General material balance equation for oil reservoirs

Quick refresher

We treat the accumulation as one “tank” with a single average pressure and compare two states: the initial one and the current one. Every term is expressed in reservoir volumes at the current pressure.
Withdrawal: $F = N_p[B_o + (R_p - R_s)B_g] + W_pB_w$ . It is the oil with its dissolved gas, the free gas and the water. $R_p$ is the cumulative gas–oil ratio, not the current one.
The sources of energy are: the expansion of the oil with its gas, $NE_o$ ; the expansion of the gas cap, $NmE_g$ ; the expansion of the connate water together with the shrinking of the pores, $NE_{fw}$ ; and the water influx, $W_e$ . The general equation is $F = N(E_o + mE_g + E_{fw}) + W_e$ .
Above the bubble point pressure (the pressure at which gas starts to come out of the oil) only the liquids and the pores expand. The water and the pores supply up to half of the energy, so we must not neglect them. Below the bubble point pressure, the gas that has come out of the oil quickly becomes the main source.
The equation has four groups of quantities: the oil in place, the production, the water influx and the pressure. If three of them are known, we can find the fourth: the oil in place, the water influx or the future pressure.
The method knows nothing about time, or about how pressure and saturation (the share of the pore space filled with each fluid) vary across the reservoir. It “sees” only the volume that is connected to the wells. It is sensitive to errors in the average pressure and in the PVT properties.

Key relationships

F = N_p\big[B_o + (R_p - R_s)B_g\big] + W_pB_w

underground withdrawal

E_o = (B_o - B_{oi}) + (R_{si} - R_s)B_g = B_t - B_{ti}

expansion of the oil with its gas per 1 m³ of oil in place

E_g = B_{oi}\left(\dfrac{B_g}{B_{gi}} - 1\right)

expansion of the gas cap per 1 m³ of oil in place for $m = 1$

E_{fw} = (1 + m)B_{oi}\,\dfrac{c_wS_w + c_f}{1 - S_w}\,\Delta p

expansion of the connate water and shrinking of the pore volume

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

general material balance equation

Common pitfalls

Putting the current gas–oil ratio in place of the cumulative one, $R_p$ .
Neglecting the compressibility of the water and the pores above the bubble point pressure. The oil in place is then too high by a factor of one and a half to two.
Taking the pressure of one well instead of the average pressure of the accumulation brought to one common depth.
Calculating the oil in place from the first months of production, when the pressure drop is about the same size as the error in measuring it.
Taking the oil in place from material balance as the full oil in place. Isolated blocks are not included in the calculation.

After this lesson you can

derive the general material balance equation by setting the underground withdrawal equal to the expansion of the fluids, the shrinking of the pore volume and the water influx
explain what each term means physically and say when it is small
simplify the equation for an undersaturated accumulation (one whose pressure is still high enough to keep all the gas dissolved in the oil), for solution gas drive (the gas that comes out of the oil pushes it to the wells), and for an accumulation with a gas cap or with water influx
estimate the oil in place from production, pressure and PVT data (how the properties of the fluids change with pressure), and use the growth of the calculated oil in place to recognise a source of energy that has not been counted
calculate the water influx and the drive indices when the oil in place is known
list the assumptions of the method, the requirements for the data and the common mistakes

The idea of the method: the accumulation as one tank

Material balance does not follow how fluid flows through the rock. It compares the accumulation at two moments: before production starts, at the initial pressure $p_i$ , and now, when the average reservoir pressure has fallen to $p$ . In doing so we treat the accumulation as one vessel, a “tank” (the tank model). The pressure is the same everywhere in this tank, and the oil, gas and water are in equilibrium with one another.

The pores of the accumulation are always full. When oil, gas and water are taken out, the fluids that stay behind fill the space that is freed: as the pressure falls, they expand. At the same time the pores themselves shrink a little, and water enters the accumulation from the aquifer. Nothing else can happen to the space in the pores. So the volume taken out always equals the sum of these changes.

What happens in the accumulation when the pressure falls from the initial to the current value

Component: Oil with dissolved gas
What happens to it: above the bubble point pressure it expands; below it, it releases gas, and the gas expands
Term: $NE_o$

Component: Gas cap
What happens to it: expands
Term: $NmE_g$

Component: Connate water
What happens to it: expands
Term: part of $NE_{fw}$

Component: Pore space
What happens to it: shrinks under the weight of the rocks above
Term: part of $NE_{fw}$

Component: Aquifer
What happens to it: water enters the accumulation
Term: $W_e$

Component	What happens to it	Term
Oil with dissolved gas	above the bubble point pressure it expands; below it, it releases gas, and the gas expands	$NE_o$
Gas cap	expands	$NmE_g$
Connate water	expands	part of $NE_{fw}$
Pore space	shrinks under the weight of the rocks above	part of $NE_{fw}$
Aquifer	water enters the accumulation	$W_e$

The symbols are introduced below. All terms are volumes at reservoir conditions at the current pressure $p$ .

Strictly speaking, it is mass that is conserved, not volume. But it is convenient to express the mass of oil and gas through volumes at the surface: cubic metres of stock-tank oil (the oil left after the gas has come out at the surface) and of gas at standard conditions (0.101325 MPa and 20 °C). We then move to reservoir volumes with the formation volume factors, the numbers that show how much room one surface cubic metre takes underground (Black oil model: solution gas–oil ratio and formation volume factors). So the equation is written in volumes, but at the same pressure $p$ in every term.

The left side: underground withdrawal

Suppose that by the time the pressure has fallen to $p$ we have produced $N_p$ cubic metres of stock-tank oil, $G_p$ cubic metres of gas and $W_p$ cubic metres of water, all measured at the surface. What volume would these fluids occupy in the reservoir at pressure $p$ ? This volume is called the underground withdrawal.

Oil. Each cubic metre of stock-tank oil, together with the gas dissolved in it, takes up $B_o$ cubic metres in the reservoir. With the oil, $N_pR_s$ cubic metres of gas reached the surface: this is the gas that was dissolved in the oil at pressure $p$ .
Free gas. The total gas produced is $G_p = N_pR_p$ , where $R_p$ is the cumulative gas–oil ratio. The difference $N_p(R_p - R_s)$ could not have dissolved in the oil produced at pressure $p$ . In the reservoir it is free gas (gas that is not dissolved in the oil), with a volume of $N_p(R_p - R_s)B_g$ .
Water. The water produced takes up $W_pB_w$ in the reservoir.

