Line-source solution and the pressure function

Quick refresher

In an unbounded reservoir, the pressure depends not on distance and time separately but on one combination of the two, $x = r^2/(4\eta t)$ . This is the square of the ratio of the distance to the radius of investigation (how far the pressure disturbance has travelled).
The solution is $p_i - p = \frac{qB\mu}{4\pi kh}\,E_1(x)$ , where $E_1(x) = -\mathrm{Ei}(-x)$ is the exponential integral. It is large for small $x$ and practically zero for $x > 5$ .
When $x < 0.01$ , we can replace the function with a logarithm: $E_1(x) \approx -\ln(1.781x)$ , with an error of less than 0.25%. At the wall of a well this condition holds within seconds after start-up. A hundred metres away it holds only after weeks or months.
At the well, $p_D = \tfrac12(\ln t_D + 0.809) + S$ , where $S$ is the skin factor (a number that shows how much the zone right around the well helps or hinders flow). Every time the time grows tenfold, the pressure drop grows by the same amount.
The solution holds until the pressure disturbance reaches the boundaries. For a well at the centre of a circle, that is roughly until $t_D/r_{eD}^2 = 0.25$ . After that, pseudo-steady-state or steady-state flow takes over.
If the bottomhole pressure is fixed at the well instead of the rate, the rate falls with time roughly as $1/p_D(t_D)$ .

Key relationships

p_i - p(r,t) = \dfrac{qB\mu}{4\pi kh}\,E_1\!\left(\dfrac{\phi\mu c_t r^2}{4kt}\right)

line-source solution (SI)

E_1(x) \approx -\ln(1.781\,x)\quad \text{for}\ x < 0.01

logarithmic approximation

p_D(t_D) = \tfrac12\left(\ln t_D + 0.809\right) + S

pressure function of a well in an infinite reservoir, $t_D > 25$

p_i - p = 0.933\,\dfrac{qB\mu}{kh}\,E_1\!\left(\dfrac{\phi\mu c_t r^2}{0.01421\,kt}\right)

metric oilfield units: m³/d, mPa·s, mD, m, 1/MPa, h, MPa

q_D(t_D) \approx \dfrac{2}{\ln t_D + 0.809}

dimensionless rate at constant bottomhole pressure

Common pitfalls

Using the logarithmic formula far from the well. For $x > 0.56$ it gives a negative drop, that is, a “rise” in pressure.
Confusing signs and notation. $E_1(x)$ and $-\mathrm{Ei}(-x)$ are the same positive number. Euler's constant is 0.5772, but the flow formulas contain its exponential, 1.781.
Extending the logarithmic line back to the first moments after start-up. There the well can no longer be treated as a line.
Using the solution for an infinite reservoir after the disturbance has reached the boundaries.
Putting time in days or distance in feet into the argument with the constant 0.01421, which is made for hours and metres.

After this lesson you can

set up the problem of a well started at a constant rate in an unbounded reservoir, and explain why the well is replaced by a line source
use the exponential integral to calculate the pressure at a given point of the reservoir at a given moment, in metric and in oilfield units
use the logarithmic approximation, and check whether it is valid at the well and away from it
state the time limits of the solution, and choose the right formula for the pressure function $p_D(t_D)$ for each period
explain how the solution at constant bottomhole pressure differs from the solution at constant rate, and where it is needed

The problem: a well in an unbounded reservoir

A well is started at a constant rate in a reservoir where the pressure is the same everywhere and equal to the initial pressure. What will the pressure be at any point of the reservoir at any moment, as long as the disturbance has not reached the boundaries? This is a problem for the diffusivity equation with three conditions (the initial and boundary conditions). First, the pressure is $p_i$ throughout the reservoir before start-up. Second, the rate at the well is constant. Third, the pressure is still $p_i$ at infinite distance.

Strictly, the condition at the well is set at its wall, at $r = r_w$ . But a well has a radius of only about 0.1 m, while the pressure changes over distances of tens or hundreds of metres. So we replace the well by a line of zero radius, through which the same rate is withdrawn. This is the line source. For a producing well, “line sink” would be more accurate, but the term is used for both production and injection.

Constant-rate condition for the line source

\lim_{r\to 0}\left(r\,\frac{\partial p}{\partial r}\right) = \frac{qB\mu}{2\pi kh}

where:

$p(r, t)$ pressure at distance $r$ from the well axis at time $t$ , Pa
$q,\ B$ rate at standard conditions and formation volume factor, m³/s, m³/m³
$k,\ h,\ \mu$ permeability, reservoir thickness, fluid viscosity, m², m, Pa·s

Units: SI

Holds when

The rate is constant from the moment of start-up.
The well penetrates the full thickness of the reservoir, and the inflow is uniform over the thickness.

This is the same condition as at the wall of a real well, only moved to its axis: the whole flow rate passes through a cylinder of any small radius around the axis. The well radius drops out of the problem.

One variable instead of two

The line-source problem has no length scale of its own: the well radius has dropped out, and the reservoir is infinite. From the distance $r$ , the time $t$ and the hydraulic diffusivity $\eta$ (how fast pressure spreads through the rock), we can make only one dimensionless combination (a number without units): $r^2/(\eta t)$ . So the pressure depends not on distance and time separately, but only on this combination. It is called the Boltzmann variable:

Similarity variable (Boltzmann variable)

x = \frac{r^2}{4\eta t} = \frac{\phi\mu c_t\,r^2}{4kt} = \left(\frac{r}{r_i}\right)^{2},\qquad r_i = \sqrt{4\eta t}

where:

$x$ dimensionless variable on which the solution depends, —
$r,\ t$ distance from the well axis and time since start-up, m, s
$\eta$ hydraulic diffusivity $k/(\phi\mu c_t)$ , m²/s
$r_i$ radius of investigation at time $t$ (how far the disturbance has spread), m

Units: SI

Holds when

An unbounded homogeneous reservoir, a line source and a constant rate from start-up.

