Straight-line material balance (Havlena–Odeh method)

Quick refresher

The balance in short notation: $F = N(E_o + mE_g + E_{fw}) + W_e$ . On the left is the withdrawal, calculated from the production. On the right are the unknowns $N$ , $m$ and $W_e$ , multiplied by the expansions, which are calculated from the PVT properties at the measured pressure.
Each date of a pressure measurement is a point on the chart. We choose the coordinates so that, if the model of the accumulation is right, the points lie on a straight line. Its slope and intercept are the quantities we want. A systematic drift of the points away from the line is a sign that the model is incomplete or the data are wrong.
Accumulation without a gas cap: the apparent oil in place $F/(E_o + E_{fw})$ is constant if there is no water influx, and grows with time if water comes in. A constant value still does not prove that the accumulation is closed. A gas cap that has not been counted, or a small aquifer above the bubble point pressure, gives a constant but too high oil in place.
With a gas cap of unknown size, the chart of $F/E_o$ against $E_g/E_o$ gives the intercept $N$ and the slope $mN$ . But $E_o$ and $E_g$ grow almost in proportion as the pressure falls, so the points crowd into a narrow band, and the line cannot tell $N$ and $m$ apart well. The size of the cap is usually taken from geological data, and the chart serves as a check.
With an aquifer, we express the influx through an aquifer model: $W_e$ equals the parameter of the model multiplied by a function of the pressure history. Then $F/E_t$ plotted against this function divided by $E_t$ is a straight line, with intercept $N$ and a slope equal to the aquifer parameter, if the model is right. A bend in the chart means a wrong model, not an inaccurate coefficient.
The drive indices are the shares of the reservoir volume of the oil and gas produced that were supplied by the expansion of the oil, the gas cap, the water and pores, and the water influx. Their sum is one when the point lies on the line. How far the sum is from one is the relative deviation of the point.

Key relationships

F = N\,(E_o + m\,E_g + E_{fw}) + W_e

material balance equation of an oil accumulation in short notation

\frac{F}{E_o + E_{fw}} = N + \frac{W_e}{E_o + E_{fw}}

apparent oil in place of an accumulation without a gas cap: constant if there is no water influx

\frac{F}{E_o + E_{fw0}} = N + mN\,\frac{E_g + E_{fw0}}{E_o + E_{fw0}}

accumulation with a gas cap and no water influx; $E_{fw} = (1 + m)\,E_{fw0}$

\frac{F}{E_t} = N + J_a\,\frac{1}{E_t}\int_0^t (p_i - p)\dd t

accumulation with a steady-state aquifer; $E_t = E_o + mE_g + E_{fw}$

\frac{NE_o}{H} + \frac{NmE_g}{H} + \frac{NE_{fw}}{H} + \frac{W_e - W_pB_w}{H} = 1

drive indices; $H$ is the reservoir volume of the oil and gas produced

Common pitfalls

Taking a straight line as proof of the model. A line follows from a right model, but lines of other models can pass through the same points.
Finding the oil in place and the gas cap size together from a short history with inaccurate pressure measurements. Pressure errors of a few hundredths of a megapascal change the result by tens of per cent.
Neglecting the connate water and the pores above the bubble point pressure. The oil in place will then be too high by a factor of one and a half to two.
Adjusting the aquifer coefficient when the points lie on a curve. The coefficient changes the slope, and only a different aquifer model can remove a bend.
Trusting the first points. While the pressure has fallen only a little, its error is about the same size as the fall, and the coordinates are divided by a small expansion.

After this lesson you can

write the material balance equation in the short notation $F$ , $E_o$ , $E_g$ , $E_{fw}$ , $W_e$ and turn it into the equation of a straight line for an accumulation without a gas cap, with a gas cap, and with an aquifer
tell a closed accumulation from one with water influx, and recognise a gas cap that has not been counted, from a chart of apparent oil in place against cumulative production
find the oil in place and the size of the gas cap from the chart, and explain why finding both together needs very accurate pressure measurements
find the oil in place and the aquifer productivity index (the water that enters per day for each unit of pressure drop) from the straight line of the steady-state aquifer model, and recognise a wrong aquifer model from the shape of the chart
calculate the drive indices from the development history and explain why they add up to one

Short notation and the idea of the line

The general material balance equation is derived in the lesson General material balance equation for oil reservoirs: the underground withdrawal equals the expansion of the oil with its dissolved gas, of the gas cap, and of the connate water and pores, plus the water influx. Here we need it in short notation, where the separate terms are combined into a few groups.

Material balance equation in short notation

F = N\,(E_o + m\,E_g + E_{fw}) + W_e

where:

$F$ underground withdrawal of the oil, gas and water at the current average pressure, m³
$N$ oil in place at standard conditions, m³
$E_o$ expansion of the oil with the gas dissolved in it, per 1 m³ of oil in place, m³/m³
$E_g$ expansion of the gas cap per 1 m³ of oil in place, for $m = 1$ , m³/m³
$E_{fw}$ expansion of the connate water and shrinking of the pore volume, per 1 m³ of oil in place, m³/m³
$m$ gas cap ratio, —
$W_e$ cumulative water influx at reservoir conditions, m³

Units: consistent units: volumes in m³ (oil in place and production at standard conditions, 0.101325 MPa and 20 °C, everything else at reservoir conditions); in oilfield units, STB, scf and rb

Holds when

the assumptions of the general equation: the accumulation is one “tank” with a single average pressure, the temperature is constant, and the fluid properties are described by the black oil model (the oil is treated as two components, stock-tank oil and gas)
there is no injection; injected volumes $W_{inj}B_w + G_{inj}B_g$ are subtracted from $F$

On the left is what we calculate from the measurements of production and pressure. On the right are the unknowns ( $N$ , $m$ , $W_e$ ), multiplied by quantities that we calculate from the PVT properties at the same pressure. The method is to divide the equation by a suitable group, so that we get a straight line with the unknowns in its slope and intercept.

