Superposition principle

Quick refresher

The diffusivity equation for a slightly compressible fluid is linear: a solution multiplied by a number, and a sum of solutions, are also solutions. So the pressure drop is proportional to the rate, and the drops from different sources add up.
Adding in space: the pressure drop at a point equals the sum of the drops from all the wells, each with its own rate, distance and time of work. The skin is added only for the well in which we calculate the pressure.
Adding in time: a change of the rate from $q_{j-1}$ to $q_j$ is equivalent to starting one more well at the same point with the rate $q_j - q_{j-1}$ . A shut-in is the start of an imaginary injection well with the former rate.
The skin terms add up to $q_nS$ : the pressure drop across the skin is set by the current rate, while the whole earlier history of the rate acts through the reservoir.
A pressure record at several rates is processed on a plot with the superposition time function: the points of all the rate steps fall on one straight line. Its slope gives the flow capacity $kh$ , and its position gives the skin.
A no-flow boundary is replaced by an image well with the same rate history at twice the distance to the boundary. A constant-pressure boundary is replaced by an image with the opposite sign.

Key relationships

p_i - p = \sum_{j} \dfrac{q_jB\mu}{4\pi kh}\,E_1\!\left(\dfrac{\phi\mu c_t\,r_j^2}{4k\,(t - t_j)}\right)

pressure at a point from several wells (SI); for a producing well at its own wellbore, a logarithm and the skin are used

p_i - p_{wf} = \dfrac{B\mu}{2\pi kh}\sum_{j=1}^{n}\left(q_j - q_{j-1}\right)\left[p_D\big(t - t_{j-1}\big) + S\right]

stepwise change of the rate; $q_0 = 0$ , $p_D$ is the pressure function without skin

X_n = \sum_{j=1}^{n}\dfrac{q_j - q_{j-1}}{q_n}\,\lg\left(t - t_{j-1}\right)

superposition time function ( $\lg$ is the base-10 logarithm); $(p_i - p_{wf})/q_n$ depends linearly on $X_n$

kh = 2.149\,\dfrac{B\mu}{m'}

flow capacity from the slope $m'$ of the multi-rate plot (MPa per 1 m³/d per log cycle)

\Delta p_{\text{im}} = \dfrac{qB\mu}{4\pi kh}\,E_1\!\left(\dfrac{\phi\mu c_t\,(2L)^2}{4kt}\right)

addition from a sealing fault at a distance $L$ (SI)

Common pitfalls

Adding the skin of a neighbouring well to its contribution. It changes the pressure only at its own wellbore.
Counting the time of all the rate steps from the start of production, instead of from the moment of each rate change.
Using the logarithmic approximation for a neighbouring well or an image. At distances of hundreds of metres it holds only after months and years.
Adding solutions where the problem is nonlinear: gas at large pressure changes, gas coming out of the oil at the wellbore, pressure-dependent permeability, a skin that grows with the rate.
Thinking that the pressure around a well starts to rise right after it is shut in. In a neighbouring well it keeps falling for a long time.

After this lesson you can

explain why the solutions of the diffusivity equation can be added, and name the conditions under which this stops being true
calculate the pressure in a producing well, and at a point of the reservoir, when several wells that started at different times are working, taking the skin (the effect of the zone right around the well) into account correctly
represent the rate history as steps, and calculate the bottomhole pressure when the rate changes, and the pressure in a neighbouring well after a shut-in
plot a graph with the superposition time function, and find the flow capacity and the skin from a pressure record at several rates, without shutting the well in
replace a no-flow boundary or a constant-pressure boundary by an imaginary image well, and estimate when the boundary shows up in the pressure of the well

Why solutions can be added

In the lesson Line-source solution and the pressure function we found the solution for one well: it starts at a constant rate in a reservoir where the pressure is the same everywhere. A real history is more complicated. The rate is changed with a choke (a valve that limits the flow), the well is shut in for a test and started again, and new wells are drilled nearby and put on production. We do not need to look for a new solution of the equation for each such case. It is enough to add up the solutions we already know.

The reason lies in the structure of the diffusivity equation (Diffusivity equation). The pressure and its derivatives enter it only to the first power: there are no products of them, and no coefficients that depend on the pressure itself. Such an equation is called linear, and it has two properties. A solution multiplied by a number is also a solution: twice the rate gives twice the pressure drop. A sum of solutions is also a solution: if two sources act in the reservoir, the pressure change from them equals the sum of the changes from each. Together these properties make up the superposition principle.

Linearity of the diffusivity equation

\nabla^2 \Delta p = \frac{\phi\mu c_t}{k}\,\frac{\partial\,\Delta p}{\partial t}\quad\Rightarrow\quad \Delta p = a_1\,\Delta p_1 + a_2\,\Delta p_2\ \ \text{— also a solution}

where:

$\Delta p$ pressure drop relative to the initial pressure, $p_i - p(x, y, t)$ , at the point of the reservoir with coordinates $x$ , $y$ at the time $t$ , Pa
$\nabla^2$ Laplace operator over the coordinates in the plane of the reservoir, $\partial^2/\partial x^2 + \partial^2/\partial y^2$ (the sum of the second derivatives, a measure of the curvature of the pressure profile), 1/m²
$\Delta p_1,\ \Delta p_2$ two solutions of the same equation, for example from two wells or from two rate steps of one well, Pa
$a_1,\ a_2$ constant factors (ratios of rates), —
$\phi,\ \mu,\ c_t,\ k$ porosity, viscosity, total compressibility, permeability, —, Pa·s, 1/Pa, m²

Units: SI (any consistent system of units)

Holds when

The fluid is single-phase and slightly compressible. The porosity, permeability, viscosity and compressibility do not depend on pressure.
Darcy's law applies: the filtration velocity is proportional to the pressure gradient, with no inertial losses.
The conditions at the wells and at the boundaries are linear too: given rates, constant pressures or no-flow boundaries.
Before the events begin, the pressure in the whole reservoir is the same, $p = p_i$ . If the reservoir was already disturbed, the solution that describes this earlier disturbance is added to the sum.

We add not the pressures themselves but their changes relative to the initial state. Each term is the solution for one source: one well at a constant rate that started at its own moment. If there is one source but its rate changes, the terms are the rate steps.

Several wells: adding in space

Suppose several wells work in the reservoir. Each creates its own drawdown cone (the region where the pressure has dropped), and the cones overlap. The pressure drop at any point (in a producing well, in a shut-in observation well, or simply at a point of the reservoir) equals the sum of the drops that each well would give if it worked alone.

