Wellbore storage · MunaiHub

Quick refresher

The wellbore is a buffer between the wellhead and the reservoir. After start-up it gives out the fluid stored in it. After shut-in at the wellhead it takes in the inflow from the reservoir. The sandface flow rate $q_{sf}$ (the inflow from the reservoir at the bottom of the well) catches up with the rate at the wellhead only gradually.
The wellbore storage coefficient $C = c_{wb}V_{wb}$ is the compressibility of the fluid in the wellbore times the volume of the wellbore below the closing valve, in m³/MPa. Free gas (gas that is not dissolved in the oil) in the wellbore makes $C$ tens of times larger. Shutting in at the bottom of the well makes it tens or hundreds of times smaller.
While only the wellbore is working, the pressure changes linearly with time: $\Delta p = qBt/(24C)$ . On a log–log plot this is a straight line with a slope of 1. It gives $C$ and nothing else: these points contain neither the permeability nor the skin.
Then comes a transition section about two logarithmic cycles of time long (two tenfold steps of time). Only after it comes the semilog straight line of radial flow (flow towards the well from all sides).
The classic estimate for the start of the line is $t_D \approx (60 + 3.5S)\,C_D$ . By this time the inflow from the reservoir reaches 99% of the rate, but the slope of the curve is still too high. The pressure derivative (the slope of the pressure curve on a semilog plot) gets within 10% of the radial-flow level only 2–4 times later.
The duration of wellbore storage is proportional to $C$ and $\mu/(kh)$ , grows with the skin, and does not depend on the rate.

Key relationships

C = c_{wb}\,V_{wb}

wellbore storage coefficient of a wellbore filled with one fluid, m³/MPa

(q_{sf} - q)\,B = 24\,C\,\dfrac{\dd p_w}{\dd t}

wellbore balance: m³/d, m³/MPa, MPa, hours (the constant is also 24 in oilfield units)

C_D = \dfrac{C}{2\pi\,\phi\,c_t\,h\,r_w^2}

dimensionless coefficient; in oilfield units the numerator has a factor 5.615

\Delta p = \dfrac{qB}{24\,C}\,t,\qquad p_D = \dfrac{t_D}{C_D}

section of pure wellbore storage: unit slope

t \approx 44.8\,(60 + 3.5S)\,\dfrac{\mu\,C}{kh}

start of the semilog straight line, h (mPa·s, m³/MPa, mD·m); in oilfield units the constant is 3390

Common pitfalls

Drawing the semilog line through the points of the transition section. It is steeper than the true line, and the permeability comes out too low.
Estimating $C$ from the volume of the wellbore alone and forgetting free gas. The real coefficient can be tens of times larger than the calculated one.
Setting the moment of start-up or shut-in inaccurately. On a log–log plot the early points then move, and the unit slope is distorted.
Mixing up the coefficient $C$ (m³/MPa), the dimensionless $C_D$ and the volume of the wellbore itself.
Using the formulas with a constant $C$ when a liquid level moves in the wellbore, or when gas and liquid redistribute.

After this lesson you can

explain why, after start-up or shut-in at the wellhead, the inflow from the reservoir lags behind the rate at the wellhead, and how this distorts the drawdown and build-up curves
calculate the wellbore storage coefficient $C$ from the volume of the wellbore and the compressibility of its contents, and convert it to the dimensionless $C_D$ in metric and oilfield units
recognise the section of pure wellbore storage by its straight line on ordinary axes and by its unit slope (a straight line of slope 1) on a log–log plot (both axes logarithmic), and find $C$ from it
estimate when wellbore storage ends and the semilog straight line begins, and explain how accurate this estimate is
name ways to shorten wellbore storage, and the conditions in which the model with a constant $C$ does not work

The wellbore as a buffer between the wellhead and the reservoir

The theory of the semilog straight line assumes that the rate changes right at the bottom of the well. At start-up the inflow from the reservoir at once becomes equal to $q$ , and at shut-in it at once becomes zero (Pressure drawdown analysis, Pressure build-up analysis). In reality the valve is at the wellhead. Between it and the reservoir there are two or three kilometres of pipe filled with a compressible fluid. Because of this, a pressure record has an early section in which the simple model does not work. This lesson is about that section.

Start-up. Before start-up, the reservoir pressure keeps the fluid in the wellbore compressed. When we open the valve, the fluid from the wellbore itself goes to the surface first. The pressure in the wellbore falls, the fluid expands, and the “extra” volume leaves through the wellhead. The reservoir joins in gradually, as the difference between the reservoir pressure and the bottomhole pressure grows. The rate at the wellhead is already $q$ . But the sandface flow rate $q_{sf}$ , that is, the inflow from the reservoir into the wellbore, only grows from zero to $q$ .

Shut-in. The picture is the mirror image. We close the valve, and the rate at the wellhead is zero. But the reservoir keeps supplying fluid. It compresses the contents of the wellbore, and the bottomhole pressure rises. This afterflow (the inflow that goes on after shut-in) fades as the pressure in the wellbore comes close to the reservoir pressure.

SchematicThe wellbore as a tank between the reservoir and the valve at the wellhead. Top: the first minutes after start-up. The rate at the wellhead is already

q

, but it comes mostly from the expansion of the fluid in the wellbore, and the inflow from the reservoir

q_{sf}

is small. Bottom: the first minutes after shut-in. The valve is closed, but the reservoir still supplies fluid and compresses the contents of the wellbore. The zigzag is the resistance of the reservoir and the skin, which the inflow has to pass. A sketch, not to scale.

In both cases the bottomhole pressure changes more slowly than in an ideal well. On a drawdown curve (the record after start-up) the early points lie above the semilog straight line. On a build-up curve (the record after shut-in) they lie below it. If we measure the pressure change $\Delta p$ from the moment of start-up or shut-in, the picture is the same: at first $\Delta p$ is smaller than a well without wellbore storage would give.

Wellbore storage changes neither the reservoir nor the diffusivity equation (the equation for how pressure spreads through the rock). What changes is the condition at the bottom of the well. The inflow from the reservoir is no longer fixed in advance. It depends on how fast the pressure in the wellbore changes. So the early section of the record carries information about the wellbore, and it says nothing about the reservoir.

The wellbore storage coefficient

How much fluid the wellbore gives out when the pressure falls depends on two quantities: the volume of the wellbore that is connected with the reservoir, and the compressibility of what fills it. Their product is the wellbore storage coefficient.

Storage coefficient of a wellbore filled with one fluid

C = \frac{\Delta V_{wb}}{\Delta p_w} = c_{wb}\,V_{wb}

where:

$C$ wellbore storage coefficient: the volume of fluid that the wellbore gives out or takes in when the bottomhole pressure changes by one unit, m³/Pa
$\Delta V_{wb}$ the volume by which the contents of the wellbore expand or shrink (the index $wb$ stands for wellbore), m³
$\Delta p_w$ change of the bottomhole pressure, Pa
$c_{wb}$ compressibility of the fluid in the wellbore, 1/Pa
$V_{wb}$ volume of the wellbore that is connected with the reservoir: from the closing valve to the reservoir, m³

Units: SI

Holds when

The wellbore is entirely filled with one fluid, with no free liquid surface in it.
The compressibility is taken as constant. The pressure change is small compared with the pressure itself (for gas this is a rough assumption).
The pressure changes by the same amount along the whole wellbore: the weight of the fluid column and the friction losses stay as they were. In reality the fluid that has entered makes the column heavier, and $C$ is smaller. In a wellbore with liquid it is 1–4% smaller. In the gas well of the table below it is about 15% smaller. In an oil well with free gas in the tubing (the pipe through which the fluid flows up) it is almost half as large (example below).
We ignore the elasticity of the pipes (their stretching under pressure) and the change of temperature.

The coefficient grows with both the volume and the compressibility. The volume is set by the design of the well and by the place where we close it. It is the whole string of pipes when we close at the wellhead, and only the part below the valve when we close at the bottom. The compressibility is set by the contents. For water and oil it is of the order of $10^{-3}$ MPa⁻¹. For gas it is about $1/p$ , that is, tens and hundreds of times larger.

