Reservoir simulation in practice: case studies

Quick refresher

A study begins with the decision. The question “how much oil will the option give” is worse than “which option is more profitable under all plausible descriptions of the reservoir”. The decision sets the measure of comparison, the level of detail the model needs and the uncertainties that must be tested.
At the appraisal stage (when we are still checking what the discovery holds) the decisions are the largest and the data the scarcest. So we run not one model but a set of realisations. We rank them by the result that matters for the decision: the realisation that is average by oil in place may be far from average by production.
Sensitivity runs that change one parameter at a time are cheap, but they do not see the combined action of the parameters. Running all the combinations quickly becomes unmanageable, so we need a design of runs (a planned set of runs).
The increment from an intervention is the difference between the runs “with” and “without” for a sector or a field. The production of a new well includes oil that the neighbouring wells would have produced anyway. This is acceleration, not increment, and it is worth less.
In gas injection projects, the gas-handling capacity, not the reservoir, often limits production: the oil rate cannot exceed $Q_g/R$ . The forecast of the producing gas–oil ratio is one of the main results of the model.
In a mature field, a history-matched model shows where the oil has been left. We compare an intervention with the continuation of development at the same injection volumes. After a few years we check the forecast against the actual outcome.

Key relationships

\Delta Q = Q_{\text{with}} - Q_{\text{without}}

incremental production of a sector or field from an intervention

Q = \dfrac{q_i}{D},\qquad Q_\text{disc} = \dfrac{q_i}{D + j},\quad j = \ln(1 + i)

production and discounted production under exponential decline, over the whole life

q_o = \min\left(q_{\text{pot}},\ \dfrac{Q_g}{R}\right)

oil rate with a limit on gas-handling capacity

\rho_S = 1 - \dfrac{6\sum d^2}{n\,(n^2 - 1)}

Spearman's rank correlation of two rankings of $n$ realisations

Common pitfalls

Building a single “most likely” model and presenting its calculation as the forecast.
Choosing the low, medium and high models by oil in place when the decision depends on production.
Evaluating an infill well by its own production.
Calculating a forecast without the limits of the surface facilities — for gas, liquid, injection.
Isolating a thief zone without checking whether it communicates with its neighbours.

After this lesson you can

formulate the question of a study from the decision to be made, and choose the measure by which the options are compared
work with a set of geological realisations (alternative versions of the geology): rank them, and choose the cases to run by the result that matters for the decision
plan sensitivity runs and understand how many are needed
evaluate infill drilling and other interventions by the incremental production of the sector (the extra production), separating the increment from acceleration
allow, in a gas injection project, for the rise in the producing gas–oil ratio and for the limits of the surface facilities
use a history-matched model of a mature field to find the remaining oil and choose interventions, and check forecasts against the actual outcome

A study starts with the decision

A study with a reservoir simulation model is not “building a model of the field”. It is answering a question on which a decision depends. The sequence of work is familiar from the lesson on the reservoir simulation model: data, model, history matching, forecasts, economics. But the quality of a study is settled before the first run, by how precisely we pose the question.

Compare two questions. One is “how much oil will an infill well produce”. The other is “will an infill well increase the value of the sector, taking into account its cost and the fact that neighbouring wells would have produced part of its oil”. Any calculation answers the first, and the answer is almost certainly too high. The second needs “with” and “without” runs, economics (Net present value (NPV), the value of future income in today's money) and a range over the uncertain parameters — and only it supports the decision. Along with the question, we write down the measure for comparing the options and the threshold at which the decision changes. Then it is clear which uncertainties we must test by runs and which do not affect the decision.

What the model decides at different stages

Stage: Appraisal
Typical decisions: development concept, number and type of wells, capacities, whether injection is needed
What the model must show: the range of production over the realisations of the geology; which uncertainties change the decision

Stage: Development, production plateau (the period of steady, maximum production)
Typical decisions: drilling sequence, flood pattern, gas injection, well operating regimes
What the model must show: profiles by well, water and gas breakthroughs, capacity limits

Stage: Mature field
Typical decisions: infill drilling, converting wells to injection, isolating thief zones, redistributing injection, enhanced oil recovery methods
What the model must show: where the oil has been left; the increment of each intervention against the continuation of development

Stage: Late stage
Typical decisions: extending operation, stopping injection, abandonment
What the model must show: production as costs fall; when production will no longer pay for the costs

Stage	Typical decisions	What the model must show
Appraisal	development concept, number and type of wells, capacities, whether injection is needed	the range of production over the realisations of the geology; which uncertainties change the decision
Development, production plateau (the period of steady, maximum production)	drilling sequence, flood pattern, gas injection, well operating regimes	profiles by well, water and gas breakthroughs, capacity limits
Mature field	infill drilling, converting wells to injection, isolating thief zones, redistributing injection, enhanced oil recovery methods	where the oil has been left; the increment of each intervention against the continuation of development
Late stage	extending operation, stopping injection, abandonment	production as costs fall; when production will no longer pay for the costs

The lesson on field development says more about the stages of the life of a field.