Underground withdrawal

F = N_p\big[B_o + (R_p - R_s)\,B_g\big] + W_pB_w

where:

$F$ volume of the oil, gas and water produced, converted to reservoir conditions at pressure $p$ , m³
$N_p,\ W_p$ cumulative production of oil and water at standard conditions, m³
$R_p$ cumulative gas–oil ratio, $R_p = G_p/N_p$ , m³/m³
$R_s,\ B_o,\ B_g$ solution gas–oil ratio and formation volume factors of the oil and the gas at the current average pressure, m³/m³
$B_w$ formation volume factor of the water; if it is unknown, take it as 1, m³/m³

Units: m³; in oilfield units, reservoir barrels (rb), with $N_p$ in barrels of stock-tank oil (STB), $R$ in standard cubic feet per barrel (scf/STB) and $B_g$ in rb/scf

Holds when

the properties are taken at the same average reservoir pressure $p$
all the gas produced is counted, including gas burned at a flare and gas used for the field's own needs; gas injected into the wells for gas lift (to help the oil rise) is subtracted from the production
above the bubble point pressure $R_p = R_s = R_{si}$ , and the free-gas term is zero

This is the space in the pores that the produced fluids have left empty. It is also the space they would fill if all of them were put back into the accumulation at the present pressure. With the total formation volume factor, the same quantity is written as $N_p[B_t + (R_p - R_{si})B_g] + W_pB_w$ . The two forms are equal because $B_t = B_o + (R_{si} - R_s)B_g$ .

The right side: what fills the space that was freed

Now imagine that we lower the pressure in the accumulation from $p_i$ to $p$ without taking anything out. Let us see how much the volume of each fluid grows and how much the pore volume shrinks. This “extra” volume does not fit in the pores. It is exactly the amount that was taken out.

Oil and its dissolved gas. The oil in place $N$ is the volume of stock-tank oil that all the oil in the accumulation would give. In the reservoir it first took up $NB_{oi}$ . At pressure $p$ the liquid takes up $NB_o$ , and the gas that has come out of it takes up $N(R_{si} - R_s)B_g$ . Together this is $NB_t$ , where $B_t$ is the total formation volume factor.

Expansion of the oil and its dissolved gas

E_o = (B_o - B_{oi}) + (R_{si} - R_s)\,B_g = B_t - B_{ti}

where:

$E_o$ increase in the reservoir volume of the oil together with the gas that has come out of it, per 1 m³ of oil in place, m³/m³
$B_{oi},\ B_o$ oil formation volume factor at the initial and the current pressure, m³/m³
$R_{si},\ R_s$ solution gas–oil ratio at the initial and the current pressure, m³/m³
$B_g$ gas formation volume factor at the current pressure, m³/m³
$B_t,\ B_{ti}$ total formation volume factor at the current and the initial pressure; $B_{ti} = B_{oi}$ , m³/m³

Units: m³/m³; in oilfield units rb/STB, with $B_g$ in rb/scf and $R_s$ in scf/STB

Holds when

$B_o$ , $R_s$ and $B_g$ belong to the average reservoir pressure and to the laboratory way of releasing the gas that is closest to what happens in the reservoir (Laboratory PVT tests)
the composition and properties of the oil are the same throughout the accumulation
gas that has come out of the oil has either stayed in the accumulation or been counted in the production; the left side of the equation takes care of this

Above the bubble point pressure the second term is zero, and only the weak expansion of the liquid remains: $E_o = B_o - B_{oi}$ . Below the bubble point pressure the oil itself shrinks ( $B_o$ gets smaller and soon becomes less than $B_{oi}$ ), but the gas that has come out takes up much more space, and $E_o$ grows quickly.

Gas cap. Its size is given by the ratio $m$ of the initial reservoir volume of the gas in the cap to the initial reservoir volume of the oil (the gas cap ratio). The volume of the cap is $mNB_{oi}$ . In standard cubic metres this is $mNB_{oi}/B_{gi}$ , and at pressure $p$ the same gas takes up $mNB_{oi}B_g/B_{gi}$ .

Expansion of the gas cap

E_g = B_{oi}\left(\frac{B_g}{B_{gi}} - 1\right)

where:

$E_g$ expansion of the cap per 1 m³ of oil in place for $m = 1$ ; the full expansion of the cap is $NmE_g$ , m³/m³
$m$ ratio of the initial reservoir volume of the gas cap to the initial reservoir volume of the oil, —
$B_{gi},\ B_g$ gas formation volume factor at the initial and the current pressure, m³/m³
$B_{oi}$ initial oil formation volume factor, m³/m³

Units: m³/m³

Holds when

the gas of the cap and the gas coming out of the oil are described by one formation volume factor $B_g$
the cap is at the same average pressure as the oil zone
condensate dropping out of the gas of the cap is neglected

$B_g$ is roughly inversely proportional to pressure. So the cap expands by a fraction close to $\Delta p/p$ : a pressure drop of 5% increases its volume by about 5%. For the same drop the liquids and the pores give only fractions of a per cent. That is why even a small gas cap noticeably slows the fall in pressure.

Connate water and pores. Oil takes up the fraction $1 - S_w$ of the pore volume. So the pore volume of the oil zone is $NB_{oi}/(1 - S_w)$ , and together with the gas cap it is $(1 + m)NB_{oi}/(1 - S_w)$ . When the pressure falls by $\Delta p = p_i - p$ , the connate water expands by the fraction $c_w\Delta p$ of its volume. The pore volume shrinks by the fraction $c_f\Delta p$ . The fluid pressure that props up the grains from inside falls, and the rock is compacted by the weight of the rocks above (Rock compressibility and the effect of stress on core properties). Both changes take space away from the oil and gas, so they act like expansion.

Expansion of the connate water and shrinking of the pore volume

E_{fw} = (1 + m)\,B_{oi}\,\frac{c_wS_w + c_f}{1 - S_w}\,\Delta p

where:

$E_{fw}$ volume released by the connate water and the pores, per 1 m³ of oil in place, m³/m³
$c_w,\ c_f$ compressibility of the water and of the pore volume (the fraction by which a volume changes when the pressure changes by one unit), 1/MPa
$S_w$ connate water saturation, fraction
$\Delta p$ fall in the average reservoir pressure from its initial value, $p_i - p$ , MPa
$m,\ B_{oi}$ gas cap ratio and initial oil formation volume factor, —, m³/m³

Units: consistent units of pressure and compressibility: MPa and 1/MPa, or psi and 1/psi

Holds when

the compressibilities are constant over the pressure range considered
$c_f$ is the compressibility of the pore volume, measured against the fall in pore pressure at reservoir conditions, not the compressibility of the rock as a whole
the water saturation and the pore compressibility in the gas cap are the same as in the oil zone; otherwise the terms for the cap and for the oil zone are written separately

This term is small in size, but above the bubble point pressure it is comparable with the expansion of the oil itself. When there is a gas cap, or when the pressure is well below the bubble point pressure, this term is small beside the expansion of the gas. Then the factor $(1 + m)$ is sometimes left out.