In other unit systems (2)

Metric oilfield units

C\,[\text{m}^3/\text{MPa}] = c_{wb}\,[\text{MPa}^{-1}]\cdot V_{wb}\,[\text{m}^3]

$c_{wb}$ in 1/MPa, $V_{wb}$ in m³, $C$ in m³/MPa.

There is no constant. The number shows how many cubic metres the wellbore gives out when the bottomhole pressure falls by 1 MPa.

Oilfield units

C\,[\text{bbl/psi}] = c_{wb}\,[\text{psi}^{-1}]\cdot V_{wb}\,[\text{bbl}]

$c_{wb}$ in 1/psi, $V_{wb}$ in barrels, $C$ in bbl/psi.

Conversion: 1 m³/MPa = 6.2898/145.04 = 0.0434 bbl/psi; 1 bbl/psi = 23.06 m³/MPa.

The same wellbore with different contents

Contents of the wellbore: Water (injection well)
Compressibility, MPa⁻¹: 0.00045
$C$ , m³/MPa: 0.0037
$C_D$: 17

Contents of the wellbore: Oil above the bubble point pressure
Compressibility, MPa⁻¹: 0.0016
$C$ , m³/MPa: 0.0132
$C_D$: 60

Contents of the wellbore: Oil and free gas: a quarter of the tubing volume, 7 MPa
Compressibility, MPa⁻¹: 0.042
$C$ , m³/MPa: 0.17
$C_D$: 783

Contents of the wellbore: Gas at 20 MPa and 80 °C (gas well)
Compressibility, MPa⁻¹: 0.046
$C$ , m³/MPa: 0.38
$C_D$: —

Contents of the wellbore: Oil below the shut-in valve at the packer (0.71 m³)
Compressibility, MPa⁻¹: 0.0016
$C$ , m³/MPa: 0.0011
$C_D$: 5

Contents of the wellbore	Compressibility, MPa⁻¹	$C$ , m³/MPa	$C_D$
Water (injection well)	0.00045	0.0037	17
Oil above the bubble point pressure	0.0016	0.0132	60
Oil and free gas: a quarter of the tubing volume, 7 MPa	0.042	0.17	783
Gas at 20 MPa and 80 °C (gas well)	0.046	0.38	—
Oil below the shut-in valve at the packer (0.71 m³)	0.0016	0.0011	5

The wellbore of the lesson's examples: tubing (the string of pipes that carries the fluid up the well) with an inside diameter of 62 mm down to a packer (a seal between the tubing and the casing) at a depth of 2500 m, and 40 m of casing below the packer, 8.26 m³ in all. The gas compressibility is for a gas with a relative density of 0.65 (Gas formation volume factor, compressibility and viscosity). In the third row it refers to the mixture in the tubing, and oil stays below the packer. In the third row $C$ includes the weight of the oil column (see the example below), and in the other rows it is simply $c_{wb}V_{wb}$ . The dimensionless coefficient $C_D$ (formula below) is calculated for the oil reservoir of the examples. For the gas well we do not give it, because its denominator contains the compressibility of a gas-saturated reservoir. The data are assumed for illustration.

Worked exampleWellbore storage coefficient from the design of the well

Given (assumed data)

Tubing: inside diameter; length to the packer: 62 mm; 2500 m
Casing below the packer: inside diameter; length: 150.5 mm; 40 m
Compressibility of the oil in the wellbore $c_o$: $1.6\cdot10^{-3}$ MPa⁻¹ (the pressure is above the bubble point everywhere)
Compressibility of the gas at 7 MPa and 40 °C: 0.163 MPa⁻¹ (for an ideal gas $1/p$ = 0.143)
Density of the oil in the wellbore: 820 kg/m³
Reservoir: $\phi$ = 0.20; $c_t = 1.5\cdot10^{-3}$ MPa⁻¹; $h$ = 10 m; $r_w$ = 0.108 m

Solution

Volume of the tubing: $\pi\cdot0.062^2/4\cdot2500 = 7.55$ m³. Volume below the packer: $\pi\cdot0.1505^2/4\cdot40 = 0.71$ m³. In all, $V_{wb} = 8.26$ m³.
The wellbore is full of oil and we close it at the wellhead: $C = 1.6\cdot10^{-3}\cdot8.26 = 0.0132$ m³/MPa. When the bottomhole pressure falls by 1 MPa, the wellbore gives out about 13 litres of oil.
The same in oilfield units: $C = 0.0132\cdot0.0434 = 5.73\cdot10^{-4}$ bbl/psi.
Dimensionless coefficient (the formula is in the next section): $2\pi\phi c_thr_w^2 = 2.199\cdot10^{-4}$ m³/MPa, $C_D = 0.0132/(2.199\cdot10^{-4}) = 60$ .
Check in oilfield units: $c_t = 1.0342\cdot10^{-5}$ psi⁻¹, $h = 32.81$ ft, $r_w = 0.3543$ ft. The denominator: $\phi c_thr_w^2 = 0.20\cdot1.0342\cdot10^{-5}\cdot32.81\cdot0.3543^2 = 8.52\cdot10^{-6}$ ; $C_D = 0.894\cdot5.73\cdot10^{-4}/(8.52\cdot10^{-6}) = 60$ , the same number.
Free gas takes a quarter of the tubing volume. The compressibility of the mixture: $0.25\cdot0.163 + 0.75\cdot0.0016 = 0.04195$ MPa⁻¹. If the weight of the column did not change, the coefficient would be $C_g = 0.04195\cdot7.55 + 0.0016\cdot0.71 = 0.318$ m³/MPa.
But the oil inflow does not only compress the gas. It also makes the column in the tubing heavier: every cubic metre that enters raises the bottomhole pressure by $10^{-6}\cdot9.81\cdot820/(3.019\cdot10^{-3}) = 2.665$ MPa. The compression of the gas and the weight of the column act in series (closed tubing with free gas): $1/C = 1/0.318 + 2.665 = 5.810$ , $C = 0.172$ m³/MPa. This is 13 times as much as without gas, but almost half of the estimate from the compressibility alone. Closed 62 mm tubing with oil never gives more than $1/2.665 = 0.375$ m³/MPa, however much gas there is in it.
Shut-in with a valve at the packer: only the oil below it is compressed, $C = 1.6\cdot10^{-3}\cdot0.71 = 0.00114$ m³/MPa. This is 11.6 times less than for a wellbore full of oil.

Answer: For a wellbore full of oil,

C \approx 0.0132

m³/MPa. With gas in a quarter of the tubing volume, it is about 0.172 m³/MPa. With shut-in at the packer, it is about 0.00114 m³/MPa.

What it means. The design is the same, but the coefficient changes by a factor of about 150, depending on what is in the wellbore and where we close it. A calculation from the volume of the wellbore without gas gives a lower estimate. We find the real value from the pressure record itself (example below), because we do not know the share of gas in the wellbore beforehand. The data in this example are assumed for illustration.

If a liquid level moves in the wellbore

In pumped wells, and in wells whose liquid does not reach the wellhead, the wellbore is not full up to the wellhead. After shut-in, the inflow from the reservoir raises the liquid level, and the bottomhole pressure rises together with the weight of the column. The wellbore then stores fluid not by compression but by the rise of the level. The coefficient is $C = A/(\rho g)$ , where $A$ is the area of the cross-section in which the level moves and $\rho$ is the density of the liquid. Take the annulus (the space between the casing and the tubing) between a casing with an inside diameter of 150.5 mm and tubing of 73 mm, and oil with a density of 850 kg/m³. Then $C$ is about 1.6 m³/MPa, more than a hundred times as much as the compression of the oil in the same wellbore gives. The lesson Nonideal wellbore storage, well clean-up and changing skin covers such records (liquid-level recovery curves) and other cases of a changing $C$ .

The wellbore balance and the dimensionless coefficient

We write a volume balance for the wellbore, as for a tank. In a short time $\dd t$ , the volume $q_{sf}B\,\dd t$ enters the wellbore from the reservoir, and the volume $qB\,\dd t$ leaves through the wellhead. Both are at reservoir conditions. The difference goes to change the stock of fluid in the wellbore: if the pressure rose by $\dd p_w$ , the wellbore took in an extra $C\,\dd p_w$ .

