Unsteady-state water influx: the van Everdingen–Hurst method

Quick refresher

The van Everdingen–Hurst problem is the diffusivity equation for an aquifer in which the accumulation plays the role of the well. At the contact, of radius $r_o$ , the pressure has fallen by $\Delta p$ and stays there. The outer boundary $r_a$ is closed or infinitely far. The answer is the cumulative influx $W_e = U\Delta p\,W_D(t_D, r_{aD})$ .
The influx is the elastic expansion of the part of the aquifer that the pressure drop has reached. The dimensionless influx $W_D$ is the volume of this “funnel” in dimensionless form, so it grows together with the funnel.
At first $W_D \approx 2\sqrt{t_D/\pi}$ : four times the time gives only about twice as much water. A closed aquifer reaches the limit $(r_{aD}^2 - 1)/2$ , its elastic store. An infinite one gives water without a limit, but more and more slowly.
A changing pressure is represented as steps. A step $\Delta p_j$ applied at time $t_j$ acts independently of the others, and its contribution by time $t_n$ is $U\Delta p_jW_D(t_{Dn} - t_{Dj})$ . For each new date all the steps are recalculated, because the time since each of them has grown.
The drops are chosen so that the stepped curve follows the average pressures of the intervals: $\Delta p_0 = (p_0 - p_1)/2$ , $\Delta p_j = (p_{j-1} - p_{j+1})/2$ . We call them half drops, because each step takes half of the pressure fall in the interval before it and half of the fall in the interval after it. This rule understates the influx, most of all in the first steps; shorter steps reduce the error.
In the material balance we compare the influx from the aquifer with the influx from the balance, $F - NE_t$ . For a forecast we choose the pressure at the end of the period at which the two estimates agree.

Key relationships

W_e = U\,\Delta p\,W_D(t_D,\,r_{aD}),\qquad U = 2\pi f\phi c_t h r_o^2

influx for a constant pressure drop at the contact; $U$ in m³/MPa

t_D = 3.553\cdot10^{-3}\,\dfrac{k\,t}{\phi\,\mu_w c_t\,r_o^2}

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

W_e(t_n) = U\sum_{j=0}^{n-1}\Delta p_j\,W_D(t_{Dn} - t_{Dj})

superposition of the pressure steps at the contact

\Delta p_0 = \tfrac12(p_0 - p_1),\qquad \Delta p_j = \tfrac12(p_{j-1} - p_{j+1})

drops from the average pressures of the intervals; $p_0 = p_i$

W_D \to \tfrac12\,(r_{aD}^2 - 1),\qquad U\Delta p\,\tfrac12(r_{aD}^2 - 1) = (c_w + c_f)\,W\Delta p

the limit of a closed aquifer is its elastic store

Common pitfalls

Applying the whole drop of an interval at its start (the influx is then too high) or at its end (too low) instead of two halves.
Counting the time of all the steps from the start of development instead of from the moment of each step, or not recalculating the old steps for the new date.
Taking $W_D$ of an infinite aquifer after the pressure drop has reached the boundary of a closed one. The influx will be too high.
Mixing up the constants of the dimensionless time for hours, days and years, or putting the radius of a well in place of the radius of the accumulation $r_o$ .
Taking a mismatch with the balance in the first years for an error of the aquifer model. Often it is the error of steps that are too long.

After this lesson you can

state the problem of a radial aquifer with a constant pressure drop at the contact, and explain the meaning of the dimensionless time, the dimensionless influx and the aquifer constant $U$
calculate $U$ and $t_D$ in metric and oilfield units, and find the influx for a constant pressure drop for an infinite and for a closed aquifer
represent a changing contact pressure as steps, choose the drops from the average pressures of the intervals, and calculate the cumulative influx by superposition (adding up the responses to the steps)
estimate in which direction and by how much the step calculation is wrong, and choose the length of the step
check the aquifer model against the material balance, and forecast the pressure for the next period by trial and error, until the influx from the balance and the influx from the aquifer agree

The problem of a radial aquifer

In the lesson Aquifer models: water influx the unsteady-state model was described in general terms: the pressure drop spreads gradually from the accumulation through the aquifer, and the influx is set by how much of the aquifer has already “felt” the drop. Here we give the model itself and show how it is used for a real pressure history.

The accumulation is represented as a circle of radius $r_o$ (for an accumulation of another shape we take a circle of the same area). The aquifer is a ring of the same thickness from $r_o$ to the outer radius $r_a$ , or a sector of a ring if the aquifer touches only the fraction $f$ of the perimeter (geometry of the aquifer). At the moment $t = 0$ the pressure at the contact falls by $\Delta p$ and does not change afterwards. Water starts to flow towards the accumulation, and the pressure drop spreads through the aquifer in the same way as around a well that has just started to produce. The only difference is that the whole accumulation plays the role of the wellbore (Diffusivity equation).

This is a problem with a given pressure at the inner boundary: the constant terminal pressure solution (CTP). In well testing, the paired problem is more often solved: with a given rate, where the pressure is calculated (the constant terminal rate solution). For water influx the first setting is natural: the contact pressure is known from measurements, and the influx has to be found. The second one will be needed in the step-by-step Carter–Tracy method (Simplified water influx models: Fetkovich and Carter–Tracy).

The van Everdingen–Hurst problem in dimensionless form

\frac{1}{r_D}\frac{\partial}{\partial r_D}\!\left(r_D\,\frac{\partial P}{\partial r_D}\right) = \frac{\partial P}{\partial t_D},\qquad P(1, t_D) = 1,\quad P(r_D, 0) = 0,\quad \left.\frac{\partial P}{\partial r_D}\right|_{r_D = r_{aD}} = 0

where:

$P$ fraction of the pressure drop at the contact reached at a point of the aquifer, $(p_i - p)/\Delta p$ , —
$r_D$ distance from the centre of the accumulation in radii of the accumulation, $r/r_o$ , —
$t_D$ dimensionless time of the aquifer, $kt/(\phi\mu_wc_tr_o^2)$ , —
$r_{aD}$ outer radius of the aquifer in radii of the accumulation, $r_a/r_o$ , —

Units: dimensionless form

Holds when

a uniform aquifer of constant thickness, full of water; the compressibilities are small and constant, so the equation is linear
the flow is radial, with no crossflow through the top and bottom
the outer boundary is impermeable; for an infinite aquifer the condition on it is replaced by $P \to 0$ as $r_D \to \infty$

No properties of the aquifer are left in the problem: the solution depends only on $t_D$ and $r_{aD}$ . It is found once, in the Laplace domain (a mathematical method that makes such equations easier to solve; see the advanced section) or numerically, and then converted to any aquifer.

The rate of influx is the flow of water through the contact, by Darcy's law. In dimensionless form it is the slope of the pressure profile at the contact, $q_D = -\partial P/\partial r_D$ at $r_D = 1$ . It is the same dimensionless rate as for a well with a constant bottomhole pressure. Its integral over the dimensionless time is the dimensionless influx $W_D$ . Going back to dimensional quantities (the derivation is in the advanced section), we get the main formula of the method.