Adding in space: the pressure at a point from several wells

p_i - p(x, y, t) = \sum_{j=1}^{N} \frac{q_jB\mu}{4\pi kh}\,E_1\!\left(\frac{\phi\mu c_t\,r_j^2}{4k\,(t - t_j)}\right)

where:

$p(x, y, t)$ pressure at a point of the reservoir at the time $t$ , Pa
$p_i$ initial reservoir pressure, Pa
$N$ number of wells, —
$q_j$ rate of well $j$ at standard conditions; negative for an injection well, m³/s
$r_j$ distance from well $j$ to the point, m
$t_j$ moment well $j$ starts; while $t \le t_j$ , its term is zero, s
$E_1$ exponential integral (a standard mathematical function that describes how the pressure drop depends on the distance from a well and on time), —
$B,\ \mu,\ k,\ h,\ \phi,\ c_t$ formation volume factor, viscosity, permeability, thickness, porosity, total compressibility, m³/m³, Pa·s, m², m, —, 1/Pa

Units: SI

Holds when

The reservoir is uniform and unbounded; each well works at a constant rate from the moment it starts.
The linearity conditions of the previous section hold.
For a producing well, at the point of its own wellbore ( $r = r_w$ ), the function is replaced by a logarithm and the skin of that well is added. The skins of the other wells are not in the sum.

The contribution of each well is set by its rate and by how far the point is from its drawdown cone: the argument of the function is the square of the ratio of the distance to the radius of investigation of that well. Close wells that have worked for a long time give a large contribution. Far wells that started recently give a small one.

In other unit systems (2)

Metric oilfield units

p_i - p = 0.933\,\frac{B\mu}{kh}\sum_{j} q_j\,E_1\!\left(\frac{\phi\mu c_t\,r_j^2}{0.01421\,k\,(t - t_j)}\right)

$q_j$ in m³/d at standard conditions, $\mu$ in mPa·s, $k$ in mD, $h$ and $r_j$ 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{B\mu}{kh}\sum_{j} q_j\,E_1\!\left(\frac{948\,\phi\mu c_t\,r_j^2}{k\,(t - t_j)}\right)

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

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

For the well in which we calculate the pressure, the distance equals its radius $r_w$ . There the argument of the function becomes smaller than 0.01 within seconds, and the function is replaced by a logarithm (logarithmic approximation): $E_1 \approx \ln t_D + 0.809$ . We add $2S$ to it, twice the skin of this well, because the factor in front of the bracket is half of the “unit” $qB\mu/(2\pi kh)$ . For neighbouring wells hundreds of metres away, the argument is small only after months and years of work, so for them we need the function $E_1$ itself.

What goes into the sum for the bottomhole pressure of a well

Source: the well itself
Distance: $r_w$ (logarithm)
Time counted from: its start
Skin: its skin

Source: a neighbouring producer
Distance: between the wells ( $E_1$ )
Time counted from: the start of the neighbour
Skin: not included

Source: a neighbouring injector
Distance: between the wells ( $E_1$ )
Time counted from: the start of injection
Skin: not included; the contribution has a minus sign

Source: an image beyond a boundary
Distance: to the image ( $E_1$ )
Time counted from: as for the real well
Skin: not included

Source	Distance	Time counted from	Skin
the well itself	$r_w$ (logarithm)	its start	its skin
a neighbouring producer	between the wells ( $E_1$ )	the start of the neighbour	not included
a neighbouring injector	between the wells ( $E_1$ )	the start of injection	not included; the contribution has a minus sign
an image beyond a boundary	to the image ( $E_1$ )	as for the real well	not included

If the rate of any well changed, its contribution is itself split into steps; see the next section.

Worked exampleBottomhole pressure of a well with a neighbouring well taken into account

Given (assumed data)

Reservoir (assumed data): $k$ = 40 mD; $h$ = 12 m; $\phi$ = 0.18; $\mu$ = 1.6 mPa·s; $B$ = 1.25; $c_t = 1.6\cdot10^{-3}$ MPa⁻¹; $p_i$ = 25 MPa; bubble point pressure 12 MPa
Well P-1: works for 480 h (20 days) at 80 m³/d; skin 1
Well P-2: 450 m from P-1; works for 192 h (8 days) at 60 m³/d; $r_w$ = 0.108 m; $S$ = 2

Solution

Hydraulic diffusivity: $\eta = 3.553\cdot10^{-3}\,k/(\phi\mu c_t) = 308.4$ m²/h. The factor $0.933\,B\mu/(kh) = 0.003888$ MPa per 1 m³/d of rate.
Own contribution of P-2: $t_D = \eta t/r_w^2 = 5.08\cdot10^{6}$ ; $\ln t_D + 0.809 + 2S = 20.25$ ; pressure drop $60\cdot0.003888\cdot20.25 = 4.724$ MPa.
Contribution of P-1 at the bottom of P-2: $x = r^2/(4\eta t) = 450^2/(4\cdot308.4\cdot480) = 0.3420$ . This is far larger than 0.01, so the logarithm does not apply and we need the function itself: $E_1(0.3420) = 0.8106$ . Pressure drop $80\cdot0.003888\cdot0.8106 = 0.252$ MPa. The skin of P-1 is not included.
Bottomhole pressure of P-2: $p_{wf} = 25 - 4.724 - 0.252 = 20.02$ MPa. The neighbouring well gives 5.1% of the pressure drop.
If we used the logarithm for P-1, we would get $-\ln(1.781\cdot0.3420) = 0.4958$ instead of 0.8106. The contribution of the neighbour would be too low by 39%.
The effect the other way round: at the bottom of P-1, well P-2 gives $x = 450^2/(4\cdot308.4\cdot192) = 0.855$ , $E_1 = 0.2816$ , and a pressure drop of $60\cdot0.003888\cdot0.2816 = 0.066$ MPa.

Answer: The bottomhole pressure of P-2 is about 20.02 MPa. Of the pressure drop of 4.98 MPa, the neighbouring well P-1 gives 0.252 MPa.

What it means. The contribution of the neighbouring well is smaller than the well's own, but it is not negligible. If we leave it out when we test P-2, part of the pressure drop will be put down to P-2 itself, and the permeability and the skin from the record will be wrong. The wells do not affect each other equally: P-2 works for less time and at a lower rate, so at the bottom of P-1 it gives an addition almost four times smaller. The pressure everywhere is above the bubble point, so the single-phase model applies. The data in this example are assumed for illustration.

The interference test is built on the same addition. The rate is changed in one well, and the pressure is recorded in a neighbouring shut-in well. The response in it is one term of the sum. From the time of its arrival and its size, we find the flow capacity and the hydraulic diffusivity of the reservoir between the wells (pressure in a neighbouring well). If other wells work at the same time, their contributions are calculated and subtracted from the record, again by superposition.

A changing rate: adding in time

Now there is one well, but its rate changes in steps: $q_1$ from the start, $q_2$ from the moment $t_1$ , $q_3$ from the moment $t_2$ , and so on. The method is the same. Each change of the rate is treated as the start of one more well at the same point, with a rate equal to the increase: the first “well” keeps working at the rate $q_1$ , the second adds $q_2 - q_1$ , the third adds $q_3 - q_2$ . After $t_2$ the sum is exactly $q_3$ . A decrease of the rate is the addition of a negative rate, that is, an imaginary injection well.

