Aquifer models: water influx

Quick refresher

The accumulation and the aquifer form one hydrodynamic system (fluid flows between them). When the pressure at the contact falls, the water and the pores of the aquifer expand elastically, and water enters the accumulation. Recharge of the aquifer from the surface is rare.
The largest possible influx for a pressure drop of $\Delta p$ throughout the aquifer is its elastic store $(c_w + c_f)W\Delta p$ , where $W$ is the pore volume of the aquifer. Compressibility is small, so only aquifers many times larger than the accumulation give a noticeable influx.
The pressure drop spreads from the contact according to the diffusivity equation. The larger the aquifer and the smaller its hydraulic diffusivity $k/(\phi\mu_wc_t)$ (the speed at which a pressure change travels), the longer it takes to give up its store. A small aquifer equalises in weeks or months, a large one in years or decades.
The small-aquifer model: the pressure throughout the aquifer equals the pressure at the contact, $W_e = (c_w + c_f)W(p_i - p)$ , and there is no time in it. Above the bubble point pressure it acts as an extra compressibility of the accumulation.
The steady-state model: the rate of influx is proportional to the pressure drop at the contact, $J_a(p_i - p)$ . The pressure at the far boundary is constant, and the influx keeps accumulating as long as the pressure at the contact is lowered. The coefficient $J_a$ is found from Darcy's law for the aquifer or from the development history.
The pseudo-steady-state model treats a finite aquifer as a tank behind a resistance. The unsteady-state model solves the diffusivity equation for the aquifer. The two simple models are limiting cases of them: with a very small resistance we get the tank, and with a very large volume we get the steady-state model.

Key relationships

W_e = (c_w + c_f)\,W\,(p_i - p)

small-aquifer model; for a pressure drop throughout the aquifer, this is its elastic store

W = \pi\,(r_a^2 - r_o^2)\,h\,\phi\,f

pore volume of a radial aquifer; $f$ is the fraction of the perimeter of the accumulation that it touches

\frac{\dd W_e}{\dd t} = J_a\,(p_i - p),\qquad W_e = J_a\int_0^t (p_i - p)\dd t

steady-state model

J_a = \frac{0.536\,f\,k\,h}{\mu_w\ln(r_a/r_o)}

productivity index, m³/(d·MPa): $k$ in mD, $h$ in m, $\mu_w$ in mPa·s; $r_a$ is the radius of the constant-pressure boundary

t_D = \frac{k\,t}{\phi\,\mu_w\,c_t\,r_o^2}

dimensionless time of the aquifer (SI); $c_t = c_w + c_f$

Common pitfalls

Assuming that the aquifer gives up its elastic store at once. A large aquifer gives it up over decades, and the small-aquifer model overstates the early influx many times.
Taking the compressibility of the water alone, without that of the pores. The elastic store is then understated almost threefold.
Confusing the pore volume of the aquifer with the volume of the rock, or forgetting the fraction $f$ of the perimeter that the aquifer touches.
Extending into the future a steady-state coefficient fitted to the first years. For a closed aquifer the influx weakens with time.
Putting the pressure in the wells into the model in place of the pressure at the contact with the aquifer.

After this lesson you can

explain where the energy of an aquifer comes from and why the water influx lags behind the pressure drop at the contact
calculate the elastic store of an aquifer (the most water it can give up by expanding) and the influx from the small-aquifer model, and estimate how it changes the performance of the accumulation
calculate the productivity index of an aquifer (how much water enters per day for each unit of pressure drop) from its properties, and the cumulative influx from the steady-state model for a given pressure history
describe the ideas of the pseudo-steady-state and unsteady-state models, and choose the class of model from the size and the hydraulic diffusivity of the aquifer (how fast a pressure change travels through it)
explain how the parameters of a model are refined from the development history, and why the fitted model is seldom the only possible one

Where the water influx comes from

Oil and gas came into the trap (the structure that holds the accumulation) when the rock was already full of water, and they pushed the water out only from the upper part of the formation (Traps and timing). So an accumulation is almost always surrounded by the same water: edge water, beyond the oil boundary, or bottom water, under the accumulation. Together they form one hydrodynamic system (water drive).

When the accumulation is developed, the pressure at its contact with the aquifer falls. The water in the aquifer is compressed to the reservoir pressure. When the pressure falls, it expands a little, and the pores of the aquifer shrink a little under the weight of the rocks above (Rock compressibility and the effect of stress on core properties). Both effects push water into the accumulation. Sometimes there is also recharge from outside: the formation may reach the surface at a higher level than the accumulation, and water may enter there at the same rate at which it is withdrawn. But usually the aquifer is cut off by faults, or the formation thins out to nothing. Then the influx is the elastic expansion of the aquifer itself.

SchematicThe aquifer and the accumulation (a diagram, not to scale). Left, in plan view: the aquifer touches the part of the perimeter of the accumulation that spans an angle

\theta

, that is, the fraction

f = \theta/360°

; the rest of the perimeter is closed by a fault.

r_o

is the radius of the accumulation (the initial contact), and

r_a

is the outer radius of the aquifer. Right, in cross-section: edge water comes along the formation from the flanks, and bottom water underlies the accumulation.

