Material balance for gas reservoirs

Quick refresher

In a closed accumulation the pore volume occupied by gas is constant: the gas that remains fills the same space that all the gas filled at the start, $GB_{gi} = (G - G_p)B_g$ .
Since $B_g$ is proportional to $Z/p$ , we get the straight line $p/Z = (p_i/Z_i)(1 - G_p/G)$ . On the $p/Z$ axis it cuts off the initial value $p_i/Z_i$ , and on the production axis it cuts off the gas in place $G$ .
We must plot exactly $p/Z$ . A straight line drawn through the pressure gives a gas in place that is too low by tens of per cent for a deep accumulation, and too high for a shallow one.
Under volumetric depletion drive (the gas is produced only by its own expansion) the gas recovery factor is $1 - (p_a/Z_a)/(p_i/Z_i)$ . It depends only on the initial and the abandonment pressure. For the lesson's accumulation, with an abandonment pressure of 3 MPa, it is 86%.
Water influx holds up the pressure: the $p/Z$ points lie above the straight line of a closed accumulation, and a line through the early points overestimates the gas in place. The sign is a growth of the apparent gas in place from one measurement to the next.
Water traps gas at high pressure, so under an active water drive the gas recovery factor is lower than under depletion drive. The condensate that was produced was gas in the reservoir, so it is added to the production as its gas equivalent.

Key relationships

G\,B_{gi} = (G - G_p)\,B_g

balance of a closed gas accumulation (volumes at reservoir conditions)

\dfrac{p}{Z} = \dfrac{p_i}{Z_i}\left(1 - \dfrac{G_p}{G}\right)

the $p/Z$ line; pressures are absolute, $Z$ is at reservoir temperature

\dfrac{G_p}{G} = 1 - \dfrac{p_a/Z_a}{p_i/Z_i}

gas recovery factor of a closed accumulation at the abandonment pressure $p_a$

G_pB_g + W_pB_w = G\,(B_g - B_{gi}) + W_e

balance with water influx: the withdrawal equals the expansion of the gas plus the water influx

G_a = \dfrac{G_pB_g + W_pB_w}{B_g - B_{gi}} = G + \dfrac{W_e}{B_g - B_{gi}}

apparent gas in place: constant for a closed accumulation, growing with water influx

Common pitfalls

Plotting the pressure instead of $p/Z$ .
Thinking that because the $p/Z$ points fell on a straight line, there is no water influx. Under water drive the early points also fall almost on a straight line, only a flatter one.
Taking the pressure of one well, or of an unfinished build-up, instead of the average pressure over the accumulation brought to one datum depth.
Leaving out of the production the condensate, the gas used on site (fuel) and the losses.
Taking $Z$ at a temperature other than the reservoir temperature, or converting the production and the gas in place to different standard conditions.

After this lesson you can

write the material balance of a closed gas accumulation and derive from it the straight line of $p/Z$ against cumulative production
find the gas in place from pressure measurements and production, and explain why the plot uses $p/Z$ and not the pressure
calculate the pressure after a given withdrawal, and the gas recovery factor for a given abandonment pressure
write the balance of an accumulation with water influx, recognise water drive from the growth of the apparent gas in place (the amount of gas the data seem to show), and estimate the volume of water that has come in
estimate the gas recovery factor under water drive, taking trapped gas into account
convert produced condensate to its gas equivalent, and name the assumptions and limits of the method

A gas accumulation as a tank of gas

The material balance compares the accumulation at two moments, before production and now, and does not follow how the fluid flows through the reservoir (the idea of the method). The accumulation is treated as one “tank” with an average reservoir pressure $p$ (in the lessons on well tests it is written $\bar p$ ; average reservoir pressure). For a gas accumulation the balance is especially simple: there is one mobile fluid in the pores, and there is also only one main source of energy, the expansion of the gas itself.

Gas is incomparably more compressible than anything else in the reservoir. For the gas of the lesson Gas formation volume factor, compressibility and viscosity, at 20 MPa the compressibility is about 0.046 MPa⁻¹: a pressure fall of 1 MPa increases the volume of the gas by almost 5%. The connate water and the pores together release only about 0.1% of the pore volume occupied by gas. So, as a first approximation, the pore volume occupied by gas is taken as constant.

The assumption of a single pressure holds better for gas than for oil. The viscosity of gas is tens of times lower, and pressure disturbances spread through the accumulation fast. For the lesson's gas, the product $\phi\mu c_t$ is 4–5 times smaller than for the oil reservoir of the lesson Flow regimes, and at the same permeability the pressure evens out that many times faster. This convenience has a flip side: the pressure looks “right” throughout the accumulation, even when water is holding it up.

Symbols and units

Quantity: $G$ , gas initially in place
Metric: sm³
Oilfield: scf

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

Quantity: $B_g$ , $B_{gi}$ , gas formation volume factor, current and initial
Metric: m³/m³
Oilfield: rb/scf or ft³/scf

Quantity: $p$ , $p_i$ , average reservoir pressure, current and initial
Metric: MPa (abs.)
Oilfield: psia

Quantity: $Z$ , $Z_i$ , gas deviation factor at $p$ and $p_i$
Metric: —
Oilfield: —

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

Quantity: $W_p$ , produced water; $B_w$ , its formation volume factor
Metric: sm³; m³/m³
Oilfield: STB; rb/STB

Quantity	Metric	Oilfield
$G$ , gas initially in place	sm³	scf
$G_p$ , cumulative gas production	sm³	scf
$B_g$ , $B_{gi}$ , gas formation volume factor, current and initial	m³/m³	rb/scf or ft³/scf
$p$ , $p_i$ , average reservoir pressure, current and initial	MPa (abs.)	psia
$Z$ , $Z_i$ , gas deviation factor at $p$ and $p_i$	—	—
$W_e$ , water influx at reservoir conditions	m³	rb
$W_p$ , produced water; $B_w$ , its formation volume factor	sm³; m³/m³	STB; rb/STB

“sm³” is a standard cubic metre, a cubic metre measured at standard conditions: in the lessons of the library these are 0.101325 MPa and 20 °C (standard conditions). scf is a standard cubic foot, rb is a barrel at reservoir conditions, and STB is a barrel at standard conditions. If $B_w$ is not known, it is taken as 1.