Influx for a constant pressure drop at the contact

W_e = U\,\Delta p\,W_D(t_D,\,r_{aD}),\qquad U = 2\pi f\phi c_t h r_o^2,\qquad t_D = \frac{k\,t}{\phi\,\mu_w c_t\,r_o^2}

where:

$W_e$ cumulative water influx at reservoir conditions by time $t$ , m³
$U$ aquifer constant, m³/MPa
$\Delta p$ pressure drop at the contact, constant from the moment $t = 0$ , MPa
$W_D$ dimensionless influx: a function of $t_D$ and $r_{aD} = r_a/r_o$ , —
$f$ fraction of the perimeter of the accumulation that the aquifer touches, $\theta/360°$ , —
$\phi,\ c_t,\ h$ porosity, total compressibility $c_w + c_f$ and thickness of the aquifer, —, 1/MPa (1/Pa in $t_D$ ), m
$r_o$ radius of the accumulation (the initial contact with the water), m
$k,\ \mu_w$ permeability of the aquifer and viscosity of the reservoir water, m², Pa·s
$t$ time since the pressure drop, s

Units: $W_e$ and $U$ in one system of volume and pressure (m³ and MPa); $t_D$ in SI

Holds when

the problem above: a uniform radial aquifer, with the contact pressure falling in one step and not changing afterwards
the pressure at the contact is taken as equal to the average pressure of the accumulation, as in the material balance
the aquifer is full of water, $c_t = c_w + c_f$ , and the compressibilities are constant

The formula separates three factors. $U$ is the capacity of the aquifer on the scale of the accumulation: how much water it gives up for each megapascal of pressure drop. $\Delta p$ is the pressure drop that drives the influx. $W_D$ is how the response develops in time, the same for all aquifers with the same $r_{aD}$ . Permeability and viscosity set the speed of the response, and they enter only $t_D$ . Porosity, compressibility, thickness and dimensions set the store, and they enter $U$ and $r_{aD}$ . The fraction of the perimeter $f$ changes only $U$ .

In other unit systems (2)

Metric oilfield units

U\,[\text{m}^3/\text{MPa}] = 2\pi f\phi c_t h r_o^2,\qquad t_D = 3.553\cdot10^{-3}\,\frac{k\,t}{\phi\,\mu_w c_t\,r_o^2}

$c_t$ in 1/MPa, $h$ and $r_o$ in m, $k$ in mD, $t$ in hours, $\mu_w$ in mPa·s. With time in days the multiplier is 0.0853, and with time in years (8766 h) it is 31.15.

There is no constant in the formula for $U$ . $3.553\cdot10^{-3}$ is the same multiplier as for the hydraulic diffusivity in the lesson Diffusivity equation: the conversion of mD to m², hours to seconds, mPa·s to Pa·s and 1/MPa to 1/Pa.

Oilfield units

U\,[\text{bbl/psi}] = 1.119\,f\phi c_t h r_o^2,\qquad t_D = 2.637\cdot10^{-4}\,\frac{k\,t}{\phi\,\mu_w c_t\,r_o^2}

$c_t$ in 1/psi, $h$ and $r_o$ in ft, $k$ in mD, $t$ in hours, $\mu_w$ in cP. With time in days the multiplier is $6.33\cdot10^{-3}$ .

$1.119 = 2\pi/5.615$ converts cubic feet to barrels; $2.637\cdot10^{-4}$ is the diffusivity multiplier in these units.

Influx as the expansion of the pressure-drop funnel

W_e = c_t\int_{V_p}(p_i - p)\dd V_p = U\,\Delta p\int_1^{r_{aD}} P\,r_D\dd r_D,\qquad W_D = \int_1^{r_{aD}} P\,r_D\dd r_D

where:

$V_p$ pore volume of the aquifer; $\dd V_p = 2\pi f h\phi\,r\dd r$ , m³
$p_i - p$ pressure drop at a point of the aquifer, MPa
$P,\ r_D$ fraction of the pressure drop and the dimensionless distance, as in the problem above, —

Units: m³, MPa, 1/MPa

Holds when

there is only water in the aquifer; $c_t$ is constant; there is no recharge from outside

The water that has entered the accumulation is the elastic expansion of the water and the pores where the pressure has already fallen. $W_D$ is the volume of the pressure-drop “funnel” in dimensionless form: as long as the funnel grows, the influx grows. When the funnel fills the whole closed aquifer ( $P = 1$ everywhere), the integral equals $(r_{aD}^2 - 1)/2$ , and the influx reaches the elastic store of the aquifer $(c_w + c_f)W\Delta p$ . On the chart of pressure profiles in the lesson on aquifer models, $W_D$ is the area under the curve, weighted by the distance.

Dimensionless influx of an infinite and of a closed aquifer

The solution of the problem is a family of curves $W_D(t_D)$ , one for each radius ratio $r_{aD}$ (the chart and the table). Their course is explained by how the pressure-drop funnel grows.

CalculatedThe dimensionless influx

W_D

for a constant pressure drop at the contact, for closed aquifers with

r_{aD}

= 2, 5 and 10 and for an infinite aquifer (logarithmic axes; calculated from the diffusivity equation). The dashed line is the early asymptote (the straight-line limit of the curves at the start)

2\sqrt{t_D/\pi}

The early period. For $t_D \ll 1$ the funnel is a thin layer at the contact. Its curvature does not matter yet, and the aquifer behaves like one with a flat boundary: $W_D \approx 2\sqrt{t_D/\pi}$ . Four times the time gives only twice the influx. Later the funnel covers wider and wider rings, the growth speeds up, and by $t_D = 1$ the asymptote understates $W_D$ by more than a quarter.
The infinite aquifer. $W_D$ grows without a limit, but more and more slowly: for large $t_D$ roughly as $2t_D/\ln t_D$ . The rate of influx for a constant $\Delta p$ falls, because the water has to be brought from a greater and greater distance.
The closed aquifer. Until the pressure drop reaches the outer boundary, its curve coincides with that of the infinite aquifer. The difference becomes noticeable at about $t_D \approx (r_{aD} - 1)^2/4$ (the time to reach the boundary). Then $W_D$ levels off at the limit $(r_{aD}^2 - 1)/2$ , the elastic store. An aquifer with $r_{aD} = 2$ gives nine tenths of the limit by $t_D \approx 1.2$ , and one with $r_{aD} = 10$ only by $t_D \approx 190$ .