Volume balance in the wellbore

(q_{sf} - q)\,B = C\,\frac{\dd p_w}{\dd t}

where:

$q_{sf}$ sandface flow rate: the inflow from the reservoir into the wellbore, converted to standard conditions, m³/s
$q$ rate at the wellhead 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
$C$ wellbore storage coefficient, m³/Pa
$p_w$ bottomhole pressure, Pa
$t$ time, s

Units: SI

Holds when

The coefficient $C$ is constant: the contents of the wellbore and their compressibility do not change.
We treat the wellbore as a single tank: the pressure changes by the same amount everywhere in it.
The rate at the wellhead is given (constant or zero), and the sandface rate is the unknown.

After start-up the pressure falls, so its rate of change is negative and $q_{sf} < q$ : the wellbore supplies part of the rate. After shut-in at the wellhead, $q = 0$ , the pressure rises, and $q_{sf} = (C/B)\,\dd p_w/\dd t > 0$ . This is the afterflow. In both cases the difference of the rates is proportional to the speed at which the pressure changes. As soon as the pressure starts to change slowly, the wellbore stops having an effect.

In other unit systems (2)

Metric oilfield units

(q_{sf} - q)\,B = 24\,C\,\frac{\dd p_w}{\dd t}

$q$ and $q_{sf}$ in m³/d, $C$ in m³/MPa, $p_w$ in MPa, $t$ in h.

24 is the number of hours in a day: the rate is measured per day, and the time in hours.

Oilfield units

(q_{sf} - q)\,B = 24\,C\,\frac{\dd p_w}{\dd t}

$q$ and $q_{sf}$ in STB/d (stock-tank barrels of oil per day), $C$ in bbl/psi, $p_w$ in psi, $t$ in h.

The same constant 24: the volumes are in barrels on both sides of the equation.

To compare different wells, we make the coefficient dimensionless, in the same way as time and pressure (dimensionless variables).

Dimensionless wellbore storage coefficient

C_D = \frac{C}{2\pi\,\phi\,c_t\,h\,r_w^2}

where:

$C_D$ dimensionless wellbore storage coefficient, —
$C$ wellbore storage coefficient, m³/MPa
$\phi$ porosity of the reservoir, fraction
$c_t$ total compressibility of the reservoir and the fluids in it, 1/MPa
$h$ net thickness of the reservoir, m
$r_w$ wellbore radius, m

Units: metric oilfield units or SI: no constant is needed if $C$ and $c_t$ are given per the same unit of pressure

Holds when

The same as for the coefficient $C$ .
$\phi$ , $c_t$ and $h$ refer to the reservoir, not to the wellbore.

The denominator is twice the volume of fluid that a cylinder of rock the size of the well itself would give out when the pressure falls by one unit. Here $\pi r_w^2h$ is the volume of the cylinder, and $\phi c_t$ is the storativity of the rock (its ability to store fluid by elastic expansion). So $C_D$ is half the ratio of the “capacity” of the wellbore to the capacity of the rock whose place it took. With this definition, the wellbore balance in dimensionless variables reads $q_{sf}/q = 1 - C_D\,\dd p_D/\dd t_D$ .

In other unit systems (1)

Oilfield units

C_D = \frac{5.615\,C}{2\pi\,\phi\,c_t\,h\,r_w^2} = \frac{0.894\,C}{\phi\,c_t\,h\,r_w^2}

$C$ in bbl/psi, $c_t$ in 1/psi, $h$ and $r_w$ in ft.

5.615 is the number of cubic feet in a barrel: the numerator is in barrels, and the denominator is in cubic feet; $0.894 = 5.615/(2\pi)$ .

For the reservoir of the examples ( $\phi$ = 0.20; $c_t = 1.5\cdot10^{-3}$ MPa⁻¹; $h$ = 10 m; $r_w$ = 0.108 m) the denominator is $2.2\cdot10^{-4}$ m³/MPa. A wellbore full of oil gives $C_D \approx 60$ . The wellbore with gas from the lesson's record ( $C$ = 0.31 m³/MPa, example below) gives about 1400, and the section below the valve at the packer gives about 5. In oil wells closed at the wellhead, the measured $C_D$ is often in the thousands. The reason is free gas in the wellbore.

CalculatedCalculated share of the rate change that the wellbore provides after start-up or shut-in, for the well of the examples at three values of

C

. The time scale is logarithmic. After start-up this share of the rate comes from the wellbore. After shut-in, this much of the former rate still arrives from the reservoir. The dots mark the classic estimate of the start of the line: by that time the wellbore accounts for about 1%.

The start of the record: a straight line on ordinary axes and the unit slope

In the very first moments the inflow from the reservoir has not had time to change. After start-up $q_{sf} \approx 0$ , and after shut-in $q_{sf} \approx q$ . The left-hand side of the wellbore balance is then constant, and the pressure changes at a constant speed.

Section of pure wellbore storage

\Delta p = \frac{qB}{C}\,t

where:

$\Delta p$ change of the bottomhole pressure since start-up or shut-in: $p_i - p_{wf}$ after start-up, $p_{ws} - p_{wf}(\Delta t = 0)$ after shut-in, Pa
$q$ change of the rate at the wellhead: the rate after start-up or the rate before shut-in, at standard conditions, m³/s
$B$ formation volume factor of the fluid, m³/m³
$C$ wellbore storage coefficient, m³/Pa
$t$ time since start-up or shut-in (after shut-in, $\Delta t$ ), s
$p_D,\ t_D,\ C_D$ in dimensionless form: the change of the bottomhole pressure $2\pi kh\,\Delta p/(qB\mu)$ , the time $kt/(\phi\mu c_tr_w^2)$ and the wellbore storage coefficient; $k$ , $h$ , $\mu$ are the permeability, the thickness of the reservoir and the viscosity, —

Units: SI

Holds when

The inflow from the reservoir has not changed yet: this is the section of pure wellbore storage.
The coefficient $C$ is constant.
Before start-up or shut-in, the pressure in the well changed far more slowly than after it.

The pressure changes linearly with time, and only the wellbore sets the speed. It is like a closed tank, where the $c_{wb}V_{wb}$ of the wellbore stands in place of the $c_tV_p$ of the reservoir (the rate of pressure decline in a closed area). The formula contains neither the permeability nor the skin: from these points we can learn nothing about the reservoir. But we find $C$ from them.

In other unit systems (3)

Metric oilfield units

\Delta p = \frac{qB}{24\,C}\,t,\qquad C = \frac{qB\,t}{24\,\Delta p}

$q$ in m³/d, $C$ in m³/MPa, $\Delta p$ in MPa, $t$ in h.

24 is the number of hours in a day.

Oilfield units

\Delta p = \frac{qB}{24\,C}\,t,\qquad C = \frac{qB\,t}{24\,\Delta p}

$q$ in STB/d, $C$ in bbl/psi, $\Delta p$ in psi, $t$ in h.

The same constant 24.

Dimensionless variables

p_D = \frac{t_D}{C_D},\qquad \frac{t_D}{C_D} = 0.02232\,\frac{kh\,t}{\mu\,C}

in the second formula $k$ is in mD, $h$ in m, $t$ in h, $\mu$ in mPa·s, $C$ in m³/MPa.

$0.02232 = 2\pi\cdot3.553\cdot10^{-3}$ ; in oilfield units the factor is $2\pi\cdot2.637\cdot10^{-4}/5.615 = 0.000295$ . The porosity, the compressibility of the reservoir and the wellbore radius cancel.

Take the logarithm: $\lg\Delta p = \lg t + \lg[qB/(24C)]$ . On a plot with two logarithmic scales this is a straight line with a slope of exactly 1. It is the unit-slope line: the pressure grows by the same factor as the time. The larger $C$ is, the lower this line lies. On ordinary axes of pressure against time, the same section is a straight line that starts at the beginning of the record.

