Nonideal wellbore storage, well clean-up and changing skin

Quick refresher

The formula $C = c_{wb}V_{wb}$ holds for a wellbore filled with one phase. In a closed tubing with free gas, the inflow both compresses the gas and makes the column heavier: $1/C = 1/C_g + \rho g/A_t$ . So $C$ does not exceed $A_t/(\rho g)$ , and it changes together with the pressure.
After shut-in at the wellhead, the gas rises and the liquid settles. The volume of the gas hardly changes, the pressure in it grows, and the bottomhole pressure can rise above the reservoir pressure. A “hump” appears on the record, and the pressure derivative becomes negative.
With increasing wellbore storage, the early unit slope (the first straight line of slope 1 on a log–log plot) gives a small apparent coefficient $C_a$ . With decreasing wellbore storage, it gives a large one, and the derivative plateau comes earlier than the early section promises. In both cases, estimates of duration from $C_a$ are wrong.
Empirical models, the exponential one (Fair) and the one with the error function (Hegeman, Hallford and Joseph), add to the wellbore balance a redistribution pressure $p_\phi$ with an amplitude $A$ and a time $t_\phi$ . They describe the shape of the record well, but they do not predict the parameters: they fit them.
During clean-up the skin falls while the drawdown test is running. The pressure falls more slowly, and at a constant rate it can even rise. The permeability and the skin from such a drawdown test are too high by factors of several. A build-up gives the right values, and the skin versus time comes from the drawdown points together with the permeability and $p^*$ from the build-up.
The best remedy for nonideal wellbore storage is shut-in at the bottom of the well. It removes the redistribution of phases and the movement of the level, and it cuts wellbore storage to minutes.

Key relationships

\dfrac{1}{C} = \dfrac{1}{C_g} + \dfrac{\rho g}{A_t}

coefficient of a closed wellbore with free gas: the gas and the weight of the column act in series

q_{sf}B = 24\,C\left(\dfrac{\dd p_{ws}}{\dd\Delta t} - \dfrac{\dd p_\phi}{\dd\Delta t}\right)

wellbore balance with the redistribution pressure: m³/d, m³/MPa, MPa, h

\dfrac{1}{C_a} = \dfrac{1}{C} + \dfrac{24A}{qB\,t_\phi}

apparent coefficient in the first minutes (exponential model)

p_c \approx \dfrac{\sum p_iV_{g,i}}{\sum V_{g,i}},\qquad p_{wsc} = p_c + \rho g h_l

pressure of the gas pocket (the gas collected at the top) and the bottomhole pressure of an isolated wellbore after the gas and liquid have separated

\Delta p' = m_e\left(1 + 2t\,\dfrac{\dd S}{\dd t}\right)

derivative of a drawdown test with a changing skin

S(t) = 1.151\left[\dfrac{p^* - p_{wf}(t)}{m} - \lg\dfrac{k\,t}{\phi\mu c_tr_w^2} + 2.098\right]

skin versus time from the drawdown points: MPa, mD, h, mPa·s, 1/MPa, m

Common pitfalls

Taking the peak of the build-up “hump” for the reservoir pressure, or drawing the semilog line through the points before and after it.
Taking the coefficient from the early unit slope as the measure of all of wellbore storage. When $C$ changes, it gives only the initial value $C_a$ .
Explaining a dip of the derivative by dual porosity or a layered reservoir without checking where the well is shut in and what is in the wellbore.
Calculating the permeability from the drawdown test of a new well while it is cleaning up.
Fitting an empirical wellbore storage model to a record that has no radial flow: the fit will find the parameters of the wellbore, not of the reservoir.

After this lesson you can

explain why the wellbore storage coefficient changes during a record, calculate the coefficient of a closed tubing with free gas, and tell increasing from decreasing wellbore storage
estimate, from the profile of pressure and gas content in the tubing, how far the pressure can rise when the gas redistributes after shut-in and how long this takes, and recognise the “hump” on a record
write the wellbore balance with the redistribution pressure and the empirical models of changing wellbore storage, and understand what fitting them gives and does not give
calculate the instantaneous wellbore storage coefficient from a measurement of the afterflow, the coefficient of a wellbore with a moving liquid level, and the correction for water rising under the gauge
show how well clean-up distorts the permeability and the skin found from a drawdown test, and calculate the skin as a function of time from the drawdown points and the results of a build-up
name the signs of each effect on the diagnostic plot (the log–log plot of the pressure change and its derivative), and the ways to check and remove them, above all shut-in at the bottom of the well

When the wellbore storage coefficient changes

The wellbore storage model of the lesson Wellbore storage rests on three assumptions: the wellbore is filled with one phase, the compressibility of the contents is constant, and the weight of the column does not change when the inflow comes. Then the coefficient $C$ is a single number for the whole record. These assumptions break down in several kinds of wells. One is an oil well with free gas in the tubing that is shut in at the wellhead. Another is a pumped well where the liquid level stands below the wellhead. Others are a gas well, and an injection well that “goes on vacuum” after shut-in (the water level in the tubing falls, and the pressure at the wellhead drops below atmospheric). In them the coefficient changes during the record, and the early section no longer reflects a single value of $C$ .

What changes the wellbore storage coefficient

Effect: Gas in the wellbore is compressed: its compressibility is close to $1/p$ and falls as the pressure rises
Where it occurs: gas wells; oil wells with free gas in the tubing
How $C$ changes: decreases

Effect: Gas rises and liquid settles (redistribution of phases)
Where it occurs: oil wells with gas in the tubing, shut in at the wellhead
How $C$ changes: early $C$ is small, then it grows; a pressure “hump” is possible

Effect: The liquid level rises
Where it occurs: pumped wells and wells whose liquid does not reach the wellhead
How $C$ changes: large; it decreases when the annulus (the space between the casing and the tubing) is closed

Effect: The level falls in an injection well
Where it occurs: the reservoir pressure is below the weight of a water column up to the wellhead
How $C$ changes: grows sharply when the wellhead pressure falls to atmospheric

Effect: Gas lift: the column drains down, and the valves and the annulus let gas through
Where it occurs: gas-lift wells (wells lifted by gas injected into the annulus) after the gas injection stops
How $C$ changes: unpredictable: humps and kinks

Effect	Where it occurs	How $C$ changes
Gas in the wellbore is compressed: its compressibility is close to $1/p$ and falls as the pressure rises	gas wells; oil wells with free gas in the tubing	decreases
Gas rises and liquid settles (redistribution of phases)	oil wells with gas in the tubing, shut in at the wellhead	early $C$ is small, then it grows; a pressure “hump” is possible
The liquid level rises	pumped wells and wells whose liquid does not reach the wellhead	large; it decreases when the annulus (the space between the casing and the tubing) is closed
The level falls in an injection well	the reservoir pressure is below the weight of a water column up to the wellhead	grows sharply when the wellhead pressure falls to atmospheric
Gas lift: the column drains down, and the valves and the annulus let gas through	gas-lift wells (wells lifted by gas injected into the annulus) after the gas injection stops	unpredictable: humps and kinks

The rise of water between the gauge and the perforated interval changes not $C$ but the correction of the pressure to the depth of the reservoir. It is covered below together with the levels.

Sometimes the wellbore is “stiffer” in the first minutes than later: a small amount of fluid is enough to change the pressure. Then the early unit-slope section gives a small coefficient, and later the pressure grows more slowly than along this line. We call this increasing wellbore storage. The general name for such cases is nonideal wellbore storage. If it is the other way round, we call it decreasing wellbore storage. In both cases the early unit slope gives not the “true” coefficient but the initial, apparent coefficient $C_a$ . Estimates built on it are wrong for such a record. One example is the time of the start of the semilog straight line (how long wellbore storage lasts).

We start with the most frequent case: free gas in a closed tubing. After shut-in, the inflow from the reservoir does two things. It compresses the gas, which frees room for the liquid. And it makes the column heavier, because there is more liquid in it. The wellhead pressure grows only from the compression of the gas. The bottomhole pressure also grows by the weight of the added liquid. These two “springs” are in series, so the wellbore storage coefficient at the bottom is smaller than each of them:

Storage coefficient of a closed wellbore with free gas

\frac{1}{C} = \frac{1}{C_g} + \frac{\rho g}{A_t},\qquad C_g = \sum_i \frac{V_{g,i}}{p_i} + c_lV_l

where:

$C$ wellbore storage coefficient at the depth of the gauge, m³/Pa
$C_g$ the part of the coefficient that comes from the compression of the contents: the volume that the gas and the liquid free up when the pressure rises by one unit (for an ideal gas the compressibility is $1/p$ ), m³/Pa
$V_{g,i},\ p_i$ volume of free gas and its absolute pressure in the $i$ -th interval of the wellbore, m³, Pa
$c_l,\ V_l$ compressibility and volume of the liquid in the wellbore, 1/Pa, m³
$\rho$ density of the liquid in the wellbore, kg/m³
$g$ acceleration of gravity, 9.81, m/s²
$A_t$ area of the cross-section through which the column rises: the inside cross-section of the tubing, m²

Units: SI

Holds when

The wellhead is closed, and the tubing is separated from the annulus by a packer.
The gas is ideal, and the temperature of each portion of gas is constant. The liquid is almost incompressible.
The gas stays at its own depth or is collected in a pocket under the wellhead. We ignore friction in a shut-in well.

The second term is the rise of the bottomhole pressure from each cubic metre of liquid that has entered the tubing. So in a closed tubing $C$ cannot exceed $A_t/(\rho g)$ , however much gas there is in it. For tubing with an inside diameter of 62 mm and oil of density 820 kg/m³, the limit is 0.375 m³/MPa. In a wellbore filled with one liquid, $C_g$ is hundredths of a cubic metre per megapascal, and the correction for the weight changes $C$ by only a few per cent. The formula $C = c_{wb}V_{wb}$ stays a good approximation. With free gas the weight of the column becomes the main term, and an estimate from the compressibility of the contents alone overestimates $C$ .

