves of the lesson are calculated). The ratio of the slope in ordinary coordinates to the limit equals $\Pi_x\Pi_y$ , where $\Pi_x = 1 + 2\sum_{n\ge1}\e^{-n^2\pi^2\eta t/L^2}\cos^2(n\pi x_w/L)$ , and $\Pi_y$ is the same for the width. Each term is its own “mode” of equalisation. The slowest is the first term of $\Pi_x$ , and this gives the formula of the section on the transition period. If the well is at the middle of the length, $\cos(\pi/2) = 0$ , and in $\Pi_x$ the first remaining term is the one with $n = 2$ , which decays four times faster. Only the mode along the width can be slower than it, if the well is shifted from the long axis.

The average pressure over the area in such a model equals the material balance at any moment, $2\pi t_{DA}$ in dimensionless form. This is a good check of any numerical calculation. For an area of any shape, several wells and a heterogeneous formation, we solve the same problem numerically, and fit the record with a model (Well testing: purposes and types of test).

Connections to other topics

Pseudo-steady-state flow and the shape factor — pseudo-steady-state flow, the shape factor and the time to reach it.
Pressure drawdown analysis — the semilog analysis of the same record: $kh$ , skin, $p_{1\text{h}}$ .
Diagnosis with the pressure derivative — the unit slope of the derivative as a sign of depletion.
Build-up in a bounded reservoir and average reservoir pressure — the average pressure from a build-up, and the material balance from repeated tests.
Well testing: purposes and types of test — the kinds of tests and the place of an extended well test among them.
Several boundaries: parallel faults and wedge-shaped blocks — linear flow in a long block, and faults that cross.
A sealing fault near the well: the method of images — the distance to the nearest boundary from the derivative step.
Material balance for gas reservoirs — a gas accumulation: the $p/Z$ line instead of the pressure line.
Rock compressibility and the effect of stress on core properties — pore compressibility, the main source of error of the volume.
Volumetric estimation of hydrocarbons in place — how the connected volume is compared with the volumetric estimate of hydrocarbons in place.

Check yourself

1
How does pseudo-steady-state flow of a closed area look on the diagnostic plot of a long drawdown test?
1. A.A plateau of the derivative
2. B.A derivative with unit slope at the end of the record
3. C.A fall of the derivative to zero
4. D.A derivative with slope ½
Show the answer
Correct answer: B. A derivative with unit slope at the end of the record
In pseudo-steady-state flow, $\Delta p' = t\,\dd\Delta p/\dd t = m^*t$ , a line with slope 1. A plateau is radial flow. Slope ½ is linear flow (for example, along a long block). A fall of the derivative in a drawdown test is a constant-pressure boundary. The $\Delta p$ itself does not show unit slope, because of the constant term.
2
The late points of a drawdown test lie on a line with a slope of 0.0024 MPa/h. The rate is 120 m³/d, $B$ = 1.2, $c_t = 1.5\cdot10^{-3}$ MPa⁻¹. Find the pore volume.
Show the answer
$V_p = qB/(24\,c_tm^*) = 120\cdot1.2/(24\cdot1.5\cdot10^{-3}\cdot0.0024) = 144/(8.64\cdot10^{-5}) \approx 1.67$ million m³. Before this, we check on the derivative that pseudo-steady-state flow has really started.
3
The total compressibility was taken 20% larger than the true one. How does this affect the pore volume that we find?
1. A.The volume is understated by about 17%
2. B.It has no effect: the slope is measured
3. C.The volume is understated by 40%
4. D.The volume is overstated by 20%
Show the answer
Correct answer: A. The volume is understated by about 17%
$V_p = qB/(c_tm^*)$ is inversely proportional to $c_t$ . With $1.2c_t$ we get $V_p/1.2 \approx 0.83V_p$ , that is, 17% less. That is why we measure the pore compressibility on core, and carry its uncertainty into the range of the volume.
4
The slope of the semilog line is $m_e$ = 0.05 MPa per unit of natural logarithm, and the intercept of the depletion line $p_{int}$ is found with an error of 0.01 MPa. How wrong can $C_A$ be?
Show the answer
$C_A$ is proportional to $\exp[(p_{int} - p_1)/m_e]$ . So an error of 0.01 MPa changes it by a factor of $\e^{0.01/0.05} = \e^{0.2} \approx 1.22$ , that is, by 22%. The shape pseudo-skin changes only by 0.1. So for a forecast of productivity this accuracy is acceptable, and for a choice between shapes it is not.
5
Edge water flows towards the block. What will happen to the slope in ordinary coordinates at the end of a long drawdown test?
1. A.It will become negative right after start-up
2. B.It will stay constant, as in a closed area
3. C.It will decrease: water partly replaces the production
4. D.It will grow
Show the answer
Correct answer: C. It will decrease: water partly replaces the production
The water inflow partly replaces the production, the pressure falls more and more slowly, and there is no line of constant slope. The volume calculated from a late pair of points will be overstated, and it will grow with the length of the record. This is a sign of support from outside, not of a large block.
6
A well stands on the axis of a long block 1800 m long, 450 m from the short side; $\eta$ = 1000 m²/h. Estimate after how many hours the slope in ordinary coordinates comes within 1% of its limit.
Show the answer
$t_{pss} \approx \frac{L^2}{\pi^2\eta}\ln\big(200\cos^2\frac{\pi x_w}{L}\big) = \frac{1800^2}{9.87\cdot1000}\ln(200\cos^2 45^\circ) = 328\cdot\ln 100 \approx 328\cdot4.61 \approx 1510$ h, about 63 days. To measure the slope reliably, we need a record about as long again.

Updated: 7 October 2026