Lumped multi-tank material balance model

Quick refresher

A tank is a volume inside which the pressure is the same. We divide an accumulation into tanks where the pressure difference is concentrated at faults and tight partings. Inside the blocks the pressure must equalise quickly, in a time much shorter than the time constant of the link between the blocks (the time that the pressures of two blocks need to level out).
The balance of a tank: its storage capacity multiplied by the rate of change of pressure equals the net inflow from the neighbours minus the withdrawal. The storage capacity $C = c_tV_p$ (the capacitance) is how much fluid the tank gives up when its pressure falls by 1 MPa. The transmissibility $T$ of a link is the crossflow per 1 MPa of pressure difference.
The crossflows between tanks add up to zero. So the average pressure of all the tanks, weighted by storage capacity, behaves like the pressure of a single tank. A single tank is not wrong on average. It is wrong in every block.
$N$ connected tanks have $N - 1$ time constants. The worst-determined parameters are those of a far block behind a weak link: the pressure history must be comparable with the longest time constant.
We find the transmissibilities and storage capacities from the block pressures by least squares (the fit that makes the sum of the squared differences between model and data as small as possible). The result is ranges, and the parameters are correlated: a larger storage capacity of a far block can partly be replaced by a smaller transmissibility.
A well with good productivity in a small block quickly loses rate, and in the end it produces no more than the fault lets through. Stimulation (a treatment that raises the productivity of the well) raises noticeably only the initial rate. Pressure maintenance in the block itself helps.

Key relationships

C_j\,\dfrac{\dd p_j}{\dd t} = -q_j + \sum_k T_{jk}\,(p_k - p_j)

balance of tank $j$ ; $C_j = c_tV_{p,j}$

\sum_j C_j\,\dfrac{\dd p_j}{\dd t} = -\sum_j q_j

the sum of the balances: the crossflows cancel, and we get the balance of a single tank

q_\infty = \dfrac{J\,T}{J + T}\,(p_{\text{nb}} - p_{wf}),\quad \tau = \dfrac{C}{J + T}

the limiting rate of a well in a small block, and the time it takes to settle

\dfrac{\Delta p_2}{a\,t} = 1 - \dfrac{1 - \e^{-x}}{x},\quad x = \dfrac{t\,T}{C_2}

pressure decline in a block behind a fault when the neighbouring block declines evenly

Common pitfalls

Taking the pressure of the wells of some blocks for the average of the accumulation and using it for a single tank.
Dividing an accumulation into tanks where the pressure inside a block does not equalise faster than between blocks.
Fitting all the volumes and transmissibilities at once to a few readings: there are more parameters than independent pieces of information in the data.
Placing an injection well where it is easiest to drill, without checking whether that block is connected with the producing wells.
Carrying the fitted transmissibilities into a grid model without recalculating them for its cells.

After this lesson you can

write the balance of a tank with crossflow to its neighbours, assemble the system for several tanks, and explain why the sum of the equations gives the material balance of a single tank
justify dividing an accumulation into tanks: along faults and partings, and by comparing the time for pressure to equalise inside a block with the time between blocks
describe wells in the model, with either a given rate or a given bottomhole pressure (the pressure at the bottom of the well), and explain why the rate of a well in a small block is limited not by its productivity but by the transmissibility of the fault
fit the transmissibilities and storage capacities of the tanks to the block pressures by least squares, and judge which parameters the data determine and which they do not
explain when several tanks are better than one, and where the model is inferior to a reservoir simulation model

From one tank to several

The material balance of a single tank (the idea of the method) assumes that the pressure is the same over the whole accumulation. The more slowly the pressure equalises, the worse this assumption is (assumptions of the balance). In an accumulation cut up by faults and tight partings, the pressure does not change smoothly. It changes by steps at these boundaries. Inside a block it is almost the same, but between blocks it differs by megapascals (blocks and faults). So the structure of the accumulation itself suggests a multi-tank model: each block or unit is a tank, and each fault or parting is a link that lets a limited amount of fluid through (a finite transmissibility).

We test the assumption that the pressure inside a tank is the same by comparing two times. Inside a block the pressure equalises in about the time that the flow needs to reach the pseudo-steady-state regime (the regime in which the pressure falls at the same rate everywhere). In dimensionless time $t_{DA}$ (time scaled to a pure number) this is from 0.1 to 1 or more, depending on the shape of the block and the layout of the wells (stabilisation time). Between blocks the time is set by the time constant of the link, $\tau = C_1C_2/\big((C_1 + C_2)\,T\big)$ (time constant of a barrier). Tanks are suitable if the first time is much shorter than the second and shorter than the intervals between pressure readings.

SchematicAn accumulation of three blocks separated by faults (top, plan view) and its multi-tank model (bottom): tanks with storage capacities

C

, linked by transmissibilities

T

, and the withdrawals of the wells

q

. A diagram, not to scale.

Worked exampleAre the blocks suitable as tanks

Given (assumed data)

Block A: 4 × 2.5 km, permeability 200 mD, porosity 0.2; oil 1.2 mPa·s; $c_t = 2.1\cdot10^{-3}$ MPa⁻¹
Pore volumes from the maps: A: 40 million m³, neighbouring block B: 25 million m³
Fault between them: transmissibility estimated from its zone, about 40 m³/(d·MPa)
Pressure readings: once every six months

Solution

Equalisation inside block A: $t = t_{DA}\,\phi\mu c_tA/(0.0853\,k)$ . At $t_{DA} = 0.1$ : $0.1\cdot 0.2\cdot 1.2\cdot 0.0021\cdot 10^7/(0.0853\cdot 200) = 29.5$ d. At $t_{DA} = 0.3$ it is 88.6 d.
Storage capacities: $C_A = 0.0021\cdot 40\cdot 10^6 = 84{,}000$ and $C_B = 52{,}500$ m³/MPa.
Time constant of the link: $84{,}000\cdot 52{,}500/\big((84{,}000 + 52{,}500)\cdot 40\big) = 808$ d, or 2.2 years.
Ratio of the times: $808/88.6 = 9$ and $808/29.5 = 27$ .