In other unit systems (2)

Metric oilfield units

\frac{1}{C} = \frac{1}{C_g} + 10^{-6}\,\frac{9.81\,\rho}{A_t}

$C$ and $C_g$ in m³/MPa; $p_i$ in MPa; $V$ in m³; $c_l$ in 1/MPa; $\rho$ in kg/m³; $A_t$ in m².

$10^{-6}$ converts pascals to megapascals: the second term is the rise of the bottomhole pressure in MPa per cubic metre of liquid that has entered.

Oilfield units

\frac{1}{C} = \frac{1}{C_g} + \frac{5.615\,\rho}{144\,A_t}

$C$ and $C_g$ in bbl/psi; $V$ in bbl; $p$ in psia (absolute pressure in psi); $\rho$ in lb/ft³; $A_t$ in ft².

$\rho/144$ is the pressure gradient of the column in psi/ft; 5.615 ft³ per barrel converts the volume of the inflow into the height of the column.

The lesson's flowing well at the moment of shut-in: the profile in the tubing (assumed data)

Tubing interval, m: 0–720
Mean pressure, MPa: 2.9
Gas share: 0.45
Gas volume, m³: 0.9782

Tubing interval, m: 720–1440
Mean pressure, MPa: 6.8
Gas share: 0.25
Gas volume, m³: 0.5434

Tubing interval, m: 1440–2160
Mean pressure, MPa: 11.7
Gas share: 0.10
Gas volume, m³: 0.2174

Tubing interval, m: 2160–2880
Mean pressure, MPa: 17.1
Gas share: 0.03
Gas volume, m³: 0.0652

Tubing interval, m	Mean pressure, MPa	Gas share	Gas volume, m³
0–720	2.9	0.45	0.9782
720–1440	6.8	0.25	0.5434
1440–2160	11.7	0.10	0.2174
2160–2880	17.1	0.03	0.0652

A flowing well is a well that flows by the pressure of the reservoir alone. The tubing has an inside diameter of 62 mm down to a packer at 2880 m, and there are 20 m of casing with an inside diameter of 150.5 mm below the packer. The gauge is at 2900 m, at the middle of the perforated interval. The wellhead pressure is 1.3 MPa and the bottomhole pressure is 20.21 MPa. All pressures are absolute. The oil in the wellbore has a density of 820 kg/m³ and a compressibility of $1.6\cdot10^{-3}$ MPa⁻¹. The gas share is the volume fraction of free gas in the pipe. Programs that calculate the flow up the tubing (lift programs) give such a profile (Multiphase flow in tubing). The gas volume is calculated in the example below.

Worked exampleCoefficient of a closed tubing with gas at the moment of shut-in and after the gas and oil have separated

Given (assumed data)

Tubing profile: the table above: four intervals of 720 m each
Tubing cross-section $A_t$: $\pi\cdot0.062^2/4 = 3.019\cdot10^{-3}$ m²
Liquid: oil 820 kg/m³, $c_l = 1.6\cdot10^{-3}$ MPa⁻¹; below the packer 0.3558 m³
Gas after it has separated: a pocket under the wellhead at the pressure $p_c$ = 5.648 MPa (calculated in the next section)

Solution

Volume of one interval: $3.019\cdot10^{-3}\cdot720 = 2.1737$ m³; the gas in the intervals is 0.9782, 0.5434, 0.2174 and 0.0652 m³, in all $V_g$ = 1.8042 m³. The liquid in the tubing is $4\cdot2.1737 - 1.8042 = 6.8906$ m³, and with the casing below the packer $V_l$ = 7.2464 m³.
The gas at the moment of shut-in: $0.9782/2.9 + 0.5434/6.8 + 0.2174/11.7 + 0.0652/17.1 = 0.4396$ m³/MPa; the liquid: $1.6\cdot10^{-3}\cdot7.2464 = 0.0116$ m³/MPa. The sum is $C_g$ = 0.4512 m³/MPa.
The weight of the column: $10^{-6}\cdot9.81\cdot820/(3.019\cdot10^{-3}) = 2.665$ MPa per cubic metre. The coefficient at the bottom: $1/C = 1/0.4512 + 2.665 = 4.881$ , $C$ = 0.205 m³/MPa. This is 2.2 times less than the compressibility of the contents alone would give.
After the gas has separated, all of it is in the pocket: $C_g = 1.8042/5.648 + 0.0116 = 0.3310$ m³/MPa; $1/C = 3.021 + 2.665 = 5.686$ , $C$ = 0.176 m³/MPa.
For comparison, the same tubing filled with oil alone: $C_g = 1.6\cdot10^{-3}\cdot9.051 = 0.01448$ m³/MPa, and with the correction for the weight, $C$ = 0.01394 m³/MPa, which is smaller by only 3.7%.

Answer: At the moment of shut-in,

C \approx 0.205

m³/MPa. After the gas separates, it is about 0.176 m³/MPa. Both values are below the limit

A_t/(\rho g)

= 0.375 m³/MPa.

What it means. The coefficient changes already because the gas collects in a pocket at a higher pressure. Then it keeps decreasing: the inflow compresses the pocket, the pressure in it rises, and the volume that the pocket frees for each unit of pressure rise, $V_g/p_c$ , falls. So wellbore storage in such a well decreases on its own, and the redistribution of the phases acts on top of this (the next section). An estimate from the compressibility of the contents alone, without the weight of the column, is wrong here by a factor of two. The data in this example are assumed for illustration.

We can measure the instantaneous wellbore storage coefficient. Suppose we record the afterflow together with the pressure, with a downhole flowmeter (a spinner, a small propeller that turns with the flow) a little above the perforated interval (production logging). Then the wellbore balance gives $C$ at each moment:

Instantaneous wellbore storage coefficient from the afterflow

C(\Delta t) = \frac{q_{sf}\,B}{\dd p_{ws}/\dd\Delta t}

where:

$C(\Delta t)$ wellbore storage coefficient at the time $\Delta t$ after shut-in, m³/Pa
$q_{sf}$ sandface flow rate: the afterflow at standard conditions, m³/s
$B$ formation volume factor of the fluid (its volume in the reservoir per unit volume at the surface), m³/m³
$\dd p_{ws}/\dd\Delta t$ the rate of growth of the bottomhole pressure: the derivative with respect to time, not to its logarithm: $\Delta p'/\Delta t$ , Pa/s

Units: SI

Holds when

The well is shut in at the wellhead, and the whole inflow goes to fill the wellbore.
The flowmeter is between the reservoir and the gauge, and it measures the whole inflow.

This is the wellbore balance from the lesson Wellbore storage, solved for $C$ . For a constant $C$ the ratio is the same along the whole afterflow section. If it changes, the coefficient changes too.

In other unit systems (1)

Metric oilfield units

C(\Delta t) = \frac{q_{sf}\,B}{24\,\dd p_{ws}/\dd\Delta t}

$q_{sf}$ in m³/d, $\dd p_{ws}/\dd\Delta t$ in MPa/h, $C$ in m³/MPa.

24 is the number of hours in a day.

Worked exampleDecreasing wellbore storage from a flowmeter record

Given (assumed data)

Well: another well of the same reservoir ( $k$ = 25 mD, $h$ = 12 m, $\mu$ = 2.5 mPa·s, $S$ = 5), 20 m³/d, $B$ = 1.12; at shut-in there is a lot of gas at low pressure in the tubing
The first point of the record: after 18 s the pressure rose by 0.0155 MPa
Flowmeter and pressure: 0.05 h: 18.95 m³/d and 2.858 MPa/h; 0.5 h: 12.14 and 2.383; 1 h: 6.36 and 1.610; 2 h: 1.83 m³/d and 0.475 MPa/h

Solution

The early unit slope: $C_a = qB\Delta t/(24\Delta p) = 20\cdot1.12\cdot0.005/(24\cdot0.0155) = 0.301$ m³/MPa.
The coefficient from the flowmeter: $C = q_{sf}B/(24\,\dd p/\dd t)$ . At 0.05 h, $18.95\cdot1.12/(24\cdot2.858) = 0.309$ ; at 0.5 h, 0.238; at 1 h, 0.184; at 2 h, 0.180 m³/MPa.
In the first hour the coefficient fell by a factor of 1.7, and after that it hardly changes: the wellbore storage is decreasing.
The classic estimate of the start of the line, $t \approx 44.8\,(60 + 3.5S)\,\mu C/(kh)$ , with the early $C_a$ : $44.8\cdot77.5\cdot2.5\cdot0.301/300 = 8.7$ h; with the late $C$ = 0.18 m³/MPa, 5.2 h.
By the model from which the record was calculated, the derivative gets within 10% of the level without wellbore storage after 19 h. A well with a constant $C$ = 0.301 would get there after 30 h, and with a constant 0.18, after 17 h.

Answer: The coefficient decreases from 0.309 to 0.180 m³/MPa in the first two hours. The derivative plateau comes after about 19 h, earlier than the early unit slope promises (30 h).

What it means. The flowmeter shows what a pressure record alone shows only indirectly: the early section belongs to a “soft” wellbore, and the late section to a “stiff” one. If we plan the duration from the early $C_a$ , we get a margin of about one and a half times. If we fit a type curve with a constant $C$ to the record, the transition section will not match any curve (Type-curve matching). Gas at low pressure in the wellbore is the cause of the decrease: as the pressure rises, the volume that it frees for each unit of pressure rise, $V_g/p$ , falls fast. The data were calculated with the error-function model (the section on empirical models). The data in this example are assumed for illustration.

CalculatedCalculated build-up derivatives of the lesson's well with changing wellbore storage (solid lines) and with a constant coefficient equal to the early apparent

C_a

(dashed). Top: increasing wellbore storage with gas redistribution. The hump is higher, then the derivative becomes negative and breaks off on the logarithmic plot (a logarithmic axis cannot show a negative value). Bottom: decreasing wellbore storage. After the early section the derivative rises above the dashed one and falls to the plateau before it. The horizontal line is the level of radial flow.

