History matching of aquifer parameters

Quick refresher

We fit the type of aquifer model and its parameters. For a radial unsteady-state aquifer these are the aquifer constant $U$ , the radius ratio $r_{aD}$ and the scale of the dimensionless time, which is set by the permeability of the aquifer. What is known is the history of the contact pressure, the production, the PVT properties, and often the oil in place from the volumetric method.
We compare the model either with the influx from the balance, $W_e = F - NE_t$ (if the oil in place is known), or with the measured pressure. The influx from the balance is inaccurate in the first years, because for a small pressure drop the expansion is a small difference.
If the model is right, the apparent constant $U_n = W_{e,n}/w_n$ is the same on all dates, and the Havlena–Odeh points lie on a straight line. If $U_n$ grows with time, the influx of the model grows more slowly than the real one; as a rule the aquifer of the model is too small or too slow. If it falls, the model gives water faster; as a rule its aquifer is too large or too fast.
The fit is not unique. A smaller permeability of the aquifer is made up for by a larger constant $U$ , and until the pressure drop has reached the outer boundary, the history says nothing about the size of the aquifer. Models that are equally accurate in the past differ in the forecast.
A fit is plausible if the parameters lie within the geological limits ( $U$ from the porosity, thickness, compressibility and the fraction of the perimeter; permeability from core; size from maps), the oil in place agrees with the volumetric estimate, and the model predicts data that were not available when it was fitted.
The parameters of step-by-step models are effective: a fitted Fetkovich aquifer makes up for the missing transient period (the early time while the pressure drop is still spreading through the aquifer) and can turn out to be much smaller than the real one. Its storage capacity $c_tW$ (the water it gives up per megapascal of pressure drop) cannot be taken as the volume of the aquifer.

Key relationships

W_{e,n} = F_n - N E_{t,n}

influx from the balance, for a known oil in place

U_n = \frac{W_{e,n}}{w_n},\qquad w_n = \sum_j \Delta p_j\,W_D(t_{Dn} - t_{Dj})

apparent aquifer constant; $w$ is the influx function, MPa

\frac{F}{E_t} = N + U\,\frac{w}{E_t}

Havlena–Odeh line with an unsteady-state influx function

U = 2\pi f\phi c_t h r_o^2 \le 2\pi\phi c_t h r_o^2

geological limit of the aquifer constant ( $f \le 1$ )

M = \frac1n\sum_k\left(\frac{p_k^{\text{meas}} - p_k^{\text{calc}}}{\sigma}\right)^2 \le 1

agreement with the pressure history within the measurement error $\sigma$

Common pitfalls

Taking a straight line, or a small mismatch, as proof that the model we found is the only one.
Fitting the parameters with steps that are too long. The error of the step passes into the parameters of the aquifer.
Using an infinite aquifer in the forecast when the history is shorter than the time in which the pressure drop reaches the boundary.
Taking the capacity and the productivity index of the aquifer fitted by the Fetkovich method for its volume and conductivity.
Fitting the oil in place and the aquifer at the same time and not comparing the oil in place with the volumetric estimate.

After this lesson you can

list what is known and what is fitted when matching an aquifer, and choose what to compare the model with: the influx from the balance or the pressure
test an aquifer model by the constancy of the apparent aquifer constant, if the oil in place is known
find the oil in place and the aquifer constant from the straight line of the Havlena–Odeh method with an unsteady-state influx function, and recognise a wrong aquifer size from the shape of the chart
fit the parameters to the pressure history and find all the models that agree with the data within the measurement error
apply the plausibility checks: the geological limits of the parameters, the agreement of the oil in place with the volumetric estimate (the figure worked out from maps and rock properties), and a forecast check on data that were not used in the fit
explain why the parameters of the Fetkovich and Carter–Tracy models fitted to the history are effective numbers (they are adjusted to make that model fit) and cannot be carried into other models as properties of the aquifer

What is known and what is fitted

An aquifer is almost always described by a model (Aquifer models: water influx). There are few wells beyond the oil boundary, and the size, the permeability and even the shape of the aquifer are known only roughly. When the accumulation is put on production, the aquifer “answers” for the first time: the pressure falls more slowly than in a closed accumulation, because water is coming in. The model is fitted to this answer.

What is fitted for a radial unsteady-state aquifer

Parameter: Constant $U = 2\pi f\phi c_th r_o^2$
What it sets: how much water the aquifer gives per 1 MPa of pressure drop at a given $W_D$
What limits it: porosity, thickness, compressibility, fraction of the perimeter $f$

Parameter: Radius ratio $r_{aD}$
What it sets: when the influx starts to weaken and how large the elastic store is (the most water the aquifer can give up)
What limits it: maps, seismic surveys, regional geology

Parameter: Permeability of the aquifer $k$
What it sets: the time scale: how fast the pressure drop spreads through the aquifer
What limits it: core, tests of wells beyond the oil boundary

Parameter: Oil in place $N$ and cap size $m$
What it sets: the energy of the accumulation itself
What limits it: volumetric estimate, geology

Parameter: Type of model
What it sets: small, steady-state, unsteady-state or linear aquifer
What limits it: shape of the charts, geology

Parameter	What it sets	What limits it
Constant $U = 2\pi f\phi c_th r_o^2$	how much water the aquifer gives per 1 MPa of pressure drop at a given $W_D$	porosity, thickness, compressibility, fraction of the perimeter $f$
Radius ratio $r_{aD}$	when the influx starts to weaken and how large the elastic store is (the most water the aquifer can give up)	maps, seismic surveys, regional geology
Permeability of the aquifer $k$	the time scale: how fast the pressure drop spreads through the aquifer	core, tests of wells beyond the oil boundary
Oil in place $N$ and cap size $m$	the energy of the accumulation itself	volumetric estimate, geology
Type of model	small, steady-state, unsteady-state or linear aquifer	shape of the charts, geology

The porosity, thickness and compressibility of the aquifer are usually taken the same as for the reservoir rock in the accumulation. The fraction of the perimeter $f$ and the permeability enter the fit through $U$ and the time scale.