Terms of the short notation

\begin{aligned} F &= N_p\big[B_o + (R_p - R_s)B_g\big] + W_pB_w\\ E_o &= (B_o - B_{oi}) + (R_{si} - R_s)\,B_g\\ E_g &= B_{oi}\left(\frac{B_g}{B_{gi}} - 1\right)\\ E_{fw} &= (1 + m)\,E_{fw0},\qquad E_{fw0} = B_{oi}\,\frac{c_wS_w + c_f}{1 - S_w}\,\Delta p\end{aligned}

where:

$N_p,\ W_p$ cumulative production of oil and water at standard conditions, m³
$R_p$ cumulative gas–oil ratio: all the gas produced per all the oil produced, m³/m³
$B_o,\ R_s,\ B_g,\ B_w$ formation volume factors of the oil, gas and water, and solution gas–oil ratio, at the current average pressure, m³/m³
$B_{oi},\ R_{si},\ B_{gi}$ the same properties at the initial pressure, m³/m³
$E_{fw0}$ expansion of the connate water and shrinking of the pore volume for $m = 0$ , that is, for the oil zone alone, m³/m³
$c_w,\ c_f,\ S_w$ compressibility of the water and of the pore volume, connate water saturation, 1/MPa, 1/MPa, —
$\Delta p$ fall in the average reservoir pressure from its initial value, $p_i - p$ , MPa

Units: $F$ in m³; the expansions in m³ per 1 m³ of oil in place; pressure and compressibilities in MPa and 1/MPa (or psi and 1/psi)

Holds when

all the properties are taken at the same average pressure $p$
the connate water saturation and the pore compressibility in the gas cap are the same as in the oil zone

The derivation of each term is in the lesson General material balance equation for oil reservoirs. One thing is new here. The term for the water and the pores is proportional to the pore volume of the whole accumulation, so it is split between the oil zone ( $E_{fw0}$ ) and the cap ( $mE_{fw0}$ ). This gives the exact grouping $F = N(E_o + E_{fw0}) + mN(E_g + E_{fw0}) + W_e$ , in which each zone has its own water and pores.

The equation links four groups of quantities: the withdrawal, the expansions, the unknown oil in place with the cap size, and the water influx. For one date this is a single equation, and it can give only one unknown, provided the others are known. But over the years of development there are many pressure measurements, and each gives its own equation with the same $N$ and $m$ . If we divide the equation by a suitable group, it becomes $y = a + bx$ , where $y$ and $x$ are calculated for each date, and the unknowns stand in the intercept $a$ and the slope $b$ .

Which coordinates to choose

Accumulation: no cap and no water
$y$ axis: $F$
$x$ axis: $E_o + E_{fw}$
Slope: $N$
Intercept: 0

Accumulation: no cap, check for influx
$y$ axis: $F/(E_o + E_{fw})$
$x$ axis: $N_p$ or time
Slope: —
Intercept: $N$

Accumulation: cap of known size
$y$ axis: $F$
$x$ axis: $E_o + mE_g + E_{fw}$
Slope: $N$
Intercept: 0

Accumulation: cap of unknown size
$y$ axis: $F/E_o$
$x$ axis: $E_g/E_o$
Slope: $mN$
Intercept: $N$

Accumulation: aquifer
$y$ axis: $F/E_t$
$x$ axis: influx function divided by $E_t$
Slope: aquifer parameter
Intercept: $N$

Accumulation	$y$ axis	$x$ axis	Slope	Intercept
no cap and no water	$F$	$E_o + E_{fw}$	$N$	0
no cap, check for influx	$F/(E_o + E_{fw})$	$N_p$ or time	—	$N$
cap of known size	$F$	$E_o + mE_g + E_{fw}$	$N$	0
cap of unknown size	$F/E_o$	$E_g/E_o$	$mN$	$N$
aquifer	$F/E_t$	influx function divided by $E_t$	aquifer parameter	$N$

$E_t = E_o + mE_g + E_{fw}$ is the total expansion per 1 m³ of oil in place. In the row for a cap of unknown size the connate water and the pores are left out; the section on the gas cap shows how to include them. The influx function, used in the row for an aquifer, is the part of the influx that we calculate from the pressure history (see the section on the aquifer).

Order of work

1
Collect the data for the dates of measurement
The average reservoir pressure brought to one depth, the cumulative production of oil, gas and water, and the injection.
2
Find the fluid properties
$B_o$ , $R_s$ , $B_g$ at each pressure, from PVT studies adjusted to the conditions in the field separators (the surface vessels that split gas from oil) (Black oil model: solution gas–oil ratio and formation volume factors).
3
Calculate the withdrawal and the expansions
$F$ , $E_o$ , $E_g$ and $E_{fw}$ for each date.
4
Choose the coordinates
According to the sources of energy we expect, using the table above.
5
Plot the chart and judge its shape
A straight line means that the model agrees with the history, and the slope and intercept give the parameters. A bend or a scatter means that the model is incomplete or the data are wrong.
6
Check the conclusion
Compare the oil in place with the volumetric estimate, the cap size with the geological data, and the water influx with the wells starting to produce water. Calculate the drive indices.

Accumulation without a gas cap: is there water influx?

If there is no gas cap ( $m = 0$ , and then $E_{fw} = E_{fw0}$ ), the equation becomes $F = N(E_o + E_{fw}) + W_e$ . For a closed accumulation $W_e = 0$ , and the points “ $F$ against $E_o + E_{fw}$ ” lie on a straight line through the origin with slope $N$ . An example with two dates is in the lesson on material balance. It is more convenient, however, to divide both sides by the expansion.

Apparent oil in place of an accumulation without a gas cap

\frac{F}{E_o + E_{fw}} = N + \frac{W_e}{E_o + E_{fw}}

where:

$F/(E_o + E_{fw})$ apparent oil in place: the oil in place calculated from one date, on the assumption that there is no water influx, m³
$N$ oil in place, m³
$W_e$ cumulative water influx by the same date, m³
$E_o,\ E_{fw}$ expansion of the oil with its gas, and of the connate water with the pores, per 1 m³ of oil in place, m³/m³

Units: m³ (in oilfield units, STB)

Holds when

there is no gas cap; there is no injection, or it has been subtracted from $F$
the pressure has fallen enough for its error to be small compared with $\Delta p$

We calculate the left side for each date and plot it against cumulative production or time. If there is no water, the second term is zero, and the points lie on a horizontal line at the level $N$ . If water comes in, a term that grows with the withdrawal is added to the oil in place, and the curve rises. Extending the curve back to zero production points to $N$ .