A closed accumulation: the balance and the p/Z line

An accumulation into which no water comes is called closed, and its drive is volumetric depletion drive (a volumetric reservoir). The pore volume occupied by gas does not change in it. At the start this volume held all the gas $G$ ; now it holds the gas that is left, $G - G_p$ . The volume of the gas in the reservoir is its amount in standard cubic metres multiplied by the formation volume factor at reservoir pressure (the ratio of the volume of the gas in the reservoir to its volume at the surface).

Material balance of a closed gas accumulation

G\,B_{gi} = (G - G_p)\,B_g

where:

$G$ gas initially in place at standard conditions (GIIP), m³
$G_p$ cumulative gas production at standard conditions, m³
$B_{gi}$ gas formation volume factor at the initial reservoir pressure $p_i$ , m³/m³
$B_g$ gas formation volume factor at the current average reservoir pressure $p$ , m³/m³

Units: both sides are volumes at reservoir conditions, in m³; in oilfield units they are reservoir barrels, with $G$ in scf and $B_g$ in rb/scf

Holds when

There is no water influx, and no water is produced.
The pore volume occupied by gas is constant: the expansion of the connate water and the decrease of the pore volume are neglected (the correction is in the advanced section).
The reservoir temperature is constant, and the pressure is the same throughout the accumulation.
The gas in the reservoir stays single-phase, and $G_p$ includes all the gas produced.

On the left is the space that the gas filled at the start. On the right is the same space filled by the remaining gas at a lower pressure. The share of gas produced is set only by the ratio of the formation volume factors: $G_p/G = 1 - B_{gi}/B_g$ .

The gas formation volume factor is proportional to $ZT/p$ (gas formation volume factor). The reservoir temperature is constant, and constant factors cancel on both sides of the equation, so in place of $B_g$ we can simply put $Z/p$ :

The p/Z line for a closed gas accumulation

\frac{p}{Z} = \frac{p_i}{Z_i}\left(1 - \frac{G_p}{G}\right)

where:

$p,\ p_i$ current and initial average reservoir pressure, absolute, MPa
$Z,\ Z_i$ gas deviation factor at reservoir temperature and pressures $p$ and $p_i$ , —
$G_p,\ G$ cumulative production and gas initially in place, m³

Units: any units of pressure and gas volume: the formula contains only ratios

Holds when

The same as for the balance of a closed accumulation.
$Z$ matches the composition of the reservoir gas and does not change because the composition changes during development.

In the coordinates of $p/Z$ against cumulative production, the measurements fall on a straight line (p/Z plot). At $G_p = 0$ it gives $p_i/Z_i$ , and it reaches zero at $G_p = G$ : if all the gas could be withdrawn, the pressure would fall to zero. The slope of the line is $-p_i/(Z_iG)$ : the larger the gas in place, the flatter the line. In Russian-language practice this technique is called the pressure decline method.

In other unit systems (1)

Gas in place from one measurement

G = \frac{G_p}{1 - \dfrac{p/Z}{p_i/Z_i}}

$G$ comes out in the same units as $G_p$ .

There are no constants. A calculation from one point is sensitive to a pressure error, especially while the pressure has fallen only a little. A line through several measurements is more reliable.

Reservoir pressure measurements of a closed accumulation (assumed data)

Measurement: initial
$G_p$ , billion m³: 0.00
$p$ , MPa: 20.00
$Z$: 0.8687
$p/Z$ , MPa: 23.022

Measurement: 1st
$G_p$ , billion m³: 0.32
$p$ , MPa: 18.64
$Z$: 0.8646
$p/Z$ , MPa: 21.559

Measurement: 2nd
$G_p$ , billion m³: 0.71
$p$ , MPa: 16.91
$Z$: 0.8620
$p/Z$ , MPa: 19.617

Measurement: 3rd
$G_p$ , billion m³: 1.05
$p$ , MPa: 15.62
$Z$: 0.8622
$p/Z$ , MPa: 18.117

Measurement: 4th
$G_p$ , billion m³: 1.46
$p$ , MPa: 13.95
$Z$: 0.8652
$p/Z$ , MPa: 16.123

Measurement	$G_p$ , billion m³	$p$ , MPa	$Z$	$p/Z$ , MPa
initial	0.00	20.00	0.8687	23.022
1st	0.32	18.64	0.8646	21.559
2nd	0.71	16.91	0.8620	19.617
3rd	1.05	15.62	0.8622	18.117
4th	1.46	13.95	0.8652	16.123

The accumulation is the one from the example of the lesson Gas formation volume factor, compressibility and viscosity: 20 MPa, 80 °C, gas with a specific gravity of 0.65 (density relative to air). The pressures were calculated from the balance equation for a gas in place of 4.89 billion m³, with an assumed measurement error of up to 0.05 MPa added. $Z$ comes from the generalised correlation for this gas (Ideal and real gas: the gas deviation factor (Z)).

Calculated

p/Z

plot for the measurements of the table. The straight line drawn through the five points is extended to the production axis, where the gas in place is read. The marked point matches an abandonment pressure of 3 MPa (

p/Z

= 3.13 MPa): the production at this point is the recoverable gas under depletion drive.

Worked exampleGas in place from the p/Z plot

Given (assumed data)

Measurements: the table above: five values of $G_p$ and $p/Z$
Gas in place by the volumetric method: 4.89 billion m³ (Gas formation volume factor, compressibility and viscosity)

Solution

For each measurement we find $Z$ at reservoir temperature and calculate $p/Z$ . For example, for the last one: $p = 13.95$ MPa, $Z = 0.8652$ , $p/Z = 16.123$ MPa.
We draw the line by the least-squares method (the line that fits all the points best): $p/Z = 23.033 - 4.7252\,G_p$ , where $G_p$ is in billion m³.
The gas in place is the production at which the line reaches zero: $G = 23.033/4.7252 = 4.87$ billion m³.
The intercept on the $p/Z$ axis equals $p_i/Z_i$ : 23.033 MPa. It corresponds to a pressure of 20.01 MPa, so the line reproduces the measured initial pressure of 20.00 MPa.
A check from the two extreme measurements: $G = \dfrac{1.46}{1 - 16.123/23.022} = 4.87$ billion m³.
In oilfield units: $4.87\cdot34.8 = 169$ billion scf (1 m³ at 20 °C and 0.101325 MPa is about 34.8 scf at 60 °F and 14.7 psia).

Answer: The gas in place from the material balance is about 4.87 billion m³; the volumetric method gave 4.89 billion m³.