CalculatedCalculated example from the lesson: the rate of well P-3 is raised twice, so that it is 40, then 70, then 100 m³/d (top). Below is the fall of the bottomhole pressure: the contributions of the three steps (dashed lines), each counted from its own moment, and their sum (solid line). Skin 3. Wellbore storage (the effect of the fluid held in the wellbore itself) is not taken into account, so at the moments of a rate change the pressure changes almost in a jump.

Adding in time: the bottomhole pressure for a stepwise change of the rate

p_i - p_{wf}(t) = \frac{B\mu}{2\pi kh}\sum_{j=1}^{n}\left(q_j - q_{j-1}\right)\left[p_D\!\left(\frac{k\,(t - t_{j-1})}{\phi\mu c_t r_w^2}\right) + S\right]

where:

$p_{wf}(t)$ bottomhole pressure at the time $t$ , Pa
$p_i$ initial reservoir pressure (the pressure before the first start), Pa
$q_j$ rate in step $j$ at standard conditions; $q_0 = 0$ , m³/s
$t_{j-1}$ moment step $j$ begins; $t_0 = 0$ , s
$n$ number of the current rate step: $t_{n-1} < t$ , —
$p_D$ dimensionless pressure at the wall of the well at a constant rate, without skin; in an infinite reservoir $\tfrac12(\ln t_D + 0.809)$ , —
$S$ skin factor, —
$B,\ \mu,\ k,\ h,\ \phi,\ c_t,\ r_w$ formation volume factor, viscosity, permeability, thickness, porosity, total compressibility, radius of the well, m³/m³, Pa·s, m², m, —, 1/Pa, m

Units: SI

Holds when

The linearity conditions hold. The skin does not depend on the rate and does not change with time.
The sandface rate (the rate at the bottom of the well) changes instantly: the effect of the wellbore is not taken into account. It is taken into account by putting the solution with wellbore storage and skin in place of $p_D + S$ (Wellbore storage).
In an infinite reservoir the logarithmic form of $p_D$ is correct if more than a few seconds have passed after each rate change. In a bounded reservoir we take the solution for a bounded reservoir.

Each rate change sends its own disturbance into the reservoir, and all of them keep acting at the same time. The skin terms add up to $q_nS$ : the sum of the rate increases equals the current rate. So the pressure drop across the skin “remembers” only the current rate, while the whole earlier history acts through the reservoir.

In other unit systems (2)

Metric oilfield units, logarithmic approximation

p_i - p_{wf} = 0.933\,\frac{B\mu}{kh}\left[\sum_{j=1}^{n}\left(q_j - q_{j-1}\right)\left(\ln\frac{3.553\cdot10^{-3}\,k\,(t - t_{j-1})}{\phi\mu c_t r_w^2} + 0.809\right) + 2q_nS\right]

$q_j$ in m³/d, $\mu$ in mPa·s, $k$ in mD, $h$ and $r_w$ in m, $c_t$ in 1/MPa, $t$ in h, pressure in MPa; the logarithm is natural.

$0.933 = 1/(4\pi\cdot0.0853)$ ; $3.553\cdot10^{-3}$ converts time in hours into dimensionless time; $0.809 = \ln(4/\gamma)$ , $\gamma = 1.781$ .

Oilfield units, logarithmic approximation

p_i - p_{wf} = 70.6\,\frac{B\mu}{kh}\left[\sum_{j=1}^{n}\left(q_j - q_{j-1}\right)\left(\ln\frac{2.637\cdot10^{-4}\,k\,(t - t_{j-1})}{\phi\mu c_t r_w^2} + 0.809\right) + 2q_nS\right]

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

$70.6 = 1/(4\pi\cdot1.127\cdot10^{-3})$ ; $2.637\cdot10^{-4}$ is the dimensionless-time factor in these units.

Worked exampleBottomhole pressure at three rates

Given (assumed data)

Reservoir: the same as in the previous example
Well P-3: $r_w$ = 0.108 m; $S$ = 3
Rate steps: 40 m³/d from 0 to 24 h; 70 m³/d from 24 to 48 h; 100 m³/d after 48 h

Solution

Dimensionless time per 1 h: $\eta/r_w^2 = 26{,}442$ . The pressure “unit” per 1 m³/d: $B\mu/(0.536\,kh) = 0.007777$ MPa.
The pressure function without skin, $p_D = \tfrac12(\ln t_D + 0.809)$ : at 24 h it is 7.085, at 48 h 7.431, at 72 h 7.634, and at 96 h 7.778.
End of the first rate step, 24 h: $\Delta p = 0.007777\cdot40\cdot(7.085 + 3) = 3.137$ MPa; $p_{wf}$ = 21.863 MPa.
End of the second rate step, 48 h: the first step has worked for 48 h, the second for 24 h. $\Delta p = 0.007777\,[40\,(7.431 + 3) + 30\,(7.085 + 3)] = 5.598$ MPa; $p_{wf}$ = 19.402 MPa.
At 96 h the steps have worked for 96, 72 and 48 h. $\Delta p = 0.007777\,[40\cdot7.778 + 30\cdot7.634 + 30\cdot7.431 + 100\cdot3] = 8.268$ MPa; $p_{wf}$ = 16.732 MPa.
Of this, the skin gives $100\cdot0.007777\cdot3 = 2.33$ MPa, the same as at a constant rate of 100 m³/d.

Answer: The bottomhole pressure is 21.863 MPa at the end of the first rate step, 19.402 MPa at the end of the second, and 16.732 MPa 96 h after the start.

What it means. By 48 h after the last rate change, the history has little effect on the pressure. At a constant rate of 100 m³/d from the very start, the pressure would be 16.62 MPa, only 0.11 MPa lower. The pressure function grows as the logarithm of time, and the contributions of the steps that have worked for 96, 72 and 48 h differ little. The pressure drop across the skin at once matches the new rate. Right after a rate change the history matters much more, as the next example shows. The pressure stays above the bubble point of 12 MPa. The data in this example are assumed for illustration.

Worked exampleIf we forget the rate history

Given (assumed data)

Well: in the same reservoir, $S$ = 3, $r_w$ = 0.108 m
History: 744 h (31 days) at 100 m³/d, then the rate is cut to 30 m³/d
Question: the pressure drop at the bottom of the well 24 h after the change of the rate

Solution

Superposition: the step of 100 m³/d has worked for 768 h, and the step of $-70$ m³/d for 24 h. $\Delta p = 0.007777\,[100\,(8.818 + 3) - 70\,(7.085 + 3)] = 3.70$ MPa.
First mistake: assuming that the well worked at 30 m³/d for all the 768 h. Then $\Delta p = 2.76$ MPa.
Second mistake: counting the time from the change of the rate, as if the reservoir had been untouched before it. Then $\Delta p = 2.35$ MPa.

Answer: The pressure drop is 3.70 MPa. Without the history we would get 2.76 or 2.35 MPa, which is 25–36% less.