Dimensionless influx

W_D

of a radial aquifer

$t_D$: 0.1
$r_{aD} = 2$: 0.4043
$r_{aD} = 5$: 0.4043
$r_{aD} = 10$: 0.4043
infinite: 0.4043

$t_D$: 0.3
$r_{aD} = 2$: 0.7509
$r_{aD} = 5$: 0.7564
$r_{aD} = 10$: 0.7564
infinite: 0.7564

$t_D$: 1
$r_{aD} = 2$: 1.295
$r_{aD} = 5$: 1.568
$r_{aD} = 10$: 1.568
infinite: 1.568

$t_D$: 3
$r_{aD} = 2$: 1.495
$r_{aD} = 5$: 3.198
$r_{aD} = 10$: 3.200
infinite: 3.200

$t_D$: 10
$r_{aD} = 2$: 1.500
$r_{aD} = 5$: 6.967
$r_{aD} = 10$: 7.401
infinite: 7.402

$t_D$: 30
$r_{aD} = 2$: 1.500
$r_{aD} = 5$: 10.98
$r_{aD} = 10$: 16.56
infinite: 16.74

$t_D$: 100
$r_{aD} = 2$: 1.500
$r_{aD} = 5$: 12.00
$r_{aD} = 10$: 35.44
infinite: 43.03

$t_D$: 300
$r_{aD} = 2$: 1.500
$r_{aD} = 5$: 12.00
$r_{aD} = 10$: 48.26
infinite: 105.7

$t_D$: 1000
$r_{aD} = 2$: 1.500
$r_{aD} = 5$: 12.00
$r_{aD} = 10$: 49.50
infinite: 292.6

$t_D$: limit
$r_{aD} = 2$: 1.5
$r_{aD} = 5$: 12
$r_{aD} = 10$: 49.5
infinite: —

$t_D$	$r_{aD} = 2$	$r_{aD} = 5$	$r_{aD} = 10$	infinite
0.1	0.4043	0.4043	0.4043	0.4043
0.3	0.7509	0.7564	0.7564	0.7564
1	1.295	1.568	1.568	1.568
3	1.495	3.198	3.200	3.200
10	1.500	6.967	7.401	7.402
30	1.500	10.98	16.56	16.74
100	1.500	12.00	35.44	43.03
300	1.500	12.00	48.26	105.7
1000	1.500	12.00	49.50	292.6
limit	1.5	12	49.5	—

Calculated from the exact solution of the diffusivity equation (by the inverse Laplace transform from the advanced section); all the digits shown are correct. Between the rows we interpolate on the logarithms of $t_D$ and $W_D$ . The values for the infinite aquifer also suit a closed one until the pressure drop has reached its boundary.

Worked exampleInflux after a single pressure step

Given (assumed data)

Aquifer: the large aquifer of the lesson Aquifer models: water influx: around the whole accumulation ( $f = 1$ ), closed at a radius of 16 km; radius of the accumulation $r_o = 1595.8$ m; thickness 10 m, porosity 0.20, permeability 50 mD, water viscosity 0.4 mPa·s, $c_t = 1.25\cdot10^{-3}$ MPa⁻¹
Pressure drop: the pressure at the contact has fallen by 1 MPa and does not change afterwards

Solution

The aquifer constant: $U = 2\pi\cdot1\cdot0.20\cdot1.25\cdot10^{-3}\cdot10\cdot1595.8^2 = 40{,}000$ m³/MPa. This is exactly twice the product of $c_t$ and the pore volume of the accumulation itself, 16.0 million m³, because the thickness and the porosity are the same in the accumulation and in the aquifer.
A check with the oilfield form: $c_t = 8.618\cdot10^{-6}$ psi⁻¹, $h = 32.81$ ft, $r_o = 5235$ ft, and $U = 1.119\cdot0.20\cdot8.618\cdot10^{-6}\cdot32.81\cdot5235^2 = 1734$ bbl/psi. A direct conversion gives the same: $40{,}000\cdot6.290/145.04 = 1735$ bbl/psi.
Dimensionless time per hour: $3.553\cdot10^{-3}\cdot50/(0.20\cdot0.4\cdot1.25\cdot10^{-3}\cdot1595.8^2) = 6.976\cdot10^{-4}$ ; per year (8766 h), 6.115. After half a year, 2 years and 8 years, $t_D$ = 3.06, 12.23 and 48.92. The radius ratio is $r_{aD} = 16{,}000/1595.8 = 10.03$ .
The dimensionless influx from the solution of the diffusivity equation: for the closed aquifer $W_D$ = 3.241, 8.565 and 23.348; for the infinite one, 3.241, 8.566 and 24.422.
The influx $W_e = U\cdot1\cdot W_D$ is 129.6, 342.6 and 933.9 thousand m³. For the infinite aquifer after 8 years it is 976.9 thousand m³, which is 4.6% more.
The elastic store of the closed aquifer per 1 MPa: the limit is $W_D = (r_{aD}^2 - 1)/2 = 49.77$ , and $U\cdot49.77 = 1991$ thousand m³. In 8 years the aquifer has given up 47% of the store.
If a fault closed 40% of the perimeter ( $f = 0.6$ ), $U$ and the influx would fall in the same proportion: 560.4 thousand m³ in 8 years. $t_D$ and $W_D$ would stay the same.

Answer: For each megapascal of pressure drop the aquifer gives 129.6 thousand m³ of water in half a year, 342.6 in 2 years and 933.9 thousand m³ in 8 years.

What it means. The time grew four times, and the influx only 2.6 times. This is more than twice (which is what $W_D \propto \sqrt{t_D}$ would give), because the funnel in a ring grows faster than at a flat boundary, but it is far from four times. During the first two years the boundary of the aquifer is not yet visible, and a closed aquifer cannot be told from an infinite one. By year 8 it lags behind by 4.6% and has given up less than half of its elastic store. So the outer radius of the aquifer cannot be found from the first few years of the history (History matching of aquifer parameters). The data in this example are assumed for illustration.

A changing pressure: superposition of steps

In reality the pressure at the contact does not fall in one step. It falls throughout the development history. The diffusivity equation is linear, so the responses to different pressure changes add up (Superposition principle). We represent the pressure history as steps: at the moments $t_0 = 0,\ t_1,\ t_2, \dots$ the contact pressure falls by $\Delta p_0,\ \Delta p_1,\ \Delta p_2, \dots$ and does not change in between. Each step sends its own wave of pressure drop into the aquifer, and each wave causes its own influx, as if there were no others.

CalculatedThe calculated pressure drop in this lesson's large aquifer 2.5 years after the start of development: the waves of the first three steps of the history (dashed lines, each labelled with the drop at the contact and the time since the step) and their sum. The early waves have gone far, the later ones are still near the contact but larger in size.

Influx for a stepwise pressure drop at the contact

W_e(t_n) = U\sum_{j=0}^{n-1}\Delta p_j\,W_D\big(t_{Dn} - t_{Dj}\big)

where:

$W_e(t_n)$ cumulative influx by time $t_n$ , m³
$\Delta p_j$ drop of the $j$ -th step, applied at time $t_j$ ; the sum of the drops is the pressure drop from the initial value, MPa
$t_{Dn} - t_{Dj}$ dimensionless time that has passed since the $j$ -th step, —
$U,\ W_D$ aquifer constant and dimensionless influx, as above, m³/MPa, —

Units: m³, MPa

Holds when

the conditions of linearity: the properties of the water and the aquifer do not depend on pressure
the contact pressure is constant between the moments of the steps; how well the steps replace a smooth history depends on the choice of the drops and of the step (the next section)
all the steps are counted from one initial state: before development the pressure everywhere in the aquifer equals $p_i$

Each step acts from its own moment and keeps giving influx until the elastic store of the aquifer is used up. So for each new date the sum is recalculated completely: the time since each step has grown, and its $W_D$ is larger. After $n$ steps the calculation contains $n$ terms, and the whole history contains $n(n+1)/2$ values of $W_D$ . This is where the step-by-step methods save work (Simplified water influx models: Fetkovich and Carter–Tracy).