What it means. The difference of 0.3% is within the error of the pressure measurements. The agreement of two independent estimates says that the accumulation is closed and that all of it is connected to the wells. By the last measurement 30% of the gas in place has been produced, and the pressure has fallen by 6.05 MPa: this is already enough for a confident estimate. After the first year, when the pressure had fallen by only 1.36 MPa, a measurement error of 0.05 MPa would change the gas in place by 4%. The data in this example are assumed for illustration.

Why p/Z and not pressure

If the gas were ideal, $Z$ would equal one, and the pressure itself would be a straight line against production. For a real gas, $Z$ changes with pressure, and the plot of pressure against production is curved. The trouble is that the curvature is small and cannot be seen from a few early points, while the extension of such a “line” to zero pressure gives the wrong gas in place.

CalculatedCalculated pressure of a closed accumulation as a fraction of the initial pressure, against the fraction of the gas in place produced, for three accumulations with the same gas. The value of

p/Z

as a fraction of its initial value falls for all three on one straight line (dashed), while the pressure follows its own curve: below the line in a deep accumulation, above it in a shallow one.

Gas in place from a “pressure against production” line drawn through the initial point and the point of 20% production

Initial conditions: 10 MPa, 60 °C
$Z_i$: 0.855
$Z$ after 20% production: 0.875
Gas in place from the line, % of the true value: 110

Initial conditions: 20 MPa, 80 °C
$Z_i$: 0.869
$Z$ after 20% production: 0.862
Gas in place from the line, % of the true value: 97

Initial conditions: 40 MPa, 120 °C
$Z_i$: 1.073
$Z$ after 20% production: 0.975
Gas in place from the line, % of the true value: 73

Initial conditions	$Z_i$	$Z$ after 20% production	Gas in place from the line, % of the true value
10 MPa, 60 °C	0.855	0.875	110
20 MPa, 80 °C	0.869	0.862	97
40 MPa, 120 °C	1.073	0.975	73

The calculation is for the lesson's gas (specific gravity 0.65). The $p/Z$ line returns the true gas in place in all three cases: 100%.

In a deep accumulation at 40 MPa, $Z$ decreases as the pressure falls. The pressure falls faster than $p/Z$ , and a line through the pressure underestimates the gas in place by more than a quarter. In a shallow accumulation at 10 MPa, $Z$ grows as the pressure falls, and the gas in place is overestimated by a tenth. For the lesson's accumulation the error is only 3%, but this is a coincidence: its initial pressure is close to the minimum of $Z$ .

Pressure forecast and gas recovery factor under depletion drive

Once the line is established, it also works the other way: for any future cumulative production we read $p/Z$ from it, and from it we find the pressure. The step back from $p/Z$ to $p$ needs trial and error, because $Z$ itself depends on the pressure. Two or three approximations are usually enough (see the advanced section).

Production ends not when the gas has run out, but when the reservoir pressure is no longer enough to deliver gas into the pipeline at an acceptable rate and compression cost. This pressure is called the abandonment pressure. It sets the gas recovery factor.

Gas recovery factor of a closed accumulation

\frac{G_p}{G} = 1 - \frac{p_a/Z_a}{p_i/Z_i}

where:

$G_p/G$ gas recovery factor: the share of the gas in place that is produced by the end of development, fraction
$p_a,\ Z_a$ abandonment pressure and the gas deviation factor at it, MPa, —
$p_i,\ Z_i$ initial reservoir pressure and the gas deviation factor at it, MPa, —

Units: a dimensionless share; pressures are absolute

Holds when

Depletion drive: there is no water influx.
The whole accumulation is brought down to one pressure $p_a$ by the end of development.

The gas recovery factor depends neither on the size of the accumulation nor on the withdrawal rate, only on how low the reservoir pressure can be brought. Each lowering of the abandonment pressure, for example by installing booster compressors (machines that push the gas into the pipeline when the reservoir pressure is low), adds to the recovery factor almost linearly.

Worked examplePressure after a given withdrawal, and recoverable gas

Given (assumed data)

The line from the previous example: $p/Z = 23.033 - 4.7252\,G_p$ ( $G_p$ in billion m³); $G = 4.87$ billion m³
Produced by the last measurement: 1.46 billion m³
Plan: 1.0 million m³/d for another 6 years
Abandonment pressure: 3.0 MPa; with booster compressors 1.5 MPa

Solution

Production in 6 years: $1.0\cdot10^{6}\cdot365.25\cdot6 = 2.1915$ billion m³; the cumulative production is 3.6515 billion m³.
From the line: $p/Z = 23.033 - 4.7252\cdot3.6515 = 5.779$ MPa.
The pressure by trial and error. We try $Z = 0.93$ : $p = 5.779\cdot0.93 = 5.37$ MPa. At this pressure $Z = 0.930$ , so the approximation has converged. The reservoir pressure after 6 years is about 5.4 MPa.
Recoverable gas at an abandonment pressure of 3.0 MPa: $Z_a = 0.9591$ , $p_a/Z_a = 3.128$ MPa; $G_p = (23.033 - 3.128)/4.7252 = 4.21$ billion m³. The gas recovery factor is $1 - 3.128/23.033 = 0.864$ .
If compressors allow the pressure to be lowered to 1.5 MPa ( $p_a/Z_a = 1.532$ MPa): $G_p = 4.55$ billion m³, and the recovery factor is 0.933. The gain is about 340 million m³.

Answer: After 6 years the reservoir pressure will fall to about 5.4 MPa. Down to a pressure of 3 MPa the accumulation will give about 4.21 billion m³, which is 86% of the gas in place.

What it means. The six years come from the withdrawal plan, not from the balance. The balance says what the pressure will be after a production of 3.6515 billion m³, whenever it takes place. Whether the wells can deliver 1.0 million m³/d at a reservoir pressure of 5.4 MPa is a separate calculation of their productivity and of the pressure losses in the wellbore and at the surface (Well inflow: the well performance diagram, Nodal analysis of the production system). The data in this example are assumed for illustration.

