Supercharging and permeability from formation-tester pretests

Quick refresher

A well with overbalance (mud pressure above formation pressure) works as a slow injection well (a well that pushes fluid into the formation). The filtrate flows through the mudcake at a rate that is set mostly by the cake. The formation must carry this flow away, and the more resistance it has, the higher the pressure at the wall rises.
The excess is proportional to the filtration rate and to the viscosity of the filtrate, and inversely proportional to the permeability, with a small correction that depends on the logarithm of time. In a layer ten times less permeable, it is almost ten times larger. The overbalance sets its upper limit.
A supercharged point looks valid: the pressure has stabilised. Three signs give it away. One is a low mobility. Another is a shift to the right of the line, which grows as the mobility falls (in proportion to the inverse mobility). The third is an increased scatter of the points.
The drawdown mobility reflects the few centimetres of rock at the probe. In a layered rock it lies between the spherical and the horizontal permeabilities. A build-up under spherical flow gives the spherical permeability $(k_h^2k_v)^{1/3}$ (a mix of the horizontal and vertical permeabilities), over tens of centimetres to metres of rock.
Tool storage (the compressible liquid in the tool's line) distorts the build-up long after the record looks stable. A slope taken from points that are too early gives a permeability that is several times too low.
An observation probe above the sink probe (the probe that draws the fluid) gives $k_h$ from the level of its response, $k_s$ from the slope of the response in time, and $k_v$ from the two together. A dual-packer module (two packers that seal off a stretch of the wall) gives radial flow to the interval (flow that converges on it from the sides).

Key relationships

\Delta p_s = \Delta p_{\text{over}}\,\dfrac{R_{\text{res}}}{R_{\text{mc}} + R_{\text{res}}}

supercharging as a pressure divider: the mudcake and the formation are connected in series

\Delta p_s \approx \dfrac{u_2 r_w \mu_f}{2k_f}\left[\dfrac{u_1}{u_2}\ln\dfrac{T_d + t_s}{t_s} + \ln\dfrac{k_f t_s}{\phi\mu_f c_t r_w^2} + 0.809\right]

the model of filtrate injection, SI

p(\Delta t) = p^* - m_s\left(\dfrac{1}{\sqrt{\Delta t}} - \dfrac{1}{\sqrt{t_p + \Delta t}}\right)

spherical build-up after a pretest of duration $t_p$

m_s = \dfrac{q\mu\sqrt{\phi\mu c_t}}{4\pi^{3/2}k_s^{3/2}},\quad k_s = (k_h^2k_v)^{1/3}

the slope of the spherical line and the spherical permeability, SI

Common pitfalls

Rejecting only the records that have not stabilised. A supercharged record has stabilised.
Thinking that the extrapolated pressure $p^*$ removes supercharging. It equals the pressure at the wall, together with the excess.
Taking the slope of the spherical build-up from the first minutes. The tool storage is still at work there, and the permeability comes out several times too low.
Comparing the drawdown mobility with the permeability from core or from a well test without allowing for the filtrate, the anisotropy and the scale.

After this lesson you can

explain the mechanism of supercharging through the resistances of the mudcake and of the formation in series, and calculate the excess pressure with the model of filtrate injection
estimate how supercharging distorts pressures, gradients and the position of contacts, and recognise supercharged measurements by their mobility, by their scatter and by the link between the deviations and the inverse mobility
choose a way to deal with supercharging: rejecting the points, using them as limits, an approximate correction, or a repeat measurement
find the spherical permeability and the extrapolated pressure from the build-up after a pretest, and explain why they differ from the drawdown mobility
relate the drawdown, spherical, horizontal and vertical permeabilities in an anisotropic rock, and explain what an observation probe and a dual-packer module give
allow for tool storage and for the limits of the method: the resolution of the gauge, the bubble point pressure (the pressure at which gas starts to come out of the oil), the invaded zone and heterogeneity

Why the pressure at the wall is higher than the formation pressure

While the hole is open and the mud pressure is higher than the formation pressure, the well works as a very slow injection well. The filtrate goes into permeable rocks in three stages (dynamics of filtration). The first is spurt loss: an instant initial loss of filtrate before the cake has formed. The second is dynamic filtration during circulation (while the mud is pumped round the well). The flow of mud washes away the outer layer of the cake, so the thickness of the cake, and with it the filtration rate, settle at a constant value. The third is static filtration after circulation stops. The rate at once falls noticeably, and then it goes on decreasing slowly as the cake grows. Logging and testing take place in the static period.

The pressure difference between the mud and the formation is shared by two resistances connected in series: the mudcake, and the rock behind it. This is a pressure divider, as in an electric circuit. Usually the resistance of the cake is larger by orders of magnitude (many powers of ten). Almost the whole difference falls across the cake, and the pressure at the wall is almost the same as the formation pressure. But the resistance of the rock grows as its permeability falls, and in a tight layer its share becomes noticeable.

Supercharging as a pressure divider

\Delta p_s = p_s - p_{\text{res}} = \Delta p_{\text{over}}\,\frac{R_{\text{res}}}{R_{\text{mc}} + R_{\text{res}}},\qquad R_{\text{mc}} = \frac{\mu_f\,h_{mc}}{k_{mc}}

where:

$\Delta p_s$ supercharging: the excess of the pressure at the wall $p_s$ over the formation pressure $p_{\text{res}}$ , Pa
$\Delta p_{\text{over}}$ overbalance: the mud pressure minus the formation pressure, Pa
$R_{\text{mc}}$ resistance of the mudcake per unit area of the wall, Pa·s/m
$R_{\text{res}}$ resistance of the rock per unit area of the wall: the excess pressure divided by the filtration rate $u$ (next section), Pa·s/m
$\mu_f,\ h_{mc},\ k_{mc}$ viscosity of the filtrate, and thickness and permeability of the cake, Pa·s, m, m²

Units: SI

Holds when

The flow through the cake is steady: its thickness and permeability are constant (dynamic equilibrium, or slow growth in the static period).
The resistance of the rock changes slowly with time (logarithmically), so the formula is applied to the moment of the measurement.

While $R_{\text{res}} \ll R_{\text{mc}}$ , the cake sets the filtration rate, $u \approx \Delta p_{\text{over}}/R_{\text{mc}}$ . The supercharging is then this rate multiplied by the resistance of the rock, so it is inversely proportional to the permeability. When the two resistances are comparable, the flow falls and the supercharging grows more slowly. It can never exceed the overbalance.