Redistribution of gas and liquid: the hump on a build-up

While the well produces, the gas is spread through the tubing in bubbles. There is little of it at the bottom, and towards the wellhead the share of gas grows together with its volume. When we shut the well in at the wellhead, the bubbles start to rise through the oil, and the oil drains down. After a few hours the gas has collected in a pocket under the wellhead. This is the phase redistribution. The oil is almost incompressible, and the volume of the tubing is constant, so the total volume of the gas hardly changes either. But a bubble that has risen from a depth where the pressure was 17 MPa into a zone where it is 3 MPa cannot expand as it would in an open pipe. The gas “carries” the high pressure of the depth up to the wellhead, and the pressure in the whole wellbore rises. This is the same effect as the rise of a gas kick (an influx of gas) in a shut-in well during drilling (gas in the well), except that here the gas is our own, released from the oil.

SchematicA sketch: the tubing of the lesson's well at the moment of shut-in and after the gas and oil have separated, if the wellbore were isolated from the reservoir. The volume of the gas is the same in both cases, 1.8 m³. The pressure in the pocket equals the mean pressure of the gas before it separated, and at the bottom the weight of the oil column is added to it. Not to scale.

How far can the pressure rise? Imagine that the wellbore is isolated from the reservoir (a very large skin): the gas has risen, the oil has settled, and there is no inflow. The gas pocket takes the same volume that the dispersed gas took. Its pressure at the same temperature is the mean pressure of the gas before it separated, with each interval counted in proportion to its gas volume. The bottomhole pressure is the pressure of the pocket plus the weight of the oil column below it:

Pressure in a closed wellbore after the gas and the liquid have separated

p_c = \frac{\sum_i p_i V_{g,i}}{\sum_i V_{g,i}},\qquad p_{wsc} = p_c + \rho g h_l,\qquad A \approx p_{wsc} - p_{wf}

where:

$p_c$ pressure of the gas pocket after the gas and liquid have separated, Pa
$p_i,\ V_{g,i}$ mean absolute pressure and volume of free gas in the $i$ -th interval of the wellbore at the moment of shut-in, Pa, m³
$p_{wsc}$ pressure at the depth of the gauge in the isolated wellbore after the gas and liquid have separated, Pa
$h_l$ height of the liquid column above the gauge after the gas and liquid have separated, m
$\rho,\ g$ density of the liquid and acceleration of gravity, kg/m³, m/s²
$A$ amplitude of the redistribution pressure: by how much the separating of gas and liquid could raise the pressure over and above the inflow, Pa
$p_{wf}$ bottomhole pressure at the moment of shut-in, Pa

Units: SI; the pressures are absolute

Holds when

The wellbore is isolated from the reservoir. This is an upper estimate for a well that manages to take part of the liquid back.
The gas is ideal and does not dissolve in the oil while the phases separate, and its temperature does not change.
We ignore the weight of the gas pocket and the friction (the weight of the pocket would raise the estimate by tenths of a megapascal).

The oil weighs as much as it did: it has only moved down. So the rise of the pressure at the bottom, compared with the producing well, is almost the difference between the mean pressure of the gas and the wellhead pressure. The gas that was “below and under pressure” has ended up at the top. The deeper the well, the more gas at depth, and the lower the wellhead pressure, the larger $A$ .

How long this takes is set by the rise speed of the bubbles relative to the oil at rest, usually of the order of 0.1–0.3 m/s (Multiphase flow in tubing). The time for the gas from the depth $z$ to reach the pocket is $z/v_s$ . Averaged over the volume of the gas, it gives the order of the redistribution time $t_\phi$ .

Worked exampleHow much the redistribution of gas can give in the lesson's well

Given (assumed data)

Tubing profile at shut-in: the table in the previous section; 0.9782, 0.5434, 0.2174 and 0.0652 m³ of gas in the intervals with middles at 360, 1080, 1800 and 2520 m
Wellbore: tubing down to 2880 m, $A_t = 3.019\cdot10^{-3}$ m², 20 m of casing below the packer; the gauge at 2900 m
Pressures: at the bottom at the moment of shut-in 20.21 MPa, initial reservoir pressure 25.0 MPa
Rise speed of the bubbles $v_s$: 0.2 m/s (assumed for the example)

Solution

Pressure of the pocket: $(2.9\cdot0.9782 + 6.8\cdot0.5434 + 11.7\cdot0.2174 + 17.1\cdot0.0652)/1.8042 = 10.1904/1.8042 = 5.648$ MPa.
The oil column: in the tubing $2880 - 1.8042/(3.019\cdot10^{-3}) = 2282.4$ m, and with the casing below the packer $h_l$ = 2302.4 m. Its weight is $10^{-6}\cdot820\cdot9.81\cdot2302.4 = 18.521$ MPa.
Pressure of the isolated wellbore after the gas and liquid have separated: $p_{wsc} = 5.648 + 18.521 = 24.169$ MPa. The amplitude: $A \approx 24.169 - 20.21 = 3.96$ MPa; the drawdown at the moment of shut-in is $25.0 - 20.21 = 4.79$ MPa.
The mean depth of the gas, weighted by volume: $(360\cdot0.9782 + 1080\cdot0.5434 + 1800\cdot0.2174 + 2520\cdot0.0652)/1.8042 = 828$ m. The rise from it takes $828/0.2 = 4140$ s, that is, 1.15 h. From the depth of the packer it takes $2880/0.2 = 14{,}400$ s, or 4 h.

Answer: The redistribution of the gas can raise the pressure in the wellbore by about 3.96 MPa in a time from one to four hours. This is comparable with the drawdown (4.79 MPa).

What it means. When the possible rise is comparable with the drawdown, in a few hours the pressure in the wellbore can catch up with the pressure in the reservoir at the well and pass it. Then the liquid goes back into the reservoir, and a hump appears on the build-up. This is an upper estimate: part of the rise is “eaten” by the inflow, which keeps compressing the gas. For the model in the next example, the amplitude is rounded to 4.0 MPa, the redistribution time is taken as 2 h (the middle of the range we found), and the wellbore storage coefficient is 0.18 m³/MPa, from the previous example. The data in this example are assumed for illustration.

Shut-in at the wellhead. With shut-in at the bottom, the gas stays above the valve, and the gauge does not “see” it. Redistribution is an effect of a build-up with shut-in at the wellhead.
A deep well. The longer the column, the more pressure the gas carries up, and the longer it takes to rise.
An almost liquid column. There is little gas in the tubing, but it is spread over a long length at a noticeable pressure, and the wellhead pressure is low. Then the mean “transfer” of pressure by the gas is large compared with the drawdown.
A large skin and a low flow capacity. The wellbore is poorly connected with the reservoir, and the extra pressure has no time to go into the reservoir, so the rise shows itself more fully.
Only after shut-in. At start-up the gas leaves with the flow. So drawdown tests have no humps of this kind: wellbore storage at start-up and at shut-in can be different.

How it looks on a record. In the first minutes the pressure rises faster than at a constant $C$ , because the rise from the rising gas is added to the inflow. Then the pressure in the wellbore overtakes the pressure in the reservoir at the well, and some of the liquid goes back. The pressure passes a maximum, a hump, and for some time it falls. In the theory of how pressure spreads through the reservoir (the diffusion problem), with a constant rate at the bottom of the well, this is impossible: after shut-in the pressure can only rise. So any fall of the pressure at the start of a build-up is a sign of an effect in the wellbore, not in the reservoir.

CalculatedCalculated build-up of the lesson's well with gas redistribution (solid line) and without it, at the same coefficient 0.18 m³/MPa (dashed). Top: the pressure rise on ordinary axes, with the pressure axis starting at 3 MPa: a maximum at 2.75 h, a minimum at 7.4 h. Bottom: the rate of the pressure rise on an ordinary (linear) scale. Between the maximum and the minimum it is negative.

Build-up of the lesson's well with gas redistribution (assumed data)

$\Delta t$: 18 s
$\Delta p$ , MPa: 0.0357
Without redistribution, MPa: 0.0258

$\Delta t$: 3 min
$\Delta p$ , MPa: 0.3417
Without redistribution, MPa: 0.2474

$\Delta t$: 15 min
$\Delta p$ , MPa: 1.4364
Without redistribution, MPa: 1.0554

$\Delta t$: 30 min
$\Delta p$ , MPa: 2.3670
Without redistribution, MPa: 1.7719

$\Delta t$: 1 h
$\Delta p$ , MPa: 3.3763
Without redistribution, MPa: 2.6256

$\Delta t$: 2 h
$\Delta p$ , MPa: 3.9688
Without redistribution, MPa: 3.3144

$\Delta t$: 2.75 h
$\Delta p$ , MPa: 4.0239
Without redistribution, MPa: 3.5119

$\Delta t$: 4 h
$\Delta p$ , MPa: 3.9813
Without redistribution, MPa: 3.6651

$\Delta t$: 6 h
$\Delta p$ , MPa: 3.9203
Without redistribution, MPa: 3.7780

$\Delta t$: 7.4 h
$\Delta p$ , MPa: 3.9106
Without redistribution, MPa: 3.8266

$\Delta t$: 10 h
$\Delta p$ , MPa: 3.9270
Without redistribution, MPa: 3.8908

$\Delta t$: 16 h
$\Delta p$ , MPa: 3.9955
Without redistribution, MPa: 3.9836

$\Delta t$: 24 h
$\Delta p$ , MPa: 4.0661
Without redistribution, MPa: 4.0594

$\Delta t$: 48 h
$\Delta p$ , MPa: 4.1864
Without redistribution, MPa: 4.1834

$\Delta t$: 100 h
$\Delta p$ , MPa: 4.3100
Without redistribution, MPa: 4.3086

$\Delta t$	$\Delta p$ , MPa	Without redistribution, MPa
18 s	0.0357	0.0258
3 min	0.3417	0.2474
15 min	1.4364	1.0554
30 min	2.3670	1.7719
1 h	3.3763	2.6256
2 h	3.9688	3.3144
2.75 h	4.0239	3.5119
4 h	3.9813	3.6651
6 h	3.9203	3.7780
7.4 h	3.9106	3.8266
10 h	3.9270	3.8908
16 h	3.9955	3.9836
24 h	4.0661	4.0594
48 h	4.1864	4.1834
100 h	4.3100	4.3086

$\Delta p = p_{ws} - p_{wf}(\Delta t = 0)$ , $p_{wf}(\Delta t = 0)$ = 20.205 MPa. Producing time before shut-in 1500 h at 20 m³/d; reservoir: $k$ = 25 mD, $h$ = 12 m, $\phi$ = 0.18, $\mu$ = 2.5 mPa·s, $c_t = 1.4\cdot10^{-3}$ MPa⁻¹, $B$ = 1.12, $r_w$ = 0.108 m, $S$ = 5. The model: a homogeneous infinite reservoir, $C$ = 0.18 m³/MPa, the exponential model of redistribution with $A$ = 4.0 MPa and $t_\phi$ = 2 h (the section below); the values are rounded to 0.0001 MPa.