How to choose the steps: the average pressures of the intervals

The pressure is measured at separate moments, usually once a quarter, half a year or a year, and between the measurements it falls smoothly. The stepped curve should follow this smooth one as closely as possible. If the whole drop of an interval is applied at its start, water starts to come in earlier than it really does, and the influx is overstated. If it is applied at the end, the influx is understated. Usually we split the drop in two: half of the drop of the interval is applied at its start and half at its end. We call these half drops. Then, in each interval, the stepped curve follows the average pressure of its two ends (see the chart).

Step drops from the average pressures of the intervals

\Delta p_0 = \frac{p_0 - p_1}{2},\qquad \Delta p_j = \frac{p_{j-1} - p_{j+1}}{2}\quad (j \ge 1),\qquad \sum_{j=0}^{n-1}\Delta p_j = p_0 - \frac{p_{n-1} + p_n}{2}

where:

$p_0 = p_i$ initial pressure at the contact, MPa
$p_j$ pressure at the contact, measured (or averaged) at time $t_j$ , MPa
$\Delta p_j$ drop of the step applied at time $t_j$ , MPa

Units: MPa

Holds when

if the intervals between measurements differ, the formula is the same, since the stepped curve still follows the average pressures of the intervals, but the longest intervals bring the largest error
the pressure in an interval changes smoothly, without sharp jumps (the start-up or shutdown of many wells)

The drop applied at time $t_j$ is the difference between the average pressures of the neighbouring intervals: half of the pressure fall in the previous interval plus half of the fall in the next one. By time $t_n$ the sum of the drops equals the fall not to $p_n$ but to the average pressure of the last interval.

CalculatedThe contact pressure of this lesson's assumed accumulation from yearly measurements (dots; between them the pressure fell along a straight line), and its stepped representation: the steps are the average pressures of the intervals, and the jumps are at the moments of the measurements.

Error of the influx for this lesson's history with different choices of steps, % of the exact solution: the whole drop of an interval at its start or at its end, and half drops with a yearly and with a quarterly step

Year: 1
whole at start: 67.6
whole at end: −100.0
halves, yearly: −16.2
halves, quarterly: −2.20

Year: 2
whole at start: 39.6
whole at end: −52.1
halves, yearly: −6.3
halves, quarterly: −0.83

Year: 4
whole at start: 18.2
whole at end: −20.7
halves, yearly: −1.2
halves, quarterly: −0.15

Year: 8
whole at start: 7.4
whole at end: −8.0
halves, yearly: −0.3
halves, quarterly: −0.03

Year	whole at start	whole at end	halves, yearly	halves, quarterly
1	67.6	−100.0	−16.2	−2.20
2	39.6	−52.1	−6.3	−0.83
4	18.2	−20.7	−1.2	−0.15
8	7.4	−8.0	−0.3	−0.03

The exact solution is the diffusivity equation for a pressure that changes along a straight line between the measurements. The quarterly steps are built from the same yearly measurements, with the pressure between them taken along a straight line.

The rule of half drops is much more accurate than the two simple variants, but it too understates the influx, most of all in the first steps. In the first year the yearly steps give an influx one sixth smaller than the exact one, by the second year 6% smaller, and by the eighth year only a quarter of a per cent smaller. Quarterly steps reduce the error of the first year seven times. The cause is the shape of the $W_D$ curve, explained in the advanced block.

More detail:Why half drops understate the influx

If the pressure in an interval falls steadily, the exact influx from this fall equals the drop of the interval multiplied by the average value of $W_D$ over all the moments inside the interval. The rule of half drops replaces it with the average of the two end values, which is the trapezoid rule. The curve $W_D(t_D)$ bends over (its slope, the rate of influx, falls with time), and for such a curve the trapezoid always lies under the curve, so the influx is understated.

This matters most in the first interval, where $W_D$ grows most steeply. In the early period, for $W_D \propto \sqrt{t_D}$ , an interval of steady pressure fall gives an exact influx proportional to $\tfrac23$ of the value of $W_D$ at its end, while the rule of half drops gives $\tfrac12$ . That is an error of $-25$ % in the first step, whatever its length. Later the $W_D$ curve is closer to a straight line, and the error of the first step is smaller: about $-16$ % in this lesson's example, where $t_D \approx 6$ in a year. In the old intervals $W_D$ is almost straight, and the error is small. So the relative error is set above all by the number of steps since the start of the history: the shorter the steps, the more of them have already accumulated by any date.

If the whole drop of an interval is applied in its middle, the error has the opposite sign and is smaller (about +4% in the first year for this lesson's history): the value in the middle of a curve that bends over is larger than its average. We can also take the steady fall within the intervals into account exactly, through the integral of $W_D$ ; this is how the exact solution in the table above was calculated. In practice it is simpler to make the step shorter.

Example: the influx from the pressure history

Assumed development history: the contact pressure and the cumulative production at the end of the year

Year: 0
$p$ , MPa: 30.00
$N_p$ , thousand m³: 0
$W_p$ , thousand m³: 0
$B_o$ , m³/m³: 1.3243

Year: 1
$p$ , MPa: 28.47
$N_p$ , thousand m³: 183
$W_p$ , thousand m³: 0
$B_o$ , m³/m³: 1.3276

Year: 2
$p$ , MPa: 26.52
$N_p$ , thousand m³: 584
$W_p$ , thousand m³: 8
$B_o$ , m³/m³: 1.3317

Year: 3
$p$ , MPa: 24.79
$N_p$ , thousand m³: 1132
$W_p$ , thousand m³: 37
$B_o$ , m³/m³: 1.3354

Year: 4
$p$ , MPa: 23.73
$N_p$ , thousand m³: 1717
$W_p$ , thousand m³: 102
$B_o$ , m³/m³: 1.3377

Year: 5
$p$ , MPa: 22.87
$N_p$ , thousand m³: 2301
$W_p$ , thousand m³: 213
$B_o$ , m³/m³: 1.3395

Year: 6
$p$ , MPa: 22.37
$N_p$ , thousand m³: 2849
$W_p$ , thousand m³: 368
$B_o$ , m³/m³: 1.3406

Year: 7
$p$ , MPa: 21.89
$N_p$ , thousand m³: 3360
$W_p$ , thousand m³: 567
$B_o$ , m³/m³: 1.3416

Year: 8
$p$ , MPa: 21.44
$N_p$ , thousand m³: 3835
$W_p$ , thousand m³: 811
$B_o$ , m³/m³: 1.3426

Year	$p$ , MPa	$N_p$ , thousand m³	$W_p$ , thousand m³	$B_o$ , m³/m³
0	30.00	0	0	1.3243
1	28.47	183	0	1.3276
2	26.52	584	8	1.3317
3	24.79	1132	37	1.3354
4	23.73	1717	102	1.3377
5	22.87	2301	213	1.3395
6	22.37	2849	368	1.3406
7	21.89	3360	567	1.3416
8	21.44	3835	811	1.3426

The oil zone is the same as in the lessons General material balance equation for oil reservoirs and Aquifer models: water influx (oil in place 9.06 million m³, initial pressure 30 MPa, bubble point pressure 18 MPa), but this is a different version of its development, with the large aquifer of the lesson Aquifer models: water influx. $N_p$ is the oil at standard conditions, $W_p$ is the water produced, and $B_o$ is the oil formation volume factor at the pressure of the measurement. Between the measurements the pressure fell along a straight line.