What the water influx depends on

What: Pore volume of the aquifer $W$
How it matters: sets the elastic store: how much water the aquifer can give for each megapascal of pressure drop

What: Compressibility of water and pores $c_w + c_f$
How it matters: the same; for water and cemented rock together it is of the order of $1\cdot10^{-3}$ MPa⁻¹

What: Conductivity $kh/\mu_w$
How it matters: speed: how fast the water reaches the accumulation and how far the pressure drop spreads in the same time

What: Fraction of the perimeter $f$
How it matters: the part of the outline of the accumulation through which water enters

What: Outer boundary
How it matters: a closed aquifer gives only its elastic store, while an aquifer with recharge can give water for as long as we like

What	How it matters
Pore volume of the aquifer $W$	sets the elastic store: how much water the aquifer can give for each megapascal of pressure drop
Compressibility of water and pores $c_w + c_f$	the same; for water and cemented rock together it is of the order of $1\cdot10^{-3}$ MPa⁻¹
Conductivity $kh/\mu_w$	speed: how fast the water reaches the accumulation and how far the pressure drop spreads in the same time
Fraction of the perimeter $f$	the part of the outline of the accumulation through which water enters
Outer boundary	a closed aquifer gives only its elastic store, while an aquifer with recharge can give water for as long as we like

Elastic store of an aquifer

W_{e,\max} = (c_w + c_f)\,W\,\Delta p,\qquad W = \pi\,(r_a^2 - r_o^2)\,h\,\phi\,f

where:

$W_{e,\max}$ largest influx for a pressure drop of $\Delta p$ throughout the aquifer, m³
$W$ pore volume of the aquifer, which is also the initial volume of water in it (often written $W_i$ in the literature), m³
$c_w,\ c_f$ compressibility of the water and of the pore volume of the aquifer (the fraction by which a volume changes for a unit change in pressure), 1/MPa
$\Delta p$ fall in pressure from the initial value, $p_i - p$ , MPa
$r_o,\ r_a$ radius of the accumulation (the initial contact with the water) and the outer radius of the aquifer, m
$h,\ \phi$ thickness and porosity of the aquifer, m, —
$f$ fraction of the perimeter of the accumulation that the aquifer touches, $\theta/360°$ , —

Units: consistent units: m³, MPa and 1/MPa

Holds when

the aquifer is full of water, and the compressibilities are constant
the geometry is radial, a sector of a ring with the angle $\theta$ ; for a strip-shaped aquifer of width $w$ and length $L$ the volume is $wLh\phi$
the pressure drop has had time to reach all parts of the aquifer

This is an upper limit: the aquifer gives this much water only when the pressure throughout it has fallen to the pressure at the contact. Each megapascal of pressure drop releases about one thousandth of the pore volume of the aquifer. So only aquifers many times larger than the accumulation give noticeable support to the pressure (sources of reservoir energy).

In other unit systems (1)

Oilfield units

W_{e,\max}\,[\text{bbl}] = 0.1781\,(c_w + c_f)\,\pi\,(r_a^2 - r_o^2)\,h\,\phi\,f\,\Delta p

$r$ and $h$ in feet, $c$ in 1/psi, $\Delta p$ in psi.

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

The small-aquifer model

If the aquifer is small and the rock lets water through easily, the pressure drop crosses it quickly, in a time that is short compared with the intervals between the pressure measurements in the accumulation. Then we can take the pressure throughout the aquifer to be equal to the pressure at the contact at every moment. This is the small-aquifer model, also called the pot aquifer or “tank” model.

Influx from a small aquifer

W_e = (c_w + c_f)\,W\,(p_i - p)

where:

$W_e$ cumulative water influx at reservoir conditions, m³
$W$ pore volume of the aquifer, m³
$c_w + c_f$ compressibility of the water and the pores of the aquifer, 1/MPa
$p_i,\ p$ initial and current pressure at the contact, MPa

Units: m³, MPa, 1/MPa (or barrels, psi and 1/psi)

Holds when

the pressure throughout the aquifer equals the pressure at the contact: the aquifer equalises faster than the pressure in the accumulation changes
there is no recharge from outside
the pressure at the contact is taken as equal to the average pressure of the accumulation

The influx depends only on the pressure drop: the aquifer responds at once, and there is no time in the model. In the material balance it looks like one more elastic term, $F = N(E_o + E_{fw}) + (c_w + c_f)W\Delta p$ . That is, it simply increases the compressibility of the accumulation (General material balance equation for oil reservoirs).

