Performance prediction for solution gas drive reservoirs

Quick refresher

Below the bubble point pressure the balance equation links two unknowns to the pressure: the cumulative production of oil $N_p$ and of gas $G_p$ . The second equation is the instantaneous gas–oil ratio from Darcy's law. The third is the average oil saturation, through which the gas–oil ratio depends on $N_p$ .
The instantaneous gas–oil ratio is the sum of the gas dissolved in the oil produced ( $R_s$ ) and the free gas that came with it. The free gas is proportional to the ratio of the relative permeabilities $k_{rg}/k_{ro}$ (how well gas flows compared with oil in the same rock), multiplied by a large factor $\mu_oB_o/(\mu_gB_g)$ . The gas is tens of times less viscous than the oil, and in the reservoir it is compressed tens to hundreds of times.
$R$ is the ratio of the current gas and oil rates, $R_p$ is the ratio of all the gas produced to all the oil produced, and $R_s$ is the gas that the oil still holds at the current pressure. The balance equation uses $R_p$ , and Darcy's law uses $R$ .
Tarner's method: for each pressure step we choose the oil production for which two amounts of gas agree. One is the gas that the balance needs. The other is the gas given up by the wells according to the gas–oil ratio (the average over the two ends of the step). Muskat's method is the same system in differential form, and Tracy's form is the same balance written with functions of pressure.
The forecast is only as good as the ratio $k_{rg}/k_{ro}$ and the PVT properties. We first test the model against the history. The field ratio of relative permeabilities is usually higher than the laboratory one, so a forecast made with laboratory curves comes out too optimistic.
Time comes from the productivity of the wells. It is proportional to the oil mobility $k_{ro}/(\mu_oB_o)$ (how easily the oil moves), which falls together with the pressure. The time of a step is the oil produced in the step divided by the average rate.

Key relationships

R = R_s + \frac{k_{rg}}{k_{ro}}\cdot\frac{\mu_oB_o}{\mu_gB_g}

instantaneous gas–oil ratio

R_p = \frac{1}{N_p}\int_0^{N_p} R\,\dd N_p

cumulative gas–oil ratio through the instantaneous one

G_p = \frac{N E_t - N_p\,(B_o - R_sB_g)}{B_g}

gas that the balance of a closed accumulation without a cap needs; $E_t = E_o + E_{fw}$

S_o = \frac{(N - N_p)\,B_o}{V_p}

average oil saturation; $V_p$ is the current pore volume

\Delta t = \frac{\Delta N_p}{\bar q_o}

moving from production to time

Common pitfalls

Calculating the gas of a step from the gas–oil ratio at the start of the step only, or taking too large pressure steps where the gas–oil ratio grows fast.
Putting the current gas–oil ratio into the balance in place of the cumulative one.
Taking laboratory relative permeability curves without a check against the history. The gas–oil ratio will come out too low and the oil recovery too high.
Dropping the expansion of the connate water and the shrinking of the pore volume when development started well above the bubble point pressure.
Building the forecast on the pressure of one well instead of the volume-average pressure of the accumulation.

After this lesson you can

explain why material balance alone is not enough for a forecast below the bubble point pressure, and which equations complete it
derive the equation of the instantaneous gas–oil ratio from Darcy's law, use it to explain how the gas–oil ratio changes, and not confuse the instantaneous and cumulative gas–oil ratios with the solution gas–oil ratio
calculate the average oil and gas saturation from the cumulative production, including the expansion of the connate water and the shrinking of the pore volume
carry out a forecast step with Tarner's method and build a forecast of pressure, production and gas–oil ratio down to the final pressure
check the input data against the development history, including finding the field gas–oil relative permeability ratio, and average the pressure over the drainage areas of the wells
turn the forecast of “production against pressure” into a forecast in time by using the productivity of the wells

What the balance is missing below the bubble point pressure

Take a closed accumulation without a gas cap and without an aquifer, the same one as in the lesson General material balance equation for oil reservoirs: oil in place 9.06 million m³, initial pressure 30 MPa, bubble point pressure 18 MPa. While the pressure is above the bubble point pressure, no gas comes out, the gas–oil ratio equals the initial solution gas–oil ratio, and the balance equation gives the cumulative production unambiguously at any pressure. Down to 18 MPa, compressibility drive (drive by the expansion of the liquids and the rock) gives about 302 thousand m³ of oil, which is 3.3% of the oil in place.

Below the bubble point pressure free gas appears in the underground withdrawal (the reservoir volume of everything produced), $N_p(R_p - R_s)B_g$ , and the cumulative gas–oil ratio $R_p$ is not known in advance. Let us rewrite the balance so that the gas stands on its own:

Gas that the balance needs

N_pB_o + (G_p - N_pR_s)\,B_g = N E_t\quad\Longrightarrow\quad G_p = \frac{N E_t - N_p\,(B_o - R_sB_g)}{B_g}

where:

$G_p$ cumulative gas production at standard conditions, $G_p = N_pR_p$ , m³
$N_p$ cumulative oil production at standard conditions, m³
$N$ oil in place, m³
$E_t$ expansion per 1 m³ of oil in place from the initial to the current pressure: $E_o + E_{fw}$ from the lesson General material balance equation for oil reservoirs, m³/m³
$B_o,\ B_g,\ R_s$ formation volume factors of the oil and the gas, and the solution gas–oil ratio, at the current average pressure, m³/m³

Units: consistent units: m³ (oil and gas at standard conditions: 0.101325 MPa and 20 °C), $B_g$ in m³/m³; in oilfield units, STB, scf, rb/STB, rb/scf

Holds when

the accumulation is closed: no gas cap, no water influx or water production, no injection
the pressure is the same throughout the volume (the “tank”), the phases are in equilibrium, and the properties are those of the black oil model at the average pressure
$E_t$ is counted from the initial pressure of 30 MPa. If the accumulation started development above the bubble point pressure, $E_t$ includes the elastic expansion of the oil, the connate water and the pores gained there.

For each pressure this is a straight line in the coordinates $(N_p, G_p)$ : the more gas the wells carry away, the less oil the accumulation gives for the same pressure drop. The balance does not know which point of this line will be realised. That is decided by the mobility of the gas in the reservoir.

Unknowns of the forecast and the equations that link them

Quantity: $N_p$ and $G_p$ at a given pressure
What gives it: material balance: one equation for two unknowns

Quantity: Instantaneous gas–oil ratio $R$
What gives it: Darcy's law for gas and oil

Quantity: Average saturation $S_o$ , $S_g$
What gives it: balance of pore volumes: the oil that is left in the current pore volume

Quantity: $k_{rg}/k_{ro}$ at this saturation
What gives it: relative permeability curves, either from the laboratory or corrected against the history

Quantity: Time
What gives it: well rates: productivity, and the limits set by the bottomhole pressure and by the production plateau

Quantity	What gives it
$N_p$ and $G_p$ at a given pressure	material balance: one equation for two unknowns
Instantaneous gas–oil ratio $R$	Darcy's law for gas and oil
Average saturation $S_o$ , $S_g$	balance of pore volumes: the oil that is left in the current pore volume
$k_{rg}/k_{ro}$ at this saturation	relative permeability curves, either from the laboratory or corrected against the history
Time	well rates: productivity, and the limits set by the bottomhole pressure and by the production plateau

The instantaneous gas–oil ratio

Below the bubble point pressure both oil and free gas flow to the well. Each phase obeys Darcy's law with its own effective permeability, $k\,k_{ro}$ or $k\,k_{rg}$ (Relative permeability). If we neglect capillary pressure (the pressure difference between gas and oil that surface forces in the pores create), the two phases have the same pressure drop. Then, in the ratio of the reservoir rates of gas and oil, both the pressure drop and the geometry of the inflow cancel: $q_g/q_o = (k_{rg}/\mu_g)/(k_{ro}/\mu_o)$ .

At the surface, one cubic metre of reservoir oil gives $1/B_o$ cubic metres of stock-tank oil and $R_s/B_o$ cubic metres of dissolved gas, and one cubic metre of free gas gives $1/B_g$ cubic metres of gas. The ratio of all the gas to the oil at the surface is the instantaneous gas–oil ratio:

Instantaneous gas–oil ratio

R = R_s + \frac{k_{rg}}{k_{ro}}\cdot\frac{\mu_o\,B_o}{\mu_g\,B_g}

where:

$R$ instantaneous (current) gas–oil ratio: the ratio of the gas rate to the oil rate, both at standard conditions, m³/m³
$R_s$ solution gas–oil ratio of the oil at the current pressure: the dissolved gas, m³/m³
$k_{rg},\ k_{ro}$ relative permeabilities to gas and to oil at the current gas saturation, —
$\mu_o,\ \mu_g$ viscosities of the oil and the gas at reservoir conditions, mPa·s
$B_o,\ B_g$ formation volume factors of the oil and the gas at the current pressure, m³/m³

Units: any consistent system: in metric units $R$ and $R_s$ in m³/m³, $B_g$ in m³/m³; in oilfield units $R$ and $R_s$ in scf/STB, $B_o$ in rb/STB, $B_g$ in rb/scf; the viscosities in the same units

Holds when

gas and oil flow to the well under the same pressure drop (capillary pressure is neglected)
the relative permeabilities are taken at the average gas saturation of the drainage area, and in a tank forecast at that of the whole accumulation
all the gas in the production is either dissolved in the oil produced or free gas of the same reservoir: there is no breakthrough of gas from a gas cap or from neighbouring formations

The first term is the gas that the oil brought in solution, and the second is the free gas. For ordinary oils the factor $\mu_oB_o/(\mu_gB_g)$ is thousands or tens of thousands: the gas is tens of times less viscous, and in the reservoir it is compressed tens to hundreds of times. So even a very small ratio $k_{rg}/k_{ro}$ gives a large flow of free gas.