Worked exampleCumulative influx by year

Given (assumed data)

History: the table above: the pressure at the contact at the end of each year; the initial pressure is 30 MPa
Aquifer: from the previous example: $U = 40{,}000$ m³/MPa, $t_D$ = 6.115 per year, $r_{aD}$ = 10.03

Solution

The drops of the steps from the average pressures of the intervals: at the start $\Delta p_0 = (30 - 28.47)/2 = 0.765$ MPa; after a year $\Delta p_1 = (30 - 26.52)/2 = 1.74$ ; after two years $\Delta p_2 = (28.47 - 24.79)/2 = 1.84$ ; then 1.395, 0.96, 0.68, 0.49 and 0.465 MPa. Their sum, 8.335 MPa, equals $30 - (21.89 + 21.44)/2$ .
The dimensionless influx 1, 2, …, 8 years after a step ( $t_D$ = 6.115; 12.23; …; 48.92): $W_D$ = 5.217; 8.565; 11.536; 14.267; 16.796; 19.143; 21.323; 23.348.
End of year 1: one step acts: $W_e = 40{,}000\cdot0.765\cdot5.217 = 159.6\cdot10^3$ m³.
End of year 2: two steps act. The first has now acted for two years, and its contribution has been recalculated: $W_e = 40{,}000\cdot(0.765\cdot8.565 + 1.74\cdot5.217) = 625.2\cdot10^3$ m³.
End of year 8: eight steps (the table after the example). The sum of the products $\Delta p_jW_D$ is 141.78 MPa, and $W_e = 40{,}000\cdot141.78 = 5671.2\cdot10^3$ m³.
The same for each year; the results are in the table of the next section.

Answer: In eight years the aquifer has given about 5.67 million m³ of water. The steps applied one and two years after the start give 51% of this sum, and the very first step gives 13%.

What it means. The influx on any date depends on the whole pressure history. By year 8 the steps of the first years, when the pressure fell fastest, give more than half of the sum, although they are long past: the aquifer is still expanding in response to them. The last step gives only 1.7%, because it has acted for only a year. The data in this example are assumed for illustration.

Terms of the sum at the end of year 8

Step: at the start
$\Delta p_j$ , MPa: 0.765
$t_{D8} - t_{Dj}$: 48.92
$W_D$: 23.3484
$\Delta p_jW_D$ , MPa: 17.861

Step: after 1 year
$\Delta p_j$ , MPa: 1.740
$t_{D8} - t_{Dj}$: 42.81
$W_D$: 21.3235
$\Delta p_jW_D$ , MPa: 37.103

Step: after 2 years
$\Delta p_j$ , MPa: 1.840
$t_{D8} - t_{Dj}$: 36.69
$W_D$: 19.1433
$\Delta p_jW_D$ , MPa: 35.224

Step: after 3 years
$\Delta p_j$ , MPa: 1.395
$t_{D8} - t_{Dj}$: 30.58
$W_D$: 16.7957
$\Delta p_jW_D$ , MPa: 23.430

Step: after 4 years
$\Delta p_j$ , MPa: 0.960
$t_{D8} - t_{Dj}$: 24.46
$W_D$: 14.2665
$\Delta p_jW_D$ , MPa: 13.696

Step: after 5 years
$\Delta p_j$ , MPa: 0.680
$t_{D8} - t_{Dj}$: 18.35
$W_D$: 11.5364
$\Delta p_jW_D$ , MPa: 7.845

Step: after 6 years
$\Delta p_j$ , MPa: 0.490
$t_{D8} - t_{Dj}$: 12.23
$W_D$: 8.5649
$\Delta p_jW_D$ , MPa: 4.197

Step: after 7 years
$\Delta p_j$ , MPa: 0.465
$t_{D8} - t_{Dj}$: 6.12
$W_D$: 5.2168
$\Delta p_jW_D$ , MPa: 2.426

Step	$\Delta p_j$ , MPa	$t_{D8} - t_{Dj}$	$W_D$	$\Delta p_jW_D$ , MPa
at the start	0.765	48.92	23.3484	17.861
after 1 year	1.740	42.81	21.3235	37.103
after 2 years	1.840	36.69	19.1433	35.224
after 3 years	1.395	30.58	16.7957	23.430
after 4 years	0.960	24.46	14.2665	13.696
after 5 years	0.680	18.35	11.5364	7.845
after 6 years	0.490	12.23	8.5649	4.197
after 7 years	0.465	6.12	5.2168	2.426

The sum of the last column is the multiplier of $U$ in the superposition formula. The steps are counted from the start of development; for example, the step “after 2 years” has acted for 6 years by the end of year 8.

Influx in the material balance: check and forecast

Material balance gives a second, independent estimate of the influx. If the oil in place $N$ is known, $W_e = F - NE_t$ , where $F$ is the underground withdrawal (the reservoir volume of the fluids produced) and $E_t$ is the expansion of the oil, the connate water and the pores per 1 m³ of oil in place (General material balance equation for oil reservoirs). If the chosen aquifer is right, the two estimates agree on all dates. If the oil in place is unknown, we find it together with $U$ from the straight line of the Havlena–Odeh method, in which the influx function is the sum $\sum\Delta p_jW_D$ (accumulation with an aquifer). If neither the size nor the conductivity of the aquifer is known, we fit them to the history, which is the subject of the lesson History matching of aquifer parameters.

Worked exampleChecking the aquifer with the material balance

Given (assumed data)

Oil in place and properties: $N$ = 9.06 million m³; $B_{oi}$ = 1.3243 and $B_w$ = 1.02 m³/m³; $(c_wS_w + c_f)/(1 - S_w) = 1.2167\cdot10^{-3}$ MPa⁻¹, as in the lesson General material balance equation for oil reservoirs; the pressure stays above the bubble point pressure all the time
History and influx from the aquifer: the history table and the previous example

Solution

Year 8, $p$ = 21.44 MPa, $B_o$ = 1.3426. Withdrawal: $F = N_pB_o + W_pB_w = 3835\cdot1.3426 + 811\cdot1.02 = 5976.1$ thousand m³.
Expansion per 1 m³ of oil in place: $E_o = 1.3426 - 1.3243 = 0.0183$ ; $E_{fw} = 1.3243\cdot1.2167\cdot10^{-3}\cdot(30 - 21.44) = 0.013792$ ; $E_t = 0.032092$ , and $NE_t = 9060\cdot0.032092 = 290.8$ thousand m³.
Influx from the balance: $W_e = 5976.1 - 290.8 = 5685.3$ thousand m³; from superposition, 5671.2 thousand m³, which is 0.25% less.
Year 1, $p$ = 28.47 MPa, $B_o$ = 1.3276: $F = 183\cdot1.3276 = 242.95$ thousand m³; $E_t = (1.3276 - 1.3243) + 1.3243\cdot1.2167\cdot10^{-3}\cdot1.53 = 0.005765$ ; $W_e = 242.95 - 9060\cdot0.005765 = 190.7$ thousand m³. Superposition gave 159.6 thousand m³, which is 16% less.
With quarterly steps the difference from the balance is −2.3% in year 1 and −0.02% by year 8 (the table below).