Worked exampleWhat we can and cannot read from a record with a hump

Given (assumed data)

Record: the table above
Well: $q$ = 20 m³/d, $B$ = 1.12, $\mu$ = 2.5 mPa·s, $h$ = 12 m
Derivative at the late points: from 24 to 48 h, 0.173 MPa on average

Solution

The early unit slope from the first point: $C_a = 20\cdot1.12\cdot0.005/(24\cdot0.0357) = 0.131$ m³/MPa. This is less than the estimates from the design give (0.18–0.20). The wellbore storage is increasing.
The hump: a maximum of 4.0239 MPa at 2.75 h, then a fall to 3.9106 MPa at 7.4 h, by 0.113 MPa. The pressure regains the level of the maximum only by 19 h.
A line through the points at 1 and 2 h would give a slope of $(3.9688 - 3.3763)/\lg 2 = 1.968$ MPa per cycle and a permeability of $2.149\cdot20\cdot1.12\cdot2.5/(1.968\cdot12) = 5.1$ mD.
Only the late points see the reservoir. From the derivative level: $kh = 0.933\cdot20\cdot1.12\cdot2.5/0.173 = 302$ mD·m, $k$ = 25.2 mD.

Answer: The hump takes the first 19 h of the record. We get the permeability only from the points after a day, about 25.2 mD. A line through the early points would have underestimated it by a factor of 5.

What it means. The peak of the hump (4.0239 MPa above the shut-in pressure) is not the reservoir pressure: after a day the pressure is above it by only 0.042 MPa, and after 100 h by 0.29 MPa, and it keeps rising. We calculate the skin from such a record at the points on the plateau, after the hump. It makes no sense to match the early part to type curves with a constant $C$ . The model from which the record was calculated has a true permeability of 25 mD. The data in this example are assumed for illustration.

The small details of a record show best on an ordinary plot, not on a logarithmic one. Plot the pressure against time on an enlarged scale. Plot also the rate of the pressure rise $\dd p_{ws}/\dd\Delta t$ (the derivative with respect to time, not to its logarithm) on a linear scale, where negative values are visible too. We cannot draw a negative logarithmic derivative on a log–log plot: on it a hump looks like a break in the curve. A record of the wellhead pressure helps too. In redistribution it rises much more than in ordinary inflow, because the gas pocket collects exactly there.

Empirical models of changing wellbore storage

An exact calculation of the redistribution needs a model of two-phase flow (gas and liquid moving together) in the wellbore, connected with a model of the reservoir. In practice we do it more simply. We add to the wellbore balance a “redistribution pressure” $p_\phi(\Delta t)$ , a source of pressure that acts by itself, independently of the inflow:

Wellbore balance with the redistribution pressure

q_{sf}B = C\left(\frac{\dd p_{ws}}{\dd\Delta t} - \frac{\dd p_\phi}{\dd\Delta t}\right)

where:

$q_{sf}$ inflow from the reservoir into the wellbore after shut-in, at standard conditions, m³/s
$C$ wellbore storage coefficient after the redistribution is over, m³/Pa
$p_{ws}$ pressure in the wellbore at the depth of the gauge, Pa
$p_\phi$ redistribution pressure: how much the pressure in an isolated wellbore would have risen by the time $\Delta t$ ; $p_\phi(0) = 0$ , Pa
$B,\ \Delta t$ formation volume factor; time since shut-in, m³/m³, s

Units: SI

Holds when

Shut-in at the wellhead; the rate at the wellhead after shut-in is zero.
$C$ is constant, and everything that changes the behaviour of the wellbore is collected in the function $p_\phi$ .
$p_\phi$ is a given function of time that does not depend on the pressure in the reservoir: the model is empirical.

If $p_\phi$ grows, the pressure in the wellbore rises even without inflow, and when it overtakes the reservoir pressure, $q_{sf}$ becomes negative: the liquid goes into the reservoir. If $p_\phi$ is negative and decreasing, the wellbore takes in more inflow than a constant $C$ would allow, and the wellbore storage is decreasing.

In other unit systems (2)

Metric oilfield units

q_{sf}B = 24\,C\left(\frac{\dd p_{ws}}{\dd\Delta t} - \frac{\dd p_\phi}{\dd\Delta t}\right)

$q_{sf}$ in m³/d; $C$ in m³/MPa; pressures in MPa; $\Delta t$ in h.

24 is the number of hours in a day, as in the wellbore balance of the lesson Wellbore storage.

Oilfield units

q_{sf}B = 24\,C\left(\frac{\dd p_{ws}}{\dd\Delta t} - \frac{\dd p_\phi}{\dd\Delta t}\right)

$q_{sf}$ in STB/d; $C$ in bbl/psi; pressures in psi; $\Delta t$ in h.

The same constant 24.

Two empirical forms of the redistribution pressure

p_\phi = A\left(1 - \e^{-\Delta t/t_\phi}\right)\quad\text{or}\quad p_\phi = A\,\operatorname{erf}\frac{\Delta t}{t_\phi}

where:

$A$ amplitude: the value to which $p_\phi$ tends; above zero the pressure rises (gas redistribution, humps), below zero the wellbore storage decreases, Pa
$t_\phi$ redistribution time; for the exponential form, 63% of the change happens in this time, s
$\operatorname{erf}$ the error function (a smooth S-shaped curve), $\operatorname{erf}x = \frac{2}{\sqrt\pi}\int_0^x \e^{-u^2}\dd u$ ; at $x = 1$ it equals 0.843, —

Units: any consistent units: $A$ in units of pressure, $t_\phi$ in units of time

Holds when

The forms were chosen to fit observations, and were not derived from the physics of the wellbore.
We find $A$ and $t_\phi$ by fitting the model to the record. They tell us only roughly about the wellbore (the amplitude is of the order of the estimate for an isolated wellbore, and the time is of the order of the rise time of the bubbles).

The exponential form was proposed by W. B. Fair for the humps from gas redistribution. The error-function form (proposed by Hegeman, Hallford and Joseph) rises and levels off more sharply. With a negative amplitude it describes decreasing wellbore storage better. At the first moment both forms give the greatest rate of growth of $p_\phi$ : $A/t_\phi$ for the exponential form and $2A/(\sqrt\pi\,t_\phi)$ for the second.

In the first minutes after shut-in the inflow is still almost equal to the rate before shut-in, $q_{sf} \approx q$ , and the balance gives a rate of pressure rise $qB/(24C) + \dd p_\phi/\dd\Delta t$ . So the early unit slope gives an apparent coefficient:

Apparent coefficient at the start of the record

\frac{1}{C_a} = \frac{1}{C} + \frac{24A}{qB\,t_\phi}\quad\text{(exponential form)},\qquad \frac{1}{C_a} = \frac{1}{C} + \frac{48A}{\sqrt\pi\,qB\,t_\phi}\quad\text{(error function)}

where:

$C_a$ the coefficient that the early unit slope would give, m³/MPa
$C$ coefficient after the redistribution, m³/MPa
$A,\ t_\phi$ amplitude and time of the redistribution, MPa, h
$q,\ B$ rate before shut-in at standard conditions and formation volume factor, m³/d, m³/m³

Units: metric oilfield: m³/MPa, MPa, h, m³/d

Holds when

The start of the record, while the inflow is still almost equal to the rate before shut-in.
$t_\phi$ is much larger than the time of the early readings.

For $A > 0$ the apparent coefficient is smaller than the true one: at the start the pressure grows faster than the inflow alone would give. For $A < 0$ it is larger. With this relation we find $C$ when we know $C_a$ from the unit slope and $A$ and $t_\phi$ from the shape of the hump. And the other way round, we use $C$ from the design of the wellbore to check whether the fitted model agrees with the early points.

Worked exampleParameters of the model and the early points of the record

Given (assumed data)

Model of the record with a hump: $C$ = 0.18 m³/MPa, $A$ = 4.0 MPa, $t_\phi$ = 2 h; $q$ = 20 m³/d, $B$ = 1.12
Model of the record with decreasing wellbore storage: $C$ = 0.18 m³/MPa, error function, $A$ = −0.92 MPa, $t_\phi$ = 0.5 h

Solution

Hump: $1/C_a = 1/0.18 + 24\cdot4.0/(22.4\cdot2) = 5.556 + 2.143 = 7.699$ , $C_a$ = 0.130 m³/MPa. The first point of the record gave 0.131.
Decrease: $48\cdot(-0.92)/(\sqrt\pi\cdot22.4\cdot0.5) = -2.225$ ; $1/C_a = 5.556 + (-2.225) = 3.331$ , $C_a$ = 0.300 m³/MPa. The first point of the record gave 0.301.
The greatest rate of redistribution at the first moment: for the hump $4.0/2 = 2.0$ MPa/h, against the rate from the inflow of $22.4/(24\cdot0.18) = 5.19$ MPa/h.

Answer: The early points of both records agree with the models:

C_a

≈ 0.130 m³/MPa for the hump and 0.300 m³/MPa for decreasing wellbore storage, at the same

C

= 0.18 m³/MPa.

What it means. Three parameters, $C$ , $A$ and $t_\phi$ , are easy to adjust to almost any early part of a record. So a match between a model and the data does not yet prove the mechanism. A useful check is a physical one. Does $A$ agree with the estimate for an isolated wellbore? Does $t_\phi$ agree with the rise time of the bubbles? Does $C$ agree with the design and the contents of the wellbore? Does the wellhead pressure rise as a gas pocket requires? The data in this example are assumed for illustration.