Answer: Inside a block the pressure equalises in one to three months, and between blocks in 2.2 years. The blocks can be treated as tanks if we compare the model with pressures averaged over months, and not with readings from the first weeks after a rate change.

What it means. A margin of 9 to 27 times is enough for readings every six months. But right after a well is started up or shut in, the pressure in the block is not uniform, and the readings of those weeks are compared not with the average pressure of the tank but with the pressure near the well (which pressure to compare with which). If the block were low in permeability, 2 mD instead of 200, the equalisation inside it would take 8 years. We would have to divide it into several tanks or calculate it with a grid model. The data in this example are assumed for illustration.

Equations: the balance of each tank

For each tank the balance is the same as for a single tank, except that crossflows to the neighbours are added to the production and injection. Above the bubble point pressure there is no free gas, and the elastic expansion is proportional to the pressure drop (undersaturated accumulation). So the equations are linear:

Material balance of a tank in the multi-tank model

C_j\,\frac{\dd p_j}{\dd t} = -q_j + \sum_k T_{jk}\,(p_k - p_j),\qquad C_j = c_t\,V_{p,j}

where:

$p_j$ average pressure of tank $j$ , brought to one datum depth (a chosen reference depth), MPa
$C_j$ storage capacity (capacitance) of the tank: the volume that it gives up when its pressure falls by 1 MPa, m³/MPa
$c_t,\ V_{p,j}$ total compressibility of the rock and the fluids, and the pore volume of the tank, 1/MPa, m³
$q_j$ net withdrawal of the wells of the tank (production minus injection) at reservoir conditions, m³/d
$T_{jk}$ transmissibility of the link between tanks $j$ and $k$ : the crossflow per 1 MPa of pressure difference, $T_{jk} = T_{kj}$ . For a section of a fault it is $0.0853\,(k_f/l_f)\,A_f/\mu$ , m³/(d·MPa)
$t$ time, d

Units: metric oilfield: MPa, m³, m³/d, days

Holds when

the pressure inside a tank is the same: it equalises much faster than between tanks
single-phase undersaturated oil or water, with constant compressibility and viscosity; below the bubble point pressure the storage capacity depends on pressure, and the equations are nonlinear (see the advanced section)
the crossflow through a link is proportional to the pressure difference of the tanks: the flow through it is steady-state, and the storage capacity of the link itself is ignored

The pressure of a tank changes at a rate equal to the net inflow, with the crossflows, divided by its storage capacity. Crossflow goes from the tank with the higher pressure, so the links smooth out the differences. The lower the transmissibility, the longer a pressure difference lasts. This is the same equation as for a cell of a grid model (cell equation). There the same quantity is called the connection coefficient of the cells and is expressed through their geometric transmissibility, and here the “cells” are natural blocks.

We express the transmissibility of a fault through the transmissibility of its zone, $k_f/l_f$ (the permeability of the fault zone divided by its thickness), and the area open to crossflow (transmissibility of a fault section). We express the transmissibility of a tight parting between units through its permeability, thickness and area (balance of two units).

Sum of the balances. Each crossflow enters the equations of two tanks with opposite signs, and it cancels when we add them: $\sum_j C_j\,\dd p_j/\dd t = -\sum_j q_j$ . The pressure averaged with the storage capacities as weights, $\bar p = \sum_j C_jp_j/\sum_j C_j$ , behaves exactly like the pressure of a single tank with the total storage capacity (depletion of a closed area). So a single tank is not wrong on average, provided that the average pressure is a true average over the volume. It is wrong when the “average” is taken over the wells and the wells stand in only some of the blocks.

How many time constants the system has. The pressure difference of two tanks settles with a single time constant (exact solution for two units). $N$ tanks have $N - 1$ such times. The short ones equalise neighbours that are well connected, and the long ones equalise far blocks behind weak links. This gives the main rule of fitting: the parameter that belongs to the longest time can be determined only if the pressure history is comparable with that time (see the advanced section).

Edge water and boundaries. We describe an aquifer by one more tank, with a large storage capacity and its own transmissibility of the link with the accumulation. This is the Fetkovich model (the aquifer as a tank behind a resistance). A block whose pressure is held by injection is described by a tank at constant pressure.

Wells in tanks

The withdrawal of a tank is the sum of the rates of its wells. When we fit to the history, we take them from field records, which are the most reliable development data. In a forecast we calculate the rate with the productivity index (the rate per unit of pressure drop between the reservoir and the well) from the pressure of the tank: $q = J\,(p_j - p_{wf})$ (inflow in the pseudo-steady-state regime). The rate of a well that is open to several units (each unit is a tank) is shared out between them by their productivities and pressures. A shut-in well of this kind passes fluid between the units (commingled production and crossflow through the wellbore). The formulas of that lesson show that the transmissibility of this link through the wellbore is $J_1J_2/(J_1 + J_2)$ , like two resistances in series. In the model we switch it on for the time of the shut-ins.