A horizontal line at the level of the volumetric estimate. The accumulation is closed, and all its oil is connected to the wells.
A rise. Water is coming in from the aquifer. With a large aquifer the curve keeps rising. With a small one the rise slows down, because the elastic store of the aquifer (the water it can give up by expanding) is used up. Below the bubble point pressure the expansion of the oil with its gas grows faster than the influx, and the curve can even turn downward (Aquifer models: water influx).
An almost horizontal line well above the volumetric estimate. There is a source of energy that changes with pressure in about the same way as the expansion of the oil itself. It may be a gas cap that has not been counted (for this lesson's oil, such a curve slowly falls), or, above the bubble point pressure, a small aquifer. Or the volumetric estimate is too low.
A steady fall around the volumetric estimate and below it cannot be explained by a source of energy. Usually it comes from errors in the pressure measurements or in the production records, or from fluids flowing out into neighbouring layers.

CalculatedThe apparent oil in place

F/(E_o + E_{fw})

, as a fraction of the volumetric estimate, against the fraction of the oil in place that has been produced, for three assumed accumulations in this lesson: a closed one (the dots are the examples of the lesson on material balance), one with an aquifer, and one with a gas cap that was “forgotten” in the calculation. Water influx lifts the curve. A gas cap that was not counted keeps it almost horizontal, but 2.6 to 2.9 times above one.

Accumulation with a gas cap

In an accumulation with a gas cap, the oil at the gas–oil contact (the boundary between the gas and the oil) is saturated with gas, and the initial pressure there equals the bubble point pressure. There are two unknowns, the oil in place $N$ and the gas cap ratio $m$ , and there are two ways to build a straight line.

Accumulation with a gas cap of known size

F = N\,\big[(E_o + E_{fw0}) + m\,(E_g + E_{fw0})\big]

where:

$F$ underground withdrawal, m³
$N$ oil in place, m³
$m$ gas cap ratio, known from geological data, —
$E_o,\ E_g,\ E_{fw0}$ expansion of the oil with its gas, of the cap gas (for $m = 1$ ) and of the connate water with the pores (for $m = 0$ ), per 1 m³ of oil in place, m³/m³

Units: m³

Holds when

there is no water influx
one formation volume factor $B_g$ describes both the cap gas and the gas that came out of the oil; the cap is at the same average pressure as the oil zone

If $m$ is right, the points “ $F$ against the square bracket” lie on a straight line through the origin with slope $N$ . The oil in place calculated for each date then does not depend on the date. If $m$ is wrong, the points drift systematically away from the line: the oil in place calculated for each date drifts. If we neglect the water and the pores, we get the form $F = N(E_o + mE_g)$ .

The size of the cap is found from geological data: from the positions of the gas–oil and oil–water contacts, the maps of net thickness (the thickness of the productive rock) and well logs. $m$ is the ratio of the initial reservoir volume of the cap gas to the initial reservoir volume of the oil. In standard cubic metres, the cap holds $G = mNB_{oi}/B_{gi}$ of gas. If there are no reliable geological data, we can try to find both unknowns from the balance itself, by dividing both its sides by $E_o + E_{fw0}$ .

Accumulation with a gas cap of unknown size

\frac{F}{E_o + E_{fw0}} = N + mN\,\frac{E_g + E_{fw0}}{E_o + E_{fw0}}

where:

$F/(E_o + E_{fw0})$ ordinate (vertical coordinate) of the point, m³
$(E_g + E_{fw0})/(E_o + E_{fw0})$ abscissa (horizontal coordinate) of the point, —
$N$ intercept on the vertical axis: the oil in place, m³
$mN$ slope: the initial reservoir volume of the cap divided by $B_{oi}$ , m³

Units: m³ (in oilfield units, STB)

Holds when

there is no water influx
the pressure drop is large compared with the error of its measurement

If we neglect the water and the pores, this is the familiar “ $F/E_o$ against $E_g/E_o$ ” form. The intercept is the oil in place, and the slope divided by the intercept is the cap size $m$ . If the points do not lie on a straight line, water is most likely entering the accumulation.

In other unit systems (1)

In terms of the cap gas at standard conditions

\frac{F}{E_o} = N + G\,\frac{B_g - B_{gi}}{E_o}

$G = mNB_{oi}/B_{gi}$ is the cap gas, in m³ at standard conditions; $B_g - B_{gi}$ is the expansion of 1 m³ of this gas, m³/m³; the connate water and the pores are left out.

There are no constants: $mNE_g = mNB_{oi}(B_g/B_{gi} - 1) = G\,(B_g - B_{gi})$ . The slope in this form is $G$ , not $mN$ . It is important not to mix the two definitions of the gas expansion.

Why N and m are hard to tell apart

The expansion of the oil with its gas and the expansion of the cap gas grow in about the same way as the pressure falls. Both depend mainly on the expansion of gas: the gas that came out of the oil and the cap gas. So the ratio $E_g/E_o$ changes little over the years of development, and the points crowd into a narrow band along the $x$ axis. We find the intercept $N$ by extending the line to $x = 0$ , which is several times farther than the width of the band, and small errors in the pressure shift it strongly.
What is found reliably is something else: the value of $F/(E_o + E_{fw0})$ in the middle of the cloud of points, that is, the sum $N(1 + m\bar x)$ : the oil in place together with the “weight” of the cap. So in practice the size of the cap is taken from geological data, and the balance checks whether the development history agrees with them.

Development history of an assumed accumulation with a gas cap

Year: Year 1
$p$ , MPa: 17.47
$N_p$ , thousand m³: 150
$R_p$ , m³/m³: 112

Year: Year 2
$p$ , MPa: 16.81
$N_p$ , thousand m³: 340
$R_p$ , m³/m³: 118

Year: Year 3
$p$ , MPa: 16.04
$N_p$ , thousand m³: 560
$R_p$ , m³/m³: 128

Year: Year 4
$p$ , MPa: 15.21
$N_p$ , thousand m³: 790
$R_p$ , m³/m³: 140

Year: Year 5
$p$ , MPa: 14.32
$N_p$ , thousand m³: 1020
$R_p$ , m³/m³: 155

Year: Year 6
$p$ , MPa: 13.44
$N_p$ , thousand m³: 1240
$R_p$ , m³/m³: 170

Year	$p$ , MPa	$N_p$ , thousand m³	$R_p$ , m³/m³
Year 1	17.47	150	112
Year 2	16.81	340	118
Year 3	16.04	560	128
Year 4	15.21	790	140
Year 5	14.32	1020	155
Year 6	13.44	1240	170

The same oil as in the lesson Black oil model: solution gas–oil ratio and formation volume factors: bubble point pressure 18 MPa, initial solution gas–oil ratio 110 m³/m³. The initial pressure at the gas–oil contact equals the bubble point pressure; there is no water influx and no water is produced. $R_p$ is the cumulative gas–oil ratio.