Appraisal: a set of geological realisations instead of one model

Between the wells, the properties of the reservoir are unknown. Geostatistical modelling (modelling that uses the statistics of the data) builds not one picture but many equally probable realisations. Each one agrees with the wells and with our ideas about the structure of the reservoir: the sizes of the sand bodies, how well they connect, how much the properties vary (Reservoir architecture and performance). But between the wells the realisations differ. The difference between them is the uncertainty that the wells do not remove. On top of it come scenario uncertainties: the position of the contacts, the transmissibility of faults and the size of the aquifer.

A full calculation of hundreds of realisations is usually too expensive. So we rank the realisations by a fast measure and run only a few in full: a low, a medium and a high one, or, for large sets, those close to P90, P50 and P10 (the values that 90%, 50% and 10% of the realisations exceed; reporting a probabilistic estimate). The measure can be static (oil in place, connected pore volume) or dynamic (a fast streamline calculation, Types of reservoir simulation model, or a simplified displacement calculation). The main question is which measure to rank by.

Nine realisations of a hypothetical field

Realisation: 1
$V_p$ , million m³: 9.6
$V_{DP}$: 0.55
Oil in place, million m³: 5.990
Recoverable oil, million m³: 3.295

Realisation: 2
$V_p$ , million m³: 10.4
$V_{DP}$: 0.80
Oil in place, million m³: 6.490
Recoverable oil, million m³: 2.815

Realisation: 3
$V_p$ , million m³: 11.0
$V_{DP}$: 0.62
Oil in place, million m³: 6.864
Recoverable oil, million m³: 3.673

Realisation: 4
$V_p$ , million m³: 11.8
$V_{DP}$: 0.85
Oil in place, million m³: 7.363
Recoverable oil, million m³: 2.709

Realisation: 5
$V_p$ , million m³: 12.3
$V_{DP}$: 0.68
Oil in place, million m³: 7.675
Recoverable oil, million m³: 3.952

Realisation: 6
$V_p$ , million m³: 12.9
$V_{DP}$: 0.78
Oil in place, million m³: 8.050
Recoverable oil, million m³: 3.648

Realisation: 7
$V_p$ , million m³: 13.4
$V_{DP}$: 0.58
Oil in place, million m³: 8.362
Recoverable oil, million m³: 4.553

Realisation: 8
$V_p$ , million m³: 14.2
$V_{DP}$: 0.83
Oil in place, million m³: 8.861
Recoverable oil, million m³: 3.524

Realisation: 9
$V_p$ , million m³: 15.0
$V_{DP}$: 0.72
Oil in place, million m³: 9.360
Recoverable oil, million m³: 4.640

Realisation	$V_p$ , million m³	$V_{DP}$	Oil in place, million m³	Recoverable oil, million m³
1	9.6	0.55	5.990	3.295
2	10.4	0.80	6.490	2.815
3	11.0	0.62	6.864	3.673
4	11.8	0.85	7.363	2.709
5	12.3	0.68	7.675	3.952
6	12.9	0.78	8.050	3.648
7	13.4	0.58	8.362	4.553
8	14.2	0.83	8.861	3.524
9	15.0	0.72	9.360	4.640

$V_p$ is the pore volume, and $V_{DP}$ is the Dykstra–Parsons heterogeneity coefficient (Reservoir heterogeneity and permeability averaging). Oil in place is $V_p(1 - S_{wi})/B_o$ at $S_{wi}$ = 0.22 and $B_o$ = 1.25. Recoverable oil is the oil produced by waterflooding a layered log-normal section (no crossflow between the layers; log-normal section) with oil of 3.0 mPa·s, up to a water cut of 98%, multiplied by the areal sweep efficiency (the share of the area that the water reaches), 0.85, the same for all realisations. The data are assumed for illustration.