The equation explains how the gas–oil ratio changes, as described in the lesson on drive mechanisms. Above the bubble point pressure there is no free gas, and $R = R_{si}$ . Just below it, gas comes out, but while its saturation is less than the critical gas saturation $S_{gc}$ (the saturation at which gas starts to move), it does not move: $k_{rg} = 0$ and $R = R_s$ , a little less than the initial value. Then the gas becomes mobile, $k_{rg}/k_{ro}$ grows by orders of magnitude (by factors of ten), and the gas–oil ratio rises several times over. Later the rise stops. At low pressure $B_g$ grows quickly, and the same reservoir volume of gas gives less and less gas at the surface. The gas–oil ratio passes a maximum and falls.

Worked exampleGas–oil ratio at 14 MPa

Given (assumed data)

Average gas saturation: $S_g = 0.0795$ (from the forecast below)
Relative permeability curves: $k_{ro} = 0.80\left(\dfrac{0.55 - S_g}{0.55}\right)^{2}$ ; $k_{rg} = 0.30\left(\dfrac{S_g - 0.03}{0.52}\right)^{1.5}$
Properties at 14 MPa: $R_s = 87.73$ m³/m³; $B_o = 1.2893$ ; $B_g = 0.007693$ m³/m³; $\mu_o = 1.175$ ; $\mu_g = 0.01700$ mPa·s

Solution

Relative permeabilities: $k_{ro} = 0.80\cdot(0.4705/0.55)^2 = 0.5854$ ; $k_{rg} = 0.30\cdot(0.0495/0.52)^{1.5} = 0.00881$ ; the ratio is $k_{rg}/k_{ro} = 0.015050$ .
The factor: $\dfrac{\mu_oB_o}{\mu_gB_g} = \dfrac{1.175\cdot1.2893}{0.01700\cdot0.007693} = 11{,}584$ .
Free gas: $0.015050\cdot11{,}584 = 174.3$ m³/m³; the gas–oil ratio is $R = 87.73 + 174.3 = 262.1$ m³/m³.
Volumes in the reservoir: for each 1 m³ of reservoir oil, $(k_{rg}/k_{ro})(\mu_o/\mu_g) = 1.04$ m³ of reservoir gas flows to the wells.

Answer: The gas–oil ratio at 14 MPa is about 262.1 m³/m³, which is 2.4 times the initial solution gas–oil ratio. 67% of the gas at the surface is free gas from the reservoir.

What it means. The permeability to gas here is only one and a half per cent of the permeability to oil. Yet in volume at reservoir conditions almost as much gas flows to the wells as oil, because the gas is 69 times less viscous. At the surface each cubic metre of this gas expands by another factor of $1/B_g \approx 130$ . This is how the energy of the drive is “spent”: the gas leaves the reservoir faster than the oil that it is supposed to push out. The data in this example are assumed for illustration.

Do not confuse the instantaneous gas–oil ratio with two other quantities. The solution gas–oil ratio $R_s$ is how much gas the oil still holds in solution at the current pressure; below the bubble point pressure it only decreases. The cumulative gas–oil ratio $R_p$ is all the gas produced divided by all the oil produced. It enters the balance equation and is linked to the instantaneous ratio by an integral over the production history.

Cumulative gas–oil ratio through the instantaneous one

R_p = \frac{G_p}{N_p} = \frac{1}{N_p}\int_0^{N_p} R\,\dd N_p \approx \frac{1}{N_p}\sum_i \bar R_i\,\Delta N_{p,i}

where:

$R_p$ cumulative gas–oil ratio, m³/m³
$\bar R_i$ average instantaneous gas–oil ratio over the $i$ -th interval of production, m³/m³
$\Delta N_{p,i}$ oil produced over the $i$ -th interval, m³

Units: m³/m³ (scf/STB)

Holds when

gas used for gas lift and injected gas are excluded from the production figures
in the step-by-step calculation $\bar R_i$ is the average of the gas–oil ratios at the two ends of the interval, so the intervals are taken shorter where $R$ changes fast

$R_p$ is the accumulated “memory” of the gas–oil ratio. When the instantaneous ratio grows, the cumulative one lags behind it, because it also includes all the early production with a low gas–oil ratio.

Average saturation from the balance

The relative permeabilities depend on saturation, and the balance gives the saturation too. The accumulation is a “tank”, and the saturation in it is an average. In the pore volume there remain the oil $(N - N_p)B_o$ , the connate water and the free gas, which fills the rest. For an accumulation that has passed the elastic stage, the pore volume has already shrunk a little and the connate water has expanded. These corrections are needed because the balance contains $E_{fw}$ (the balance of the occupied space).

Average saturations of a closed accumulation

S_o = \frac{(N - N_p)\,B_o}{V_p},\qquad V_p = \frac{N B_{oi}}{1 - S_{wi}}\,(1 - c_f\Delta p),\qquad S_w = \frac{S_{wi}\,(1 + c_w\Delta p)}{1 - c_f\Delta p},\qquad S_g = 1 - S_o - S_w

where:

$S_o,\ S_w,\ S_g$ average oil, water and gas saturations at the current pressure, fraction
$V_p$ current pore volume of the accumulation, m³
$S_{wi}$ initial connate water saturation, fraction
$B_{oi}$ oil formation volume factor at the initial pressure, m³/m³
$c_w,\ c_f$ compressibility of the water and of the pore volume, 1/MPa
$\Delta p$ fall in pressure from the initial value, $p_i - p$ , MPa

Units: fractions; volumes in m³, pressure and compressibilities in MPa and 1/MPa

Holds when

the saturation is the same throughout the accumulation: the gas does not collect at the top in a secondary gas cap
the accumulation is closed, and no water is produced
the compressibilities are constant over the pressure range considered

The oil saturation falls for two reasons: oil is produced, and the oil that is left shrinks as it gives up gas. Free gas takes the space that is freed, and its saturation sets $k_{rg}/k_{ro}$ .

In other unit systems (1)

Without the compressibility of the water and the pores

S_o = \left(1 - \frac{N_p}{N}\right)\frac{B_o}{B_{oi}}\,(1 - S_{wi})

The same quantities; $S_w = S_{wi}$ and the pore volume is constant.

There are no constants. If the oil in place and the production are counted from the bubble point pressure, $B_{ob}$ stands in place of $B_{oi}$ .

Worked exampleSaturations at 14 MPa: with and without the corrections

Given (assumed data)

Production to 14 MPa: $N_p = 908.8$ thousand m³; oil in place $N = 9.06$ million m³
Properties: $B_{oi} = 1.3243$ ; $B_o = 1.2893$ m³/m³; $S_{wi} = 0.25$ ; $c_w = 0.45\cdot10^{-3}$ , $c_f = 0.8\cdot10^{-3}$ MPa⁻¹; $\Delta p = 16$ MPa

Solution

Pore volume: $\dfrac{9.06\cdot1.3243}{0.75}\cdot(1 - 0.0128) = 15.793$ million m³.
Oil: $S_o = (9.06 - 0.9088)\cdot1.2893/15.793 = 0.6654$ .
Water: $S_w = 0.25\cdot1.0072/0.9872 = 0.2551$ ; gas: $S_g = 1 - 0.6654 - 0.2551 = 0.0795$ .
Without the corrections: $S_o = (1 - 0.1003)\cdot\dfrac{1.2893}{1.3243}\cdot0.75 = 0.6569$ , $S_g = 1 - 0.6569 - 0.25 = 0.0931$ .
With the curves of the example above: $k_{rg}/k_{ro} = 0.01505$ with the corrections and 0.02297 without them.

Answer: The average gas saturation at 14 MPa is 7.95%. Without the water and the pores we would get 9.31%, and the ratio

k_{rg}/k_{ro}

would be 1.5 times larger.