An accumulation with water influx

If an aquifer lies under the accumulation or around it, then when the pressure falls, water enters the accumulation (water influx). The water occupies part of the pores that the gas filled before, and the remaining gas gets a smaller volume, so the pressure falls more slowly. The water that has entered, $W_e$ , and the water produced, $W_p$ , appear in the balance:

Material balance of a gas accumulation with water influx

G\,B_{gi} = (G - G_p)\,B_g + W_e - W_pB_w

where:

$W_e$ cumulative water influx from the aquifer at reservoir conditions, m³
$W_p,\ B_w$ cumulative water production at standard conditions, and its formation volume factor, m³, m³/m³
$G,\ G_p,\ B_g,\ B_{gi}$ as in the balance of a closed accumulation, m³, m³, m³/m³, m³/m³
$p,\ Z,\ p_i,\ Z_i$ in the form with $p/Z$ : the current and initial average reservoir pressure (absolute) and the gas deviation factors at them, as on the $p/Z$ line, MPa, —, MPa, —

Units: m³ at reservoir conditions; in oilfield units, reservoir barrels

Holds when

The expansion of the connate water and the decrease of the pore volume are neglected.
The pressure is the same throughout the accumulation, including the part flooded by water.
$W_e$ is the whole volume of water that has entered the initial gas-filled volume, no matter how much gas it displaced and how much it bypassed.

The initial gas-filled pore volume is now shared by the gas that is left and the water that has come in and has not been produced. The equation agrees with the general equation of the lesson General material balance equation for oil reservoirs if we keep only the gas in it: the water influx, as there, is expressed in reservoir cubic metres.

In other unit systems (2)

The withdrawal equals the expansion of the gas plus the water influx

G_pB_g + W_pB_w = G\,(B_g - B_{gi}) + W_e

The same quantities. On the left is the reservoir volume of the withdrawal; on the right are the expansion of the gas of the accumulation and the water influx.

Through p/Z

\frac{p}{Z} = \frac{p_i}{Z_i}\cdot\frac{1 - G_p/G}{1 - (W_e - W_pB_w)/(G\,B_{gi})}

Pressures in any identical units; $(W_e - W_pB_w)/(GB_{gi})$ is the share of the initial gas-filled volume that water has taken.

When $W_e = W_pB_w$ the denominator equals one, and the formula turns into the line of a closed accumulation.

CalculatedCalculated

p/Z

plots for the lesson's accumulation at the same withdrawal rate: a closed accumulation (a straight line), moderate water influx and strong water influx. The curves with water influx stop where the water has swept 85% of the volume of the accumulation and the wells start to produce water. The dashed straight line is drawn through the first four measurements of the accumulation with moderate water influx, and it “shows” a gas in place of 6.2 billion m³ instead of 4.9.

The danger is that with water influx the early points also fall almost on a straight line, only a flatter one. The aquifer responds with a delay, and produced water appears late: the wells open the upper part of the accumulation, and the water moves towards them as a slow, even front. The reason is the mobility ratio of water to gas, $M' = (k_{rw}'/\mu_w)/(k_{rg}'/\mu_g)$ (how easily water moves compared with gas). It is only hundredths, because the viscosity of water is tens of times higher than that of gas, and its end-point relative permeability (the value at the end of the displacement) is lower (Relative permeability). The first sign of water is not that the wells start to produce water, but a pressure that is “too good”.

The apparent gas in place helps to recognise water influx. It is the gas in place calculated from each measurement on the assumption that the accumulation is closed. For a closed accumulation it is the same from one measurement to the next. With water influx it grows.

Apparent gas in place

G_a = \frac{G_pB_g + W_pB_w}{B_g - B_{gi}} = G + \frac{W_e}{B_g - B_{gi}}

where:

$G_a$ apparent gas in place: the gas in place calculated from one measurement without taking water influx into account, m³
$B_g - B_{gi}$ expansion of one standard cubic metre of the gas of the accumulation when the pressure falls from $p_i$ to $p$ , m³/m³
$G_p,\ W_p,\ B_w,\ W_e,\ G$ as in the balance with water influx, m³, m³, m³/m³, m³, m³

Units: m³ at standard conditions

Holds when

The compressibility of the pores and the connate water is neglected.
The pressure measurements are accurate enough. When the pressure has fallen only a little, the difference $B_g - B_{gi}$ is small, and the error is large.

The first fraction is what we can calculate from the data. The second equality shows what it means. If there is no water, $G_a = G$ at any date. If water comes in, a term is added to the gas in place that grows with time. When no water is produced, the first fraction is the same as the gas in place from one measurement on the $p/Z$ plot. In the literature the difference $B_g - B_{gi}$ is written $E_g$ . Here it is written out in full, because in the lesson General material balance equation for oil reservoirs this symbol stands for the expansion of the gas cap per cubic metre of oil in place.

History of an accumulation with an aquifer (assumed data)

Date: start
$G_p$ , billion m³: 0.000
$p$ , MPa: 20.00
$p/Z$ , MPa: 23.02
$G_a$ , billion m³: —

Date: year 1
$G_p$ , billion m³: 0.365
$p$ , MPa: 18.55
$p/Z$ , MPa: 21.46
$G_a$ , billion m³: 5.4

Date: year 2
$G_p$ , billion m³: 0.731
$p$ , MPa: 17.37
$p/Z$ , MPa: 20.14
$G_a$ , billion m³: 5.8

Date: year 3
$G_p$ , billion m³: 1.096
$p$ , MPa: 16.36
$p/Z$ , MPa: 18.98
$G_a$ , billion m³: 6.2

Date: year 4
$G_p$ , billion m³: 1.461
$p$ , MPa: 15.45
$p/Z$ , MPa: 17.92
$G_a$ , billion m³: 6.6

Date: year 5
$G_p$ , billion m³: 1.826
$p$ , MPa: 14.62
$p/Z$ , MPa: 16.93
$G_a$ , billion m³: 6.9

Date	$G_p$ , billion m³	$p$ , MPa	$p/Z$ , MPa	$G_a$ , billion m³
start	0.000	20.00	23.02	—
year 1	0.365	18.55	21.46	5.4
year 2	0.731	17.37	20.14	5.8
year 3	1.096	16.36	18.98	6.2
year 4	1.461	15.45	17.92	6.6
year 5	1.826	14.62	16.93	6.9

The same accumulation, connected with an aquifer; the withdrawal is 1.0 million m³/d, and no water is produced. The rate of water influx is taken, as in the examples of the lessons Reservoir drive mechanisms and General material balance equation for oil reservoirs, as proportional to the pressure fall; the coefficient here is 700 m³/(d·MPa) at reservoir conditions. The apparent gas in place $G_a$ is calculated in the example below.

CalculatedApparent gas in place for three calculated histories of the lesson's accumulation. For the closed accumulation it is constant and equal to the true gas in place. With water influx it grows from the very first measurement, the faster the stronger the aquifer. The points are the values from the table.

