Material Balance Equation

For Oil Reservoirs

Oil Material Balance Equation

we

wc

fwcw

gi

g

oi

gssioiooiiwpgspop

BW

pS

CSCm

B

Bm

B

BRRBBBNBWBRRBN

111

Nomenclature

Np = cumulative oil produced at reservoir pressure p, STB

Ni = initial oil in place, STB

m = oii

gii

BN

BG

Rs = ratio of gas in place to oil in place under standard conditions,

SCF/STB Gi = initial gas in place, SCF

We = cumulative water influx, STB

Wp = cumulative water produced, STB

Rp = cumulative producing gas oil ratio, SCF/STB = Gp/Np

Gp = cumulative gas produced, SCF

Nomenclature

p = (pi-p), psia, where pi is the initial pressure

Rs = solution gas ratio, SCF/STB

Bo = oil formation volume factor, bbl/STB

Bg = gas formation volume factor, bbl/SCF = .005z T

p

Bw = water formation volume factor, bbl/STB

Cw = compressibility of water, psi-1

Cf = compressibility of formation, psi-1

Swc = connate water saturation, fraction

subscript i = initial conditions (For example, Boi = initial formation volume factor, bbl/STB)

Oil Material Balance Equation

In words, the oil material balance equation says that any production we obtain is due to change in volume of reservoir rock and fluids and/or displacement by encroaching water from an adjoining aquifer.

Left Hand Side

Production from the reservoir at reservoir conditions We measure these terms

NpBo – Oil produced at reservoir conditions, (res. bbl) Note that Bo includes changes in oil volume due to

gas going into solution.

wpgspop BWBRRBN

Left hand Side

Np(Rp Rs)Bg – Amount of free gas produced at reservoir conditions.Note – Rp is total produced gas (free +

dissolved) per barrel; Rp is gas dissolved at reservoir conditions that is produced.

The effect of oil volume changes due to solution gas, Rs is included in Bo

WpBw – Amount of water produced (or injected)

Right Hand Side

Accounts for expansion of oil and free gas in the reservoir, as well as influx of water and change in reservoir pore volume.

gssioioi BRRBBN

Expansion of oil + dissolved gas

Right Hand Side

Expansion of gas in the gas-cap

Expansion of rock and connate water

Natural Water influx

giigigi

goii BGBG

B

BmBN

1

pS

CSCBGBNp

S

CSCmBN

wc

fwcwgiioii

wc

fwcwoii

111

weBW

Initial Reservoir Conditions

R eservo ir C o nditio ns

O il

G as

N i S T B

N iR si scf

G i scf

S urface C o nditio ns

{• Original Reservoir Oil = NiBoi res bbl.

• Original Reservoir Gas = GiBgi res bbl. = m NiBoi res bbl

Initial Reservoir Conditions

Reservoir Pore Volume = Vp bbl Pore Volume occupied by water = VpSwc bbl Hydrocarbon Pore Volume (HCPV) = Vp (1 – Swc) bbl But HCPV = NiBoi + m NiBoi = (1+m) NiBoi bbl So Vp = (1+m) NiBoi / (1 – Swc) bbl Free gas volume = mNiBoi bbl So initial gas saturation = mNiBoi/ Vp

= mNiBoi (1 – Swc) /((1+m) NiBoi )

Initial oil saturation = NiBoi/ Vp

= NiBoi (1 – Swc) /((1+m) NiBoi )

After Production of Oil and Gas

R eservo ir C o nditio ns

O il

G as

N i S T B

N iR si scf

G i scf

S urface C o nditio ns

N p S T B

N pR p scf

(N i - N p)B o bblco ntaining

(N i - N p)R s scf gas

(G i+N iR si - N pR p

-(N i - N p)R s)B g

bbl gas

• Free Gas in Reservoir = (mNiBoi/Bgi+NiRsi - NpRp

- (Ni – Np)Rs)Bg bbl.

After Production of Oil and Gas

Gas Saturation = Free Gas Volume Pore Volume (mNiBoi/Bgi+NiRsi - NpRp - (Ni – Np)Rs)Bg [(1+m)

NiBoi / (1 – Swc) ]

Oil Saturation = Oil Volume Pore Volume (Ni – Np)Bo (1+m) NiBoi / (1 – Swc)

Example

For a solution gas drive reservoir, calculate the

original oil in place if the following information is

given. Assume water and rock compressibility

are negligible.

p = 2,000 psia Bo = 1.22 bbl/STB

Rs = 350 SCF/STB

z = 0.80 Rsi = 600 SCF/STB

Boi = 1.3 bbl/STB

Np = 20.0 MMSTB

T = 150o F Rp = 900 SCF/STB

Solution

Solution Gas drive reservoir implies:Negligible water influx.No initial gas cap.

gssioioi

wpgspop

BRRBBN

BWBRRBN

Solution

Calculate Bg

bbl/SCF10625.7

000,2

4601508.005.

005.

4

p

zTBg

Solution

Calculate reservoir production

Calculate PVT dependent terms on RHS

bbl res. 10279.3

010625.735090022.110207

46

wpgspop BWBRRBN

bbl/STB res. 1106.