A well in a small block: the fault limits the rate

Sometimes a productive well stands in a small block that is connected to the rest of the accumulation by a fault that passes fluid poorly. At constant bottomhole pressure, the rate of the well is described by one tank with one link and the well. The initial rate is set by the productivity of the well. The final rate is set by a chain of two “resistances” in series, the fault and the well.

Rate of a well in a small block behind a fault

\begin{aligned} q(t) &= q_\infty + (q_0 - q_\infty)\,\e^{-t/\tau},\qquad \tau = \frac{C}{J + T}, \\ q_0 &= J\,(p_{\text{nb}} - p_{wf}),\qquad q_\infty = \frac{J\,T}{J + T}\,(p_{\text{nb}} - p_{wf}) \end{aligned}

where:

$q,\ q_0,\ q_\infty$ rate of the well at reservoir conditions: current, initial and limiting, m³/d
$J$ productivity index of the well, m³/(d·MPa)
$T$ transmissibility of the fault between the block and its neighbour, m³/(d·MPa)
$C$ storage capacity of the block, $c_tV_p$ , m³/MPa
$p_{\text{nb}}$ pressure of the neighbouring block: that block is large, or its pressure is held by injection, MPa
$p_{wf}$ bottomhole pressure of the well, MPa
$\tau$ time constant of the decline of the rate, d

Units: metric oilfield

Holds when

the block is a tank; the pressure of the neighbouring block is constant; at the start the pressure of the block equals that of its neighbour
the bottomhole pressure and the productivity are constant; the flow is single-phase, above the bubble point pressure

When $J \gg T$ , the limiting rate is close to $T\,(p_{\text{nb}} - p_{wf})$ and hardly depends on the productivity of the well. The change from the initial rate to the limiting one takes several $\tau$ . After the well is shut in, the block recovers more slowly, with the constant $C/T$ , because it is fed only through the fault.

Worked exampleA well in a small block, and stimulation

Given (assumed data)

Block: pore volume 3.0 million m³; $c_t = 2.1\cdot10^{-3}$ MPa⁻¹
Fault: transmissibility 8 m³/(d·MPa); injection holds the pressure of the neighbouring block at 28.0 MPa
Well: $J = 40$ m³/(d·MPa), bottomhole pressure 20.0 MPa; after stimulation $J = 80$

Solution

Storage capacity of the block: $C = 0.0021\cdot 3.0\cdot 10^6 = 6300$ m³/MPa. Initial rate: $q_0 = 40\cdot(28.0 - 20.0) = 320$ m³/d.
Limiting rate: $q_\infty = 40\cdot 8/48\cdot 8.0 = 53.33$ m³/d. Time constant: $\tau = 6300/48 = 131.25$ d.
After a year: $\e^{-365/131.25} = 0.0620$ and $q = 53.33 + (320 - 53.33)\cdot 0.0620 = 69.9$ m³/d. Produced in the year: $53.33\cdot 365 + (320 - 53.33)\cdot 131.25\cdot(1 - 0.0620) = 52.3$ thousand m³.
With $J = 80$ : $q_0 = 640$ , $q_\infty = 80\cdot 8/88\cdot 8.0 = 58.2$ m³/d and $\tau = 6300/88 = 71.6$ d. In the year it produces 62.6 thousand m³.
After a shut-in the pressure of the block recovers to 28.0 MPa with the constant $C/T = 6300/8 = 788$ d, which is more than two years.

Answer: The rate falls from 320 to 69.9 m³/d in a year and tends to 53.33. Stimulation doubles the initial rate, but it raises the limiting rate only to 58.2 m³/d, and the production in a year by only 20%.

What it means. The rate of such a well is limited not by the well but by the fault: in the end it produces as much as the fault lets through. Suppose the bottomhole pressure cannot be lowered below some limit: the well stops flowing on its own, or the pump can no longer deliver. Then the well “dies”. After a long shut-in, when the block has been recharged through the fault, it works again. This is how wells in strongly fragmented accumulations come to work on and off. What helps is not stimulation but pressure maintenance in the block itself: an injection well in it, or a well that is open to both blocks. The data in this example are assumed for illustration.

CalculatedThe rate of the well in the small block of the example, for productivities of 40 and 80 m³/(d·MPa). Stimulation doubles the initial rate, but after only four months this well produces less, because it has used up the elastic reserve of the block faster. After a year both curves approach almost the same limits, which are set by the fault (dotted).

Fitting the model to the block pressures

The model works in two modes. In the forward mode the storage capacities and transmissibilities are given, and we calculate the pressures from the schedule of withdrawals. This is a forecast. In the inverse mode the block pressures are known from measurements, and we adjust the storage capacities and transmissibilities until the calculation agrees with them within the error. This is history matching (History matching).

Which pressures are suitable. We need the average pressures of the blocks, brought to one datum depth. They come from the pressure build-up in shut-in wells (Build-up in a bounded reservoir and average reservoir pressure), from tester surveys in new wells (Pressure measurements in producing reservoirs: barriers and accuracy) and from permanent gauges. The most valuable reading is the first well in a block that has not yet been put on production. Its pressure shows at once whether the block communicates with the working ones (see the example below). We take the withdrawals and injection of each block from field records. We divide the production of a well that is open to several tanks by the results of production logging (measurements of the flow of each unit in a working well).