CalculatedCalculated excess of pressure over the formation pressure against the distance from the wall, for layers of 30 and 0.3 mD at the same filtration rate (the model of the next section, with the parameters of the example). Almost the whole overbalance, about 1.9 MPa, is lost across the cake and is not shown on the plot.

The raised pressure does not occupy a thin shell at the wall. It covers a region tens of metres across: with distance it falls only logarithmically (very slowly), like the pressure around an injection well. So the build-up after a pretest, which covers tens of centimetres to metres of rock, lies wholly within this region and sees the same excess.

How much: the model of filtrate injection

We find the resistance of the rock by treating the well as an injection well. Filtrate leaves through each square metre of the wall at a rate $u$ . For each metre of formation thickness, this is a flow of $2\pi r_w u$ . First, during circulation, the rate $u_1$ holds for a time $T_d$ . Then, in the static period, it is smaller, $u_2$ . Each stage is radial injection at a constant rate. The line-source solution (the pressure around an injection well treated as a thin line, logarithmic approximation) and superposition of the two rates (adding up their effects) give the pressure at the wall.

Supercharging by the model of filtrate injection

\Delta p_s = \frac{u_2\,r_w\,\mu_f}{2k_f}\left[\frac{u_1}{u_2}\ln\frac{T_d + t_s}{t_s} + \ln\frac{k_f\,t_s}{\phi\,\mu_f\,c_t\,r_w^2} + 0.809\right]

where:

$u_1,\ u_2$ filtration rates through unit area of the wall during circulation and after it stops (the volume of filtrate per unit time per unit area), m/s
$T_d$ duration of dynamic filtration before circulation stops, s
$t_s$ time of static filtration up to the measurement, s
$r_w$ radius of the hole, m
$k_f$ permeability of the rock to the filtrate: in a water-bearing layer the permeability, in an oil-bearing layer the effective permeability of the flushed zone to the filtrate, m²
$\mu_f,\ \phi,\ c_t$ viscosity of the filtrate, and porosity and total compressibility of the rock at the hole, Pa·s, —, 1/Pa
$0.809$ $\ln(4/\gamma)$ , where $\gamma = 1.781$ , as in the logarithmic approximation of the line-source solution, —

Units: SI; natural logarithm

Holds when

The cake sets the filtration rate: the supercharging is small compared with the overbalance (otherwise use the pressure divider above). In each stage the rate is constant.
The flow of filtrate is single-phase (one fluid), in a uniform rock. The logarithmic approximation at the wall holds a few minutes after injection starts, even in a layer of 0.03 mD, long before logging.
The filtrate spreads away from the wall horizontally (radially), as in a uniform formation. A thin tight layer between permeable parts partly drains into its neighbours in the vertical direction, and the supercharging in it is smaller than calculated.

The factor before the bracket sets the scale: it contains the filtration rate, the viscosity of the filtrate and the resistance of the rock, $1/k_f$ . The bracket changes slowly. Its first term, the memory of dynamic filtration, falls with the time of rest (the time since circulation stopped). The second term grows as the logarithm of time. So the supercharging is almost inversely proportional to the permeability, and it changes little during logging.

In other unit systems (1)

Metric units

\Delta p_s\,[\text{kPa}] = 140.7\,\frac{u_2\,r_w\,\mu_f}{k_f}\left[\frac{u_1}{u_2}\ln\frac{T_d + t_s}{t_s} + \ln\frac{k_f\,t_s}{\phi\,\mu_f\,c_t\,r_w^2} - 4.831\right]

$u$ in mm/h, $r_w$ in m, $\mu_f$ in mPa·s, $k_f$ in mD, $T_d$ and $t_s$ in h, $c_t$ in 1/MPa.

$140.7 = (10^{-3}/3600)\cdot 10^{-3}/(2\cdot 0.9869\cdot 10^{-15})/1000$ ; $-4.831 = \ln(0.9869\cdot10^{-15}\cdot 3600/10^{-9}) + 0.809$ . A rate of 0.3 mm/h is 0.05 cm³ of filtrate a minute through 100 cm² of wall.

Worked exampleSupercharging of water-bearing layers of 30, 3 and 0.3 mD

Given (assumed data)

Filtration: 0.72 mm/h during circulation for 30 h, then 0.30 mm/h; the measurement is made 10 h after circulation stops
Hole and filtrate: $r_w$ = 0.108 m (drill bit of 215.9 mm); water-based filtrate, 0.40 mPa·s
Rock at the hole: $\phi$ = 0.18; $c_t = 1.0\cdot10^{-3}$ MPa⁻¹
Overbalance: 1.9 MPa

Solution

Memory of circulation: $\dfrac{0.72}{0.30}\ln\dfrac{30 + 10}{10} = 2.4\cdot 1.3863 = 3.327$ .
The denominator under the logarithm: $0.18\cdot 0.40\cdot 1.0\cdot10^{-3}\cdot 0.108^2 = 8.40\cdot10^{-7}$ . For 30 mD: $\ln(30\cdot 10/(8.40\cdot10^{-7})) - 4.831 = 14.86$ ; the bracket is 18.19.
The factor: $140.7\cdot 0.30\cdot 0.108\cdot 0.40/30 = 0.0608$ kPa; the supercharging is $0.0608\cdot 18.19 = 1.11$ kPa.
For 3 mD: the factor is 0.608 kPa, the bracket is $3.327 + 12.56 = 15.89$ , and the supercharging is 9.7 kPa. For 0.3 mD: the factor is 6.08 kPa, the bracket is 13.58, and the supercharging is 82.6 kPa.
Checking the assumption: even for 0.3 mD the supercharging is 4.3% of the overbalance. By the pressure divider it is $82.6/(1 + 82.6/1900) = 79.2$ kPa, so the correction is small.

Answer: The supercharging is about 1.11 kPa in the 30 mD layer, 9.7 kPa in the 3 mD layer and 82.6 kPa in the 0.3 mD layer.

What it means. With a scatter of about 1 kPa in valid points, the supercharging of the 30 mD layer is almost unnoticeable. For the 3 mD layer it is already ten times the scatter. For the 0.3 mD layer it is tens of kilopascals, which means metres of error in the contact. Rules of thumb of the type “supercharging matters below such-and-such a permeability” depend on the mud: with a filtration rate five times smaller, all the numbers are five times smaller. The static filtration rate in a well is almost never measured. So the model serves to understand the mechanism and the orders of magnitude, and supercharging itself is recognised from the survey data (below). The data in this example are assumed for illustration.