Worked exampleWater influx from the pressure history

Given (assumed data)

History: the table above; the calculation is shown for the fifth year
Gas in place by the volumetric method $G$: 4.89 billion m³; gas-filled pore volume 25.92 million m³
Initial conditions: $p_i = 20$ MPa; $B_{gi} = 0.005302$ m³/m³
Water production: none

Solution

The formation volume factor in the fifth year: $p = 14.62$ MPa, $Z = 0.8636$ , $B_g = B_{gi}\cdot\dfrac{p_i/Z_i}{p/Z} = 0.005302\cdot\dfrac{23.02}{16.93} = 0.007209$ m³/m³.
The withdrawal at reservoir conditions: $G_pB_g = 1.826\cdot10^{9}\cdot0.007209 = 13.16$ million m³.
The expansion of a cubic metre of gas: $B_g - B_{gi} = 0.001907$ m³/m³.
The apparent gas in place: $G_a = 13.16\cdot10^{6}/0.001907 = 6.90$ billion m³. Year by year it grows: 5.4, 6.2 and 6.9 billion m³ in the first, third and fifth year. The accumulation is not closed.
With a gas in place of 4.89 billion m³, the expansion of the gas gave $4.89\cdot10^{9}\cdot0.001907 = 9.33$ million m³. The rest is water: $W_e = 13.16 - 9.33 = 3.8$ million m³, which is 15% of the gas-filled pore volume.
A closed accumulation after the same withdrawal would have a pressure of 12.55 MPa, which is 2.1 MPa lower than the measured one.

Answer: In five years about 3.8 million m³ of water has come into the accumulation. The apparent gas in place has grown from 5.4 to 6.9 billion m³, against a true 4.89.

What it means. A line drawn through all six $p/Z$ points of this table would give 6.9 billion m³, which is 41% more than the true gas in place, although the points depart from it by no more than 0.3 MPa. The bend (the end points above the line, the middle ones below it) can be seen only when the pressure measurements are accurate. The conclusion about water influx rests on an independent estimate of the gas in place. When there is none, the gas in place and the water influx must be found together by fitting an aquifer model (Aquifer models: water influx, Straight-line material balance (Havlena–Odeh method)). The data in this example are assumed for illustration.

Gas recovery factor under water drive

For oil, an active aquifer is a blessing: it maintains the pressure and displaces the oil. For gas it is different. As water moves through the pores, it leaves trapped gas behind it: the residual gas saturation $S_{gr}$ , roughly 20 to 40% of the pore volume. This gas is trapped at the pressure that the accumulation had when the water arrived, and with a strong influx that pressure is high: each cubic metre of pore space keeps many standard cubic metres of gas.

The second loss is the gas in those parts of the accumulation that the water has not reached by the time the wells start to produce water. The share of the accumulation's volume that the water has swept is called the volumetric sweep efficiency $E_V$ . This is the whole volumetric sweep. In the lessons on waterflooding the same letter stands for the vertical sweep (over the thickness), and the volumetric sweep is written as the product $E_AE_V$ (recovery factor in waterflooding). In the swept part, trapped gas is left. In the unswept part, there is gas at its initial saturation. All of it is at the abandonment pressure $p_a$ .

Gas recovery factor under water drive

\frac{G_p}{G} = 1 - \frac{p_a/Z_a}{p_i/Z_i}\left[E_V\,\frac{S_{gr}}{1 - S_w} + (1 - E_V)\right]

where:

$E_V$ sweep efficiency: the share of the accumulation volume swept by water by the end of development, fraction
$S_{gr}$ residual gas saturation behind the water front, fraction
$S_w$ connate water saturation, fraction
$p_a,\ Z_a$ reservoir pressure at the end of development and the gas deviation factor at it, MPa, —
$p_i,\ Z_i,\ G_p,\ G$ initial pressure and gas deviation factor, cumulative production and gas in place, MPa, —, m³, m³
$W_e,\ W_p,\ B_w,\ B_{gi}$ in the second form: the water influx, the water production and its formation volume factor, and the initial gas formation volume factor, as in the balance with water influx, m³, m³, m³/m³, m³/m³

Units: a dimensionless share

Holds when

At the end of development, all the remaining gas is at one pressure $p_a$ .
In the swept part the gas saturation equals $S_{gr}$ , and in the unswept part it equals the initial value, $1 - S_w$ .
The compressibility of the pores and the connate water is neglected.

The square bracket is the share of the initial gas-filled volume in which gas has been left. At $E_V = 0$ it equals one, and the formula turns into the gas recovery factor of a closed accumulation. This is the same balance with water influx, only the volume of water that has come in is expressed through the sweep (the second form). The two levers of the recovery factor, the abandonment pressure and the sweep, pull in opposite directions: a strong aquifer sweeps the accumulation well, but keeps the pressure high.

In other unit systems (2)

The volume of water left in the accumulation, through the sweep

W_e - W_pB_w = E_V\,G\,B_{gi}\,\frac{1 - S_{gr} - S_w}{1 - S_w}

$W_e$ in m³ at reservoir conditions; $GB_{gi}$ is the initial gas-filled pore volume, in m³.

There are no constants: in the swept part, the water takes the share of the pores that the displaced gas has freed.

Trapped gas as a share of the gas in place

\frac{G_{trap}}{G} = E_V\,\frac{S_{gr}}{1 - S_w}\cdot\frac{p_a/Z_a}{p_i/Z_i}

$G_{trap}$ is the gas left behind the water front, in standard m³; the rest are dimensionless shares and pressures in identical units.

There are no constants: the pore volume with trapped gas, $E_V\,GB_{gi}\,S_{gr}/(1 - S_w)$ , is divided by the formation volume factor at the pressure $p_a$ .

Worked exampleHow much gas is left behind the water front

Given (assumed data)

The lesson's accumulation: $p_i/Z_i = 23.02$ MPa; $S_w = 0.25$
Residual gas saturation; sweep by the time the wells start to produce water: $S_{gr} = 0.30$ ; $E_V = 0.85$
Pressure at that moment (the calculated curves of the figure above): moderate water influx: 10.5 MPa; strong: 17.6 MPa

Solution

The share of the volume where gas has been left: $0.85\cdot0.30/0.75 + 0.15 = 0.49$ .
Moderate influx: $p_a/Z_a = 11.91$ MPa; the gas recovery factor is $1 - (11.91/23.02)\cdot0.49 = 0.746$ .
Strong influx: $p_a/Z_a = 20.40$ MPa; the gas recovery factor is $1 - (20.40/23.02)\cdot0.49 = 0.566$ .
Trapped gas with strong influx: $0.85\cdot0.40\cdot20.40/23.02 = 0.30$ of the gas in place; with moderate influx, 0.18.
The volume of water that has come in is the same in both cases: $0.85\cdot25.92\cdot(1 - 0.30 - 0.25)/0.75 = 13.2$ million m³.