What it means. A month of work at a high rate left a deep drawdown cone in the reservoir, and after the rate is cut it fills up slowly. The pressure in the well rises, but stays much lower than in a well that has worked all the time at the lower rate. So a test that starts after a rate change cannot be processed as an ordinary drawdown or build-up test counted from that moment: we need the whole rate history. The data in this example are assumed for illustration.

A shut-in is the start of an injection well

A shut-in is a special case of a rate change: the rate changes from $q$ to zero. At the moment of the shut-in, a well with the rate $-q$ is “started” at the same point, that is, an injection well with the same rate. For the bottomhole pressure this gives the build-up formula, on which the Horner and MDH methods are built (Pressure build-up analysis). If the well worked at several rates before the shut-in, the formula only gets longer:

Pressure build-up after work at several rates

p_i - p_{ws}(\Delta t) = \frac{B\mu}{2\pi kh}\left[\sum_{j=1}^{n}\left(q_j - q_{j-1}\right)p_D\big(t_n - t_{j-1} + \Delta t\big) - q_n\,p_D(\Delta t)\right]

where:

$p_{ws}(\Delta t)$ bottomhole pressure the time $\Delta t$ after the shut-in, Pa
$t_n$ moment of the shut-in (the end of the $n$ -th, last rate step), s
$q_j,\ t_{j-1}$ rates of the steps and the moments they begin, $q_0 = 0$ , $t_0 = 0$ , m³/s, s
$q_n$ rate before the shut-in, m³/s
$p_D(\cdot)$ pressure function of the well without skin; the argument is the time converted to dimensionless time, —
$p_i,\ B,\ \mu,\ k,\ h$ initial pressure, formation volume factor, viscosity, permeability, thickness, Pa, m³/m³, Pa·s, m², m

Units: SI

Holds when

The conditions of adding in time hold; the effect of the wellbore after the shut-in is not taken into account.
For a single rate step ( $n = 1$ , $t_1 = t_p$ ) the formula coincides with the one from which the Horner equation is derived.

The skin terms cancel: the sum of the rate increases equals $q_n$ , and the shut-in subtracts exactly $q_nS$ . So the build-up pressures contain no skin, and we find the skin from the pressure at the moment of the shut-in. To process such a record we use the same superposition time function as in the next section, or we simplify the history by an equivalent producing time (Equivalent producing time and synthetic flowing time).

The most unexpected consequence is seen not in the shut-in well but in a neighbouring one. The disturbance from the shut-in, a rise of pressure, spreads through the reservoir as gradually as the fall after the start did. As long as it has not reached the neighbouring point, the pressure there keeps falling, although the well is already shut in.

CalculatedCalculated pressure drop in an observation well 200 m from well P-4, which worked for 72 h at 100 m³/d and was then shut in. The upper dashed curve is the contribution of the well if it had kept working. The lower one is the contribution of the imaginary injection well from the moment of the shut-in (negative). The solid curve is their sum. The moment of the shut-in and the largest pressure drop, 15 h after it, are marked.

Worked examplePressure in an observation well after a neighbouring well is shut in

Given (assumed data)

Active well P-4 (the well whose rate is changed): 100 m³/d for 72 h, then shut in
Observation well: 200 m from P-4, not working itself
Reservoir: the same; $\eta$ = 308 m²/h

Solution

The factor: $0.933\,qB\mu/(kh) = 0.389$ MPa; $r^2/(4\eta) = 32.4$ h, so the argument of the function is $x = 32.4/t$ with $t$ in hours.
At the moment of the shut-in, 72 h: $x = 0.450$ , $E_1 = 0.625$ , pressure drop 0.243 MPa.
After the shut-in: $\Delta p(t) = 0.389\,[E_1(32.4/t) - E_1(32.4/(t - 72))]$ . The derivative of $E_1(a/t)$ with respect to time is $\e^{-a/t}/t$ , so the pressure stops falling when $\e^{-a/t}/t = \e^{-a/(t - 72)}/(t - 72)$ . By trial, $t = 87.4$ h, which is 15 h after the shut-in.
The largest drop: $0.389\cdot[E_1(0.371) - E_1(2.108)] = 0.277$ MPa, which is 0.034 MPa more than at the moment of the shut-in.
After that the pressure rises: 144 h after the start the drop is 0.195 MPa, and after 400 h it is 0.071 MPa.

Answer: The pressure in the observation well keeps falling for another 15 h after the neighbouring well is shut in, down to 0.277 MPa below the initial pressure, and only then starts to rise.

What it means. The delay is longer the farther the observation well is and the lower the hydraulic diffusivity of the reservoir. The pulse test is built on this: the active well is started and shut in by turns, and from the delay and the size of the response in the neighbouring well we find the hydraulic diffusivity and the flow capacity of the reservoir between them. The same effect matters in the analysis of a build-up in a reservoir already developed with many wells. The pressure in the shut-in well rises, while disturbances from the neighbouring wells keep arriving (Testing in a developed reservoir: Slider's method). The data in this example are assumed for illustration.

A record at several rates: the superposition time function

The formula for adding in time can be turned round: from a record of the bottomhole pressure at several rates, we can find the flow capacity and the skin. This is needed when a well cannot be shut in, or when shutting it in is too costly. An example is clean-up of a new well on several chokes with a downhole gauge. We put the logarithmic approximation into the formula and divide both sides by the current rate $q_n$ . All the logarithms of time collect into one quantity, the superposition time function:

The superposition time function and the straight line of a multi-rate record

\frac{p_i - p_{wf}(t)}{q_n} = m'\,X_n(t) + b',\qquad X_n(t) = \sum_{j=1}^{n}\frac{q_j - q_{j-1}}{q_n}\,\lg\left(t - t_{j-1}\right)

where:

$q_n$ rate of the current, $n$ -th step at standard conditions, m³/d
$X_n(t)$ superposition time function: a weighted sum of the base-10 logarithms ( $\lg$ ) of the times elapsed since each rate change, —
$q_j,\ t_{j-1}$ rates of the steps and the moments they begin, $q_0 = 0$ , $t_0 = 0$ , m³/d, h
$m'$ slope of the line: the change of $(p_i - p_{wf})/q_n$ when $X_n$ grows by one (per log cycle), MPa/(m³/d)
$b'$ value of $(p_i - p_{wf})/q_n$ on the line at $X_n = 0$ , MPa/(m³/d)
$p_i,\ p_{wf}$ the pressure before the first start, and the bottomhole pressure, MPa
$k,\ h,\ B,\ \mu,\ \phi,\ c_t,\ r_w,\ S$ in the forms below: permeability, thickness, formation volume factor, viscosity, porosity, total compressibility, radius of the well, skin, mD, m, m³/m³, mPa·s, —, 1/MPa, m, —

Units: metric oilfield: pressure in MPa, rate in m³/d, time in h

Holds when

The reservoir is infinite and uniform. After each rate change, enough time has passed for the effect of the wellbore to end.
The rates and the moments of the rate changes are known exactly. Before the first start the pressure in the reservoir had evened out and equals $p_i$ .
The skin is constant and does not depend on the rate.