Worked exampleThe average by oil in place is not the average by production

Given (assumed data)

Set: nine equally probable realisations (the table above)
What is compared: the order of the realisations by oil in place and by recoverable oil

Solution

Oil in place, $0.624\,V_p$ , runs from 5.990 million m³ (realisation 1) to 9.360 million m³ (realisation 9). The order by oil in place follows the numbers, and the median (the fifth of nine) is realisation 5, 7.675 million m³.
Recoverable oil runs from 2.709 million m³ (realisation 4) to 4.640 million m³ (realisation 9). The order is different: 4, 2, 1, 8, 6, 3, 5, 7, 9.
The median by recoverable oil is realisation 6, 3.648 million m³. Realisation 5, the median by oil in place, is seventh in this order: 3.952 million m³, which is 8% more.
The high case by oil in place (eighth in the ranking) is realisation 8. Its recoverable oil is 3.524 million m³, which puts it fourth. The high case by oil is realisation 7, 4.553 million m³.
Spearman's rank correlation measures how well the two orders agree. The sum of the squared differences of the ranks is $\sum d^2 = 44$ , so $\rho_S = 1 - 6\cdot44/(9\cdot80) = 0.63$ .

Answer: Realisation 5, the median by oil in place, gives 3.952 million m³ of recoverable oil, which is 8% more than realisation 6, the median by oil (3.648 million m³). The rank correlation of the two orders is 0.63.

What it means. The pore volume decides oil in place, but recoverable oil also depends on heterogeneity. Realisation 8 has a lot of oil in place and realisation 4 a moderate amount, but both have a high $V_{DP}$ (see the table). Water passes quickly through the best layers, and less oil is displaced before the limiting water cut is reached. A model chosen as the median by oil in place would give, for design, a production forecast that is 8% too high. We rank the realisations by what matters for the decision: recoverable oil, the length of the plateau, water production. For different measures, a different realisation may turn out to be the median. The data in this example are assumed for illustration.

CalculatedOil in place and recoverable oil of the nine realisations (the numbers are the realisation numbers); the dashed lines are the medians of the two rankings: the fifth of nine by oil in place and by recoverable oil.

We need a set of realisations for the decision itself too. We run each development option on all the selected realisations, and we compare the options by the distribution of the result, not by a single calculation (Expected monetary value and preference theory). A robust decision stays sensible across the whole range. If there is none, flexibility and additional data become valuable: an appraisal well, an extended well test. The methods of the lesson on decision trees calculate what such information is worth.

How many runs are needed: sensitivity and designs of runs

Besides the geology, the fluid properties, the relative permeabilities, the capillary pressure, the vertical permeability, the aquifer and the transmissibility of faults are all uncertain, and the result also depends on the grid (limitations of models). The first step is one-at-a-time runs: a base case and, for each parameter, a low and a high value, with the others at their base values. We show the result with spider and tornado diagrams (charts that rank the parameters by their effect; how they are built and read). They pick out the few parameters that really affect the decision.

Number of runs for

k

uncertain parameters

Parameters $k$: 2
One at a time: $2k + 1$: 5
All combinations of two levels: $2^k$: 4
All combinations of three levels: $3^k$: 9

Parameters $k$: 4
One at a time: $2k + 1$: 9
All combinations of two levels: $2^k$: 16
All combinations of three levels: $3^k$: 81

Parameters $k$: 5
One at a time: $2k + 1$: 11
All combinations of two levels: $2^k$: 32
All combinations of three levels: $3^k$: 243

Parameters $k$: 8
One at a time: $2k + 1$: 17
All combinations of two levels: $2^k$: 256
All combinations of three levels: $3^k$: 6561

Parameters $k$	One at a time: $2k + 1$	All combinations of two levels: $2^k$	All combinations of three levels: $3^k$
2	5	4	9
4	9	16	81
5	11	32	243
8	17	256	6561

A run of a full-scale model takes hours, so all combinations are out of reach for five or more parameters.

One-at-a-time runs do not see the combined action. Suppose poor vertical permeability is dangerous only together with a weak aquifer. In one-at-a-time runs, both parameters will look moderate. So after we select the 3–5 main parameters, we build a design of runs in which the parameters change together. We fit a fast approximating (proxy) formula to the results. Then we build the distribution of the result from it by Monte Carlo simulation (many random trials). We do the same in assisted history matching (manual and assisted history matching).

Waterflooding and infill drilling: increment or acceleration

We compare options for developing a waterflood — infill drilling, converting wells to injection, new injection wells, changes in injection — by “with” and “without” runs. These are two runs of the same model that differ only in the intervention. Economics calculates the incremental cash flow in the same way. The incremental production is the difference in the production of the sector or field, not the production of the new well.