What it means. The corrections are small, about one per cent of the pore volume. But the gas saturation itself is small, and an error of 17% in it changes the gas mobility noticeably: near the critical gas saturation $k_{rg}$ rises very steeply. Neglecting the compressibility of the water and the pores overstates the gas saturation and the gas–oil ratio, which makes the forecast pessimistic. For an accumulation that was saturated with gas from the very beginning, the difference would be much smaller. The data in this example are assumed for illustration.

Tarner's method: one forecast step

The forecast goes from the bubble point pressure downward in pressure steps. In each step one quantity is unknown: the cumulative oil production at the end of the step, $N_{p,k}$ . Everything else is calculated from it. The gas produced in the step can be found in two independent ways. From the balance: this is the amount of gas that had to leave so that the volumes fit at the new pressure. From the gas–oil ratio: this is the amount of gas that the wells brought up together with the oil of the step. The right oil production is the one for which the two ways give the same result.

Step condition in Tarner's method

\underbrace{G_p(p_k, N_{p,k}) - G_{p,j}}_{\text{from the balance}} = \underbrace{\frac{R_j + R_k}{2}\,\big(N_{p,k} - N_{p,j}\big)}_{\text{from the gas–oil ratio}}

where:

$j,\ k$ start and end of the pressure step ( $p_k < p_j$ ), —
$G_p(p_k, N_{p,k})$ cumulative gas production that the balance needs at pressure $p_k$ and oil production $N_{p,k}$ (the equation above), m³
$G_{p,j},\ N_{p,j},\ R_j$ cumulative gas and oil and the instantaneous gas–oil ratio at the start of the step; they are known, m³, m³, m³/m³
$R_k$ instantaneous gas–oil ratio at the end of the step: from $N_{p,k}$ we get the saturation, and from it $k_{rg}/k_{ro}$ , m³/m³

Units: m³ of gas at standard conditions

Holds when

within a step the gas–oil ratio changes linearly with the oil production (the trapezoid rule), so the pressure steps are taken small, a fraction of a megapascal where $R$ grows fast
all the assumptions of the tank balance and of the gas–oil ratio equation

As $N_{p,k}$ grows, the left side falls (the more oil the accumulation gives, the less gas it had to give), and the right side grows. So the solution is unique, and we find it by trial: two trials and an interpolation between them.

Order of a step

1
Set the pressure at the end of the step
For example, 0.25–1 MPa lower than at the start of the step. The faster the gas–oil ratio changes, the shorter the step.
2
Guess the oil production
$N_{p,k}$ , for example by continuing the trend of the previous steps.
3
Gas from the balance
$G_p(p_k, N_{p,k})$ and the gain in gas over the step.
4
Gas–oil ratio at the end of the step
From $N_{p,k}$ the average saturations, from the gas saturation $k_{rg}/k_{ro}$ , and from these $R_k$ .
5
Gas from the gas–oil ratio
The average gas–oil ratio of the step multiplied by the oil production of the step.
6
Compare and refine
If the balance needs more gas than the wells gave, increase the oil production; if less, decrease it. Two trials on opposite sides of the solution and a linear interpolation usually give the answer to within a fraction of a per cent.
7
Move to the next step
The $N_{p,k}$ , $G_{p,k}$ and $R_k$ that we found become the start of the next step.

Properties of the assumed oil below the bubble point pressure

$p$ , MPa: 18
$R_s$ , m³/m³: 110.00
$B_o$ , m³/m³: 1.3500
$B_g$ , m³/m³: 0.005983

$p$ , MPa: 16
$R_s$ , m³/m³: 98.94
$B_o$ , m³/m³: 1.3198
$B_g$ , m³/m³: 0.006731

$p$ , MPa: 14
$R_s$ , m³/m³: 87.73
$B_o$ , m³/m³: 1.2893
$B_g$ , m³/m³: 0.007693

$p$ , MPa: 13
$R_s$ , m³/m³: 82.07
$B_o$ , m³/m³: 1.2738
$B_g$ , m³/m³: 0.008285

$p$ , MPa: 12
$R_s$ , m³/m³: 76.37
$B_o$ , m³/m³: 1.2583
$B_g$ , m³/m³: 0.008975

$p$ , MPa: 10
$R_s$ , m³/m³: 64.81
$B_o$ , m³/m³: 1.2268
$B_g$ , m³/m³: 0.010770

$p$ , MPa: 8
$R_s$ , m³/m³: 53.02
$B_o$ , m³/m³: 1.1946
$B_g$ , m³/m³: 0.013462

$p$ , MPa: 6
$R_s$ , m³/m³: 40.92
$B_o$ , m³/m³: 1.1616
$B_g$ , m³/m³: 0.017950

$p$ , MPa: 5
$R_s$ , m³/m³: 34.73
$B_o$ , m³/m³: 1.1447
$B_g$ , m³/m³: 0.021540

$p$ , MPa	$R_s$ , m³/m³	$B_o$ , m³/m³	$B_g$ , m³/m³
18	110.00	1.3500	0.005983
16	98.94	1.3198	0.006731
14	87.73	1.2893	0.007693
13	82.07	1.2738	0.008285
12	76.37	1.2583	0.008975
10	64.81	1.2268	0.010770
8	53.02	1.1946	0.013462
6	40.92	1.1616	0.017950
5	34.73	1.1447	0.021540

The oil of the lesson Black oil model: solution gas–oil ratio and formation volume factors: bubble point pressure 18 MPa at 85 °C; $B_g$ is for $Z = 0.87$ and standard conditions of 0.101325 MPa and 20 °C.

Viscosities and the factor of the gas–oil ratio equation

$p$ , MPa: 18
$\mu_o$ , mPa·s: 1.000
$\mu_g$ , mPa·s: 0.01912
$\mu_oB_o/(\mu_gB_g)$: 11,801

$p$ , MPa: 16
$\mu_o$ , mPa·s: 1.079
$\mu_g$ , mPa·s: 0.01800
$\mu_oB_o/(\mu_gB_g)$: 11,754

$p$ , MPa: 14
$\mu_o$ , mPa·s: 1.175
$\mu_g$ , mPa·s: 0.01700
$\mu_oB_o/(\mu_gB_g)$: 11,584

$p$ , MPa: 13
$\mu_o$ , mPa·s: 1.232
$\mu_g$ , mPa·s: 0.01653
$\mu_oB_o/(\mu_gB_g)$: 11,459

$p$ , MPa: 12
$\mu_o$ , mPa·s: 1.296
$\mu_g$ , mPa·s: 0.01609
$\mu_oB_o/(\mu_gB_g)$: 11,293

$p$ , MPa: 10
$\mu_o$ , mPa·s: 1.451
$\mu_g$ , mPa·s: 0.01529
$\mu_oB_o/(\mu_gB_g)$: 10,810

$p$ , MPa: 8
$\mu_o$ , mPa·s: 1.660
$\mu_g$ , mPa·s: 0.01457
$\mu_oB_o/(\mu_gB_g)$: 10,110

$p$ , MPa: 6
$\mu_o$ , mPa·s: 1.960
$\mu_g$ , mPa·s: 0.01394
$\mu_oB_o/(\mu_gB_g)$: 9099

$p$ , MPa: 5
$\mu_o$ , mPa·s: 2.166
$\mu_g$ , mPa·s: 0.01366
$\mu_oB_o/(\mu_gB_g)$: 8427

$p$ , MPa	$\mu_o$ , mPa·s	$\mu_g$ , mPa·s	$\mu_oB_o/(\mu_gB_g)$
18	1.000	0.01912	11,801
16	1.079	0.01800	11,754
14	1.175	0.01700	11,584
13	1.232	0.01653	11,459
12	1.296	0.01609	11,293
10	1.451	0.01529	10,810
8	1.660	0.01457	10,110
6	1.960	0.01394	9099
5	2.166	0.01366	8427

The data are assumed. The viscosity of the oil grows as gas leaves it: from 1.00 mPa·s at the bubble point pressure to 2.17 mPa·s at 5 MPa. The gas viscosity is for gas of relative density 0.75 at 85 °C. The gas that comes out in the reservoir at these pressures is lighter than all the gas dissolved in the oil.

