500 gas material balance basic concepts g=10 bscfpetro/faculty/kelly/grm-chap1-matbal.pdf · gas...

Gas Material Balance Basic Concepts

• Estimate OGIP

• Predict recovery

• Identify drive

mechanism

GRM-Engler-09

0

500

1000

1500

2000

2500

3000

3500

4000

0 1 2 3 4 5 6 7 8 9 10 11 12

Cumulative gas produced,Bscf

p/z

, p

sia

(p/z)i

(p/z)a

Gpa=8.5 BscfG=10 Bscf

Last measured

data point

extrapolate

Gas Material Balance Basic Concepts

GRM-Engler-09

0

500

1000

1500

2000

2500

3000

3500

4000

0 1 2 3 4 5 6 7 8 9 10 11 12

Cumulative gas produced,Bscf

p/z

, p

sia

(p/z)i

(p/z)a

Gpa=8.5 BscfG=10 Bscf

Last measured

data point

extrapolate

G

pG1

izip

z

p

Gas property, f(T,P,g)

Measured

pressure Cumulative gas

Production @ P

ipiz

az

ap1

gaB

giB1volRF

Gas Material Balance Advanced Topics

GRM-Engler-09

• Nonlinear gas material balance

Water drive gas reservoirs

(additional pressure support)

Abnormally pressured gas reservoirs

(rock compressibility)

Low permeability gas reservoirs (measured pressures don’t achieve ave. press.)

Gas Material Balance Advanced Topics

GRM-Engler-09

• Comprehensive gas material balance

eWwBinjWwBpWgB

615.5swRpWinjGpG

G

iz/p

iz

p

)pip()p(ec1z

p

Geopressured

component

Water drive

component Gas injection

Gas in solution

p/z

Gp/G

(p/z)i

(p/z)a

Depletion drive

water drive

strength

(p/z)a

0 100 50

Gas Material Balance Water Drive Reservoirs

GRM-Engler-09

giBgB

pWwB615.5eWgBpGG

Cumulative water influx, rcf Cumulative water

production, stb

How to determine

We?


GRM-Engler-09

giS

grS

ip

iz

az

ap

giS

grS

gaB

giB

wdRF

1

1

Gas saturation

Recovery (water drive) < Recovery (depletion)

45 to 75% >75%

vE

vE

giS

grS

gaB

giB

vEwdRF1

1

Volumetric sweep

efficiency

Method accounts for

pressure gradients within

the invaded zone due to

relative permeability

effects resulting from

trapped residual gas

rt

ra

ro reservoir

aquifer

Invaded

zone

Original

Reservoir

boundary

Current

Reservoir

boundary

Water

influx

Modified Gas Material Balance

Prediction of water influx and reservoir pressure

(Schafer, et al, 1993)

pWwBeWgBgtBtGgiBgBGgBpG )()(

GRM-Engler-09 Trapped gas volume


Input data

Calculate

ro

Guess

pok

Calculate

We1

Calculate

rt

Calculate

pt

Calculate

pave

Calculate

Gt

Solve for

We 2

We1-We

2 < tol N Y

Gpn+1 = Gp

n +DGp

Update

Pok+1

Given Gpn

*

*

GRM-Engler-09


Comparison of reservoir performance from simulation,

Conventional and modified material balance methods (Hower and Jones, 1991)

GRM-Engler-09


Modified Gas Material Balance - example

0

500

1000

1500

2000

2500

3000

3500

4000

4500

0 500 1000 1500 2000 2500

Gp, mscf

p/z

,psia

G=2.16 Bscf

measured

We = 831 mstb

Sgr = 24%

GRM-Engler-09

F/E

t,stb

Intercept=G

tE

wBeW

We correct

We too small

We too large


GRM-Engler-09

tE

wBeWG

tE

F

Linearized Gas Material Balance

wBpWgBpGF

giBgBgE

pipwiS1

fcwcwiS*giBcfE

F = total net reservoir voidage

Et = Eg + Ecf = total expansion

Eg = expansion of gas in reservoir

Ecf = connate water and formation expansion * Bgi


GRM-Engler-09

Linearized Gas Material Balance - example

y = 1.0602x + 2.1074

R2 = 0.9843

3.0

3.2

3.4

3.6

3.8

4.0

4.2

4.4

4.6

4.8

5.0

1.0 1.5 2.0 2.5 3.0

We/Et

F/E

t

G = 2.107 Bscf

484mstb

1,427mstb

We = 831 mstb

p/z

Gp

(p/z)i

Gas expansion

Gas expansion

+

Formation compaction

+

Water expansion

Overestimate of G

Gas Material Balance Abnormally pressured gas reservoirs

GRM-Engler-09

pi

pe

c1

G

pG

1

iz

ip

z

p

wiS1

fc

wiS

wc

ec

Where,

GRM-Engler-09


Method 1 : Assume formation compressibility is known and constant

Volumetric (geopressured)

y = -92.336x + 6532.2

R2 = 0.9979

0

1000

2000

3000

4000

5000

6000

7000

0 20 40 60 80 100

Gp, Bcf

mo

difie

d p

/z, p

sia

G=70.7

pGvs

)wi

S1(

)pi

p(f

cwi

Sw

c

1z

p

Plotting function

Conventional

overestimates G

by > 25%!