Incremental production and acceleration from an intervention

\Delta Q = Q_{\text{with}} - Q_{\text{without}},\qquad Q_{\text{acc}} = Q_{\text{well}} - \Delta Q

where:

$\Delta Q$ incremental production of the sector from the intervention, m³
$Q_{\text{with}},\ Q_{\text{without}}$ production of the sector over the same period in the runs with and without the intervention; $Q_{\text{without}}$ is the baseline production, m³
$Q_{\text{well}}$ the new well's own production, m³
$Q_{\text{acc}}$ the part of the production of the new well that the neighbouring wells would have produced later — acceleration of production, m³

Units: any identical units of oil volume at standard conditions

Holds when

Both runs use the same model and the same operating regimes of the other wells, apart from the changes that the intervention itself causes.
The period of comparison is the same for both runs and long enough for the acceleration to cancel out: over a short period, part of the acceleration looks like an increment.

A new well shares oil with its neighbours. In a swept zone it intercepts oil that they would have produced later. In a zone that the flood bypasses, it opens up new oil. Only a model that links the wells through the reservoir can tell one from the other.

Worked exampleTwo locations for an infill well

Given (assumed data)

Runs of the sector model (reduced to an exponential decline $q = q_i e^{-Dt}$ ): without a new well: $q_i$ = 146.0 thousand m³/yr, $D$ = 0.120 1/yr
Option A — between producers in the swept zone: sector: 171.6 thousand m³/yr and 0.137 1/yr; the well itself: 25.6 thousand m³/yr and 0.120 1/yr
Option B — in the zone bypassed by the flood: sector: 164.3 thousand m³/yr and 0.117 1/yr; the well itself: 18.3 thousand m³/yr and 0.095 1/yr
Economics (assumed): discount rate 10% a year; net income of US$140 per m³ of oil (US$0.14 million per thousand m³); the well costs US$3.5 million

Solution

Production over the whole life, $Q = q_i/D$ : without the well 1216.7, with A — 1252.6, with B — 1404.3 thousand m³. Increment of the sector: A — 35.9, B — 187.6 thousand m³.
Own production of the wells: A — $25.6/0.120 = 213.3$ , B — $18.3/0.095 = 192.6$ thousand m³. Of the production of A, 177.4 thousand m³ (83%) is acceleration, because the neighbours would have produced this oil without well A. For B the acceleration is 3%.
Discounted production (production in which later oil counts for less) is $q_i/(D + j)$ , with $j = \ln1.1 = 0.0953$ (exponential decline). Without the well it is 678.1, with A 738.7, with B 773.9 thousand m³. The increment is 60.6 and 95.8 thousand m³.
Net present value of the increment (US$ million): A — $60.6\cdot0.14 - 3.5 = 5.0$ , B — $95.8\cdot0.14 - 3.5 = 9.9$ .
Now the evaluation by the wells' own production. The discounted production of A is $25.6/(0.120 + 0.0953) = 118.9$ , and of B $18.3/(0.095 + 0.0953) = 96.2$ thousand m³. The “value” of A is 13.1, and of B 10.0 US$ million. Option A would look the better one, and its value would be overstated by a factor of 2.6.

Answer: By the increment of the sector, option B is better: 187.6 thousand m³ of new oil and US$9.9 million against 35.9 thousand m³ and US$5.0 million for A. By the wells' own production the order would be the reverse.

What it means. Well A, in the swept zone, intercepts oil that the neighbours would have produced later. The increment of the sector over the whole life is small, and the value of the option comes mainly from acceleration: we get the oil earlier, and discounting values this, but less than new oil. Well B opens up oil that the flood does not reach, so almost all of its production is increment. Reducing the runs to an exponential is a simplification for the example. In a real study, we compare the yearly profiles up to the economic limit or the end of the licence. The lesson Acceleration projects and incremental analysis covers the methods of evaluating such projects. The data in this example are assumed for illustration.

CalculatedCumulative incremental production of the sector: option A first runs ahead of the run without the well and then gives back most of the gain; option B accumulates new oil.

Gas injection: gas–oil ratio and capacity

A gas injection study answers several questions at once. Should we inject gas or water (how gas differs from water)? Where should we inject, into the crest or into the gas cap (free gas on top of the oil), and at what rate, so that the front stays stable (gravity stabilisation)? Do we need a compositional model (Types of reservoir simulation model)? Are the grid layers thin enough to show a gas tongue at the top (how many cells are needed)? Often one more question turns out to be decisive: how much gas the surface facilities can take.