Water influx and injection. If the accumulation is connected to an aquifer, water flows into it when the pressure falls. The cumulative volume of this water at reservoir conditions is $W_e$ (water influx). It depends on the size and the conductivity of the aquifer (how easily water flows through it), and on how long ago and by how much the pressure was lowered (Aquifer models: water influx). Water and gas injected from the surface are counted in the same way: $W_{inj}B_w$ and $G_{inj}B_g$ are added to the right side.

The general equation

What remains is to set the withdrawal equal to the sum of the sources.

General material balance equation of an oil accumulation

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

where:

$F$ underground withdrawal of oil, gas and water, m³
$N$ oil in place at standard conditions, m³
$E_o,\ E_g,\ E_{fw}$ expansion of the oil with its gas, of the gas cap, and of the connate water and pores, per 1 m³ of oil in place, m³/m³
$m$ gas cap ratio, —
$W_e$ cumulative water influx from the aquifer at reservoir conditions, m³

Units: consistent units: volumes in m³, gas in standard m³ (sm³), pressure in MPa, compressibilities in 1/MPa; in oilfield units STB, scf, rb, psi and 1/psi, with no conversion constants

Holds when

the accumulation is one “tank”: the pressure is the same throughout the volume and the phases are in equilibrium
the reservoir temperature is constant
the fluid properties are described by the black oil model (oil taken as two components, stock-tank oil and gas) and belong to the average reservoir pressure
there is no injection; injected volumes, $W_{inj}B_w + G_{inj}B_g$ , are added to the right side
time does not enter the equation: it links states, not rates

On the left is everything that was taken out; on the right is everything that made room for it. In the lesson Reservoir drive mechanisms the same equation is written with $B_t$ .

In other unit systems (1)

Expanded form

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

The same quantities and units. The lines of the right side are the oil with its gas, the gas cap, the connate water with the pores, and the water influx.

Oil in place from material balance

N = \frac{F - W_e}{E_o + m\,E_g + E_{fw}}

where:

$N$ oil in place, m³
$F,\ W_e$ underground withdrawal and cumulative water influx, m³
$E_o,\ E_g,\ E_{fw},\ m$ as in the general equation, m³/m³, —

Units: consistent units, as in the general equation

Holds when

the water influx $W_e$ and the gas cap ratio $m$ are known or equal to zero
the pressure drop is large compared with the error in finding it

The numerator is the part of the withdrawal that the accumulation itself supplied. The denominator is how much reservoir volume one cubic metre of oil in place frees for the pressure drop reached. It is a small difference between two close numbers, the formation volume factors at two pressures. So the result is sensitive to errors in the pressure and in the factors themselves.

Symbols and units

Quantity: $N$ , $N_p$ : oil in place and cumulative oil production
Metric: sm³
Oilfield: STB

Quantity: $G_p$ : cumulative gas production
Metric: sm³
Oilfield: scf

Quantity: $R_s$ , $R_{si}$ , $R_p$ : solution gas–oil ratio and cumulative gas–oil ratio
Metric: m³/m³
Oilfield: scf/STB

Quantity: $B_o$ , $B_t$ : oil formation volume factors
Metric: m³/m³
Oilfield: rb/STB

Quantity: $B_g$ : gas formation volume factor
Metric: m³/m³
Oilfield: rb/scf

Quantity: $W_p$ : water produced
Metric: sm³
Oilfield: STB

Quantity: $F$ , $W_e$ : withdrawal and water influx at reservoir conditions
Metric: m³
Oilfield: rb

Quantity: $c_w$ , $c_f$ : compressibilities
Metric: 1/MPa
Oilfield: 1/psi

Quantity	Metric	Oilfield
$N$ , $N_p$ : oil in place and cumulative oil production	sm³	STB
$G_p$ : cumulative gas production	sm³	scf
$R_s$ , $R_{si}$ , $R_p$ : solution gas–oil ratio and cumulative gas–oil ratio	m³/m³	scf/STB
$B_o$ , $B_t$ : oil formation volume factors	m³/m³	rb/STB
$B_g$ : gas formation volume factor	m³/m³	rb/scf
$W_p$ : water produced	sm³	STB
$F$ , $W_e$ : withdrawal and water influx at reservoir conditions	m³	rb
$c_w$ , $c_f$ : compressibilities	1/MPa	1/psi

“sm³” is a cubic metre at standard conditions. STB is a barrel of stock-tank oil, rb is a barrel at reservoir conditions, and scf is a standard cubic foot. Pressures are absolute (measured from zero, not from atmospheric pressure). The equation uses their difference and the fluid properties at the average pressure.

Special cases: what is left of the equation

In a particular accumulation some terms are zero or small, and the equation becomes simpler.

Simplifications of the general equation

Accumulation: Undersaturated, no water influx
What drops out: $m = 0$ , $W_e = 0$ ; no free gas: $R_p = R_s = R_{si}$
What remains: expansion of the oil, the connate water and the pores

Accumulation: On solution gas drive
What drops out: $m = 0$ , $W_e = 0$
What remains: expansion of the oil with the gas that has come out; the water and the pores are a small addition

Accumulation: With a gas cap
What drops out: $W_e = 0$
What remains: the oil with its gas, and the cap; the water and the pores are usually small

Accumulation: With an aquifer
What drops out: nothing
What remains: all the terms and the water influx

Accumulation	What drops out	What remains
Undersaturated, no water influx	$m = 0$ , $W_e = 0$ ; no free gas: $R_p = R_s = R_{si}$	expansion of the oil, the connate water and the pores
On solution gas drive	$m = 0$ , $W_e = 0$	expansion of the oil with the gas that has come out; the water and the pores are a small addition
With a gas cap	$W_e = 0$	the oil with its gas, and the cap; the water and the pores are usually small
With an aquifer	nothing	all the terms and the water influx

Water can also be produced from an accumulation without an aquifer, when part of the accumulation's own water becomes mobile. Then $W_p \ne 0$ while $W_e = 0$ .