What the data determine. The block pressures respond not to each quantity separately, but to combinations of them. The total pressure decline sets the total storage capacity of the connected blocks. The pressure difference of neighbouring blocks is the crossflow divided by the transmissibility, and the crossflow depends on the storage capacities behind the link. A larger storage capacity of a far block can partly be replaced by a smaller transmissibility. Two things help to separate them: the transient periods (after wells start up and shut in, and when injection begins), where the time constants show themselves, and independent estimates of the volumes from the maps. So we fit only a few parameters, as many as the data can determine, and take the rest from the geology.

Worked exampleThe first well in a neighbouring block: does the fault hold back the flow

Given (assumed data)

Block 1: on production for 500 days; its pressure from permanent gauges falls almost evenly, by 2.4 kPa/d
Block 2: behind the fault, no wells; pore volume from the maps 18 million m³ (±20%); $c_t = 2.1\cdot10^{-3}$ MPa⁻¹
Reading: a new well in block 2 showed on day 500 a pressure 0.30 MPa below the initial one; the error is ±0.02 MPa (the same gauge and the same datum depth as in the initial survey)

Solution

Block 2 is a tank fed only through the fault: $C_2\,\dd p_2/\dd t = T\,(p_1 - p_2)$ . If $p_1$ falls evenly at the rate $a$ , the decline in block 2 is $\Delta p_2 = a\,\big[t - \tau\,(1 - \e^{-t/\tau})\big]$ , with $\tau = C_2/T$ . We divide by the decline in block 1: $\Delta p_2/(a\,t) = 1 - (1 - \e^{-x})/x$ , where $x = t/\tau$ .
The decline in block 1 is $0.0024\cdot 500 = 1.2$ MPa. The ratio is $0.30/1.2 = 0.25$ .
We solve the equation $1 - (1 - \e^{-x})/x = 0.25$ by trial: $x = 0.606$ . Check: $\e^{-0.606} = 0.5455$ and $1 - (1 - 0.5455)/0.606 = 0.2500$ .
Time constant: $\tau = 500/0.606 = 825$ d. Storage capacity: $C_2 = 0.0021\cdot 18\cdot 10^6 = 37{,}800$ m³/MPa. Transmissibility: $T = 37{,}800/825 = 45.8$ m³/(d·MPa).
The reading error of ±0.02 MPa gives $T$ from 42 to 50. The volume error of ±20% adds another ±20%, because the transmissibility is proportional to the storage capacity that we assume. A decline of less than 0.04 MPa (two errors) would mean $T < 5$ : the fault hardly passes any fluid. A decline equal to the 1.2 MPa of block 1 would mean that the fault does not hinder the flow.

Answer: The fault partly passes fluid:

T \approx 46

m³/(d·MPa) (±22% with the volume error included), and the time constant of block 2 is about 2.3 years.

What it means. A single reading in a new well answers a question that seismic data cannot answer: does the fault hold back the flow? The answer is not “yes or no” but a number with an error, and almost all of the error comes from the volume of the block. We check the transmissibility that we found with the next readings. With it, the model must predict how the pressure in block 2 will go on falling. The calculation assumes that block 2 is fed only through this fault. The pressure of block 1 is taken from the gauges, so its own links with other blocks do not affect the result. The data in this example are assumed for illustration.

Pressure readings in blocks A, B and C (day — pressure, MPa)

Block: A
Readings: 200 — 28.72; 400 — 27.31; 600 — 26.26; 800 — 24.82; 1000 — 23.73; 1200 — 22.36
How obtained: pressure build-up in the production wells

Block: B
Readings: 400 — 29.68; 600 — 28.14; 800 — 26.99; 1000 — 25.72; 1200 — 24.42
How obtained: on day 400, a formation tester in the first well of the block, then pressure build-up

Block: C
Readings: 900 — 29.71; 1050 — 29.67; 1200 — 29.52
How obtained: a formation tester in a new well, then a permanent gauge

Block	Readings	How obtained
A	200 — 28.72; 400 — 27.31; 600 — 26.26; 800 — 24.82; 1000 — 23.73; 1200 — 22.36	pressure build-up in the production wells
B	400 — 29.68; 600 — 28.14; 800 — 26.99; 1000 — 25.72; 1200 — 24.42	on day 400, a formation tester in the first well of the block, then pressure build-up
C	900 — 29.71; 1050 — 29.67; 1200 — 29.52	a formation tester in a new well, then a permanent gauge

The pressures are block averages brought to the datum depth. The error of each is ±0.10 MPa. The initial pressure of all the blocks is 30.0 MPa. The data are assumed for illustration.

Worked exampleFault transmissibilities and the storage capacity of the far block from the pressures

Given (assumed data)

Blocks: A, B and C in a row, with faults A–B and B–C; pore volumes from the maps 40, 25 and 30 million m³; $c_t = 2.1\cdot10^{-3}$ MPa⁻¹
Withdrawals at reservoir conditions: A: 600 m³/d from the start of production; B: 300 m³/d from day 400; there are no producing wells in C
Readings: the table above
What is fitted: $T_{AB}$ , $T_{BC}$ and the storage capacity of block C (63,000 m³/MPa from the map, but the block has only one well, an observation well); the storage capacities of A and B are taken from the maps