For a single rate step $X_1 = \lg t$ , and the formula turns into the semilog straight line of the drawdown test (a test in which the well produces at a constant rate and the pressure falls), divided by the rate (Pressure drawdown analysis). For several rates the points of all the rate steps fall on one straight line: the function $X_n$ takes the whole rate history into account. The slope depends only on the flow capacity, and the position of the line depends on the skin as well.

In other unit systems (2)

Metric oilfield units

kh = 2.149\,\frac{B\mu}{m'},\qquad S = 1.151\left[\frac{b'}{m'} - \lg\frac{k}{\phi\mu c_t r_w^2} + 2.098\right]

$m'$ and $b'$ in MPa/(m³/d), $\mu$ in mPa·s, $k$ in mD, $h$ and $r_w$ in m, $c_t$ in 1/MPa, time in h; $kh$ in mD·m.

$2.149 = \ln 10/(4\pi\cdot0.0853)$ ; $1.151 = \ln 10/2$ ; $2.098 = -\lg(3.553\cdot10^{-3}) - 0.809/\ln 10$ : the same as in the drawdown analysis.

Oilfield units

kh = 162.6\,\frac{B\mu}{m'},\qquad S = 1.151\left[\frac{b'}{m'} - \lg\frac{k}{\phi\mu c_t r_w^2} + 3.23\right]

$m'$ and $b'$ in psi/(STB/d), $\mu$ in cP, $k$ in mD, $h$ and $r_w$ in ft, $c_t$ in 1/psi, time in h; $kh$ in mD·ft.

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

Below is an assumed record of well P-3, worked at the three rates of the example above. A gauge at the bottom of the well recorded the pressure all the time. The table gives a selection of the readings and the plot coordinates calculated from them.

Bottomhole pressure of well P-3 at three rates (assumed data)

$t$ , h: 0.25
$q_n$ , m³/d: 40
$p_{wf}$ , MPa: 22.606
$X_n$: −0.602
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 59.85

$t$ , h: 1
$q_n$ , m³/d: 40
$p_{wf}$ , MPa: 22.365
$X_n$: 0.000
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 65.88

$t$ , h: 3
$q_n$ , m³/d: 40
$p_{wf}$ , MPa: 22.189
$X_n$: 0.477
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 70.28

$t$ , h: 8
$q_n$ , m³/d: 40
$p_{wf}$ , MPa: 22.035
$X_n$: 0.903
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 74.13

$t$ , h: 24
$q_n$ , m³/d: 40
$p_{wf}$ , MPa: 21.863
$X_n$: 1.380
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 78.43

$t$ , h: 24.25
$q_n$ , m³/d: 70
$p_{wf}$ , MPa: 20.066
$X_n$: 0.533
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 70.49

$t$ , h: 25
$q_n$ , m³/d: 70
$p_{wf}$ , MPa: 19.881
$X_n$: 0.799
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 73.13

$t$ , h: 27
$q_n$ , m³/d: 70
$p_{wf}$ , MPa: 19.737
$X_n$: 1.022
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 75.19

$t$ , h: 32
$q_n$ , m³/d: 70
$p_{wf}$ , MPa: 19.594
$X_n$: 1.247
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 77.23

$t$ , h: 48
$q_n$ , m³/d: 70
$p_{wf}$ , MPa: 19.403
$X_n$: 1.552
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 79.96

$t$ , h: 48.25
$q_n$ , m³/d: 100
$p_{wf}$ , MPa: 17.605
$X_n$: 0.908
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 73.95

$t$ , h: 49
$q_n$ , m³/d: 100
$p_{wf}$ , MPa: 17.418
$X_n$: 1.095
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 75.82

$t$ , h: 51
$q_n$ , m³/d: 100
$p_{wf}$ , MPa: 17.271
$X_n$: 1.256
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 77.29

$t$ , h: 56
$q_n$ , m³/d: 100
$p_{wf}$ , MPa: 17.121
$X_n$: 1.422
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 78.79

$t$ , h: 72
$q_n$ , m³/d: 100
$p_{wf}$ , MPa: 16.906
$X_n$: 1.661
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 80.94

$t$ , h: 96
$q_n$ , m³/d: 100
$p_{wf}$ , MPa: 16.733
$X_n$: 1.854
$(p_i - p_{wf})/q_n$ , kPa/(m³/d): 82.67

$t$ , h	$q_n$ , m³/d	$p_{wf}$ , MPa	$X_n$	$(p_i - p_{wf})/q_n$ , kPa/(m³/d)
0.25	40	22.606	−0.602	59.85
1	40	22.365	0.000	65.88
3	40	22.189	0.477	70.28
8	40	22.035	0.903	74.13
24	40	21.863	1.380	78.43
24.25	70	20.066	0.533	70.49
25	70	19.881	0.799	73.13
27	70	19.737	1.022	75.19
32	70	19.594	1.247	77.23
48	70	19.403	1.552	79.96
48.25	100	17.605	0.908	73.95
49	100	17.418	1.095	75.82
51	100	17.271	1.256	77.29
56	100	17.121	1.422	78.79
72	100	16.906	1.661	80.94
96	100	16.733	1.854	82.67

Rate steps: 40 m³/d from 0 h, 70 m³/d from 24 h, 100 m³/d from 48 h; $p_i$ = 25 MPa. The data were calculated with a model of a uniform infinite reservoir ( $k$ = 40 mD, $S$ = 3) with wellbore storage ( $C$ = 0.02 m³/MPa) and rounded to 0.001 MPa.

CalculatedPlot of the multi-rate record for the data of the table: the horizontal axis is the superposition time function, the vertical axis is the pressure drop divided by the current rate. The points of all three rate steps fall on one straight line. Only the readings taken 15 min after the start and after a rate change lie a little below it, when the effect of the wellbore is still felt. The line is drawn in the example below.

Worked exampleFlow capacity and skin without shutting the well in

Given (assumed data)

Reservoir and well: $h$ = 12 m; $\phi$ = 0.18; $\mu$ = 1.6 mPa·s; $B$ = 1.25; $c_t = 1.6\cdot10^{-3}$ MPa⁻¹; $r_w$ = 0.108 m
Rate steps: 40, 70 and 100 m³/d from 0, 24 and 48 h
Points for the line: 10 readings taken no earlier than 3 h after a rate change

Solution

The superposition function, for example for 56 h (third rate step): $X_3 = \tfrac{40}{100}\lg 56 + \tfrac{30}{100}\lg 32 + \tfrac{30}{100}\lg 8 = 1.422$ .
A least-squares line (the line that fits the points best): slope $m' = 9.000$ kPa/(m³/d), that is 0.00900 MPa/(m³/d); intercept at $X_n = 0$ : $b' = 65.99$ kPa/(m³/d).
Flow capacity: $kh = 2.149\cdot1.25\cdot1.6/0.00900 = 478$ mD·m; permeability $k = 478/12 = 39.8$ mD.
Skin: $b'/m' = 7.332$ ; $\lg\dfrac{k}{\phi\mu c_tr_w^2} = 6.869$ ; $S = 1.151\,(7.332 - 6.869 + 2.098) = 2.95$ .
What a naive analysis would give. Suppose we process the third rate step as an ordinary drawdown test, counting time from 48 h. On the same points, the slope of $p_{wf}$ against $\lg(t - 48)$ comes out as 0.444 MPa per cycle, instead of $2.149\cdot100\cdot1.25\cdot1.6/(40\cdot12) = 0.895$ , and the permeability comes out as 81 mD.