Worked exampleOil in place and the size of the gas cap

Given (assumed data)

History: the table above: pressure, cumulative oil production and cumulative gas–oil ratio by year
Initial properties: $B_{oi} = 1.35$ m³/m³; $R_{si} = 110$ m³/m³; $B_{gi} = 0.005983$ m³/m³
Water and pores: $S_w = 0.25$ ; $c_w = 0.45\cdot10^{-3}$ and $c_f = 0.8\cdot10^{-3}$ MPa⁻¹, so $(c_wS_w + c_f)/(1 - S_w) = 1.2167\cdot10^{-3}$ MPa⁻¹
Geological data: the volumetric estimate of the oil zone is about 6.5 million m³; the cap size from the contacts and the maps is $m \approx 0.5$

Solution

Year 6, $p = 13.44$ MPa. The properties at this pressure: $R_s = 84.57$ m³/m³, $B_o = 1.2806$ m³/m³, $B_g = 0.008013$ m³/m³. Withdrawal: $F = 1240\cdot\big[1.2806 + (170 - 84.57)\cdot0.008013\big] = 2437$ thousand m³.
Expansions per cubic metre of oil in place: $E_o = (1.2806 - 1.35) + (110 - 84.57)\cdot0.008013 = 0.1344$ ; $E_g = 1.35\cdot(0.008013/0.005983 - 1) = 0.4580$ ; $E_{fw0} = 1.35\cdot1.2167\cdot10^{-3}\cdot(18 - 13.44) = 0.00749$ m³/m³.
If we assume there is no cap: $F/(E_o + E_{fw0}) = 17.2$ million m³, which is 2.6 times the volumetric estimate for the oil zone. Without a gas cap, the history cannot be explained.
With the cap from the geological data, $m = 0.5$ : $N = F/\big[(E_o + E_{fw0}) + 0.5\,(E_g + E_{fw0})\big] = 6.50$ million m³. Over the six dates it ranges from 6.47 to 6.50 million m³ (the table after the example). The oil in place hardly depends on the date, and it agrees with the volumetric estimate.
Fitting both unknowns together, without geological data. For each date we calculate $x = (E_g + E_{fw0})/(E_o + E_{fw0})$ and $y = F/(E_o + E_{fw0})$ . For year 6, $x = 3.280$ and $y = 17.17$ million m³. The straight line through the six points (by least squares, which makes the sum of the squared distances of the points from the line as small as possible) has intercept $N = 7.1$ million m³ and slope $mN = 3.07$ million m³, so $m = 0.43$ .
The quality of the two descriptions: the points deviate from the fitted line by no more than 0.16%, and from the line with $N = 6.5$ million m³ and $m = 0.5$ by no more than 0.5%. Both descriptions are within the accuracy of the input data.
Sensitivity: if the pressures were known only to 0.1 MPa (17.5; 16.8; 16.0; 15.2; 14.3; 13.4 MPa), the same fit would give $N = 1.6$ million m³ and $m = 2.9$ . The first point shifts the result most: the pressure drop of 0.53 MPa became 0.5 MPa, an error of 6%, and its coordinates are divided by the smallest expansion. But even without it the fit gives $N = 7.6$ million m³ and $m = 0.38$ .

Answer: With a cap size of 0.5 from the geological data, the oil in place is about 6.5 million m³, and the history confirms it. Without geological data, the balance is described equally well by the pair 6.5 million m³ and 0.5, and by the pair 7.1 million m³ and 0.43.

What it means. The abscissas of the six points change only from 3.835 to 3.280, because the expansion of the oil and the expansion of the cap gas grow almost in proportion. We find the intercept by extending the line about 6 times farther than the width of the cloud of points, so a small change in the slope shifts it noticeably. What is found reliably is the average ordinate of the cloud, $N(1 + m\bar x) \approx 18.0$ million m³: both pairs give it to within a fraction of a per cent. That is why the size of the cap is taken from geological data, and the balance checks that they agree with the history. $N$ and $m$ can be separated by the balance alone only with a long history, a large pressure drop and very accurate pressure measurements. The data in this example are assumed for illustration: the history was calculated with the balance equation for 6.5 million m³ of oil and $m = 0.5$ , and the pressures were rounded to 0.01 MPa.

CalculatedSix measurements of an accumulation with a gas cap in the coordinates

x = (E_g + E_{fw0})/(E_o + E_{fw0})

y = F/(E_o + E_{fw0})

, and two straight lines that describe them to within a fraction of a per cent: one fitted to the points and one that matches the geological data (

m = 0.5

). The points occupy a narrow band of abscissas, and the intercepts, for which the chart is drawn, lie far outside it.

Oil in place for each date at different cap sizes, million m³

Year: Year 1
$m = 0.3$: 8.78
$m = 0.5$: 6.47
$m = 0.7$: 5.12

Year: Year 2
$m = 0.3$: 8.75
$m = 0.5$: 6.47
$m = 0.7$: 5.13

Year: Year 3
$m = 0.3$: 8.75
$m = 0.5$: 6.50
$m = 0.7$: 5.16

Year: Year 4
$m = 0.3$: 8.72
$m = 0.5$: 6.50
$m = 0.7$: 5.18

Year: Year 5
$m = 0.3$: 8.68
$m = 0.5$: 6.49
$m = 0.7$: 5.19

Year: Year 6
$m = 0.3$: 8.65
$m = 0.5$: 6.50
$m = 0.7$: 5.21

Year	$m = 0.3$	$m = 0.5$	$m = 0.7$
Year 1	8.78	6.47	5.12
Year 2	8.75	6.47	5.13
Year 3	8.75	6.50	5.16
Year 4	8.72	6.50	5.18
Year 5	8.68	6.49	5.19
Year 6	8.65	6.50	5.21

$N = F/\big[(E_o + E_{fw0}) + m(E_g + E_{fw0})\big]$ for each date. If $m$ is right, the oil in place does not depend on the date. If $m$ is too low, it decreases from date to date, and if $m$ is too high, it grows. The direction of the drift here is set by this lesson's oil, whose $E_g/E_o$ falls as the pressure falls. For another oil it may be the opposite, but the warning sign is the same: the result depends on the date.