What the models give. A description of the early part of the record, so that it does not interfere with the fit of the whole reservoir model, and a consistent estimate of $C$ , if the record has radial flow.
What they do not give. Predictions: we cannot calculate the parameters beforehand from the physics of the wellbore, and we cannot plan a test with the model. Uniqueness: with three free parameters of the wellbore, there are many suitable combinations.
When they are useless. If a hump or a transition section hides radial flow, no wellbore model will bring back the permeability and the skin: they are simply not in the record. Then we repeat the test with shut-in at the bottom of the well.
How to treat the early points in a fit. If the wellbore model clearly does not fit, we leave the early points out of the objective function (the measure of misfit that the fit makes as small as possible) or give them a small weight. Otherwise they, being numerous on the logarithmic time scale, would distort the reservoir properties that we find (Type-curve matching).

The liquid level in the wellbore

In a pumped well, and in a well that does not flow by itself, the liquid stands below the wellhead. After shut-in the inflow raises the level, in the annulus of a pumped well or in the tubing. The wellbore then stores fluid by the rise of the column: $C = A/(\rho g)$ (the wellbore storage coefficient). A record of this rise is called a liquid-level recovery curve. An echo sounder (an instrument that finds the level by the echo of a sound pulse) measures the level. From the level and the gas pressure in the annulus we calculate the bottomhole pressure (dynamic fluid level). If the annulus is closed at the wellhead, there is a gas pocket above the level, and the inflow compresses it too. This is the same series pair of “springs” as in a closed tubing:

Wellbore storage coefficient for a rising level under a gas pocket

\frac{1}{C} = \frac{p_g}{V_g} + \frac{\rho g}{A},\qquad p_gV_g = \text{const}

where:

$p_g,\ V_g$ absolute pressure and volume of the gas above the level, Pa, m³
$A$ area of the cross-section in which the level rises (the annulus or the tubing), m²
$\rho$ density of the liquid, kg/m³

Units: SI; in metric oilfield units the second term is $10^{-6}\cdot9.81\,\rho/A$ MPa/m³

Holds when

The gas above the level is ideal, and its mass and temperature do not change (no gas comes out of the oil).
The liquid is incompressible, and there is no friction.

If the annulus is open to a line at constant pressure, the gas leaves and the first term disappears: $C = A/(\rho g)$ . If it is closed, then as the level rises the volume of the pocket gets smaller, the pressure in it grows, and the first term grows as $p_g^2$ . The wellbore storage decreases, and it decreases faster the closer the level is to the wellhead.

Worked exampleA liquid-level recovery curve of a pumped well with the annulus open and closed

Given (assumed data)

Annulus: casing with an inside diameter of 150.5 mm, tubing of 73 mm
At shut-in: dynamic level at a depth of 1400 m, gas pressure in the annulus 0.6 MPa (absolute)
Liquid: oil 850 kg/m³
Reservoir and well: the same as for the lesson's well: $k$ = 25 mD, $h$ = 12 m, $\mu$ = 2.5 mPa·s, $S$ = 5

Solution

Cross-section of the annulus: $\pi(0.1505^2 - 0.073^2)/4 = 0.0136$ m²; the term for the rising level is $10^{-6}\cdot9.81\cdot850/0.0136 = 0.6131$ MPa/m³. The annulus is open: $C = 1/0.6131 = 1.631$ m³/MPa.
The annulus is closed, at the start: the gas is $0.0136\cdot1400 = 19.040$ m³, $p_g/V_g = 0.6/19.040 = 0.0315$ ; $C = 1/(0.0315 + 0.6131) = 1.551$ m³/MPa.
The level has risen by 700 m: the gas is compressed twofold, $p_g$ = 1.2 MPa, $V_g$ = 9.520 m³, $p_g/V_g$ = 0.1261; $C$ = 1.353 m³/MPa. After another 350 m: $p_g$ = 2.4 MPa, $V_g$ = 4.760 m³; $C$ = 0.895 m³/MPa.
The start of the semilog straight line by the classic estimate at $C$ = 1.631 m³/MPa: $44.8\cdot77.5\cdot2.5\cdot1.631/300 = 47$ h, that is, 2.0 days.

Answer: The rise of the level gives

C

of about 1.631 m³/MPa. This is 8–9 times more than for the closed tubing of the lesson's flowing well. With the annulus closed, the coefficient falls to 0.895 m³/MPa as the level rises.

What it means. By the classic estimate, the line in such a level recovery curve starts only after two days. The derivative reaches the plateau even 2–4 times later, after 4–8 days (how long wellbore storage lasts). So the level recovery curves of pumped wells more often give the productivity and the reservoir pressure, from the late points and a model fit, than the permeability from a line. A closed annulus shortens wellbore storage, but it makes it decreasing. For the analysis we need a record of the gas pressure in the annulus: without it, the conversion of the level to the bottomhole pressure is wrong. The data in this example are assumed for illustration.

A falling level in an injection well. If the reservoir pressure is less than the weight of a column of water from the bottom to the wellhead, then after the injection stops, the wellhead pressure falls to atmospheric. After that the water keeps going into the reservoir, and the water level in the tubing falls below the wellhead, so the pressure at the wellhead drops below atmospheric: the well “goes on vacuum”. While the wellbore is full, the coefficient is the compression of the water, $c_wV_{wb}$ : thousandths of a cubic metre per megapascal. When the level starts to go down, it becomes $A/(\rho_wg)$ . For 62 mm tubing this is 0.31 m³/MPa, tens of times more. On the diagnostic plot of the pressure falloff after injection (the fall of pressure after the injection stops) we see a first unit slope, then a dip of the derivative, and then a second unit-slope section. This section is shifted to the right by as many times as the coefficient has grown. Taking this dip for dual porosity or for a change of permeability is a frequent mistake.

Water below the gauge. The gauge usually stands above the perforated interval, and we convert its pressure to the depth of the reservoir with the density of the fluid in the wellbore. If the well produces water, after shut-in the water settles, and the water level rises from the bottom to the gauge. While it rises, the correction changes. The pressure at the gauge grows more slowly than at the bottom, by $(\rho_w - \rho_o)g$ for each metre of rise of the water. When the water has reached the gauge, the correction is constant again, but now with water. Two gauges a few metres apart help us to recognise this: the difference of their readings is the density of the fluid between them.

Worked exampleWater rising below the gauge

Given (assumed data)

Gauge: 45 m above the middle of the perforated interval (vertically)
Densities in the wellbore: oil 760 kg/m³, water 1080 kg/m³
Second gauge: 3.0 m below the first
Slope of the semilog straight line: 0.40 MPa per cycle

Solution

The correction to the depth of the reservoir with oil: $10^{-6}\cdot760\cdot9.81\cdot45 = 0.3355$ MPa; with water: $10^{-6}\cdot1080\cdot9.81\cdot45 = 0.4768$ MPa. While the water rises to the gauge, the correction changes by 0.1413 MPa.
This is 35% of the slope of the line. If the water rises over a few hours, the derivative will show a dip of the same order.
The difference of the readings of the two gauges: with oil between them, $10^{-6}\cdot760\cdot9.81\cdot3.0 = 0.0224$ MPa; with water, 0.0318 MPa; the difference is 9.4 kPa.

Answer: The rise of water distorts the pressure converted to the reservoir by 0.1413 MPa. A pair of gauges shows the moment when the water has reached them, by a jump in the difference of the readings of 9.4 kPa.

What it means. Quartz gauges distinguish a pressure change of a fraction of a kilopascal. So a jump of a few kilopascals is clearly visible, although each of the gauges reads about 20 MPa. The closer the gauge is to the perforated interval, the smaller the distortion. When we run instruments into a well that produces water, we should consider this beforehand. The data in this example are assumed for illustration.

Gas-lift wells. After the gas injection stops, the column in the tubing gets heavier, and its weight can exceed the reservoir pressure. The liquid drains into the reservoir, and the pressure passes a hump of the same kind as in gas redistribution. Leaking gas-lift valves and a connection of the tubing with the annulus bring the annulus with its gas into the wellbore, and the coefficient changes in jumps (Gas lift operation and optimisation). Such records can hardly be interpreted without shut-in at the bottom of the well.

Shut-in at the bottom and the test programme

Almost all the effects of this lesson arise in the wellbore above the perforated interval. Shut-in at the bottom of the well cuts this wellbore off. Below the valve there is only a small volume of liquid, the gauge hangs below the valve, and neither the redistribution of gas, nor the movement of a level, nor a connection with the annulus changes the pressure under the valve. A valve with a timer and its own power supply is lowered on a wire (wireline) and seated in a landing nipple (a seat machined in the tubing; Wireline operations). It closes by itself after a set time. After the record, the pressures above and below it are equalised, and the tool is pulled out.

To run the valve, we usually shut the well in at the wellhead for a short time, and then start it again. This short shut-in stays in the rate history, and its trace is visible on the future build-up: the derivative is a little lowered. Superposition shows by how much (Superposition principle):

The largest distortion of the derivative plateau after a short shut-in

\varepsilon_{\max} = \frac{\tau}{\left(\sqrt{t_f + \tau} + \sqrt{t_f}\right)^2} \approx \frac{\tau}{4\,t_f}

where:

$\varepsilon_{\max}$ the largest relative lowering of the build-up derivative compared with the radial-flow level, —
$\tau$ duration of the short shut-in at the wellhead for running the valve, h
$t_f$ time that the well produces after the short shut-in before the valve closes, h

Units: dimensionless; $\tau$ and $t_f$ in the same units

Holds when

Before the short shut-in, the well produced for a long time at the same rate, and it returned to it again.
Around the well there is radial flow; the wellbore storage after the short shut-in and the restart is over.

The short shut-in adds to the build-up the term $m_e\ln[(\Delta t + t_f + \tau)/(\Delta t + t_f)]$ . Its derivative with respect to $\ln\Delta t$ is negative, and the largest in absolute value at $\Delta t = \sqrt{(t_f + \tau)t_f}$ . The approximate form holds for $\tau \ll t_f$ : for the distortion to be less than 5%, the restarted flow must last at least five times longer than the short shut-in.