Solution

The model is three tank balances for the chain A–B–C, with $C_A = 84{,}000$ and $C_B = 52{,}500$ m³/MPa. We calculate the pressures numerically, with a time step of one day.
The guesses “the faults hardly pass fluid” ( $T = 1$ ) and “the faults are open” ( $T = 200$ ) give $M = 37$ and $M = 187$ ( $M$ is the objective function, a number that measures how far the model is from the readings). The mismatches are many times larger than the error.
The minimum of the objective function $M$ by least squares is at $T_{AB} = 38.5$ and $T_{BC} = 13.8$ m³/(d·MPa), $C_C = 64.3$ thousand m³/MPa. Here $M = 0.47$ : the mismatches are at the level of the reading error.
Errors from the curvature of $M$ near the minimum: $T_{AB}$ is ±8%, $T_{BC}$ is ±23% and $C_C$ is ±28%. The correlation coefficient of the estimates of $T_{BC}$ and $C_C$ (it shows how strongly the two estimates trade off against each other) is 0.80.
Calculation for day 1200: A is 22.41, B is 24.46 and C is 29.50 MPa, against the measured 22.36, 24.42 and 29.52.
A single tank for all the readings: the best storage capacity is 211 thousand m³/MPa, but $M = 486$ . On day 1200 it gives $30.0 - 960{,}000/211{,}390 = 25.46$ MPa, which is 3.10 MPa above the pressure of block A and 4.06 MPa below the pressure of block C.

Answer: Fault A–B has a transmissibility of about 38.5, and fault B–C about 13.8 m³/(d·MPa). The storage capacity of block C from the pressures agrees with the map. A single tank cannot describe these data.

What it means. The parameters of the far block C are determined worst. Its slow time constant, 3723 d, is three times longer than the history, and its storage capacity and the transmissibility of the fault B–C partly replace each other. In terms of the fault zone (from transmissibility to the model), $k_f/l_f = T\mu/(0.0853\,A_f)$ is about 0.011 and 0.0039 mD/m for an open area of 50,000 m². The readings were generated by this same model, with transmissibilities of 40 and 15 m³/(d·MPa), a storage capacity of C of 63,000 m³/MPa, and random errors of a few hundredths of a megapascal. The fit recovered them within the error. If the storage capacities of A and B were fitted as well, the uncertainty would grow. The pressures determine above all the sum of the storage capacities and the ratios of the storage capacities to the transmissibilities. The data in this example are assumed for illustration.

CalculatedThe block pressures from the example: the readings with an error of ±0.10 MPa, and the calculation with the fitted model. The dashed line is the best single tank. It runs between the blocks and does not describe any of them.

When several tanks are better than one

Suppose we treat the accumulation as a single tank, using the pressure of the working wells. If these wells stand in only some of the blocks, the calculation sees only the volume that has had time to respond. The apparent storage capacity (the withdrawal so far divided by the pressure drop), $Q/(p_i - p)$ , then grows with time, as if edge water were feeding the accumulation or new volumes were joining in (limitations of the balance). In the example above, from the pressure of block A it would be 89, 116 and 126 thousand m³/MPa on days 400, 800 and 1200, while the total storage capacity of the three blocks is about 201 thousand m³/MPa. Pressure readings on the other side of the faults, and the geology, help to tell far blocks from an aquifer. The growth itself is a reason to build a model of several tanks.

In a gas accumulation the same picture shows on the $p/Z$ plot (pressure divided by the gas deviation factor $Z$ ; the p/Z straight line). With the pressure of the wells of the nearest block, the points first lie on a steep straight line that belongs to this block only. When the far block starts to give up its gas, the plot flattens. A straight line through the early points underestimates the gas in place, and the gas of the far blocks shows up later, the worse the link is. A multi-tank model of a gas accumulation writes the same balance for each tank in the $p/Z$ form, and the crossflow between tanks in standard cubic metres (the volume at standard surface conditions). The equations become nonlinear, but they are solved numerically in the same way.

The main benefit of several tanks is a forecast by block. A single tank says how the average pressure will fall, and it is “supported” by injection in the same way at any point. Several tanks show in which block the wells will lose pressure first, and which injection will reach them.

Worked exampleWhere to inject water

Given (assumed data)

Model: the one fitted in the previous example; the same withdrawals
Options: injection of 600 m³/d from day 1300, into an existing well of block C, into a new well of block B, or into a new well of block A
Constraint: the pressure in block A must not fall below the bubble point pressure of 18 MPa

Solution

Without injection the pressure in block A falls to 18 MPa on day 1921. By day 3000 it is 11.7 MPa.
Injection into C: block A reaches 18 MPa on day 1927, and by day 3000 it is 12.2 MPa. The pressure in block C itself rises to 39.75 MPa, which is 9.75 MPa above the initial pressure.
Injection into B: 18 MPa is reached on day 2129, which is 208 days later than without injection. By day 3000 it is 15.5 MPa.
Injection into A: by day 3000 the pressure in block A is 21.1 MPa, above the bubble point pressure for the whole period.
A single tank (storage capacity 211 thousand m³/MPa) does not tell the injection points apart: by day 3000 it gives 22.6 MPa with injection and 17.8 MPa without it.

Answer: Injection into block C delays the fall of the pressure in block A to the bubble point pressure by only 6 days, and injection into block B by 208 days. Injection into block A itself keeps the pressure above the bubble point pressure for the whole period.

What it means. Injection behind two weakly conducting faults builds up the pressure in its own block, and the producing wells get almost no help. In addition, the pressure in block C would reach the allowable injection pressure (the highest pressure that does not fracture the rock; Fracture gradient and leak-off testing), and the injection would have to be cut back. A single tank does not see this. At any injection point it promises the same thing: a pressure above the bubble point pressure for the whole period. A multi-tank model compares the options in seconds. We then test the chosen option with a grid model, which accounts for the displacement of oil by water and for the water cut of the wells (the share of water in the produced liquid) (voidage replacement). The data in this example are assumed for illustration.