Accumulation with an aquifer

The water influx $W_e$ is not a number but a function of time. To get a straight line, we express it through an aquifer model (Aquifer models: water influx). All simple models have the form “an aquifer parameter multiplied by a function of the pressure history at the contact”. For a small aquifer (one that equalises its pressure with the accumulation at once) the function is the pressure drop $\Delta p$ itself. For the steady-state model (the pressure at the far edge of the aquifer stays constant) it is the integral of the pressure drop over time. For the unsteady-state model (the pressure drop spreads through the aquifer gradually) it is the sum of the pressure steps, weighted by the dimensionless water influx (Unsteady-state water influx: the van Everdingen–Hurst method). We calculate the function from the pressure history, and the parameter stays unknown, like the oil in place.

Straight line for an accumulation with an aquifer

\frac{F}{E_t} = N + C\,\frac{w(t)}{E_t},\qquad E_t = E_o + m\,E_g + E_{fw}

where:

$E_t$ total expansion per 1 m³ of oil in place, for a known $m$ (for $m = 0$ it is $E_o + E_{fw}$ ), m³/m³
$C$ parameter of the aquifer model: the slope of the line, depend on the model
$w(t)$ influx function: the part of the model calculated from the pressure history, $W_e = C\,w(t)$ , depend on the model
$N,\ F$ oil in place (the intercept of the line) and the underground withdrawal, m³

Units: m³; the units of $C$ and $w$ are such that their product is in m³ at reservoir conditions

Holds when

the type of the aquifer model and its geometry are chosen in advance, and only the parameter $C$ is unknown
the pressure at the contact with the aquifer is taken as equal to the average pressure of the accumulation
the gas cap ratio is known, or there is no gas cap

If the model is right, the points lie on a straight line: the intercept is the oil in place and the slope is the aquifer parameter. If the model is wrong, no coefficient will straighten the chart. Its shape suggests in which direction to change the model.

In other unit systems (2)

Steady-state aquifer model

\frac{F}{E_t} = N + J_a\,\frac{1}{E_t}\int_0^t (p_i - p)\dd t

$J_a$ is the aquifer productivity index, in m³/(d·MPa) at reservoir conditions; the integral is in MPa·d; $p$ is the pressure at the contact.

There are no constants. We calculate the integral from the measurements by the trapezoid rule: for each interval, the average of the pressure drop at its two ends, multiplied by the length of the interval.

Small aquifer

\frac{F}{E_t} = N + (c_w + c_f)\,W\,\frac{p_i - p}{E_t}

$W$ is the pore volume of the aquifer, m³; $c_w + c_f$ is the compressibility of the water and the pores of the aquifer, 1/MPa; pressure in MPa.

There are no constants. Above the bubble point pressure $E_t$ is almost proportional to $\Delta p$ , the abscissa does not change from date to date, and such an aquifer cannot be detected with the chart.

A straight line. The aquifer model agrees with the history; the intercept is $N$ and the slope is the aquifer parameter.
Late points above the line drawn through the early ones. The real influx grows faster than the chosen function: the aquifer is larger, or lets water through more easily, than the model assumes.
Late points below the line. The influx grows more slowly. The aquifer is smaller, its pressure falls with time, and it gives up water more and more slowly.
An S-shaped curve. The geometry is wrong, for example a radial aquifer (one that surrounds the accumulation) instead of a linear one (one that lies on one side of it).
A random scatter. There are errors in the data: in the pressure, the PVT properties or the production records.
The points move “backwards” along the $x$ axis. This is normal for the small-aquifer model below the bubble point pressure: its abscissa $\Delta p/E_t$ gets smaller, because $E_t$ grows faster than $\Delta p$ . Above the bubble point pressure this abscissa is almost constant, and the points gather in a column.

Accumulation with an aquifer: coordinates of the line for the steady-state model

Year: Year 1
$F/E_t$ , million m³: 26.06
$\int(p_i - p)\dd t$ , MPa·d: 547.9
$x$ , thousand MPa·d: 48.77

Year: Year 2
$F/E_t$ , million m³: 46.47
$\int(p_i - p)\dd t$ , MPa·d: 2008.9
$x$ , thousand MPa·d: 107.10

Year: Year 3
$F/E_t$ , million m³: 70.20
$\int(p_i - p)\dd t$ , MPa·d: 4054.3
$x$ , thousand MPa·d: 174.83

Year: Year 4
$F/E_t$ , million m³: 95.44
$\int(p_i - p)\dd t$ , MPa·d: 6464.9
$x$ , thousand MPa·d: 246.95

Year	$F/E_t$ , million m³	$\int(p_i - p)\dd t$ , MPa·d	$x$ , thousand MPa·d
Year 1	26.06	547.9	48.77
Year 2	46.47	2008.9	107.10
Year 3	70.20	4054.3	174.83
Year 4	95.44	6464.9	246.95

The history is from the lesson General material balance equation for oil reservoirs: an oil zone with an initial pressure of 30 MPa, with the pressure above the bubble point pressure all the time. $x = \int(p_i - p)\dd t/E_t$ ; the integral is taken by the trapezoid rule between the yearly pressure measurements.

Worked exampleOil in place and the aquifer from one history

Given (assumed data)

History: the pressure and production over four years from the lesson General material balance equation for oil reservoirs ( $p$ = 27.0; 25.0; 23.8; 23.0 MPa, with an initial pressure of 30 MPa)
Properties: $B_{oi} = 1.3243$ ; $B_w = 1.02$ m³/m³; $(c_wS_w + c_f)/(1 - S_w) = 1.2167\cdot10^{-3}$ MPa⁻¹; no gas cap
Oil in place: taken as unknown; the volumetric estimate, 9.06 million m³, is needed only for comparison