The variable $x$ is the square of the ratio of the distance to the radius of investigation. When $x \ll 1$ , the point lies deep inside the drawdown cone (the region where the pressure has dropped). When $x \approx 1$ , it is at the edge of the cone. When $x \gg 1$ , the disturbance has not reached it yet. The pressure profiles at different times are similar to one another: if we stretch the distances in proportion to $\sqrt t$ , we get one and the same curve.

In other unit systems (2)

Metric oilfield units

x = \frac{\phi\mu c_t\,r^2}{0.01421\,k\,t}

$r$ in m, $t$ in h, $k$ in mD, $\mu$ in mPa·s, $c_t$ in 1/MPa.

$0.01421 = 4\cdot3.553\cdot10^{-3}$ , where $3.553\cdot10^{-3}$ is the diffusivity factor in these units (Diffusivity equation).

Oilfield units

x = \frac{948\,\phi\mu c_t\,r^2}{k\,t}

$r$ in ft, $t$ in h, $k$ in mD, $\mu$ in cP, $c_t$ in 1/psi.

$948 = 1/(4\cdot2.637\cdot10^{-4})$ .

Using one variable instead of two turns the partial differential equation into an ordinary differential equation, which we can integrate (the derivation is in the advanced section). The result:

Line-source solution

p_i - p(r, t) = \frac{qB\mu}{4\pi kh}\,E_1\!\left(\frac{\phi\mu c_t\,r^2}{4kt}\right),\qquad E_1(x) = -\mathrm{Ei}(-x) = \int_x^{\infty}\frac{\e^{-u}}{u}\,\dd u

where:

$p(r, t)$ pressure in the reservoir at distance $r$ from the well axis, a time $t$ after start-up, Pa
$p_i$ initial reservoir pressure, Pa
$q,\ B$ rate at standard conditions and formation volume factor, m³/s, m³/m³
$\mu,\ k,\ h$ fluid viscosity, permeability and reservoir thickness, Pa·s, m², m
$\phi,\ c_t$ porosity and total compressibility, —, 1/Pa
$E_1(x)$ exponential integral, —

Units: SI

Holds when

An unbounded homogeneous isotropic (the same permeability in every direction) reservoir of constant thickness. Before start-up the pressure is $p_i$ everywhere.
A constant rate from start-up. The well is replaced by a line source and penetrates the full thickness.
One slightly compressible fluid (these are the conditions of the diffusivity equation: Diffusivity equation).
There is no skin and no wellbore storage (the effect of the volume of fluid in the well itself). The formula describes the pressure in the reservoir, not in the well itself.

The pressure drop at any point equals half of the “unit” radial pressure drop, $qB\mu/(2\pi kh)$ , multiplied by a function of one argument. In dimensionless variables it is $p_D = \tfrac12E_1\!\big(r_D^2/(4t_D)\big)$ . The farther the point and the less time has passed, the larger the argument and the smaller the pressure drop.

In other unit systems (2)

Metric oilfield units

p_i - p = 0.933\,\frac{qB\mu}{kh}\,E_1\!\left(\frac{\phi\mu c_t\,r^2}{0.01421\,k\,t}\right)

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

$0.933 = 1/(4\pi\cdot0.0853)$ , where $0.0853$ is the constant of Darcy's law in these units; $0.01421 = 4\cdot3.553\cdot10^{-3}$ .

Oilfield units

p_i - p = 70.6\,\frac{qB\mu}{kh}\,E_1\!\left(\frac{948\,\phi\mu c_t\,r^2}{k\,t}\right)

$q$ in STB/d (barrels of oil at standard conditions per day), $\mu$ in cP, $k$ in mD, $h$ and $r$ in ft, $c_t$ in 1/psi, $t$ in h, pressure in psi.

$70.6 = 1/(4\pi\cdot1.127\cdot10^{-3})$ ; $948 = 1/(4\cdot2.637\cdot10^{-4})$ .

The exponential integral

The function $E_1(x)$ is the area under the curve $\e^{-u}/u$ from $x$ to infinity. For small $x$ it is large and behaves like a logarithm. For large $x$ it quickly falls towards zero. We can calculate it from a series, read it from tables, or use the built-in functions of software.

Series for the exponential integral

E_1(x) = -\ln x - 0.5772 + x - \frac{x^2}{2\cdot2!} + \frac{x^3}{3\cdot3!} - \dots

where:

$x$ argument, $x > 0$ , —
$0.5772$ Euler's constant, —

Units: no units

Holds when

The series converges for any $x > 0$ , but it is convenient only for $x \le 1$ . For larger $x$ , use tables or a continued fraction (advanced section).

The first two terms are the logarithmic part: $-\ln x - 0.5772 = -\ln(\gamma x)$ , where $\gamma = \e^{0.5772} = 1.781$ . The remaining terms get small quickly when $x$ is small. Do not confuse Euler's constant, 0.5772, with its exponential: in flow formulas the letter $\gamma$ stands for 1.781.