Worked exampleA small aquifer and the production down to the bubble point pressure

Given (assumed data)

Oil zone: the accumulation of the lessons Reservoir drive mechanisms and General material balance equation for oil reservoirs: pore volume 16.0 million m³ with a thickness of 10 m and a porosity of 0.20, which gives a radius $r_o = 1596$ m; oil in place 9.06 million m³; initial pressure 30 MPa, bubble point pressure 18 MPa
Properties: $B_{oi} = 1.3243$ and $B_{ob} = 1.35$ m³/m³; $(c_wS_w + c_f)/(1 - S_w) = 1.2167\cdot10^{-3}$ MPa⁻¹
Aquifer: a ring of the same thickness and porosity around the whole accumulation ( $f = 1$ ), outer radius 3.2 km; $c_w + c_f = (0.45 + 0.8)\cdot10^{-3} = 1.25\cdot10^{-3}$ MPa⁻¹

Solution

Pore volume of the aquifer: $W = \pi\cdot(3200^2 - 1596^2)\cdot10\cdot0.20 = 48.3$ million m³, which is 3.0 times the pore volume of the accumulation.
Elastic store per megapascal: $(c_w + c_f)\,W = 1.25\cdot10^{-3}\cdot48.3\cdot10^6 = 60.4$ thousand m³/MPa. For comparison, the accumulation itself frees an average of 34.0 thousand m³/MPa down to the bubble point pressure (the next step).
Pressure fall to the bubble point pressure, $\Delta p = 12$ MPa. Expansion of the accumulation: $E_o = 1.35 - 1.3243 = 0.0257$ ; $E_{fw} = 1.3243\cdot1.2167\cdot10^{-3}\cdot12 = 0.01933$ ; $N(E_o + E_{fw}) = 9.06\cdot10^6\cdot0.04503 = 408.0$ thousand m³. Influx from the aquifer: $W_e = 60.4\cdot12 = 725.1$ thousand m³.
Production down to the bubble point pressure ( $N_pB_{ob}$ equals the sum of the expansion and the influx): without the aquifer $N_p = 408.0/1.35 = 302$ thousand m³ (3.3% of the oil in place); with the aquifer $N_p = (408.0 + 725.1)/1.35 = 839$ thousand m³ (9.3%).

Answer: An aquifer only three times larger than the accumulation increases the production down to the bubble point pressure by a factor of 2.8, from 302 to 839 thousand m³.

What it means. The aquifer gave almost twice as much energy as the accumulation itself. Per unit of pore volume it is less elastic, because it has no oil, which compresses more than water. But its pore volume is three times larger. The result is right if the aquifer equalises faster than the pressure in the accumulation falls. For this aquifer the pressure drop reaches the outer boundary in about two weeks (see the example in the section on the pseudo-steady-state and unsteady-state models). So the assumption holds for a production period of several years. For an aquifer tens of times larger this is no longer so. The data in this example are assumed for illustration.

Why the influx lags behind the pressure

The small-aquifer model assumes that the pressure drop covers the whole aquifer at once. In fact it spreads gradually from the contact, in the same way as around a well that has been put on production (Flow regimes). The flow of water in the aquifer is described by the diffusivity equation, in which the accumulation plays the role of the well. The inner boundary is the contact at the radius $r_o$ , and the outer boundary is the edge of the aquifer at the radius $r_a$ (Diffusivity equation).

Until the pressure drop has reached the outer boundary, the aquifer behaves as an infinite one: only the parts near the accumulation give water, and the influx grows more and more slowly with time. When the drop has reached the boundary, what follows depends on the boundary. A closed aquifer starts to be depleted as a whole: the pressure in it falls everywhere at almost the same rate. This is the pseudo-steady-state regime (Pseudo-steady-state flow and the shape factor). An aquifer with recharge at its boundary reaches the steady state: for a constant pressure drop at the contact, the influx proceeds at a constant rate.

Dimensionless time of the aquifer and the time to reach its boundary

t_D = \frac{\eta\,t}{r_o^2} = \frac{k\,t}{\phi\,\mu_w\,c_t\,r_o^2},\qquad t_a \approx \frac{(r_a - r_o)^2}{4\eta}

where:

$t_D$ dimensionless time of the aquifer, —
$t$ time since the pressure at the contact started to fall, s
$\eta$ hydraulic diffusivity of the aquifer, $k/(\phi\mu_wc_t)$ , m²/s
$k,\ \phi$ permeability and porosity of the aquifer, m², —
$\mu_w$ viscosity of the reservoir water, Pa·s
$c_t$ total compressibility of the aquifer, $c_w + c_f$ , 1/Pa
$r_o,\ r_a$ radius of the accumulation and outer radius of the aquifer, m
$t_a$ time in which the radius of investigation $\sqrt{4\eta t}$ (how far the pressure drop has spread) goes from the contact to the outer boundary, s

Units: SI

Holds when

a uniform aquifer of constant thickness, with radial geometry
$t_a$ is an order-of-magnitude estimate: a noticeable pressure drop at the boundary starts earlier, and the equalisation of the whole aquifer comes several times later

Dimensionless time is the “clock” of the aquifer. For the same ratio $r_a/r_o$ , the same $t_D$ means the same fraction of the store given up. The time to reach the boundary grows as the square of the distance from the contact to the boundary: twice as far takes four times as long.