Solution

The withdrawal and the expansion for each year are calculated as in the lesson General material balance equation for oil reservoirs. In year 4, $F = 2498.5$ thousand m³ and $E_t = E_o + E_{fw} = 0.02618$ m³/m³. The apparent oil in place $F/E_t$ by year is 26.06, 46.47, 70.20 and 95.44 million m³. It grows, so water is entering the accumulation.
Small-aquifer model: the abscissa $\Delta p/E_t$ by year is 267, 267, 267 and 267 MPa, practically the same, while the ordinates differ almost four times. We cannot draw a straight line through such points: the influx grows not only with the pressure drop but also with time.
Steady-state model. We take the integral of the pressure drop by the trapezoid rule, with a year of 365.25 days. For the first year, $\tfrac{0 + 3}{2}\cdot365.25 = 547.9$ MPa·d. By the end of the second year, $547.9 + \tfrac{3 + 5}{2}\cdot365.25 = 2008.9$ ; by the end of the third, 4054.3; by the end of the fourth, 6464.9 MPa·d.
The abscissas $x = \int(p_i - p)\dd t/E_t$ are 48.77, 107.10, 174.83 and 246.95 thousand MPa·d. The straight line through the four points (least squares) has slope $J_a = 350.1$ m³/(d·MPa) and intercept $N = 8.98$ million m³.
A check with the two end points: $J_a = (95.44 - 26.06)\cdot10^6/\big[(246.95 - 48.77)\cdot10^3\big] = 350.1$ m³/(d·MPa); $N = 26.06 - 350.1\cdot48.77\cdot10^{-3} = 8.99$ million m³.
Water influx by the end of year 4: $W_e = J_a\int(p_i - p)\dd t = 350.1\cdot6464.9 = 2263$ thousand m³.

Answer: The oil in place is about 8.98 million m³, the aquifer productivity index is about 350 m³/(d·MPa), and in four years 2263 thousand m³ of water entered the accumulation.

What it means. The volumetric estimate is 9.06 million m³. The difference of 0.9% comes from the rounding of the input data (the pressures to 0.1 MPa, the oil formation volume factor to four digits) and from integrating with yearly trapezoids. In the lesson on material balance the water influx was found by putting in the volumetric oil in place. Here the same quantities come from one history, without a volumetric estimate, but at the price of assuming a model of the aquifer. The line confirms the steady-state model only over these four years. If the aquifer is finite, its pressure will fall with time, and the late points will lie below the line. So we redraw the chart as new measurements arrive. The data in this example are assumed for illustration: the history was calculated with a coefficient of 350 m³/(d·MPa) and rounded.

CalculatedThe four yearly points of the accumulation with an aquifer, in the coordinates of the steady-state model, and the straight line drawn through them. The intercept on the vertical axis is the oil in place, and the slope is the aquifer productivity index

J_a

Drive indices from the development history

When the model of the accumulation agrees with the history, we can split the withdrawal into the contributions of the sources of energy. We divide by the reservoir volume of the oil and gas produced, $H = N_p[B_o + (R_p - R_s)B_g]$ . From the water influx we subtract the water produced, because only the water that stayed in the accumulation takes up space in it (drive indices).

Drive indices and their sum

\underbrace{\frac{NE_o}{H}}_{\text{DDI}} + \underbrace{\frac{NmE_g}{H}}_{\text{SDI}} + \underbrace{\frac{NE_{fw}}{H}}_{\text{EDI}} + \underbrace{\frac{W_e - W_pB_w}{H}}_{\text{WDI}} = 1

where:

$\text{DDI}$ depletion drive index: the share of the expansion of the oil with the gas dissolved in it (solution gas drive, and above the bubble point pressure the elastic expansion of the oil), —
$\text{SDI}$ segregation drive index: the share of the gas cap drive, —
$\text{EDI}$ expansion drive index: the share of the expansion of the connate water and the shrinking of the pore volume, —
$\text{WDI}$ water drive index: the share of the water drive, that is, of the water that has come in, less the water produced, —
$H$ reservoir volume of the oil and gas produced, $F - W_pB_w$ , m³

Units: dimensionless fractions; all volumes in the same units

Holds when

all the quantities belong to one date
$N$ , $m$ and the water influx have been found from the chart or are given

The sum of the left side is $\big[N(E_o + mE_g + E_{fw}) + W_e - W_pB_w\big]/H$ . It equals one exactly when the point lies on the line. If a point deviates from the line by 0.5%, the sum differs from one by about the same amount. So the sum of the indices is also a check of how well the model describes each date.

Worked exampleHow the drive of two accumulations changed

Given (assumed data)

Accumulation with an aquifer: $N = 8.98$ million m³ and $J_a = 350.1$ m³/(d·MPa) from the previous example; no gas cap
Accumulation with a gas cap: $N = 6.5$ million m³, $m = 0.5$ ; no water influx

Solution

Aquifer, year 1: $H = N_pB_o = 292.8$ thousand m³. The expansion of the oil is $NE_o = 57.5$ thousand m³, the water and pores $NE_{fw} = 43.4$ thousand m³, and the water influx $J_a\cdot547.9 = 191.8$ thousand m³; no water was produced. Indices: DDI = 0.196, EDI = 0.148, WDI = 0.655; the sum is 1.000.
Year 4: $H = 2437.3$ thousand m³; the influx is 2263.4 thousand m³, and the water produced is $60\cdot1.02 = 61.2$ thousand m³. Indices: DDI = 0.055, EDI = 0.042, WDI = 0.904; the sum is 1.000.
Gas cap, year 6: $H = F = 2436.9$ thousand m³; oil $NE_o = 873.8$ , cap $NmE_g = 1488.6$ , water and pores $N(1 + m)E_{fw0} = 73.0$ thousand m³. Indices: DDI = 0.359, SDI = 0.611, EDI = 0.030; the sum is 0.999.
Gas cap, year 1: the sum of the indices is 1.005. The point of year 1 lies below the line by about 0.5%. With a pressure drop of only 0.53 MPa, rounding the pressure to 0.01 MPa is already noticeable.

Answer: In the accumulation with an aquifer, the share of aquifer water grew in four years from 66% to 90%. In the accumulation with a gas cap, the cap supplies about 61% of the withdrawal.

What it means. The indices show how the drive changes with time. The share of aquifer water grows. In the steady-state model the influx keeps accumulating as long as the pressure at the contact is lowered. The elastic expansion of the oil, the water and the pores grows only together with the pressure drop. In the accumulation with a cap, the share of the oil slowly grows, because below the bubble point pressure the expansion of the oil with the gas that comes out speeds up. The indices do not say how the water or the gas will spread through the reservoir, but they show which source of energy to save. In the first accumulation the point is not to take oil out faster than the water comes in. In the second, the point is not to waste the gas of the cap. The data in this example are assumed for illustration.

CalculatedDrive indices by year for the two accumulations in this lesson: one with an aquifer (the oil in place and the aquifer coefficient are taken from the line in the example) and one with a gas cap (

N = 6.5

million m³,

m = 0.5

). In the legend, DDI is the oil, SDI the gas cap, EDI the water with the pores, and WDI the aquifer water. The length of each bar is the sum of the indices; it differs from one by no more than half a per cent.