CalculatedThe pressure in block A from the fitted model with injection of 600 m³/d from day 1300 into different blocks, and without injection. The dashed line is the forecast of a single tank without injection. The horizontal line is the bubble point pressure of 18 MPa.

Limits of the method, and the reservoir simulation model

What a single tank, several tanks and a grid model can see

Question: Oil in place of the connected volume
Single tank: as a whole
Several tanks: by block
Grid model: by cell

Question: Communication between blocks
Single tank: none
Several tanks: transmissibilities of the links
Grid model: transmissibility multipliers

Question: Pressure at the well
Single tank: no
Several tanks: average over its block
Grid model: with the drawdown and the influence of neighbours

Question: Saturation, water front
Single tank: average
Several tanks: average by block
Grid model: by cell

Question: Time of one run
Single tank: fractions of a second
Several tanks: seconds
Grid model: minutes and hours

Question	Single tank	Several tanks	Grid model
Oil in place of the connected volume	as a whole	by block	by cell
Communication between blocks	none	transmissibilities of the links	transmissibility multipliers
Pressure at the well	no	average over its block	with the drawdown and the influence of neighbours
Saturation, water front	average	average by block	by cell
Time of one run	fractions of a second	seconds	minutes and hours

The comparison is for a single-phase problem above the bubble point pressure. Below it, and in waterflooding, a multi-tank model needs phase balances and quickly loses its simplicity.

Uniformity of the tank. Inside a tank there are no drawdown cones (zones of lowered pressure around the wells) and no fronts. The pressure at a working well is lower than the average pressure of the tank by the drawdown of the well, so we compare the model with the average pressures of the blocks.
Phases. Above the bubble point pressure the equations are linear. Below it gas comes out of solution, the storage capacity of a tank starts to depend on pressure and saturation, and the crossflow carries oil and gas in different shares. We add phase balances to each tank, and the model comes closer to a grid model.
The division is set in advance. The tanks follow the faults and partings that the geologist has chosen. With a wrong division, the fit will find parameters that explain the history but predict the future badly (non-uniqueness). We compare the options for the division by the quality of the fit and by the forecasts.
Number of parameters. Each tank adds a storage capacity, and each link adds a transmissibility. There must be noticeably fewer parameters than independent pieces of information in the data, so we tie them together with the geology. For example, a long fault across which five pairs of neighbouring blocks communicate is described by one fault-zone transmissibility $k_f/l_f$ , and the area of each section open to crossflow is taken from the maps. Five unknown transmissibilities turn into one.

We usually build the multi-tank model before the grid model, and together with it. We convert its transmissibilities into $k_f/l_f$ and set the transmissibility multipliers of the faults (factors that scale the flow between neighbouring cells) in the cells of the model, after recalculating them for the size of the cells (transmissibility multiplier, Interblock flow and transmissibility). We compare the storage capacities of the blocks with the volumes from the maps. The grid model must reproduce the same block pressures to which the multi-tank model was fitted. Perched water under a block (Perched and tilted contacts, trapped water; a pocket of water that is cut off from the main aquifer) in such a model is not an aquifer that gives pressure support. It is only a small addition to the storage capacity of its own tank.

Terms and symbols

Tank: A part of an accumulation, a block or a unit, in which we take the pressure to be the same. An element of the multi-tank model.
Storage capacity (capacitance) $C$: $c_tV_p$ , the volume that a tank gives up when its pressure falls by 1 MPa, in m³/MPa.
Transmissibility of a link $T$: The crossflow between tanks per 1 MPa of pressure difference, in m³/(d·MPa). For a fault it is found from the transmissibility of its zone $k_f/l_f$ and its area.
Time constant of a link: The time for the pressure to equalise between tanks. $N$ connected tanks have $N - 1$ such times.
Forward and inverse problem: Calculating the pressures from known storage capacities and transmissibilities, and fitting these parameters to the measured pressures.
Apparent storage capacity: $Q/(p_i - p)$ from the pressure of a part of the accumulation. Its growth points to volumes that have not been counted, or to water influx.

Common mistakes and misconceptions

Wrong

“The average pressure of the wells is the average pressure of the accumulation.”

In fact

The wells stand in only some of the blocks. The average pressure that obeys the balance of a single tank is the average over the storage capacities of all the blocks, and it needs readings in all the blocks.

Wrong

“A well in a small block has quickly lost its rate, so it needs stimulation.”

In fact

If the fault limits the rate, stimulation raises noticeably only the initial rate. In the lesson's example the limiting rate grew only from 53 to 58 m³/d. Pressure maintenance in the block itself helps.

Wrong

“The model matched the readings, so the transmissibilities have been found accurately.”

In fact

Other combinations of parameters also describe the history within the error, especially for far blocks. The result of a fit is ranges and their correlation, and we check the forecast with new readings.

Wrong

“Injection into any block supports the whole accumulation, and only its volume matters.”

In fact

Behind weakly conducting faults, injection builds up the pressure of its own block and hardly reaches the producing wells. We choose the injection point by the transmissibilities of the links.

Wrong

“The initial pressures of the blocks were the same, so the blocks communicate well and tanks are not needed.”

In fact

The match tells us that the pressures equalised over geological time. We test communication over the years of production from the response of the block pressures to production (isolated blocks).

