mine balance

APPLICATION OF MATERIAL BALANCE

PRESENTED BY:NORHANISAH BINTI ALIAS

PRESENTATION PRESENTATION OUTLINEOUTLINE IntroductionElements of material balanceComponents of Material balance equationData requirementsData gatheringAnalysis and ResultsConclusions

INTRODUCTIONINTRODUCTION

• The material balance is the total of fluid produce in the reservoir to the difference between the amount of fluid present at the initial reservoir and amount of fluid remaining at the final reservoir.

• The mass that cannot be destroyed or created. • Application areas of material balance:

- hydrocarbon initially in place (HIIP)

- reservoir drive mechanisms identification

- production prediction

- reserve estimates

- Development planning

ELEMENTS OF MATERIAL BALANCEELEMENTS OF MATERIAL BALANCE

The drive mechanisms are consists of: Solution gas drive

◦ Associated gas

Gas cap drive◦ Saturated

Natural water drive◦ Aquifer support-system

Gravity drainage drive Compaction drive

◦ High formation compressibility

ELEMENTS OF MATERIAL BALANCEELEMENTS OF MATERIAL BALANCEPropertiesProperties

Rock properties Porosity Permeability Formation

compressibility Water saturation

Fluid properties Water viscosity Oil viscosity Solution gas, Rs Water compressibility Gas FVF Oil FVF Water FVF Oil compressibility Gas gravity Oil gravity

Components of Material Balance Components of Material Balance EquationEquation

ΔP

Gas cap

Oil + originally dissolved gas

Volumes at initial pressure Volumes at the reduced pressure

AC

B

Underground withdrawal = expansion of oil + originally dissolved gas (A)+ expansion of gas cap (B)+ reduction in HCPV due to connate water expansion and decrease in the pore volume (C)+ cumulative water infux (D)

D

Np (Bo + (Rp - Rs) Bg ) + WpBw = N[(Bo – Boi) + (Rsi – Rs)Bg] + mNBoi(Bg / Bgi – 1) + (1+m) NBoil[(cwSwc+ cf)/(1-Swc)]

ΔP + WeBw

MBE in Linear FormMBE in Linear Form

F = Np (Bo + (Rp - Rs) Bg ) + WpBw (rb)

Eo = (Bo – Boi) + (Rsi – Rs)Bg (rb/stb)

Eg = Boi(Bg / Bgi – 1) (rb/stb)

Ef,w = (1+m) Boi([cwSwc+ cf])/[1-Swc] ΔP(rb/stb)

F = N (Eo + mEg + Ef,w ) + WeBw

Material Balance Equation VariablesMaterial Balance Equation Variables

Production data

Np = cumulative oil produced (stb)

Gp = cumulative gas produced (scf)

Wp = cumulative water produced (stb)

Rp = Gp/Np (scf/stb)

= cumulative produced gas oil ratio

Reservoir data

Pi = initial pressure in the reservoir (psi)

P = current pressure in the reservoir (psi)

Swc = connate water saturation (fraction)

Cf = formation compressibility (1/psi)

Material Balance Equation VariablesMaterial Balance Equation Variables

Fluid PVT data

Bgi = initial gas volume factor (ft³/scf)

Bg = gas volume factor (ft³/scf)

Boi = initial oil volume factor (rb/stb)

Bo = oil volume factor (rb/stb)

Cw = compressibility of water (1/psi)

Bw = formation volume factor of water (rb/stb)

Rsi = solution gas oil ratio at initial pressure(scf/stb)

Rs = solution gas oil ratio at current pressure (scf/stb)

OILOIL RESERVOIRRESERVOIR

There are 2 methods to solve the material balance:Analytical methodGraphical method

- Havlena-Odeh

- F/Et versus We/Et

- (F-We)/Et versus F known as Campbell

- F-We versus Et

- (F-We)/(Eo+Efw) versus Eg/(Eo+Efw)

Analytical InterpretationAnalytical Interpretation

GRAPHICAL INTERPRETATION (OIL)GRAPHICAL INTERPRETATION (OIL)

Havlena-Odeh Campbell graph

N can be obtained when the straight line touch y-axis


F/Et versus We/Et F-We versus Et


(F-We)/(Eo+Efw) versus Eg/(Eo+Efw)

Drive MechanismDrive Mechanism

Production PredictionProduction Prediction

Number of wells: 8The estimate reserve: 96.9961 MMstb

Material Balance Equation for Gas Material Balance Equation for Gas ReservoirReservoir

For Gas Reservoir,

G(Bg-Bgi) + We = GpBg +BwWp

For volumetric gas reservoir, We = 0

Assuming water production is negligible, Wp = 0

The above equation can becomes:

G(Bg-Bgi) = GpBg

Gas Reservoir without aquiferGas Reservoir without aquifer

By using the equation of real gas and substituting expression for Bg and Bgi into the equation, the following is obtained:

G(Z/P) – G(Zi/Pi) = Gp(Z/P)

Rearranging it, the equation can become

P/Z = Pi/Zi (1 – Gp/G)

Graphical Interpretation (Gas)Graphical Interpretation (Gas)

P/Z method:

For volumetric depletion

P/Z = (Pi/Zi)*(1-Gp/G)

P/Z = 0, Gp = G

Gas Reservoir With AquiferGas Reservoir With Aquifer

The relation between Gp and p/z is not linear.

Rearranging the equation, it can be obtained

GBgi = (G – Gp)Bg + We - BwWp

When the water drive, We ≠ 0

And Efw = negligible

The equation for gas reservoir with water influx,

(GpBg+WpBw)/(Bg-Bgi) = G + We/(Bg-Bgi)


Graph without water influx


Van Everdingen and Hurst Carter-Tracy


Fetkovich semi steady state

CONCLUSIONCONCLUSION

Material balance equation is a tool that helps determine the reserves, recovery factor and drive mechanism.

Can be applied to a variety of reservoirs, either with or without water influx.

THANK YOU..THANK YOU..

mine balance

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