Answer: The gas recovery factor is about 75% with moderate water influx and 57% with strong; a closed accumulation with an abandonment pressure of 3 MPa would give 86%.

What it means. The same amount of water has come in, but the losses differ. With strong influx the gas is trapped at 17.6 MPa, and 30% of the gas in place is left behind the front, against 18% with moderate influx. Hence the rule for developing gas accumulations with active water: take the gas faster than the water can arrive, so that trapping happens at the lowest possible pressure. The possibilities for this are limited by the number of wells and by the demand for gas. The calculation agrees with the balance: the production on the calculated curves by this moment is 3.65 and 2.76 billion m³. The data in this example are assumed for illustration.

Wet gas and condensate: the gas equivalent

Wet gas (gas with a lot of heavy hydrocarbons in it) is single-phase in the reservoir, but condensate (light liquid hydrocarbons) drops out of it in the separator (Types of reservoir fluids). For the balance this is the same reservoir gas, only produced as a liquid, and it must be put back into the production as gas. Fresh water that was water vapour in the reservoir is treated in the same way. The conversion goes through the number of moles: a cubic metre of liquid contains $\rho/M$ kilomoles, and each kilomole of gas takes the same volume at standard conditions.

Gas equivalent of condensate

GE = \frac{R\,T_{\text{sc}}}{p_{\text{sc}}}\cdot\frac{\rho_c}{M_c} = 24.05\,\frac{\rho_c}{M_c}

where:

$GE$ gas equivalent: the volume of gas at standard conditions that 1 m³ of liquid gives, m³/m³
$\rho_c$ density of the stabilised condensate (the liquid left after the light gases are removed) at standard conditions, kg/m³
$M_c$ molar mass of the condensate, kg/kmol
$R$ universal gas constant, 8.3145 kPa·m³/(kmol·K), kPa·m³/(kmol·K)
$T_{\text{sc}},\ p_{\text{sc}}$ standard temperature and pressure: 293.15 K and 101.325 kPa, K, kPa

Units: $\rho_c$ in kg/m³, $M_c$ in kg/kmol; the result is m³ of gas per m³ of liquid at 20 °C and 0.101325 MPa

Holds when

The vapour of the condensate at standard conditions is taken as an ideal gas ( $Z = 1$ ).
The density and molar mass refer to the liquid whose volume is counted in the production.

24.05 m³/kmol is the volume of a kilomole of ideal gas at standard conditions. The lighter the condensate, the smaller its molar mass and the more gas it “contains”. For water ( $\rho$ = 1000 kg/m³, $M$ = 18 kg/kmol) the equivalent is about 1340 m³ of vapour per cubic metre.

In other unit systems (1)

Oilfield units

GE = 133{,}000\,\frac{\gamma_c}{M_c}

$GE$ in scf of gas per barrel of condensate (STB) at 60 °F and 14.7 psia; $\gamma_c$ is the specific gravity of the condensate (relative to water); $M_c$ in lb/lb-mol.

$133{,}000 = 10.73\cdot520/14.7\cdot62.4\cdot5.615$ : the volume of a pound-mole of gas (10.73 is the gas constant in psia·ft³/(lb-mol·°R), with 520 °R and 14.7 psia) multiplied by the mass of a barrel of water (62.4 lb/ft³ and 5.615 ft³ in a barrel).

Worked exampleProduction of wet gas with condensate taken into account

Given (assumed data)

Separator gas produced: 1.40 billion m³
Stabilised condensate produced: 180 thousand m³
Condensate: density 750 kg/m³; molar mass 120 kg/kmol

Solution

Gas equivalent: $GE = 24.05\cdot750/120 = 150.3$ m³/m³.
The condensate converted to gas: $180\cdot10^{3}\cdot150.3 = 27.1$ million m³.
Production of reservoir gas: $G_p = 1.40 + 0.027 = 1.427$ billion m³, which is 1.9% more than the separator gas.
The same in oilfield units: $\gamma_c = 0.75$ ; $GE = 133{,}000\cdot0.75/120 = 831$ scf/STB.

Answer: The balance uses

G_p = 1.427

billion m³; the gas equivalent of the condensate is about 150.3 m³/m³.

What it means. If we leave the condensate out, all the points on the $p/Z$ plot shift to the left by 1.9%, and the gas in place comes out too low by the same amount. For rich gas-condensate systems the correction reaches 10% and more. The data in this example are assumed for illustration.

Data, assumptions and typical mistakes

Working procedure with the p/Z plot

1
Collect the measurements
The average reservoir pressure on several dates, and the cumulative production on the same dates: gas, condensate and water.
2
Convert the production to reservoir gas
Add to the separator gas the gas equivalent of the condensate, the gas used on site and the losses.
3
Find Z for each pressure
From the composition or the specific gravity of the gas, at reservoir temperature and absolute pressure.
4
Plot p/Z against cumulative production
Draw a straight line through all the reliable points; read $p_i/Z_i$ and the gas in place.
5
Check whether the accumulation is closed
Calculate the apparent gas in place from each measurement, and compare the gas in place with the volumetric estimate. A growth of the apparent gas in place, or a gas in place clearly larger than the volumetric one, is a sign of water influx.
6
Make the forecast
The pressure after the planned withdrawal and the recoverable gas at the abandonment pressure. With water influx, we use an aquifer model and take trapped gas into account.