Answer: From the record at three rates,

kh \approx 478

mD·m,

k \approx 39.8

mD and the skin is about 2.95, without shutting the well in.

What it means. The data were calculated with a model of permeability 40 mD and skin 3. The analysis returned the permeability with an error of 0.5% and the skin with an error of 0.05. This is a trace of the effect of the wellbore: 3 h after a rate change it has not quite finished. For a record calculated without the wellbore, the same calculation returns 40 mD and skin 3 exactly, and the rounding of the readings to 0.001 MPa hardly matters. The readings 15 min after a rate change lie below the line by 0.2–0.7 kPa/(m³/d): the inflow from the reservoir has not yet caught up with the new rate. Processing the last rate step without the history would double the permeability: the pressure in it falls more slowly, because the disturbances from the first two rates have already slowed down. The data in this example are assumed for illustration.

What this analysis needs

The main thing is accurate rates and accurate moments of the rate changes. An error in the rate of the current step goes into the flow capacity, and an error in the moment of a change distorts the superposition function for the first hours of the new step. If the rate changes smoothly, its history is split into small steps or processed by convolution (a way of adding up the effects of a continuously changing rate; see the advanced section). For a build-up after several rates, we plot $p_{ws}$ against the superposition function in which the shut-in enters as the last step with the rate $-q_n$ , and the rate before the shut-in, $q_n$ , stands in the denominator. For a single rate step this function equals the base-10 logarithm of the Horner ratio.

Reservoir boundaries and image wells

Superposition helps also when the disturbance reaches a boundary. Suppose a well works at a distance $L$ from a straight sealing fault (a fault that does not let fluid through). There is no ready-made solution for a reservoir with such a boundary, but we can build it from the solutions for an infinite reservoir. On the other side of the fault, at the same distance, we place an imaginary well with the same rate history. This is an image well. In Russian-language literature it is also called a fictitious well, and the technique is called the method of reflection. The two identical wells are symmetric about the line of the fault: at symmetric points on both sides of it the pressure is the same. So there is no pressure drop across the line, and nothing flows through it; the flow goes only along it. This is exactly the condition of a no-flow boundary.

SchematicSketch: how images replace boundaries. A no-flow boundary: an image with the same rate. A constant-pressure boundary: an image with the opposite sign (an injector). A corner between two perpendicular faults: three images. A channel between parallel faults: an infinite row of images, cut off when the contribution of the next ones becomes negligible. Not to scale.

A well near a sealing fault

p_i - p_{wf} = \frac{qB\mu}{4\pi kh}\left[\ln\frac{kt}{\phi\mu c_t r_w^2} + 0.809 + 2S + E_1\!\left(\frac{\phi\mu c_t\,(2L)^2}{4kt}\right)\right]

where:

$L$ distance from the well to the fault; $2L$ is the distance to the image, m
$E_1$ exponential integral: the contribution of the image, —
$p_{wf},\ p_i$ bottomhole and initial pressure, Pa
$q,\ B,\ \mu,\ k,\ h$ rate, formation volume factor, viscosity, permeability, thickness, m³/s, m³/m³, Pa·s, m², m
$\phi,\ c_t,\ r_w,\ S$ porosity, total compressibility, radius of the well, skin, —, 1/Pa, m, —

Units: SI; the logarithm is natural

Holds when

The fault is a straight sealing boundary through the whole thickness of the reservoir. The disturbance has not reached any other boundary.
The reservoir is uniform. Beyond the fault we imagine the same reservoir continuing.
The skin is added only for the real well. The image has the same rate history. For a changing rate, both are split into the same steps.

The last term is the contribution of the image. While the disturbance has not had time to reach the fault and “return”, it is negligible. Later it grows and eventually becomes a logarithm of time itself, so the slope of the semilog straight line doubles. How to find the distance to the fault from a pressure record is covered in the lesson A sealing fault near the well: the method of images.

In other unit systems (1)

Metric oilfield units

p_i - p_{wf} = 0.933\,\frac{qB\mu}{kh}\left[\ln\frac{3.553\cdot10^{-3}\,kt}{\phi\mu c_t r_w^2} + 0.809 + 2S + E_1\!\left(\frac{\phi\mu c_t\,(2L)^2}{0.01421\,kt}\right)\right]

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

$0.933 = 1/(4\pi\cdot0.0853)$ ; $3.553\cdot10^{-3}$ and $0.01421 = 4\cdot3.553\cdot10^{-3}$ are the factors of dimensionless time.

CalculatedCalculated pressure drop along a line through well P-5 perpendicular to the fault, 240 h after the start (the well is at the origin, the fault is at 150 m). The solid curve is the reservoir with the fault: the sum of the contributions of the well and the image. At the fault it is horizontal, because the flow across the boundary is zero. The dotted curve beyond the fault is the imaginary half of the picture with the image at 300 m. The grey dashed curve is the same reservoir without the fault.

Worked exampleWhen the well “feels” the fault

Given (assumed data)

Well P-5: 80 m³/d; $S$ = 1; $r_w$ = 0.108 m
Distance to the fault $L$: 150 m
Reservoir: the same; $\eta$ = 308 m²/h

Solution

The radius of investigation $\sqrt{4\eta t}$ reaches the fault in $L^2/(4\eta) = 18$ h. The argument of the image at that moment is $x = (2L)^2/(4\eta t) = 4$ , $E_1(4) = 0.0038$ : the addition to the pressure drop is only 0.0012 MPa.
The radius of investigation reaches the image, at a distance $2L$ , in $(2L)^2/(4\eta) = 73$ h. There $x = 1$ , $E_1(1) = 0.219$ , and the addition is 0.068 MPa, which is 1.3% of the pressure drop in the well.
After 240 h: $x = 0.304$ , $E_1 = 0.896$ , addition 0.279 MPa; the bottomhole pressure is 18.97 MPa instead of 19.25 MPa without the fault.
After 720 h: the addition is 0.56 MPa, which is 8% of the pressure drop.

Answer: A fault 150 m away becomes noticeable after about three days of work, when the radius of investigation approaches twice the distance to it (73 h). By 30 days it gives 8% of the pressure drop in the well.