It is convenient to write the influx of the model as the product of the constant $U$ and the influx function $w$ , the same function that stands on the horizontal axis in the Havlena–Odeh method (accumulation with an aquifer). For an unsteady-state aquifer, $w$ is the sum of the pressure steps weighted by the dimensionless influx (Unsteady-state water influx: the van Everdingen–Hurst method). It depends on $r_{aD}$ and on the time scale, but not on $U$ :

Influx of the unsteady-state aquifer model

W_e(t_n) = U\,w_n,\qquad w_n = \sum_{j=0}^{n-1}\Delta p_j\,W_D\big(t_{Dn} - t_{Dj};\ r_{aD}\big),\qquad t_D = \frac{k\,t}{\phi\,\mu_wc_t\,r_o^2}

where:

$W_e(t_n)$ cumulative water influx by date $t_n$ , at reservoir conditions, m³
$U$ aquifer constant, m³/MPa
$w_n$ influx function by date $t_n$ , MPa
$\Delta p_j$ pressure steps at the contact from the average pressures of the intervals (or the exact integral for a pressure that changes along a straight line between the measurements), MPa
$W_D,\ r_{aD},\ t_D$ dimensionless influx, radius ratio and dimensionless time of the aquifer, —

Units: m³, MPa; $t_D$ in SI, or with the multiplier $3.553\cdot10^{-3}$ for $k$ in mD, $t$ in hours, $\mu_w$ in mPa·s and $c_t$ in 1/MPa

Holds when

a uniform radial aquifer; the pressure at the contact equals the average pressure of the accumulation
the function $w$ is calculated with short steps, or exactly; otherwise the error of the stepped representation of the pressure passes into the fitted parameters

This separation is convenient. For a chosen $r_{aD}$ and permeability, the influx is linear in $U$ , and the constant is found at once, by division or by least squares. Only the two “nonlinear” parameters have to be found by trial.

What to compare the model with depends on what is known. If the oil in place is reliably estimated by the volumetric method, the balance gives a “measured” influx for each date, $W_e = F - NE_t$ (General material balance equation for oil reservoirs), and the model is compared with it. If the oil in place is unknown too, we find it together with the aquifer from the Havlena–Odeh straight line. Finally, we can compare the pressure itself: from the production and the aquifer model we calculate what it should have been, and compare it with the measurements. This is how it is done in material balance programs and in simulation models (History matching).

The apparent aquifer constant

Suppose the oil in place is known. For a chosen aquifer, we calculate the influx function $w_n$ for each date and divide the influx from the balance by it. If the shape of the model is right, the quotient is the same number, the constant $U$ . If it is not, the apparent aquifer constant drifts, and the direction of the drift suggests what to change.

Apparent aquifer constant

U_n = \frac{W_{e,n}}{w_n} = \frac{F_n - N E_{t,n}}{w_n}

where:

$U_n$ apparent constant for date $n$ , m³/MPa
$F_n,\ E_{t,n}$ underground withdrawal (the reservoir volume of the fluids produced) and expansion per 1 m³ of oil in place on the same date, m³, m³/m³
$N$ oil in place: known, m³
$w_n$ influx function of the chosen aquifer model, MPa

Units: m³/MPa

Holds when

the oil in place is known independently; there is no gas cap, or its size is known ( $E_t$ includes $mE_g$ )
the early dates, where the pressure drop is about the same size as the error of its measurement, get a smaller weight

If $U_n$ grows with time, the real influx grows faster than the function of the model: the real aquifer is larger, or lets water through more easily, than the model assumes. If it falls, the opposite is true: the model gives water more generously than the accumulation receives it, and the real aquifer is smaller, more closely bounded or slower than in the model. The rule is not absolute: if the permeability of a closed aquifer in the model is too high, its boundary “switches on” earlier than the real one, and after the initial fall $U_n$ grows again. So the drift tells us that the shape of the influx function is wrong, and the fit decides which parameter to change.

Apparent aquifer constant by year for three models, thousand m³/MPa

Year: 1
$W_e$ from the balance, thousand m³: 190.7
12 km: 40.06
16 km: 40.06
infinite: 40.06

Year: 2
$W_e$ from the balance, thousand m³: 668.0
12 km: 40.10
16 km: 40.05
infinite: 40.05

Year: 3
$W_e$ from the balance, thousand m³: 1372.8
12 km: 40.22
16 km: 40.00
infinite: 39.99

Year: 4
$W_e$ from the balance, thousand m³: 2187.9
12 km: 40.58
16 km: 39.99
infinite: 39.94

Year: 5
$W_e$ from the balance, thousand m³: 3057.7
12 km: 41.14
16 km: 40.00
infinite: 39.86