Above the bubble point pressure no gas comes out of the oil, and $B_o - B_{oi} \approx B_{oi}c_o\Delta p$ , where $c_o$ is the compressibility of the oil. The equation turns into the formula of compressibility drive (drive by the expansion of the liquids and the shrinking of the pores alone) from the lesson Reservoir drive mechanisms:

Material balance of a closed undersaturated accumulation

N_pB_o = N B_{oi}\,c_e\,\Delta p,\qquad c_e = c_o + \frac{c_wS_w + c_f}{1 - S_w}

where:

$c_e$ effective compressibility, based on the pore volume filled with oil; it is the same quantity as $(S_oc_o + S_wc_w + c_f)/S_o$ when $S_o = 1 - S_w$ , 1/MPa
$c_o$ average compressibility of the oil between $p_i$ and $p$ , 1/MPa
$N_p,\ N,\ B_o,\ B_{oi},\ \Delta p$ as in the general equation, m³, m³, m³/m³, m³/m³, MPa

Units: consistent units of pressure and compressibility

Holds when

the pressure is above the bubble point pressure; there is no gas cap and no water influx, and no water is produced
the compressibilities are constant from $p_i$ to $p$

The withdrawal is proportional to the pressure drop. While the accumulation is undersaturated and closed, the pressure falls almost along a straight line against the cumulative production. The slope of this line is inversely proportional to the oil in place, and this is how $N$ is estimated from the first years of production. If the pressure fall slows down, another source of energy has appeared: water influx or gas coming out of the oil.

Worked exampleOil in place of an undersaturated accumulation from the pressure drop

Given (assumed data)

Initial and current reservoir pressure: 30 and 24 MPa; bubble point pressure 18 MPa
Cumulative oil production $N_p$: 152 thousand m³; no water in the production; the gas–oil ratio is constant at 110 m³/m³
Oil formation volume factors: $B_{oi} = 1.3243$ at 30 MPa; $B_o = 1.3371$ at 24 MPa
Connate water and compressibilities: $S_w = 0.25$ ; $c_w = 0.45\cdot10^{-3}$ and $c_f = 0.8\cdot10^{-3}$ MPa⁻¹
Gas cap and water influx: none

Solution

Withdrawal. The pressure is above the bubble point pressure and there is no free gas: $F = N_pB_o = 152\cdot1.3371 = 203.24$ thousand m³.
Expansion of the oil: $E_o = B_o - B_{oi} = 1.3371 - 1.3243 = 0.0128$ m³/m³.
Water and pores: $\dfrac{c_wS_w + c_f}{1 - S_w} = \dfrac{(0.45\cdot0.25 + 0.8)\cdot10^{-3}}{0.75} = 1.2167\cdot10^{-3}$ MPa⁻¹; $E_{fw} = 1.3243\cdot1.2167\cdot10^{-3}\cdot6 = 0.009667$ m³/m³.
Oil in place: $N = \dfrac{F}{E_o + E_{fw}} = \dfrac{203.24\cdot10^3}{0.022467} = 9.05$ million m³, or 56.9 million STB.
If we forget the water and the pores: $N = F/E_o = 15.9$ million m³, which is 1.76 times more.
If the average pressure is found with an error of 0.5 MPa, that is 24.5 instead of 24 MPa ( $B_o = 1.3360$ ), we get 9.88 million m³, which is 9% more.

Answer: The oil in place is about 9.05 million m³. The volumetric estimate for this accumulation gave 9.06 million m³ (Black oil model: solution gas–oil ratio and formation volume factors).

What it means. The match with the volumetric estimate tells us that the accumulation really is closed and that all of it is connected to the wells. The two side calculations show the main sources of error. Here the water and the pores gave 43% of the whole expansion, and without them the oil in place is too high by a factor of 1.76. And an error of only 0.5 MPa in the average pressure, for a pressure drop of 6 MPa, changes the result by 9%, because the oil in place is inversely proportional to $\Delta p$ . The data in this example are assumed for illustration: the production was calculated with these same formulas for 9.06 million m³ in place and rounded.

Below the bubble point pressure, the expansion of the gas that has come out of the oil appears in $E_o$ , and it quickly becomes the main part. The chart shows the expansion per cubic metre of oil in place for the assumed oil of the lesson Black oil model: solution gas–oil ratio and formation volume factors, and the share of it that the connate water and the pores give.

CalculatedTop: the volume that one cubic metre of oil in place frees as the pressure falls from 30 MPa. It is the expansion of the oil with its gas,

E_o

, and the contribution of the connate water and pores,

E_{fw}

(the vertical axis is logarithmic). Bottom: the share of

E_{fw}

in their sum. The bubble point pressure is 18 MPa; the calculation is for this lesson's assumed accumulation, without a gas cap.

Worked exampleOil in place from data below the bubble point pressure

Given (assumed data)

Reservoir pressure: initial 30 MPa, current 14 MPa
Cumulative production: oil 909 thousand m³; cumulative gas–oil ratio 130 m³/m³; no water
Oil properties at 14 MPa: $R_s = 87.73$ m³/m³; $B_o = 1.2893$ m³/m³; $B_g = 0.007693$ m³/m³
Initial properties: $R_{si} = 110$ m³/m³; $B_{oi} = 1.3243$ m³/m³
Connate water and compressibilities: as in the previous example; no gas cap and no water influx

Solution

Withdrawal of oil with its dissolved gas: $N_pB_o = 909\cdot1.2893 = 1172.0$ thousand m³.
Free gas: the cumulative gas–oil ratio, 130, is larger than the solution gas–oil ratio, 87.73 m³/m³. At 14 MPa this difference cannot dissolve in the oil produced, so in the reservoir it takes up the volume of free gas: $N_p(R_p - R_s)B_g = 909\cdot(130 - 87.73)\cdot0.007693 = 295.6$ thousand m³. In total $F = 1467.6$ thousand m³.
Expansion of the oil with its gas. The oil itself has shrunk: $B_o - B_{oi} = -0.0350$ m³/m³. The gas that has come out: $(R_{si} - R_s)B_g = (110 - 87.73)\cdot0.007693 = 0.1713$ m³/m³. In sum, $E_o = 0.1363$ m³/m³.
Water and pores: $E_{fw} = 1.3243\cdot1.2167\cdot10^{-3}\cdot16 = 0.0258$ m³/m³, which is 16% of the whole expansion.
Oil in place: $N = \dfrac{1467.6\cdot10^3}{0.1363 + 0.0258} = 9.05$ million m³.
Without the water and the pores we would get 10.8 million m³, which is 19% more.
If the average pressure is found with an error of 0.5 MPa, that is 14.5 instead of 14 MPa ( $R_s = 90.55$ ; $B_o = 1.2969$ ; $B_g = 0.007428$ ), we get 10.2 million m³, which is 12% more.

Answer: The oil in place is about 9.05 million m³, the same as from the data above the bubble point pressure.