In other unit systems (2)

Metric oilfield units

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

$k$ in mD, $t$ in hours, $\mu_w$ in mPa·s, $c_t$ in 1/MPa, $r_o$ in m. With time in days the multiplier is 24 times larger, 0.0853.

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

t_D = 2.637\cdot10^{-4}\,\frac{k\,t}{\phi\,\mu_w\,c_t\,r_o^2}

$k$ in mD, $t$ in hours, $\mu_w$ in cP, $c_t$ in 1/psi, $r_o$ in ft.

CalculatedThe calculated pressure drop in the large aquifer (outer radius 16 km, closed), as a fraction of the drop at the contact, after 0.1, 1, 3, 10 and 30 years from the moment when the pressure at the contact fell and did not change afterwards. The aquifer and the accumulation are those of this lesson's examples; the calculation is by the diffusivity equation.

The chart shows why the influx lags behind the pressure. After a little over a month, the pressure drop is noticeable only in a ring about three kilometres wide next to the contact, and only this part of the aquifer gives water. After a year it reaches 10 km, but far from the contact it is only a few per cent of the drop at the contact. After three years it reaches the boundary. The aquifer will give up its whole elastic store only when the pressure everywhere in it has become equal to the pressure at the contact, and for a large aquifer this takes decades.

The steady-state model

Suppose the pressure at the outer boundary of the aquifer is held constant. This happens when the aquifer is connected to a recharge area (where water enters it from outside) or to a huge highly permeable body of rock. After a short transient period (while the flow is still settling) the flow of water in the aquifer becomes steady. Then, by Darcy's law, the rate of influx is proportional to the difference between the pressures at the boundary and at the contact, that is, to the pressure drop at the contact. This is the steady-state aquifer model, proposed by R. Schilthuis in 1936.

Steady-state model of water influx

\frac{\dd W_e}{\dd t} = J_a\,(p_i - p),\qquad W_e = J_a\int_0^t (p_i - p)\dd t

where:

$\dd W_e/\dd t$ rate of water influx at reservoir conditions, m³/d
$J_a$ aquifer productivity index (the influx constant), m³/(d·MPa)
$p_i,\ p$ initial and current pressure at the contact, MPa
$t$ time, d
$W_e$ cumulative influx, m³

Units: m³ at reservoir conditions, days, MPa; in oilfield units, barrels per day per psi

Holds when

the pressure at the outer boundary of the aquifer is constant and equal to the initial pressure
the flow in the aquifer is steady: the transient period is short compared with the development history
the pressure at the contact is taken as equal to the average pressure of the accumulation

The influx accumulates all the time while the pressure at the contact is lowered: for a constant pressure drop it grows linearly and without limit. The pressure in the accumulation stabilises when the rate of influx becomes equal to the underground withdrawal per day, which is the volume of the fluids produced in a day, measured at reservoir conditions (the limiting withdrawal). The integral over the pressure measurements is calculated by the trapezoid rule, as in the lesson Straight-line material balance (Havlena–Odeh method).

Productivity index of a radial aquifer (steady flow)

J_a = \frac{2\pi\,f\,k\,h}{\mu_w\,\ln(r_a/r_o)}

where:

$J_a$ productivity index of the aquifer, m³/(s·Pa)
$f$ fraction of the perimeter of the accumulation that the aquifer touches, —
$k,\ h$ permeability and thickness of the aquifer, m², m
$\mu_w$ viscosity of the reservoir water, Pa·s
$r_a$ radius at which the pressure is constant (the recharge boundary), m
$r_o$ radius of the accumulation, m

Units: SI

Holds when

a radial aquifer of constant thickness and permeability
the pressure at the radius $r_a$ is constant; the flow is steady and single-phase

This is Dupuit's formula of radial flow (radial flow to a well), in which the accumulation plays the role of the well. The coefficient depends only weakly on the distance to the recharge boundary, through a logarithm. What matters most is the conductivity of the aquifer $kh/\mu_w$ and the fraction of the perimeter $f$ .

In other unit systems (2)

Metric oilfield units

J_a\,[\text{m}^3/(\text{d}\cdot\text{MPa})] = \frac{0.536\,f\,k\,h}{\mu_w\ln(r_a/r_o)}

$k$ in mD, $h$ in m, $\mu_w$ in mPa·s.

$0.536 = 2\pi\cdot0.0853$ , the same constant as in the radial flow formula of the lesson Rock permeability.

Oilfield units

J_a\,[\text{bbl}/(\text{d}\cdot\text{psi})] = \frac{7.08\cdot10^{-3}\,f\,k\,h}{\mu_w\ln(r_a/r_o)}

$k$ in mD, $h$ in ft, $\mu_w$ in cP; $J_a$ in reservoir barrels per day per psi.

$7.08\cdot10^{-3} = 2\pi\cdot1.127\cdot10^{-3}$ .