What it means. The boundary acts with a delay: the disturbance must reach the fault and “return” to the well. If we do not allow for the fault, the faster fall of pressure in the late hours may be taken for a worsening of the reservoir or for a growing skin. If the boundary were not sealing but kept a constant pressure, the pressure in the well would eventually stabilise at 5.56 MPa below the initial pressure (the formula below). The data in this example are assumed for illustration.

If the boundary holds a constant pressure (for example, a nearby edge of an active aquifer, or a row of injection wells), the image is put in with the opposite sign: an injection well with the same rate. On the line of the boundary the contributions of the well and of the image are equal in size and opposite in sign, and the pressure there stays at its initial value. With time, fluid entering through the boundary makes up the whole withdrawal, and the pressure in the well stops falling:

A well near a constant-pressure boundary: the steady state

p_i - p_{wf} \to \frac{qB\mu}{2\pi kh}\left[\ln\frac{2L}{r_w} + S\right]\quad \text{for}\ t \gg \frac{\phi\mu c_t\,(2L)^2}{4k}

where:

$L$ distance from the well to the constant-pressure boundary, m
$p_i$ initial pressure, which is also the pressure at the boundary, Pa
$q,\ B,\ \mu,\ k,\ h,\ r_w,\ S$ rate, formation volume factor, viscosity, permeability, thickness, radius of the well, skin, m³/s, m³/m³, Pa·s, m², m, m, —
$\phi,\ c_t$ porosity and total compressibility (in the condition on time), —, 1/Pa

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

Holds when

The boundary is a straight line with the constant pressure $p_i$ through the whole thickness of the reservoir; there are no other boundaries.
The time is much longer than the time in which the disturbance reaches the image.

The difference between the logarithm of the contribution of the well and the logarithm of the contribution of the image stops depending on time. The result is the formula of steady-state inflow, in which twice the distance to the boundary plays the role of the radius of the supply contour (the outer boundary that holds the pressure constant) (Steady-state radial flow).

For other boundaries the images are placed by the same rules as the images in a system of mirrors. A well in the corner between two perpendicular faults needs three images. Between two parallel faults we need an infinite row, which is cut off when the contribution of the next images becomes negligible. A closed rectangle needs a two-dimensional lattice of images. The signs of such boundaries on pressure records are covered in the lessons A sealing fault near the well: the method of images and Several boundaries: parallel faults and wedge-shaped blocks.

Where superposition works

Superposition in methods that appear in other lessons

Task: Pressure build-up
What is added: a start at the rate $q$ and a “start” at the rate $-q$ at the moment of the shut-in
Where it is covered: Pressure build-up analysis

Task: Multi-rate record
What is added: the rate steps of one well
Where it is covered: this lesson

Task: Interference test
What is added: the contributions of several wells at the observation point
Where it is covered: Well testing: purposes and types of test

Task: Wellbore storage
What is added: a smoothly changing inflow from the reservoir
Where it is covered: Wellbore storage

Task: Faults and no-flow boundaries
What is added: the real well and its images
Where it is covered: A sealing fault near the well: the method of images

Task: Water influx from the aquifer
What is added: steps of pressure drop at the edge of the accumulation
Where it is covered: Unsteady-state water influx: the van Everdingen–Hurst method

Task: Simplified rate history
What is added: an equivalent producing time instead of steps
Where it is covered: Equivalent producing time and synthetic flowing time

Task	What is added	Where it is covered
Pressure build-up	a start at the rate $q$ and a “start” at the rate $-q$ at the moment of the shut-in	Pressure build-up analysis
Multi-rate record	the rate steps of one well	this lesson
Interference test	the contributions of several wells at the observation point	Well testing: purposes and types of test
Wellbore storage	a smoothly changing inflow from the reservoir	Wellbore storage
Faults and no-flow boundaries	the real well and its images	A sealing fault near the well: the method of images
Water influx from the aquifer	steps of pressure drop at the edge of the accumulation	Unsteady-state water influx: the van Everdingen–Hurst method
Simplified rate history	an equivalent producing time instead of steps	Equivalent producing time and synthetic flowing time

In all cases we add the solutions for a constant rate (or a constant pressure), which are solutions of one and the same linear equation.

Terms and symbols

Superposition principle: Adding the solutions of a linear equation: the pressure change from several sources equals the sum of the changes from each of them.
Adding in space: The pressure at a point from several wells. Each has its own rate, distance and time of work. The skin counts only for the well in which the pressure is calculated.
Adding in time: A change of rate treated as the start of another well at the same point, with a rate equal to the increase $q_j - q_{j-1}$ . A shut-in is the start of an injection well.
Superposition time function $X_n$: A weighted sum of the base-10 logarithms of the times elapsed since each rate change. The plot of a multi-rate record is drawn against it.
Image well: An imaginary well beyond a boundary of the reservoir whose contribution reproduces the condition at the boundary: with the same rate for a no-flow boundary, with the opposite sign for a constant-pressure boundary.
Pulse test: An interference test in which the active well is started and shut in by turns. It uses the delay and the size of the response in the neighbouring well.

Common mistakes and misconceptions

Wrong

“The skin of a neighbouring well increases the pressure drop in our well.”

In fact

At a given rate, a skin changes the pressure only at the wellbore of its own well. Only the well's own skin goes into the sum for our well.

Wrong

“After a change of rate the pressure depends only on the new rate and the time elapsed since the change.”

In fact

The disturbances from all the earlier rates keep acting. In the lesson's example, a day after the rate was cut from 100 to 30 m³/d, a calculation without the history underestimates the pressure drop by a quarter or more.

Wrong

“The well is shut in, so the pressure around it starts to rise at once.”

In fact

In its own wellbore, yes. But at a neighbouring point of the reservoir the pressure keeps falling for some time: in the lesson's example, 200 m from the well, for another 15 h.

Wrong

“For a neighbouring well, as for the well itself, we can use the logarithmic approximation.”

In fact

Only when $x = r^2/(4\eta t) < 0.01$ . For wells 450 m apart in the lesson's reservoir, this holds only after almost two years. After 20 days the logarithm underestimates the contribution of the neighbour by more than a third.

Wrong

“A fault shows up in the well as soon as the radius of investigation reaches it.”

In fact

The contribution of the image becomes noticeable when the radius of investigation approaches twice the distance to the fault: the disturbance must reach the boundary and “return”.

In depth: advanced detail

A smoothly changing rate: the Duhamel integral

If the rate changes continuously, we can split its history into smaller and smaller steps. In the limit the sum becomes an integral, called the Duhamel integral, or convolution:

The Duhamel integral (convolution) for a changing rate

p_i - p_{wf}(t) = \int_0^t \Delta p_u(t - \tau)\,\dd q(\tau)

where:

$\Delta p_u(t)$ pressure drop at the bottom of the well when the well starts at a unit constant rate (together with the skin and, if needed, the wellbore), Pa/(m³/s)
$\dd q(\tau)$ increase of the rate at the moment $\tau$ , including the jump from zero at the start, m³/s
$\tau$ the moment to which the increase of the rate refers, s

Units: SI (any consistent system of units)

Holds when

The problem is linear; before the start the pressure in the reservoir has evened out.
For a stepwise history the integral turns into the sum of the section on adding in time.