Worked exampleA test programme with shut-in at the bottom

Given (assumed data)

Short shut-in to run the valve: 2 h
The lesson's well: with shut-in at the wellhead $C$ ≈ 0.18 m³/MPa; below the valve at the packer, 0.3558 m³ of oil with $c_o = 1.6\cdot10^{-3}$ MPa⁻¹

Solution

Restart for 4 h: $\varepsilon_{\max} = 2/(\sqrt6 + \sqrt4)^2 = 0.101$ , that is, 10%. For 10 h: $2/(\sqrt{12} + \sqrt{10})^2 = 0.0455$ , 4.6%. The largest distortion comes $\sqrt{12\cdot10} = 11.0$ h after the valve closes.
The coefficient below the valve: $1.6\cdot10^{-3}\cdot0.3558 = 0.000569$ m³/MPa. This is 316 times less than with shut-in at the wellhead.
The start of the semilog straight line by the classic estimate: with shut-in at the wellhead, $44.8\cdot77.5\cdot2.5\cdot0.18/300 = 5.2$ h; with shut-in at the valve, 1.0 min. By the calculation, the derivative gets within 10% of the radial-flow level after 2.3 min.

Answer: A restart for 10 h leaves a distortion of the derivative of about 4.6%. Shut-in at the valve cuts wellbore storage from hours to minutes and removes the hump.

What it means. The price of shut-in at the bottom is wireline work and the risk that the valve will not close or will not open. The gain is a record in which the reservoir is visible from the first minutes. The rule “restart for five times longer than the short shut-in” comes from the same superposition as the Horner equation. The trace of any change in the way the well works fades faster the farther it is from the start of the record. The data in this example are assumed for illustration.

Shut in at the bottom where possible: in exploration wells with a drill stem tester (Well testing: purposes and types of test), and in producing wells with a valve on a wire.
Place the gauge as close to the perforated interval as possible, and in wells that produce water use a pair of gauges a few metres apart.
Record the pressure and the temperature at the wellhead during the build-up. They show whether a gas pocket is collecting, and whether the depth correction changes in a gas well, where the column of gas cools after shut-in.
Measure the afterflow with a flowmeter if the early section matters: it gives the instantaneous $C$ directly.
Compare the drawdown test and the build-up of the same well. Wellbore storage at start-up and at shut-in is different, and this in itself hints that $C$ is not constant.
For gas wells, use the pseudo-pressure and the pseudo-time. They allow for the fact that the compressibility and the viscosity of the gas change with the pressure, and they remove part of the “nonideality” without empirical models (real gas pseudo-pressure).

Well clean-up and a changing skin

The skin of a new well is usually not constant in the first hours and days of work. The well carries out the filtrate and the solid phase of the drilling fluid, the debris after perforation, the spent acid after a treatment, and the fracturing fluid after hydraulic fracturing. The skin falls: this is well clean-up (Sources of formation damage). The opposite also happens. Fines (fine particles) carried to the wellbore, deposits of salts (scale), or condensate that drops out of the gas near the well make the skin grow while the well produces. In both cases the assumption of a constant skin, on which the analysis of a drawdown test rests (Pressure drawdown analysis), is broken.

The skin acts at once: the pressure drop across it is proportional to the current rate and the current skin (Skin factor). So at a constant rate and radial flow, $\Delta p = m_e[\ln t_D + 0.809 + 2S(t)]$ , and the derivative with respect to $\ln t$ gets an extra term from the change of the skin:

The derivative of a drawdown test with a changing skin

\Delta p' = \frac{\dd\,\Delta p}{\dd\ln t} = m_e\left(1 + 2t\,\frac{\dd S}{\dd t}\right)

where:

$\Delta p'$ logarithmic derivative of the pressure drop $p_i - p_{wf}$ , Pa
$m_e$ plateau level of the derivative, $qB\mu/(4\pi kh)$ , Pa
$S(t)$ skin at the time $t$ , —
$t$ time since the start of production at a constant rate, s

Units: any consistent system; $t\,\dd S/\dd t$ has no units

Holds when

A constant rate, radial flow in a homogeneous reservoir, and wellbore storage is over.
The skin is a thin layer, and the pressure drop across it is proportional to the rate at that moment.

While the skin falls, the derivative is below the plateau. If $2t\,\dd S/\dd t < -1$ , it is negative: at a constant rate the pressure rises. Even a modest change of the skin changes the slope a lot. A fall of the skin by 0.5 per decimal cycle of time lowers the slope of the semilog plot by 43%, and the permeability from it comes out almost twice too high.

A build-up after such production is not affected by the clean-up. After shut-in the sandface rate is zero, there is no pressure drop across the skin, and the reservoir alone sets the pressure rise (Pressure build-up analysis). The skin enters the analysis of a build-up only through the pressure at the moment of shut-in. So the build-up gives the skin at the end of the producing period. Then the permeability and $p^*$ from the build-up let us calculate the skin at each point of the preceding drawdown test. This is the skin versus time:

The skin at the time t from a drawdown point

S(t) = \frac12\left[\frac{p^* - p_{wf}(t)}{m_e} - \ln\frac{k\,t}{\phi\mu c_tr_w^2} - 0.809\right]

where:

$p^*$ the extension of the Horner line from the build-up; for a new well, the initial reservoir pressure, Pa
$p_{wf}(t)$ bottomhole pressure at the time $t$ of production at a constant rate, Pa
$m_e$ plateau level from the build-up, $m/\ln 10$ , Pa
$k$ permeability from the build-up, m²
$\phi,\ \mu,\ c_t,\ r_w$ porosity, viscosity, total compressibility, wellbore radius, —, Pa·s, 1/Pa, m

Units: SI; natural logarithm

Holds when

Before production the pressure in the reservoir was equalised and the rate is constant; at the time $t$ wellbore storage is over.
The permeability did not change during production: only the near-wellbore zone changes.
The same formula holds also with boundaries, if $p^*$ is taken from the radial-flow line of the build-up.

This is the skin formula of a drawdown test, solved at every point and not only at 1 h. All the quantities except $p_{wf}(t)$ are taken from the build-up. We get a series of values of the skin, and we can see from it how and to what the clean-up goes.

In other unit systems (2)

Metric oilfield units

S(t) = 1.151\left[\frac{p^* - p_{wf}(t)}{m} - \lg\frac{k\,t}{\phi\mu c_tr_w^2} + 2.098\right]

$m$ is the slope per decimal cycle, MPa; pressures in MPa; $k$ in mD; $t$ in h; $\mu$ in mPa·s; $c_t$ in 1/MPa; $r_w$ in m.

$1.151 = \ln10/2$ ; $2.098 = -\lg(3.553\cdot10^{-3}) - 0.809/\ln10$ , as in the analysis of a drawdown test (Pressure drawdown analysis).

Oilfield units

S(t) = 1.151\left[\frac{p^* - p_{wf}(t)}{m} - \lg\frac{k\,t}{\phi\mu c_tr_w^2} + 3.23\right]

pressures and $m$ in psi; $k$ in mD; $t$ in h; $\mu$ in cP; $c_t$ in 1/psi; $r_w$ in ft.

$3.23 = -\lg(2.637\cdot10^{-4}) - 0.809/\ln10$ .

A new well: the first drawdown test and the build-up that follows it (assumed data)

$t$ or $\Delta t$ , h: 1
$p_{wf}$ , MPa (production): 23.358
$p_{ws}$ , MPa (shut-in): 26.207

$t$ or $\Delta t$ , h: 2
$p_{wf}$ , MPa (production): 23.288
$p_{ws}$ , MPa (shut-in): 26.328

$t$ or $\Delta t$ , h: 4
$p_{wf}$ , MPa (production): 23.250
$p_{ws}$ , MPa (shut-in): 26.442

$t$ or $\Delta t$ , h: 6
$p_{wf}$ , MPa (production): 23.240
$p_{ws}$ , MPa (shut-in): —

$t$ or $\Delta t$ , h: 8
$p_{wf}$ , MPa (production): 23.233
$p_{ws}$ , MPa (shut-in): —

$t$ or $\Delta t$ , h: 10
$p_{wf}$ , MPa (production): 23.225
$p_{ws}$ , MPa (shut-in): 26.575

$t$ or $\Delta t$ , h: 12
$p_{wf}$ , MPa (production): 23.215
$p_{ws}$ , MPa (shut-in): —

$t$ or $\Delta t$ , h: 16
$p_{wf}$ , MPa (production): 23.191
$p_{ws}$ , MPa (shut-in): —

$t$ or $\Delta t$ , h: 24
$p_{wf}$ , MPa (production): 23.139
$p_{ws}$ , MPa (shut-in): 26.672

$t$ or $\Delta t$ , h	$p_{wf}$ , MPa (production)	$p_{ws}$ , MPa (shut-in)
1	23.358	26.207
2	23.288	26.328
4	23.250	26.442
6	23.240	—
8	23.233	—
10	23.225	26.575
12	23.215	—
16	23.191	—
24	23.139	26.672

A test after perforation (shooting holes through the casing into the reservoir): the well was started and shut in with a valve at the bottom; rate 95 m³/d for 24 h, then a build-up. Reservoir: $h$ = 11 m, $\phi$ = 0.21, $\mu$ = 0.85 mPa·s, $c_t = 1.5\cdot10^{-3}$ MPa⁻¹, $B$ = 1.21, $r_w$ = 0.108 m. The data were calculated from a model of a homogeneous infinite reservoir with a skin that falls during production, and rounded to 0.001 MPa.

CalculatedCalculated drawdown test of a new well with clean-up. Top: the pressure on a semilog plot. The points are the measurements of the table, the orange curve is the calculation by the model, the dashed line is the same well with a constant skin of 2.52 (the level at the end of production), and the dark dashed line is the “line” through 1 and 10 h. Bottom: the skin versus time. The points are the calculation by the formula of the lesson, and the line is the skin put into the model.