Worked exampleProductivity index of the aquifer and the influx over four years

Given (assumed data)

Accumulation: the oil zone of the lesson General material balance equation for oil reservoirs, radius 1596 m; the pressure at the contact is 27.0, 25.0, 23.8 and 23.0 MPa at the end of years 1 to 4, with an initial pressure of 30 MPa
Aquifer: touches half of the perimeter ( $f = 0.5$ ); thickness 10 m, permeability 50 mD, water viscosity 0.4 mPa·s; at a distance of 4 km from the centre of the accumulation it is connected to a highly permeable body of rock whose pressure does not change

Solution

Productivity index: $\ln(4000/1596) = 0.9189$ ; $J_a = 0.536\cdot0.5\cdot50\cdot10/(0.4\cdot0.9189) = 364.6$ m³/(d·MPa).
The integral of the pressure drop, by the trapezoid rule between the yearly measurements, with a year of 365.25 days: by the end of years 1 to 4 it is 547.9, 2008.9, 4054.3 and 6464.9 MPa·d.
The cumulative influx $W_e = J_a\int(p_i - p)\dd t$ is 200, 732, 1478 and 2357 thousand m³. The rate of influx at the end of year 4 is $J_a\cdot7 = 2552$ m³/d.
Comparison with the balance: for an oil in place of 9.06 million m³, on the same dates the accumulation “needs” 191, 702, 1418 and 2261 thousand m³ of water (General material balance equation for oil reservoirs). The model gives 4.2–4.6% more.
For the same history, the straight line of the Havlena–Odeh method gave a coefficient of about 350 m³/(d·MPa) (Straight-line material balance (Havlena–Odeh method)). With the same geometry this corresponds to a permeability of $50\cdot350/364.6 = 48$ mD.

Answer: The productivity index of the aquifer is about 365 m³/(d·MPa). In four years the aquifer would give 2357 thousand m³ of water, which is 4% more than the balance requires.

What it means. A difference of a few per cent is smaller than the uncertainty in the properties of the aquifer. Its permeability, with almost no wells, is known at best to within tens of per cent. So a coefficient calculated from the properties of the aquifer is a plausibility check, and the working value is fitted to the history. Note that more water came in during year 4 than in the first two years together, although the pressure fell by only 0.8 MPa in that year. The influx is set not by the rate at which the pressure falls, but by how far and for how long the pressure has been lowered. The data in this example are assumed for illustration.

If an aquifer has no recharge but is just very large, the pressure far from the contact gradually falls, and the “distance to the region of constant pressure” grows with time. W. Hurst proposed in 1943 to take this into account by replacing the constant logarithm in the coefficient with a logarithm that grows with time. Then $J_a$ falls as $1/\ln(at)$ , where the constant $a$ is fitted to the history. This is an intermediate step between the steady-state and unsteady-state models, and today it is rarely used.

Pseudo-steady-state and unsteady-state models

Most aquifers have no recharge from outside, but they are not small either. They need models that take into account both the finite elastic store and the time it takes to give it up.

The pseudo-steady-state aquifer model treats the aquifer as a tank behind a resistance. The pressure in the “tank” is the average pressure of the aquifer, $\bar p_a$ . Water enters the accumulation at a rate proportional to the difference $\bar p_a - p$ , and the aquifer itself is depleted according to its own balance.

Pseudo-steady-state aquifer model

\frac{\dd W_e}{\dd t} = J\,(\bar p_a - p),\qquad W_e = (c_w + c_f)\,W\,(p_i - \bar p_a)

where:

$\bar p_a$ average pressure in the aquifer, MPa
$J$ productivity index of the aquifer for pseudo-steady-state flow, m³/(d·MPa)
$W,\ c_w + c_f$ pore volume and compressibility of the aquifer, m³, 1/MPa
$p_i,\ p$ initial pressure and pressure at the contact, MPa
$W_e$ cumulative influx, m³

Units: m³ at reservoir conditions, days, MPa

Holds when

the aquifer is closed; after a transient period the pressure in it falls everywhere at the same rate
the pressure at the contact is taken as equal to the average pressure of the accumulation

For a constant pressure drop at the contact $\Delta p$ , the two equations give $W_e = (c_w + c_f)W\Delta p\,\big(1 - \e^{-t/\tau}\big)$ with a time constant $\tau = (c_w + c_f)W/J$ : the aquifer gives up its elastic store exponentially. With a very small resistance ( $J \to \infty$ ) the model turns into the small aquifer, and with a very large volume ( $W \to \infty$ ) it turns into the steady-state model. The step-by-step calculation for a changing pressure is covered in the lesson Simplified water influx models: Fetkovich and Carter–Tracy.

The unsteady-state aquifer model does not simplify the flow in the aquifer. It solves the diffusivity equation for it, the same equation that was used to draw the chart of pressure profiles. The solution for a single step in the pressure at the contact is written in a dimensionless form that is common to all aquifers.

