Applied reservoir eng

Applied Reservoir Engineering

Applied Reservoir Engineering : Dr. Hamid Khattab

Reservoir DefinitionReservoir DefinitionReservoir DefinitionReservoir DefinitionCap rock

Res. Fluid

Reservoir rock

R iReservoir

Shallow Deep

offshare onshare offshare onshare


Reservoir rocks

Sedimentry Chemical

Sandstone Sand L.s Dolomit


Rock Properties

Porosity Saturation Permeability Capillary Wettability

Absolute EffectiveSo

Sw Sg

Absolute

Eff ti

Relative

Primary PrimaryEffective

Ratio

Secondary Seccondary


Reservoir fluidsReservoir fluids

Water Oil Gas

Salt Fresh BlackVolatile

Drey Wet Condensate

Low volatileHigh volatile Ideal

Real (non ideal)


Fluid properties

Gas Oil Water

AM T P Z C βρg AMw γg Tc PC Z Cg βg µg βw rs µw SalinityCw

ρo γo APT rs βo βt µo Co TR PR


Applied reservoir Engineering ContentsApplied reservoir Engineering Contents

1. Calculation of original hydrocarbon in placei. Volumetric methodi. Volumetric methodii. Material balance equation (MBE)

2 Determination of the reservoir drive mechanism2. Determination of the reservoir drive mechanism

– Undersaturated– Depletion– Gas cap– Water drive– Combination

3. Prediction of future reservoir performance

– Primary recoveryPrimary recovery– Secoundry recovery by : Gas injection

Water injection


Calculation of original hydrocarbon in place by Calculation of original hydrocarbon in place by Calculation of original hydrocarbon in place by Calculation of original hydrocarbon in place by volumetric methodvolumetric method

● ●6

7Well Depth

1 D1

●

●●4

32

1 D1

2 D2

3 D3

4 D4

●

●●

●

1

9

4 D4

5 D5

6 D6

7 D ●8 5 9

Scale:1:50000

7 D7

8 D8

9 D9

Location mapStructural contour map



Well Depth1 h1 G1 h1

2 h2

3 h3

4 h4

GocGas

Oil4 h4

5 h5

6 h6

7 h

Woc

Oil

Water7 h7

8 h8

9 h9

30

10

0

Isopach map



)1(. wiSBVN −= φ)1()( wiSAh −= φ )()( wiφ

wiSAhβ

φ6155

)1(43560 −= SCFBblgβ

oiβ615.5

wiSAhN φ )1(7758 − STB

STBbbloβ

acresA :

oi

wiNβφ )(

=

iSAhφ )1(7758 −

STB

SCF

fth :

fractionsSwi :,φ

gi

wiSAhGβφ )1(7758

= SCF fwi,φ


Calculation of (BV) using isopach mapCalculation of (BV) using isopach map

Area inch2C.L

( ) g p p( ) g p p

1. Trapozoidal method:

A110

Ao0 WOC

5.01 >−nn AA

A330

A220

[ ]AAAAAhBV nn +++++= − 22......222 1210

A550

A440A’GOC

[ ]

[ ]AAhn

nn

′+′

+2

2 1210

A770

A660

O76

A770


Calculation of (BV) using isopach mapCalculation of (BV) using isopach map

Area inch2C.L

( ) g p p( ) g p p

2. Pyramid or cone method

5.01 ≤−nn AA A110

Ao0 WOC

[ ]AAAAhBV .3 1010 ++=

A330

A220

[ ]AAAAh .3 2121 +++

A550

A440A’GOC

[ ] [ ]nnnn AhAAAnAh3

.3 11 ++++ −−

A770

A660

O76

A770


Calculation of (BV) using isopach mapCalculation of (BV) using isopach map( ) g p p( ) g p p

3. Simpson method

Odd number of contour lines

[ ]nn AAAAAAhBV 24......4243 13210 ++++++= −

[ ]nAh3

3′

+3


Converting map areas (inchConverting map areas (inch22) to acres) to acresConverting map areas (inchConverting map areas (inch22) to acres) to acres

Say : Scale 1 : 50000Say : Scale 1 : 50000

1 inch = 50,000 inch

acres56398(50,000)inch12

2 acres56.39843560144

( )inch 1 =×

=


Converting map areas (inchConverting map areas (inch22) to acres) to acresExample 1 :

Converting map areas (inchConverting map areas (inch22) to acres) to acres

Gi th f ll i l i t d f it fGiven the following planimetred areas of an oit ofreservoir. Calculate the original oil place (N) if φ =25%,Swi=30%, βoi=1.4 bbl/STB and map scale=1:15000

C.L : 0 10 20 30 40 50 60 70 80 86Area inch2 : 250 200 140 98 76 40 26 12 5 0


Converting map areas (inchConverting map areas (inch22) to acres) to acresConverting map areas (inchConverting map areas (inch22) to acres) to acres

Solution :

[ ]