Worked examplePermeability and skin of a new well that is cleaning up

Given (assumed data)

Record: the table above; production for 24 h at 95 m³/d, $B$ = 1.21, $\mu$ = 0.85 mPa·s, $h$ = 11 m
Reservoir: $\phi\mu c_tr_w^2 = 0.21\cdot0.85\cdot1.5\cdot10^{-3}\cdot0.108^2 = 3.123\cdot10^{-6}$
Initial pressure: 26.800 MPa (measured before start-up)

Solution

The drawdown test “as usual”: a line through 1 and 10 h, a slope of $23.358 - 23.225 = 0.133$ MPa per cycle, and a permeability of $2.149\cdot95\cdot1.21\cdot0.85/(0.133\cdot11) = 209.97/1.463 = 143.5$ mD. The skin from it: $\lg(143.5/3.123\cdot10^{-6}) = 7.662$ , $S = 1.151\,[(26.800 - 23.358)/0.133 - 7.662 + 2.098] = 23$ .
The build-up: the Horner abscissae are $\lg(26/2) = 1.11394$ and $\lg(34/10) = 0.53148$ ; the slope is $m = (26.575 - 26.328)/(1.11394 - 0.53148) = 0.4241$ MPa per cycle; $k = 209.97/(0.4241\cdot11) = 45.0$ mD; $p^* = 26.575 + 0.4241\cdot0.53148 = 26.800$ MPa, which is the initial pressure.
The skin versus time: $\lg[k/(\phi\mu c_tr_w^2)] = \lg(45.0/3.123\cdot10^{-6}) = 7.159$ . At 1 h: $1.151\,[(26.800 - 23.358)/0.4241 - 7.159 + 2.098] = 1.151\,[8.116 - 7.159 + 2.098] = 3.52$ .
At 4 h: $(26.800 - 23.250)/0.4241 = 8.371$ , $\lg4 = 0.602$ , $S = 1.151\,(8.371 - 7.159 - 0.602 + 2.098) = 3.12$ . At 24 h: $(26.800 - 23.139)/0.4241 = 8.632$ , $\lg24 = 1.380$ , $S = 2.52$ .
The derivative of the drawdown test by the model at 6 h is 0.0216 MPa, against the radial-flow level of 0.1842 MPa: 8.5 times lower.

Answer: The reservoir:

k \approx 45.0

mD. The skin fell from 3.52 to 2.52 in a day of production. The drawdown test “as usual” would have given 143.5 mD and a skin of 23: a permeability error of a factor of 3.2.

What it means. The skin changed by only a little more than one, but in the first day this almost fully offset the growth of the drawdown. The pressure between 2 and 12 h changed by only 0.073 MPa, and it looked like a “stabilisation”. This is the source of the gross errors of a drawdown test, and of the rule: we take the permeability of a new well from the build-up, and from the drawdown test we build the skin versus time. The series shows whether the clean-up has ended. Here the skin approaches a value of about 2.5: this is residual damage, which we should compare with what we expect and decide whether a treatment is needed (Formation damage and damage skin). The model from which the record was calculated has $k$ = 45 mD and a skin that falls from 3.7 to 2.5. The data in this example are assumed for illustration.

How to recognise an effect from the record

Signs of nonideal wellbore storage and of clean-up

Effect: Gas redistribution (a hump)
Pressure: the build-up passes a maximum and falls
Derivative: higher than an ordinary hump, then negative (a break)
How to check: wellhead pressure; rise time of the bubbles; shut-in at the bottom

Effect: Redistribution without a hump
Pressure: a flat step between two “lines”
Derivative: a deep dip, as in dual porosity
How to check: the same; geology

Effect: Increasing wellbore storage
Pressure: early growth faster than by the $C$ of the wellbore
Derivative: a unit slope with a small $C_a$ ; the plateau later than expected
How to check: an estimate of $C$ from the design; models with $A > 0$

Effect: Decreasing wellbore storage
Pressure: early growth slower
Derivative: a unit slope with a large $C_a$ ; the plateau earlier than expected
How to check: a flowmeter; the pressure in the annulus

Effect: A falling level when injection stops
Pressure: the wellhead pressure fell to atmospheric
Derivative: a dip and a second unit slope
How to check: a record of the wellhead pressure

Effect: Water below the gauge
Pressure: a shift of pressure by $\Delta\rho g z$
Derivative: a dip for the time that the water rises
How to check: a pair of gauges

Effect: Clean-up during a drawdown test
Pressure: the pressure “stabilises” or rises
Derivative: below the plateau, sometimes negative
How to check: a build-up and the skin versus time

Effect	Pressure	Derivative	How to check
Gas redistribution (a hump)	the build-up passes a maximum and falls	higher than an ordinary hump, then negative (a break)	wellhead pressure; rise time of the bubbles; shut-in at the bottom
Redistribution without a hump	a flat step between two “lines”	a deep dip, as in dual porosity	the same; geology
Increasing wellbore storage	early growth faster than by the $C$ of the wellbore	a unit slope with a small $C_a$ ; the plateau later than expected	an estimate of $C$ from the design; models with $A > 0$
Decreasing wellbore storage	early growth slower	a unit slope with a large $C_a$ ; the plateau earlier than expected	a flowmeter; the pressure in the annulus
A falling level when injection stops	the wellhead pressure fell to atmospheric	a dip and a second unit slope	a record of the wellhead pressure
Water below the gauge	a shift of pressure by $\Delta\rho g z$	a dip for the time that the water rises	a pair of gauges
Clean-up during a drawdown test	the pressure “stabilises” or rises	below the plateau, sometimes negative	a build-up and the skin versus time

One and the same shape of the derivative occurs for effects in the wellbore and in the reservoir. The decision is made by the combination of signs, the design of the well, the way it is shut in, and the data of other tests (false signs).

The order of checking the early part of a record

1
Where and how the well is shut in
At the wellhead or at the bottom, whether there is a packer, what is in the tubing and in the annulus, and where the gauge is relative to the perforations. With shut-in at the bottom, most of the effects of this lesson are ruled out.
2
An ordinary plot and the rate of pressure rise
Look at the pressure against time on an enlarged scale, and at $\dd p/\dd t$ on a linear scale. A fall of the pressure after shut-in is an effect of the wellbore.
3
Compare $C_a$ with the estimate
The early unit slope against the coefficient from the design and the contents of the wellbore, including the weight of the column. A strong difference is a sign of a changing $C$ .
4
Compare the drawdown test and the build-up
If wellbore storage at start-up and at shut-in is different, the coefficient is not constant. If the drawdown test of a new well is “flat” and the build-up is normal, it is clean-up.
5
The reservoir only after that
We look for dual porosity, boundaries and supply in the section after the effects of the wellbore have ended. If there is no such section, we repeat the test with shut-in at the bottom.

Terms and symbols

Nonideal wellbore storage: Wellbore storage with a coefficient that changes during the record. Models with a constant $C$ do not apply to it.
Increasing and decreasing wellbore storage: The early coefficient is smaller or larger than the late one. The first gives humps, and the second an early derivative plateau.
Apparent coefficient $C_a$: The coefficient that the early unit slope gives. When $C$ changes, it is only the initial value.
Phase redistribution: The rise of gas and the settling of liquid in a closed wellbore. It carries the pressure of the depth up to the wellhead and raises the pressure in the wellbore.
Redistribution pressure $p_\phi$: An empirical function of time in the wellbore balance, with an amplitude $A$ and a time $t_\phi$ .
Well clean-up: The fall of the skin in the first hours and days of production, when the well carries out filtrate, the solid phase and spent fluids.
Skin versus time: The skin calculated at every point of a drawdown test, with the permeability and $p^*$ from the build-up.

Common mistakes and misconceptions

Wrong

“The pressure three hours after shut-in was higher than after a day, so the reservoir is being depleted.”

In fact

In a well shut in at the wellhead with gas in the tubing, this is how a hump from phase redistribution looks. The shut-in itself cannot lower the bottomhole pressure in the first hours. The wellbore does it. A fall at the end of a long build-up in a developed field is another effect: it comes from the neighbouring wells (Testing in a developed reservoir: Slider's method).

Wrong

“The wellbore storage coefficient equals the volume of the tubing times the compressibility of the contents.”

In fact

Only for a wellbore filled with one phase. In a closed tubing with gas the weight of the column is added, $1/C = 1/C_g + \rho g/A_t$ , and the coefficient changes with the pressure.

Wrong

“A dip of the derivative after wellbore storage is dual porosity.”

In fact

The same dip comes from phase redistribution without a hump, from a falling level in an injection well, from water rising below the gauge, and from clean-up. We accept dual porosity only if geology and core confirm it, and the well was shut in at the bottom.

Wrong

“The drawdown test of a new well stabilised quickly, so the reservoir has good permeability or support.”

In fact

This is how clean-up looks: a falling skin offsets the growth of the drawdown. The permeability from such a drawdown test is too high by factors of several. We find it from the build-up.

Wrong

“The empirical model matched the record well, so its parameters describe the wellbore.”

In fact

The three parameters of the wellbore adjust to almost any early part of a record. We check them against physical estimates, and we find the reservoir from the section after the effects of the wellbore have ended.

In depth: advanced detail

Models of changing wellbore storage in Laplace space

We switch to the dimensionless variables of the lesson on wellbore storage: $p_{wD}$ is the pressure rise in units of $qB\mu/(2\pi kh)$ , then $t_D$ , $C_D$ , and $p_{\phi D} = 2\pi kh\,p_\phi/(qB\mu)$ . For a build-up after long production, the wellbore balance with the redistribution pressure takes the form $1 - q_{sfD} = C_D\left(\dd p_{wD}/\dd t_D - \dd p_{\phi D}/\dd t_D\right)$ . Here $q_{sfD}$ is the share of the change of rate that has already reached the bottom of the well. In Laplace space, the response of the reservoir with a skin to a changing inflow is a product: $\bar p_{wD} = \bar q_{sfD}\,(s\,\bar p_D^{\,*} + S)$ . We eliminate $\bar q_{sfD}$ and get:

A well with changing wellbore storage: the Laplace transform

\bar p_{wD}(s) = \frac{R(s)\left[1 + C_D\,s^2\,\bar p_{\phi D}(s)\right]}{s\left[1 + C_D\,s\,R(s)\right]},\qquad R(s) = s\,\bar p_D^{\,*}(s) + S

where:

$\bar p_{wD}(s)$ Laplace transform of the dimensionless pressure rise in the well, —
$\bar p_D^{\,*}(s)$ Laplace transform of the solution without wellbore storage and skin; for a well of finite radius, $K_0(\sqrt s)/(s^{3/2}K_1(\sqrt s))$ , —
$\bar p_{\phi D}(s)$ Laplace transform of the dimensionless redistribution pressure, —
$C_D,\ S$ dimensionless wellbore storage coefficient and skin, —

Units: dimensionless

Holds when

$p_{\phi D}(0) = 0$ , $p_{wD}(0) = 0$ .
For $p_{\phi D} \equiv 0$ the formula turns into the solution with wellbore storage and skin from the lesson Wellbore storage.