Influx for a constant pressure drop at the contact (unsteady-state model)

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

where:

$U$ 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 the dimensionless time $t_D$ and of the ratio $r_{aD} = r_a/r_o$ , —
$f,\ \phi,\ c_t,\ h,\ r_o$ fraction of the perimeter, porosity, compressibility $c_w + c_f$ and thickness of the aquifer, radius of the accumulation, —, —, 1/MPa, m, m

Units: m³, MPa, 1/MPa, m

Holds when

a uniform radial aquifer; the pressure at the contact fell by $\Delta p$ in one step and does not change afterwards
for a changing pressure the history is split into steps and the responses are added (Superposition principle)

At first $W_D \approx 2\sqrt{t_D/\pi}$ : the influx grows as the square root of time, and twice the time gives not twice the influx. For a closed aquifer $W_D$ tends to $(r_{aD}^2 - 1)/2$ , and then $U\Delta p\,W_D = (c_w + c_f)W\Delta p$ , the elastic store of the aquifer. The values of $W_D$ are found by solving the diffusivity equation. How to use them for a real pressure history is covered in the lesson Unsteady-state water influx: the van Everdingen–Hurst method.

In other unit systems (1)

Oilfield units

U\,[\text{bbl/psi}] = 1.119\,f\,\phi\,c_t\,h\,r_o^2

$c_t$ in 1/psi, $h$ and $r_o$ in ft.

$1.119 = 2\pi/5.615$ converts cubic feet to barrels.

CalculatedThe fraction of its elastic store that the aquifer gives up after the pressure at the contact has fallen and then stayed the same (time on a logarithmic scale). The solid lines are the solution of the diffusivity equation for the small (3.2 km) and the large (16 km) closed aquifers of the lesson. The dashed lines are the simple models. One is the tank (the whole store at once, for any aquifer). For the large aquifer there are two more: the pseudo-steady-state model (its resistance comes from the formula for pseudo-steady-state inflow, in which the accumulation plays the role of the well) and the steady-state model, as if its boundary were recharged.

Worked exampleHow fast an aquifer gives up its elastic store

Given (assumed data)

Aquifers: a small one (outer radius 3.2 km) and a large one (16 km), both closed, around the whole accumulation of radius 1596 m; thickness 10 m, porosity 0.20
Properties: permeability 50 mD, water viscosity 0.4 mPa·s, $c_t = c_w + c_f = 1.25\cdot10^{-3}$ MPa⁻¹
Pressure drop: the pressure at the contact has fallen by 1 MPa and does not change afterwards

Solution

Hydraulic diffusivity of the aquifer: $\eta = 3.553\cdot10^{-3}\cdot50/(0.20\cdot0.4\cdot1.25\cdot10^{-3}) = 1776.5$ m²/h, or 15.6 km² per year. Dimensionless time in a year: $t_D = 1776.5\cdot8766/1596^2 = 6.1$ .
The radius of investigation reaches the outer boundary of the small aquifer in $(3200 - 1596)^2/(4\cdot1776.5) = 362$ h, that is, in about 15 days. For the large aquifer it takes 29,198 h, about 3.3 years.
Elastic store per 1 MPa: 60.4 thousand m³ for the small aquifer and 1991 thousand m³ for the large one.
By the solution of the diffusivity equation, the small aquifer gives up 66% of its store in the first month and 95% in three months. The large one gives up 10% in the first year, that is 209 thousand m³ instead of the 1991 thousand m³ of the tank model. In 10 years it gives up 54%, and in 30 years 90%.

Answer: The small aquifer equalises in months, and for a yearly balance it can be treated as a tank. The large one gives up its store over decades: the tank model would overstate its influx in the first year about 10 times.

What it means. The outer radius is five times larger, the distance from the contact to the boundary is 9 times larger, and the time to reach the boundary is 81 times larger, because it grows as the square of this distance. So the large aquifer needs the unsteady-state model. The pseudo-steady-state model suits the period after the first few years: on the chart it approaches the exact solution after five to ten years. A steady-state model with a suitable coefficient can describe a stretch of the history, but for a closed aquifer it will overstate the influx later. The data in this example are assumed for illustration.

The unsteady-state model is the general case, and the others are its simplifications for limiting conditions. In simulation models the aquifer is set in one of two ways. It can be an analytical model (a ready set of formulas) attached to the boundary cells, usually the unsteady-state model in the Carter–Tracy approximation or the pseudo-steady-state model. Or it can be additional grid cells with a large pore volume (Building a simulation model: input data and results).