The exponential integral and its logarithmic approximation

$x$: 0.001
$E_1(x)$: 6.332
$-\ln(1.781x)$: 6.331
Error of the logarithm: 0.016%

$x$: 0.01
$E_1(x)$: 4.038
$-\ln(1.781x)$: 4.028
Error of the logarithm: 0.25%

$x$: 0.05
$E_1(x)$: 2.468
$-\ln(1.781x)$: 2.419
Error of the logarithm: 2.0%

$x$: 0.1
$E_1(x)$: 1.823
$-\ln(1.781x)$: 1.725
Error of the logarithm: 5.4%

$x$: 0.5
$E_1(x)$: 0.5598
$-\ln(1.781x)$: 0.1159
Error of the logarithm: 79%

$x$: 1
$E_1(x)$: 0.2194
$-\ln(1.781x)$: below zero
Error of the logarithm: not applicable

$x$: 2
$E_1(x)$: 0.0489
$-\ln(1.781x)$: below zero
Error of the logarithm: not applicable

$x$: 5
$E_1(x)$: 0.001148
$-\ln(1.781x)$: below zero
Error of the logarithm: not applicable

$x$	$E_1(x)$	$-\ln(1.781x)$	Error of the logarithm
0.001	6.332	6.331	0.016%
0.01	4.038	4.028	0.25%
0.05	2.468	2.419	2.0%
0.1	1.823	1.725	5.4%
0.5	0.5598	0.1159	79%
1	0.2194	below zero	not applicable
2	0.0489	below zero	not applicable
5	0.001148	below zero	not applicable

The error is the percentage by which the logarithmic approximation falls short of the exact value. For $x > 0.56$ the logarithm is negative and no longer makes sense.

CalculatedCalculated values of the function

E_1(x)

and of the logarithmic approximation

-\ln(1.781x)

. The argument scale is logarithmic. To the left of

x = 0.01

the two curves cannot be told apart. At the edge of the drawdown cone, at

x = 1

, the function equals 0.219, but the logarithm is already negative.

This gives three regions. For $x < 0.01$ we replace the function with a logarithm. For $x$ from 0.01 to about 5 we need the function itself. For $x > 5$ it is less than 0.0012: the pressure at such a point has practically not changed.

The logarithmic approximation

If we put $E_1(x) \approx -\ln(\gamma x)$ into the solution, we get a formula without special functions. This is the logarithmic approximation:

Logarithmic approximation of the line-source solution

p_i - p(r, t) = \frac{qB\mu}{4\pi kh}\left[\ln\frac{kt}{\phi\mu c_t r^2} + 0.809\right]

where:

$p(r, t)$ pressure in the reservoir at distance $r$ , a time $t$ after start-up, Pa
$p_i$ initial reservoir pressure, Pa
$q,\ B,\ \mu,\ k,\ h,\ \phi,\ c_t$ as in the line-source solution, m³/s, m³/m³, Pa·s, m², m, —, 1/Pa
$0.809$ $\ln(4/\gamma)$ , where $\gamma = 1.781$ , —

Units: SI; natural logarithm

Holds when

The conditions of the line-source solution.
$x < 0.01$ , that is, $kt/(\phi\mu c_tr^2) > 25$ : the error is then less than 0.25%. At $kt/(\phi\mu c_tr^2) = 5$ the error is about 2%.

Deep inside the drawdown cone, the pressure depends linearly on both the logarithm of time and the logarithm of distance. The profile there has the same shape as in steady-state flow, but its level sinks with time. In dimensionless form, $p_D = \tfrac12\left[\ln(t_D/r_D^2) + 0.809\right]$ .

In other unit systems (2)

Metric oilfield units (base-10 logarithm)

p_i - p = 2.149\,\frac{qB\mu}{kh}\left[\log_{10}\frac{k\,t}{\phi\mu c_t r^2} - 2.098\right]

$q$ in m³/d, $\mu$ in mPa·s, $k$ in mD, $h$ and $r$ in m, $c_t$ in 1/MPa, $t$ in h, pressure in MPa.

$2.149 = \ln 10/(4\pi\cdot0.0853)$ ; $2.098 = -\log_{10}(3.553\cdot10^{-3}) - 0.809/\ln 10$ .

Oilfield units (base-10 logarithm)

p_i - p = 162.6\,\frac{qB\mu}{kh}\left[\log_{10}\frac{k\,t}{\phi\mu c_t r^2} - 3.23\right]

$q$ in STB/d, $\mu$ in cP, $k$ in mD, $h$ and $r$ in ft, $c_t$ in 1/psi, $t$ in h, pressure in psi.

$162.6 = 70.6\cdot\ln 10$ ; $3.23 = -\log_{10}(2.637\cdot10^{-4}) - 0.809/\ln 10$ .

The condition $x < 0.01$ means that more than $25\,r^2/\eta$ of time has passed. Take the well of the examples below ( $\eta$ = 350 m²/h). At the well wall, $r$ = 0.1 m, this takes 2.6 seconds. At 10 m from the axis it takes 7 hours. At 100 m it takes about a month, and by then the disturbance has long since reached the boundaries. So at the well we use the logarithm, and for points in the reservoir we use the function $E_1$ itself.