In such projects, all or part of the produced gas goes back into the reservoir. We separate it, treat it and compress it. While only solution gas flows to the wells, the producing gas–oil ratio equals the solution gas–oil ratio. After the injected gas breaks through to the wells, the ratio rises, and with constant capacity we have to reduce the oil production.

Oil rate with a limit on gas-handling capacity

q_o = \min\left(q_{\text{pot}},\ \frac{Q_g}{R}\right)

where:

$q_o$ oil rate of the field, m³/d
$q_{\text{pot}}$ the potential of the wells — the oil rate they would deliver without the limit on gas, m³/d
$Q_g$ capacity for gas treatment and injection, m³/d (standard)
$R$ producing gas–oil ratio, m³/m³

Units: gas and oil at standard conditions

Holds when

All the produced gas passes through the limited capacity; gas used for the field's own needs and gas sold are accounted for in $Q_g$ .
The forecast gives the gas–oil ratio. When withdrawal is restricted, the gas approaches the wells more slowly, so in a model run with a gas limit on the group the gas–oil ratio rises later.

When $Q_g/R$ falls below the potential of the wells, the compressors, not the reservoir, decide production: each doubling of the gas–oil ratio halves the allowable oil rate.

Forecast of the model of a gas injection project (hypothetical)

Year: 1
Well potential, thousand m³/d: 9.0
$R$ , m³/m³: 110
Oil at 2 million m³/d of gas: 9.00
Oil at 3 million m³/d of gas: 9.00

Year: 2
Well potential, thousand m³/d: 9.0
$R$ , m³/m³: 110
Oil at 2 million m³/d of gas: 9.00
Oil at 3 million m³/d of gas: 9.00

Year: 3
Well potential, thousand m³/d: 9.0
$R$ , m³/m³: 110
Oil at 2 million m³/d of gas: 9.00
Oil at 3 million m³/d of gas: 9.00

Year: 4
Well potential, thousand m³/d: 9.0
$R$ , m³/m³: 135
Oil at 2 million m³/d of gas: 9.00
Oil at 3 million m³/d of gas: 9.00

Year: 5
Well potential, thousand m³/d: 8.6
$R$ , m³/m³: 175
Oil at 2 million m³/d of gas: 8.60
Oil at 3 million m³/d of gas: 8.60

Year: 6
Well potential, thousand m³/d: 8.2
$R$ , m³/m³: 235
Oil at 2 million m³/d of gas: 8.20
Oil at 3 million m³/d of gas: 8.20

Year: 7
Well potential, thousand m³/d: 7.8
$R$ , m³/m³: 310
Oil at 2 million m³/d of gas: 6.45
Oil at 3 million m³/d of gas: 7.80

Year: 8
Well potential, thousand m³/d: 7.4
$R$ , m³/m³: 390
Oil at 2 million m³/d of gas: 5.13
Oil at 3 million m³/d of gas: 7.40

Year: 9
Well potential, thousand m³/d: 7.0
$R$ , m³/m³: 470
Oil at 2 million m³/d of gas: 4.26
Oil at 3 million m³/d of gas: 6.38

Year: 10
Well potential, thousand m³/d: 6.6
$R$ , m³/m³: 545
Oil at 2 million m³/d of gas: 3.67
Oil at 3 million m³/d of gas: 5.50

Year: 11
Well potential, thousand m³/d: 6.2
$R$ , m³/m³: 610
Oil at 2 million m³/d of gas: 3.28
Oil at 3 million m³/d of gas: 4.92

Year: 12
Well potential, thousand m³/d: 5.8
$R$ , m³/m³: 670
Oil at 2 million m³/d of gas: 2.99
Oil at 3 million m³/d of gas: 4.48

Year	Well potential, thousand m³/d	$R$ , m³/m³	Oil at 2 million m³/d of gas	Oil at 3 million m³/d of gas
1	9.0	110	9.00	9.00
2	9.0	110	9.00	9.00
3	9.0	110	9.00	9.00
4	9.0	135	9.00	9.00
5	8.6	175	8.60	8.60
6	8.2	235	8.20	8.20
7	7.8	310	6.45	7.80
8	7.4	390	5.13	7.40
9	7.0	470	4.26	6.38
10	6.6	545	3.67	5.50
11	6.2	610	3.28	4.92
12	5.8	670	2.99	4.48

Oil rates are in thousand m³/d. The gas–oil ratio of 110 m³/m³ in the first three years is the solution gas–oil ratio of the oil. The rise from year 4 is the breakthrough of the injected gas. The data are assumed for illustration.