[ ] [ ] [ ]0500506512512101226122610

26402762982140220022502

10+×+×+×+×+×+=BV

[ ] [ ] [ ]05005036512512

31012261226

210

×+++×+++×+++

ftinch :7198 2=

acres87.3543560144

(15,000) inch 12

2 =×

=

ftinch :7198

acresBV 39.25819387.357198 =×=∴

MMSTBN 38.2504.1

)3.01(25.039.2581937758=

−×××=∴


Converting map areas (inchConverting map areas (inch22) to acres) to acresBy using Simpson method


[ ]

[ ]05056

52124262404762984140220042503

10

×+++

×+×+×+×+×+×+×+×+=BV

[ ]05053

×+++

ftinch .6.7156 2=ftacro

f.6.25670987.356.7156 =×=

)301(25062567097758 MMSTBN 94.2484.1

)3.01(25.06.2567097758=

−×××=∴

MMSTBNav 66.2492)94.24838.250( =+=∴



If th i f l 1 i i d


If the reservoir of example 1 is a gas reservoir andβg=0.001 bbl/SCF. Calculate the original gas in place

S l tiSolution :

)301(250392581937758 −××× MMSCFG 53.350001.0

)3.01(25.039.2581937758=

×××=




A gas cap has the following data : φ =25%, Swi=30%, βoi=1.3bbl/STB, βgi=0.001 bbl/SCF and map scale=1:20000

C.L : 0(WOC) 10 20 30 33(GOC) 40 50 60 70 76Area inch2 : 350 310 270 220 200 190 130 55 25 0

Calculate the original oil in place (N) and the original gas inplace (G)


Converting map areas (inchConverting map areas (inch22) to acres) to acresSolution :


[ ] [ ]310 2[ ] [ ] ftinchBVoil .92802002202322027023102350

210 2=+++×+×+=

[ ] [ ] [ ]BV 551305513010130190101902007×++++++= [ ] [ ] [ ]

[ ] [ ] ftinch

BVgas

.79.4303253625552555

310

55130551302

1301902

1902002

2=+×+++

×++++++=

acres77.6343560144

(20,000) inch 12

2 =×

=

MMSTBN 6183.1

)3.01(25.077.6392807758=

−××××=

MMSCFG 6.372001.0

)3.01(25.077.6379.43037758=

−××××=


Reservoir drive mechanismReservoir drive mechanismReservoir drive mechanismReservoir drive mechanism

Water reservoirWater reservoirP

Gas reservoir

GasGasBg

Waterwith bottom water drive

without bottom water drive

g

Oil reservoir


Oil reservoir

Undersaturated

P>Pb

OilOilOil

Waterwith bottom without bottom Saturatedwater drive

without bottom water drive

Saturated

P≤Pb

OilOil

OilOil

W

Gas

Gas

Gas

Oil WaterGas

Combination drive

Bottom water drive

Gas cap drive

Depletion drive


PVT data for gas and oil reservoirsPVT data for gas and oil reservoirsPVT data for gas and oil reservoirsPVT data for gas and oil reservoirs

Gas reservoirs

Bg

Gas reservoirs

PZTBg 00504.0=

P


PVT data for gas and oil reservoirsPVT data for gas and oil reservoirs

Saturated oil reservoirs

PVT data for gas and oil reservoirsPVT data for gas and oil reservoirs

Boi=Bti

µoB

rsi

Bt = Bo+(rsi-rs)Bg

PZTBg 00504.0=

Bo

rs

Boi= BtiP

Bg

0

1

P0 Pi


PVT data for gas and oil reservoirsPVT data for gas and oil reservoirsPVT data for gas and oil reservoirsPVT data for gas and oil reservoirs

Undersaturated oil reservoirs

saturated undersat.

P1 > Pb

Bt

µorsi=c

Bo

rs

Bg1

0


Laboratory measurment of PVT dataLaboratory measurment of PVT dataLaboratory measurment of PVT dataLaboratory measurment of PVT data

GasGas

SCFOil

P

OilOil

P

OilOil

SCF

STB

undersaturatedsaturated

Pb P > Pb Pi

P = 14.7 psi

T = 60o F


Calculation of original gas in place by MBECalculation of original gas in place by MBECalculation of original gas in place by MBECalculation of original gas in place by MBE

1. Gas reservoir without bottom water drive

pG

T∆ ( ) gip BGG −giGB

( )

ipp∠ip

gpBGG =∴

( ) gpgi BGGGB −=

1gig BB

G−

=∴ 1



1. Gas reservoir without bottom water driveExample 4 :

psiP SCFG p SCFbblgB Z G

201.60.810.00084123900

0x1060.830.000770x10-64000p

G

195.20.770.00095373700

200.20.790.00089273800 .constG ≠

Solution :

199.70.750.00107583600

Using eq. (1)



MBE as an equation of a straight liney1( ) gpgi BGGGB −=

( )BBGBG∴ 2

gpBG

G

y1

( )giggp BBGBG −=∴

Another form:

2

gig BB −

x1

( )

−=

ip p

TZpZTG

pZTG 00504.000504.0 ZGp

y2

ip ppp

−=∴ iZZGGZ 3

ppG

=∴i

p ppGG

p 3

ii PZpZ −

x2


Calculation of original gas in place by MBECalculation of original gas in place by MBECalculation of original gas in place by MBECalculation of original gas in place by MBEAnother form:

( ) gpgi BGGGB −=

( ) ZZp

i

i

Zp

( )

−=

pZGGG

pZ

pi

i 00504.000504.0

pGP

Z

y3 i

i

GZp

ppp

i

ip

Zp

GG

ZP

−=∴ 1

Gp

i

i

i

i GGZp

Zp

Zp

−=∴

at pGx3

G

0

0=Zp

pGG =

p


Calculation of original gas in place by MBECalculation of original gas in place by MBEExample 5 :

Calculation of original gas in place by MBECalculation of original gas in place by MBE

Solve the previous example using MBE as a straigh lineSolution :

p PZ ii PZpZ −gp BGgig BB − ZP pG

1244411.752.251.0680.000053900

0x10-64819― x10-52.075x10-4―x104―4000

3738965.912.663.5150.000183700

2741773.152.392.4030.000123800

x3y3x2y2y1x1

5634218.092.885.9900.000303600

From Figgers STBG 610200×=


Calculation of original gas in place by MBECalculation of original gas in place by MBE2.Gas reservoir with bottom water drive


RpGpW

T∆

∴

Assuming =0 causes an increase in G continuously



MBE as a straight line


F/

45o

∴

N

∴

/Assuming is known

33



A i i h k b d i h h f ll i


Example 6 :A gas reservoir with a known bottom water drive has the following data: =0 and

B

0x10-60.000930x10940000

We bblT years psiP SCFG p SCFbblgB

7.4900.0010772.3338002

2.2970.0009827.8539001

13.3080.00117113.8537003

18.4860.00125151.4836004

34

Calculate the original gas in place



Tyear F Eg F/Eg x109 We/Eg x109

Solution


Tyear F Eg F/Eg x10 We/Eg x101 27.2x106 0.00005 546 45.932 77.39 0.00014 553 53.043 133.20 0.00024 555 55.444 189.35 0.00032 554 54.25

F/EF/Eg

45

From Fig: G=500x109 SCF G=500x109

We/Eg


Calculation of original gas in place by MBECalculation of original gas in place by MBEGas Cap Expansion an Shrinkage

G


Gpc

gasGOC

Gpc

expansion

GOC

GOC

Oil

shrinkage

Oil

Shrinkage due to: poor planning or accident and corrosiong p p g

- Assume gas cap expansion = (G-Gpc).Bg-GBgi

Assume gas cap shrinkage = GB (G Gp )B- Assume gas cap shrinkage = GBgi - (G-Gpc)BgGpc: gas produced from the gas cap and my be = zero



Example: 7


Calculate the gas cap volume change if G=40x109 SCF

P Gpc x109 Bg

4000 0 0.00203900 4 0 00223900 4 0.00223800 7 0.00253700 10 0.00283600 13 0.00313500 17 0.0035



Solution


Assuming gas cap expansion = (G-Gpc).Bg-Ggi

Pressure Gas cap change x103 type4000 - -3900 -800 shrinkage3800 +2500 expansion3800 +2500 expansion3700 +4000 expansion3600 +3700 shrinkage3500 +5000 expansion

Shrinkage at P=3600 may be due PVT or Gp dataShrinkage at P=3600 may be due PVT or Gpc data


C l l ti f i i l il i l b MBEC l l ti f i i l il i l b MBECalculation of original oil in place by MBECalculation of original oil in place by MBE

a) Under-saturated oil reservoirs

Characteristics

P>P- P>Pb

- No free gas, no Wp

- Large volumeLimited K- Limited K

- Low flow rate- Produce by Cw and Cf


C l l ti f i i l il i l b MBEC l l ti f i i l il i l b MBE

1- Under-saturated oil reservoirs without bottom water

Calculation of original oil in place by MBECalculation of original oil in place by MBE

Np

NBoi

P P

(N-Ni)Bo

P>PPi>PbP>Pb

neglecing Cw and Cf

NBoi=(N-Np)Bo

opBNN∴ (1)

oio

p

BBN

−=∴ (1)


C l l ti f i i l il i l b MBEC l l ti f i i l il i l b MBEExample: 8

C l l t th i i l il i l i t d i d l ti C


Calculate the original oil in place assuming no water drive and neglecting Cwand Cf using the following data

P Np x106 Bo

4000 0 1.403800 1 535 1 423800 1.535 1.423600 3.696 1.453400 7.644 1.493200 9.545 1.54


C l l ti f i i l il i l b MBEC l l ti f i i l il i l b MBESolution

Pressure NpBo x106 Bo-Boi N x106


4000 - - -3800 2.179 0.02 108.953600 5 539 0 05 110 78 constN ≠3600 5.539 0.05 110.783400 11.389 0.09 126.643200 14.699 0.14 104.99

rearrange MBE as a straight line

NBoi = (N-Np)Bo

F

F = NEo

From Fig:6

NSTBxN 610110≠

Eo



o.b.p=1 psi/ftDConsidering Cw and Cf


- overburden pressure = 1 psi/ftD- rock strength = 0.5 psi/ftD

r s rv ir pr ssur 0 5 psi/ftD- reservoir pressure = 0.5 psi/ftD

o.b.p


C l l ti f i i l il i l b MBEC l l ti f i i l il i l b MBECalculation of original oil in place by MBECalculation of original oil in place by MBEConsidering Cw and Cf

NBoiVpBNNNB fwopoi ∆+−= )( ,

Pi>Pb

dpVCdVpdVp

C

VpVpVp

f

fwfw

→

∆+∆=∆

1,

(N-Np)BodVp

dpVCdVpdpV

C pfff

pf =→=

1

.