GRM-Engler-09


Method 2 [Roach (1981)] : Simultaneous solution of G and cf

Revised material

Balance eq. wi

S1

fc

wc

wiS

ipz

zi

p

pi

p

pG

G

11

ipz

zi

p

pi

p

1

Geopressured

y = 13.199x - 17.511

R2 = 0.993

0

50

100

150

0 2 4 6 8 10 12 14

x, Bscf/psi x 10-3

y, p

si-1

x1

0-6

Bscf8.75199.13

1000

slope

1G

1psi

610x5.12

wc

wiS)

wiS1(

610bx

fc

GRM-Engler-09


Method 3 [Fetkovich (1991)] : Addition of gas solubility and total water

Defines: )

wiS1(

])p(f

c)p(tw

c[M)p(f

cwi

S)p(tw

c

)p(ec

Where,

Ctw, cumulative total water compressibility

M, associated water-volume ratio given by:

water expansion due to pressure depletion

the release of solution gas in the water

1

2

rr

aqr

rh

aqh

gh/

nh

gh/

nh1

r

nnp

aqM

NNPMM

non-net pay water and

pore volumes

external water volume

found in limited aquifers

GRM-Engler-09


Method 3 [Fetkovich (1991)]

Back-calculate ce from,

pi

p

1

G

pG

1z/p

iz/p1

atedbackcalculec

0

5

10

15

20

25

30

35

40

45

50

0 2000 4000 6000 8000 10000

pressure, psia

ce, p

si- 1

ce(back) assuming OGIP

ce(p) generated from rock &

w ater properties

Compare with rock and fluid derived ce

GRM-Engler-09


Method 3 [Fetkovich (1991)]

Results

0

1000

2000

3000

4000

5000

6000

7000

0 20 40 60 80 100

Gp, Bscf

p/z

, p

sia

historical performance data

model M = 2.25

Cf = 3.2x10-6 psi-1

G = 72 Bscf

GRM-Engler-09

Gas Material Balance Low-permeability gas reservoirs

G p

p/z

G G p

p/z

G

Expanding

Radius

Rebound

Effect

GRM-Engler-09


Gp

p/z

Conventional response

G

Tight gas response

(p/z)i

(p/z)intm

1

m2 = m

1

?

Gp

p/z

Conventional response

G

Tight gas response

(p/z)i

(p/z)intm

1

m2 = m

1

?m2 = m

1

?pG

)z/p(m

D

D

m

1*

scT

scTPhcV

)wS1(Ah43560hcV

Hydrocarbon volume

Estimate area

slope

GRM-Engler-09


m

1*

iz

ipG

m

1*

scT

scTPhcV

)1(43560 wSAhhcV

Gas-in-place

Hydrocarbon volume

Estimate area

(p/z)

(p/z)i

Gp

CASE A m2

G1 G2

GRM-Engler-09


2m

1*

iz

ip

1m

1*

intz

pG

Find Gas-in-place and m2

2m

1*

scT

scTPhcV

Hydrocarbon volume

(p/z)

(p/z)i

Gp

CASE B m2

G1= G2

(p/z)int

GRM-Engler-09


• Different slope and

intercept

• Estimate using other

cases

Case B < actual < Case A

(p/z)

(p/z)i

Gp

CASE C m2

G1 G2

(p/z)int

GRM-Engler-09


P/Z vs. Cumulative Production

No. 114

y = -0.8367x + 552.88

R2 = 0.9580

200

400

600

800

1000

1200

1400

1600

0 100 200 300 400 500 600 700

Cumulative Production, mmscf

P/Z

, p

sia

Example 1.8

??? 11

gi, cp 0.0134

h, ft 40

cti, psi-1

x 10-4

5.77

gg 0.67

Tr , deg F 106

Sw, % 44

rw, ft. 0.229

Pi, psi 1131

mmscfmz

pG 660

8367.

9.552

1

1*

int

m2 = 2.12 psia/mmscf

Vhc = 7.544 mmrcf

A = 70 acres

Case B

GRM-Engler-09


0

500

1000

1500

2000

2500

3000

3500

4000

4500

0 100 200 300 400 500 600 700

Cumulative production, mmscf

flo

w r

ate

, m

scf/

mo

nth

GIP = 700 mmscf

Gp = 526 mmscf

RF = 75%

GRM-Engler-09


1

10

100

1000

0 5 10 15 20 25

time, years

pro

du

cti

on

ra

te,

msc

f/m

o

0

200

400

600

800

1000

1200

SIB

HP

, p

si

simulated

measured

Single well simulation model

• Pwf =

250 150 psia

•Model area =

86 acres

GRM-Engler-09

Gas Material Balance Low-permeability gas reservoirs Problem 4

0

200

400

600

800

1000

1200

1400

0 200 400 600 800 1000

cumulative production, mmscf

p/z

, p

sia

, % 11

gi, cp 0.0131

h, ft 67

cti, psi-1 x 10-4 6.22

gg 0.67

Tr , deg F 103

Sw, % 44

rw, ft. 0.229

Pi, psi 1045

Estimate the gas-in-place and drainage area for this well. If

cumulative production was 752 mmscf, what has been the

recovery factor?

500 gas material balance basic concepts g=10 bscfpetro/faculty/kelly/grm-chap1-matbal.pdf · gas...

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