Worked exampleWhat gas capacity to build

Given (assumed data)

Forecast of the model: the well potential and the gas–oil ratio by year — the table above
Capacity options: 2 and 3 million m³/d of gas

Solution

Year 7, $R$ = 310 m³/m³. At 2 million m³/d, the allowable oil rate is $2000/310 = 6.45$ thousand m³/d. This is less than the well potential of 7.8, so gas is the limit. At 3 million m³/d it is $3000/310 = 9.68$ thousand m³/d, so the wells are the limit.
Gas begins to restrict production at 2 million m³/d from year 7, and at 3 million m³/d from year 9.
Production over 12 years (365 d per year): 28.68 and 32.59 million m³. The larger capacity gives 3.91 million m³ more, which is 14% of the production of the smaller option. All this oil comes in the years from the 7th.

Answer: A capacity of 3 million m³/d instead of 2 million keeps 3.91 million m³ of oil over 12 years, about 14% of the production of the smaller option.

What it means. We build capacity years before the gas–oil ratio rises, so we rely on the model forecast. The forecast of the gas–oil ratio, and its range over the realisations of the geology, is therefore one of the main results of the study. It decides whether to build the capacity at once or to allow for expansion in phases. An estimate from an unchanged forecast of the gas–oil ratio overstates the losses of the smaller option, because when withdrawal is restricted the gas reaches the wells more slowly. In a real study we set the limit in the model itself. We compare the incremental production with the cost of the compressors by net present value. The data in this example are assumed for illustration.

CalculatedOil rate by year: the well potential (dashed line) and the rate allowed by gas-handling capacities of 2 and 3 million m³/d.

In Kazakhstan, the largest fields of the Pre-Caspian Basin re-inject raw gas. According to its own data for 2024–2025, the operator of Karachaganak is increasing the share of gas returned to the reservoir. It links this, among other things, with the rise in the gas–oil ratio and with extending the plateau of liquid hydrocarbon production (gas injection at the fields of Kazakhstan). The calculations of this lesson do not relate to these fields.

A mature field: where the oil has been left

In a mature field, a model matched to many years of history (History matching) becomes a tool for managing development. It shows where and in which layers the oil has been left, which parts the flood has swept, and where the injection goes. From this come the interventions: infill wells and sidetracks (new branches from existing wells) into the zones that the flood bypasses; conversion of wells to injection; redistribution of injection; isolation of swept intervals; and enhanced oil recovery methods. We compare each with the continuation of current development at the same injection volumes, and we update the model regularly as new data arrive.

A typical story: history matching cannot reproduce the early water and the rapid rise in water cut until the model contains a thin high-permeability layer — a thief zone. Injectivity and inflow profiles (measurements of how much fluid each layer takes or gives) confirm that almost all the water goes through it. Pressure measurements with a formation tester in new wells show whether the layers communicate with one another (barriers from the pressure profile). If they do not, we can isolate the thief zone in the injection wells.

Worked exampleShould we isolate a thief zone

Given (assumed data)

Sector: the section of eight layers from the lesson Displacement in heterogeneous reservoirs, with no crossflow between them (16 m); displacement in each layer is piston-like (a flat front pushes the oil ahead; the Dykstra–Parsons calculation); oil 3.0 mPa·s ( $M'$ = 1.765), $B_o$ = 1.25
Element between the rows of wells: 500 × 300 m, pore volume 442.5 thousand m³
Model after history matching: layer 4 (900 mD, 1 m) has been swept since the first months; almost all the water goes through it
Decision: after 0.25 pore volumes injected (injected volume divided by the pore volume of the element): continue as it is, or isolate layer 4 in the injection well; the comparison runs to 0.55 pore volumes injected

Solution

By the time of the decision, 0.1741 pore volumes of oil have been produced, and the water cut is 53.4%. 67% of the moveable oil (the oil that water can displace) remains in the reservoir. Most of it is in layers 3, 7, 5 and 2: 22.4, 19.1, 17.8 and 15.9% of the remainder. None of these layers has watered out yet.
Immediately after isolation there is no water in the production, because the other layers have not yet been swept. But injectivity (how easily the well takes water) falls: for the same injection rate we need a pressure drop 1.9 times greater.
After another 0.30 pore volumes injected, 0.2756 pore volumes have been produced without changes, and 0.3354 with isolation. The additional 0.0598 pore volumes are $0.0598\cdot442.5/1.25 = 21.2$ thousand m³ of oil per element.
The water cut at the end of the period is 79.9% without changes and 59.1% with isolation. For the same injection, 1.6 times more oil has been produced: $(0.3354 - 0.1741)/(0.2756 - 0.1741)$ .

