Technology ServicesReservoir & Well Performance

Material Balance:The Forgotten

ReservoirEngineering Tool

John McMullan

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Understanding the OilAnd Gas ReservoirUsing Material BalanceLafayette, LouisianaOctober 21, 2004

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http://www.cgrpttc.lsu.edu/products/matbal/

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Possible Talk Titles

Material Balance: The Forgotten Reservoir Engineering Tool

Are Traditional Material Balance Calculations Obsolete?

Material Balance: Obsolete in 2005?

Material Balance: A Quaint Reservoir Engineering Tool from the Past

Material Balance, Why Bother?

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Material Balance

Can Provide an estimate of initial HC in place– independent of geological interpretation

– can be used to verify volumetric estimates

Determines the degree of aquifer influence– understanding of the “drive mechanism”

– estimate recovery factor

Estimate of recoverable reserves

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Uses of Material Balance

As a precursor to reservoir simulation

Identify undrained hydrocarbons

Can be used as a forecasting tool in certain situations

Can be used to help evaluate operating strategies such as new wells, accelerated rate, compression

In some cases can be used to screen for enhanced recovery

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Gas Material Balance

G ≡ initial gasvolume - SCF

Bgi – bbl/SCF Vgi = G × Bgi

Vgi ≡ initial gasvolume - bbls

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Gas Material Balance

G × Bgi

Bg is a functionof new pressure

Expansion (bbls)

= Vg - Vgi

= G (Bg – Bgi)}Vg = G × Bg

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Gas Material Balance

Expansion (bbls)

= Vg - Vgi

= G (Bg – Bgi)}

Expansion mustequal production:

Gp Bg = G (Bg – Bgi)

Fix piston

Bleed off Gp SCFof gas until

pressure equalsthe same as before.

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Gas Material Balance

Vti = Vgi + Vwi (bbls)

Vgi = Vti (1-Sw) = G Bgi

Vti = G Bgi / (1-Sw)

Vwi = Vti Sw = G Bgi Sw / (1-Sw)

Vgi = G × Bgi

Vwi = G Bgi Sw / (1-Sw)

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Gas Material Balance

The change in water volume can be found:

∆Vwi = Vwi cw ∆p

Since

Vwi = Vti Sw = G Bgi Sw / (1-Sw)

Substituting:

∆Vwi = G Bgi cw ∆p Sw / (1-Sw)

(the expansion of water with a drop in pressure)

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Gas Material Balance

G × BgiExpansion (bbls)

= G (Bg – Bgi) +

G Bgi cw ∆p Sw / (1-Sw)

}Vwi =

G Bgi Sw / (1-Sw)

Vg = G × Bg

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Gas Material Balance

As before,

Expansion must equal production:

G (Bg – Bgi) + G Bgi cw ∆p Sw / (1-Sw)

= Gp Bg + WpBwFix piston

Bleed off Gp SCFof gas, Wp water until

pressure equalsthe same as before.

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Expansion (bbls)

= G (Bg – Bgi)

+

G Bgi cw ∆p Sw / (1-Sw) + We

}

Suppose while the pressure drops, weinject We reservoir barrels of water.

Expansion must equal production:

G (Bg – Bgi) + G Bgi cw ∆p Sw / (1-Sw) + We

= Gp Bg + WpBw

Gas Material Balance

We

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Gas Material BalanceFinally, consider the possibility that the actual

initial pore volume will reduce as the pressure falls:

∆Vti = Vti cf ∆p

Recall,

Vti = G Bgi / (1-Sw)

Substituting:

∆Vti = cf ∆p G Bgi / (1-Sw)

This loss in original volume results inan additional amount of expansion from

the original volume.