CalculatedCalculated supercharging against permeability for two muds: static filtration of 0.30 and 0.06 mm/h (dynamic filtration 2.4 times larger), the rest as in the example. The dashed line is the model of constant filtration rate. The solid lines allow for the pressure divider at an overbalance of 1.9 MPa.

How supercharging distorts pressures, gradients and contacts

Each supercharged point is shifted to the right by its own $\Delta p_s$ . If the supercharging is the same at all the points of a line, the line shifts in parallel. The gradient and the fluid density look believable, but the level is wrong. If the supercharging changes through the section (for example, if the rock is tighter lower down), the slope changes too. A shift of the level turns into a shift of the contact, in the same way as a pressure error does in the topic Pressure gradients, fluid contacts and capillary equilibrium.

Distortion of the gradient and of the contact by supercharging

G_{\text{app}} = G + \frac{\Delta(\Delta p_s)}{\Delta z},\qquad \delta z^* = \frac{\Delta p_{s,o} - \Delta p_{s,w}}{G_w - G_o}

where:

$G_{\text{app}},\ G$ apparent and true gradients from the points of one zone, Pa/m
$\Delta(\Delta p_s)/\Delta z$ change of the supercharging with depth over the interval of the points (the mean slope), Pa/m
$\delta z^*$ shift of the FWL that we find (plus is downward), m
$\Delta p_{s,o},\ \Delta p_{s,w}$ supercharging of the oil and water lines at the depth where they cross, Pa

Units: SI (in practice kPa, kPa/m and m)

Holds when

The lines are fitted to the supercharged points as to valid ones. The supercharging is small compared with the pressure differences on which the classification of the fluids is based.

Supercharging of the water points raises the FWL, and supercharging of the points in the oil part lowers it. The water part below the contact is often tighter than the oil part. The reason is cementation (minerals filling the pores): it could go on in the water part after the oil filled the trap (the structure that holds the oil). So the first case is more dangerous: the FWL comes out too high, and the oil in place is underestimated.

Pretests of the water-bearing part of well 2

Depth, m: 2547.5
Pressure, MPa: 28.0495
Mobility, mD/(mPa·s): 4

Depth, m: 2553.0
Pressure, MPa: 28.0952
Mobility, mD/(mPa·s): 12

Depth, m: 2559.5
Pressure, MPa: 28.1700
Mobility, mD/(mPa·s): 6

Depth, m: 2566.0
Pressure, MPa: 28.2289
Mobility, mD/(mPa·s): 25

Depth, m: 2573.5
Pressure, MPa: 28.3143
Mobility, mD/(mPa·s): 8

Depth, m	Pressure, MPa	Mobility, mD/(mPa·s)
2547.5	28.0495	4
2553.0	28.0952	12
2559.5	28.1700	6
2566.0	28.2289	25
2573.5	28.3143	8

Well 2 from the topic Pressure gradients, fluid contacts and capillary equilibrium: the oil part has six valid pretests, and the oil line is $G_o$ = 7.068 kPa/m. The well was drilled with the mud of the example above, which has a higher fluid loss than the mud of the new well, and the water-bearing part here is tighter. All the records have stabilised. The data are assumed for illustration.

Worked exampleA tight water-bearing formation: a believable line and a wrong contact

Given (assumed data)

Pretests of the water-bearing part: the table above
Oil line of well 2: $G_o$ = 7.068 kPa/m; the FWL from the regional water line is 2540.3 m ± 5 m
Regional water line: gradient 10.50 kPa/m (water density from a laboratory analysis, 1.0706 g/cm³ in the formation); 28.163 MPa at 2560.0 m

Solution

The line through five points: $G_w = 10.22$ kPa/m, density 1.042 g/cm³, which is believable formation water. The scatter is $s = 5.8$ kPa, several times larger than usual.
The intersection of this line with the oil line of well 2: 2535.5 m, which is 4.8 m higher than by the regional line.
The deviations of the points from the regional water line: 17.8, 5.7, 12.3, 2.9, 9.5 kPa, with the inverse mobilities 0.250, 0.083, 0.167, 0.040, 0.125 mPa·s/mD. The lower the mobility, the larger the deviation.
The straight line $d = c + a/(k/\mu)_d$ by least squares: $a = 72 \pm 2$ kPa·mD/(mPa·s), $c = 0.1 \pm 0.4$ kPa. The residuals (what is left of the deviations after the line) are 0.4 kPa, so the deviations are fully explained by supercharging.
The water level without supercharging is the regional line plus $c$ . Its intersection with the oil line gives FWL = 2540.3 ± 0.5 m. The error no longer comes from comparing two wells: the water level is taken from the measurements of the same well.

Answer: Supercharging raised the FWL that we found by 4.8 m. With the correction, the FWL is 2540.3 ± 0.5 m, in agreement with the new well.

What it means. Each point on its own looked valid, and the line looked believable. Only the low mobilities, the increased scatter and the link between the deviations and the inverse mobility gave the supercharging away. The correction works here because the mobilities are different, and the filtration rate is apparently the same for all the layers. Errors in the drawdown mobility weaken the link that we find: the slope $a$ comes out smaller and $c$ larger. So the correction stays incomplete, and the FWL is a little too high. A sign of this is residuals that are noticeably larger than the scatter of the measurements. If all the points are equally tight, the deviations do not depend on mobility, and supercharging cannot be told from a shift of the level. The only defence is then to reject points by mobility and to use the regional water line. The method with a known water gradient is a special case of the forced gradient technique. The data in this example are assumed for illustration.

How to recognise and allow for supercharging

Signs of supercharging

1
Mobility
Supercharging is noticeable first of all at points with a mobility below a few mD/(mPa·s). With a mud of high fluid loss, it is noticeable at higher mobilities too.
2
Position of the point
A shift to the right of the line of valid points of the same zone, larger when the mobility is lower. We check the link between the deviation and the inverse mobility on a plot.
3
Scatter and density
The scatter of the points of the line is several times larger than usual, and the apparent fluid density is outside the expected limits or does not match the sample.
4
Repeat pretests
A repeat test at the same point gives a different stabilised pressure. A repeat survey a few hours later gives a smaller excess.
5
Stabilisation time
In a tight layer the pressure takes longer to stabilise. Both the time of the tool storage and the time of the spherical “tail” are inversely proportional to the mobility (tool storage). A long build-up is a sign of the risk of supercharging.