Data, checks and limitations

The average pressure decides everything. The expansions are small differences between formation volume factors at two pressures, so an error in the pressure passes into the coordinates many times magnified. The early points are especially vulnerable, because their coordinates are divided by a small expansion. We do not drop them silently, but we do not let them decide the slope either.
PVT properties. The formation volume factors and the solution gas–oil ratio must match the way gas comes out of the oil in the reservoir and the conditions in the field separators (Laboratory PVT tests). A systematic error in $B_o$ or $R_s$ bends the chart in the same way as a wrong model.
Complete records of withdrawal and injection. We need all the gas produced and all the water. Injected water and gas are subtracted from $F$ . Gas used for gas lift is removed from the production figures.
Ambiguity. The more parameters an aquifer model has, the easier it is to “straighten” any chart, and the less the straightening means. We prefer the simplest model that agrees with the geology, and we review it with every new measurement.
No time. The chart makes the withdrawal and the pressure agree, but does not say when a given pressure will be reached. To forecast in time, we combine the balance with a calculation of well rates and water influx (Performance prediction for solution gas drive reservoirs, Aquifer models: water influx).

Terms and symbols

Havlena–Odeh method: A way of writing the material balance equation as a straight line. The slope and intercept of the line give the oil in place, the gas cap size and the aquifer parameter.
$F$ , $E_o$ , $E_g$ , $E_{fw}$: The underground withdrawal is $F$ . The expansions of the oil with its gas, of the gas cap, and of the connate water with the pores, per 1 m³ of oil in place, are $E_o$ , $E_g$ and $E_{fw}$ .
$E_{fw0}$: The expansion of the connate water and the shrinking of the pore volume for $m = 0$ . With a gas cap, $E_{fw} = (1 + m)E_{fw0}$ .
Apparent oil in place: Oil in place calculated from one date on the assumption that there are no extra sources of energy. Its growth with time points to water influx.
Influx function $w(t)$: The part of the aquifer model that is calculated from the pressure history: $W_e = C\,w(t)$ , where $C$ is the aquifer parameter.
Aquifer productivity index $J_a$: The parameter of the steady-state model: the water influx per day for each 1 MPa of pressure drop at the contact.
Drive indices (DDI, SDI, EDI, WDI): The shares of the reservoir volume of the oil and gas produced that are supplied by the expansion of the oil, the gas cap, the water with the pores, and the aquifer water.

Common mistakes and misconceptions

Wrong

“The points lie on a straight line, so the model of the accumulation is right.”

In fact

A straight line is needed, but it is not enough. Above the bubble point pressure, a small aquifer and a larger closed accumulation give the same line. The joint fit of the oil in place and the cap in this lesson's example gave two different pairs with equally good agreement. We check the conclusion against the volumetric estimate, the geology and the wells starting to produce water.

Wrong

“From the chart of F/Eo against Eg/Eo we can find both the oil in place and the gas cap size.”

In fact

Only with a long history and very accurate pressure. The abscissas of the points change little, the intercept comes from a long extrapolation, and rounding the pressure to 0.1 MPa in this lesson's example turned 6.5 million m³ into 1.6, and, without the most sensitive, first point, into 7.6. The cap size is taken from geological data, and the balance serves as a check.

Wrong

“The points have bent, so we need to pick another aquifer coefficient.”

In fact

The coefficient changes the slope of the line, not its shape. A bend means that the aquifer model itself is wrong: its size, its geometry or its type (small, steady-state, unsteady-state).

Wrong

“The first measurements are the most reliable: the accumulation is still untouched.”

In fact

The opposite is true. While the pressure has fallen only a little, its error is about the same size as the fall, and the coordinates are divided by a small expansion. The first points are the ones that most often deviate from the line more than the others.

Wrong

“The sum of the drive indices is not one, so there is an arithmetic error.”

In fact

The sum is one only if the point lies on the line. Its difference from one is the relative deviation of the point from the model, and this is a useful check.

In depth: advanced detail

The same line for a gas accumulation

For a gas accumulation with an aquifer, the withdrawal $F = G_pB_g + W_pB_w$ equals the expansion of the gas, $G(B_g - B_{gi})$ , plus the water influx. The expansion of the connate water and the pores is usually neglected (Material balance for gas reservoirs). Dividing by the expansion of one cubic metre of gas gives a straight line of the same kind.

Straight line for a gas accumulation with a steady-state aquifer

\frac{G_pB_g + W_pB_w}{B_g - B_{gi}} = G + J_a\,\frac{1}{B_g - B_{gi}}\int_0^t (p_i - p)\dd t

where:

$G,\ G_p$ gas in place (the intercept of the line) and cumulative gas production at standard conditions, m³
$B_g,\ B_{gi}$ gas formation volume factor at the current and the initial pressure, m³/m³
$W_p,\ B_w$ water produced and its formation volume factor, m³, m³/m³
$J_a$ aquifer productivity index: the slope of the line, m³/(d·MPa)
$p_i,\ p$ initial and current pressure at the contact (the average reservoir pressure), absolute, MPa

Units: m³ at standard conditions; the integral in MPa·d

Holds when

a steady-state aquifer model
the expansion of the connate water and the pores is neglected; the gas in the reservoir is single-phase

The left side is the apparent gas in place $G_a$ of the lesson on gas material balance. There it grew from one measurement to the next, which was only a sign of water influx. Here its growth is explained quantitatively, and the gas in place is found together with the aquifer parameter. The expansion of one cubic metre of gas is written out, $B_g - B_{gi}$ , because the symbol $E_g$ in the oil balance means a different quantity.

Worked exampleGas accumulation: gas in place and the aquifer together

Given (assumed data)

History: the accumulation with an aquifer from the lesson Material balance for gas reservoirs: five yearly pressure measurements, withdrawal of 1.0 million m³/d, no water produced
Initial conditions: $p_i = 20$ MPa, $B_{gi} = 0.005302$ m³/m³; the gas in place is taken as unknown

Solution

For each year, $y = G_pB_g/(B_g - B_{gi})$ is the apparent gas in place. For example, in year 5, $p = 14.62$ MPa, $B_g = 0.007210$ m³/m³ and $y = 6.90$ billion m³; in year 1 it is 5.38 billion m³.
The integral of the pressure drop by the trapezoid rule: by the end of year 5 it is 5464 MPa·d. The abscissas $x = \int(p_i - p)\dd t/(B_g - B_{gi})$ range from 0.69 to 2.86 million MPa·d.
The straight line through the five points: intercept $G = 4.91$ billion m³, slope $J_a = 695.5$ m³/(d·MPa).