In depth: advanced detail

Time constants of a chain of tanks

In matrix form (with tables of numbers called matrices) the system of tanks is written as $\mathbf C\,\dot{\mathbf p} = -\mathbf K\,\mathbf p - \mathbf q$ . Here $\mathbf C$ is the diagonal matrix of storage capacities, and $\mathbf K$ is the symmetric matrix of links. On its diagonal stands the sum of the transmissibilities of the tank, and off the diagonal stand the transmissibilities with a minus sign, so that the sum of each row is zero. The solution is a sum of modes (separate patterns of change, each with its own time). One mode is the common pressure decline of all the tanks. The other $N - 1$ decay as $\e^{-\lambda t}$ , where $\lambda$ are the eigenvalues of the matrix $\mathbf C^{-1}\mathbf K$ (the numbers that give the decay rates), and $1/\lambda$ are the time constants. For three tanks in a chain, the non-zero eigenvalues come from a quadratic equation:

Time constants of three tanks in a chain

\begin{aligned} &\lambda^2 - s_1\lambda + s_2 = 0,\qquad \tau = 1/\lambda, \\ &s_1 = \frac{T_{AB}}{C_A} + \frac{T_{AB} + T_{BC}}{C_B} + \frac{T_{BC}}{C_C},\qquad s_2 = \frac{T_{AB}\,T_{BC}\,(C_A + C_B + C_C)}{C_A\,C_B\,C_C} \end{aligned}

where:

$\lambda$ non-zero eigenvalues of $\mathbf C^{-1}\mathbf K$ , the inverse time constants, 1/d
$s_1$ trace of the matrix $\mathbf C^{-1}\mathbf K$ (the sum of the numbers on its diagonal), which is the sum of the eigenvalues, 1/d
$s_2$ the sum of its principal minors of second order (small determinants cut out along the diagonal), which is the product of the non-zero eigenvalues, 1/d²
$C_A,\ C_B,\ C_C$ storage capacities of the tanks, m³/MPa
$T_{AB},\ T_{BC}$ transmissibilities of the links; there is no A–C link, m³/(d·MPa)

Units: metric oilfield

Holds when

a chain A–B–C with no direct A–C link; the storage capacities and transmissibilities are constant.

The two roots give two times: a short one and a long one. If one link is much stronger than the other, the short time equalises the strongly connected pair, and the long time pulls the far tank towards it. A pressure history shorter than the long time determines the parameters of the far tank poorly.

Worked exampleThe two times of the field in the example

Given (assumed data)

Storage capacities: $C_A = 84{,}000$ and $C_B = 52{,}500$ m³/MPa; $C_C$ is the fitted one, 64,300 m³/MPa
Transmissibilities: fitted: $T_{AB} = 38.5$ and $T_{BC} = 13.8$ m³/(d·MPa)

Solution

$s_1 = 38.5/84{,}000 + 52.3/52{,}500 + 13.8/64{,}300 = 1.6691\cdot 10^{-3}$ 1/d.
$s_2 = 38.5\cdot 13.8\cdot 200{,}800/(84{,}000\cdot 52{,}500\cdot 64{,}300) = 3.762\cdot 10^{-7}$ 1/d².
Roots: $\lambda = \big(s_1 \pm \sqrt{s_1^2 - 4s_2}\big)/2$ . $\sqrt{s_1^2 - 4s_2} = 1.1319\cdot 10^{-3}$ , $\lambda_1 = 1.4005\cdot 10^{-3}$ and $\lambda_2 = 2.686\cdot 10^{-4}$ 1/d.
Time constants: $1/\lambda_1 = 714$ d and $1/\lambda_2 = 3723$ d.

Answer: About 2.0 and 10 years. The pair A–B equalises in two years, and block C catches up with it in ten.

What it means. The three-year history covered the short time in full and only a third of the long one. This is why the transmissibility of the fault A–B was found three times more accurately than that of B–C, and the storage capacity of block C is known only roughly. Readings in block C itself are not the only way to narrow the range. In the model of the example, two readings in block B, 165 and 365 days after the last ones, reduce the error of $T_{BC}$ from 23 to 13%. The same readings in block C hardly change it (23%). The crossflow from C affects the pressure of block B most, while the pressure of C itself changes little. The model shows which readings are more useful even before they are planned. The data in this example are assumed for illustration.

The same matrix gives a practical rule for the calculation. An explicit time step (one that uses only the values from the previous step) is limited by the shortest of the times of the system. In the fitted model of three blocks it is 714 d, so an explicit step of one day is stable with a margin. But a small tank behind a strong link has a time of a day or less. The more the times differ, the stiffer the system is (the harder it is to calculate step by step). So multi-tank models like these, as grid models do, are calculated implicitly (each step solves for all the new pressures together) or exactly, through the eigenvalues (why pressure is calculated implicitly).

Below the bubble point pressure, and gas tanks

Below the bubble point pressure we write the balance of each tank in the full form of the lesson General material balance equation for oil reservoirs. The withdrawal $F_j$ equals $N_j(E_o + E_{fw})$ plus the inflow from the neighbours. The crossflow through a link carries oil and the gas that has come out of solution, in shares set by the mobilities of the phases at the link. The storage capacity of the tank is no longer constant, and it grows as gas is released. We solve the equations in time steps, with iterations at each step.

For gas tanks the $p/Z$ form is convenient. The gas of tank $j$ , after the withdrawal $G_{p,j}$ and the crossflows, occupies the same pore volume, and $(p/Z)_j = (p/Z)_i\,\big(1 - (G_{p,j} + \text{outflow}_j)/G_j\big)$ (closed gas accumulation). The crossflow between gas tanks is roughly proportional to the difference of the squares of the pressures, as long as the product of the gas viscosity and the gas deviation factor changes little. As before, we fit the gas in place of the tanks and the transmissibilities of the links to the block pressures in time.