Below is the assumed pressure build-up record that we analyse in this lesson. The well produced for a month at 40 m³/d, and then we closed it at the wellhead. A gauge at the bottom of the well records the pressure every second.

Rise of the bottomhole pressure after shut-in at the wellhead (assumed data)

$\Delta t$: 10 s
$\Delta p$ , MPa: 0.0179
$\Delta p/\Delta t$ , MPa/h: 6.44
Inflow from the reservoir, % of $q$: 99.4

$\Delta t$: 20 s
$\Delta p$ , MPa: 0.0356
$\Delta p/\Delta t$ , MPa/h: 6.41
Inflow from the reservoir, % of $q$: 99.0

$\Delta t$: 30 s
$\Delta p$ , MPa: 0.0533
$\Delta p/\Delta t$ , MPa/h: 6.40
Inflow from the reservoir, % of $q$: 98.5

$\Delta t$: 1 min
$\Delta p$ , MPa: 0.1060
$\Delta p/\Delta t$ , MPa/h: 6.36
Inflow from the reservoir, % of $q$: 97.2

$\Delta t$: 2 min
$\Delta p$ , MPa: 0.2092
$\Delta p/\Delta t$ , MPa/h: 6.28
Inflow from the reservoir, % of $q$: 94.8

$\Delta t$: 5 min
$\Delta p$ , MPa: 0.5044
$\Delta p/\Delta t$ , MPa/h: 6.05
Inflow from the reservoir, % of $q$: 88.4

$\Delta t$: 10 min
$\Delta p$ , MPa: 0.9542
$\Delta p/\Delta t$ , MPa/h: 5.73
Inflow from the reservoir, % of $q$: 79.2

$\Delta t$: 30 min
$\Delta p$ , MPa: 2.3501
$\Delta p/\Delta t$ , MPa/h: 4.70
Inflow from the reservoir, % of $q$: 52.8

$\Delta t$: 1 h
$\Delta p$ , MPa: 3.6554
$\Delta p/\Delta t$ , MPa/h: 3.66
Inflow from the reservoir, % of $q$: 30.5

$\Delta t$: 2 h
$\Delta p$ , MPa: 4.9029
$\Delta p/\Delta t$ , MPa/h: 2.45
Inflow from the reservoir, % of $q$: 11.7

$\Delta t$: 5 h
$\Delta p$ , MPa: 5.8343
$\Delta p/\Delta t$ , MPa/h: 1.17
Inflow from the reservoir, % of $q$: 1.9

$\Delta t$: 10 h
$\Delta p$ , MPa: 6.1764
$\Delta p/\Delta t$ , MPa/h: 0.62
Inflow from the reservoir, % of $q$: 0.6

$\Delta t$: 24 h
$\Delta p$ , MPa: 6.5102
$\Delta p/\Delta t$ , MPa/h: 0.27
Inflow from the reservoir, % of $q$: 0.2

$\Delta t$: 48 h
$\Delta p$ , MPa: 6.7474
$\Delta p/\Delta t$ , MPa/h: 0.14
Inflow from the reservoir, % of $q$: 0.1

$\Delta t$	$\Delta p$ , MPa	$\Delta p/\Delta t$ , MPa/h	Inflow from the reservoir, % of $q$
10 s	0.0179	6.44	99.4
20 s	0.0356	6.41	99.0
30 s	0.0533	6.40	98.5
1 min	0.1060	6.36	97.2
2 min	0.2092	6.28	94.8
5 min	0.5044	6.05	88.4
10 min	0.9542	5.73	79.2
30 min	2.3501	4.70	52.8
1 h	3.6554	3.66	30.5
2 h	4.9029	2.45	11.7
5 h	5.8343	1.17	1.9
10 h	6.1764	0.62	0.6
24 h	6.5102	0.27	0.2
48 h	6.7474	0.14	0.1

The pressure at the moment of shut-in is 16.307 MPa; $\Delta p = p_{ws} - p_{wf}(\Delta t = 0)$ . The data were calculated from a model of a homogeneous infinite reservoir ( $k$ = 20 mD, $S$ = 3) with wellbore storage ( $C$ = 0.31 m³/MPa), and rounded to 0.0001 MPa. The last column is the afterflow, calculated from the same model.

Worked exampleWellbore storage coefficient from the first seconds of the record

Given (assumed data)

Rate before shut-in $q$ ; formation volume factor $B$: 40 m³/d; 1.2 m³/m³
First readings: 10 s — 0.0179 MPa; 20 s — 0.0356 MPa; 30 s — 0.0533 MPa
Reservoir: $\phi$ = 0.20; $c_t = 1.5\cdot10^{-3}$ MPa⁻¹; $h$ = 10 m; $r_w$ = 0.108 m
Wellbore: tubing 7.55 m³, below the packer 0.71 m³ (previous example)

Solution

Check the unit slope. The time grew threefold, from 10 to 30 s. The pressure grew by a factor of $0.0533/0.0179 = 2.978$ . The slope on the log–log plot is $\lg 2.978/\lg 3 = 0.99$ , practically 1.
Speed of the pressure rise from the first point: 10 s is 1/360 h, so $\Delta p/\Delta t = 0.0179\cdot360 = 6.444$ MPa/h.
Wellbore storage coefficient: $C = \dfrac{qB}{24\,(\Delta p/\Delta t)} = \dfrac{40\cdot1.2}{24\cdot6.444} = 0.3104$ m³/MPa.
The same in oilfield units: $q = 251.6$ STB/d, $\Delta p = 2.596$ psi; $C = \dfrac{251.6\cdot1.2}{24\cdot360\cdot2.596} = 0.01346$ bbl/psi. Converting $0.01346/0.0434$ gives back the same 0.31 m³/MPa.
Dimensionless coefficient: $2\pi\phi c_thr_w^2 = 2.199\cdot10^{-4}$ m³/MPa, $C_D = 0.3104/(2.199\cdot10^{-4}) = 1412$ .
Comparison with the design: a wellbore full of oil would give 0.0132 m³/MPa, which is 23.5 times less. So there is free gas in the wellbore.
How much gas? The inflow both compresses the gas and makes the column heavier (previous example). So first we find the part of the coefficient that comes from the compression of the contents: $C_g = 1/(1/0.3104 - 2.665) = 1.80$ m³/MPa. At 7 MPa ( $c_g$ = 0.163 MPa⁻¹) this would need $1.80/0.163 = 11$ m³ of gas. That is more than the whole volume of the tubing (7.55 m³). So there is a lot of gas, and it is mostly at low pressure, in the upper part of the tubing. The coefficient of the record is close to the limit of 0.375 m³/MPa for closed 62 mm tubing.

Answer:

C \approx 0.31

m³/MPa (0.01346 bbl/psi),

C_D \approx 1412

. There is a lot of free low-pressure gas in the tubing.

What it means. We calculated the data from a model with $C$ = 0.31 m³/MPa, and the analysis returned this value. The model holds $C$ constant. In a real well, low-pressure gas is compressed as the pressure grows, and the coefficient would be smaller by the end of the record (Nonideal wellbore storage, well clean-up and changing skin). Look at the speed of the pressure rise in the table: after only a minute it is 1% lower than the initial speed, and after 5 minutes it is 6% lower. The section of pure wellbore storage is short. We can catch it only with frequent readings from the first seconds. These points say nothing about the reservoir: the same straight line would appear at any permeability. The data in this example are assumed for illustration.

From the wellbore to the reservoir: the transition section

The unit slope ends when the inflow from the reservoir has changed noticeably. After that, the pressure depends on the wellbore, the reservoir and the skin together. This is the transition section. It ends when the sandface rate becomes equal to the rate at the wellhead. From this moment the record is the same as that of a well without wellbore storage, and the straight line starts on the semilog plot.