Answer: The gas in place is about 4.91 billion m³, against a volumetric estimate of 4.89, and the aquifer productivity index is about 696 m³/(d·MPa).

What it means. In the lesson on gas material balance the water influx was found from the volumetric gas in place. Here the gas in place and the aquifer are found together. The history there was calculated with a coefficient of 700 m³/(d·MPa) in monthly steps, so the yearly trapezoids and the rounding of the pressures to 0.01 MPa give a small difference. The data in this example are assumed for illustration.

Connections to other topics

General material balance equation for oil reservoirs — the derivation of the general equation, its terms and assumptions.
Material balance for gas reservoirs — the $p/Z$ line and the apparent gas in place of a gas accumulation.
Aquifer models: water influx — the aquifer models whose functions go on the horizontal axis.
Reservoir drive mechanisms — the sources of reservoir energy whose shares the drive indices show.
Black oil model: solution gas–oil ratio and formation volume factors — the formation volume factors and the solution gas–oil ratio from which the expansions are calculated.
Unsteady-state water influx: the van Everdingen–Hurst method — the unsteady-state influx model and its function for the chart.
History matching of aquifer parameters — fitting the size and the conductivity of the aquifer to the history.
Performance prediction for solution gas drive reservoirs — the forecast of the pressure and the gas–oil ratio from a matched balance.
History matching — the balance as a check of the simulation model.

Check yourself

1
The apparent oil in place of an accumulation without a gas cap, calculated for four years, is 9.1, 9.0, 9.1 and 9.0 million m³. The pressure stays above the bubble point pressure all the time, and the volumetric estimate is 6.0 million m³. What can this mean?
1. A.There is an error in the cumulative gas–oil ratio
2. B.The accumulation is closed and the volumetric estimate is right
3. C.Water is entering from a large aquifer
4. D.There is an extra source of energy that changes with pressure in the same way as the expansion of the oil, for example a small aquifer, or the volumetric estimate is too low
Show the answer
Correct answer: D. There is an extra source of energy that changes with pressure in the same way as the expansion of the oil, for example a small aquifer, or the volumetric estimate is too low
A constant apparent oil in place says that an extra source of energy, if there is one, is proportional to $\Delta p$ , like the expansion of undersaturated oil. A small aquifer above the bubble point pressure behaves in exactly this way: it raises the level of the line without changing its shape. A large aquifer would make the apparent oil in place grow with time. The geology of the area beyond the oil boundary helps to choose between an aquifer and an error in the volumetric estimate.
2
Two points of the chart $F/E_o$ against $E_g/E_o$ : $x = 4.0$ , $y = 15.0$ million m³ and $x = 3.0$ , $y = 12.5$ million m³. Find the oil in place and the gas cap size. How does the result change if the ordinate of the first point is 1% larger?
Show the answer
The slope is $mN = (15.0 - 12.5)/(4.0 - 3.0) = 2.5$ million m³, the intercept is $N = 12.5 - 2.5\cdot3.0 = 5.0$ million m³, and $m = 2.5/5.0 = 0.5$ . With $y_1 = 15.15$ the slope is 2.65, $N = 12.5 - 2.65\cdot3.0 = 4.55$ million m³ (9% less), and $m = 0.58$ . An error of 1% has grown nine times. When we extend the line to the vertical axis, the absolute error triples, from 0.15 to 0.45 million m³, because the intercept (at $x = 0$ ) lies three times farther from the points than the points lie from each other. And the intercept itself is three times smaller than the ordinate of the point.
3
An accumulation with an aquifer is processed with the steady-state model. The points of the first years lie on a straight line, and the points of the last years lie clearly below its continuation. What does this mean?
1. A.A gas cap has appeared in the accumulation
2. B.The real influx grows more slowly than the integral of the pressure drop: the aquifer is finite and its pressure is falling
3. C.The coefficient $J_a$ must be increased
4. D.The aquifer is larger than assumed
Show the answer
Correct answer: B. The real influx grows more slowly than the integral of the pressure drop: the aquifer is finite and its pressure is falling
The steady-state model assumes that the pressure at the outer boundary of the aquifer is constant, and that the influx is larger the longer the pressure has been lowered. A finite aquifer is depleted with time, and the real influx falls behind the model, so the late points lie lower. The coefficient only changes the slope. We need a different model: one of a finite aquifer, or the unsteady-state model.
4
On a certain date: $H = 2000$ thousand m³, the water produced is $W_pB_w = 100$ thousand m³; $NE_o = 300$ , $NmE_g = 500$ , $NE_{fw} = 60$ and $W_e = 1240$ thousand m³. Find the drive indices and their sum.
Show the answer
DDI = 300/2000 = 0.15; SDI = 500/2000 = 0.25; EDI = 60/2000 = 0.03; WDI = (1240 − 100)/2000 = 0.57. The sum is 1.00. The withdrawal $F = H + W_pB_w = 2100$ thousand m³ is exactly equal to the sum of the sources, $300 + 500 + 60 + 1240 = 2100$ thousand m³, so the point lies on the line. More than half of the withdrawal came from aquifer water, and a quarter from the gas cap.
5
Why is it hard to find the oil in place and the gas cap size together from one balance?
1. A.The expansion of the oil and the expansion of the cap gas grow almost in proportion as the pressure falls, so the points on the chart crowd into a narrow band and the intercept is found by a long extrapolation
2. B.The cap gas is not in the balance equation
3. C.The size of the cap changes with time
4. D.The balance does not work below the bubble point pressure
Show the answer
Correct answer: A. The expansion of the oil and the expansion of the cap gas grow almost in proportion as the pressure falls, so the points on the chart crowd into a narrow band and the intercept is found by a long extrapolation
The ratio $E_g/E_o$ changes little over the history: in this lesson's example, from 4.1 to 3.4 (the abscissa with the connate water and pores included goes from 3.8 to 3.3). The intercept on the vertical axis is far from the points, and pressure errors of a few hundredths of a megapascal shift it by tens of per cent. Only the combination $N(1 + m\bar x)$ is found reliably, so $m$ is taken from geological data.

Updated: 19 October 2026