Worked exampleA forecast step from 14 to 13 MPa

Given (assumed data)

Start of the step, 14 MPa (the forecast table below): $N_{p,j} = 908.8$ thousand m³; $G_{p,j} = 118.24$ million m³; $R_j = 262.2$ m³/m³
Properties at 13 MPa: $R_s = 82.07$ m³/m³; $B_o = 1.2738$ ; $B_g = 0.008285$ m³/m³; $\mu_o = 1.232$ ; $\mu_g = 0.01653$ mPa·s
Accumulation: $N = 9.06$ million m³; $B_{oi} = 1.3243$ ; $S_{wi} = 0.25$ ; $(c_wS_{wi} + c_f)/(1 - S_{wi}) = 1.2167\cdot10^{-3}$ MPa⁻¹; $\Delta p = 17$ MPa
Relative permeability curves: as in the example “Gas–oil ratio at 14 MPa”

Solution

Expansion per 1 m³ of oil in place: $E_t = (1.2738 - 1.3243) + (110 - 82.07)\cdot0.008285 + 1.3243\cdot1.2167\cdot10^{-3}\cdot17 = 0.20829$ m³/m³; $NE_t = 1.8871$ million m³. The factor is $B_o - R_sB_g = 0.5939$ m³/m³.
Pore volume at 13 MPa: $15.998\cdot(1 - 0.0136) = 15.780$ million m³; the connate water saturation is $S_w = 0.2554$ .
Trial 1: $N_{p,k} = 1000$ thousand m³. From the balance, $G_p = (1.8871 - 1.000\cdot0.5939)/0.008285 = 156.1$ million m³, a gain of 37.9 million m³. Saturations: $S_o = 8.06\cdot1.2738/15.780 = 0.6506$ , $S_g = 0.0940$ ; $k_{rg}/k_{ro} = 0.02355$ ; $R_k = 82.07 + 0.02355\cdot11{,}459 = 351.9$ m³/m³. From the gas–oil ratio: $\tfrac12(262.2 + 351.9)\cdot91.2\cdot10^3 = 28.0$ million m³. The balance needs 9.9 million m³ more gas than the wells give, so more than 1000 thousand m³ of oil must have been produced.
Trial 2: $N_{p,k} = 1050$ thousand m³. From the balance, 152.5 million m³, a gain of 34.3 million m³. $S_o = 0.6466$ , $S_g = 0.0980$ , $R_k = 383.2$ m³/m³; from the gas–oil ratio, 45.6 million m³. Now the wells have given 11.3 million m³ more than the balance allows.
Interpolation: $N_{p,k} = 1000 + 50\cdot\dfrac{9.9}{9.9 + 11.3} = 1023.3$ thousand m³. The exact solution of the step equation is 1023.8 thousand m³, with $R_k = 366.6$ m³/m³.

Answer: In the step from 14 to 13 MPa the accumulation will give about 115 thousand m³ of oil. The cumulative production will be 1023.3 thousand m³, and the gas–oil ratio will grow from 262 to 367 m³/m³.

What it means. The forecast table, calculated in steps of 0.25 MPa, gives 1024.5 thousand m³ at 13 MPa. One step of 1 MPa was wrong by only 0.07%. The error comes from the trapezoid, because the gas–oil ratio does not grow linearly with production. Note the proportion: in this step about 1.3% of the oil in place was produced, but 3.6% of the initial dissolved gas. The data in this example are assumed for illustration.

Forecast down to the final pressure

The forecast starts at the bubble point pressure, where everything is known: compressibility drive has produced $N_{pb} = NE_t(p_b)/B_{ob}$ of oil, all the gas produced was dissolved, and $R = R_p = R_{si}$ . Then, step by step, we go on down to the final pressure, at which production stops paying for itself or the wells can no longer work.

Forecast for the assumed accumulation below the bubble point pressure

$p$ , MPa: 18
$N_p/N$ , %: 3.33
$R$ , m³/m³: 110
$R_p$ , m³/m³: 110
$S_g$: 0.000

$p$ , MPa: 17
$N_p/N$ , %: 4.88
$R$ , m³/m³: 104
$R_p$ , m³/m³: 109
$S_g$: 0.019

$p$ , MPa: 16
$N_p/N$ , %: 6.70
$R$ , m³/m³: 113
$R_p$ , m³/m³: 108
$S_g$: 0.040

$p$ , MPa: 15
$N_p/N$ , %: 8.50
$R$ , m³/m³: 174
$R_p$ , m³/m³: 115
$S_g$: 0.061

$p$ , MPa: 14
$N_p/N$ , %: 10.03
$R$ , m³/m³: 262
$R_p$ , m³/m³: 130
$S_g$: 0.080

$p$ , MPa: 13
$N_p/N$ , %: 11.31
$R$ , m³/m³: 367
$R_p$ , m³/m³: 151
$S_g$: 0.096

$p$ , MPa: 12
$N_p/N$ , %: 12.39
$R$ , m³/m³: 483
$R_p$ , m³/m³: 174
$S_g$: 0.111

$p$ , MPa: 11
$N_p/N$ , %: 13.32
$R$ , m³/m³: 609
$R_p$ , m³/m³: 200
$S_g$: 0.125

$p$ , MPa: 10
$N_p/N$ , %: 14.14
$R$ , m³/m³: 741
$R_p$ , m³/m³: 228
$S_g$: 0.137

$p$ , MPa: 9
$N_p/N$ , %: 14.87
$R$ , m³/m³: 878
$R_p$ , m³/m³: 256
$S_g$: 0.150

$p$ , MPa: 8
$N_p/N$ , %: 15.54
$R$ , m³/m³: 1017
$R_p$ , m³/m³: 286
$S_g$: 0.161

$p$ , MPa: 7
$N_p/N$ , %: 16.16
$R$ , m³/m³: 1153
$R_p$ , m³/m³: 316
$S_g$: 0.173

$p$ , MPa: 6
$N_p/N$ , %: 16.74
$R$ , m³/m³: 1280
$R_p$ , m³/m³: 348
$S_g$: 0.184

$p$ , MPa: 5
$N_p/N$ , %: 17.29
$R$ , m³/m³: 1388
$R_p$ , m³/m³: 379
$S_g$: 0.195

$p$ , MPa	$N_p/N$ , %	$R$ , m³/m³	$R_p$ , m³/m³	$S_g$
18	3.33	110	110	0.000
17	4.88	104	109	0.019
16	6.70	113	108	0.040
15	8.50	174	115	0.061
14	10.03	262	130	0.080
13	11.31	367	151	0.096
12	12.39	483	174	0.111
11	13.32	609	200	0.125
10	14.14	741	228	0.137
9	14.87	878	256	0.150
8	15.54	1017	286	0.161
7	16.16	1153	316	0.173
6	16.74	1280	348	0.184
5	17.29	1388	379	0.195

Tarner's method, step 0.25 MPa; relative permeability curves matched to the history (see the section on the input data). Above 18 MPa the drive is compressibility drive; the bubble point pressure is 18 MPa, and the final pressure of the forecast is 5 MPa. The data are assumed.

CalculatedGas–oil ratios in the forecast for the assumed accumulation against the fraction of the oil in place produced: the instantaneous ratio

R

, the cumulative ratio

R_p

and the solution gas–oil ratio

R_s

of the oil at the current pressure. The labels give the reservoir pressure.

The gas becomes mobile early. The critical gas saturation, 3%, is reached already at about 16.5 MPa, and at 14 MPa the gas–oil ratio is 2.4 times the initial solution gas–oil ratio.
The pressure falls fast. From 18 to 14 MPa the accumulation gave 6.7% of the oil in place, from 14 to 10 MPa only 4.1%, and from 10 to 5 MPa 3.2%. Each following megapascal “costs” less and less oil, because a larger and larger part of the reservoir energy leaves with the gas.
In the range of the forecast the gas–oil ratio is still rising. The maximum mentioned above comes for this oil below 5 MPa, when the gas formation volume factor starts to grow faster than the ratio $k_{rg}/k_{ro}$ .

Worked exampleWhere the energy of the dissolved gas went

Given (assumed data)

Forecast at 5 MPa: $N_p = 1566.6$ thousand m³; $R_p = 379.5$ m³/m³
Oil in place and solution gas–oil ratio: $N = 9.06$ million m³; $R_{si} = 110$ m³/m³; at 5 MPa $R_s = 34.73$ m³/m³

Solution

The dissolved gas at the start was $NR_{si} = 9.06\cdot110 = 996.6$ million m³.
The gas produced is $G_p = N_pR_p = 594.5$ million m³, which is 60% of the initial amount.
The gas that stays dissolved in the oil is $(N - N_p)R_s = 260.2$ million m³, or 26%. The rest is free gas in the reservoir: 141.9 million m³ (14%).
Oil recovery: $N_p/N = 17.3$ %.

Answer: By 5 MPa, 17.3% of the oil and 60% of the dissolved gas have been produced.

What it means. Gas is the main source of energy in this drive, and more than half of it went to the surface while pushing out only a sixth of the oil. That is why pressure maintenance is so important for such accumulations: injecting water before the pressure falls below the bubble point pressure keeps the gas in solution and the energy in the reservoir (Why water is injected: pressure maintenance and oil displacement). The data in this example are assumed for illustration.