For the exponential form, $\bar p_{\phi D} = A_D/[s(1 + s\,t_{\phi D})]$ . For the error-function form, $\bar p_{\phi D} = (A_D/s)\,\e^{(s\,t_{\phi D}/2)^2}\operatorname{erfc}(s\,t_{\phi D}/2)$ . Here $A_D$ and $t_{\phi D}$ are the dimensionless amplitude and time. For large $s$ (a short time), $\bar p_{wD} \to (1/C_D + A_D/t_{\phi D})/s^2$ in the first form. This is the unit slope with the apparent coefficient $C_a$ .

The curves and data of the lesson were calculated by inverting this formula along a vertical line in the complex plane. We replace the inversion integral with a Fourier series and sum it with Euler acceleration, using 33 values of the transform per point. For this we calculate $K_0$ , $K_1$ and $\e^{z^2}\operatorname{erfc}z$ for a complex argument. To the response to the shut-in we add the superposition of the production before it. The familiar Stehfest algorithm takes only real $s$ , and it is not accurate enough here. In the transition section of wellbore storage, its derivative is wrong by about 0.2% with sixteen terms and by up to one and a half per cent with twelve. This is enough to shift the time of reaching the plateau by a per cent, and the rate of pressure rise in the third digit. An independent calculation by another program, and for the exponential form an inversion along a Talbot contour, agreed with the data of the lesson in all the digits shown. We calculate the factor $\e^{z^2}\operatorname{erfc}z$ as one function: separately, both factors go beyond the range of floating-point numbers at large $|z|$ .

Why the gas and the weight of the column add in series

Let a tubing of cross-section $A_t$ , closed at the wellhead, contain liquid and gas in any arrangement, and let the wellhead pressure be $p_t$ . A liquid at rest obeys hydrostatics, so the pressure at the bottom equals $p_t + (M_l + M_g)g/A_t$ , where $M_l$ and $M_g$ are the masses of the liquid and the gas in the pipe. Suppose a volume $\dd V$ of liquid has entered from below. The mass of the liquid has grown by $\rho\,\dd V$ , and the pressure difference between the bottom and the wellhead has grown by $\rho g\,\dd V/A_t$ .

Each portion of gas is under the same load as before the inflow: the new liquid entered below everything. So the pressure on each portion has grown by the same amount as at the wellhead, $\dd p_t$ , and its volume has decreased by $V_{g,i}\,\dd p_t/p_i$ . The sum of the decreases equals $\dd V$ : $\dd p_t = \dd V/C_g$ . Adding, we get $\dd p_w = \dd V\,(1/C_g + \rho g/A_t)$ , the formula of the main section. The same picture holds for a pocket of gas above a liquid level in the annulus, where the sum is replaced by the single term $V_g/p_g$ .

Two consequences follow. First, in a closed tubing the coefficient cannot exceed $A_t/(\rho g)$ . The large coefficients that we sometimes find from a record mean that the annulus is connected to the wellbore (no packer, or a leak in the tubing or in a gas-lift valve). Second, for a constant mass of gas, $C_g = \sum V_{g,i}/p_i$ falls as the pressure rises, so wellbore storage decreases even without any redistribution of phases.

Connections to other topics

Wellbore storage — the constant wellbore storage coefficient, the unit slope and the solution in Laplace space, which is extended here.
Diagnosis with the pressure derivative — the diagnostic plot and the false signs, among which are effects of the wellbore.
Type-curve matching — matching to type curves, and why early points with a changing $C$ interfere with it.
Pressure build-up analysis — the build-up from which we find the permeability and $p^*$ for the skin versus time.
Pressure drawdown analysis — the analysis of a drawdown test at a constant skin, and the skin formula applied here to each point.
Skin factor — the skin as a thin layer with a pressure drop proportional to the current rate.
Sources of formation damage — where the skin that the well carries out in clean-up comes from.
Multiphase flow in tubing — the profile of pressure and gas content in the tubing of a producing well, and the rise speed of bubbles.
Well control: primary barrier and kick indicators — the same effect in a shut-in well during drilling: gas rising without expanding.
Wireline operations — how valves for shut-in at the bottom are run on a wire.
Inflow at high rates: non-Darcy flow — a skin that grows with the rate, which is easy to confuse with clean-up.
Effect of producing time on build-up analysis — the distortions of a build-up that a short producing time before shut-in gives.

Check yourself

1
A build-up of an oil well shut in at the wellhead: in 3 h the pressure rose by 3.2 MPa, for the next 5 h it fell by 0.1 MPa, and only then did it start to rise slowly. What is most likely?
1. A.A neighbouring injection well was shut in
2. B.Dual porosity of the reservoir
3. C.A constant-pressure boundary near the well
4. D.Redistribution of gas and liquid in the wellbore
Show the answer
Correct answer: D. Redistribution of gas and liquid in the wellbore
After shut-in, the pressure at the bottom of a well in a homogeneous reservoir can only rise, and a constant-pressure boundary only slows the rise. The shut-in of a neighbouring injection well would act with a delay, and a few hours later it would not be followed by a rise. A fall in the first hours is an effect of the wellbore: rising gas lifted the pressure above the reservoir pressure at the well, and the liquid went back into the reservoir. We check it by the rise of the wellhead pressure and by the rise time of the bubbles, and we remove it by shutting in at the bottom.
2
A tubing with an inside diameter of 62 mm, closed at the wellhead; under the wellhead there is a gas pocket of 2.5 m³ at 4 MPa (absolute), and below it oil of 840 kg/m³. Find the wellbore storage coefficient at the bottom, and compare it with the coefficient that we would get without the weight of the column.
Show the answer
$C_g = V_g/p_g = 2.5/4 = 0.625$ m³/MPa (we neglect the compressibility of the oil). The weight of the column: $A_t = 3.019\cdot10^{-3}$ m², $10^{-6}\cdot9.81\cdot840/(3.019\cdot10^{-3}) = 2.73$ MPa/m³. $1/C = 1/0.625 + 2.73 = 1.60 + 2.73 = 4.33$ , $C \approx 0.231$ m³/MPa. This is almost three times less than the 0.625 from the compression of the gas alone. A closed tubing cannot give more than $1/2.73 = 0.366$ m³/MPa.
3
To run a downhole valve, the well was shut in at the wellhead for 3 h. How long must it then produce, so that the future build-up is lowered by the short shut-in by no more than 5%?
Show the answer
The largest lowering of the derivative is $\varepsilon = \tau/(\sqrt{t_f + \tau} + \sqrt{t_f})^2 \approx \tau/(4t_f)$ . The condition $\tau/(4t_f) \le 0.05$ gives $t_f \ge 5\tau = 15$ h. Check with the exact formula: $3/(\sqrt{18} + \sqrt{15})^2 = 3/(4.243 + 3.873)^2 = 3/65.9 = 0.046$ , that is, 4.6%.
4
A build-up after 30 h of production of a new well gave $k$ = 30 mD, $m$ = 0.35 MPa per cycle and $p^*$ = 31.00 MPa; $\phi\mu c_tr_w^2 = 3.695\cdot10^{-6}$ (metric units). On the drawdown test the pressure at 2 h was 27.20 MPa, and at 20 h it was 27.05 MPa. Find the skin at these moments. What would happen to the permeability if we calculated it from these two points of the drawdown test?
Show the answer
$\lg(30/3.695\cdot10^{-6}) = 6.910$ . At 2 h: $(31.00 - 27.20)/0.35 = 10.857$ ; $S = 1.151\,(10.857 - 6.910 - 0.301 + 2.098) = 1.151\cdot5.744 = 6.61$ . At 20 h: $(31.00 - 27.05)/0.35 = 11.286$ ; $S = 1.151\,(11.286 - 6.910 - 1.301 + 2.098) = 1.151\cdot5.173 = 5.95$ . The skin decreased by 0.66, so the well is cleaning up. The slope from the two points of the drawdown test is $27.20 - 27.05 = 0.15$ MPa per cycle instead of 0.35, and the permeability would come out $0.35/0.15 = 2.3$ times higher, about 70 mD.
5
What most reliably removes the humps from phase redistribution and shortens wellbore storage?
1. A.Increase the rate before shut-in
2. B.Record the pressure ten times more often
3. C.Raise the gauge closer to the wellhead
4. D.Shut the well in with a valve at the bottom
Show the answer
Correct answer: D. Shut the well in with a valve at the bottom
A valve at the bottom cuts the wellbore off: only a small volume of liquid is left below it, while the gas, the levels and the annulus stay above and do not affect the pressure under the valve. The rate before shut-in does not change how long the wellbore effects last. Frequent readings only show the hump in more detail. And a gauge higher up, farther from the reservoir, only adds uncertainty to the depth correction.
6
On the drawdown test of a new gas well, at a constant rate the pressure hardly changed for 12 h, and the derivative fell 5–6 times below the level that the build-up showed later. Which explanation should we check first, and how?
Show the answer
Clean-up: the skin was falling and offsetting the growth of the drawdown. We check it with the skin versus time. We find $k$ and $p^*$ from the build-up and calculate the skin at each point of the drawdown test with the formula $S(t)$ . If it decreases steadily and reaches a constant value, this is clean-up, and not supply or “stabilisation”. For gas we do the calculation in pseudo-pressures, and we tell clean-up from a non-Darcy skin by the order of the rates.

Updated: 28 October 2026