Answer: By year 8 the superposition of yearly steps and the balance agree to within 0.25%, and in year 1 only to within 16%. With quarterly steps they agree to within 2% already in year 1.

What it means. The aquifer was chosen correctly (the assumed history was calculated with it), and the difference in the first years is created only by the length of the step: yearly steps are late in responding to the fast initial pressure fall. A wrong aquifer would show itself differently: the difference would not disappear as the step is shortened, and it would grow with time. The calculation from the balance is not perfect either. For a small pressure drop the expansion is a small difference, and a pressure error of 0.1 MPa changes the influx of year 1 by 1.7% and that of year 8 by only 0.04%. The data in this example are assumed for illustration.

Cumulative influx by year, thousand m³

Year: 1
superposition, yearly step: 159.6
superposition, quarterly step: 186.3
balance $F - NE_t$: 190.7

Year: 2
superposition, yearly step: 625.2
superposition, quarterly step: 661.6
balance $F - NE_t$: 668.0

Year: 3
superposition, yearly step: 1333.1
superposition, quarterly step: 1367.7
balance $F - NE_t$: 1372.8

Year: 4
superposition, yearly step: 2161.0
superposition, quarterly step: 2185.0
balance $F - NE_t$: 2187.9

Year: 5
superposition, yearly step: 3034.2
superposition, quarterly step: 3054.6
balance $F - NE_t$: 3057.7

Year: 6
superposition, yearly step: 3919.3
superposition, quarterly step: 3933.5
balance $F - NE_t$: 3935.7

Year: 7
superposition, yearly step: 4795.3
superposition, quarterly step: 4808.8
balance $F - NE_t$: 4811.0

Year: 8
superposition, yearly step: 5671.2
superposition, quarterly step: 5684.1
balance $F - NE_t$: 5685.3

Year	superposition, yearly step	superposition, quarterly step	balance $F - NE_t$
1	159.6	186.3	190.7
2	625.2	661.6	668.0
3	1333.1	1367.7	1372.8
4	2161.0	2185.0	2187.9
5	3034.2	3054.6	3057.7
6	3919.3	3933.5	3935.7
7	4795.3	4808.8	4811.0
8	5671.2	5684.1	5685.3

The quarterly steps are built from the yearly measurements (the pressure between them along a straight line). The influx from the balance is from the history table, for an oil in place of 9.06 million m³.

CalculatedWater influx into this lesson's assumed accumulation: the exact solution for the aquifer (line), the superposition of yearly steps (dashed) and the influx from the material balance (dots); for comparison, the underground withdrawal

F

. The differences of the first years are tens of thousands of cubic metres, and they are almost invisible on the chart (see the table above).

The chart also shows the main property of such an accumulation: by year 8 the aquifer has supplied about 95% of the underground withdrawal, and the expansion of the oil, water and pores themselves only the remainder. So the pressure in it is set by the aquifer, and the forecast is built on this: the pressure at the end of the next period must be such that the influx from the aquifer and the influx that the balance needs at the planned withdrawal agree. There are two unknowns, the pressure and the influx, and we find them by trial.

Forecast of the pressure for the next period

1
Set the withdrawal
The planned cumulative production of oil, gas and water at the end of the period, from the development programme.
2
Choose a trial pressure
For example, by continuing the trend of the last measurements.
3
Calculate the influx in two ways
From the aquifer, by superposition with a new step that depends on the trial pressure; from the balance, $F - NE_t$ at the trial pressure and the planned withdrawal.
4
Refine the pressure
If the balance needs more water than the aquifer gives, the pressure is lower than the trial one, and the other way round. Two trials and a linear interpolation between them usually give the answer to within hundredths of a megapascal.
5
Move to the next period
The accepted pressure becomes a measurement from which the drops are built further. The steps of the forecast are three to six months.

Worked exampleForecast of the pressure for year 9

Given (assumed data)

Plan for year 9: cumulative production at its end: $N_p$ = 4273 thousand m³ of oil and $W_p$ = 1103 thousand m³ of water
Aquifer, history, properties: as in the previous examples; yearly steps

Solution

The eight known steps have acted a year longer by the end of year 9 than by the end of year 8, and their contribution is $U\sum\Delta p_jW_D = 6453.08$ thousand m³. The new step, $(21.89 - p)/2$ , is applied at the end of year 8 and acts for one year. Its contribution is $40{,}000\cdot5.2168\cdot(21.89 - p)/2$ m³, or $104.34\cdot(21.89 - p)$ thousand m³.
Trial $p$ = 21.2 MPa. From the aquifer: $W_e = 6453.08 + 104.34\cdot0.69 = 6525.07$ thousand m³. From the balance ( $B_o$ = 1.3431): $F = 4273\cdot1.3431 + 1103\cdot1.02 = 6864.13$ thousand m³, $E_t = 0.032979$ , $W_e = 6864.13 - 9060\cdot0.032979 = 6565.34$ thousand m³. The balance needs 40.27 thousand m³ more than the aquifer gives, so the pressure must be lower.
Trial $p$ = 20.8 MPa. From the aquifer, 6566.81 thousand m³; from the balance ( $B_o$ = 1.3440), 6555.19 thousand m³. Now the aquifer gives 11.62 thousand m³ more than is needed.
Linear interpolation: $p = 21.2 - 0.4\cdot40.27/(40.27 + 11.62) = 20.89$ MPa. A check at this pressure: the two influx estimates differ by only 0.01 thousand m³.
The same calculation with quarterly steps (the pressure inside year 9 along a straight line) gives 20.99 MPa.

Answer: For the planned withdrawal the pressure at the contact will fall by the end of year 9 to about 20.89 MPa, a further 0.55 MPa in the year.

What it means. The new step gives 104.34 thousand m³ of water for each megapascal, while the elasticity of the accumulation itself gives about 34 thousand m³ per megapascal. The pressure is indeed set by the aquifer. For the same reason the forecast is sensitive to an error in the influx: yearly steps understate the influx by about a quarter of a per cent, and the forecast with them is 0.1 MPa lower than with quarterly steps. So the steps of a forecast are taken shorter than a year, and the parameters of the aquifer are first checked against the history. The data in this example are assumed for illustration.

Practical notes

Pressure at the contact. The model takes the pressure at the contact with the aquifer, brought to one depth. Usually it is taken as equal to the average pressure of the accumulation. In a large accumulation, or one with low permeability, the pressure at the contact is higher than the average, and then we take the pressure of wells beyond or near the oil boundary.
The step. The drops are built from measurements, or from a smoothed pressure curve, with a step of three to six months. A yearly step noticeably understates the influx of the first years. For a small aquifer that responds fast, the step must be shorter than its response time. Otherwise the steps based on average pressures lag by half a step (Simplified water influx models: Fetkovich and Carter–Tracy).
Geometry. An accumulation of irregular shape is replaced by a circle of the same area, and an aquifer along part of the perimeter by the fraction $f$ . A strip-shaped aquifer has its own solution (see the advanced section). For bottom water that comes from below over the whole area there are solutions with vertical flow. They are also written through the dimensionless influx for a single pressure step, and the superposition of the steps is the same.
Ambiguity. The same influx history is often described equally well by different combinations of $U$ , $r_{aD}$ and the scale of the dimensionless time, especially while the boundary of the aquifer has not shown itself. The parameters are checked against the geology of the area beyond the oil boundary and refined with each new measurement (History matching of aquifer parameters).
Amount of calculation. At step $n$ the sum contains $n$ terms, and in a forecast it is recalculated for each trial pressure. In simulation models with hundreds of steps and many parts of the boundary this is expensive, so the step-by-step methods of Fetkovich or Carter–Tracy are used there (Simplified water influx models: Fetkovich and Carter–Tracy).