Input data and the check against the history

What the forecast needs

Data: PVT properties
What exactly: $B_o$ , $R_s$ , $B_g$ (or $Z$ ), $\mu_o$ , $\mu_g$ as functions of pressure
Where from: studies of bottomhole samples (taken at the bottom of the well), adjusted to the conditions in the field separators (Laboratory PVT tests)

Data: Production history
What exactly: $N_p$ , $G_p$ , $W_p$ and the average pressure by date
Where from: field production records and pressure measurements

Data: Relative permeabilities
What exactly: $k_{rg}/k_{ro}$ as a function of saturation, $S_{gc}$
Where from: core (Relative permeability), corrected against the history

Data: Geology
What exactly: $N$ , gas cap ratio $m$ , $S_{wi}$ , porosity
Where from: volumetric estimate (Volumetric estimation of hydrocarbons in place)

Data: Wells
What exactly: productivity, limiting bottomhole pressure, number of wells, plateau
Where from: well tests and the field development plan

Data	What exactly	Where from
PVT properties	$B_o$ , $R_s$ , $B_g$ (or $Z$ ), $\mu_o$ , $\mu_g$ as functions of pressure	studies of bottomhole samples (taken at the bottom of the well), adjusted to the conditions in the field separators (Laboratory PVT tests)
Production history	$N_p$ , $G_p$ , $W_p$ and the average pressure by date	field production records and pressure measurements
Relative permeabilities	$k_{rg}/k_{ro}$ as a function of saturation, $S_{gc}$	core (Relative permeability), corrected against the history
Geology	$N$ , gas cap ratio $m$ , $S_{wi}$ , porosity	volumetric estimate (Volumetric estimation of hydrocarbons in place)
Wells	productivity, limiting bottomhole pressure, number of wells, plateau	well tests and the field development plan

A forecast in time needs all five groups; a forecast of “production against pressure” needs the first four.

If the accumulation already has a history, we first make the model reproduce the past: a forecast that cannot repeat the known years tells us nothing reliable about the future. Most often the gas–oil ratio does not match, and the relative permeabilities are to blame. A core is a few centimetres of uniform rock, while in the reservoir the saturation is uneven. Around the wells the pressure is lower than the average, and more gas has come out; the gas moves along high-permeability layers. So a “tank” with laboratory curves usually underestimates the mobility of the gas. We can invert the gas–oil ratio equation and get the ratio of the relative permeabilities directly from the production history.

Field gas–oil relative permeability ratio

\frac{k_{rg}}{k_{ro}} = (R - R_s)\,\frac{\mu_gB_g}{\mu_oB_o}

where:

$R$ measured gas–oil ratio of the accumulation (the sum over the wells) on the date, m³/m³
$R_s,\ B_o,\ B_g,\ \mu_o,\ \mu_g$ properties at the average reservoir pressure on the same date, m³/m³, mPa·s

Units: a dimensionless ratio; $R$ and $R_s$ in the same units

Holds when

the gas in the production is only dissolved and free gas of the oil zone: no gas cap, no gas injection and no crossflow
the gas saturation on the same date is taken from the balance, from the cumulative oil production and the average pressure
the gas measurements are complete: gas burned at a flare and gas used for the field's own needs are counted

Each date of the history gives a point “gas saturation against $k_{rg}/k_{ro}$ ”. We compare it with the laboratory curve. A systematic difference means that the curve for the forecast must be corrected. The result is a “field” curve, which describes not a core sample but the accumulation as a “tank”.

History of the accumulation below the bubble point pressure

$p$ , MPa: 16
$N_p$ , thousand m³: 607
$R$ , m³/m³: 113

$p$ , MPa: 15
$N_p$ , thousand m³: 770
$R$ , m³/m³: 174

$p$ , MPa: 14
$N_p$ , thousand m³: 909
$R$ , m³/m³: 262

$p$ , MPa	$N_p$ , thousand m³	$R$ , m³/m³
16	607	113
15	770	174
14	909	262

$R$ is the instantaneous gas–oil ratio of the accumulation. The last row is the history from the lesson General material balance equation for oil reservoirs: by 14 MPa, 909 thousand m³ of oil was produced at a cumulative gas–oil ratio of 130 m³/m³. The data are assumed.

Worked exampleThe relative permeability ratio from the history

Given (assumed data)

History: the table above; in detail for 14 MPa: $N_p = 909$ thousand m³, $R = 262$ m³/m³
Properties at 14 MPa: $R_s = 87.73$ m³/m³; $\mu_oB_o/(\mu_gB_g) = 11{,}584$
Laboratory curves: $k_{rg} = 0.30\left(\dfrac{S_g - 0.06}{0.49}\right)^{2}$ ; $k_{ro}$ as in the example “Gas–oil ratio at 14 MPa”

Solution

The gas saturation from the balance (as in the example “Saturations at 14 MPa”, but for $N_p = 909$ thousand m³): $S_g = 0.0795$ .
From the history: $k_{rg}/k_{ro} = (262 - 87.73)/11{,}584 = 0.0150$ .
From the laboratory curves at the same saturation: $k_{ro} = 0.5854$ , $k_{rg} = 0.30\cdot(0.0195/0.49)^2 = 0.000475$ , and the ratio 0.00081 is 19 times smaller.
At 15 and 16 MPa ( $S_g$ = 0.061 and 0.040) the history gives 0.0069 and 0.0012, while the laboratory curve gives practically zero: the gas saturation is still below its critical value of 0.06, or only just above it.

Answer: The gas in the accumulation is more mobile than in the core: at 14 MPa the field ratio is 19 times the laboratory one, and at 15–16 MPa the gas is already flowing, although by the laboratory curve it is immobile.

What it means. The field points show that the gas starts to move at a gas saturation of about 3%, not 6%. We correct the curve for the forecast. We lower the critical gas saturation and make the first part of the gas curve steeper ( $S_{gc} = 0.03$ , an exponent of 1.5 instead of 2). We go on until the points lie on the curve and the forecast reproduces the history. These are the curves used in this lesson's forecast. The data in this example are assumed for illustration.

CalculatedThe ratio of the gas and oil relative permeabilities, in per cent, against gas saturation: the laboratory curve (dashed), the points found from the production history of the accumulation, and the curve matched to the history (the vertical axis is logarithmic).

How the input data change the forecast

Variant: curves matched to the history (base case)
$N_p/N$ at 14 MPa, %: 10.0
$R_p$ at 14 MPa: 130
$N_p/N$ at 5 MPa, %: 17.3
$R$ at 5 MPa: 1388

Variant: laboratory curves
$N_p/N$ at 14 MPa, %: 11.5
$R_p$ at 14 MPa: 103
$N_p/N$ at 5 MPa, %: 23.0
$R$ at 5 MPa: 1217

Variant: without expansion of water and pores
$N_p/N$ at 14 MPa, %: 8.3
$R_p$ at 14 MPa: 134
$N_p/N$ at 5 MPa, %: 15.2
$R$ at 5 MPa: 1503

Variant	$N_p/N$ at 14 MPa, %	$R_p$ at 14 MPa	$N_p/N$ at 5 MPa, %	$R$ at 5 MPa
curves matched to the history (base case)	10.0	130	17.3	1388
laboratory curves	11.5	103	23.0	1217
without expansion of water and pores	8.3	134	15.2	1503

The history at 14 MPa is 10.0% of the oil in place (909 thousand m³) at $R_p$ = 130 m³/m³. The laboratory curves do not reproduce it: in the model the gas stays immobile longer, and the forecast of the recovery is too high by almost six percentage points. The variant without the expansion of water and pores, on the contrary, understates the energy of the accumulation. $R_p$ and $R$ are in m³/m³.

CalculatedReservoir pressure against the fraction of the oil in place produced: the forecast with curves matched to the history, with laboratory curves, and without the expansion of the connate water and pores. The dot is the history at 14 MPa.

Which pressure to take as the average

The balance is written for a “tank” with one pressure, but the pressure is measured in individual wells. In a high-permeability accumulation with light oil the pressure equalises fast, and the measurements of different wells lie on one curve. In a less conductive one (where fluid flows less easily) each well has its own drainage area, the part of the reservoir that it drains, with its own average pressure. Where more is produced per unit of volume, the pressure is lower. The average pressure of each area is found from well tests (Build-up in a bounded reservoir and average reservoir pressure), and for the balance these pressures are averaged by pore volume.

Average pressure of the accumulation from the drainage areas

\bar p = \frac{\sum_j \bar p_j\,V_j}{\sum_j V_j}

where:

$\bar p_j$ average pressure in the drainage area of the $j$ -th well, brought to a common depth, MPa
$V_j$ pore volume of this area, m³

Units: MPa; volumes in any, but the same, units

Holds when

the drainage areas do not overlap and together make up the accumulation
the pressures belong to one date and to one reference depth

The balance conserves volumes, so the pressure is averaged with the weight of the volume: an area twice as large “weighs” twice as much. The arithmetic mean over the wells is right only for equal areas.