How to allow for it. The first remedy is rejection. Points with a low mobility are not used for the gradient and the contacts, unless their pressure is confirmed in another way. If a zone can be estimated only from such points, we use them as limits. The stabilised pressure of a supercharged point is an upper limit of the formation pressure. An unfinished build-up gives a lower limit. An approximate correction comes from the deviations linked with the inverse mobility, as in the example above. It is good enough for a rough estimate (the right order of magnitude), because the filtration rate of different layers can differ by several times. So, in the survey of the topic about the formation tester, the point at 2526.5 m is supercharged by 46 kPa. The point at 2520.0 m has a mobility six times higher and shows no excess, although at an equal filtration rate it would have been about 9 kPa. Finally, supercharging can be reduced at the planning stage. We use a smaller overbalance, a mud with a low fluid loss, and repeat measurements at suspicious depths. For a tight water-bearing formation, we take the water line from regional data (regional water line).

Permeability from a pretest: drawdown and build-up

The drawdown mobility, from the steady pressure difference during the draw by the piston, is covered in the topic about the formation tester. Almost the whole pressure difference falls across the first centimetres at the probe. So the mobility reflects the rock right at the wall, flushed by filtrate. It is sensitive to everything that is there: leftover mudcake, fine particles, and rock damaged or compacted by the tool.

The build-up after a pretest carries information about rock farther away. The disturbance from the draw spreads from the probe spherically. Far from the probe, the pressure behaves as around a point sink in an infinite medium (a point that draws fluid from all sides): a steady part plus a “tail” that decays as $1/\sqrt t$ . The build-up is a superposition (a sum of effects) of a draw at a rate $q$ for a time $t_p$ and its stopping (a shut-in as a negative rate). The steady parts cancel, and two “tails” remain. This gives the spherical time function $f_s$ , and in the coordinates “pressure against $f_s$ ” the record falls on a straight line.

Pressure build-up after a pretest under spherical flow

p(\Delta t) = p^* - m_s\,f_s(\Delta t),\qquad f_s = \frac{1}{\sqrt{\Delta t}} - \frac{1}{\sqrt{t_p + \Delta t}},\qquad m_s = \frac{q\,\mu\sqrt{\phi\,\mu\,c_t}}{4\pi^{3/2}\,k_s^{3/2}}

where:

$p(\Delta t)$ pressure at the probe a time $\Delta t$ after the piston stops, Pa
$p^*$ the pressure that the build-up approaches ( $f_s \to 0$ ): the pressure at the wall, Pa
$t_p$ duration of the draw, s
$q$ rate of the draw by the piston, m³/s
$m_s$ slope of the spherical line, Pa·s^½
$k_s$ spherical permeability $(k_h^2 k_v)^{1/3}$ , m²
$\mu,\ \phi,\ c_t$ viscosity of the fluid that flows in (usually the filtrate), and porosity and total compressibility of the rock at the hole, Pa·s, —, 1/Pa

Units: SI

Holds when

The rock is uniform (possibly anisotropic). The disturbance has moved away from the probe by more than a few hole radii. The flow to the probe is then a full sphere, because the hole is a thin impermeable tube. (For flow to a hemisphere at a flat wall, the slope would be twice as large.)
The influence of tool storage has ended (see the section on tool storage). The disturbance has not yet reached the boundaries of the formation.
One phase of known viscosity flows. With two rates of draw, $f_s$ is built by superposition, by the same rules.

The slope depends on $k_s^{3/2}$ , and on porosity and compressibility only as a square root. So their inaccuracy affects $k_s$ weakly: to the power 1/3. The intercept on the axis $f_s = 0$ gives $p^*$ , even if the build-up has not finished.

In other unit systems (1)

Metric units

m_s\,[\text{MPa·s}^{1/2}] = 45.8\,\frac{q\,[\text{cm}^3/\text{s}]\;\mu\,[\text{mPa·s}]\sqrt{\phi\,\mu\,c_t\,[\text{MPa}^{-1}]}}{k_s^{3/2}\,[\text{mD}]}

$q$ in cm³/s, $\mu$ in mPa·s, $c_t$ in 1/MPa, $k_s$ in mD, $\Delta t$ and $t_p$ in s, pressure in MPa.

$45.8 = \dfrac{10^{-6}\cdot10^{-3}\sqrt{10^{-3}\cdot10^{-6}}}{4\pi^{3/2}(0.9869\cdot10^{-15})^{3/2}}\cdot 10^{-6}$ ; hence $k_s = (45.8\,q\mu\sqrt{\phi\mu c_t}/m_s)^{2/3}$ .

Worked exampleSpherical permeability from the build-up at 2520.0 m

Given (assumed data)

Pretest: the repeat test at the point 2520.0 m from the topic Pressure measurement with the wireline formation tester: 5 cm³ at a rate of 0.2 cm³/s ( $t_p$ = 25 s); the tool stayed at the point for 6 minutes
End of the draw: 24.1254 MPa; the formation pressure at this point is 27.8141 MPa
Build-up: at 40, 80, 160 and 320 s: 27.81139; 27.81342; 27.81386; 27.81401 MPa (resolution 0.01 kPa)
Rock and filtrate: $\phi$ = 0.18; $c_t = 1.0\cdot10^{-3}$ MPa⁻¹; $\mu_f$ = 0.40 mPa·s; tool storage 0.25 cm³/MPa

Solution

Drawdown mobility: $39.02\cdot 0.2/(27.8141 - 24.1254) = 2.12$ mD/(mPa·s). The time constant of the tool storage is $0.25\cdot 39.02/2.1 = 4.65$ s.
Spherical time function: $1/\sqrt{40} - 1/\sqrt{65} = 0.03408$ . For 80, 160 and 320 s it is 0.01421, 0.00554 and 0.00206 s^−½.
The constant: $45.8\cdot 0.2\cdot 0.40\sqrt{0.18\cdot 0.40\cdot 1.0\cdot10^{-3}} = 0.03109$ .
Early points (40 and 80 s): $m_s = (27.81342 - 27.81139)/(0.03408 - 0.01421) = 0.1022$ MPa·s^½, $k_s = (0.03109/0.1022)^{2/3} = 0.45$ mD, which is a mobility of 1.1.
Late points (160 and 320 s): $m_s = 0.00015/(0.00554 - 0.00206) = 0.0431$ MPa·s^½, $k_s = 0.80$ mD, which is a mobility of 2.0. Then $p^* = 27.81401 + 0.0431\cdot 0.00206 = 27.8141$ MPa.

Answer: From the late points the spherical mobility is about 2.0 mD/(mPa·s), and from the drawdown it is 2.12. The early points would give 1.1, almost half as much.