What it means. One cubic metre of oil in place has now freed 0.162 m³ of reservoir volume, seven times more than at 24 MPa. The oil by itself takes up less space than at the start, and all the expansion comes from the gas that has come out. The estimate has not become more accurate, though. Below the bubble point pressure the expansion changes very quickly with pressure, so the same 0.5 MPa error now shifts the result by 12%. On top of that come the errors in the gas properties and in the counting of the gas produced. The contribution of the water and the pores has fallen from 43 to 16% but is still noticeable. The accumulation went through a 12 MPa pressure drop above the bubble point pressure, and the elastic expansion gained there has not disappeared. For an accumulation that is saturated from the very beginning, this contribution would be much smaller. The data in this example are assumed for illustration.

What we can find from the equation

The quantities in the equation fall into four groups: what was in the accumulation (the oil in place $N$ and the gas cap ratio $m$ ), what was taken out ( $N_p$ , $R_p$ , $W_p$ ), what came in ( $W_e$ ), and the average reservoir pressure now. This pressure sets the formation volume factors and $\Delta p$ . Any one group can be found if the other three are known.

Four tasks

1
Oil in place
If there is no water influx, or if it is known, we find $N$ from production and pressure and compare it with the volumetric estimate (Volumetric estimation of hydrocarbons in place).
2
Water influx
If the oil in place is known from the volumetric method, the equation gives $W_e$ for each date of a pressure measurement. From this history we fit the parameters of the aquifer (History matching of aquifer parameters) and decide whether injection is needed.
3
Pressure forecast
When the oil in place and the drive mechanism are known, we calculate the average pressure after a given withdrawal. This shows when artificial lift (pumps or gas lift that help the oil reach the surface) will be needed and how much oil can be produced before the pressure reaches a given final value. Below the bubble point pressure we also need a forecast of the gas–oil ratio (Performance prediction for solution gas drive reservoirs).
4
Shares of the drive mechanisms
If we divide the terms of the right side by the withdrawal of oil and gas, we get the drive indices.

The simplest sign of a source of energy that we have not counted is the behaviour of the calculated oil in place. Suppose we treat the accumulation as closed and calculate $N = F/(E_o + E_{fw})$ for every date. For a truly closed accumulation we get the same number every time. If the calculated (“apparent”) oil in place grows from date to date, something else is supporting the pressure, most often water.

Drive indices

\begin{gathered}\frac{NE_o}{H} + \frac{NmE_g}{H} + \frac{NE_{fw}}{H} + \frac{W_e - W_pB_w}{H} = 1,\\ H = N_p\big[B_o + (R_p - R_s)B_g\big]\end{gathered}

where:

$H$ reservoir volume of the oil and gas produced, m³
$NE_o/H$ share of the expansion of the oil and the dissolved gas, —
$NmE_g/H$ share of the gas cap, —
$NE_{fw}/H$ share of the elasticity of the connate water and the pores, —
$(W_e - W_pB_w)/H$ share of the aquifer water, less the water produced, —

Units: dimensionless fractions; volumes in the same units

Holds when

all the quantities belong to one date
the oil in place $N$ , the gas cap ratio $m$ and the water influx $W_e$ have already been found

Each fraction is a drive index: the share of the withdrawal of oil and gas that one mechanism supplied. We subtract the water produced from the water influx, because only the water that stayed in the accumulation takes up space in it. The indices change with time, and they show how the drive of the accumulation changes (Reservoir drive mechanisms).

Development history of an assumed accumulation with an aquifer

Year: Year 1
$p$ , MPa: 27.0
$N_p$ , thousand m³: 220
$W_p$ , thousand m³: 0
$B_o$ , m³/m³: 1.3307

Year: Year 2
$p$ , MPa: 25.0
$N_p$ , thousand m³: 649
$W_p$ , thousand m³: 5
$B_o$ , m³/m³: 1.3350

Year: Year 3
$p$ , MPa: 23.8
$N_p$ , thousand m³: 1198
$W_p$ , thousand m³: 25
$B_o$ , m³/m³: 1.3375

Year: Year 4
$p$ , MPa: 23.0
$N_p$ , thousand m³: 1820
$W_p$ , thousand m³: 60
$B_o$ , m³/m³: 1.3392

Year	$p$ , MPa	$N_p$ , thousand m³	$W_p$ , thousand m³	$B_o$ , m³/m³
Year 1	27.0	220	0	1.3307
Year 2	25.0	649	5	1.3350
Year 3	23.8	1198	25	1.3375
Year 4	23.0	1820	60	1.3392

The same oil zone as in the previous examples, but connected to an aquifer. The pressure stays above the bubble point pressure all the time. The data are assumed: the rate of water influx is taken as proportional to the pressure drop at the oil–water contact (the boundary between the oil and the aquifer water), 350 m³/(d·MPa), as in the example of the lesson Reservoir drive mechanisms.

Worked exampleWater influx and drive indices

Given (assumed data)

Production and pressure history: the table above; the calculation is shown for year 4
Oil in place by the volumetric method $N$: 9.06 million m³; no gas cap
Initial conditions: $p_i = 30$ MPa; $B_{oi} = 1.3243$ m³/m³
Connate water and compressibilities: $S_w = 0.25$ ; $c_w = 0.45\cdot10^{-3}$ and $c_f = 0.8\cdot10^{-3}$ MPa⁻¹
Water formation volume factor $B_w$: 1.02 m³/m³

Solution

Withdrawal in year 4: oil $1820\cdot1.3392 = 2437.3$ thousand m³, water $60\cdot1.02 = 61.2$ thousand m³; $F = 2498.5$ thousand m³.
Expansion per cubic metre of oil in place for $\Delta p = 7$ MPa: $E_o = 1.3392 - 1.3243 = 0.0149$ ; $E_{fw} = 1.3243\cdot1.2167\cdot10^{-3}\cdot7 = 0.01128$ ; the sum is 0.02618 m³/m³.
Check the assumption “the accumulation is closed”: $N = F/(E_o + E_{fw}) = 95.4$ million m³, ten times the volumetric estimate. The apparent oil in place grows from year to year (table below), so the accumulation is not closed.
The accumulation itself, with 9.06 million m³ in place, freed $9.06\cdot10^6\cdot0.02618 = 237.2$ thousand m³. The rest is water: $W_e = 2498.5 - 237.2 = 2261$ thousand m³.
Drive indices (we divide by the withdrawal of oil, 2437.3 thousand m³): expansion of the oil, 5.5%; water and pores, 4.2%; aquifer water less the water produced, $(2261 - 61.2)/2437.3$ , which is 90.3%. The sum is 100%.