It is hard to find the volumes of the areas from maps. It is easier to use the fact that each closed area works as a small “tank” itself. Over the same period its underground withdrawal $\Delta F_j$ and its pressure drop $\Delta p_j$ are linked through the storage capacity of the area (how much fluid it releases for each unit of pressure drop), $\Delta F_j = c_tV_j\,\Delta p_j$ . If the total compressibility $c_t$ is the same in all areas, the volume is proportional to $\Delta F_j/\Delta p_j$ :

Average pressure weighted by withdrawal and pressure drop

\bar p = \frac{\sum_j \bar p_j\,\Delta F_j/\Delta p_j}{\sum_j \Delta F_j/\Delta p_j}

where:

$\Delta F_j$ underground withdrawal from area $j$ over the period (for example, half a year), m³
$\Delta p_j$ fall in the average pressure of this area over the same period, MPa

Units: MPa; $\Delta F_j$ and $\Delta p_j$ are for the same period

Holds when

the areas are closed: there is no crossflow between them, and the pressure drop of each is determined by its own withdrawal
the total compressibility is the same in all areas, with the same fluids and saturations; below the bubble point pressure, where $c_t$ is set by the free gas, this assumption is rougher
the pressure drop over the period is clearly larger than the error in finding it

A well whose pressure falls slowly in spite of a large withdrawal drains a large volume and gets a large weight. This is the same principle as the balance of the whole accumulation, applied to each area separately.

Worked exampleAverage pressure over four wells

Given (assumed data)

Well 1: $\bar p_j = 14.6$ MPa; $\Delta F_j = 60$ thousand m³; $\Delta p_j = 0.9$ MPa
Well 2: $\bar p_j = 13.2$ MPa; $\Delta F_j = 95$ thousand m³; $\Delta p_j = 1.6$ MPa
Well 3: $\bar p_j = 15.4$ MPa; $\Delta F_j = 30$ thousand m³; $\Delta p_j = 0.5$ MPa
Well 4: $\bar p_j = 12.4$ MPa; $\Delta F_j = 70$ thousand m³; $\Delta p_j = 1.8$ MPa

Solution

The weights $\Delta F_j/\Delta p_j$ , in thousand m³/MPa: $60/0.9 = 66.67$ ; $95/1.6 = 59.38$ ; $30/0.5 = 60.00$ ; $70/1.8 = 38.89$ ; the sum is 224.94. The shares of the volume are 30, 26, 27 and 17%.
The volume average: $\bar p = (14.6\cdot66.67 + 13.2\cdot59.38 + 15.4\cdot60.00 + 12.4\cdot38.89)/224.94 = 14.06$ MPa.
For comparison: the arithmetic mean is 13.90 MPa; the average weighted by the withdrawal $\Delta F_j$ alone is 13.57 MPa.
Sensitivity of the balance: the expansion $E_t$ at 13.90 and at 14.06 MPa differs by 4%.

Answer: The average pressure of the accumulation is about 14.06 MPa, which is 0.16 MPa above the arithmetic mean over the wells.

What it means. Well 4 produces a lot, but its pressure falls faster than the others. It drains a smaller volume, and its low pressure should not pull the average down. Weighting by the withdrawal alone does the opposite and understates the average by 0.49 MPa. Below the bubble point pressure the expansion grows quickly as the pressure falls, and an error of 0.16 MPa changes it by 4%. This is as much as an error of the same size in the oil in place or in the production records. The data in this example are assumed for illustration.

From pressure to time

The balance and the gas–oil ratio tell us how much oil and gas the accumulation will give at each pressure, but not when. Time appears when we add the well rates. At each pressure step the oil production $\Delta N_p$ is known, and the rate follows from the productivity of the wells at that pressure. We divide the production by the average rate of the step.

Below the bubble point pressure the productivity of a well changes for two reasons. The reservoir pressure falls, and with it the oil mobility $k_{ro}/(\mu_oB_o)$ falls too: the gas saturation grows, and the oil loses gas and becomes more viscous. The productivity index at zero drawdown is proportional to this mobility (forecast of the inflow performance curve). This index is the oil rate per unit of pressure drop between the reservoir and the well, for a very small drop. It is written with the formula of pseudo-steady-state flow (Pseudo-steady-state flow and the shape factor), which is flow in a closed drainage area once the pressure drop has reached its edges:

Well productivity at the current oil mobility

J^* = \frac{0.536\,k\,h\,k_{ro}}{\mu_oB_o\left(\ln\dfrac{r_e}{r_w} - \dfrac34 + S\right)}

where:

$J^*$ productivity index at zero drawdown, for stock-tank oil, m³/(d·MPa)
$k,\ h$ permeability and effective thickness of the reservoir, mD, m
$k_{ro}$ relative permeability to oil at the average saturation, from the forecast, —
$\mu_o,\ B_o$ viscosity and formation volume factor of the oil at the average pressure, mPa·s, m³/m³
$r_e,\ r_w,\ S$ radius of the drainage area, radius of the well, skin factor (how much the zone right around the well helps or hinders flow), m, m, —

Units: metric oilfield units: $k$ in mD, $h$ in m, $\mu_o$ in mPa·s; $J^*$ in m³/(d·MPa)

Holds when

pseudo-steady-state flow, with the well at the centre of a round drainage area
$k_{ro}$ , $\mu_o$ and $B_o$ are taken at the average saturation and pressure of the accumulation, as in the “tank”
the skin factor and the thickness do not change

The ratio of $J^*$ in the future to $J^*$ now equals the ratio of the oil mobilities: the geometry and the permeability of the reservoir cancel.

In other unit systems (1)

Oilfield units

J^*\,[\text{STB/(d·psi)}] = \frac{7.08\cdot10^{-3}\,k\,h\,k_{ro}}{\mu_oB_o\left(\ln\dfrac{r_e}{r_w} - \dfrac34 + S\right)}

$k$ in mD, $h$ in ft, $\mu_o$ in cP, $B_o$ in rb/STB.

$7.08\cdot10^{-3} = 2\pi\cdot1.127\cdot10^{-3}$ , as in the lesson Rock permeability; in the metric form $0.536 = 2\pi\cdot0.0853$ .

Well rate and the time of a step

q_{\text{well}} = \frac{J^*\bar p}{1.8}\left(1 - 0.2\,x - 0.8\,x^2\right),\quad x = \frac{p_{wf}}{\bar p};\qquad q_o = \min\big(q_{\text{plateau}},\ n\,q_{\text{well}}\big);\qquad \Delta t = \frac{\Delta N_p}{\tfrac12\,(q_{o,j} + q_{o,k})}

where:

$q_{\text{well}}$ oil rate of one well from the Vogel curve, which links the rate of a well to its flowing pressure (Vogel's equation), m³/d
$\bar p,\ p_{wf}$ average reservoir pressure and the bottomhole flowing pressure, limited by the capability of the pump, MPa
$q_o,\ q_{\text{plateau}}$ rate of the accumulation and the level of the production plateau, m³/d
$n$ number of producing wells, —
$\Delta N_p,\ \Delta t$ oil production in a pressure step, and the duration of the step, m³, d

Units: m³/d, MPa, days

Holds when

the average pressure is below the bubble point pressure; above it the inflow performance curve is composite (Inflow performance relationship (IPR))
all the wells are identical, the pressure in the accumulation is uniform, and no new wells are drilled
the rate changes about linearly within a step, so the pressure steps are not too large

While the wells can give more than the plateau, the plan limits the production; later the reservoir does. The time of a step grows towards the end of the forecast, because each megapascal of pressure fall now happens at smaller and smaller rates.

Worked exampleHow long a step from 14 to 13 MPa takes

Given (assumed data)

Wells: 8 wells, each with an area of 1 km² ( $r_e = 564$ m), $r_w = 0.1$ m, no skin; $k = 50$ mD, $h = 10$ m; $p_{wf} = 3$ MPa; plateau 1200 m³/d
At 14 MPa: $k_{ro} = 0.5854$ ; $\mu_o = 1.175$ mPa·s; $B_o = 1.2893$
Production: 908.8 thousand m³ at 14 MPa and 1024.5 thousand m³ at 13 MPa (the forecast table); the rate of the accumulation at 13 MPa is 622 m³/d

Solution

The denominator: $\ln(564/0.1) - 0.75 = 7.888$ .
$J^* = \dfrac{0.536\cdot50\cdot10\cdot0.5854}{1.175\cdot1.2893\cdot7.888} = 13.13$ m³/(d·MPa); at the bubble point pressure it was 20.13 ( $k_{ro} = 0.80$ , $\mu_o = 1.00$ , $B_o = 1.35$ ).
Well rate: $x = 3/14 = 0.2143$ ; $q_{\text{well}} = 13.13\cdot14/1.8\cdot(1 - 0.2x - 0.8x^2) = 94.0$ m³/d; of the accumulation, $8\cdot94.0 = 752$ m³/d, which is less than the plateau.
Time of the step: $\Delta t = (1024.5 - 908.8)\cdot10^3/\big(\tfrac12(752 + 622)\big) = 168$ days, or 0.46 years.