Average reservoir pressure. It is found from the pressure build-up in shut-in wells, brought to one datum depth and averaged over the volume (Pressure build-up analysis, Build-up in a bounded reservoir and average reservoir pressure). In a high-permeability accumulation the pressures in the wells are close. In a low-permeability or block-divided one they differ, and the model of one “tank” works poorly.
Complete production accounting. $G_p$ includes all the gas that has left the reservoir: separator gas, gas used on site and gas burned off at a flare, and condensate and water as gas equivalents. Gas injected back is subtracted.
Gas properties. $Z$ is for the composition of the reservoir gas and the reservoir temperature; the gas in place and the production are brought to the same standard conditions.
Early measurements. While the pressure has fallen by only a few per cent, the error of its measurement is comparable to the fall itself, and water influx has not yet shown itself. Early estimates of the gas in place cannot be trusted.
Only the connected volume. The balance “sees” the part of the accumulation whose pressure responds to the withdrawal. If the gas in place from the balance is clearly smaller than the volumetric one, part of the accumulation is not being drained.
No time. The balance links the withdrawal with the pressure, but does not say when that withdrawal will take place. The rates are set by the productivity of the wells and by the gas gathering system.
High initial pressure. In accumulations with abnormally high reservoir pressure (overpressure) the decrease of the pore volume noticeably supports the pressure, and a simple straight line overestimates the gas in place (see the advanced section; Abnormal formation pressures: origin, prediction and recognition).

Terms and symbols

Volumetric depletion drive: The drive of a closed gas accumulation: gas is produced only by its own expansion, and the pore volume occupied by gas is constant.
The $p/Z$ plot: The dependence of the ratio of the reservoir pressure to the gas deviation factor on cumulative gas production. For a closed accumulation it is a straight line that cuts off the gas in place on the production axis.
Apparent gas in place $G_a$: The gas in place calculated from one measurement on the assumption of a closed accumulation. Its growth from one measurement to the next points to water influx.
Abandonment pressure $p_a$: The pressure at which further production becomes impossible or unprofitable. It sets the gas recovery factor.
Gas recovery factor: The share of the gas initially in place that is produced by the end of development.
Residual gas saturation $S_{gr}$: The share of the pore volume in which gas stays trapped behind the water front.
Gas equivalent: The volume of gas at standard conditions that matches a unit volume of produced condensate or water.

Common mistakes and misconceptions

Wrong

“The p/Z points fell on a straight line, so the accumulation is closed and the gas in place is determined.”

In fact

With water influx the early points also fall almost on a straight line, only a flatter one. In the lesson's example, such a line overestimates the gas in place by 41%. Whether the accumulation is closed is checked by the constancy of the apparent gas in place and by an independent volumetric estimate.

Wrong

“We can plot the pressure: Z is close to one.”

In fact

$Z$ changes with pressure, and a line through the pressure is wrong: in the lesson's examples, from minus 27% for a deep accumulation to plus 10% for a shallow one. We plot $p/Z$ .

Wrong

“Edge water raises the gas recovery factor, as it raises the oil recovery.”

In fact

Water traps gas at high pressure, and it reaches the wells before the pressure has time to fall. In the lesson's example, the gas recovery factor with strong water influx is 57%, against 86% under depletion drive.

Wrong

“Condensate is not gas, so there is no need to include it in the gas production.”

In fact

In the reservoir the condensate was part of the gas. Its production is converted to the gas equivalent and added to $G_p$ ; otherwise the gas in place will come out too low.

Wrong

“The p/Z plot shows how many years the field will last.”

In fact

There is no time in the balance. It shows what the pressure will be after a given withdrawal. The duration depends on the withdrawal rate, which is set by the productivity of the wells and by demand.

In depth: advanced detail

Compressibility of the pores and the connate water

The full balance of a gas accumulation contains the same term as that of an oil accumulation. When the pressure falls by $\Delta p = p_i - p$ , the connate water expands and the pore volume decreases, and together they take from the gas the share $(c_wS_w + c_f)\Delta p/(1 - S_w)$ of the initial gas-filled volume (expansion of water and pores).

Balance of a gas accumulation with the water and the pores taken into account

G_pB_g + W_pB_w = G\,(B_g - B_{gi}) + G\,B_{gi}\,\frac{c_wS_w + c_f}{1 - S_w}\,\Delta p + W_e

where:

$c_w,\ c_f$ compressibility of water and of the pore volume, 1/MPa
$S_w$ connate water saturation, fraction
$\Delta p$ fall of the average reservoir pressure from the initial one, $p_i - p$ , MPa
$G,\ G_p,\ B_g,\ B_{gi}$ gas in place, cumulative production and gas formation volume factors, as in the balance of a closed accumulation, m³, m³, m³/m³, m³/m³
$W_e,\ W_p,\ B_w$ water influx, water production and its formation volume factor, as in the balance with water influx, m³, m³, m³/m³
$p,\ Z,\ p_i,\ Z_i$ in the second form: the current and initial average reservoir pressure (absolute) and the gas deviation factors at them, MPa, —, MPa, —

Units: m³ at reservoir conditions; pressure and compressibilities in consistent units

Holds when

The compressibilities are constant over the pressure range considered.
$c_f$ is the compressibility of the pore volume with respect to the fall of the pore pressure (Rock compressibility and the effect of stress on core properties).

The term with water and pores behaves as one more, small, source of energy. For a closed accumulation, it is no longer $p/Z$ that falls on the straight line, but $p/Z$ with a correction factor (the second form). Without the correction the pressure seems “too high”, and the gas in place is overestimated, just as with water influx.

In other unit systems (1)

Closed accumulation, through p/Z

\frac{p}{Z}\left[1 - \frac{c_wS_w + c_f}{1 - S_w}\,\Delta p\right] = \frac{p_i}{Z_i}\left(1 - \frac{G_p}{G}\right)

Pressures in MPa, compressibilities in 1/MPa (or psi and 1/psi).

There are no constants. The factor in square brackets is the share of the initial gas-filled volume that is left to the gas after the expansion of the connate water and the decrease of the pore volume.

Worked exampleHow large the correction is

Given (assumed data)

The lesson's accumulation: $S_w = 0.25$ ; $c_w = 0.45\cdot10^{-3}$ and $c_f = 0.8\cdot10^{-3}$ MPa⁻¹
Withdrawal: 1.6 billion m³ out of 4.89 billion m³

Solution

The coefficient: $(0.45\cdot0.25 + 0.8)\cdot10^{-3}/0.75 = 1.2167\cdot10^{-3}$ MPa⁻¹.
Without water and pores, the pressure after such a withdrawal would be 13.43 MPa. With them it is 13.53 MPa, which is 0.10 MPa higher.
The correction factor: $1 - 1.2167\cdot10^{-3}\cdot(20 - 13.53) = 0.9921$ .
The gas in place from this point without the correction: $\dfrac{1.6}{1 - 15.610/23.022} = 4.97$ billion m³, which is 1.7% more than the true value. With the correction: $\dfrac{1.6}{1 - 15.487/23.022} = 4.89$ billion m³.