Answer: In four years about 2261 thousand m³ of water entered the accumulation. Water drive supplied 90.3% of the withdrawal.

What it means. A closed accumulation would have given only 177 thousand m³ of oil as the pressure fell to 23 MPa, but 1820 thousand m³ was produced. Water took up the rest of the space. Note that the conclusion about water influx rests on the oil in place from the volumetric method. Without it, the same history could also be explained by a very large closed accumulation. Three things help to tell these cases apart: the growth of the apparent oil in place from year to year, the geology, and the wells starting to produce water. The data in this example are assumed for illustration.

Results by year

Year: Year 1
$F$ , thousand m³: 292.8
$E_o + E_{fw}$: 0.01123
Apparent oil in place, million m³: 26.1
$W_e$ , thousand m³: 191

Year: Year 2
$F$ , thousand m³: 871.5
$E_o + E_{fw}$: 0.01876
Apparent oil in place, million m³: 46.5
$W_e$ , thousand m³: 702

Year: Year 3
$F$ , thousand m³: 1627.8
$E_o + E_{fw}$: 0.02319
Apparent oil in place, million m³: 70.2
$W_e$ , thousand m³: 1418

Year: Year 4
$F$ , thousand m³: 2498.5
$E_o + E_{fw}$: 0.02618
Apparent oil in place, million m³: 95.4
$W_e$ , thousand m³: 2261

Year	$F$ , thousand m³	$E_o + E_{fw}$	Apparent oil in place, million m³	$W_e$ , thousand m³
Year 1	292.8	0.01123	26.1	191
Year 2	871.5	0.01876	46.5	702
Year 3	1627.8	0.02319	70.2	1418
Year 4	2498.5	0.02618	95.4	2261

The apparent oil in place is the result of a calculation that assumes there is no water influx. The water influx $W_e$ is calculated for 9.06 million m³ of oil in place.

CalculatedThe balance for the two examples in this lesson. In each pair, the upper bar is the underground withdrawal and the lower bar shows the sources that freed space for it; the two bars of a pair have equal lengths. At the top is the closed accumulation at 14 MPa; at the bottom is the accumulation with an aquifer in year 4. The shares are taken relative to the whole withdrawal

F

, so they differ slightly from the drive indices: those are taken relative to the withdrawal of oil and gas, and use the water influx less the water produced.

Assumptions and input data

The equation itself is exact, because it is a conservation law. What makes it approximate are the assumptions about the accumulation and the quality of the data.

One pressure. The whole accumulation is given one average pressure, although around the wells it is lower and far from them it is higher. The average is found from measurements in shut-in wells: we bring them to one depth and average them by volume (average reservoir pressure, Build-up in a bounded reservoir and average reservoir pressure).
Fast pressure equalisation. The “tank” assumption works better the faster a pressure change spreads through the accumulation. The speed is measured by the hydraulic diffusivity $k/(\phi\mu c_t)$ , made of the permeability $k$ , porosity $\phi$ , viscosity $\mu$ and total compressibility $c_t$ : the larger it is, the faster the spread (Flow regimes). The assumption holds well for a reservoir with high permeability (the rock lets fluid through easily) and thin, low-viscosity oil. It holds badly for a reservoir with low permeability or one broken into blocks.
Constant temperature. The reservoir is large and holds a lot of heat, so the process is taken as isothermal (at constant temperature). The exceptions are thermal methods and the injection of large volumes of cold water.
Representative fluid properties. The formation volume factors and the solution gas–oil ratio must match two things: the way gas comes out of the oil in the reservoir, and the conditions in the field separators, the surface vessels that split gas from oil (How to read a PVT report). One sample does not always represent the whole accumulation.
Reliable production records. We need the cumulative volumes not only of oil but also of gas and water. In old fields it is the gas and the water that were often recorded worst.

Where the input data come from

Quantity: $N_p$ , $G_p$ , $W_p$
Source: field production records
What to watch for: injection is counted separately; the gas of gas lift is subtracted

Quantity: $p$
Source: reservoir pressure measurements in shut-in wells
What to watch for: bringing to one depth, averaging by volume

Quantity: $B_o$ , $R_s$ , $B_g$
Source: PVT studies of bottomhole samples
What to watch for: conversion to the separation conditions

Quantity: $S_w$
Source: well logs and cores
What to watch for: average by pore volume

Quantity: $c_w$
Source: measurements or correlations
What to watch for: usually about $0.45\cdot10^{-3}$ MPa⁻¹ ( $3\cdot10^{-6}$ psi⁻¹)

Quantity: $c_f$
Source: tests on core samples
What to watch for: a handbook value can be wrong by several times, especially for weak rocks

Quantity: $N$ , $m$
Source: volumetric estimate of the oil in place
What to watch for: or they are the unknowns

Quantity: $W_e$
Source: aquifer model
What to watch for: or it is the unknown

Quantity	Source	What to watch for
$N_p$ , $G_p$ , $W_p$	field production records	injection is counted separately; the gas of gas lift is subtracted
$p$	reservoir pressure measurements in shut-in wells	bringing to one depth, averaging by volume
$B_o$ , $R_s$ , $B_g$	PVT studies of bottomhole samples	conversion to the separation conditions
$S_w$	well logs and cores	average by pore volume
$c_w$	measurements or correlations	usually about $0.45\cdot10^{-3}$ MPa⁻¹ ( $3\cdot10^{-6}$ psi⁻¹)
$c_f$	tests on core samples	a handbook value can be wrong by several times, especially for weak rocks
$N$ , $m$	volumetric estimate of the oil in place	or they are the unknowns
$W_e$	aquifer model	or it is the unknown

The more quantities we have to find from one equation, the less certain the answer. Different combinations of oil in place, gas cap size and water influx can explain the same pressure history.

Limitations and common mistakes

No time. The equation tells us what the pressure will be after a given withdrawal, but not when this will happen. Time enters when the balance is combined with the calculation of well rates (Well inflow: the well performance diagram) and of water influx.
No space. The model is zero-dimensional (it treats the whole accumulation as a single point). It gives the average saturations of the accumulation, but does not say where the gas that has come out, or the water that has come in, is located. If the oil properties change a lot across the area or with depth (the bubble point pressure, for example), a single set of PVT properties describes the accumulation badly.
Only the connected volume. The calculation includes only the part of the accumulation where the pressure responds to the withdrawal. Material balance does not “see” isolated blocks and very tight zones, so its oil in place is usually not larger than the volumetric one. The difference shows how much of the accumulation is not being drained.
Small differences. Both the withdrawal and the expansion are small compared with the oil in place. While the pressure has fallen only a little, the error in measuring it is comparable with the fall itself, and early estimates cannot be trusted.
Weakly falling pressure. With an active aquifer or a large gas cap the pressure hardly changes, and the equation cannot tell the options apart well.
Ambiguity. A large closed accumulation and a smaller one with water influx can give a similar pressure history. Independent data help to tell them apart: the volumetric estimate, the geology of the area beyond the oil boundary, and the wells starting to produce water.