Answer: The pressure fall from 14 to 13 MPa takes about 168 days. In that time the rate of the accumulation falls from 752 to 622 m³/d.

What it means. By 14 MPa the productivity of the wells has fallen by 35% compared with the bubble point pressure, although the permeability of the reservoir has not changed: the oil has lost mobility. So under this drive the rates fall faster than the pressure. The data in this example are assumed for illustration.

Forecast in time

$\bar p$ , MPa: 30
$N_p/N$ , %: 0.0
Oil, m³/d: 1200
Gas, thousand m³/d: 132
Time, years: 0.0

$\bar p$ , MPa: 18
$N_p/N$ , %: 3.3
Oil, m³/d: 1200
Gas, thousand m³/d: 132
Time, years: 0.7

$\bar p$ , MPa: 16
$N_p/N$ , %: 6.7
Oil, m³/d: 1089
Gas, thousand m³/d: 123
Time, years: 1.4

$\bar p$ , MPa: 14
$N_p/N$ , %: 10.0
Oil, m³/d: 752
Gas, thousand m³/d: 197
Time, years: 2.3

$\bar p$ , MPa: 12
$N_p/N$ , %: 12.4
Oil, m³/d: 510
Gas, thousand m³/d: 247
Time, years: 3.2

$\bar p$ , MPa: 10
$N_p/N$ , %: 14.1
Oil, m³/d: 332
Gas, thousand m³/d: 246
Time, years: 4.3

$\bar p$ , MPa: 8
$N_p/N$ , %: 15.5
Oil, m³/d: 198
Gas, thousand m³/d: 201
Time, years: 5.6

$\bar p$ , MPa: 6
$N_p/N$ , %: 16.7
Oil, m³/d: 99
Gas, thousand m³/d: 126
Time, years: 7.7

$\bar p$ , MPa: 5
$N_p/N$ , %: 17.3
Oil, m³/d: 60
Gas, thousand m³/d: 83
Time, years: 9.4

$\bar p$ , MPa	$N_p/N$ , %	Oil, m³/d	Gas, thousand m³/d	Time, years
30	0.0	1200	132	0.0
18	3.3	1200	132	0.7
16	6.7	1089	123	1.4
14	10.0	752	197	2.3
12	12.4	510	247	3.2
10	14.1	332	246	4.3
8	15.5	198	201	5.6
6	16.7	99	126	7.7
5	17.3	60	83	9.4

Eight wells, plateau 1200 m³/d, bottomhole pressure 3 MPa. Above 18 MPa the drive is compressibility drive, and the wells can give more than the plateau. Gas is the gas rate at the surface, $Rq_o$ . The data are assumed.

CalculatedOil rate of the assumed accumulation in time: a production plateau while the wells can hold it, then a decline. The labels give the average reservoir pressure.

Terms and symbols

Instantaneous gas–oil ratio $R$: The ratio of the current gas and oil rates at standard conditions: the dissolved gas plus the free gas of the reservoir.
Cumulative gas–oil ratio $R_p$: All the gas produced, divided by all the oil produced. It enters the balance equation.
Ratio $k_{rg}/k_{ro}$: The ratio of the relative permeabilities to gas and to oil. Together with the factor $\mu_oB_o/(\mu_gB_g)$ , it sets the share of free gas in the production.
Tarner's method: A step-by-step forecast in which, at each pressure step, the oil production is chosen so that the gas from the balance and the gas from the gas–oil ratio agree.
Muskat's method: The same system in differential form: an equation for the change of the oil saturation with pressure.
Field relative permeabilities: The ratio $k_{rg}/k_{ro}$ found from the history of the gas–oil ratio. It describes the accumulation as a “tank”, not a core sample.
Secondary gas cap: Gas that has come out of the oil and collected at the top. It lowers the gas–oil ratio of the wells and raises the recovery.

Common mistakes and misconceptions

Wrong

“A forecast with laboratory relative permeability curves is reliable, because the core was studied carefully.”

In fact

A laboratory curve describes a sample, while a tank forecast describes an accumulation with uneven saturation. In this lesson's example the laboratory curves did not repeat the history, and overstated the recovery at 5 MPa by almost six percentage points. We first check the curves against the history of the gas–oil ratio.

Wrong

“The method is iterative, so the pressure step can be anything.”

In fact

The iterations refine the solution of the step equation, but the gas of the step is still calculated by the trapezoid rule. Where the gas–oil ratio grows more and more steeply with production, the trapezoid on a large step overstates the gas of the step, and the oil production comes out too low. In this lesson's example, steps of 2 MPa understated the production at 6 MPa by almost one per cent, and steps of 0.25 MPa by hundredths of a per cent.

Wrong

“Below the bubble point pressure the compressibility of the water and the pores no longer matters.”

In fact

For an accumulation that started development 12 MPa above the bubble point pressure, the expansion of the water and the pores gained there has not gone anywhere. Without it, the recovery at 5 MPa in this lesson's example is understated, from 17.3 to 15.2%, and the gas saturation is overstated.

Wrong

“The gas–oil ratio has grown several times, so there is more gas in the reservoir.”

In fact

There is less and less gas in the reservoir. The gas–oil ratio grows because the free gas has become mobile and outruns the oil. In this lesson's example, at a gas saturation of 8% its permeability is only 1.5% of the permeability to oil, but because of its low viscosity almost as much of it flows.

Wrong

“If the rate of the accumulation is 1200 m³/d now, in five years it will be a little lower.”

In fact

Under solution gas drive the rate falls faster than the pressure, because both the reservoir pressure and the oil mobility decrease. In this lesson's example the plateau lasts a little more than a year, and by year 5 the rate is almost four times lower.

In depth: advanced detail

Tracy's form: the balance through functions of pressure

Let us divide the balance equation of a closed accumulation by the expansion $E_t$ . All the fluid properties then gather in two functions of pressure, which can be calculated in advance from the PVT data:

Balance in Tracy's functions

N = N_p\,\Phi_n + G_p\,\Phi_g,\qquad \Phi_n = \frac{B_o - R_sB_g}{E_t},\qquad \Phi_g = \frac{B_g}{E_t}

where:

$\Phi_n,\ \Phi_g$ functions of pressure for the oil and the gas, —, m³/m³
$E_t$ expansion per 1 m³ of oil in place, $E_o + E_{fw}$ (with a gas cap, $+\,mE_g$ ), m³/m³
$N,\ N_p,\ G_p$ oil in place, and cumulative oil and gas, m³

Units: m³; $\Phi_g$ is per 1 m³ of gas at standard conditions

Holds when

the accumulation is closed, with no water; water influx is added as the term $-(W_e - W_pB_w)\Phi_w$ with $\Phi_w = 1/E_t$

The functions depend only on pressure. Above the bubble point pressure $E_t$ is small, and they are very large; below it they fall fast. In a step from $j$ to $k$ with the average gas–oil ratio of the step $\bar R$ , the balance immediately gives $\Delta N_p = (N - N_{p,j}\Phi_n - G_{p,j}\Phi_g)/(\Phi_n + \bar R\,\Phi_g)$ , where the functions are taken at $p_k$ . What we choose now is not the oil production but the gas–oil ratio at the end of the step.

Worked exampleThe same step from 14 to 13 MPa in Tracy's form

Given (assumed data)

At 13 MPa: $E_t = 0.20829$ ; $B_o - R_sB_g = 0.5939$ ; $B_g = 0.008285$ m³/m³
Start of the step: $N_{p,j} = 908.8$ thousand m³; $G_{p,j} = 118.24$ million m³; $R_j = 262.2$ ; the gas–oil ratio at the end of the step from Tarner's method is 366.6 m³/m³

Solution

The functions: $\Phi_n = 0.5939/0.20829 = 2.8513$ ; $\Phi_g = 0.008285/0.20829 = 0.039776$ m³/m³.
The numerator: $9060 - 908.8\cdot2.8513 - 118{,}240\cdot0.039776 = 1765.6$ thousand m³.
The denominator: $2.8513 + \tfrac12(262.2 + 366.6)\cdot0.039776 = 15.3569$ .
$\Delta N_p = 1765.6/15.3569 = 115$ thousand m³.

Answer:

\Delta N_p \approx 115

thousand m³, the same as Tarner's method gave: two forms of one balance.

What it means. Tracy's form is convenient for calculation: the functions of pressure are built once, and a step comes down to one formula and a refinement of the gas–oil ratio. The data in this example are assumed for illustration.