Answer: Isolating the swept layer gives 0.0598 pore volumes, that is 21.2 thousand m³ of oil per element, for the same injection. The water cut at the end of the period is 59.1% instead of 79.9%.

What it means. We again compare the intervention with the continuation of development at the same injection. Isolation works only because the layers do not communicate. With crossflow, the water would go around the isolated interval inside the reservoir (what can be done about a high-permeability layer). The price is an almost twofold rise in the pressure drop between the injection and production wells. If the injection pressure runs up against the formation fracture pressure, we will have to reduce the rate, and the gain in time will shrink. Two thirds of the moveable oil is still in the reservoir, and the flood will not sweep the tightest layers, 5 and 7, in a reasonable time. They need other interventions. The layered calculation replaces the full model here. For a section without crossflow it gives the main result, although it does not account for the transition zone behind the front in each layer. The data in this example are assumed for illustration.

CalculatedWater cut of the production of the element against injection: as it is and with layer 4 isolated after 0.25 pore volumes injected. The steps are water breakthroughs through individual layers.

After 0.57 pore volumes, the curve with isolation rises higher. The water that used to go into layer 4 has sped up the fronts in the other layers, so the next breakthrough came earlier. But more oil has been produced for the same injection all the way, both after 0.57 and after two pore volumes have been injected. So we compare the options by oil production, injection and water production over the period, not by the water cut at one moment.

Checking the forecast against the actual outcome

A few years after the decision, we compare the forecast with what actually happened. This is a look-back. It differs from the hindcast on withheld data in history matching (non-uniqueness of history matching). There we test the model on a past that was already known. Here we test it on a future that no one knew when the forecast was made.

The actual outcome is inside the forecast range — the model and the estimate of uncertainty work, and the new data can narrow the range.
The actual outcome is at the edge of the range or outside it — something essential is wrong in the model (connectivity, the aquifer or the displacement mechanism), or the range was too narrow. We revise the model and check the decisions made with it again.
Deviations in one direction across many projects — a systematic bias of the estimates, often towards optimism: underestimated heterogeneity, start-up delays, downtime. We take it into account in later forecasts.

We record and analyse not only failures. If a forecast came true, it is useful to understand why: perhaps errors of different signs cancelled out, and next time we will not be so lucky. Such an analysis is the cheapest way to improve later studies.

Terms and symbols

Simulation study: Work in which we build or update a model to answer the question behind a specific decision.
Geostatistical realisation: One of the equally probable pictures of the properties of the reservoir between wells. It agrees with the well data.
Ranking of realisations: Ordering the realisations by a fast measure, so that we can run a few representative ones in full.
“With” and “without” runs: Two runs of the same model that differ only in the intervention. Their difference is the increment.
Baseline production: The production forecast without the intervention. We compare production with the intervention against it.
Acceleration of production: The part of the production of a new facility that the existing wells would have produced later.
Thief zone: A thin high-permeability layer through which most of the injected water goes.
Look-back: A comparison of the forecast with the actual production a few years after the decision.

Common mistakes and misconceptions

Wrong

“Let us choose the median realisation by oil in place and design from it.”

In fact

Oil in place is not production. In the example of the lesson, the realisation that is the median by oil in place gave 8% more oil than the median by recoverable oil. We rank realisations by the measure that matters for the decision.

Wrong

“The new well will produce 213 thousand m³ — that is the effect of drilling.”

In fact

The effect is the increment in the production of the sector in the “with” and “without” runs. In the example of the lesson, of the 213 thousand m³ of well A only 36 thousand m³ was increment. The rest is acceleration: oil that the neighbouring wells would have produced anyway.

Wrong

“The production forecast is set by the reservoir and the wells.”

In fact

Often the surface facilities decide it: the capacity for gas, liquid and injection. In a gas injection project, the allowable oil rate is the gas-handling capacity divided by the gas–oil ratio.

Wrong

“A thief zone should always be isolated.”

In fact

Isolation at the well works only if the layers do not communicate. Otherwise the water goes around the isolated interval inside the reservoir. We check communication with pressure measurements through the section, and we check the gain with a “with” and “without” run that allows for the rise in the injection pressure.

Wrong

“One-at-a-time runs showed that no parameter is dangerous — so the risk is small.”

In fact

One-at-a-time runs do not see the combined action of the parameters. We change the main parameters together, according to a design of runs.