Terms and symbols

Aquifer constant $U$: It equals $2\pi f\phi c_th r_o^2$ , in m³/MPa. This is the influx per 1 MPa of pressure drop at $W_D = 1$ . It does not depend on permeability.
Dimensionless influx $W_D$: The cumulative influx after a unit step of the contact pressure, in units of $U$ . It is a function of $t_D$ and $r_{aD}$ .
Dimensionless time of the aquifer $t_D$: It equals $kt/(\phi\mu_wc_tr_o^2)$ : time on the scale $r_o^2/\eta$ , where $\eta$ is the hydraulic diffusivity of the aquifer.
$r_{aD}$: The ratio of the outer radius of the aquifer to the radius of the accumulation. It sets the limit $(r_{aD}^2 - 1)/2$ and the time to reach the boundary.
Pressure step: An instantaneous drop of the contact pressure by $\Delta p_j$ at time $t_j$ . A smooth history is represented as a sum of steps.
Half drops (drops from the average pressures of the intervals): The drop of a step is $\Delta p_j = (p_{j-1} - p_{j+1})/2$ . The stepped curve then follows the average pressures of the intervals between the measurements.

Common mistakes and misconceptions

Wrong

“For the same pressure drop, in four times the time the aquifer will give four times as much water.”

In fact

The influx grows more slowly than time: at first as $\sqrt{t}$ , later a little faster, but it still lags behind. In this lesson's example the influx in 2 years is 2.6 times that in half a year, not four times.

Wrong

“The pressure steps of past years have already done their work; for a new date it is enough to add the contribution of the new step.”

In fact

Each step keeps giving influx until the aquifer has used up its store. In this lesson's example, by year 8 more than half of the influx comes from the steps applied one and two years after the start. The sum is recalculated completely.

Wrong

“A difference from the balance in the first years proves that the aquifer was chosen wrongly.”

In fact

First we check the step: the rule of half drops with a yearly step understates the influx of the first year by 16%, and with a quarterly step by 2%. An error of the aquifer model shows itself as a difference that does not disappear as the step is shortened and that grows with time.

Wrong

“If the boundary of the aquifer is not visible in the history, we can ignore it in the forecast.”

In fact

After the pressure drop has reached the boundary, a closed aquifer gives less and less water than an infinite one. In this lesson's example, for a constant pressure drop the difference by year 8 is about 5%, and it keeps growing. A forecast with an infinite aquifer comes out too optimistic.

In depth: advanced detail

Derivation of the influx formula

By Darcy's law the flow of water through the contact, which is the side surface of a sector of a cylinder of radius $r_o$ , height $h$ and angle $2\pi f$ , is proportional to the pressure gradient at the contact. Let us put in $p = p_i - \Delta p\,P$ and $r = r_or_D$ :

Rate of influx and dimensionless rate

q_w = \frac{2\pi f k h\,r_o}{\mu_w}\left.\frac{\partial p}{\partial r}\right|_{r_o} = \frac{2\pi f k h\,\Delta p}{\mu_w}\,q_D(t_D),\qquad q_D = -\left.\frac{\partial P}{\partial r_D}\right|_{r_D = 1}

where:

$q_w$ rate of water influx at reservoir conditions, m³/s
$q_D$ dimensionless rate of influx, —
$k,\ h,\ \mu_w,\ f$ permeability and thickness of the aquifer, viscosity of the water, fraction of the perimeter, m², m, Pa·s, —

Units: SI

Holds when

a constant pressure drop $\Delta p$ at the contact from the moment $t = 0$

The rate of influx is proportional to the conductivity of the aquifer $kh/\mu_w$ and to the pressure drop, and its course in time is set by the dimensionless rate, which falls: at first as $1/\sqrt{\pi t_D}$ , and for an infinite aquifer later roughly as $2/\ln t_D$ .

The cumulative influx is the integral of the rate over time. We change time to dimensionless time, $\dd t = (\phi\mu_wc_tr_o^2/k)\dd t_D$ , and the permeability and the viscosity cancel:

From the rate of influx to the cumulative influx

W_e = \int_0^t q_w\dd t = \frac{2\pi fkh\,\Delta p}{\mu_w}\cdot\frac{\phi\mu_wc_tr_o^2}{k}\int_0^{t_D}q_D\dd t_D = 2\pi f\phi c_thr_o^2\,\Delta p\,W_D

where:

$W_D = \int_0^{t_D}q_D\dd t_D$ dimensionless influx, —
$\phi,\ c_t,\ r_o$ porosity and total compressibility of the aquifer, radius of the accumulation, —, 1/Pa, m

Units: SI

Holds when

the same as for the problem in dimensionless form

The conductivity decides how fast the influx goes, but not how much water the aquifer gives by a given dimensionless time: it is “hidden” in the time scale.

The balance of the aquifer gives the same result. Multiply the equation of the problem by $r_D$ and integrate from 1 to $r_{aD}$ . The left side turns into the flow through the boundaries: there is none at the outer boundary, and at the contact it equals $q_D$ . The right side turns into the rate of change of the integral $\int P\,r_D\dd r_D$ . So $\int_0^{t_D}q_D\dd t_D$ and the integral of the pressure-drop funnel coincide: the cumulative influx equals the expansion of the aquifer (the formula “Influx as the expansion of the pressure-drop funnel” in the main text).

Solution in the Laplace domain

The Laplace transform over time (a mathematical tool that turns the time derivative into a simple factor) turns the equation of the problem into an ordinary Bessel equation in $r_D$ , and the condition $P(1) = 1$ into $\bar P(1) = 1/s$ . Its solution is expressed through the modified Bessel functions $I_0, I_1, K_0, K_1$ (special functions that appear in problems with radial symmetry), and the transform of the dimensionless influx is obtained in explicit form:

Laplace transform of the dimensionless influx

\bar W_D(s) = \frac{K_1(\sqrt s)\,I_1(r_{aD}\sqrt s) - I_1(\sqrt s)\,K_1(r_{aD}\sqrt s)}{s^{3/2}\big[K_0(\sqrt s)\,I_1(r_{aD}\sqrt s) + I_0(\sqrt s)\,K_1(r_{aD}\sqrt s)\big]}\;\xrightarrow{\;r_{aD}\to\infty\;}\;\frac{K_1(\sqrt s)}{s^{3/2}\,K_0(\sqrt s)}

where:

$\bar W_D(s)$ Laplace transform of $W_D(t_D)$ , —
$s$ variable of the transform (conjugate to $t_D$ ), —
$I_\nu,\ K_\nu$ modified Bessel functions of the first and second kind of order $\nu$ , —

Units: dimensionless

Holds when

a closed outer boundary at $r_{aD}$ ; on the right is the infinite aquifer

The transform is inverted numerically (for example, with the Stehfest algorithm, a standard numerical method for this, as in the lesson on wellbore storage), or the solution is expanded in a series of eigenfunctions (the natural pressure patterns of the aquifer). In this lesson the values of $W_D$ were found the first way, by inverting this transform with the Stehfest algorithm, and checked against the series expansion. The transform of the constant-rate solution $\bar p_D(s)$ is linked to it by a simple relation, $\bar W_D = 1/\big(s^3\bar p_D\big)$ : the two problems are two views of one solution (derivation of the Carter–Tracy formula).