Three sections at the start of the record

Section: Pure wellbore storage
What happens at the bottom of the well: the inflow from the reservoir has not changed yet
What it looks like: a straight line on ordinary axes; slope 1 on the log–log plot
What we get: $C$

Section: Transition
What happens at the bottom of the well: the inflow changes, and the skin pressure drop builds up
What it looks like: the curve leaves the slope 1 and flattens; the pressure derivative passes a maximum
What we get: $C$ , $kh$ and $S$ , only by fitting a model to the whole record (Type-curve matching)

Section: Radial flow
What happens at the bottom of the well: $q_{sf} = q$ after start-up, $q_{sf} = 0$ after shut-in
What it looks like: a semilog straight line; the derivative is horizontal
What we get: $kh$ , $S$ , reservoir pressure

Section	What happens at the bottom of the well	What it looks like	What we get
Pure wellbore storage	the inflow from the reservoir has not changed yet	a straight line on ordinary axes; slope 1 on the log–log plot	$C$
Transition	the inflow changes, and the skin pressure drop builds up	the curve leaves the slope 1 and flattens; the pressure derivative passes a maximum	$C$ , $kh$ and $S$ , only by fitting a model to the whole record (Type-curve matching)
Radial flow	$q_{sf} = q$ after start-up, $q_{sf} = 0$ after shut-in	a semilog straight line; the derivative is horizontal	$kh$ , $S$ , reservoir pressure

CalculatedCalculated pressure build-up after the shut-in of the well of the examples at three values of

C

, and without wellbore storage (dashed curve). Both scales are logarithmic. Each curve starts on its own unit-slope line and, after about two cycles, merges with the common curve of the reservoir. The dots are the readings from the table.

The skin switches on together with the inflow. The pressure drop across the skin is proportional to the rate at the bottom of the well (Skin factor). While only the wellbore works, almost nothing flows through the near-wellbore zone, and there is no extra pressure drop. It builds up in the transition section. So the skin makes the transition longer: the wellbore has to “discharge” through a larger resistance.

End of the unit-slope section

\frac{t_D}{C_D} \approx \alpha\,\ln\left(3\alpha\,C_D\,\e^{2S}\right)

where:

$t_D/C_D$ dimensionless time divided by the dimensionless wellbore storage coefficient; equal to $0.02232\,kh\,t/(\mu C)$ in metric oilfield units, —
$\alpha$ allowed relative deviation of the pressure from the unit-slope line: 0.01, 0.05 or 0.10, —
$C_D$ dimensionless wellbore storage coefficient, —
$S$ skin factor, —

Units: dimensionless

Holds when

An approximate formula for $C_D\e^{2S} > 10^3$ .
A constant $C$ and a homogeneous reservoir.
For $C_D$ from $10^2$ to $10^4$ and $S$ from 0 to 10, it differs by no more than 15% from the calculation with the lesson's model.

The larger the wellbore and the skin, the longer, in units of $t_D/C_D$ , the unit slope lasts. But it always ends early: in this range of $C_D$ and $S$ , the pressure leaves the line by 5% at $t_D/C_D$ between 0.1 and 1.5. In the record of the example ( $C_D\e^{2S} \approx 5.7\cdot10^5$ ) the formula gives 0.57, that is, 3.5 min. By the model, the deviation reaches 5% after 4 min.

The transition section is a trap. On a semilog plot it is smooth, and it is easy to draw a “straight line” through its points. This line is steeper than the true one, because here the pressure is still catching up with the curve of the reservoir.

CalculatedThe same calculated curves on a semilog plot. All three reach the same radial-flow line with a slope of 0.77 MPa per cycle, but at different times. The dashed line is the false line through the points at 2 and 10 h of the record with gas in the wellbore. Its slope is 1.82 MPa per cycle.

How long wellbore storage lasts

Together, the wellbore and the reservoir behave like a tank that fills or empties through a resistance. The tank is $C$ . The resistance is the pressure drop that we need to push a unit rate through the reservoir and the skin. It equals $\mu\,p_D/(2\pi kh)$ , where $p_D$ is the dimensionless pressure drop of a well without wellbore storage (the pressure function). Multiply the two and the result is a time.

Time constant of the wellbore

\tau = \frac{\mu\,C\,p_D}{2\pi kh},\qquad p_D = \tfrac12\left(\ln t_D + 0.809\right) + S

where:

$\tau$ time constant: in this time the difference between the sandface rate and the wellhead rate falls by a factor of about $\e \approx 2.7$ , s
$C$ wellbore storage coefficient, m³/Pa
$p_D$ dimensionless pressure drop of a well without wellbore storage at the time we look at; usually between 5 and 15, —
$\mu,\ k,\ h$ viscosity, permeability, thickness of the reservoir, Pa·s, m², m
$t_D,\ S$ dimensionless time and skin factor, —

Units: SI

Holds when

A rough model: the reservoir and the skin are replaced by a constant resistance, though in reality $p_D$ slowly grows with time.
It is good for an estimate of the order of magnitude and for the first two or three $\tau$ . The “tail” of a real transition is longer than the exponential.

In this model the wellbore's share of the rate falls as $\e^{-t/\tau}$ . The time constant is proportional to $C$ and inversely proportional to the transmissibility $kh/\mu$ (the flow capacity divided by the viscosity). The skin enters $p_D$ as a term and makes the transition longer. There is no rate in the formula: if the rate is twice as large, all the pressure drops double, and the times stay the same.

In other unit systems (1)

Metric oilfield units

\tau = 44.8\,\frac{\mu\,C\,p_D}{kh}

$\tau$ in h, $\mu$ in mPa·s, $C$ in m³/MPa, $k$ in mD, $h$ in m.

$44.8 = 24/0.536 = 1/(2\pi\cdot3.553\cdot10^{-3})$ , where 0.536 is the constant of the Dupuit equation in these units.

An exact calculation of the transition uses the solution with wellbore storage and skin (advanced section). A comparison of the calculated curves with and without wellbore storage gave a simple rule that we use when we plan tests:

Estimate of the start of the semilog straight line

t_D \approx (60 + 3.5\,S)\,C_D

where:

$t_D$ dimensionless time of the start of the line, $kt/(\phi\mu c_tr_w^2)$ , —
$C_D$ dimensionless wellbore storage coefficient, —
$S$ skin factor, —
$t,\ C$ in the dimensional forms: the time of the start of the line and the wellbore storage coefficient, h; m³/MPa or bbl/psi
$\mu,\ k,\ h$ in the dimensional forms: viscosity of the fluid, permeability and thickness of the reservoir, mPa·s or cP; mD; m or ft

Units: dimensionless

Holds when

A guide, not an exact limit: it comes from a comparison of calculated curves. It applies for $C_D\e^{2S} > 10^3$ .
A constant $C$ ; until this time the reservoir behaves as an infinite one.
For a build-up: if the shut-in time is much shorter than the producing time of the well.

By this time the inflow from the reservoir differs from the rate at the wellhead by about 1%, and the pressure differs by less than 2% from the pressure of a well without wellbore storage. But the slope of the curve is still too high. In the calculations for the well of the examples, the logarithmic derivative (the slope of the curve on a semilog plot) at this moment is 1.2–1.4 times its level in radial flow. With a skin of about 10 it is one and a half to two times that level. It gets within 10% of this level after a time 2–4 times longer than the classic estimate, and within 5% after about twice as long again.

In other unit systems (2)

Metric oilfield units

t \approx 44.8\,(60 + 3.5\,S)\,\frac{\mu\,C}{kh}

$t$ in h, $\mu$ in mPa·s, $C$ in m³/MPa, $k$ in mD, $h$ in m.

$44.8 = 1/0.02232$ is the inverse of the factor in $t_D/C_D$ .

Oilfield units

t \approx 3390\,(60 + 3.5\,S)\,\frac{\mu\,C}{kh}

$t$ in h, $\mu$ in cP, $C$ in bbl/psi, $k$ in mD, $h$ in ft.

$3390 = 1/0.000295$ . If we multiply out the brackets and round, we get the well-known form $(200{,}000 + 12{,}000\,S)\,\mu C/(kh)$ .