Year: 6
$W_e$ from the balance, thousand m³: 3935.7
12 km: 41.83
16 km: 40.00
infinite: 39.70

Year: 7
$W_e$ from the balance, thousand m³: 4811.0
12 km: 42.64
16 km: 40.00
infinite: 39.47

Year: 8
$W_e$ from the balance, thousand m³: 5685.3
12 km: 43.50
16 km: 40.00
infinite: 39.16

Year	$W_e$ from the balance, thousand m³	12 km	16 km	infinite
1	190.7	40.06	40.06	40.06
2	668.0	40.10	40.05	40.05
3	1372.8	40.22	40.00	39.99
4	2187.9	40.58	39.99	39.94
5	3057.7	41.14	40.00	39.86
6	3935.7	41.83	40.00	39.70
7	4811.0	42.64	40.00	39.47
8	5685.3	43.50	40.00	39.16

The history is from the lesson Unsteady-state water influx: the van Everdingen–Hurst method; the influx from the balance is for an oil in place of 9.06 million m³. The models are closed radial aquifers with outer radii of 12 and 16 km, and an infinite one, with a permeability of 50 mD. The influx function is exact for a pressure that changes along a straight line between the measurements.

Worked exampleWhich aquifer keeps the constant

Given (assumed data)

Influx from the balance: year 1: 190.7 thousand m³; year 8: 5685.3 thousand m³ (the table above)
Influx function $w$ , MPa: year 1: 4.76 for all three models; year 8: 130.69 (12 km), 142.14 (16 km), 145.18 (infinite)

Solution

Year 1: $U_1 = 190.7/4.76 = 40.06$ thousand m³/MPa. The models cannot be told apart: in a year the boundary of the aquifer has hardly affected the influx yet, and even an aquifer of 8 km differs by only 0.4%.
Year 8, 12 km: $5685.3/130.69 = 43.50$ ; 16 km: $5685.3/142.14 = 40.00$ ; infinite: $5685.3/145.18 = 39.16$ thousand m³/MPa.
Change over eight years: 8.6%, −0.2% and −2.3%. For an aquifer of radius 8 km (not in the table) it goes from 40.23 to 63.89, a change of 59%.

Answer: The aquifer of radius 16 km keeps the constant: 40.00 thousand m³/MPa on all dates. For the 8 and 12 km aquifers it grows, and for the infinite one it slowly falls.

What it means. The growth means that the model aquifer “runs out of breath” earlier than the real one: the pressure drop has reached its boundary, and the influx of the model has slowed down, while water is still entering the accumulation. The fall for the infinite aquifer is the opposite: the model goes on enlarging the “funnel” beyond the real boundary. But note the scale: the infinite aquifer differs from the right one by only 2% over eight years, and with real errors in the pressure, the PVT properties and the production records such a drift is hard to notice. This is the first sign of the ambiguity discussed below. The data in this example are assumed for illustration: the history was calculated with an aquifer of 16 km, and the example shows how well the fit recovers it.

CalculatedThe apparent aquifer constant by year for four models with a permeability of 50 mD: closed aquifers of radius 8, 12 and 16 km, and an infinite one. It is constant only for the 16 km aquifer, with which the assumed history was calculated.

Oil in place and the aquifer together: the Havlena–Odeh line

If the oil in place is unknown, we divide the balance equation $F = NE_t + U\,w$ by $E_t$ and get a straight line. Its intercept on the vertical axis is the oil in place, and its slope is the aquifer constant (the line for an accumulation with an aquifer). For an unsteady-state aquifer the abscissa depends on the chosen $r_{aD}$ and permeability, and the fit comes down to a search for the values at which the points lie on a straight line.

Havlena–Odeh line with an unsteady-state influx function

\frac{F}{E_t} = N + U\,\frac{w}{E_t},\qquad E_t = E_o + mE_g + E_{fw}

where:

$F/E_t$ ordinate: the apparent oil in place, m³
$w/E_t$ abscissa: the influx function divided by the expansion, MPa
$N$ intercept on the vertical axis: the oil in place, m³
$U$ slope: the aquifer constant, m³/MPa

Units: m³, MPa, m³/MPa

Holds when

there is no gas cap, or its size $m$ is known
the early points, where $E_t$ is small, are unreliable and must not decide the slope

A bend in the chart says that the shape of the influx function is wrong. Late points above the line mean that the function of the model grows more slowly than the real influx. As a rule the aquifer of the model is small (its boundary “switched on” in the model earlier than in the reservoir) or slow. Points below the line mean the opposite: as a rule the aquifer of the model is large or fast. An S-shaped curve means that the geometry is wrong, for example a linear aquifer instead of a radial one. A random scatter means errors in the data.

Coordinates of the line for the 16 km, 50 mD aquifer

Year: 1
$x = w/E_t$ , thousand MPa: 0.826
$y = F/E_t$ , million m³: 42.14

Year: 2
$x = w/E_t$ , thousand MPa: 1.282
$y = F/E_t$ , million m³: 60.42

Year: 3
$x = w/E_t$ , thousand MPa: 1.760
$y = F/E_t$ , million m³: 79.48

Year: 4
$x = w/E_t$ , thousand MPa: 2.328
$y = F/E_t$ , million m³: 102.15

Year: 5
$x = w/E_t$ , thousand MPa: 2.864
$y = F/E_t$ , million m³: 123.63

Year: 6
$x = w/E_t$ , thousand MPa: 3.441
$y = F/E_t$ , million m³: 146.70

Year: 7
$x = w/E_t$ , thousand MPa: 3.960
$y = F/E_t$ , million m³: 167.49

Year: 8
$x = w/E_t$ , thousand MPa: 4.429
$y = F/E_t$ , million m³: 186.22

Year	$x = w/E_t$ , thousand MPa	$y = F/E_t$ , million m³
1	0.826	42.14
2	1.282	60.42
3	1.760	79.48
4	2.328	102.15
5	2.864	123.63
6	3.441	146.70
7	3.960	167.49
8	4.429	186.22

$F$ and $E_t$ are from the history of the lesson Unsteady-state water influx: the van Everdingen–Hurst method; $E_t = E_o + E_{fw}$ , with no gas cap.