Connections to other topics

General material balance equation for oil reservoirs — the balance of a single tank, from which the multi-tank model starts.
Material balance for gas reservoirs — the p/Z plot, and how far blocks show up on it.
Pressure measurements in producing reservoirs: barriers and accuracy — two units with crossflow, the time constant of a barrier, and faults from repeat surveys.
Partially communicating faults — fault transmissibility from a well test, and carrying it into the model.
Reservoir architecture and performance — isolated blocks and their equalisation time.
Pseudo-steady-state flow and the shape factor — depletion of a closed area, and the time to reach the regime.
History matching — the objective function, non-uniqueness and testing a match.
Simplified water influx models: Fetkovich and Carter–Tracy — an aquifer as one more tank behind a resistance.
Pressure equation for single-phase flow — the same balance equation for the cells of a grid model.
Multiple-zone completions: commingled and segregated production — the rates of the formations in a commingled completion, and crossflow through the wellbore.
Perched and tilted contacts, trapped water — closed volumes of water, and blocks linked by windows in faults.

Check yourself

1
Three blocks are connected by faults, and production comes only from the first one. How does the pressure averaged by storage capacity over all three blocks change?
1. A.It does not change until the crossflows settle
2. B.In the same way as in a single tank with their total storage capacity
3. C.In the same way as the pressure of the first block
4. D.More slowly, because crossflows feed the first block
Show the answer
Correct answer: B. In the same way as in a single tank with their total storage capacity
Each crossflow leaves one tank and enters another, so the crossflows cancel in the sum of the balances: $\sum C_j\,\dd p_j/\dd t = -q$ . The pressure averaged by storage capacity falls as in a single tank. The crossflows only share the decline out between the blocks.
2
A block with a pore volume of 2 million m³ ( $c_t = 2\cdot10^{-3}$ MPa⁻¹) is connected to a large supported block by a fault with $T = 5$ m³/(d·MPa). The well in it has $J = 30$ m³/(d·MPa), and its bottomhole pressure is 10 MPa below the pressure of the neighbouring block. Find the initial and limiting rates and the time constant.
Show the answer
$C = 0.002\cdot 2\cdot10^6 = 4000$ m³/MPa. $q_0 = 30\cdot 10 = 300$ m³/d. $q_\infty = 30\cdot 5/35\cdot 10 \approx 42.9$ m³/d, limited by the fault. $\tau = 4000/35 \approx 114$ d, so within a year the rate almost reaches the limit. After a shut-in the block recovers with the constant $4000/5 = 800$ d.
3
The first well in a block behind a fault, two years after production began in the neighbouring block, showed the initial pressure, although the neighbouring block is depleted by 3 MPa. What does this mean?
1. A.The gauge was wrong by 3 MPa
2. B.The blocks communicate well
3. C.Over a time scale of years the fault hardly passes any fluid (or another source that we have not counted feeds the block)
4. D.The block pressures equalised over geological time, so the connection is good
Show the answer
Correct answer: C. Over a time scale of years the fault hardly passes any fluid (or another source that we have not counted feeds the block)
With a good connection the pressure behind the fault would have fallen almost as much as in the neighbouring block. A ratio of declines close to zero means, by the formula for the first well, a very small transmissibility: the time constant is much longer than two years. Another explanation is strong support of the block by an aquifer. We test this with the geology and with the behaviour of the pressure later. The equality of the initial pressures tells us nothing about communication on the scale of production.
4
Why is it hard to determine separately, from the block pressures, the storage capacity of a far block and the transmissibility of the fault leading to it, and what helps to separate them?
Show the answer
As long as the history is shorter than the long time constant, the pressures of the neighbouring block respond mainly to their combination: a larger storage capacity can partly be replaced by a smaller transmissibility (in the lesson's example the correlation coefficient of the estimates is about 0.8). Several things help to separate them: transient periods after start-ups, shut-ins and the beginning of injection; a longer history; and an independent estimate of the volume of the far block from the maps. New readings are also not necessarily most useful in the far block itself. In the lesson's example two readings in its neighbour, block B, improve the transmissibility of the fault B–C almost twofold, while the same readings in C hardly improve it.
5
The producing wells stand in block A. Block B is behind a fault, and block C is behind a second fault. Where will injection of the same volume support the pressure in block A most?
1. A.Into block A itself
2. B.Into block B
3. C.Into block C
4. D.It makes no difference: only the total injection volume matters
Show the answer
Correct answer: A. Into block A itself
The more faults there are between the injection well and the producing wells, the larger the part of the injection that is “taken” by the pressure rise on its own side of each fault. In the lesson's example injection into C delayed the fall of the pressure in A to the bubble point pressure by a few days, injection into B by more than half a year, and injection into A kept it above the bubble point pressure for the whole forecast period. The volume matters for the average pressure, and for a block the place matters.
6
The apparent storage capacity, calculated from the pressure of the working block as for a single tank, grows from reading to reading. Name two possible causes and say how to tell them apart.
Show the answer
The first cause is that neighbouring volumes join the working block across the faults. The second is water influx from the aquifer. We tell them apart by pressure readings in the neighbouring blocks (has the pressure fallen there?), by the geology (is there reservoir rock behind the faults, and how large a volume?), by the appearance of water in the wells, and by the character of the growth. The neighbouring blocks are limited, and the apparent storage capacity in the end levels off at their sum.

Updated: 28 October 2026