Answer: For the lesson's accumulation, the water and pores change the estimate of the gas in place by 1.7%.

What it means. With ordinary pore compressibility and moderate pressures, the correction is small compared with the error of the measurements, and it is neglected. In accumulations with abnormally high pressure and weakly cemented rocks, the pore compressibility is several times larger, and the pressure fall is measured in tens of megapascals. There the early $p/Z$ points run noticeably flatter than the late ones, and a line through them gives a gas in place that is too high by tens of per cent. The data in this example are assumed for illustration.

How to go from p/Z to pressure

The plot gives $p/Z$ , but well calculations need the pressure. The equation $p = (p/Z)\cdot Z(p)$ is solved by successive approximations: we assume a value of $Z$ , find $p$ , use it to refine $Z$ , and repeat. Because $Z$ changes slowly with pressure, the approximations converge in two or three steps. Another way is to calculate in advance a table of $p/Z$ for a range of pressures at reservoir temperature, and find the pressure by interpolation.

Worked examplePressure from a value of p/Z

Given (assumed data)

Value from the plot: $p/Z = 10.0$ MPa
Gas and temperature: the lesson's gas, 80 °C

Solution

First approximation, $Z = 1$ : $p = 10.0$ MPa. At this pressure $Z = 0.8855$ .
Second: $p = 10.0\cdot0.8855 = 8.855$ MPa; $Z = 0.8946$ .
Third: $p = 10.0\cdot0.8946 = 8.946$ MPa; $Z = 0.8938$ , a change in the fourth digit.

Answer:

p \approx 8.94

MPa at

Z \approx 0.8938

What it means. The same technique was used in the example with the pressure forecast. If we start not from one but from the $Z$ of the nearest measurement, one step is enough. The data in this example are assumed for illustration.

Connections to other topics

Gas formation volume factor, compressibility and viscosity — the formation volume factor and compressibility of gas, on which the balance is built.
Ideal and real gas: the gas deviation factor (Z) — how the gas deviation factor $Z$ is found for each point of the plot.
General material balance equation for oil reservoirs — the general material balance equation and its assumptions.
Reservoir drive mechanisms — the depletion and water drives of gas accumulations among the other drives.
Volumetric estimation of hydrocarbons in place — the volumetric estimate of the gas in place, against which the result of the balance is checked.
Aquifer models: water influx — how the water influx is calculated, which in this lesson was found from the balance.
Straight-line material balance (Havlena–Odeh method) — finding the gas in place and the aquifer parameters together from a straight line.
Types of reservoir fluids — dry gas, wet gas and gas condensate: when the gas in the reservoir stays single-phase.
Build-up in a bounded reservoir and average reservoir pressure — where the average reservoir pressure for each point comes from.
Probabilistic volumetrics by Monte Carlo simulation — the uncertainty of the volumetric estimate, which the balance narrows.

Check yourself

1
On the p/Z plot of a closed gas accumulation, the straight line cuts off a segment of 12 billion m³ on the cumulative production axis. What is this quantity?
1. A.The production at the moment of the last measurement
2. B.The gas initially in place that is connected with the wells
3. C.The recoverable gas
4. D.The pore volume of the accumulation
Show the answer
Correct answer: B. The gas initially in place that is connected with the wells
The line reaches $p/Z = 0$ at $G_p = G$ : if the gas could be withdrawn completely, the pressure would fall to zero. This is the gas initially in place of that part of the accumulation whose pressure responds to the withdrawal. The recoverable gas is smaller: it is read at the level of the abandonment pressure.
2
A closed gas accumulation has $p_i/Z_i = 30$ MPa. After a withdrawal of 2.0 billion m³, $p/Z = 24$ MPa is measured. Find the gas in place and the gas recovery factor for an abandonment value of $p_a/Z_a = 4$ MPa.
Show the answer
Gas in place: $G = \dfrac{2.0}{1 - 24/30} = 10$ billion m³. Gas recovery factor: $1 - 4/30 = 0.867$ , that is 8.67 billion m³ of recoverable gas.
3
Why do we plot p/Z and not the reservoir pressure?
1. A. $Z$ accounts for water influx
2. B.It is easier to choose the scale
3. C.The amount of gas in a constant pore volume is proportional to $p/Z$ , not to $p$ : $Z$ changes with pressure
4. D.The pressure cannot be measured accurately
Show the answer
Correct answer: C. The amount of gas in a constant pore volume is proportional to $p/Z$ , not to $p$ : $Z$ changes with pressure
From the equation of state of a real gas, the number of moles in a given volume at constant temperature is proportional to $p/Z$ . So it is exactly $p/Z$ that falls linearly with production. The pressure by itself follows a curve, and its extension from the early points gives the wrong gas in place.
4
The apparent gas in place calculated from three successive measurements is 12, 15 and 19 billion m³. What does this mean?
1. A.The gas behaves as an ideal gas
2. B.The estimate is being refined; we should trust the last value
3. C.The production is being overstated in the accounting
4. D.Water is coming into the accumulation, and the pressure falls more slowly than in a closed one
Show the answer
Correct answer: D. Water is coming into the accumulation, and the pressure falls more slowly than in a closed one
$G_a = G + W_e/(B_g - B_{gi})$ : for a closed accumulation the second term is zero, and the apparent gas in place is constant. A steady growth is a sign of water influx (or of another source of energy that has not been counted, such as compaction of the rock). The true gas in place is smaller than any of the three values.
5
The gas in place by the volumetric method is 8.0 billion m³, and $B_{gi} = 0.0045$ m³/m³. At the date of the measurement 2.0 billion m³ of gas has been produced, no water has been produced, and $B_g = 0.0056$ m³/m³. How much water has come into the accumulation?
Show the answer
Withdrawal: $G_pB_g = 2.0\cdot10^{9}\cdot0.0056 = 11.2$ million m³. Expansion of the gas: $G(B_g - B_{gi}) = 8.0\cdot10^{9}\cdot0.0011 = 8.8$ million m³. Water influx: $W_e = 11.2 - 8.8 = 2.4$ million m³ at reservoir conditions.
6
50 thousand m³ of condensate has been produced, with a density of 780 kg/m³ and a molar mass of 130 kg/kmol. What volume of gas must be added to the production in the material balance?
Show the answer
$GE = 24.05\cdot780/130 = 144.3$ m³/m³. The addition is $50\cdot10^{3}\cdot144.3 \approx 7.2$ million m³ of gas at standard conditions.

Updated: 19 October 2026