Muskat's differential form

In a closed accumulation with a constant pore volume, let us write the amounts of oil and gas per unit of pore volume: the oil is $S_o/B_o$ , and the gas is $S_oR_s/B_o + S_g/B_g$ . For a small fall in pressure, the gas–oil ratio equals the loss of gas divided by the loss of oil. By Darcy's law the same ratio equals the instantaneous gas–oil ratio. Setting these two expressions equal, we get an equation for the oil saturation:

Muskat's equation

\frac{\dd S_o}{\dd p} = \frac{S_o\,X + S_o\,\dfrac{k_{rg}}{k_{ro}}\,Y + (1 - S_o - S_w)\,Z}{1 + \dfrac{k_{rg}}{k_{ro}}\cdot\dfrac{\mu_o}{\mu_g}},\qquad X = \frac{B_g}{B_o}\frac{\dd R_s}{\dd p},\quad Y = \frac{\mu_o}{\mu_g}\cdot\frac{1}{B_o}\frac{\dd B_o}{\dd p},\quad Z = B_g\,\frac{\dd(1/B_g)}{\dd p}

where:

$S_o,\ S_w$ oil saturation and connate water saturation (constant), —
$X,\ Y,\ Z$ functions of pressure from the PVT data: gas coming out of solution, shrinkage of the oil, expansion of the free gas, 1/MPa
$k_{rg}/k_{ro}$ ratio of the relative permeabilities at the current saturation, —

Units: 1/MPa; all the functions are positive below the bubble point pressure, so the oil saturation decreases together with the pressure

Holds when

the pore volume and the connate water saturation are constant (the compressibility of the water and the pores is not counted), which suits an accumulation that started development at the bubble point pressure
the saturation is uniform, and the gas does not separate into a cap
the derivatives of the properties (their rates of change with pressure) are taken from smoothed PVT data, because random scatter in the points is magnified in them

The numerator contains three ways in which a fall in pressure frees space: gas comes out of solution ( $X$ ), the oil shrinks ( $Y$ , with the weight of the gas mobility), and gas that has already come out expands ( $Z$ ). The denominator shows what part of the freed space leaves with the flow of gas. The equation is integrated numerically from the bubble point pressure downward, and the recovery is found from the saturation equation without corrections.

Muskat's method and Tarner's method for the same accumulation, without the compressibility of the water and the pores,

N_p/N

, %

$p$ , MPa: 14
Muskat, step 0.01 MPa: 8.28
Tarner, step 0.25 MPa: 8.27

$p$ , MPa: 10
Muskat, step 0.01 MPa: 12.23
Tarner, step 0.25 MPa: 12.22

$p$ , MPa: 5
Muskat, step 0.01 MPa: 15.24
Tarner, step 0.25 MPa: 15.23

$p$ , MPa	Muskat, step 0.01 MPa	Tarner, step 0.25 MPa
14	8.28	8.27
10	12.23	12.22
5	15.24	15.23

Both calculations are for the variant “without expansion of water and pores” from the table of variants. With equal assumptions the methods agree to within hundredths of a per cent. Counting the compressibility in Muskat's equation would need extra terms, so for accumulations that have passed the elastic stage Tarner's method is more convenient.

Accumulation with a gas cap

If the accumulation has a gas cap of relative size $m$ , the expansion of the cap is added to the balance, and otherwise Tarner's scheme does not change:

Gas that the balance of an accumulation with a gas cap needs

G_p = \frac{N\,(E_o + mE_g + E_{fw}) - N_p\,(B_o - R_sB_g)}{B_g}

where:

$m,\ E_g$ gas cap ratio and the expansion of the cap per 1 m³ of oil in place for $m = 1$ , as in the lesson General material balance equation for oil reservoirs, —, m³/m³
$E_{fw}$ expansion of the water and the pores including the cap, $(1 + m)E_{fw0}$ , m³/m³

Units: m³ of gas at standard conditions

Holds when

one $B_g$ describes both the cap gas and the gas that came out of the oil; the cap is at the same average pressure
the cap gas does not break through to the wells; if it does, the gas–oil ratio equation of the oil zone no longer describes the production, and the cap gas is counted separately

The cap supports the pressure, and for the same gas withdrawal the pressure falls more slowly. But gas cap drive is above all the displacement of oil by gas expanding from above, and its efficiency is set by the rate of withdrawal and the gravity separation of the phases (the gas rises, the oil sinks; Segregated flow and gravity displacement, Displacement of oil by gas). The tank model conveys only the energy part.

Connections to other topics

General material balance equation for oil reservoirs — the balance equation, its terms $E_o$ and $E_{fw}$ , and the same assumed accumulation.
Straight-line material balance (Havlena–Odeh method) — checking the oil in place and the sources of energy against the history before the forecast.
Relative permeability — the curves for gas and oil, and the critical gas saturation.
Reservoir drive mechanisms — solution gas drive and the course of the gas–oil ratio, qualitatively.
Black oil model: solution gas–oil ratio and formation volume factors — $B_o$ , $R_s$ , $B_g$ of the assumed oil.
Inflow performance relationship (IPR) — the Vogel curve and the recalculation of productivity to a future pressure.
Pseudo-steady-state flow and the shape factor — the formula for the productivity of a well in a closed area.
Build-up in a bounded reservoir and average reservoir pressure — how to find the average pressure of the drainage area of a well.
History matching — history matching of a simulation model: the same principle on a larger scale.
Why water is injected: pressure maintenance and oil displacement — pressure maintenance by water injection.

Check yourself

1
The wells of the accumulation give a gas–oil ratio of 480 m³/m³ at a pressure at which the solution gas–oil ratio of the oil is 75 m³/m³. What share of the gas at the surface is free gas from the reservoir?
1. A.About 16%
2. B.About 84%
3. C.It cannot be said without the viscosities
4. D.About 50%
Show the answer
Correct answer: B. About 84%
Each cubic metre of oil carries 75 m³ of dissolved gas, and the remaining $480 - 75 = 405$ m³ is free gas: $405/480 \approx 0.84$ . The viscosities are needed to calculate the gas–oil ratio from the saturation, but not to split the measured gas into dissolved and free gas.
2
At 11 MPa the accumulation gives a gas–oil ratio of 450 m³/m³. At this pressure $R_s = 75$ m³/m³, $B_o = 1.25$ , $B_g = 0.0095$ m³/m³, $\mu_o = 1.4$ and $\mu_g = 0.016$ mPa·s. Find the field ratio $k_{rg}/k_{ro}$ .
Show the answer
The factor is $\mu_oB_o/(\mu_gB_g) = 1.4\cdot1.25/(0.016\cdot0.0095) = 1.75/0.000152 \approx 11{,}510$ . The ratio is $k_{rg}/k_{ro} = (450 - 75)/11{,}510 \approx 0.033$ . We compare it with the laboratory curve at the gas saturation calculated from the balance for the same date.
3
The field points of $k_{rg}/k_{ro}$ lie an order of magnitude (ten times) above the laboratory curve. What will the forecast be if we keep the laboratory curve?
1. A.Optimistic: the gas–oil ratio is too low, the pressure falls more slowly and the recovery is too high
2. B.Pessimistic: the gas–oil ratio is too high and the recovery too low
3. C.It will not change: the balance does not depend on the relative permeabilities
4. D.Optimistic only for gas
Show the answer
Correct answer: A. Optimistic: the gas–oil ratio is too low, the pressure falls more slowly and the recovery is too high
With the laboratory curve the gas in the model is less mobile than in the accumulation. The wells carry away less gas, and at each step the balance “allows” more oil. The forecast underestimates the gas–oil ratio and overestimates the recovery, by almost six percentage points at 5 MPa in this lesson's example.
4
In a step of Tarner's method, for a trial oil production the gas from the balance turned out to be larger than the gas calculated from the gas–oil ratio. How should the trial value be changed, and why?
Show the answer
Increase it. The balance says that for this oil production the wells must have carried away more gas than the gas–oil ratio gives, for the volumes to fit. If more oil is produced, the balance will need less gas (the oil itself has freed space), and the gas from the gas–oil ratio will be larger (more oil, and a higher gas saturation). The solution is where the two estimates are equal.
5
In a pressure step the accumulation gives 90 thousand m³ of oil. The rate is 700 m³/d at the start of the step and 560 m³/d at the end. How long does the step last?
Show the answer
The average rate is $(700 + 560)/2 = 630$ m³/d; $\Delta t = 90{,}000/630 \approx 143$ days, which is about 0.39 years. If the rate changes strongly and nonlinearly within a step, the pressure step is split into smaller ones.
6
An accumulation started development 10 MPa above the bubble point pressure. An engineer forecasts it below the bubble point pressure and drops the compressibility of the water and the pores. In which direction will the forecast be wrong?
1. A.The recovery is overstated
2. B.There is no error: below the bubble point pressure only the gas works
3. C.The recovery is understated
4. D.The error is only in time
Show the answer
Correct answer: C. The recovery is understated
The expansion of the water and the pores gained in the elastic stage stays in the balance. Without it the accumulation “has” less energy and gives less oil at each pressure, and the calculated gas saturation is higher, so the gas–oil ratio starts to rise earlier. In this lesson's example the recovery at 5 MPa would be 15.2% instead of 17.3%.

Updated: 19 October 2026