Worked exampleOil in place and the aquifer constant from the line

Given (assumed data)

Points: the table above (16 km aquifer, 50 mD)
Volumetric estimate of the oil in place: 9.06 million m³, for comparison only

Solution

The slope from the second and eighth points: $U = \dfrac{(186.22 - 60.42)\cdot10^6}{(4.429 - 1.282)\cdot10^3} = 39{,}975$ m³/MPa; the intercept is $N = 60.42\cdot10^6 - 39{,}975\cdot1.282\cdot10^3 = 9.17$ million m³.
The straight line through all eight points (least squares): $N = 9.11$ million m³, $U = 40.0$ thousand m³/MPa; the points deviate from it by no more than 0.06%.
The same calculation for the two other aquifers is in the table after the example: the 12 km aquifer gives an intercept of 3.2 million m³, and the infinite one 10.56 million m³.

Answer: The 16 km aquifer gives a straight line with an oil in place of 9.11 million m³ and a constant of 40.0 thousand m³/MPa, in agreement with the volumetric estimate.

What it means. The 12 km aquifer gives itself away at once: the points are bent, and the intercept, 3.2 million m³, is almost three times smaller than the volumetric estimate. The infinite aquifer is harder. Its points deviate from the line by no more than 1.6%, so it cannot be rejected by the shape of the chart alone, but the oil in place comes out 17% larger than the volumetric one. What decides is an independent estimate of the oil in place, and this is why it is so valuable in fitting an aquifer. The data in this example are assumed for illustration.

Havlena–Odeh line for three aquifers with a permeability of 50 mD

Aquifer: 12 km
$N$ , million m³: 3.19
$U$ , thousand m³/MPa: 44.1
Largest deviation of a point, %: 5.99

Aquifer: 16 km
$N$ , million m³: 9.11
$U$ , thousand m³/MPa: 40.0
Largest deviation of a point, %: 0.06

Aquifer: infinite
$N$ , million m³: 10.56
$U$ , thousand m³/MPa: 39.1
Largest deviation of a point, %: 1.64

Aquifer	$N$ , million m³	$U$ , thousand m³/MPa	Largest deviation of a point, %
12 km	3.19	44.1	5.99
16 km	9.11	40.0	0.06
infinite	10.56	39.1	1.64

The volumetric estimate is 9.06 million m³. The deviation is the relative vertical distance of a point from the fitted line.

Fitting to the pressure

The influx from the balance is a calculated quantity. In the first years its error is large, because the expansion of the accumulation is a small difference. What is actually measured is the pressure. So it is more reliable to compare the model with the pressure. For each candidate aquifer we calculate from the production the pressure at the end of each year. We use the same trial as for the pressure forecast in the lesson on the van Everdingen–Hurst method. Then we compare the result with the measurements.

Measure of agreement with the pressure history

M = \frac1n\sum_{k=1}^{n}\left(\frac{p_k^{\text{meas}} - p_k^{\text{calc}}}{\sigma}\right)^{2},\qquad M \le 1\ \Longleftrightarrow\ \text{RMS} = \sqrt{\frac1n\sum_{k}\big(p_k^{\text{meas}} - p_k^{\text{calc}}\big)^2} \le \sigma

where:

$M$ misfit function (a number that shows how far the calculation is from the measurements), —
$p_k^{\text{meas}},\ p_k^{\text{calc}}$ measured and calculated pressure at the contact on date $k$ , MPa
$\sigma$ error in finding the average pressure (standard deviation), MPa
$\text{RMS}$ root-mean-square (RMS) deviation of the calculation from the measurements, MPa

Units: $M$ is dimensionless; pressures in MPa

Holds when

the measurement errors are independent and equal in size
each candidate aquifer is fitted so that $M$ is the smallest for its $r_{aD}$ and permeability; the constant $U$ is then fitted automatically

This is the same measure as in the lesson History matching: $M \approx 1$ means that the calculation differs from the measurements by their error. All the models with $M \le 1$ agree with the history equally well, and we cannot choose between them by the pressure history.

Fit to eight years of pressure: the constant

U

, thousand m³/MPa, and the RMS deviation of the pressure, MPa

Aquifer: 12 km
25 mD: 67.5 / 0.082
40 mD: 49.7 / 0.190
50 mD: 43.5 / 0.281
62.5 mD: 38.5 / 0.391

Aquifer: 16 km
25 mD: 66.3 / 0.053
40 mD: 46.9 / 0.030
50 mD: 40.0 / 0.003
62.5 mD: 34.3 / 0.048

Aquifer: infinite
25 mD: 66.2 / 0.059
40 mD: 46.4 / 0.087
50 mD: 39.1 / 0.110
62.5 mD: 32.9 / 0.134

Aquifer	25 mD	40 mD	50 mD	62.5 mD
12 km	67.5 / 0.082	49.7 / 0.190	43.5 / 0.281	38.5 / 0.391
16 km	66.3 / 0.053	46.9 / 0.030	40.0 / 0.003	34.3 / 0.048
infinite	66.2 / 0.059	46.4 / 0.087	39.1 / 0.110	32.9 / 0.134

In each cell $U$ is fitted to the pressure, with an oil in place of 9.06 million m³. The permeability sets the time scale of the aquifer: 6.115 units of $t_D$ per year at 50 mD, and in proportion at the others.