In depth: advanced detail

A fast ranking measure and how to check it

Ranking by a fast measure works only if the fast measure orders the realisations in the same way as the full calculation. We check this: we run about ten realisations from different parts of the set in full, and calculate the rank correlation of the fast measure with the full result. In the example of the lesson, the rank correlation of oil in place with recoverable oil is about 0.63, so oil in place is unsuitable for choosing the low, medium and high realisations by production. A fast displacement calculation that accounts for heterogeneity agrees better with the full calculation. If the correlation is still moderate, we run more realisations at the edges of the set, where a ranking error is most costly.

Connections to other topics

What a reservoir simulation model is and why it is needed — what a reservoir simulation model is and how it is used.
History matching — history matching is the basis of studies of mature fields.
Building a simulation model: input data and results — how the input data of a model are organised and checked.
Geological and technical uncertainty of a project — uncertainty over the stages of a project and the P90–P10 ranges.
Sensitivity analysis and scenarios — spider and tornado diagrams, scenarios.
Acceleration projects and incremental analysis — how projects that accelerate production are evaluated.
Displacement of oil by gas — the physics of displacing oil by gas and the stability of the front.
Displacement in heterogeneous reservoirs — waterflooding of a layered reservoir and thief zones.
Decision trees and the value of information — the value of additional information.
Reservoir engineering: tasks across the life of a field — the stages of the life of a field and project documents.

Check yourself

1
Of twenty realisations, the median one by oil in place was chosen for design. What is wrong with this, if the decision is the number of producing wells and the capacity of the facility?
1. A.Nothing: the median by oil in place is the median by every measure
2. B.A median cannot be determined from twenty realisations
3. C.The maximum by oil in place should have been chosen
4. D.The median by oil in place may be far from the median by production and by the length of the plateau, on which the decision depends
Show the answer
Correct answer: D. The median by oil in place may be far from the median by production and by the length of the plateau, on which the decision depends
Production depends not only on the pore volume but also on heterogeneity and connectivity. In the example of the lesson, the realisation that is the median by oil in place came seventh of nine by recoverable oil and gave a forecast 8% above the median. We rank realisations by the measure on which the decision depends.
2
Without a new well the sector would have produced 1.20 million m³ of oil, with it 1.26 million m³. The new well's own production is 0.35 million m³. What are the increment and the acceleration?
Show the answer
The increment is the difference in the production of the sector: $1.26 - 1.20 = 0.06$ million m³. The acceleration is $0.35 - 0.06 = 0.29$ million m³. This is oil that the neighbouring wells would have produced later. If we evaluate the well by its 0.35 million m³, we overstate the effect almost six times in volume. In value the overstatement is smaller, because acceleration also has a price.
3
The capacity for gas treatment and injection is 2.5 million m³/d, the gas–oil ratio has risen to 400 m³/m³, and the wells could deliver 8 thousand m³/d of oil. What oil rate is possible?
Show the answer
$2{,}500{,}000/400 = 6250$ m³/d. This is less than the potential of the wells, so gas restricts production to about 6.25 thousand m³/d. If the gas–oil ratio rose to 800 m³/m³, the allowable rate would halve.
4
When will isolating a high-permeability thief zone in an injection well have an effect?
1. A.Only if the thief zone has not yet been swept
2. B.Only if the thief zone is separated from the neighbouring layers by impermeable barriers
3. C.Always: the water will stop going into the thief zone
4. D.Only with gas injection
Show the answer
Correct answer: B. Only if the thief zone is separated from the neighbouring layers by impermeable barriers
If the layers communicate, water that did not enter the thief zone at the well will cross into it inside the reservoir, a few metres from the wellbore, and the effect will be small. We check communication with pressure measurements through the section and with inflow and injectivity profiles, and we check the gain with a “with” and “without” run.
5
A model has six uncertain parameters. How many runs are needed for one-at-a-time runs and for all combinations of three levels?
1. A.13 and 216
2. B.12 and 64
3. C.13 and 729
4. D.7 and 216
Show the answer
Correct answer: C. 13 and 729
One at a time: the base run and two per parameter, $2\cdot6 + 1 = 13$ . All combinations of three levels: $3^6 = 729$ . So after we select the main parameters, we build a design of runs and an approximating formula rather than try everything.
6
Four years after the decision to waterflood, the actual production turned out to be below the lower bound of the forecast range. What should we do?
Show the answer
This is a sign that something essential is wrong in the model (connectivity, the aquifer, the displacement mechanism), or that the range was too narrow. We revise the model with the new data, including history matching to the actual history. We recalculate the forecast and check whether the decisions made and planned remain sensible. We record the cause of the deviation, so that later studies can take it into account.

Updated: 17 October 2026