What it means. The rock in this model is uniform and isotropic (its permeability is the same in all directions), and there is no damage at the probe. So the drawdown and the build-up must give the same result, and they do, if we take the build-up late enough. The early points lie 9–17 time constants of the tool storage after the piston stops. The record there already seems stabilised, but the inflow that squeezes the liquid in the line still distorts the slope. The late pressure changes, on the other hand, are only 0.15 kPa, about fifteen steps of the gauge resolution. So the window of usable points is narrow on both the early and the late side. The extrapolated pressure agreed with the formation pressure, because there is no supercharging at this point. During the drawdown the pressure at the probe falls to about 24.1 MPa. This is 3.3 MPa below the bubble point pressure of the oil of this accumulation (27.4 MPa). The model is single-phase and does not allow for gas that may come out of the residual oil at the probe. That gas would raise the compressibility and distort the start of the build-up. The data in this example are assumed for illustration.

CalculatedBuild-up after the pretest at 2520.0 m in the coordinates “pressure against spherical time function” (the model with a tool storage of 0.25 cm³/MPa). The vertical axis shows the excess pressure over 27.810 MPa. The dots are the measurements from the example. The grey dashed line is the spherical line of the same rock without tool storage. The orange line is the line through the two early points.

Spherical and radial flow, anisotropy

In a layered rock the vertical permeability is lower than the horizontal one, and spherical flow turns into ellipsoidal flow (flow in the shape of a stretched sphere). Suppose we stretch the vertical coordinate by a factor $\sqrt{k_s/k_v}$ and compress the horizontal ones by a factor $\sqrt{k_h/k_s}$ . Then the rock becomes isotropic (the same permeability in every direction), with a permeability $k_s = (k_h^2k_v)^{1/3}$ , and the volumes and the flow rates do not change. So, far from the probe, where the shape of the probe itself does not matter, the build-up “sees” just the spherical permeability. The drawdown, however, is set by the flow in the first centimetres at the round opening of the probe on the wall, and there anisotropy acts differently. The drawdown mobility lies between $k_s$ and $k_h$ . As anisotropy grows, it moves away from $k_h$ more slowly than $k_s$ does (advanced section).

Drawdown and spherical permeability in an anisotropic rock (as fractions of

k_h

)

$A = k_h/k_v$: 1
$k_d/k_h$: 1.00
$k_s/k_h$: 1.00
$k_d/k_s$: 1.00

$A = k_h/k_v$: 3
$k_d/k_h$: 0.77
$k_s/k_h$: 0.69
$k_d/k_s$: 1.12

$A = k_h/k_v$: 10
$k_d/k_h$: 0.61
$k_s/k_h$: 0.46
$k_d/k_s$: 1.31

$A = k_h/k_v$: 30
$k_d/k_h$: 0.51
$k_s/k_h$: 0.32
$k_d/k_s$: 1.57

$A = k_h/k_v$: 100
$k_d/k_h$: 0.43
$k_s/k_h$: 0.22
$k_d/k_s$: 1.97

$A = k_h/k_v$	$k_d/k_h$	$k_s/k_h$	$k_d/k_s$
1	1.00	1.00	1.00
3	0.77	0.69	1.12
10	0.61	0.46	1.31
30	0.51	0.32	1.57
100	0.43	0.22	1.97

$k_s/k_h = A^{-1/3}$ ; $k_d/k_h$ is from the model of a round probe on a flat impermeable wall (advanced section). Solutions with an ellipsoidal sink instead of an opening give values of $k_d/k_h$ about 10% smaller.

This gives the usual order of the estimates in a layered uniform rock: $k_v < k_s < k_d < k_h$ . The ratio $k_d/k_s$ grows slowly with anisotropy. So the pair “drawdown and build-up” can estimate anisotropy only roughly. An error of 10% in one of the estimates, at $A$ of about 10, changes the anisotropy that we find by about a factor of two. It is more reliable to measure the response of the rock at a distance.

Radial flow. If the probe stands in a thin layer between impermeable layers, the disturbance reaches them, and spherical flow gives way to radial (cylindrical) flow through the whole thickness of the layer. On the diagnostic plot (pressure change and its derivative against time, on logarithmic axes) the derivative reaches a plateau, a flat part (Diagnosis with the pressure derivative). The semilog straight line gives the $k_h h$ of the layer (permeability times thickness). The dual-packer module seals off a section of the wall about a metre long and at once creates radial flow to this interval. Spherical flow follows and, if the layer is bounded, late radial flow. These are the same flow regimes as for a partially penetrating well (a well that opens only part of the layer), but at the scale of a metre (flow regimes). The large volume between the packers also gives a large storage. So such tests are run with a pump for tens of minutes.

Observation probe. In a multi-probe tool, an observation probe stands above the sink probe at a distance $\Delta z$ (usually about 0.7 m) along the same line of the tool. Sometimes there is one more on the opposite side of the hole. A pump draws fluid through the sink at a constant rate, and the observation probe records how the disturbance reaches it. This is a vertical interference test. Neither damage at the sink nor its storage affects the response.

Response of an observation probe at a constant draw rate

\Delta p_{\text{obs}}(t) = \frac{q\mu}{4\pi k_h\,\Delta z}\,\operatorname{erfc}\frac{\Delta z}{\sqrt{4\eta_v t}} \;\approx\; \frac{q\mu}{4\pi k_h\,\Delta z} - \frac{m_s}{\sqrt t},\qquad \eta_v = \frac{k_v}{\phi\mu c_t}

where:

$\Delta p_{\text{obs}}$ pressure drop at the observation probe, Pa
$\Delta z$ vertical distance between the probes, m
$\eta_v$ vertical hydraulic diffusivity (how fast a pressure change spreads in the vertical direction), m²/s
$m_s$ the same slope as in the spherical build-up: $q\mu\sqrt{\phi\mu c_t}/(4\pi^{3/2}k_s^{3/2})$ , Pa·s^½
$q,\ \mu,\ \phi,\ c_t,\ k_h,\ k_v$ pump rate, viscosity, porosity, compressibility, and horizontal and vertical permeabilities, m³/s, Pa·s, —, 1/Pa, m², m²

Units: SI; the approximate form holds when $\Delta z/\sqrt{4\eta_v t} < 0.15$

Holds when

A point sink in a uniform anisotropic medium; the hole is not allowed for. The hole slightly increases the response (by no more than a factor of two, the limit of a flat wall), and then the first formula underestimates $k_h$ . In processing real data, the constant is taken from a model of the tool with a hole.
The rate at the sink is constant from the start; the flow is single-phase.