Worked exampleWhen radial flow starts with three ways of shutting in

Given (assumed data)

Reservoir and well: $k$ = 20 mD; $h$ = 10 m; $\mu$ = 1.5 mPa·s; $S$ = 3
Shut-in at the packer: $C$ = 0.00114 m³/MPa
Shut-in at the wellhead, oil in the wellbore: $C$ = 0.0132 m³/MPa
Shut-in at the wellhead, gas in the tubing (the lesson's record): $C$ = 0.31 m³/MPa

Solution

The common factor: $44.8\,\mu/(kh) = 44.8\cdot1.5/200 = 0.336$ h·MPa/m³; $60 + 3.5\cdot3 = 70.5$ .
Start of the line by the classic estimate: $t = 0.336\cdot70.5\cdot C$ . At the packer, 1.6 min; wellbore with oil, 19 min; wellbore with gas, 7.3 h.
Check of the last value in oilfield units: $C = 0.31\cdot0.0434 = 0.01345$ bbl/psi, $kh = 200\cdot3.2808 = 656.2$ mD·ft; $t = 3390\cdot70.5\cdot1.5\cdot0.01345/656.2 = 7.3$ h.
Time constant for the wellbore with gas: for a well without wellbore storage, after one hour of production, $p_D = 8.16$ , so $\tau = 0.336\cdot8.16\cdot0.31 = 0.85$ h. One hour after shut-in, $\e^{-1/0.85} = 0.31$ of the former rate should still flow into the wellbore, that is, 31%. In the table of the lesson it is 30.5%.
On the calculated curve, the pressure derivative gets within 10% of the radial-flow level after 3 min, 0.76 h and 22 h. This is 1.9–3.0 times later than the classic estimate. It gets within 5% after about twice as long again. For a build-up, these times are right if we take the derivative against the equivalent time, which allows for how long the well produced before shut-in (Effect of producing time on build-up analysis).
To record at least one logarithmic cycle of the line after that, the time must be 10 times longer: about 0.5 h, 8 h and 220 h (9 days).

Answer: Radial flow starts after minutes when we shut in at the packer, after less than an hour when the wellbore is full of oil, and only after a day if there is gas in the wellbore.

What it means. The reservoir is the same in all three cases, and only $C$ changes. If we shut in a wellbore with gas at the wellhead, a full record would need nine days of shut-in. In that time the boundaries and the neighbouring wells would have time to affect the pressure. Shut-in at the packer cuts the record to half an hour. The exponential estimate with the time constant describes the first half of the transition well, and it underestimates the “tail”: after 5 h it gives 0.3% of the former rate, while the model gives 2%. The data in this example are assumed for illustration.

There is also an older rule of thumb, the “rule of one and a half cycles”: the line starts about one and a half logarithmic cycles after the end of the unit-slope section. One and a half cycles means multiplying the time by 32. In the record of the example, the unit slope ends after 4–9 min (a deviation of 5–10%), and the rule gives 2–5 h. Both estimates tell us when the pressure comes up to the line, but not when its slope settles. For planning, they are good with a margin of a few times. On a finished record we find the start of radial flow from the pressure derivative.

How to shorten wellbore storage and how to allow for it

Shut the well in at the bottom. A shut-in valve at the packer leaves tenths of a cubic metre of liquid below it instead of the whole wellbore. In the lesson's example $C$ becomes almost 300 times smaller than for a wellbore with gas, and radial flow starts after minutes. A drill stem test (a test of a formation through the drill pipe) works this way (Well testing: purposes and types of test). For producing wells there are valves that we lower on a wire or a cable (downhole shut-in).
Isolate the annulus. Without a packer (a seal between the tubing and the casing), the annulus (the space between the tubing and the casing) also becomes part of the “wellbore”. Its volume is usually several times the volume of the tubing, and gas often collects at the top of it.
Keep one phase in the wellbore (liquid only, with no free gas). If the wellhead pressure is above the bubble point pressure (the pressure at which gas starts to come out of the oil), there is no free gas in the wellbore, and $C$ is small and constant. This is not always possible, but we should remember it when we choose how the well works before shut-in.
Record the start often and exactly. Readings every second in the first minutes, and an exactly fixed moment of start-up or shut-in, let us see the unit slope and find $C$ .
Measure the sandface rate. If we record the flow rate with a downhole flowmeter together with the pressure, we can allow for the changing inflow by superposition (Superposition principle), and the early data are no longer lost.
Calculate before the test. We put the expected $C$ , $kh$ and $S$ into the formulas of this lesson or into a calculation model. Then we check whether radial flow has time to appear and whether the planned duration is enough.

Sometimes there is no radial-flow section in the record at all: wellbore storage has not ended yet, and the reservoir boundaries or the neighbouring wells are already having an effect. On a semilog plot such a record still looks like a smooth curve with “almost straight” stretches, and this is one of the main reasons for wrong interpretations. So modern analysis does not throw away the early points. It fits a model that describes the whole record at once (wellbore, skin, reservoir and boundaries), and it checks the result for common sense (Type-curve matching, Diagnosis with the pressure derivative).

When a constant

C

is not enough

What happens in the wellbore: After shut-in, gas rises and liquid settles
How it differs from the model: the pressure in the wellbore is also raised by the rising gas, and not only by the inflow from the reservoir
Sign in the record: a “hump” on the build-up curve

What happens in the wellbore: A liquid level rises or falls
How it differs from the model: the wellbore stores fluid by the rise of the level: $C$ is much larger, and it changes when the level reaches the wellhead
Sign in the record: a very long unit-slope section, then a sharp change in the course of the curve

What happens in the wellbore: Gas comes out of the oil or dissolves in it
How it differs from the model: the compressibility of the contents of the wellbore, and with it $C$ , changes with the pressure
Sign in the record: the initial slope on the log–log plot differs from 1

What happens in the wellbore: The pressure in a gas well changes severalfold
How it differs from the model: the gas compressibility is about $1/p$ , and it changes with the pressure
Sign in the record: different values of $C$ at start-up and at shut-in

What happens in the wellbore	How it differs from the model	Sign in the record
After shut-in, gas rises and liquid settles	the pressure in the wellbore is also raised by the rising gas, and not only by the inflow from the reservoir	a “hump” on the build-up curve
A liquid level rises or falls	the wellbore stores fluid by the rise of the level: $C$ is much larger, and it changes when the level reaches the wellhead	a very long unit-slope section, then a sharp change in the course of the curve
Gas comes out of the oil or dissolves in it	the compressibility of the contents of the wellbore, and with it $C$ , changes with the pressure	the initial slope on the log–log plot differs from 1
The pressure in a gas well changes severalfold	the gas compressibility is about $1/p$ , and it changes with the pressure	different values of $C$ at start-up and at shut-in

These cases are covered in the lesson Nonideal wellbore storage, well clean-up and changing skin.

What else distorts the early section

Cause: Hydraulic fracture
How it differs from the simple model: the flow near the wellbore is linear, not radial
Where it is covered: Inflow to a hydraulically fractured well

Cause: Well clean-up
How it differs from the simple model: the skin falls during the test
Where it is covered: Nonideal wellbore storage, well clean-up and changing skin

Cause: Partial penetration
How it differs from the simple model: the flow converges to the opened interval
Where it is covered: Partial penetration and well deviation

Cause: A wide zone of changed permeability
How it differs from the simple model: the zone can no longer be treated as a thin layer with a skin
Where it is covered: Skin factor

Cause	How it differs from the simple model	Where it is covered
Hydraulic fracture	the flow near the wellbore is linear, not radial	Inflow to a hydraulically fractured well
Well clean-up	the skin falls during the test	Nonideal wellbore storage, well clean-up and changing skin
Partial penetration	the flow converges to the opened interval	Partial penetration and well deviation
A wide zone of changed permeability	the zone can no longer be treated as a thin layer with a skin	Skin factor

These effects overlap with wellbore storage, and it partly hides them. The shorter wellbore storage is, the more we see on the early section.