Worked examplePressure in the reservoir 10 hours after start-up

Given (assumed data)

Well and reservoir (lesson Flow regimes): $q$ = 100 m³/d; $B$ = 1.3; $\mu$ = 1.2 mPa·s; $k$ = 50 mD; $h$ = 15 m
Hydraulic diffusivity $\eta$: 350.4 m²/h ( $\phi$ = 0.20; $c_t = 2.1125\cdot10^{-3}$ MPa⁻¹)
Time after start-up: 10 h; the disturbance has not yet reached the boundaries
Points: the well wall ( $r_w$ = 0.1 m), 30 m and 100 m from the axis

Solution

Factor in front of the function: $0.933\,\dfrac{qB\mu}{kh} = 0.933\cdot\dfrac{156}{750} = 0.194$ MPa.
Argument $x = r^2/(4\eta t)$ with $4\eta t = 14{,}016$ m²: $7.135\cdot 10^{-7}$ at the well wall, 0.0642 at 30 m, 0.713 at 100 m.
Well wall: $x \ll 0.01$ , so the logarithm is valid: $E_1 \approx -\ln(1.781x) = 13.58$ . Pressure drop: $0.194\cdot13.58 = 2.63$ MPa.
30 m: $E_1(0.0642) = 2.232$ , pressure drop 0.433 MPa. The logarithm would give 0.421 MPa, which is 2.8% less.
100 m: $E_1(0.713) = 0.364$ , pressure drop 0.071 MPa. Here the logarithm gives −0.047 MPa, a meaningless “rise” in pressure.
The radius of investigation at this moment: $r_i = \sqrt{4\eta t} = 118$ m. At this distance $x = 1$ , and the pressure has fallen by $0.194\cdot0.219 = 0.043$ MPa.

Answer: After 10 h the pressure has fallen by 2.63 MPa at the well wall, by 0.433 MPa at 30 m and by 0.071 MPa at 100 m from it.

What it means. The first 30 m account for more than 80% of the pressure drop: the drawdown cone is steep at the well and gentle at its edge. The numbers agree with the “10 h” profile on the graph of the lesson Flow regimes. That profile was calculated numerically for a bounded region. The disturbance has not yet reached the boundary, so the region behaves like an infinite reservoir. The logarithmic approximation is reliable only deep inside the cone. At 30 m it is off by a few per cent, and at 100 m it cannot be used. The data in this example are assumed for illustration.

Pressure at the well: the pressure function

In well testing, the pressure is measured in the well itself, that is, at $r = r_w$ , or $r_D = 1$ . There the argument $x = 1/(4t_D)$ falls below 0.01 within seconds after start-up, and the solution reduces to a logarithm of time. If we add the pressure drop across the skin (Skin factor), we get the relationship on which all semilog methods are built:

Pressure function of a well in an infinite reservoir

p_D(t_D) = \frac{2\pi kh\,(p_i - p_{wf})}{qB\mu} = \frac12\left(\ln t_D + 0.809\right) + S,\qquad t_D = \frac{kt}{\phi\mu c_t r_w^2}

where:

$p_D(t_D)$ dimensionless drop in bottomhole pressure, —
$p_{wf}$ bottomhole flowing pressure of the producing well a time $t$ after start-up, Pa
$t_D$ dimensionless time based on the wellbore radius, —
$S$ skin factor, —
$r_w$ wellbore radius, m
$p_i,\ q,\ B,\ \mu$ initial pressure, rate, formation volume factor, viscosity, Pa, m³/s, m³/m³, Pa·s
$k,\ h,\ \phi,\ c_t$ permeability, thickness, porosity, total compressibility, m², m, —, 1/Pa

Units: SI; in metric oilfield units $t_D = 3.553\cdot10^{-3}\,kt/(\phi\mu c_tr_w^2)$ with $t$ in hours

Holds when

An unbounded reservoir and a constant rate from start-up.
$t_D > 25$ (error about 2%) or $t_D > 100$ (about 0.5%): for an ordinary well that is only seconds.
The skin is an extra pressure drop in a thin zone near the wellbore. It appears together with the flow.
The effect of wellbore storage (the fluid volume in the well itself) has ended (Wellbore storage).

Each tenfold increase in time raises $p_D$ by $\tfrac12\ln 10 = 1.151$ , and the skin shifts the whole relationship by a constant amount. The same formula can be read as the Dupuit equation (the steady-state radial flow equation) with a growing radius: $p_D = \ln(r_d/r_w) + S$ , where $r_d = \sqrt{4\eta t/\gamma} \approx 0.75\,r_i$ is the radius of drainage (quasi-steady-state profile). The form with units, and how to find permeability and skin from measurements, are in the lesson Pressure drawdown analysis.

Dimensionless pressure at the well wall from three formulas

$t_D$: 1
Finite wellbore radius: 0.802
Line source: 0.522
Logarithm: 0.405
Error of the logarithm: −50%

$t_D$: 5
Finite wellbore radius: 1.362
Line source: 1.234
Logarithm: 1.209
Error of the logarithm: −11%

$t_D$: 10
Finite wellbore radius: 1.651
Line source: 1.568
Logarithm: 1.556
Error of the logarithm: −5.8%

$t_D$: 25
Finite wellbore radius: 2.062
Line source: 2.019
Logarithm: 2.014
Error of the logarithm: −2.3%

$t_D$: 100
Finite wellbore radius: 2.723
Line source: 2.708
Logarithm: 2.707
Error of the logarithm: −0.58%

$t_D$: 1000
Finite wellbore radius: 3.861
Line source: 3.859
Logarithm: 3.858
Error of the logarithm: −0.056%

$t_D$	Finite wellbore radius	Line source	Logarithm	Error of the logarithm
1	0.802	0.522	0.405	−50%
5	1.362	1.234	1.209	−11%
10	1.651	1.568	1.556	−5.8%
25	2.062	2.019	2.014	−2.3%
100	2.723	2.708	2.707	−0.58%
1000	3.861	3.859	3.858	−0.056%

The skin is zero. “Finite wellbore radius” is the solution with the constant-rate condition set at the wall, $r = r_w$ . It was calculated by numerical inversion of the Laplace transform (a mathematical method; see the advanced section). For the well of the examples, $t_D = 25$ is reached 2.6 s after start-up and $t_D = 100$ after 10 s.