Worked exampleWhich aquifers agree with the history and what they forecast

Given (assumed data)

Fit: the table above; error of the average pressure $\sigma$ = 0.1 MPa
Plan for years 9 to 15: liquid 2000 m³/d, with the water cut (the share of water in the liquid) rising from 40 to 67% (year 9 is the plan of the lesson Unsteady-state water influx: the van Everdingen–Hurst method)
Geology of the aquifer: porosity 0.18–0.22; thickness 9–11 m; $c_t$ = (1.0–1.5)·10⁻³ MPa⁻¹; permeability from core 30–80 mD; $r_o$ = 1596 m

Solution

The condition RMS deviation ≤ 0.1 MPa is met by 7 candidates out of twelve (the table after the example). Their pressure forecast for the end of year 15 ranges from 18.24 to 20.74 MPa.
The largest constant that the geology allows ( $f = 1$ , upper limits): $U = 2\pi\cdot0.22\cdot1.5\cdot10^{-3}\cdot11\cdot1596^2 = 58.1$ thousand m³/MPa. The three candidates with a permeability of 25 mD need 66–68 thousand m³/MPa and also contradict the core data, so they are rejected.
4 candidates remain: the 16 km aquifer with a permeability of 40–62.5 mD, and the infinite one with 40 mD. The forecast for year 15 ranges from 18.24 to 20.74 MPa.
If the regional geology says that the formation beyond the oil boundary ends within 15–20 km (at a fault, or where the reservoir rock changes into shale), the infinite aquifer also drops out. The forecast then narrows to 18.24–19.34 MPa.

Answer: A pressure history with an error of 0.1 MPa leaves a spread of 2.5 MPa in the forecast for year 15. The geological limits and the information on the size of the aquifer narrow it to 1.1 MPa. The right value in this assumed example is 18.85 MPa.

What it means. Eight years of history is not little, and yet it did not fix the aquifer uniquely. A permeability twice as small, or a quarter larger, is easily made up for by the constant, and an infinite aquifer with a smaller permeability is hardly different from a closed one. What settled the matter was information from outside: core, the limits of porosity and thickness, and maps. The data in this example are assumed for illustration.

Candidates that agree with the pressure history (RMS deviation ≤ 0.1 MPa), and their forecast

Aquifer: 12 km
$k$ , mD: 25.0
$U$ , thousand m³/MPa: 67.5
RMS deviation, MPa: 0.082
$p$ in year 15, MPa: 18.55

Aquifer: 16 km
$k$ , mD: 25.0
$U$ , thousand m³/MPa: 66.3
RMS deviation, MPa: 0.053
$p$ in year 15, MPa: 19.99

Aquifer: 16 km
$k$ , mD: 40.0
$U$ , thousand m³/MPa: 46.9
RMS deviation, MPa: 0.030
$p$ in year 15, MPa: 19.34

Aquifer: 16 km
$k$ , mD: 50.0
$U$ , thousand m³/MPa: 40.0
RMS deviation, MPa: 0.003
$p$ in year 15, MPa: 18.85

Aquifer: 16 km
$k$ , mD: 62.5
$U$ , thousand m³/MPa: 34.3
RMS deviation, MPa: 0.048
$p$ in year 15, MPa: 18.24

Aquifer: infinite
$k$ , mD: 25.0
$U$ , thousand m³/MPa: 66.2
RMS deviation, MPa: 0.059
$p$ in year 15, MPa: 20.51

Aquifer: infinite
$k$ , mD: 40.0
$U$ , thousand m³/MPa: 46.4
RMS deviation, MPa: 0.087
$p$ in year 15, MPa: 20.74

Aquifer	$k$ , mD	$U$ , thousand m³/MPa	RMS deviation, MPa	$p$ in year 15, MPa
12 km	25.0	67.5	0.082	18.55
16 km	25.0	66.3	0.053	19.99
16 km	40.0	46.9	0.030	19.34
16 km	50.0	40.0	0.003	18.85
16 km	62.5	34.3	0.048	18.24
infinite	25.0	66.2	0.059	20.51
infinite	40.0	46.4	0.087	20.74

The forecast is for the plan of years 9 to 15, made in the same way as the fit. The pressure stays above the bubble point pressure of 18 MPa all the time.

CalculatedThe pressure at the contact from year 6 to year 15: the last measurements (dots) and the calculation for the seven candidate aquifers that agree with eight years of history within 0.1 MPa. In the past the candidates coincide, and by year 15 they differ by 2.5 MPa.