P>Pb

∆Vp,,w

V

dpVCdVpdpdVp

VC www

w

ww =→= .1

dpVSCdVpVSVVVS pwwwpwwp

ww =→=→=



dpVCSCdVp pfwwwf +=∴ )(,

SNBVpSVpNB

w

oiwoi −

=→−=)1(

)1(

dpNBSCSC

dVp oifww

wf −

+=∴ )

1(,

pNBSCSC

BNNNB

S

oifww

opoi

w

∆+

+−=∴ )1

()(

1

S oiw

opoi −1



BNN op

Calculation of original oil in place by MBECalculation of original oil in place by MBEConsidering Cw and Cf

BB

pBSCSC

BBN

oiw

fwwoio

op

∆−+

+−=∴

)1

(

BNBNN

pBCBBpBBBC

opop

oiooiooi

oioo

==∴

∆=−→∆−

=Q

SSwhere

pBSCSC

SSCpB

SCSC

CN

wo

oiw

fww

w

oooi

w

fwwo

−=

∆−+

+−

∆−+

+∴

1

]11

[]1

[

pBS

CSCSCBN

Noi

fwwoo

op

wo

∆++

=∴]

1[

pCBBN

N

S

eoi

op

w

∆=

−1

(2)



PPP i −=∆Pi

Pi

VVdPdV

VC

oio −=

1

.1

PBBBC

oi

oioo ∆

−=

Pi

BBPPVV

V

oio

i

i

oi

−−−

=

)(1

.1

Voi

Vo PBoio

oi ∆=

)(.

)salinity and ,,(

)(

sw

f

rTPfC

fC

=

= φ

From the following charts


C l l ti f i i l il i l b MBEC l l ti f i i l il i l b MBECalculation of original oil in place by MBECalculation of original oil in place by MBE


C l l ti f i i l il i l b MBEC l l ti f i i l il i l b MBEExample: 9

l l ( ) d h ff f d


SolutionSolve example (8) considering the effect of Cw and Cf

R1(Fig 2)r (Fig 1)C (Fig 4)∆P=(Pi P)PBBC oio −=

182.9x10-6――4000

R1(Fig.2)rsf(Fig.1)Cwp(Fig.4)∆P=(Pi-P)P pBC

oio ∆=

0.8516.82.958.9284003600

17.22.937.143x10-52003800

15.23.0012.5008003200

162.9810.7146003400


C l l ti f i i l il i l b MBEC l l ti f i i l il i l b MBEContinue


Cw=CwpxR2R2 (Fig.3)rs= rsf x R1Pw

fwwoo

SCSCSC

−++

1

7.725x10-53.3111.1314.623800

―3.30x10-61.415.34000

13.5693.2891.10413.602400

9.5703.2471.1114.283600

13.1433.171.0912.923200


C l l ti f i i l il i l b MBEC l l ti f i i l il i l b MBEContinue

BN


pCBBN

Neoi

op

∆=

P NpBo NBoiCe∆P N

4000 ― ― ―

3800 2.179x016 0.0218 108.2x106

3600 5.359 0.0536 107.9

N ≠ C

3400 11.389 0.1131 106.5

3200 14 699 0 1470 105 13200 14.699 0.1470 105.1



Use MBE as a straight line as follows:


PCNBBN eoiop ∆=BNF

Plot the fig

oNEF = opBNF =

610100×=NPlot the fig. 10100×=N

STBN 610100×=PCBE ∆PCBE eoio ∆=


U d t t d il i ith b tt tU d t t d il i ith b tt tUndersaturated oil reservoir with bottom waterUndersaturated oil reservoir with bottom water

pNpW pW

(N-Np)BoNBoi

P>PbPi>Pb

Assuming (We) is known and neglect Cw+Cf

( ) ( )BwWBNNNB −+−= ( ) ( )( )

i

wpeop

wpeopoi

BBBwWBN

N

BwWBNNNB

−

−−=∴

+=

oio BB

Assuming We=0 will cuse an increase in (N)



Example 11 :

Using the following data in the undersaturated oil reservoir with a known (We), neglecting Cw & Cf calculate (N): wp= 0