Each small increase of the rate starts its own response, and the integral adds them. In Laplace images (the Laplace transform, a mathematical change of variable), a convolution turns into a product. This is how the solution with wellbore storage and skin was found in the lesson on wellbore storage. The inverse problem is to recover, from a record of pressure and rate, the response to a unit constant rate, as if the whole test were one long drawdown test. It is called deconvolution. It is solved numerically with regularisation (a stabilising technique), and it is very sensitive to errors in the rates and in the initial pressure.

Where superposition is approximate

Gas. The viscosity and compressibility of gas depend on pressure, and the equation for pressure is nonlinear. Changing to pseudo-pressure makes it almost linear in the coordinates. For large pressure changes the term with the time derivative remains nonlinear, and pseudo-time is introduced as well. For moderate pressure drops, superposition in pseudo-pressures works well.
Two-phase flow at the wellbore. When the bottomhole pressure is below the bubble point, gas comes out at the well, the mobility of the oil falls and depends on the pressure history. Adding is then only a rough approximation, and the pressure is calculated with a numerical model (What a reservoir simulation model is and why it is needed).
Inertial losses. The extra pressure drop grows as the square of the rate, that is, the skin is $S + Dq$ . In adding in time, the term with the skin stops being linear in the rate. In practice the skin of each rate step is taken at its own rate (Inflow at high rates: non-Darcy flow).
Pressure-dependent permeability. In fractured and weakly cemented rocks the permeability falls when the pressure falls, and the “unit” of the pressure drop changes with time (Rock compressibility and the effect of stress on core properties).
A changing skin and wellbore storage. Well clean-up, and the redistribution of gas and liquid in the wellbore, break the condition that $S$ and $C$ are constant (Nonideal wellbore storage, well clean-up and changing skin).
Non-uniform initial pressure. If the reservoir was already disturbed by the work of wells before the test, the continuation of this earlier disturbance, the pressure trend, is added to the sum. Without it the build-up or the drawdown will be distorted.

Connections to other topics

Line-source solution and the pressure function — the solution that is added up; the function $E_1$ and the logarithmic approximation.
Diffusivity equation — the linearity of the equation is the basis of superposition, and the conditions under which it holds.
Pressure build-up analysis — pressure build-up as the sum of a start and an imaginary injection well; the Horner equation.
Pressure drawdown analysis — the semilog straight line of a single rate step is a special case of the plot with the superposition function.
Wellbore storage — the smoothly changing inflow from the reservoir after a rate change, and its treatment by convolution.
Well testing: purposes and types of test — interference tests and multi-rate records among the types of well tests.
A sealing fault near the well: the method of images — the analysis of a record near a fault by the method of images: the doubling of the slope and the distance to the boundary.
Equivalent producing time and synthetic flowing time — how the rate history is simplified with an equivalent producing time.
Unsteady-state water influx: the van Everdingen–Hurst method — superposition of pressure steps at the edge of the accumulation to calculate water influx.

Check yourself

1
Well A works at a rate of 50 m³/d, and well B, 300 m from it, works at 80 m³/d. The skin of A is 5, and the skin of B is 0. Which skins go into the calculation of the bottomhole pressure of well A?
1. A.Only the skin of A
2. B.The skins of both wells
3. C.Only the skin of B
4. D.None: the skin does not enter superposition
Show the answer
Correct answer: A. Only the skin of A
A skin is a pressure drop in a thin zone at the wellbore of its own well. At a given rate of well B, its skin changes the pressure only in its own wellbore, and at the point of well A the contribution of B is set by the reservoir, that is, by the function $E_1$ . The own skin of A must be included.
2
A well worked for 10 h at 60 m³/d, and then the rate was raised to 90 m³/d. Find the bottomhole pressure drop 30 h after the start, if the pressure “unit” $B\mu/(0.536\,kh)$ = 0.005 MPa per 1 m³/d, the dimensionless time per 1 h is 20,000, and the skin is 2.
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There are two steps: 60 m³/d has worked for 30 h, and the increase of 30 m³/d for 20 h. $p_D(30\ \text{h}) = \tfrac12(\ln 600{,}000 + 0.809) = 7.06$ ; $p_D(20\ \text{h}) = \tfrac12(\ln 400{,}000 + 0.809) = 6.85$ . $\Delta p = 0.005\,[60\,(7.06 + 2) + 30\,(6.85 + 2)] = 0.005\,(543.6 + 265.5) \approx 4.05$ MPa. The skin gave $0.005\cdot90\cdot2 = 0.9$ MPa, at the current rate.
3
A well was shut in after a month of work. What happens in the first hours after the shut-in to the pressure in an observation well 300 m from it?
1. A.It does not change until the disturbance arrives, and then falls
2. B.It starts to rise at once
3. C.It keeps falling, and then starts to rise
4. D.It returns instantly to the initial value
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Correct answer: C. It keeps falling, and then starts to rise
A shut-in is the start of an imaginary injection well. Its contribution at a distant point is at first negligible, while the contribution of the well that was working keeps growing. The pressure in the observation well falls until the rate of growth of the second contribution equals that of the first, and only then does it begin to recover.
4
A well is 100 m from a sealing fault. How many times later does the fault show up if it is 200 m away?
1. A.Four times as late
2. B.A fault at 200 m will never show up
3. C.Twice as late
4. D.At the same time
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Correct answer: A. Four times as late
The contribution of the image depends on the argument $x = \phi\mu c_t(2L)^2/(4kt)$ , that is, on the ratio $L^2/t$ . The same effect at twice the $L$ comes at four times the time.
5
Why does the skin enter the formula for adding in time only as $q_nS$ , although it is written in every term?
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Each step carries a pressure drop across the skin proportional to its increase of rate: $(q_j - q_{j-1})S$ . The sum of the increases equals the current rate: $(q_1 - 0) + (q_2 - q_1) + \dots + (q_n - q_{n-1}) = q_n$ . The pressure drop across the skin is set by the current flow through the near-wellbore zone, not by the history. At a shut-in ( $q = 0$ ) it disappears, so build-up pressures contain no skin.
6
Can we calculate by superposition the bottomhole pressure of a gas well, in which the pressure falls by half, by adding solutions for pressure?
1. A.No, superposition is forbidden for gas
2. B.Yes, if the skin is zero
3. C.Only roughly: the equation for gas is nonlinear, and we add solutions in pseudo-pressures
4. D.Yes, superposition is always valid
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Correct answer: C. Only roughly: the equation for gas is nonlinear, and we add solutions in pseudo-pressures
The viscosity and compressibility of gas change with pressure, and the equation for pressure is nonlinear. Changing to pseudo-pressure makes it almost linear, and adding in pseudo-pressures works well for moderate pressure changes. For large changes we add pseudo-time or calculate numerically.

Updated: 5 October 2026