How to choose a model

Classes of aquifer models

Model: Small aquifer (“tank”)
Main assumption: the pressure throughout the aquifer equals the pressure at the contact
Where it is suitable: small permeable aquifers that equalise in weeks or months
Parameters: $W$

Model: Steady-state
Main assumption: the pressure at the outer boundary is constant
Where it is suitable: aquifers with recharge; empirically, over a limited stretch of the history
Parameters: $J_a$

Model: Pseudo-steady-state
Main assumption: the aquifer is depleted like a tank behind a resistance
Where it is suitable: finite aquifers after the first years of development
Parameters: $W$ , $J$

Model: Unsteady-state
Main assumption: the diffusivity equation for the aquifer
Where it is suitable: the general case, above all large aquifers
Parameters: $U$ , $r_a/r_o$ , $\eta$

Model	Main assumption	Where it is suitable	Parameters
Small aquifer (“tank”)	the pressure throughout the aquifer equals the pressure at the contact	small permeable aquifers that equalise in weeks or months	$W$
Steady-state	the pressure at the outer boundary is constant	aquifers with recharge; empirically, over a limited stretch of the history	$J_a$
Pseudo-steady-state	the aquifer is depleted like a tank behind a resistance	finite aquifers after the first years of development	$W$ , $J$
Unsteady-state	the diffusivity equation for the aquifer	the general case, above all large aquifers	$U$ , $r_a/r_o$ , $\eta$

The fraction of the perimeter $f$ enters every model through the volume or the coefficient. If the influx is small compared with other sources of energy (a gas cap, dissolved gas), the choice of model has little effect on the result, and the steady-state model is usually enough.

How a model is fitted to the history

1
Get the influx from the balance
If the oil in place is known, $W_e = F - N(E_o + mE_g + E_{fw})$ for each date (General material balance equation for oil reservoirs). If it is unknown, find the influx together with the oil in place from a straight line (Straight-line material balance (Havlena–Odeh method)).
2
Start with the simplest model
Estimate from the geology the size and the conductivity of the aquifer and its response time. An aquifer that equalises quickly is a tank, an aquifer with recharge is a steady-state model, and otherwise we use the unsteady-state or the pseudo-steady-state model.
3
Fit the parameters
Choose them so that the model reproduces the influx on all dates. A systematic mismatch means a wrong model, not an inaccurate parameter.
4
Check that the result is plausible
Compare the fitted volume and conductivity of the aquifer with the geological data: the extent of the formation beyond the oil boundary, its thickness and its permeability.
5
Update
With every new pressure measurement. A model found from a short history is seldom the only possible one (History matching of aquifer parameters).

Terms and symbols

Aquifer: The water-filled part of the formation that is connected to the accumulation, so that fluid can flow between them. When the pressure falls, it gives water to the accumulation.
Elastic store of an aquifer: The largest influx for a pressure drop of $\Delta p$ throughout the aquifer: $(c_w + c_f)W\Delta p$ .
Small-aquifer model: The pressure throughout the aquifer equals the pressure at the contact. The influx is proportional to the pressure drop and does not depend on time.
Steady-state model: The rate of influx is proportional to the pressure drop at the contact. The pressure at the outer boundary is constant.
Pseudo-steady-state model: A finite aquifer treated as a tank behind a resistance. It gives up its elastic store exponentially.
Unsteady-state model: The solution of the diffusivity equation for the aquifer. The influx depends on the whole pressure history.
$J_a$ , $U$ , $f$: The productivity index of the aquifer, the aquifer constant, and the fraction of the perimeter of the accumulation that the aquifer touches.

Common mistakes and misconceptions

Wrong

“The aquifer is huge, so the pressure in the accumulation will not fall.”

In fact

The volume gives a store of energy, but not the speed at which it is given up. In this lesson's example a large aquifer gives up only a tenth of its elastic store in the first year. If the withdrawal is faster than the influx, the pressure falls (Reservoir drive mechanisms).

Wrong

“The influx is calculated from the compressibility of water.”

In fact

We need the sum of the compressibilities of the water and the pores. For cemented rock $c_f$ is often larger than $c_w$ , and without it the elastic store is understated almost threefold.

Wrong

“The steady-state model described the history, so the forecast will be right.”

In fact

A fitted coefficient is right for the stretch on which it was fitted. For a closed aquifer the influx weakens with time, and the steady-state model starts to overstate it.

Wrong

“The tank model suits any aquifer; we only need to fit the volume.”

In fact

A tank does not know about time: it “discharges” a large aquifer instantly. A volume fitted to early data will turn out much too small and will grow with every new measurement, which is a sure sign that the aquifer is not a tank. First we check the response time of the aquifer.

In depth: advanced detail

The exponential of the pseudo-steady-state model and its limits

From the balance of the aquifer, the average pressure in it is $\bar p_a = p_i - W_e/\big[(c_w + c_f)W\big]$ . Let us put it into the influx equation for a constant pressure at the contact, $p = p_i - \Delta p$ :

Response of a pseudo-steady-state aquifer to a constant pressure drop

\frac{\dd W_e}{\dd t} = J\left[\Delta p - \frac{W_e}{(c_w + c_f)W}\right]\;\Rightarrow\; W_e = (c_w + c_f)\,W\,\Delta p\left(1 - \e^{-t/\tau}\right),\quad \tau = \frac{(c_w + c_f)\,W}{J}

where:

$\tau$ time constant of the aquifer, d
$J$ productivity index of the aquifer for pseudo-steady-state flow, m³/(d·MPa)
$W,\ c_w + c_f,\ \Delta p$ pore volume and compressibility of the aquifer, and the pressure drop at the contact, m³, 1/MPa, MPa

Units: m³, days, MPa

Holds when

the pressure drop at the contact is constant; at $t = 0$ the aquifer is untouched: $W_e = 0$

At the start, for $t \ll \tau$ , $1 - \e^{-t/\tau} \approx t/\tau$ , and $W_e \approx J\Delta p\,t$ , which is the steady-state model with the coefficient $J$ . For $t \gg \tau$ the influx tends to the elastic store, which is the tank model. The larger the aquifer and the smaller its conductivity, the larger $\tau$ is, and the longer the aquifer behaves as a steady-state one, still “not noticing” that it is limited. For this lesson's large aquifer, $\tau$ is about 13 years.