Why the fit is not unique

The constant and the time scale. A smaller permeability slows down the response of the aquifer, but a larger constant $U$ increases its “weight”. Over a limited stretch of history these two changes almost balance each other. In the table of the fit to pressure the constant changes almost twofold, while the agreement with the history stays within the error.
The invisible boundary. Until the pressure drop has reached the outer boundary of the aquifer, its size does not affect the history. The time to reach the boundary grows as the square of the distance to it (why the influx lags behind the pressure): for a 16 km aquifer in this accumulation, about three years. A history shorter than that cannot tell it from an infinite one.
Oil in place and the aquifer. If the oil in place is fitted too, extra influx can be “hidden” in the oil in place, and the other way round: the infinite aquifer gave an almost straight line with an oil in place 17% larger than the real one.
Errors in the data. The early pressures, the PVT properties and the production records pass into the parameters. The rougher the data, the wider the range of models that agree with them.

The most convincing check is the hindcast (a test on data kept back from the fit): we fit the model to part of the history, then forecast the rest and compare it with measurements that were not used in the fit. A model that has correctly guessed a “future” that has already come deserves more trust than a model that has only repeated the past.

Worked exampleFitting to four years and checking against the next four

Given (assumed data)

Fit: to the pressure of the first four years; oil in place 9.06 million m³
Candidates: the 16 km aquifer with a permeability of 50, 40 and 62.5 mD, and an infinite one with 50 mD
Pressure error: $\sigma$ = 0.1 MPa

Solution

The fit: for the 16 km aquifer with 50 mD, $U = 40.0$ thousand m³/MPa and an RMS deviation of 0.003 MPa; for the infinite one, $U = 39.9$ and 0.005 MPa; for 40 and 62.5 mD, 47.1 and 34.0 thousand m³/MPa and 0.012 and 0.012 MPa. Over four years all four candidates agree with the history.
The forecast for years 5 to 8 and its comparison with the measurements are in the table after the example. By year 8 the infinite aquifer overstates the pressure by 0.39 MPa, the 62.5 mD candidate understates it by 0.17 MPa, and the 40 mD candidate overstates it by 0.09 MPa.
The RMS deviation of the forecast for years 5 to 8: 0.003, 0.235, 0.048 and 0.103 MPa.

Answer: The new data reject the infinite aquifer (forecast RMS deviation 0.235 MPa) and put the 62.5 mD candidate on the edge. The 16 km aquifer with 40–50 mD passes the check.

What it means. The four years of history almost coincide with the time in which the pressure drop reaches the boundary of the 16 km aquifer, so the boundary is hardly visible in them. The next years already show it: the infinite aquifer goes on giving water that the real aquifer does not have, and its forecast departs more and more from the measurements. A model that has passed such a check can be trusted more than a model that merely describes the whole history well, because a fit to the whole history leaves no data for a check. The data in this example are assumed for illustration.

Difference between the forecast and the measurements, MPa (calculated minus measured), for the candidates fitted to the first four years

Year: 5
16 km, 50 mD: 0.00
infinite, 50 mD: 0.04
16 km, 40 mD: −0.02
16 km, 62.5 mD: −0.01

Year: 6
16 km, 50 mD: 0.00
infinite, 50 mD: 0.11
16 km, 40 mD: 0.00
16 km, 62.5 mD: −0.05

Year: 7
16 km, 50 mD: 0.00
infinite, 50 mD: 0.23
16 km, 40 mD: 0.04
16 km, 62.5 mD: −0.11

Year: 8
16 km, 50 mD: 0.00
infinite, 50 mD: 0.39
16 km, 40 mD: 0.09
16 km, 62.5 mD: −0.17

Year	infinite, 50 mD	16 km, 40 mD	16 km, 62.5 mD
5	0.04	−0.02	−0.01
6	0.11	0.00	−0.05
7	0.23	0.04	−0.11
8	0.39	0.09	−0.17

A positive difference means that the model overstates the pressure, that is, the influx.

CalculatedThe difference between the calculation and the pressure measurements for four candidate aquifers fitted to the first four years: up to year 4 it is the fit, after it the forecast. The band is the measurement error ±0.1 MPa.

What makes a fit plausible

Checks after the fit

1
Parameters within the geological limits
The constant $U$ is not larger than $2\pi\phi c_th r_o^2$ for plausible porosity, thickness and compressibility ( $f \le 1$ ). The permeability is within the limits of core data and of tests of wells beyond the oil boundary. The size is within the limits of maps and seismic data. A parameter that has “drifted” out of its limits is a sign of a wrong type of model or of an error in the data.
2
The oil in place agrees with the volumetric estimate
If the oil in place was fitted together with the aquifer, a difference from the volumetric estimate that is larger than its error needs an explanation, not acceptance.
3
The differences are random
The residuals (the differences “calculated minus measured”) must have no trend. A trend is a sign of a wrong shape of the model, not of an inaccurate parameter.
4
A check by forecast
Part of the history is kept for a check. After a fit to the whole history, the model is checked against every new measurement.
5
Agreement with the water seen in the wells
Where and when water reached the wells, and how the oil–water contact rose according to well logs: these facts are not part of the balance, and so they check independently where the water came from and how much came.
6
The spread of the forecast
If several models agree, the forecast is given as a range over them, and decisions (for example, on water injection) are tested for their sensitivity to the choice of model.

Step-by-step models in a fit

In simulation models the aquifer is usually set by the step-by-step Fetkovich or Carter–Tracy method (Simplified water influx models: Fetkovich and Carter–Tracy), and its parameters are fitted to the history in the same way. But each method has its own error, and the fit “absorbs” it: the parameters adjust themselves so that the method, with its error, reproduces the history. So the fitted parameters belong to the pair “aquifer + method”, not to the aquifer.