Terms and symbols

Wellbore storage: The effect that makes the sandface rate change only gradually after the well is started or shut in at the wellhead: the wellbore gives out or takes in fluid.
Wellbore storage coefficient $C$: The volume of fluid that the wellbore gives out or takes in when the bottomhole pressure changes by one unit. For a wellbore filled with one fluid it is $c_{wb}V_{wb}$ . Units: m³/MPa or bbl/psi.
Dimensionless coefficient $C_D$: $C/(2\pi\phi c_thr_w^2)$ : the wellbore storage coefficient divided by twice the volume of fluid that a cylinder of rock the size of the well gives out when the pressure falls by one unit.
Sandface flow rate $q_{sf}$: The inflow from the reservoir into the wellbore. After start-up or shut-in it differs from the rate at the wellhead while the contents of the wellbore are being compressed or are expanding.
Afterflow: The sandface rate after the well is closed at the wellhead. It fades as the pressure in the wellbore grows.
Unit-slope line: The start of the record on a log–log plot, where $\Delta p$ is proportional to time. It is the sign of pure wellbore storage.
Transition section: The interval between the end of the unit slope and the start of radial flow. The pressure depends on the wellbore, the reservoir and the skin together.

Common mistakes and misconceptions

Wrong

“We can process the early part of the record as radial flow if the points lie on a straight line.”

In fact

In the transition section the points also lie on a “straight line”, but it is steeper than the true one. In the lesson's example the line through 2 and 10 h underestimates the permeability by a factor of 2.4.

Wrong

“The wellbore storage coefficient is the volume of the wellbore.”

In fact

It is the volume multiplied by the compressibility of the contents. For the same volume, gas in the wellbore makes $C$ tens of times larger, and a moving liquid level makes it larger still.

Wrong

“The higher the rate, the longer wellbore storage lasts.”

In fact

The duration does not depend on the rate. It is set by $C$ , the transmissibility $kh/\mu$ and the skin. The rate changes only the size of the pressure drops.

Wrong

“As soon as the points leave the unit slope, the semilog straight line has started.”

In fact

Between them lies the transition section, about two logarithmic cycles long. In the lesson's example the unit slope ends after minutes, and radial flow starts only after a day.

Wrong

“We can estimate the permeability from the unit-slope section.”

In fact

The equation of this section contains no reservoir properties: it gives only $C$ . The permeability and the skin appear in the record together with the inflow from the reservoir.

In depth: advanced detail

The condition at the bottom of the well and the solution in Laplace space

In dimensionless variables the inflow from the reservoir equals the pressure gradient at the wall of the well: $q_{sf}/q = -r_D\,\partial p_D/\partial r_D$ at $r_D = 1$ (the condition of constant rate is the special case where this quantity equals 1). We put the wellbore balance into it and get a new condition at the bottom of the well. The pressure drop across the skin is proportional to the same inflow.

Condition at the bottom of the well for a well with wellbore storage and skin

\begin{aligned} &\left.r_D\,\frac{\partial p_D}{\partial r_D}\right|_{r_D = 1} = -1 + C_D\,\frac{\dd p_{wD}}{\dd t_D}\\[4pt] &p_{wD} = p_D(1, t_D) - S\left.r_D\,\frac{\partial p_D}{\partial r_D}\right|_{r_D = 1}\end{aligned}

where:

$p_D(r_D, t_D)$ dimensionless pressure drop in the reservoir, —
$p_{wD}(t_D)$ dimensionless pressure drop in the well itself (behind the layer with the skin), —
$C_D,\ S$ dimensionless wellbore storage coefficient and skin factor, —
$r_D,\ t_D$ dimensionless distance and time, —

Units: dimensionless; $p_D$ and $p_{wD}$ are related to the rate at the wellhead

Holds when

The rate at the wellhead is constant from start-up, and $C_D$ and $S$ are constant.
The skin is an infinitely thin layer, and the pressure drop across it is proportional to the inflow from the reservoir at that moment.
The other conditions are as in the problem of starting a well (Diffusivity equation).

For $C_D = 0$ the first equality turns into the condition of constant rate. For large $C_D$ and a short time, the right-hand side is close to zero: the reservoir “does not notice” the start-up. The condition is linear, so we can still add solutions. For a build-up this means that the pressure rise after shut-in repeats the start-up curve while the shut-in time is short compared with the producing time.

The simplest way to solve the problem with such a condition is in Laplace space (we solve the equation after the Laplace transform, in terms of a variable $s$ instead of time). Let $\bar p_D^{\,*}(s)$ be the Laplace transform of the solution of the same problem without wellbore storage and without skin, that is, at a constant rate at the bottom of the well. A changing inflow $q_{sf}(t)$ is a sequence of small steps of the rate, and we find the response to it by superposition (Superposition principle). In Laplace space, superposition becomes multiplication. Together with the wellbore balance we get two equalities: $\bar p_{wD} = \bar q_D\,(s\,\bar p_D^{\,*} + S)$ and $\bar q_D = 1/s - C_D\,s\,\bar p_{wD}$ , where $\bar q_D$ is the Laplace transform of the share $q_{sf}/q$ . We eliminate $\bar q_D$ and get:

A well with wellbore storage and skin: the Laplace transform

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

where:

$\bar p_{wD}(s)$ Laplace transform of the dimensionless pressure in a well with wellbore storage and skin, —
$\bar p_D^{\,*}(s)$ Laplace transform of the solution at a constant rate at the bottom of the well, without wellbore storage and skin: $K_0(\sqrt{s})/s$ for a line source, $K_0(\sqrt{s})/(s^{3/2}K_1(\sqrt{s}))$ for a well of finite radius, —
$s$ variable of the Laplace transform, which corresponds to the dimensionless time $t_D$ , —
$K_0,\ K_1$ modified Bessel functions of the second kind, —
$C_D,\ S$ dimensionless wellbore storage coefficient and skin factor, —

Units: dimensionless

Holds when

$C_D$ and $S$ are constant; the rate at the wellhead is constant from start-up.
The formula holds for any solution $\bar p_D^{\,*}$ , including that of a bounded reservoir: the wellbore and the skin change only the condition at the bottom of the well.

Check the limits. For large $s$ (a short time) the expression tends to $1/(C_Ds^2)$ , and this is the Laplace transform of the function $t_D/C_D$ , the unit slope. For small $s$ (a long time) the wellbore term vanishes, and what is left is $\bar p_D^{\,*} + S/s$ , the solution without wellbore storage and with a skin. If we replace the reservoir and the skin with a constant resistance $R$ , that is, put $s\,\bar p_D^{\,*} + S = R$ , the inversion gives $p_{wD} = R\,(1 - \e^{-t_D/(C_DR)})$ . This is an exponential with the time constant from the main section.

For a real reservoir the inverse transform cannot be written as a formula, so we do it numerically, with a computer. All the curves and data of this lesson were calculated in this way. For $\bar p_D^{\,*}$ we took the solution for a well of finite radius (the Laplace transform).

Numerical inversion: the Stehfest algorithm

The Stehfest algorithm recovers the value of a function at a given time from the values of its Laplace transform at several points on the real axis. We need no integrals in the complex plane, only a weighted sum.

The Stehfest algorithm

f(t) \approx \frac{\ln 2}{t}\sum_{i=1}^{N} V_i\,\bar f\!\left(\frac{i\,\ln 2}{t}\right)

where:

$f(t)$ the function of time that we are looking for, —
$\bar f(s)$ its Laplace transform, $\int_0^\infty f(t)\,\e^{-st}\,\dd t$ , —
$N$ number of terms; even, —
$V_i$ weight coefficients that depend only on $N$ , —

Units: dimensionless; $t$ and $s$ are in consistent units

Holds when

The function $f(t)$ is smooth, without jumps or oscillations near the time $t$ .
We usually take $N$ from 8 to 16. With fewer terms the accuracy is low. With more terms, huge weights of different signs eat up the significant digits.

For each time we calculate the transform at $N$ points $s_i = i\ln 2/t$ , and only there. The curves of this lesson are calculated with sixteen terms. The pressure comes out with four or five correct digits, and its logarithmic derivative in the transition section with three.

Weight coefficients of the Stehfest algorithm

V_i = (-1)^{N/2 + i}\sum_{k = \lfloor (i+1)/2 \rfloor}^{\min(i,\,N/2)} \frac{k^{N/2}\,(2k)!}{(N/2 - k)!\;k!\;(k-1)!\;(i-k)!\;(2k-i)!}

where:

$V_i$ weight of the $i$ -th term, $i = 1 \dots N$ , —
$N$ even number of terms, —
$k$ summation index; the lower limit is the integer part of $(i+1)/2$ , —

Units: dimensionless

Holds when

$N$ is even. The sum of all the weights is zero, which is a handy check of the calculation.