CalculatedCalculated dimensionless pressure at the well wall in the first moments after start-up, from three formulas: the finite-wellbore-radius solution, the line-source solution and the logarithmic approximation. The time scale is logarithmic. By

t_D = 25

the three curves come together. To the left of that, the logarithmic line gives too small a pressure drop, and for

t_D < 0.45

it goes below zero.

The practical conclusion has two parts. First, at the well the logarithmic formula is accurate within seconds after start-up. This is long before the effect of wellbore storage ends, and that effect is what actually governs the first minutes and hours of a pressure record. Second, do not extend the logarithmic line back to the moment of start-up itself: at small $t_D$ the true pressure drop is larger than this line gives.

Where the solution stops working: the reservoir boundaries

The solution for an infinite reservoir is valid until the disturbance reaches the boundaries. Take a well at the centre of a circular region of radius $r_e$ . Its bottomhole pressure departs from the infinite-reservoir solution when the radius of investigation becomes equal to the radius of the region, that is, at $t_D/r_{eD}^2 \approx 0.25$ (time to the boundary). The exact solution for a closed circle (at $r_e/r_w$ = 3000) differs from it by 0.2% at $t_D/r_{eD}^2 = 0.25$ , by 0.5% at 0.3 and by more than 1% at 0.4. After that a different formula applies. Which one depends on the condition at the boundary.

The pressure function of a well in different periods

Period: The first moments
Condition: $t_D \ll 1$
$p_D$ at the well: $2\sqrt{t_D/\pi}$

Period: Infinite reservoir
Condition: $t_D > 25$ and $t_D/r_{eD}^2 < 0.25$
$p_D$ at the well: $\tfrac12(\ln t_D + 0.809)$

Period: Closed circle, pseudo-steady-state regime
Condition: $t_D/r_{eD}^2 > 0.3$
$p_D$ at the well: $2t_D/r_{eD}^2 + \ln r_{eD} - \tfrac34$

Period: Circle with constant pressure at the boundary
Condition: $t_D/r_{eD}^2 > 0.5$
$p_D$ at the well: $\ln r_{eD}$

Period	Condition	$p_D$ at the well
The first moments	$t_D \ll 1$	$2\sqrt{t_D/\pi}$
Infinite reservoir	$t_D > 25$ and $t_D/r_{eD}^2 < 0.25$	$\tfrac12(\ln t_D + 0.809)$
Closed circle, pseudo-steady-state regime	$t_D/r_{eD}^2 > 0.3$	$2t_D/r_{eD}^2 + \ln r_{eD} - \tfrac34$
Circle with constant pressure at the boundary	$t_D/r_{eD}^2 > 0.5$	$\ln r_{eD}$

Here $r_{eD} = r_e/r_w$ . The skin is added to each formula. For a well at the centre of a circle, the transition from the second row to the third is short. For elongated regions and for wells close to one boundary it is stretched out (shape factor).

CalculatedCalculated pressure function of a well at the centre of a closed circle, for three values of

r_{eD} = r_e/r_w

, and in an infinite reservoir (dashed straight line). The horizontal axis is the base-10 logarithm of dimensionless time: 6 means

t_D = 10^6

. Each curve follows the common straight line as long as

t_D/r_{eD}^2

is less than 0.25–0.3, and then turns upward: pseudo-steady-state depletion begins. If the area of the region grows a hundredfold, the departure comes a hundred times later.

The line-source solution is useful even after the disturbance has reached the boundaries. The diffusivity equation is linear, so we can add solutions together. The pressure from several wells is the sum of their contributions. A change of rate is treated as starting one more well at the same point. A sealing fault is replaced by an imaginary image well on the other side of it (Superposition principle, Pressure build-up analysis, A sealing fault near the well: the method of images).

Pressure in a neighbouring well

We need the formula with the function $E_1$ above all where the distance is large. One case is an interference test: the rate is changed in one well and the pressure is recorded in another. Another is estimating how wells affect each other. The logarithmic approximation is not suitable for such distances (see the example above).

CalculatedCalculated pressure drop at points 50, 150 and 250 m from the well of the examples, after it is started at a rate of 100 m³/d. The reservoir is taken to be unbounded. Both scales are logarithmic. The dots on the curves mark the moment when the radius of investigation reaches that distance: by then the pressure has already fallen by 0.043 MPa. The dashed line is a drop of 0.005 MPa, which an accurate gauge can clearly detect.

Worked exampleWhen the response reaches a neighbouring well

Given (assumed data)

Active well: the well of the previous example: the factor in front of the function is 0.194 MPa, $\eta$ = 350.4 m²/h
Observation well: shut in (not producing), 150 m from the active well
Clearly detectable pressure change: 0.005 MPa (gauge resolution 0.0001 MPa)

Solution

The required value of the function: $E_1(x) = 0.005/0.194 = 0.0258$ .
From a table, or by trial and error, $x = 2.47$ (as a check, $E_1(2.5) = 0.0249$ ).
Time: $t = \dfrac{r^2}{4\eta x} = \dfrac{150^2}{4\cdot350.4\cdot2.47} = 6.5$ h.
The radius of investigation reaches 150 m later, after $r^2/(4\eta) = 16$ h. By that moment the pressure in the observation well has already fallen by 0.043 MPa.
A coarse gauge that can tell apart only 0.05 MPa would “see” the response only after 18 h ( $E_1 = 0.258$ , $x = 0.91$ ).