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Gas General MaterialBalance Equation

G (Bg – Bgi) + G Bgi cw ∆p Sw / (1-Sw) + We + cf ∆p G Bgi / (1-Sw)

= Gp Bg + WpBw

GasProduction

WaterProduction+=

GasExpansion

WaterExpansion

WaterInflux

FormationExpansion+++

Note that all terms are a function of pressure

Equation can not be directly solved

An iterative approach is required for solution

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Oil General MaterialBalance Equation

N (Bt – Bti) + N m Bti (Bg - Bgi ) + (N Bti + N m Bti ) cw ∆p Sw / (1-Sw)

+ cf ∆p (N Bti + N m Bti ) / (1-Sw) + We + WI BwI + GI BgI

= Np Bt + Np (Rp – Rsoi) Bg + WpBw

Bgi

OilExpansion

WaterExpansion+ +Gas Cap

Expansion

WaterInflux

FormationExpansion+ + Water

Injection+ GasInjection+

Free GasProduction

WaterProduction++Oil & Dissolved

Gas Production=

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Gas Material Balance as aStraight Line

G (Bg – Bgi) + Bgi cw Sw + cf ∆p = Gp Bg + WpBw - We

1-Sw

⎡⎢⎣

⎤⎥⎦⎩

⎧⎨

⎫⎬⎭

Xg ≡ (Bg – Bgi) + Bgi cw Sw + cf ∆p1-Sw

⎡⎢⎣

⎤⎥⎦

Yg ≡ Gp Bg + WpBw - We

Yg = G Xg

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Yg - rb

Xg – rb/SCF

m = G

Yg - rb

Xg – rb/SCF

m = G

Gas Material Balance as aStraight Line

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Water Influx

Aquifers come in all shapes and sizes– Aquifers can be extremely large relative to the reservoir

size, even infinite “acting”.

– Aquifers can be small, even neglected.

– Aquifer productivity can be either high or low (relative to the withdrawal rates from the reservoir).

Aquifers can be hydraulically connected to more than one reservoir.

Aquifers can even be connected to the surface.

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Steady-State and Semisteady-state Aquifer Models

dWedt = qaq = k’ × (Paq – P)

qaq = instantaneous aquifer flow rate (rb/day)

k’ = aquifer influx constant (rb/day/psi)

Paq = average aquifer pressure (psi)

P = average reservoir pressure (psi)

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Steady-State and Semisteady-state Aquifer Models

k’ is similar to the “Productivity Index” often usedto describe an individual well’s productivity.

Using a similar definition:

qaq = Jaq × (Paq – Pr)

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Steady-State and Semisteady-state Aquifer Models

If the pore volume (Vaq), compressibilities, and averageaquifer pressure are known, the total water influx atany point in time can be estimated by:

We = Vaq × cavg × (Pi – Paq)

Recall: Cavg = SoCo + SwCw + SgSg + Cf

So for aquifers, Cavg = Cw + Cf

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Steady-State and Semisteady-state Aquifer Models

Define a term Wei that we will refer to as the“maximum encroachable water”:

Wei = Vaq × cavg × (Pi – Paq)

0

Wei = Vaq × cavg × Pi

In words, this is the volume of water that will flowfrom an aquifer if it’s pressure is lowered to zero.

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Steady-State and Semisteady-state Aquifer Models

The average aquifer pressure at any point in timecan then be estimated by:

Paq = Pi × (1 – We / Wei)

qaq = Jaq × (Paq – Pr)

This equation, along with the previously shown equation below, form the basis for steady-state and semisteady-

state aquifer models.

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Steady-State and Semisteady-state Aquifer Models

“Pot” aquifer– defined as an aquifer where the aquifer and reservoir pressure

remain (nearly) equal as the reservoir depletes

– this implies a small aquifer with high productivity

“Schilthuis” steady-state aquifer– aquifer is extremely large and consequently, the aquifer pressure

can be assumed to remain constant

“Fetkovitch” semisteady-state aquifer– aquifer rate and pressure are assumed to both change with time as

described by the previous equations

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Pot AquiferDefinition: P = Paq

Recall:Paq = Pi × (1 – We / Wei)

Then:

P = Pi × (1 – We / Wei)

Solving for We yields:

We = Wei × (1 – P / Pi)

We can be directly substituted into any ofthe material balance equations.

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Steady-state Aquifer

Definition: Paq = Pi

Recall:

qaq = Jaq × (Paq – P)

Then:

qaq = Jaq × (Pi – P)

(∆We )n = Jaq × Pi –⎡⎣ ⎦

⎤Pn + Pn-1

2× ∆t

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Fetkovich AnalyticalAquifer

Recall “encroachable” water was defined by:

W ei = V aq × P i × c aq

AquiferPore

Volume

InitialPressure

AquiferCompressibility

Aquifer pressure at any point in time is given by:

P aq = P i × (1 - W e / W ei )

Cumulative Water Influx

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Fetkovich AnalyticalAquifer

Finally, the influx rate at any point in time is given by:

q aq = J aq × ( P aq - P res ) =

AquiferProductivity

Index

ReservoirPressure

dWe

dt

Note: A large J and small W ei models a pot aquifer. A large W ei models an infinite aquifer.