Terms and symbols

Material balance: The underground withdrawal equals the sum of the expansion of the fluids, the shrinking of the pore volume and the water influx.
Underground withdrawal $F$: The volume of the oil, gas and water produced, converted to reservoir conditions at the current pressure.
Cumulative gas–oil ratio $R_p$: All the gas produced, divided by all the oil produced since the start of development.
$E_o$ , $E_g$ , $E_{fw}$: The expansion 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.
Gas cap ratio $m$: The ratio of the initial reservoir volume of the gas in the cap to the initial reservoir volume of the oil.
Water influx $W_e$: The cumulative volume of water that has entered the accumulation from the aquifer, at reservoir conditions.
Apparent oil in place: Oil in place calculated on the assumption that the accumulation is closed. If it grows with time, it points to a source of energy that has not been counted.
Drive index: The share of the withdrawal of oil and gas that one mechanism supplies. The indices add up to one.

Common mistakes and misconceptions

Wrong

“Material balance gives the whole oil in place.”

In fact

It gives the oil in place of the part of the accumulation that is connected to the wells by pressure. Isolated blocks are not included. Comparing it with the volumetric estimate shows how much of the accumulation is being drained.

Wrong

“Water and rock are almost incompressible, so we can ignore them.”

In fact

Above the bubble point pressure the oil is only a little more compressible. In this lesson's example the water and the pores gave 43% of the expansion, and without them the oil in place is too high by a factor of 1.76. This term becomes small only beside the expansion of free gas.

Wrong

“We put the gas–oil ratio measured today into the equation.”

In fact

What is needed is the cumulative gas–oil ratio $R_p$ : all the gas produced per all the oil produced. The current gas–oil ratio describes today's inflow, not the whole history of withdrawal.

Wrong

“The calculated oil in place grows from one measurement to the next, so the estimate is getting more accurate.”

In fact

A steady growth of the apparent oil in place is a sign of a source of energy that has not been counted: water influx, a gas cap, or new parts of the accumulation joining in. For a closed accumulation the value must not depend on the date.

Wrong

“Material balance shows when the pressure will fall to the bubble point pressure.”

In fact

It shows after how much production this will happen, but not when. Only a calculation of the well rates and of the water influx gives time.

In depth: advanced detail

The same balance through occupied space and average saturations

The equation can also be derived in a different way. Instead of following the expansions, we follow what occupies the initial oil-filled space $NB_{oi}$ after the pressure has fallen to $p$ . In that space are the oil that is left, the gas that came out of the oil and was not produced, the gas that invaded from the expanded gas cap, and the water that came in. In addition, some of the space was taken by the expanded connate water and the shrunken pores.

Balance of the occupied space

\begin{aligned} NB_{oi} &= (N - N_p)B_o + \big[NR_{si} - (N - N_p)R_s - G_{ps}\big]B_g\\ &+ \big[(G - G_{pc})B_g - GB_{gi}\big] + (W_e - W_pB_w) + NE_{fw}\end{aligned}

where:

$(N - N_p)B_o$ reservoir volume of the oil that is left, m³
$G_{ps},\ G_{pc}$ gas produced that came out of the oil, and gas produced from the cap; $G_{ps} + G_{pc} = G_p = N_pR_p$ , sm³
$G$ initial volume of the gas of the cap at standard conditions, $G = mNB_{oi}/B_{gi}$ , sm³
$W_e - W_pB_w$ water that entered the accumulation and stayed in it, m³
$NE_{fw}$ space taken by the expansion of the connate water and the shrinking of the pore volume, m³

Units: m³ at reservoir conditions

Holds when

the same as for the general equation
the split of the produced gas into gas of the cap and gas that came out of the oil is conventional: only their sum enters the final equation

The square brackets of the first line hold the free gas that stayed in the accumulation: the initial dissolved gas, minus the gas dissolved in the oil that is left, minus the gas produced. If we move the terms and add $G_{ps}$ and $G_{pc}$ , we get the general equation $F = N(E_o + mE_g + E_{fw}) + W_e$ . The two derivations are the same bookkeeping written from different sides.

This form gives the average saturations at once. For a closed accumulation without a gas cap, the pore volume at pressure $p$ is $V_p = \dfrac{NB_{oi}}{1 - S_w}(1 - c_f\Delta p)$ . The oil takes up $(N - N_p)B_o$ , the connate water takes $S_w(1 + c_w\Delta p)$ of the initial pore volume, and the rest is free gas.

Worked exampleAverage saturations at 14 MPa

Given (assumed data)

Accumulation: closed, from the example “Oil in place from data below the bubble point pressure”: $N = 9.05$ million m³, $N_p = 909$ thousand m³
Properties: $B_{oi} = 1.3243$ ; $B_o = 1.2893$ m³/m³; $S_w = 0.25$ ; $c_w = 0.45\cdot10^{-3}$ , $c_f = 0.8\cdot10^{-3}$ MPa⁻¹; $\Delta p = 16$ MPa

Solution

Initial pore volume: $NB_{oi}/(1 - S_w) = 15.99$ million m³. At 14 MPa it is smaller by $c_f\Delta p = 1.28$ %, so $V_p = 15.78$ million m³.
Oil: $(N - N_p)B_o = 10.50$ million m³, $S_o = 0.665$ .
Water: $S_w = 0.25\cdot(1 + c_w\Delta p)/(1 - c_f\Delta p) = 0.255$ .
Gas: $S_g = 1 - S_o - S_w = 0.080$ .

Answer: The average gas saturation of the accumulation is about 8%, and the oil saturation has fallen from 75 to 66.5%.

What it means. This gas saturation is usually already above the critical gas saturation (the level at which gas starts to flow): the gas is mobile, and the gas–oil ratio rises. This agrees with the production data. The cumulative gas–oil ratio, 130 m³/m³, is larger than the initial solution gas–oil ratio, 110 m³/m³, which means that free gas was produced together with the oil. The balance gives only the average over the accumulation. It does not know whether the gas has collected at the top or is spread through the reservoir. From the average saturations and the relative permeability curves we forecast the gas–oil ratio (Performance prediction for solution gas drive reservoirs). The data in this example are assumed for illustration.