Connections to other topics

Reservoir drive mechanisms — water drive among the other sources of reservoir energy.
General material balance equation for oil reservoirs — water influx in the general balance equation, and its calculation from the volumetric oil in place.
Straight-line material balance (Havlena–Odeh method) — finding the oil in place and the aquifer parameters together from the history.
Diffusivity equation — the equation that describes how the pressure drop spreads through the aquifer.
Flow regimes — the unsteady, pseudo-steady and steady-state regimes, for an aquifer just as for a well.
Unsteady-state water influx: the van Everdingen–Hurst method — the unsteady-state model for a real pressure history.
Simplified water influx models: Fetkovich and Carter–Tracy — simplified step-by-step influx models.
History matching of aquifer parameters — how the size and the conductivity of the aquifer are fitted to the history.
Material balance for gas reservoirs — water influx into a gas accumulation, and trapped gas.

Check yourself

1
What most often supplies the water influx into an accumulation?
1. A.The elastic expansion of the water of the aquifer and the shrinking of its pores when the pressure falls
2. B.Recharge of the aquifer with water from the surface
3. C.Capillary imbibition (the soaking up of water by the rock)
4. D.Gravity sinking of water under the accumulation
Show the answer
Correct answer: A. The elastic expansion of the water of the aquifer and the shrinking of its pores when the pressure falls
Recharge from the surface needs the layer to reach the surface and water to enter there at the same rate at which it is withdrawn, and this is rare. Usually the aquifer is cut off by faults, or the formation thins out to nothing. Then the influx is the elastic expansion of the water and the shrinking of the pores of the aquifer: $(c_w + c_f)W\Delta p$ in the limit.
2
The pore volume of an aquifer is 200 million m³, and $c_w + c_f = 1.2\cdot10^{-3}$ MPa⁻¹. What is the largest influx for a pressure drop of 5 MPa, and is it always reached?
Show the answer
$W_{e,\max} = 1.2\cdot10^{-3}\cdot200\cdot10^6\cdot5 = 1.2$ million m³. The aquifer gives this much water only when the pressure drop has reached all its parts and has equalised. If the aquifer is large or has low permeability, over the years of development it gives only a part of its store, and the tank model overstates the influx here.
3
A steady-state model, $J_a = 500$ m³/(d·MPa). The pressure at the contact fell at once by 4 MPa and stays so for two years (730.5 days). How much water came in, and what is the rate of influx? What changes if the pressure fell steadily from zero to 4 MPa over the same two years?
Show the answer
The rate is $500\cdot4 = 2000$ m³/d, and the influx is $500\cdot4\cdot730.5 \approx 1.46$ million m³. With a steady fall, the average pressure drop over the two years is 2 MPa, and the influx is half as large, about 0.73 million m³, although the final pressure is the same. The steady-state influx depends on the whole pressure history.
4
An aquifer is 15 times larger than the accumulation in radius, its permeability is low, and the development history is 5 years. Which model should we choose?
1. A.Small aquifer: it will give up its whole elastic store anyway
2. B.Steady-state, with a coefficient from Darcy's law out to the outer boundary
3. C.Unsteady-state: the pressure drop has not yet reached the boundary, and the influx depends on time
4. D.No model is needed: the influx is small
Show the answer
Correct answer: C. Unsteady-state: the pressure drop has not yet reached the boundary, and the influx depends on time
The time to reach the boundary grows as the square of the distance to it, and is inversely proportional to the hydraulic diffusivity. For a large aquifer of low permeability it is decades. While the aquifer behaves as an infinite one, the influx is set by the diffusivity equation, so we need the unsteady-state model. The tank would overstate the influx many times, and the steady-state model, with a recharge boundary where there is none, would overstate it with time too.
5
The coefficient of a steady-state model was fitted to the first five years and describes them well. After ten years the influx from the model turned out to be clearly larger than the balance gives. Why, and how would this look on a chart of the Havlena–Odeh method?
Show the answer
The aquifer is most likely closed. Its pressure far from the contact falls with time, and the real influx grows more slowly than the integral of the pressure drop at the contact. On a chart of $F/E_t$ against $\int(p_i - p)\dd t/E_t$ the late points would lie below the line drawn through the early ones. We need a model of a finite aquifer, the pseudo-steady-state or the unsteady-state one.

Updated: 5 October 2026