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Aquifer Productivity Index

Source: Applied Petroleum Reservoir EngineeringCraft, Hawkins, and Terry

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Fetkovich AnalyticalAquifer

Through algebraic manipulation and integration, the waterinflux for a constant drop in pressure for a time “t” becomes:

We = ( pi – p ) (1 - e-J pi t / Wei)Weipi

Fetkovich showed that this equation can be applied in adifference form without the need for superposition tomodel a system with a continuously falling pressure:

∆Wen = ( pn-1 – pRn ) (1 - e -J pi ∆tn / Wei)Weipi

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Aquifer Boundary Pressure

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Some CommentsRegarding Aquifer Models

Steady-state and semisteady-state models are often thought to be less accurate than unsteady-state models (like the often used Hurst and van Everdingen model)

None of the analytical aquifer models directly consider the growing water invaded zone and its impact on aquifer productivity or MB (see SPE papers by Al-Hashim & Bass and Lutes et al)

Truly “rigorous” treatment of aquifer influx requires reservoir simulation

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Material Balance CalculationsTwo Approaches

“Traditional” XY Plot– Uses observed pressures and production in GMBE and

aquifer model to calculate N or G

– Some iteration may be required to estimate gas cap volume, aquifer properties, etc.

– Sparse pressure data, erratic production rates a problem

Alternate method– Uses observed production, aquifer model, and assumed

values of N or G in GMBE to calculate pressure

– Iterate until calculated and observed pressures agree

– Excellent for investigating sensitivities

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Gas Example No. 1

Moderate Water Drive

Project Based on 60-100 BCF Volumetric Estimate

Unconstrained MB Analysis Suggested Reservoir Was Depleted

Subsequent Well Was Dry Hole

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Gas Example No. 2

Weak Water Drive

P/Z Suggested 860 BCF

Downdip Water Production Suggested Limited Water Influx

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Gas Example No. 2 (cont.)

Material Balance 725 BCF

HC Pore Volume 450 MMBbls

Water Influx 100 MMBbls

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Gas Example No. 3

Strong Water Drive

“Classic” Rate Sensitive Reservoir

Used in Field to Establish Production Priority

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Gas Example No. 4

Example of “Pot” Aquifer

Observe the Sensitivity to Formation Compressibility (49.7 to 58.5 BCF)

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Oil Example No. 1

Initially Undersaturated, Moderate Water Drive

Observe the Extreme Sensitivity to Formation Compressibility (11 to 30 MMBO)

Note Culled Points

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Oil Example No. 2

Initially Undersaturated, Weak Water Drive

MB Analysis Reveals Pressure Behavior That Could Not Be Matched

Subsequent Simulation Study Was a Failure

Anomalous Behavior Was Later Determined to be the Result of Several Casing Leaks

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Oil Example No. 3

Small Offshore, Above BP Pressure

N = 35 MMBO from “XY” Plot

N of 66.5 MMBO Agrees with Volumetric Estimate of 65 MMBO

Dominated by Aquifer InfluxFrom Dake, Exercise 3.4The Practice of Reservoir Engineering

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Oil Example No. 4

Venezuela, Water-drive, Water Injection, Gas-cap Expansion, Solution Gas Drive

N = 27 MMBO with “XY” Plot and Fixed Gas Cap Size

N = 34 MMBO Allowing Gas Cap Size to Vary

Note Culled Points

From Havelana and Odeh, JPT, July 1964

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Material Balance Compared To Reservoir Simulation

Reservoir Simulation– expensive– time consuming– requires geologic

description– driven with single

phase– ability to forecast– determines location

and distribution of unswept HC’s

Material Balance– cheap– fast– independent of

geology– uses production of all

phases in calculations– limited forecasting– can determine the existence of unswept HC’s

Technology ServicesReservoir & Well Performance

Material Balance:The Forgotten

ReservoirEngineering Tool

John McMullan