Worked exampleA Fetkovich aquifer fitted to eight years

Given (assumed data)

Target of the fit: the influx from the balance over eight years (the table in the section on the apparent constant); monthly steps of the method
Aquifer from the geology: storage capacity $c_tW$ = 1.99 million m³/MPa, productivity index $J$ = 431 m³/(d·MPa), from the lesson Simplified water influx models: Fetkovich and Carter–Tracy

Solution

With the geological parameters the Fetkovich method lags behind the influx from the balance in the first years (it has no transient period): the RMS deviation of the influx is 232 thousand m³.
Fitting both parameters: $c_tW$ = 1.21 million m³/MPa, $J$ = 552 m³/(d·MPa), time constant 6.0 years; the RMS deviation of the influx is 47 thousand m³, a much better agreement.
The fitted capacity is 39% smaller than the geological one, and the productivity index is 28% larger. The “aquifer” has become smaller and more conductive, so as to give water faster in the first years.
The pressure forecast for the plan: in year 12, 18.37 MPa against 19.93 for the real aquifer. In year 13 the fitted Fetkovich aquifer brings the pressure down to the bubble point pressure.
For comparison, take the Carter–Tracy method with the real $r_{aD}$ and permeability, and fit a single constant. With a monthly step it gives $U$ = 41.4 thousand m³/MPa (3.4% more than the real one). With a yearly step it gives 39.1, and the agreement is 1.5 times worse.

Answer: The Fetkovich method describes the influx of the eight years of history with an aquifer 39% smaller than the real one. But it repeats the pressure worse than the measurement error, and its forecast for year 12 is 1.6 MPa more pessimistic.

What it means. The fit clearly improved the description of the influx of the past years, which is what it was made for. But against the pressure, which is what is really measured, the fitted model differs with an RMS deviation of 0.25 MPa, more than the measurement error of 0.1 MPa. Its forecast for year 9 is already 0.7 MPa lower than the pressure from the real aquifer. The method has no transient period, and the fit makes up for it only in part. Its capacity is not the volume of the aquifer, and its productivity index is not the conductivity: these are effective numbers. Carried into a van Everdingen–Hurst model, or into an estimate of the water store, they would give a wrong answer. The data in this example are assumed for illustration.

Terms and symbols

Influx function $w$: The part of the aquifer model that is calculated from the pressure history: $W_e = U\,w$ . For an unsteady-state aquifer it is the sum of the pressure steps weighted by the dimensionless influx.
Apparent aquifer constant $U_n$: The influx from the balance divided by the influx function of the model on the same date. It is constant if the shape of the model is right.
Time scale of the aquifer: The multiplier that turns years into dimensionless time. It is proportional to the permeability of the aquifer and inversely proportional to $\phi\mu_wc_tr_o^2$ .
Hindcast: A test in which the model is fitted to part of the history, the rest is forecast, and the forecast is compared with the measurements.
Effective parameters: Parameters of a simplified model that are fitted to the history. They make up for the errors of the model and are not equal to the physical properties of the aquifer.
Non-uniqueness of the fit: A property of the inverse problem: different sets of parameters agree with the history equally well, within the error of the data.

Common mistakes and misconceptions

Wrong

“The model repeats eight years of history perfectly, so it is right.”

In fact

In this lesson's example, seven candidates out of twelve repeat the history to within 0.1 MPa, and their pressure forecasts for year 15 differ by 2.5 MPa. Agreement with the history is a necessary condition, not a proof.

Wrong

“Let the aquifer be infinite: it is safer, since the size is unknown anyway.”

In fact

An infinite aquifer in a forecast overstates the influx after the pressure drop has reached the real boundary. In this lesson's example, the infinite aquifer fitted to four years overstated the pressure by 0.39 MPa by year 8, and by 2.2 MPa by year 15.

Wrong

“The apparent constant is growing, so we must increase the aquifer constant.”

In fact

The constant changes only the level, while a drift is a sign of a wrong shape of the influx function: the size of the aquifer, the time scale or the geometry. But first we check the calculation step, because long steps create a drift by themselves.

Wrong

“The capacity of an aquifer in the Fetkovich method is its elastic store from the geology.”

In fact

The fitted capacity is an effective quantity. In this lesson's example it is 39% smaller than the geological one, because it makes up for the lack of a transient period in the method.

In depth: advanced detail

Small aquifer and gas cap

For an aquifer that equalises faster than the interval between measurements, the influx function is the pressure drop itself, $w = \Delta p$ , and the constant is the elastic store $(c_w + c_f)W$ (the small-aquifer model). Below the bubble point pressure the expansion of the accumulation grows faster than $\Delta p$ , the abscissa $\Delta p/E_t$ gets smaller, and the points on the chart go “backwards”. Above the bubble point pressure they gather in a column, and by the balance such an aquifer cannot be told from a larger closed accumulation (apparent oil in place).

If the accumulation has a gas cap, $E_t$ includes $mE_g$ , and there are three unknowns: the oil in place, the cap size and the aquifer. It is practically impossible to separate them with one balance (accumulation with a gas cap). The cap size is taken from geological data, and we fit the aquifer and, if needed, the oil in place. An error in $m$ passes into the parameters of the aquifer: an overstated cap “takes” part of the energy, and the fitted aquifer comes out weaker than the real one.