The equation as a straight line

For a closed accumulation without a gas cap, the general equation takes the form $F = N\,(E_o + E_{fw})$ . If we calculate the withdrawal $F$ and the expansion $E_o + E_{fw}$ for each date of a pressure measurement and plot the points on a chart, they lie on a straight line through the origin. The slope of this line equals the oil in place $N$ .

Worked exampleTwo dates of one accumulation

Given (assumed data)

Pressure 24 MPa: $F = 203.24$ thousand m³; $E_o + E_{fw} = 0.022467$ m³/m³
Pressure 14 MPa: $F = 1467.6$ thousand m³; $E_o + E_{fw} = 0.16210$ m³/m³

Solution

The slope of the line from the origin to the first point: $203.24\cdot10^3/0.022467 = 9.05$ million m³.
To the second point: $1467.6\cdot10^3/0.16210 = 9.05$ million m³.

Answer: Both points lie on one straight line with a slope of about 9.05 million m³.

What it means. The withdrawal grew seven times, the expansion per cubic metre of oil in place also grew seven times, and their ratio stayed the same: the accumulation is closed. If water had supported the pressure, the second point would lie above the line. A check on all the dates at once is more reliable than a calculation from one point: it shows both the quality of the data and the moment when a new source of energy appears. The lesson Straight-line material balance (Havlena–Odeh method) explains how such charts give the oil in place, the gas cap size and the water influx. The data in this example are assumed for illustration.

Connections to other topics

Black oil model: solution gas–oil ratio and formation volume factors — the formation volume factors and the solution gas–oil ratio, which turn standard volumes into reservoir volumes.
Reservoir drive mechanisms — the sources of reservoir energy that the equation brings together in one balance.
Straight-line material balance (Havlena–Odeh method) — the linear form of the equation: oil in place, gas cap and water influx from the development history.
Rock compressibility and the effect of stress on core properties — where the pore compressibility comes from and why it must be measured.
Aquifer models: water influx — how water influx is calculated.
Performance prediction for solution gas drive reservoirs — the forecast of the pressure and the gas–oil ratio below the bubble point pressure.
Material balance for gas reservoirs — the same balance for a gas accumulation.
Volumetric estimation of hydrocarbons in place — the volumetric estimate of the oil in place, with which the result is compared.
History matching — the balance as an independent check of the simulation model.

Check yourself

1
The pressure in the accumulation is below the bubble point pressure. Which gas–oil ratio enters the underground withdrawal $F$ ?
1. A.The current gas–oil ratio of the wells
2. B.The cumulative one: all the gas produced divided by all the oil produced
3. C.The initial solution gas–oil ratio $R_{si}$
4. D.The solution gas–oil ratio at the current pressure $R_s$
Show the answer
Correct answer: B. The cumulative one: all the gas produced divided by all the oil produced
The balance counts everything taken out since the start of development, so it needs the cumulative gas–oil ratio $R_p = G_p/N_p$ . The solution gas–oil ratio $R_s$ at the current pressure also enters the formula, but for a different purpose. It shows which part of the gas produced was dissolved in the oil in the reservoir and which part was free gas.
2
The accumulation is undersaturated, there is no gas cap and no aquifer, and no water is produced. Which terms remain on the right side of the equation?
1. A.The expansion of the oil, the expansion of the connate water and the shrinking of the pore volume
2. B.Only the expansion of the oil
3. C.The expansion of the gas that has come out
4. D.The water influx
Show the answer
Correct answer: A. The expansion of the oil, the expansion of the connate water and the shrinking of the pore volume
Above the bubble point pressure no gas comes out, and there is no cap or water influx. What remains is $N(B_o - B_{oi})$ and $NE_{fw}$ . The contribution of the water and the pores is comparable with the expansion of the oil, so it cannot be dropped.
3
A closed undersaturated accumulation: $N_p = 80$ thousand m³, $B_{oi} = 1.250$ , $B_o = 1.262$ m³/m³, the pressure has fallen by 5 MPa; $S_w = 0.20$ , $c_w = 0.45\cdot10^{-3}$ , $c_f = 0.6\cdot10^{-3}$ MPa⁻¹. Estimate the oil in place.
Show the answer
$F = 80\cdot1.262 = 100.96$ thousand m³. $E_o = 0.012$ . $(c_wS_w + c_f)/(1 - S_w) = (0.09 + 0.6)\cdot10^{-3}/0.8 = 0.8625\cdot10^{-3}$ MPa⁻¹, $E_{fw} = 1.250\cdot0.8625\cdot10^{-3}\cdot5 = 0.00539$ . $N = 100.96\cdot10^3/0.01739 \approx 5.8$ million m³. Without the water and the pores we would get 8.4 million m³.
4
The oil in place calculated on the assumption of a closed accumulation grows from year to year: 12, 19, 27 million m³. What does this mean?
1. A.There is an error in the oil formation volume factor
2. B.The pressure has fallen below the bubble point pressure
3. C.The estimate is getting more accurate, and we should trust the last value
4. D.The accumulation is getting energy from a source that has not been counted, most likely the aquifer
Show the answer
Correct answer: D. The accumulation is getting energy from a source that has not been counted, most likely the aquifer
For a closed accumulation the ratio of the withdrawal to the expansion does not depend on the date. A steady growth means that the pressure falls more slowly than it should: water (or an unrecognised gas cap that is expanding) takes the place of the oil produced. In this case the oil in place is taken from an independent estimate or found together with the water influx.
5
Why is the oil in place from material balance usually not larger than the volumetric one, and what does a large difference between them tell us?
Show the answer
Material balance “sees” only the volume where the pressure responds to the withdrawal. Isolated fault blocks, lenses (small separate bodies of reservoir rock) and very tight zones are not included in it, while the volumetric method counts them together with the rest. A large difference means that a noticeable part of the accumulation is not drained by the existing wells (or that the maps and the volumetric parameters are wrong). This is a reason to look for undrained zones and plan drilling.
6
On a certain date the withdrawal of oil and gas at reservoir conditions is 1000 thousand m³, and the water produced is $W_pB_w = 50$ thousand m³. The expansion of the oil with its gas gave 150 thousand m³, the water and the pores 50 thousand m³, and the water influx 850 thousand m³. Find the drive indices.
Show the answer
Check the balance: $F = 1000 + 50 = 1050$ ; the sources: $150 + 50 + 850 = 1050$ thousand m³. The indices are found by dividing by the withdrawal of oil and gas: expansion of the oil, $150/1000 = 0.15$ ; water and pores, $0.05$ ; aquifer water less the water produced, $(850 - 50)/1000 = 0.80$ . The sum is one. The drive is water drive.

Updated: 28 October 2026