The level that the response approaches depends only on the horizontal permeability. A change of the vertical permeability stretches the path and eases the flow by the same factor, so the two effects cancel. The speed with which the response approaches this level is set by the vertical hydraulic diffusivity. At late times the response is a straight line in the coordinates “ $\Delta p_{\text{obs}}$ against $1/\sqrt t$ ”. The intercept on the axis gives $k_h$ , the slope gives $k_s$ , and together they give $k_v = k_s^3/k_h^2$ .

In other unit systems (1)

Metric units

\Delta p_{\text{obs},\infty}\,[\text{kPa}] = 80.6\,\frac{q\,[\text{cm}^3/\text{s}]\;\mu\,[\text{mPa·s}]}{k_h\,[\text{mD}]\;\Delta z\,[\text{m}]}

$q$ in cm³/s, $\mu$ in mPa·s, $k_h$ in mD, $\Delta z$ in m. The slope $m_s$ comes from the metric form of the spherical build-up (in MPa·s^½).

$80.6 = 10^{-6}\cdot10^{-3}/(4\pi\cdot 0.9869\cdot10^{-15})/1000$ .

CalculatedCalculated response of an observation probe 0.70 m above the sink at a draw of 1 cm³/s, for two values of the vertical permeability at

k_h

= 12 mD. The level that the curves approach is the same. The vertical permeability decides how fast the response gets there.

Worked exampleHorizontal and vertical permeability from an observation probe

Given (assumed data)

Test: a water-bearing layer; a pump draws 1.0 cm³/s through the sink probe; the observation probe is 0.70 m above
Fluid and rock: water, 0.40 mPa·s; $\phi$ = 0.20; $c_t = 1.0\cdot10^{-3}$ MPa⁻¹
Response of the observation probe: 450, 900 and 1800 s after the start: 3.20, 3.39 and 3.52 kPa

Solution

The values of $1/\sqrt t$ (the horizontal axis of the plot) are 0.04714, 0.03333 and 0.02357 s^−½. A check of the condition: $\Delta z/\sqrt{4\eta_v t}$ at $k_v$ of about 1 mD and 450 s is about 0.15.
A line through the end points: the slope is $(3.52 - 3.20)/(0.04714 - 0.02357) = 13.58$ kPa·s^½. The level is $3.52 + 13.58\cdot 0.02357 = 3.840$ kPa. The middle point on the line is 3.39 kPa, and the measurement is 3.39.
Horizontal permeability: $k_h = 80.6\cdot 1.0\cdot 0.40/(0.70\cdot 3.840) = 12.0$ mD.
Spherical permeability: $k_s = \left(45.8\cdot 1.0\cdot 0.40\sqrt{0.20\cdot 0.40\cdot 10^{-3}}/0.01358\right)^{2/3} = 5.26$ mD.
Vertical permeability: $k_v = k_s^3/k_h^2 = 1.01$ mD. The anisotropy is $k_h/k_v = 11.9$ .

Answer:

k_h \approx 12.0

mD,

k_v \approx 1.01

mD; the anisotropy is about 12.

What it means. The sink probe in this layer would give about 7.1 mD from the drawdown and about 5.26 from the build-up. Both estimates lie between $k_v$ and $k_h$ , and only the observation probe separates them. The numbers refer to a volume of a metre to a metre and a half around the probes. If there is a thin tight layer between the probes, the response and the “vertical permeability” reflect exactly that layer. This is valuable for forecasting coning (the drawing of water or gas towards the well) and vertical displacement (Water and gas coning). The data in this example are assumed for illustration.

Tool storage and the limits of the method

Between the probe and the gauge are the line and the chamber with liquid. Its compression and expansion, the tool storage, separate what the piston does from what happens in the rock (the storage model). During the drawdown this is the time constant $\tau = SK/(k/\mu)$ . If the piston works for less than a few $\tau$ , the drawdown has no time to become steady, and the mobility is overstated. During the build-up the effect of storage is less obvious and lasts longer. After the piston stops, the rock still squeezes the liquid in the line, and the liquid gives up part of its volume with a delay. The spherical formula, however, assumes that the whole volume left the rock during the time $t_p$ . So the straight line on the spherical plot appears only after tens of $\tau$ . The slope from earlier points is too steep, and the permeability comes out too low.

Mobility from the slope of the spherical plot in different windows (the model with tool storage)

Point and storage: 2520.0 m; 0.25 cm³/MPa
Window after stopping: 40–160 s
Mobility from the slope: 1.26
True: 2.10

Point and storage: 2520.0 m; 0.25 cm³/MPa
Window after stopping: 160–640 s
Mobility from the slope: 1.97
True: 2.10

Point and storage: 2520.0 m; 0.05 cm³/MPa
Window after stopping: 40–160 s
Mobility from the slope: 2.01
True: 2.10

Point and storage: 2526.5 m; 0.25 cm³/MPa
Window after stopping: 60–240 s
Mobility from the slope: 0.013
True: 0.35

Point and storage: 2526.5 m; 0.25 cm³/MPa
Window after stopping: 960–3840 s
Mobility from the slope: 0.327
True: 0.35

Point and storage	Window after stopping	Mobility from the slope	True
2520.0 m; 0.25 cm³/MPa	40–160 s	1.26	2.10
2520.0 m; 0.25 cm³/MPa	160–640 s	1.97	2.10
2520.0 m; 0.05 cm³/MPa	40–160 s	2.01	2.10
2526.5 m; 0.25 cm³/MPa	60–240 s	0.013	0.35
2526.5 m; 0.25 cm³/MPa	960–3840 s	0.327	0.35

Mobilities are in mD/(mPa·s). The point 2520.0 m is the pretest of the example. The point 2526.5 m is the tight supercharged layer of the survey in the topic on the formation tester: a draw of 2 cm³ at a rate of 0.05 cm³/s. The model is a spherical sink that gives the same drawdown as the standard probe, with the storage of the line.

This is one of the reasons why, in practice, the permeability from the build-up is often noticeably lower than from the drawdown. In tight layers the tool is withdrawn before the storage stops distorting the slope. Tool designers reduce the storage by making the volume of the line between the probe and the gauge smaller. The analysis is done from the late points, or with a model in which the storage is allowed for explicitly. The upper limit of the method is set by the resolution of the gauge: the build-up signal falls as $k^{-3/2}$ . For a pretest of 5 cm³ at a rate of 0.5 cm³/s, the pressure change from $\Delta t = t_p$ to $10\,t_p$ exceeds twenty steps of a quartz gauge (0.01 kPa) only up to 11 mD. With a step of 0.7 kPa (a strain gauge) it exceeds twenty steps only up to 0.6 mD. The drawdown works in much more permeable rocks.