P Np Bo We

4000 ―x106 1.40 ―x106

3800 2.334 1.45 1.135

3600 5 362 1 42 2 4163600 5.362 1.42 2.416

3400 10.033 1.49 3.561

3200 12 682 1 54 4 8323200 12.682 1.54 4.832



Solution :Solution :( )

oio

wpeop

BBBwWBN

N−

−−=

P NpBo Bo-Boi N

4000 106 1064000 ―x106 ― ―x106

3800 3.314 0.02 108.5

3600 7 775 0 05 107 1 N ≠ C3600 7.775 0.05 107.1

3400 14.950 0.09 126.5

3200 19 531 0 14 105 0

N ≠ C

3200 19.531 0.14 105.0


U d t t d il i ith b tt tU d t t d il i ith b tt t

EFRearrange MBE as a straight line

Undersaturated oil reservoir with bottom waterUndersaturated oil reservoir with bottom water

oE

o45[ ] eiowpop WBBNBWBN +−=+ 0

WENF +=

110=N

eo WENF +=

oeo EWNEF +=∴

oe EW

[ ]ioo BBE 0−= oe EWp oEFop BNF =[ ]ioo 0

48 32155 57 7750 05360056.75165.73.3140.023800― x10-6―― x10-6―4000

p p

34.51139.519.5310.14320039.56166.414.9800.09340048.32155.57.7750.053600


Undersaturated oil reservoir with bottom waterUndersaturated oil reservoir with bottom waterExample 11 :

Solve examole (10) considering Cw and Cf effect

Solution :So ut onCw, Co, Cf and Ce are the same as example (9)

Peoi

e

o

e

CBW

EW

∆

=PeoiCB ∆Peoi

oP

o CBBN

EF

∆

=P∆eCP

45.07145.060.05364009.570360052.06 x106152.01 x1060.02182007.7853800

――――― x10-54000

32.87132.860.147080013.143320031.26131.250.113960013.568340045.07145.060.05364009.5703600

FoE

F

o45

Pi

ee

CBW

EW

∆

=Pi

oP

CBBN

EF

∆

=Plot vs

610100 ×=Noe EW

Peoio CBE ∆Peoio CBE ∆

As in Fig. 610100×=N


B S t t d il iB S t t d il iB. Saturated oil reservoirsB. Saturated oil reservoirs

1 D l ti d i i1. Depletion drive reservoirs

Characteristics

bPP ≤• b0=• pW

rapidlyincreasesRp•

FRlow .•



pGpN


T∆oiNB ( ) op BNN −

p

Free gas

( )bi PP≤

Free gas

p

( ) gasfreeBNNNB opoi +−=

( ) SCFRNrNNNrgasfree ppspsi −−−=

( ) ( )[ ] gppspsiopoi BRNrNNNrBNNNB −−−+−=∴

( )[ ]( )[ ]( ) gssioio

gspop

BrrBBBrRBN

N−+−

−+=∴



Example 12 :


Calucaltion (N) for a depletion drive reservoir has the following data : Swi=30%

91 506140 0012731 4236743 873800

― x1067180.0010411.492718― x1064000

NrsBgBoRPNPP

ion

96.014000.0022001.28630776.443400

96.025100.0016271.35519375.263600

91.506140.0012731.4236743.873800

Solu

t

96.014000.0022001.28630776.443400

As shown N ≠ const., so rearrange MBE as a straight line



( )[ ] ( )[ ]gssioiogspop BrrBBNBrRBN −+−=−+


oENF =

Solution :

P F Eo

4000 0 106 0

Solution

F

4000 0x106 0

3800 5.802 0.0634

3600 19 339 0 2014

61096×=N

3600 19.339 0.2014

3400 46.124 0.4804

6 oESTBNFigFrom 61096: ×=

o



( ) gssioiop BrrBBNFR

−+−


( )( ) gspo

gssioiop

BrRBNFR

−+==.

( )PRPfFR &. = ( )PRPfFR &.

PRFR 1. ∝

To increase R.F:

• Working over high producing GOR wellsWorking over high producing GOR wells• Shut-in ,, ,, ,, ,, ,,• Reduce (q) of ,, ,, ,, ,,

R i j t f d d• Reinject some of gas produced


Calculation of original oil in place by MBECalculation of original oil in place by MBEExample 13 :

For example 12 at P=3400 psi calculate: S and R F without Gi and


Solution :

For example 12, at P 3400 psi calculate: Sg and R.F without Gi and with Gi=60 Gp

gasfreeS =

( )[ ]i BRNrNNNrgasfree −−−=

volumeporeS g =

( )[ ] gppspsi BRNrNNNrgasfree

( )bbls6

6

66

1005.280022.030771044.64061044.6967181096

×=×

××

−××−−××=

30771044.6

( ) bblsS

NBvolumeporew

oi 66

1062.2043.01

492.11096)1(

×=−××

=−

= ( )w )(

%7.13137.01062.204

1005.286

6

==××

=∴ gS



( ) gssioio BrrBBFR

−+−


( )( ) gspo

gssioioGwithout BrRB

FRi −+=.

( ) 0022.0406718492.1286.1 ×−+− ( )( )%7.6067.0

0022.04063077286.1==

×−+=

( )( )

gssioioGwith BrRB

BrrBBFR

i −+

−+−=%60.