10625.73506003.122.1 4

gssioio BRRBB

Solution

Solve for initial oil in place

gssioio

wpgspopi BRRBB

BWBRRBNN

MMSTB2961106.

10279.3 7

iN

Condensed Notation

The material balance equation is lengthy; cumbersome to work with.

Introduce shorthand notation to facilitate manipulation.

Total Production of oil, water and gas:

wpgspop BWBRRBNF

Condensed Notation

Reservoir fluid expansion terms (on a per STB basis)Expansion of oil and dissolved gas

Expansion of gas-cap gas

Expansion of rock and connate water

gssioioo BRRBBE

1

gi

goig B

BmBmE

pS

cScBmEm

wc

fwcwoiwf

111 ,

Note

The fluid expansion terms Eo, Eg, and Ef,w are composed only of fluid PVT properties and connate water saturation.At reservoir conditions (Temperature

constant), they are functions of reservoir pressure only.

Material Balance Equation

In terms of our shorthand notation

Simplifications:Solution Gas Drive Reservoir: We = 0, m = 0

F Ni Eo mEg 1 m E f , w WeBw

wfoi EENF ,

Solution Gas Drive

Production (F) is measured.Eo and Ef,w are determined from PVT and

rock properties.Material balance is a straight line

equationPlot of F versus Eo + Ef,w is a straight line

with intercept 0 and slope Ni

Solution Gas Drive

F

E o(0 ,0 )

Slope = Ni

Gas Cap Drive – No Water Influx

In this case, the material balance equation can be simplified to

Straight line form:

wfgoi EmmEENF ,1

wfo

wfgii

wfo

wfgwfoi

EE

EEmNN

EE

F

EEmEENF

,

,

,

,,

Gas Cap drive Plot

(0 ,0 )

wfo EE

F

,

wfo

wfg

EE

EE

,

,

} iN

S lope = m N i

Material Balance Equations

For Gas Reservoirs

Problem

Suppose we had a tank of gas buried underground Fixed known

temperature Pressure known Tank Volume

unknown

Gasp i

TG i

Problem (Cont’d)

Suppose we remove Gp scf gas Pressure falls to a

new measured value

Temperature constant

Can we determine the original scf of gas in the tank?

GaspT

G i - G p

G p

Solution

From real gas law:

Original number of moles of gas in the tank, ni

Solving for original volume of the tank

znRTpV

R

G

R

G

RTz

Vpn ii

i

ii 0283.0

520

7.14

giii

ii BGp

TzGV 0283.0

Solution (Cont’d)

Number of moles removed from tank

Number of moles left in the tank

R

Gn p

r 0283.0

R

GGn pi 0283.0left

Solution (Cont’d)

Gas left occupies the entire tank volume, so

Tank volume

RT

R

GGzpV

RTznpV

pi

0283.0

left

pigpi GGBGGp

zTV 0283.0

Gas Material Balance – Volumetric Reservoir

We have two expressions for tank volumeMust be equal

i

ii

g

giip

igipig

zp

zp

GB

BGG

GBGGBV

11

Straight line plot

For a volumetric gas reservoir, a plot of Gp versus p/z will be a straight line of slope –(Gi/(pi/zi)) and intercept Gi

In practice, people plot p/z versus Gp and extrapolate to p/z = 0

p/z versus Gp

pG

zp

i

iz

p

iG0

0

Problem

Suppose when we remove Gp scf gas, WeBw res. bbl of water encroached Pressure falls to a new

measured value Temperature constant

Can we determine the original scf of gas in the tank?

GaspT

G i - G p

G p

W eB w

Solution

Original volume of gas in tank

Final Volume of gas in tank

Material Balance with water influx

gii BGV

pigpiwe GGBGGp

zTBWV 0283.0

pigwegii GGBBWBG

p/z versus Gp with water influx

pG

zp

i

iz

p

iG0

0

E ffec t o f wate r inf lux

Production of Gas

Processes that determine gas production: Expansion of gas Water Influx Expansion of rock and connate water

In most cases expansion of rock and connate water is small compared to gas expansion.

In “abnormally” pressured gas reservoirs, this term may be significant.

Gas Material Balance

General form

Gas formation volume factor

wewc

fwcwgigigiwpgp BWp

S

cScBBBGBWBG

1

/scfft 0283.0 3

p

zTBg

Gas Material Balance - p/z

It is customary to express the gas material balance in terms of p/z

If there is no water influx and formation and rock compressibility are negligible

Plot of p/z vs. Gp is a straight line Intercepts the x-axis at Gi

piii

i GGzG

p

z

p

Abnormally Pressured Reservoirs

Normal pressure gradients for gas reservoirs are in the range of 0.4-0.5 psia/ft of depth

Abnormally pressured reservoirs have gradients of 0.7-1.0 psia/ft of depth > 300 abnormally pressured gas reservoirs

offshore Gulf Coast; gradients > 0.65 with depths over 10000 ft.

P/z for Abnormally Pressured Reservoirs