What limits the estimates of permeability from a pretest

Factor: Volume investigated
Drawdown: centimetres at the probe
Build-up: tens of centimetres to metres; it grows as $\sqrt{\eta\,\Delta t}$

Factor: Damage, compaction or splitting of the rock by the probe
Drawdown: strong: damage and compaction make the result too low, splitting makes it too high
Build-up: almost no effect

Factor: Anisotropy
Drawdown: between $k_s$ and $k_h$
Build-up: $k_s = (k_h^2k_v)^{1/3}$

Factor: Invaded zone
Drawdown: filtrate with residual oil
Build-up: may go beyond the invaded zone: the pores hold a mixture of fluids, and the viscosity is unclear

Factor: Tool storage
Drawdown: overstates the result if the draw is shorter than a few $\tau$
Build-up: understates the result from early points, sometimes by several times

Factor: Gauge resolution
Drawdown: in permeable rocks the drawdown is small but can be measured; the noise of the tool itself is more of a problem
Build-up: the limit is a few to tens of mD

Factor: Bubble point pressure
Drawdown: a deep drawdown may release gas in the line
Build-up: a distorted build-up

Factor	Drawdown	Build-up
Volume investigated	centimetres at the probe	tens of centimetres to metres; it grows as $\sqrt{\eta\,\Delta t}$
Damage, compaction or splitting of the rock by the probe	strong: damage and compaction make the result too low, splitting makes it too high	almost no effect
Anisotropy	between $k_s$ and $k_h$	$k_s = (k_h^2k_v)^{1/3}$
Invaded zone	filtrate with residual oil	may go beyond the invaded zone: the pores hold a mixture of fluids, and the viscosity is unclear
Tool storage	overstates the result if the draw is shorter than a few $\tau$	understates the result from early points, sometimes by several times
Gauge resolution	in permeable rocks the drawdown is small but can be measured; the noise of the tool itself is more of a problem	the limit is a few to tens of mD
Bubble point pressure	a deep drawdown may release gas in the line	a distorted build-up

The directions of the effects are for the usual cases. The comparison of estimates from pretests with core and with well tests is covered in the topic Comparing formation-tester, core and well-test data.

Terms and symbols

Supercharging $\Delta p_s$: The excess of the pressure at the wall over the formation pressure, caused by filtrate that the formation cannot carry away in time. It grows as permeability falls.
Dynamic and static filtration: The loss of filtrate during circulation (the cake is in equilibrium, and the rate is constant) and after circulation stops (the rate is lower and slowly decreases).
Spherical time function $f_s$: It equals $1/\sqrt{\Delta t} - 1/\sqrt{t_p + \Delta t}$ . In the coordinates “pressure against $f_s$ ” a spherical build-up is a straight line.
Spherical permeability $k_s$: The permeability that a build-up under spherical flow measures, $(k_h^2k_v)^{1/3}$ .
$p^*$: The pressure that the build-up approaches: the pressure at the wall, together with the supercharging.
Vertical interference test: A test with a multi-probe tool: the draw is through one probe, and the response is recorded by another at some vertical distance. It gives $k_h$ and $k_v$ .

Common mistakes and misconceptions

Wrong

“The pressure has stabilised, so it is the formation pressure.”

In fact

In a tight layer the pressure stabilises at the level of the pressure at the wall, which is higher than the formation pressure by the supercharging. We check the mobility and the position of the point among its neighbours.

Wrong

“Supercharging happens only at a mobility below 1 mD/(mPa·s).”

In fact

Supercharging is proportional to the filtration rate through the cake. With a mud of high fluid loss it is noticeable even at a few mD/(mPa·s). The limit depends on the mud, on the overbalance and on how accurately we need the contacts.

Wrong

“The build-up covers a larger volume, so its permeability is always more accurate than the drawdown.”

In fact

The build-up is distorted by tool storage and by heterogeneity (the rock changing from place to place). In an anisotropic rock it gives the spherical permeability, not the horizontal one. A slope from early points gives a permeability that is several times too low.

Wrong

“If the observation probe showed almost nothing, the rock between the probes is tight in the vertical direction.”

In fact

In a uniform rock the level of the response is set by the horizontal permeability, and the vertical permeability decides how fast this level is reached. A small level means a high $k_h$ or a small rate. A slow rise means a low $k_v$ . No response at all suggests a continuous impermeable thin layer between the probes.

In depth: advanced detail

Where the spherical line comes from

For a point sink with a constant rate $q$ in an infinite isotropic medium, the pressure drop at a distance $r$ is $\dfrac{q\mu}{4\pi k r}\operatorname{erfc}\dfrac{r}{2\sqrt{\eta t}}$ , where $\eta = k/(\phi\mu c_t)$ and erfc is the complementary error function. For a small argument, $\operatorname{erfc}x \approx 1 - 2x/\sqrt\pi$ . At the sink, where $r$ is small, the drop is the steady part $q\mu/(4\pi k r)$ minus a “tail” $\dfrac{q\mu}{4\pi k}\dfrac{1}{\sqrt{\pi\eta t}}$ , which no longer depends on $r$ .

In a build-up we add to the draw that starts at $t = 0$ an injection of the same rate from the moment $t_p$ . The steady parts cancel, and the “tails” give $p^* - p = \dfrac{q\mu}{4\pi^{3/2} k\sqrt\eta}\left(\dfrac{1}{\sqrt{\Delta t}} - \dfrac{1}{\sqrt{t_p + \Delta t}}\right)$ . We put in $\sqrt\eta = \sqrt{k/(\phi\mu c_t)}$ and get the slope $m_s = q\mu\sqrt{\phi\mu c_t}/(4\pi^{3/2}k^{3/2})$ . In an anisotropic medium, after the coordinates are stretched, $k$ is replaced by $k_s$ .

Why a full sphere and not a hemisphere at the wall? The sink stands on the wall of the hole, and at small $r$ the wall does act as a flat screen. But we analyse the build-up when the disturbance has gone tens of centimetres away. For it, a hole with a radius of about 0.1 m is a thin impermeable tube that hardly blocks the space. So the late “tail” is spherical. To take a hemisphere would double the slope in the formula and overstate $k_s$ by a factor of $2^{2/3} \approx 1.6$ . For a single draw with $t_p \ll \Delta t$ , the function is $f_s \approx t_p/(2\Delta t^{3/2})$ . The build-up is faster the higher $k_s$ is, and the time in which the pressure comes within a given small amount of $p^*$ is inversely proportional to the mobility.