Answer: A response of 0.005 MPa will arrive at the well 150 m away after about 6.5 h — 2.5 times sooner than the radius of investigation “reaches” it.

What it means. The disturbance has no sharp front: the pressure far away starts to change at once, only very weakly. When the response is noticed depends on how sensitive the gauge is. The radius of investigation is a conventional boundary, where the pressure drop equals 0.11 of a “unit” $qB\mu/(2\pi kh)$ . In an interference test, the arrival time and the size of the response give the hydraulic diffusivity and the flow capacity (permeability times thickness) of the reservoir between the wells. The data in this example are assumed for illustration.

If the bottomhole pressure is fixed instead of the rate

So far we have fixed a constant rate at the well, and the pressure came out of the calculation. It can be the other way round: the well works at a constant bottomhole pressure. An example is a flowing well (one that flows on its own, with no pump) with a constant wellhead pressure. Another is a well with a pump that holds the fluid level. Then the pressure does not fall; the rate does. The drawdown cone widens, and the same pressure difference acts over a longer and longer distance. The solution of this problem is also known. The dimensionless rate $q_D(t_D)$ has been tabulated, and for large $t_D$ it is expressed approximately through the pressure function:

Rate at constant bottomhole pressure

q(t) = \frac{2\pi kh\,(p_i - p_{wf})}{B\mu}\,q_D(t_D),\qquad q_D(t_D) \approx \frac{1}{p_D(t_D)} = \frac{2}{\ln t_D + 0.809}

where:

$q(t)$ rate at standard conditions a time $t$ after start-up, m³/s
$p_i - p_{wf}$ constant difference between the initial reservoir pressure and the bottomhole pressure, Pa
$q_D(t_D)$ dimensionless rate, —
$p_D(t_D)$ pressure function of the well at constant rate, —
$k,\ h,\ B,\ \mu$ permeability, thickness, formation volume factor, viscosity, m², m, m³/m³, Pa·s

Units: SI; in metric oilfield units the factor $2\pi$ is replaced by the constant 0.536

Holds when

An unbounded reservoir, a constant bottomhole pressure from start-up, and no skin. With skin, $q_D \approx 1/(p_D + S)$ .
The approximation $q_D \approx 1/p_D$ overestimates the rate: by about 2% at $t_D = 10^4$ and by less than 1% at $t_D > 10^6$ .

The rate decreases in inverse proportion to the logarithm of time: slowly, but without stopping. In transient flow the ratio of rate to drawdown is higher than after pseudo-steady-state flow is reached, and it keeps falling. The same set-up (constant pressure at the inner boundary) is used for water influx from an aquifer into an accumulation. There we calculate the cumulative influx, the total volume of water that has flowed in (Unsteady-state water influx: the van Everdingen–Hurst method).

Worked exampleHow the rate falls at constant drawdown

Given (assumed data)

Reservoir and well: the same: $k$ = 50 mD, $h$ = 15 m, $\mu$ = 1.2 mPa·s, $B$ = 1.3; in 1 h $t_D$ = 35,039
Difference between the initial and bottomhole pressures: 3.0 MPa, constant from start-up

Solution

Factor: $\dfrac{0.536\,kh\,\Delta p}{B\mu} = \dfrac{0.536\cdot750\cdot3.0}{1.56} = 773$ m³/d.
After 1 h: $p_D = 5.637$ , and $q \approx 773/5.637 = 137$ m³/d. Exact solution: $q_D = 0.1749$ , $q = 135$ m³/d.
After one day ( $t_D = 840{,}947$ ): $p_D = 7.226$ , approximately 107 m³/d, exactly 106 m³/d.
After 64 h, when the disturbance reaches the boundary of a region of radius 300 m: approximately 100 m³/d, exactly 99 m³/d.

Answer: At a constant drawdown of 3 MPa the rate falls from 135 m³/d an hour after start-up to 99 m³/d after 64 h. The approximation

1/p_D

overestimates it by 1.5%.

What it means. The rate fell by a quarter, but not because of depletion: a negligible fraction of the hydrocarbons in place has been produced in that time. It fell because the drawdown cone has widened. In a permeable reservoir such a decline lasts only a few days, until the boundaries make themselves felt. In a tight reservoir it is stretched over months and years, and we cannot forecast production from a single measurement of productivity (duration of the regimes). The data in this example are assumed for illustration.

Terms and symbols

Line source: An idealisation of a well: a line of zero radius through which a specified rate is withdrawn (or injected).
Exponential integral $E_1(x)$: It equals $\int_x^\infty \e^{-u}/u\,\dd u$ , the same as $-\mathrm{Ei}(-x)$ . It sets the shape of the drawdown cone in an infinite reservoir.
Boltzmann variable $x$: The quantity $r^2/(4\eta t)$ , the square of the ratio of the distance to the radius of investigation. It is the only argument of the solution.
Logarithmic approximation: Replacing $E_1(x)$ by $-\ln(1.781x)$ for $x < 0.01$ .
Constant $\gamma$: The number 1.781, the exponential of Euler's constant 0.5772. Here $\ln(4/\gamma) = 0.809$ .
Pressure function $p_D(t_D)$: The dimensionless drop in the bottomhole pressure of a well producing at a constant rate. In an infinite reservoir it is $\tfrac12(\ln t_D + 0.809) + S$ .
Dimensionless rate $q_D(t_D)$: The rate of a well at constant bottomhole pressure, in units of $2\pi kh\,\Delta p/(B\mu)$ .