( )( ) 002204063077402861

0022.0406718492.1286.1×−×+×−+−

=

( ) gspo BrRB +

( )%49.151549.0

0022.040630774.0286.1==

××+


2 Gas Cap reservoir2 Gas Cap reservoir2. Gas Cap reservoir2. Gas Cap reservoir

Characteristics

• P falls slowly• No Wp• High GOR for high structure wells• R.F > R.Fdepletion

• Ultimate R.F ∝ Kv, gas cap size, 1/µo, 1/qoUltimate R.F ∝ Kv, gas cap size, 1/µo, 1/qo


Calculation of OOIP by MBECalculation of OOIP by MBEyypNpG

giGB

(N-Np)Bo

free gas

oiNB

gi

oi

gi

NBB

m =

( ) gasfreeBNNNBGB opoigi +−=+

P>PbiP

[ ] ( ) ppspsi RNrNNGNrgasfree −−−+=

( )[ ]gspop BrRBNN

−+=∴

( ) ( )giggi

oigssioio BB

BBmBrrBB

N−+−+−

∴

( ) ( )oi BRNNNmNBNBNNNBNB

+++∴ ( ) ( ) gppspgi

oisiopoioi BRNrNN

BNrpBNNNBmNB

−−−++−=+∴

This equation contains two unknown (m and N)


Calculation of OOIP by MBECalculation of OOIP by MBEyy

Rearrange MBE to give a straight line equation

( )[ ] ( )[ ] ( )gigoi

gssioiogspop BBBmNBBrrBBNBrRBN −+−+−=−+ gg

giggpp B

EF

go GENEF +=oE

G

o

g

o EE

GNEF

+=∴NN

og EE



Example 14 :

Calculate (N) and (m) for the following gas cap reservoir

P Np Rp Bo rs Bg

4000 ―x106 510 1.2511 510 0.000873900 3.295 1050 1.2353 477 0.000923800 5 905 1060 1 2222 450 0 000963800 5.905 1060 1.2222 450 0.000963700 8.852 1160 1.2122 425 0.001013600 11.503 1235 1.2022 401 0.001073500 14.513 1265 1.1922 375 0.001133400 17.730 1300 1.1822 352 0.00120



Solution : +=∴ g

EEGNE

F

P F Eo Eg F/Eo Eg/Eo

+∴oo EGNE

4000 ―x106 0 0 ―x106 ―

3900 5.807 0.0145 0.00005 398.8 0.0034

3800 10.671 0.0287 0.00009 371.8 0.0031

3700 17.302 0.0469 0.00014 368.5 0.0029

3600 24.094 0.0677 0.00020 355.7 0.0028

3500 31.898 0.09268 0.00026 340.6 0.0027

3400 41 130 0 1207 0 00033 340 7 0 00273400 41.130 0.1207 0.00033 340.7 0.0027



FoE

910826 ×=G

From Fig.

N = 115 x 106 STB

610115 ×=N 10115 ×Nog EE

2511110115 6 ×××mmNB00087.0

2511.11011510826 9 ×××==×=m

BmNBG

gi

oi

50∴ 5.0=∴m



Another solution

Assume several values of (m) until the straight line going through the origin as follows:

go GENEF +=

mNBg

gi

oio E

BmNBNE +=

+= g

gi

oio E

BmBENF



oi EmBE +FP

m = 0.6m = 0.5m = 0.4

ggi

oio E

BE +

0 1060 0930 08110 6713800

0.0570.0510.0435.8073900

0000x1064000

0 2400 2110 18324 0943600

0.1670.1470.12717.3023700

0.1060.0930.08110.6713800

0.4050.3580.31141.1303400

0.3180.2440.24331.8983500

0.2400.2110.18324.0943600



F

From Fig.

m = 0 5m = 0.5

N = 115 x 106 STB

oi EmBE + ggi

oio E

BE +


3. Water drive reservoirs3. Water drive reservoirs

Edge water Bottom water

Finite Infinite Finite Infinite

Oil Oil

WaterW W

Water


3. Water drive reservoirs3. Water drive reservoirs

Characteristics

-P decline very gradually-Wp high for lower structure wells-Low GOR-R.F > R.Fgac cap > R.Fdepletion



(N-Np)Bo

NBoi free gas

( ) ( ) gasfreeBwWBNNNB wpeopoi +−+−=

( ) ppspsi RNrNNNrgasfree −−−=

( ) ( ) ( )[ ]

( )[ ] ( )BwWBrRBN +

( ) ( ) ( )[ ] gppspsiwpeopoi BRNrNNNrBwWBNNNB −−−+−+−=∴

( )[ ] ( )( ) gssioio

wpegspop

BrrBBBwWBrRBN

N−+−

−−−+=∴



Rearrange MBE as an equation of a straight line:

( )[ ] ( )[ ] egssioiowpgspop WBrrBBNBwBrRBN +−+−=+−+∴

eo WENF +=

o

e

o EWN

EF

+=∴oo



Example 15 :

Calculate (N) for the following bpttom water drive reservoir of known (We) value:

P Np Bo Rs Rp Bg We

4000 0x106 1.40 700 700 0.0010 0x106

3900 3.385 1.38 680 780 0.0013 3.912

3800 10.660 1.36 660 890 0.0016 13.635

3700 19.580 1.34 630 1050 0.0019 23.265

3600 27.518 1.32 600 1190 0.0022 44.044



Solution :o

e

o EWNE

F +=

P F Eo F/Eo We/Eo

4000 ―x106 ― ―x106 ―x106

3900 5.111 0.006 851.89 6523800 18.420 0.024 767.52 5683700 41.862 0.073 573.45 373.53700 41.862 0.073 573.45 373.53600 72.042 0.140 514.38 314.6

oEF

o

o45From Fig.