Drawdown at a round probe in an anisotropic rock

Let the opening of the probe be a circle of radius $r_p$ on a flat impermeable wall, with the same pressure all over it. In an isotropic rock such a disc passes a steady flow $q = 4 r_p\,(k/\mu)\,\Delta p$ . This is a classical result of potential theory. It is half as much as for the same disc in an unbounded medium, because the flow comes from one side only.

In an anisotropic rock we stretch the coordinates so that the medium becomes isotropic, with a permeability $k_s$ . The wall stays a plane, and the circle becomes an ellipse with semi-axes $r_pA^{1/3}$ in the vertical direction and $r_pA^{-1/6}$ in the horizontal one, where $A = k_h/k_v$ . The conductance of an elliptical disc (the flow it passes for a given pressure difference) is proportional to its major semi-axis. It is inversely proportional to the complete elliptic integral $K(m)$ with the parameter $m = 1 - (b/a)^2 = 1 - 1/A$ . After we return to the original coordinates we get $k_d/k_h = \pi/(2K(m))$ : it is one at $A = 1$ , 0.61 at $A = 10$ and 0.43 at $A = 100$ . The drawdown loses the horizontal permeability only logarithmically slowly, while the spherical permeability falls as $A^{-1/3}$ .

The model is idealised: the wall is flat, and the pressure is the same over the whole opening. Solutions in which the opening is replaced by an ellipsoidal sink give values of $k_d/k_h$ about 10% smaller at the same anisotropy. The direction and the order of magnitude of the effect do not change: $k_s < k_d < k_h$ .

Connections to other topics

Pressure measurement with the wireline formation tester — pretest, drawdown mobility, classes of measurements and tool storage.
Pressure gradients, fluid contacts and capillary equilibrium — how a pressure error turns into a contact error, and the regional water line.
Borehole environment: invaded zone and mudcake — the mudcake, filtration and the invaded zone.
Line-source solution and the pressure function — the line-source solution on which the model of filtrate injection is built.
Partial penetration and well deviation — spherical flow and spherical permeability at the scale of a well.
Comparing formation-tester, core and well-test data — comparison of permeability from pretests, core and well tests.
Diagnosis with the pressure derivative — diagnosing flow regimes: a slope of −½ in spherical flow, a plateau in radial flow.

Check yourself

1
Two water-bearing layers in one well: 40 and 2 mD, with the same mud and the same time of measurement. About how many times do their supercharging values differ?
1. A.About 4.5 times ( $\sqrt{20}$ )
2. B.About 15–20 times, larger in the tight layer
3. C.They are the same
4. D.About 400 times
Show the answer
Correct answer: B. About 15–20 times, larger in the tight layer
The factor before the bracket is inversely proportional to the permeability, so it differs by 20 times. The bracket changes only logarithmically: for the tight layer it is a little smaller. With the parameters of the lesson's example the brackets are about 15.5 and 18.5, and the ratio is $20\cdot 15.5/18.5 \approx 17$ .
2
Find the supercharging of a 1.0 mD layer by the lesson's model: the factor is $140.7\cdot 0.30\cdot 0.108\cdot 0.40/1.0$ kPa and the bracket is 14.8.
Show the answer
The factor is $140.7\cdot 0.30\cdot 0.108\cdot 0.40 = 1.82$ kPa; the supercharging is $1.82\cdot 14.8 \approx 27$ kPa. With a difference between the water and oil gradients of 3.4 kPa/m, such a supercharging of a water-bearing point would move the FWL up by almost 8 m.
3
The stabilised pressures of five water-bearing pretests deviate from the water line found from a water analysis by 3, 12, 6, 20 and 9 kPa. Their mobilities are 30, 8, 15, 5 and 10 mD/(mPa·s). What is this most likely to be?
1. A.A lost packer seal
2. B.Supercharging of tight layers
3. C.Five separate water-bearing systems
4. D.An error in the density of the water found by analysis
Show the answer
Correct answer: B. Supercharging of tight layers
The deviation grows with the inverse mobility almost in proportion: $3\cdot 30 \approx 12\cdot 8 \approx 6\cdot 15 \approx 20\cdot 5 \approx 9\cdot 10 \approx 90$ – $100$ kPa·mD/(mPa·s). An error in the water density would give deviations that grow with depth, not with mobility. A lost seal would give the mud pressure.
4
On a spherical plot, a line through the points 30–60 s after the piston stopped gave a mobility of 0.6 mD/(mPa·s), and the drawdown gave 2.4. The time constant of the storage is about 4 s. Which conclusion is better?
Show the answer
The window of 30–60 s is only 8–15 time constants of the storage. The inflow that squeezes the liquid in the line still distorts the slope, and the permeability from it is too low. First we check the late points (a few minutes later), or we analyse the build-up with a model that includes the storage. Only if the difference remains do we look for anisotropy, heterogeneity or damage at the probe. And damage would have made the drawdown estimate the lower one.
5
An observation probe 0.7 m above the sink, at a draw of 2 cm³/s, showed a limiting pressure drop of 5.0 kPa; the viscosity is 0.5 mPa·s. Find the horizontal permeability in the idealised model, and say in which direction the hole, which the model leaves out, shifts it.
Show the answer
$k_h = 80.6\cdot 2\cdot 0.5/(0.7\cdot 5.0) = 23$ mD. The hole is an impermeable obstacle. It increases the response compared with an infinite medium, so the idealised formula underestimates $k_h$ (by no more than a factor of two, the limit of a flat wall).
6
Why is the extrapolated pressure $p^*$ of a spherical build-up not suitable for correcting supercharging?
1. A.It cannot be found until the build-up has finished
2. B.It equals the pressure at the wall, and supercharging is exactly the excess of the pressure at the wall over the formation pressure
3. C.It is always lower than the formation pressure
Show the answer
Correct answer: B. It equals the pressure at the wall, and supercharging is exactly the excess of the pressure at the wall over the formation pressure
The build-up after a pretest is the response of the rock a few centimetres to metres from the probe, inside the region where filtrate has already raised the pressure. The line approaches the pressure at the wall, and the supercharging is fully inside $p^*$ . $p^*$ helps when the build-up had no time to finish, but it does not separate supercharging from the formation pressure.

Updated: 16 October 2026