Common mistakes and misconceptions

Wrong

“The logarithmic formula is valid at any distance from the well.”

In fact

It holds only for $x = r^2/(4\eta t) < 0.01$ . In the lesson's example, after 10 h it is off by 3% at 30 m from the well, and at 100 m it gives a negative pressure drop.

Wrong

“The well radius enters the formula for the pressure in the reservoir.”

In fact

The line-source solution has no well radius in it: the pressure at distance $r$ depends on the rate, the reservoir properties and the time. The radius appears only when the solution is applied to the wall of the well itself.

Wrong

“Euler's constant is 1.781.”

In fact

Euler's constant is 0.5772. The flow formulas contain its exponential, $\gamma = \e^{0.5772} = 1.781$ , and the constant 0.809 is $\ln(4/\gamma)$ .

Wrong

“The solution for an infinite reservoir is suitable only for very large accumulations.”

In fact

It is suitable for any reservoir, as long as the disturbance has not reached the boundaries. In the region of radius 300 m used in the examples, that is the first 64 hours.

Wrong

“If the bottomhole pressure is held constant, the rate will also be constant.”

In fact

In transient flow the rate falls roughly as $1/(\ln t_D + 0.809)$ : the drawdown cone widens, and the same pressure difference acts over a greater distance.

In depth: advanced detail

Deriving the solution with the Boltzmann substitution

We look for the pressure as a function of one variable, $x = r^2/(4\eta t)$ . The chain rule (the rule for differentiating a function of a function) gives $r\,\partial p/\partial r = 2x\,\dd p/\dd x$ and $\partial p/\partial t = -(x/t)\,\dd p/\dd x$ . Putting these into the diffusivity equation gives an ordinary differential equation:

The diffusivity equation in the Boltzmann variable

x\,\frac{\dd^2 p}{\dd x^2} + (1 + x)\,\frac{\dd p}{\dd x} = 0\quad\Rightarrow\quad \frac{\dd p}{\dd x} = C\,\frac{\e^{-x}}{x}

where:

$p(x)$ pressure as a function of the Boltzmann variable, Pa
$C$ constant of integration, Pa

Units: SI

Holds when

The conditions of the line-source solution: the initial condition and both boundary conditions can be written with the single variable $x$ .

This is a first-order equation for the derivative $\dd p/\dd x$ , and we can check its solution by putting it back into the equation. The condition on the well axis, $2x\,\dd p/\dd x \to qB\mu/(2\pi kh)$ as $x \to 0$ , gives $C = qB\mu/(4\pi kh)$ .

What remains is to integrate from $x$ to infinity, where the pressure equals the initial pressure: $p_i - p = C\int_x^\infty \e^{-u}/u\,\dd u = C\,E_1(x)$ . The initial condition is satisfied automatically: as $t \to 0$ , the variable $x$ goes to infinity at every point with $r > 0$ . The substitution works because the problem has no well radius and no boundary radius, that is, no length that $r$ could be compared with.

How to calculate the exponential integral

For $x \le 1$ the series from the main section is enough. Its power terms alternate in sign and get smaller, so the error is no greater than the first term we leave out. At $x = 0.2$ , three power terms are enough. At $x = 1$ , you need six of them for four correct digits. For large $x$ the series converges slowly, so the function is calculated by a continued fraction (a fraction nested inside another fraction, again and again) or taken from tables. This is how the values in the tables and examples of this lesson were found. A two-sided bound (an upper and a lower limit) is useful for a quick check:

Two-sided bound for the exponential integral

\frac12\,\e^{-x}\ln\left(1 + \frac{2}{x}\right) < E_1(x) < \e^{-x}\ln\left(1 + \frac{1}{x}\right)

where:

$x$ argument, $x > 0$ , —

Units: no units

Holds when

Valid for any $x > 0$ .

At $x = 1$ the bounds are 0.202 and 0.255, and the exact value is 0.219. At $x = 5$ they are 0.00113 and 0.00123, and the exact value is 0.00115. The bound helps you notice a gross error, such as a wrong sign of the argument or a base-10 logarithm used instead of a natural one.

A well of finite radius and the first moments

If the constant-rate condition is set not on the axis but at the well wall, at $r_D = 1$ , the solution is written with Bessel functions (special mathematical functions). It can be written compactly only in Laplace space (after a mathematical transformation of the time variable). For the dimensionless pressure at the wall:

Finite wellbore radius: the Laplace transform

\bar p_D(s) = \frac{K_0(\sqrt{s})}{s^{3/2}\,K_1(\sqrt{s})}

where:

$\bar p_D(s)$ Laplace transform of the function $p_D(t_D)$ at the well wall, —
$s$ variable of the Laplace transform, —
$K_0,\ K_1$ modified Bessel functions of the second kind, —

Units: no units

Holds when

An unbounded reservoir, a constant rate at the well wall, no skin and no wellbore storage.

We go back to the function of time numerically, for example with the Stehfest algorithm. This is how the “finite wellbore radius” column in the table of the main section was found. When $t_D \ll 1$ , the solution behaves as $2\sqrt{t_D/\pi}$ . As long as the disturbance has moved from the wall by no more than a fraction of the well radius, the curvature of the wall is not felt, and the flow to the wall is linear. When $t_D > 100$ , the solution differs from the logarithmic formula by less than 0.6%.