Connections to other topics

Unsteady-state water influx: the van Everdingen–Hurst method — the unsteady-state aquifer model and the influx function that are fitted here.
Simplified water influx models: Fetkovich and Carter–Tracy — step-by-step methods and their errors, which pass into the fitted parameters.
Straight-line material balance (Havlena–Odeh method) — the material balance line and the signs of a wrong aquifer model.
Aquifer models: water influx — the classes of aquifer models and the time for the pressure drop to reach the boundary.
General material balance equation for oil reservoirs — water influx from the balance, for a known oil in place.
History matching — the misfit function and the non-uniqueness of matching a simulation model.
Material balance for gas reservoirs — the same fitting for a gas accumulation with an aquifer.
Building a simulation model: input data and results — analytical aquifers in a simulation model.

Check yourself

1
With the chosen aquifer model, the apparent constant has grown over ten years from 30 to 45 thousand m³/MPa. What does this mean?
1. A.There is an error in the oil in place
2. B.The real influx grows faster than the function of the model: usually the aquifer is larger, or lets water through more easily, than assumed, and the shape of the model must be changed
3. C.The aquifer is smaller than assumed
4. D.We should take the constant of 45 thousand m³/MPa, because it is more accurate, being the last one
Show the answer
Correct answer: B. The real influx grows faster than the function of the model: usually the aquifer is larger, or lets water through more easily, than assumed, and the shape of the model must be changed
The constant sets only the level of the influx. Its drift means that the influx function of the model grows differently from the real influx: a growth of $U_n$ means that the model “runs out of breath” earlier than the accumulation. We change the size of the aquifer, the time scale or the geometry. But first we check whether the drift was created by long calculation steps.
2
From the history, an aquifer constant of $U$ = 40 thousand m³/MPa was fitted. The radius of the accumulation is 2.0 km, the porosity of the aquifer is 0.22, the thickness is 12 m, and $c_t = 1.1\cdot10^{-3}$ MPa⁻¹. What fraction $f$ of the perimeter of the accumulation does this correspond to, and is it plausible?
Show the answer
For $f = 1$ : $U = 2\pi\cdot0.22\cdot1.1\cdot10^{-3}\cdot12\cdot(2000)^2 = 2\pi\cdot11{,}616 \approx 73$ thousand m³/MPa. The fitted constant corresponds to $f = 40/73 \approx 0.55$ : the aquifer touches about half of the perimeter. This is plausible if the geology allows part of the boundary to be closed by a fault, or by the formation thinning out to nothing. If the maps show the accumulation surrounded by water on all sides, we look for the cause, for example an overstated permeability in the time scale, or an error in the oil in place.
3
Two aquifer models, a closed one of 15 km and an infinite one, repeat three years of history equally well. What will help to tell them apart?
1. A.A calculation with shorter steps
2. B.A fit with more parameters
3. C.Nothing: the models are equivalent
4. D.The pressure of the following years, when the pressure drop reaches the outer boundary, and independent information on the size of the aquifer
Show the answer
Correct answer: D. The pressure of the following years, when the pressure drop reaches the outer boundary, and independent information on the size of the aquifer
Until the pressure drop has reached the boundary, the size of the aquifer does not affect the history, and no processing of these data will show it. Only time can tell the models apart (new measurements, once the boundary “switches on”), or information from outside: maps, seismic surveys, the pressure in wells beyond the oil boundary. Until then the forecast is given as a range over both models.
4
Why does fitting the aquifer parameters with yearly steps and half drops bias the result, and in which direction?
Show the answer
Yearly steps understate the influx, most of all in the first years (choosing the steps). To reproduce the balance with an understated influx function, the fit raises the aquifer constant, or adjusts the size and the time scale so as to speed up the early influx. In this lesson's history the apparent constant of the right aquifer with yearly steps falls from 47.8 to 40.1 thousand m³/MPa. This is a false drift that is easily taken for a sign that the aquifer in the model is too large.
5
The Fetkovich method fitted to the history gave an aquifer capacity of $c_tW$ = 1.2 million m³/MPa. Can we find the volume of water in the aquifer from it by dividing by $c_t$ ?
1. A.Yes, if the calculation step was monthly
2. B.Yes, it is its elastic store
3. C.No, because the capacity depends on pressure
4. D.No: the fitted capacity is an effective quantity of the method that makes up for its error, and it can differ noticeably from the geological one
Show the answer
Correct answer: D. No: the fitted capacity is an effective quantity of the method that makes up for its error, and it can differ noticeably from the geological one
The Fetkovich method does not describe the transient period, and in a fit the capacity and the productivity index adjust to make up for it: in this lesson's example the fitted capacity is 39% smaller than the geological one. The volume of the aquifer found from it would be understated. We estimate the volume from the geology, and use the fitted numbers only inside the same method.
6
A Havlena–Odeh line with an unsteady-state influx function came out almost ideal. But the oil in place from its intercept is 12 million m³, while the volumetric estimate is 9 million m³ (with an error of about 10%). What should we do?
Show the answer
Do not accept the result. A straight line is needed but is not enough: over a limited history different aquifers give almost straight lines, and the difference goes into the intercept. We should check the aquifer models that give an oil in place within the volumetric estimate (another size or time scale), and the data that affect the balance: the average pressure, the PVT properties and the production records. If the oil in place still does not agree, we revise the volumetric estimate, but now as a separate hypothesis with a geological justification.

Updated: 7 October 2026