2- material balance equation
DESCRIPTION
Rumus Material BalanceTRANSCRIPT
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Material Balance Equation
For Oil Reservoirs
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Oil Material Balance Equation
we
wc
fwcw
gi
g
oi
gssioiooiiwpgspop
BW
pS
CSCm
B
Bm
B
BRRBBBNBWBRRBN
111
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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
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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)
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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.
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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
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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)
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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
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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
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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
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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 )
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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.
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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)
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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
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Solution
Solution Gas drive reservoir implies:Negligible water influx.No initial gas cap.
gssioioi
wpgspop
BRRBBN
BWBRRBN
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Solution
Calculate Bg
bbl/SCF10625.7
000,2
4601508.005.
005.
4
p
zTBg
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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
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Solution
Solve for initial oil in place
gssioio
wpgspopi BRRBB
BWBRRBNN
MMSTB2961106.
10279.3 7
iN
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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
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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 ,
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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.
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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 ,
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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
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Solution Gas Drive
F
E o(0 ,0 )
Slope = Ni
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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
,
,
,
,,
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Gas Cap drive Plot
(0 ,0 )
wfo EE
F
,
wfo
wfg
EE
EE
,
,
} iN
S lope = m N i
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Material Balance Equations
For Gas Reservoirs
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Problem
Suppose we had a tank of gas buried underground Fixed known
temperature Pressure known Tank Volume
unknown
Gasp i
TG i
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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
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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
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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
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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
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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
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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
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p/z versus Gp
pG
zp
i
iz
p
iG0
0
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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
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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
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p/z versus Gp with water influx
pG
zp
i
iz
p
iG0
0
E ffec t o f wate r inf lux
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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.
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Gas Material Balance
General form
Gas formation volume factor
wewc
fwcwgigigiwpgp BWp
S
cScBBBGBWBG
1
/scfft 0283.0 3
p
zTBg
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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
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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.
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P/z for Abnormally Pressured Reservoirs