610200 ×=Noe EW

N = 200 x 106


4. Combination drive reservoir4. Combination drive reservoir

Characteristics:

I W f l t t ll-Increase Wp from low structure wells-Increase GOR from high structure wells-Relativity rapid decline of Py p f-R.F > R.Fwater influx



free gasiGB giGB

(N-Np)Booi

gi

NBGB

m =

oiNB

gi

Pi

( ) ( ) gasfreeBwWBNNGBNB wpeopgioi +−+−=+

( ) RNNNNGf

P<PiPi

( ) ppspsi RNrNNNrGgasfree −−−+=

( ) ( )( )[ ]

wpeopoioi BwWBNNmNBNB −+−=+∴

( )[ ] ( )BwWBrRBN −−−+

( )[ ] gppspsi BRNrNNNr −−−+

( )[ ] ( )( ) ( )gig

gi

oigssioio

wpegspop

BBBmBBrrBB

BwWBrRBNN

−+−+−

+=∴



This equation includes 3 unknown (We, m & N)Rearange this equation as a straight line equation

( )[ ] ( )[ ] ( ) egigoi

gssioiowpgspop WBBBmBNBrrBBNBwBrRBN +

−+−+−=+−+∴

g q g q

( )[ ] [ ] ( )gggi

gpgpp B

egoi

o WEBmBENF +

+= g

giB

e

mBWNmB

F+=∴

ggi

oio E

BmBE

F

+o45

ggi

oiog

gi

oio E

BmBEE

BmBE ++

45

NIf We is assumed to be known and m is calculated by geological dat. N can be obtained

ggi

oio

e

EBmBE

W

+

N


Calculation of OOIP by MBECalculation of OOIP by MBE

Example 16 :

yy

Calculate the original oil in place (N)for the following combination drive reservoir assuming that m=0.5 and values of (We) are given:

P Np Bo rs Rp Bg We

4000 0x106 1 351 600 600 0 00100 0x1064000 0x106 1.351 600 600 0.00100 0x106

3800 4.942 1.336 567 1140 0.00105 0.515

3600 8.869 1.322 540 1150 0.00109 1.097

3400 17.154 1.301 491 1325 0.00120 3.011


Calculation of OOIP by MBECalculation of OOIP by MBE

Solution :

yy

――――0x1064000

EoFPg

gi

oio E

BmBE

F

+ ggi

oio

e

EBmBE

W

+ggi

oio E

BmBE +

13 95183 950 21590 080839 715340011.29181.290.09720.036417.6223600

9.66x106179.66x1060.05330.01969.57638000x104000

13.95183.950.21590.080839.7153400

gi

oio E

BmBE

F

+

F F giB o45From Fig.

STBN 610170×=

ggi

oio

e

EBmBE

W

+

610170 ×=N


Uses of MBECalculation of (N), (G) and (We) Prediction of future performance

Difficulties of its applicationDifficulties of its application

Lackof PVT dataAssume constant gas compositionProduction data (NP, GP and WP)Pi and We calculations

Limitation of MBE applicationThick formationHigh permeabilityHigh permeabilityHomogeneous formationLow oil viscosity

N ti t d iNo active water driveNo large gas cap


S l ti f PVT d t f MBE ppli ti sSelection of PVT data for MBE applications

Depletion drive flash

Gas cap drive differential

C bi ti (fl h diff ) rsCombination (flash + diff.)Water drive flashLow volatile oil differential

rs

High volatile oil flash

Moderate volatile (flash + diff.) pp

difflash ff


Water in flux

Due to: Cw, Cf and artesian flow

We

Oil

Bottom water Edge water Linear flux

OilOil

waterWW

water


Flow regimes

Steady state semi-steady state Unsteady state

Outer boundary condition

Infinite Limited Infinite Limited


Steady state water influx

- Open external boundary- ∆P/∆r = C with time- qe=qw=C with time- Strong We

- Steady state equation (Darcy law)y q ( y )

pe

qw qe

pw

rrw rer


Hydraulic analogHydraulic analog

( )∝∆∝

PPdtdWPq Pi

( )( )

( )∑ ∆

−=−∝

PPkWPPkdtdWPPdtdW

ie

iePw

( )∑ ∆−= tPPkW ie x

screen sandq

constantinflux water :k( ) curve (Pust)under area :∑ ∆− tPPi

C l l f KCalculation of K:Water influx rate = oil rate + gas rate + water prod. rate

dWdNdNdW )()( PPkBdtdWBrR

dtdNB

dtdN

dtdW

iwP

gspP

oPe −=+−+=


Example :C l l t K i th f ll i d t P 3500 i P 3340 Calculate K using the following data: Pi=3500 psi, P=3340 (Bo=0.00082 bbl/SCF), Qw=0, Bw=1.1 bbl/STB and Qo=13300 STB/day