The weights alternate in sign and grow quickly with $N$ : for $N = 8$ the largest in absolute value is about $1.9\cdot10^4$ , and for $N = 12$ it is already $8\cdot10^6$ . So we must calculate the transform with high accuracy.

Weights of the Stehfest algorithm for

N = 8

$i$: 1
$V_i$: −0.333

$i$: 2
$V_i$: 48.333

$i$: 3
$V_i$: −906

$i$: 4
$V_i$: 5464.667

$i$: 5
$V_i$: −14,376.667

$i$: 6
$V_i$: 18,730

$i$: 7
$V_i$: −11,946.667

$i$: 8
$V_i$: 2986.667

$i$	$V_i$
1	−0.333
2	48.333
3	−906
4	5464.667
5	−14,376.667
6	18,730
7	−11,946.667
8	2986.667

Exact values: −1/3, 145/3, −906, 16,394/3, −43,130/3, 18,730, −35,840/3, 8960/3.

Worked exampleChecking the algorithm and calculating a point of the curve with wellbore storage

Given (assumed data)

A function with a known answer: $\bar f(s) = 1/(s + 1)$ , the original is $f(t) = \e^{-t}$
The lesson's well with gas in the wellbore: $C_D$ = 1410; $S$ = 3

Solution

At $t = 1$ and $N = 8$ the points are $s_i = i\ln 2$ : 0.693; 1.386; … ; 5.545. The sum is $\sum V_i/(s_i + 1) = 0.5297$ , and the result is $\ln 2\cdot0.5297 = 0.3672$ .
The exact value is $\e^{-1} = 0.3679$ : the error is 0.2%. With $N = 12$ we get 0.36787, and the error is 0.003%.
Now the curve with wellbore storage. At the time $t_D = C_D$ , the unit-slope line would give $p_{wD} = 1$ . Inverting the formula with wellbore storage and skin (sixteen terms) gives $p_{wD} = 0.925$ , which is 7.5% less.
In real units for the lesson's record: $t_D/C_D = 1$ corresponds to 6.2 min, and $p_{wD} = 0.925$ is a pressure rise of 0.62 MPa.

Answer: The algorithm reproduces

\e^{-1}

with an error of 0.2% with eight terms and 0.003% with twelve. For the lesson's well at

t_D/C_D = 1

, the pressure is 7.5% below the unit-slope line.

What it means. A check on a function with a known original is a necessary step before we trust a calculation of curves. The value on the unit-slope line at $t_D/C_D = 1$ agrees with the table of the lesson: between the readings at 5 and 10 min the deviation from the line grows from 6 to 11%. The data in this example are assumed for illustration.

Connections to other topics

Pressure drawdown analysis — the semilog straight line after start-up; wellbore storage lifts the early points above it.
Pressure build-up analysis — afterflow pulls the early points below the Horner line.
Well testing: purposes and types of test — the early, middle and late sections of a record; shut-in at the bottom of the well in a drill stem test.
Skin factor — the pressure drop across the skin appears together with the inflow from the reservoir, and it makes the transition section longer.
Line-source solution and the pressure function — the solution without wellbore storage, which the record reaches after the transition section, and its Laplace transform.
Diffusivity equation — the condition at the wall of the well and the dimensionless variables in which the coefficient $C_D$ is written.
Gas formation volume factor, compressibility and viscosity — the compressibility of gas, which is why free gas in the wellbore raises $C$ so much.
Type-curve matching — how we process the whole record, the early section included, with a family of calculated curves for different $C_D$ and $S$ .
Diagnosis with the pressure derivative — how the pressure derivative shows the end of wellbore storage and the start of radial flow.
Nonideal wellbore storage, well clean-up and changing skin — a changing wellbore storage coefficient: liquid level, redistribution of phases, well clean-up.

Check yourself

1
On a log–log plot the first points of the record lie on a straight line with slope 1. What can we find from this section?
1. A.The distance to the reservoir boundary
2. B.The wellbore storage coefficient
3. C.The permeability of the reservoir
4. D.The skin factor
Show the answer
Correct answer: B. The wellbore storage coefficient
On the unit-slope section $\Delta p = qBt/(24C)$ : the pressure changes because the fluid in the wellbore is compressed or expands, and the inflow from the reservoir has not changed yet. The formula contains neither the permeability nor the skin, so these points give only $C$ .
2
An injection well is filled with water up to the wellhead. The volume of the wellbore is 12 m³, and the compressibility of water is $0.45\cdot10^{-3}$ MPa⁻¹. The reservoir: $\phi$ = 0.25; $c_t = 1.2\cdot10^{-3}$ MPa⁻¹; $h$ = 8 m; $r_w$ = 0.1 m. Find $C$ and $C_D$ .
Show the answer
$C = 0.45\cdot10^{-3}\cdot12 = 0.0054$ m³/MPa. The denominator: $2\pi\cdot0.25\cdot1.2\cdot10^{-3}\cdot8\cdot0.1^2 = 1.51\cdot10^{-4}$ m³/MPa. $C_D = 0.0054/(1.51\cdot10^{-4}) \approx 36$ . The value is small: a wellbore filled with one liquid acts for a short time.
3
After a well is started at 60 m³/d ( $B$ = 1.1), the points of the first minutes lie on a unit-slope line. After 0.02 h the pressure has fallen by 0.5 MPa. What is the wellbore storage coefficient?
Show the answer
$C = \dfrac{qB\,t}{24\,\Delta p} = \dfrac{60\cdot1.1\cdot0.02}{24\cdot0.5} = 0.11$ m³/MPa. For a wellbore of about 10 m³ filled with oil, we would get about 0.016 m³/MPa. A value seven times larger points to free gas in the wellbore.
4
The test was repeated at twice the rate. How will the time at which wellbore storage ends change (the properties of the fluid in the wellbore are the same)?
1. A.It will not change
2. B.It will halve
3. C.It will grow fourfold
4. D.It will double
Show the answer
Correct answer: A. It will not change
The rate is not in the estimate $t \approx 44.8\,(60 + 3.5S)\,\mu C/(kh)$ . When the rate doubles, all the pressure drops double, and the times stay the same. The duration is set by $C$ , the transmissibility and the skin.
5
For the well from the lesson Pressure drawdown analysis ( $k$ = 85 mD, $h$ = 14 m, $\mu$ = 0.9 mPa·s, $S$ = 3, $C$ = 0.06 m³/MPa), estimate the start of the semilog straight line by the classic formula, and compare it with what the record showed.
Show the answer
$60 + 3.5\cdot3 = 70.5$ ; $kh = 85\cdot14 = 1190$ mD·m; $t \approx 44.8\cdot70.5\cdot0.9\cdot0.06/1190 \approx 0.14$ h, that is, about 9 min. In that record the pressure derivative levelled out only by 2 h. There is no contradiction. The classic estimate shows when the pressure comes within 1–2% of the line, and the slope settles later. By the calculation for this well, the derivative gets within 10% of the radial-flow level after about 0.4 h, and within 2% only by 2 h.
6
Why does shutting in at a valve at the packer shorten wellbore storage more than the smaller volume alone would suggest?
1. A.The valve raises the reservoir pressure
2. B.The valve increases the permeability of the near-wellbore zone
3. C.Shutting in at the bottom reduces the skin
4. D.Only liquid is left below the valve, and the gas that has collected in the tubing is cut off
Show the answer
Correct answer: D. Only liquid is left below the valve, and the gas that has collected in the tubing is cut off
The coefficient $C$ is the volume times the compressibility. A valve at the packer reduces the volume ten times or more, and at the same time it cuts off the tubing, where the free gas with its high compressibility is. In the lesson's example $C$ falls from 0.31 to 0.0011 m³/MPa, almost 300 times, although the volume fell only 12 times.

Updated: 3 October 2026