Linear aquifer

If the aquifer touches one side of the accumulation as a strip of width $w$ and length $L$ (for example, along a long fault or in a narrow graben, a block of rock that has dropped between two faults), the flow in it is linear. The method is the same, and only the scales change: the length is taken along the strip, not as the radius of the accumulation.

Influx from a linear aquifer

W_e = U_L\,\Delta p\,W_D(t_D),\qquad U_L = w\,h\,L\,\phi\,c_t,\qquad t_D = \frac{k\,t}{\phi\,\mu_wc_t\,L^2};\qquad W_D = 2\sqrt{t_D/\pi}\ \ \text{(infinite strip)}

where:

$U_L$ constant of the linear aquifer: its elastic store per 1 MPa, m³/MPa
$w,\ h,\ L$ width of the strip along the contact, thickness, and length of the aquifer from the contact to the far boundary, m
$t_D$ dimensionless time, referred to the length of the strip, —

Units: m³, MPa, 1/MPa, m; $t_D$ in SI (the metric form has the multiplier $3.553\cdot10^{-3}$ for time in hours)

Holds when

a uniform strip of constant cross-section; the flow is perpendicular to the contact
for an infinite strip $L$ cancels: $W_e = 2wh\phi c_t\Delta p\sqrt{\eta t/\pi}$

A linear aquifer has no curvature, so the early-period formula of the radial aquifer is exact for it for as long as the pressure drop has not reached the far boundary. A closed strip levels off at the limit $W_D = 1$ , that is, at the elastic store $U_L\Delta p$ . For the same contact area and the same properties, a radial aquifer in time gives more water than a linear one. The reason is that the volume of its rings grows with the distance from the contact, while the cross-section of a strip is constant.

In other unit systems (1)

Oilfield units

U_L\,[\text{bbl/psi}] = 0.1781\,w\,h\,L\,\phi\,c_t

$w$ , $h$ , $L$ in ft, $c_t$ in 1/psi; $t_D$ with the multiplier $2.637\cdot10^{-4}$ for time in hours.

$0.1781 = 1/5.615$ converts cubic feet to barrels.

Connections to other topics

Aquifer models: water influx — the classes of influx models: the unsteady-state model is the general case, and the small-aquifer and steady-state models are its limits.
Superposition principle — adding the responses to steps, just as for a change of well rate.
Diffusivity equation — the equation solved for the aquifer, and the dimensionless variables.
Line-source solution and the pressure function — a well with a constant bottomhole pressure is the same problem as water influx into an accumulation.
Simplified water influx models: Fetkovich and Carter–Tracy — step-by-step methods without superposition, and their errors on the same history.
Straight-line material balance (Havlena–Odeh method) — the straight line of material balance with the influx function $\sum\Delta p_jW_D$ .
General material balance equation for oil reservoirs — water influx in the general balance equation.
History matching of aquifer parameters — fitting the constant, the size and the time scale of the aquifer to the history.
Material balance for gas reservoirs — water influx into a gas accumulation is calculated in the same way.

Check yourself

1
The pressure at the contact has fallen by a constant amount. In the early period ( $t_D \ll 1$ ) four times as much time has passed. How has the cumulative influx changed?
1. A.It has grown 16 times
2. B.It has grown four times
3. C.It has grown about twice
4. D.It has not changed: the aquifer has already given up its store
Show the answer
Correct answer: C. It has grown about twice
In the early period $W_D \approx 2\sqrt{t_D/\pi}$ , so the influx is proportional to the square root of time: $\sqrt4 = 2$ . Later the growth is a little faster than the square root (the curvature of the contact matters), but still slower than time.
2
An aquifer touches half of the perimeter of the accumulation ( $f = 0.5$ ); porosity 0.25, $c_t = 1.0\cdot10^{-3}$ MPa⁻¹, thickness 8 m, radius of the accumulation 2000 m. Find the aquifer constant and the influx if the pressure at the contact has fallen by 2 MPa and stays there, and $W_D$ is 6.0 by that time.
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$U = 2\pi\cdot0.5\cdot0.25\cdot1.0\cdot10^{-3}\cdot8\cdot2000^2 = 2\pi\cdot4000 \approx 25{,}130$ m³/MPa. The influx is $W_e = U\Delta pW_D = 25{,}130\cdot2\cdot6.0 \approx 302$ thousand m³ at reservoir conditions.
3
The pressure at the contact from measurements at equal intervals is 25.0, 24.4, 23.6 and 23.0 MPa. What drop of a step is applied at the moment of the second measurement ( $t_2$ , pressure 23.6 MPa) by the rule of the average pressures of the intervals?
1. A.0.70 MPa
2. B.0.30 MPa
3. C.0.60 MPa
4. D.1.40 MPa
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Correct answer: A. 0.70 MPa
$\Delta p_2 = (p_1 - p_3)/2 = (24.4 - 23.0)/2 = 0.70$ MPa. This is the difference between the average pressures of the neighbouring intervals: $(24.4 + 23.6)/2 = 24.0$ and $(23.6 + 23.0)/2 = 23.3$ MPa.
4
Why, when calculating the influx for a new date, can we not simply add the contribution of the new step to the result of the previous date?
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The old steps keep acting, and the time since each of them has grown by the length of the step, and so has its $W_D$ . The influx on the new date is the sum over all the steps with the new values $W_D(t_{Dn} - t_{Dj})$ . So the calculation reworks the whole history, and the amount of work grows as the square of the number of steps. The step-by-step methods avoid this by keeping only one or two numbers about the past.
5
With yearly steps the influx in the first two years is 6–16% smaller than from the material balance, and from year 5 on they agree to within 1%. What is the most likely cause?
1. A.The aquifer is connected to a recharge area
2. B.The aquifer is smaller than assumed in the model
3. C.The error of the rule of half drops with a long step: the calculation should be repeated with a shorter step
4. D.The oil in place is overstated
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Correct answer: C. The error of the rule of half drops with a long step: the calculation should be repeated with a shorter step
The rule of half drops understates the influx, most of all in the first steps, and the error falls quickly with the number of steps. This is exactly the picture in this lesson's example. A wrong size of the aquifer would show itself differently: the difference would grow with time, once the pressure drop reached the boundary, and would not disappear as the step is shortened.
6
A closed aquifer with $r_{aD} = 3$ . What is the limit of $W_D$ , and what does reaching it mean?
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$(r_{aD}^2 - 1)/2 = (9 - 1)/2 = 4$ . Then $U\Delta p\cdot4$ equals the elastic store of the aquifer $(c_w + c_f)W\Delta p$ : the pressure throughout the aquifer has become equal to the pressure at the contact. After that the aquifer behaves like a small “tank”: new water comes in only if the pressure at the contact falls again.

Updated: 7 October 2026