Solution :

psidaybblk

daybbldtdWe

//130)33403500(

20800/20800000082.0)700900(133004.113300

==∴

=+×−×+×=

)33403500( −

Pit1 t2 t3 t4∆t1 ∆t2 ∆t3 ∆t4

A1 A2 A3 A4

Calculation of ( )∑ ∆− tPPi

P1

A1 A2 A3 A4( )( )

11

4321

2tPP

AAAAtPP

i

i

∆−

=

+++=∆−∑

P2

P3

( ) ( )

( ) ( )32

221

2

tPPPP

tPPPP

ii

ii

∆−+−

+

∆−+−

+

P3P4

( ) ( )4

43

3

2

2

tPPPP

t

ii ∆−+−

+

∆+


Example :

The pressure history of a steady-state water drive reservoir is given as follows:

Tdays : 0 100 200 300 400Ppsi : 3500 3450 3410 3380 3340

If k=130 bbl/day/psi, calculate We at 100, 200,300 & 400 dayse , , y


100 100 100 100 t

Solution : 5090

P

t

( ) bblsWe100 000,325010034503500130

−

−=

120160

P

( )

We

e

200

100

12350001002

90501002

50130

,2

=

+

+×=

We3

200

16012012090905050

102606100212090100

29050100

250130

+++

×=

+

++

+×=

bblsWe3

200 1044201002

160120100212090100

29050100

250130 ×=

+

++

++

+×=


SemiSemi--steadysteady--state water influxstate water influxSemiSemi steadysteady state water influxstate water influx

As the water drains from the aquifer, the aquifer radius (re) increases with time, there for (re/rw) is replaced by a time increases with time, there for (re/rw) is replaced by a time

dependent function (re/rw)→at

PPCPPCPPkhdW × − )()()(10087 3

PPCdWatnPPC

rrnPPC

rrnPPkh

dtdW i

we

we

we

wee −→

−=

−×=∴

)()(

)()()(

)()(1008.7

lllµ

PPatnPPC

dtdW ie −

=∴

)()(

)(l

tatnPPCW i

e ∆−

=∴ ∑ )()(

l


The two unknown constants (a and C) are determined as:

(at)lnCdtdW

PP

e

i 1)(

)(=

−

)()(dtdWPP

e

i −

tan lnlCCdtdW

PP

e

i 11)(

)(+=

−∴

)( We

1C1

Plott this equation as a straight line:

tlnanl

C1

Gives slop = and intercept = C1

CC


Example 18:Example 18:Using the following data calculate (a) and (c)

Tmonth P We MBE ∆We (Wen+1-Wen-1) ∆We/ ∆t (Pi-P)0 3793 0x103 0 0 03 3788 4.0 12.4 136 56 3774 24.8 35.5 389 199 3748 75.5 73.6 806 4512 3709 172 116.8 1279 8415 3680 309 154 1687 113ti

on

15 3680 309 154 1687 11318 3643 480 197 2158 15021 3595 703 249 2727 198

Solu

t

24 3547 978 291 3187 24627 3518 1286 319 3494 27530 3485 1616 351 3844 30833 3437 1987 386 4228 35636 3416 2388 407 4458 377


tmonth tdays ∆We/ ∆t (Pi-P) Ln t (Pi-P)/ dWe/ dt0 0 0 0 ― ―6 182.5 389 19 5.207 0.049

12 365 1279 84 5.900 0.06612 365 1279 84 5.900 0.06618 547.5 2158 150 6.305 0.07024 780 3187 246 6.593 0.07730 912 5 3844 308 6 816 0 08130 912.5 3844 308 6.816 0.081

)()(dtdWPPi −002.01

=C

From Fig.)( dtdWe

∴C = 50

Using any point in the straight line

C

002.01=

C

Using any point in the straight line

a = 0.064

∑ − PPW i50

tln

∑=∴ln(0.064t)

W ie 50


Example 18:Example 18:Using data of example (18) calculate the cumulative water influx (We) after 39 months (1186.25 days) where the pressure equals 3379 psi

Solution :

PPPP dt

taPP

taPPWW ii

ee

−+

−×+= 250

2

39

1

363639 lnln

[ ]34163793337937933 −− [ ]109525.11862)1095064.0(

34163793)25.1186064.0(

3379379350102388 3 ××

×

+×

×+×=lnln

33 33 10508.420102388 ×+×=

bbls3102809×=


UnsteadyUnsteady--state water influxstate water influxUnsteadyUnsteady state water influxstate water influx

- P and q = C with time- q = 0 at re, q=qmax at rw

Closed extended boundryrw

- Closed extended boundry- We due to Cw and Cf


Hydraulic analogHydraulic analogHydraulic analogHydraulic analog

Pi

P2

P1

Pw

qx qscreen sand sand sand


Physical analogPhysical analogPhysical analogPhysical analog

Date post:	15-Jul-2015
Category:	Engineering
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