In practice, the first moments of a pressure record are governed not by the well radius but by the volume of the wellbore. While the liquid or gas in the wellbore is expanding, the inflow from the reservoir lags behind the rate at the wellhead (Wellbore storage). The finite-wellbore-radius solution is really needed when the “well” is large. This is how water influx into an accumulation is calculated: the inner boundary is the contour of the accumulation itself (Unsteady-state water influx: the van Everdingen–Hurst method).

Connections to other topics

Diffusivity equation — the equation whose solution is the formula of this lesson, and its assumptions.
Flow regimes — the radius of investigation, and the time the disturbance takes to reach the boundaries.
Pressure drawdown analysis — permeability and skin from the slope and position of the logarithmic straight line.
Pressure build-up analysis — pressure build-up as the sum of two constant-rate solutions.
Superposition principle — several wells, changes of rate and faults as sums of line-source solutions.
Skin factor — the extra pressure drop at the well wall, which is added to the pressure function.
Pseudo-steady-state flow and the shape factor — what replaces the infinite-reservoir solution in a closed region.
Well testing: purposes and types of test — interference tests and the radius of investigation when planning tests.
Unsteady-state water influx: the van Everdingen–Hurst method — the solution at constant pressure on the inner boundary, for aquifers.

Check yourself

1
A point in the reservoir is at a distance equal to the radius of investigation at a given moment. What are the argument $x$ and the dimensionless pressure drop $p_D$ there?
1. A. $x = 0.01$ , $p_D \approx 2$
2. B. $x = 1$ , $p_D \approx 0.11$
3. C. $x = 1$ , $p_D = 0$
4. D. $x = 4$ , $p_D \approx 0.002$
Show the answer
Correct answer: B. $x = 1$ , $p_D \approx 0.11$
By definition $x = (r/r_i)^2$ , so at $r = r_i$ the argument equals 1. Then $p_D = \tfrac12E_1(1) = \tfrac12\cdot0.219 \approx 0.11$ : at the “edge” of the cone the pressure has already fallen, by about one ninth of a “unit” of radial pressure drop. It has not stayed at its initial value.
2
A well produces at 80 m³/d ( $B$ = 1.2; $\mu$ = 1.0 mPa·s) in a reservoir with $k$ = 100 mD, $h$ = 10 m, $\phi$ = 0.20, $c_t = 1.5\cdot10^{-3}$ MPa⁻¹. By how much will the pressure fall at a point 100 m from it after 10 h?
Show the answer
Factor: $0.933\cdot80\cdot1.2\cdot1.0/(100\cdot10) = 0.0896$ MPa. Argument: numerator $\phi\mu c_tr^2 = 0.20\cdot1.0\cdot1.5\cdot10^{-3}\cdot100^2 = 3.0$ , denominator $0.01421\cdot100\cdot10 = 14.21$ , so $x = 0.211$ . This is greater than 0.01, so we need the function itself: $E_1(0.211) \approx 1.18$ (from the series: $-\ln 0.211 - 0.577 + 0.211 - 0.011 = 1.18$ ). Pressure drop: $0.0896\cdot1.18 \approx 0.106$ MPa.
3
How soon after start-up can we use the logarithmic formula instead of the function $E_1$ at the wall of an ordinary well?
1. A.After a few days
2. B.Within seconds
3. C.After a few hours
4. D.Only after pseudo-steady-state flow is reached
Show the answer
Correct answer: B. Within seconds
We need $t_D > 25$ . One “unit” of dimensionless time, $r_w^2/\eta$ , is a fraction of a second: for the well of the examples, $t_D = 25$ is reached after 2.6 s. Hours and days are needed only for points tens or hundreds of metres from the well.
4
A well produces at a constant rate in an infinite reservoir. By how much does the dimensionless drop in bottomhole pressure $p_D$ increase between 10 and 100 h, and between 100 and 1000 h?
1. A.By 1.151 and by 11.51
2. B.It depends on the skin
3. C.By 1.151 in both cases
4. D.By 2.303 in both cases
Show the answer
Correct answer: C. By 1.151 in both cases
$p_D = \tfrac12(\ln t_D + 0.809) + S$ : every tenfold increase in time adds $\tfrac12\ln 10 = 1.151$ , whatever time you start from and whatever the skin, because the skin enters as a constant term. That is why a plot of pressure against the logarithm of time is a straight line.
5
Why does the well radius not enter the formula for the pressure in the reservoir at distance $r$ from the well, but does enter the formula for the bottomhole pressure?
Show the answer
In the line-source solution the well is a line. The pressure in the reservoir depends only on the rate, the reservoir properties, the distance and the time. The bottomhole pressure is the same solution at the point $r = r_w$ : the well radius appears as the distance at which we calculate the pressure. The smaller the radius, the closer to the axis we take the point, and the greater the pressure drop.
6
A well has been started and the bottomhole pressure is held constant. How does its rate change while the disturbance has not reached the boundaries?
1. A.It falls in inverse proportion to time
2. B.It rises: the drawdown cone covers an ever greater volume
3. C.It stays constant
4. D.It falls roughly in inverse proportion to the logarithm of time
Show the answer
Correct answer: D. It falls roughly in inverse proportion to the logarithm of time
The dimensionless rate is $q_D \approx 2/(\ln t_D + 0.809)$ : the cone widens, the pressure difference is spread over a longer distance, the pressure gradient at the wall gets smaller, and the rate falls slowly. A linear decline, or a decline in inverse proportion to time, is not typical of radial inflow in an infinite reservoir.

Updated: 2 October 2026