Applied Reservoir Lectures

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Applied Reservoir Engineering

Ifh-

Applied Reservoir Engineering : Dr. Hamid Khattab

Reservoir Definition Cap rock Res. Fluid Reservoir rock

R Reservoir i

Shallow

offshare

Deep

onshare

offshare

onshare

Ifiis

Applied Reservoir Engineering : Dr. Hamid Khattab

Reservoir rocks

Sedimentry

Sandstone

Chemical

Sand

L.s

Dolomit

Hi

Applied Reservoir Engineering : Dr. Hamid Khattab I

Rock Properties

Absolute

i

Saturation

Porosity

Effective

So

Absolute Sw

Capillary

Permeability

Relative

Sg Eff ti Effective

Primary

Secondary

Primary

Seccondary

Ratio

Wettability

n

Applied Reservoir Engineering : Dr. Hamid Khattab Reservoir fluids

Water

Salt

Oil

Fresh

Black

Gas

Volatile

Drey

Wet

Low volatile High volatile

Ideal

Real (non ideal)

Condensate

Hi

Applied Reservoir Engineering : Dr. Hamid Khattab

Fluid properties

Oil

Gas

Water r

ρg AMw γg Tc PC Z

TR P R

Cg βg µg

ρo γo APT rs

f

βw rs µw Cw Salinity

βo βt µo Co



IfApplied Reservoir Engineering : Dr. Hamid Khattab

Applied reservoir Engineering Contents 1. Calculation of original hydrocarbon in place i. Volumetric method ii. Material balance equation (MBE) 2 Determination of the reservoir drive mechanism 2. – Undersaturated – Depletion – Gas cap – Water drive – Combination 3. Prediction of future reservoir performance – Primary recovery – Secoundry recovery by : Gas injection Water injection

4i'\

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original hydrocarbon in place by volumetric method

-N

\ \

✓

/ / / /ÿ

I ✓

✓

/ ✓

✓

I

s /

I

l

/ / /

2

/ / /

D2

3

D3

4

D4

5

D5

6

D6

7

D7

8

D8

9

D9

/ / /

sc*-:

\

I I

\ \ *v

/

/

/

\

\

D1

I I I

/

\

V

1

\

s

\

Depth

l

s /

Well

V

/ /

l

-/

OG

y

/

I /

\

y

/

Structural contour map

6

●

●7 ●4

3● 1●

●

●5

8

Scale:1:50000

Location map

●2 ●

9

1W-. Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original hydrocarbon in place by volumetric method

/ /

oo ✓

t

i

/

I I

✓

I

Q

/

O

/

V Yo

\

\

/

40,

\

10 0

Isopach map

Depth

1

h1

2

h2

3

h3

4

h4

5

h5

6

h6

7

h7

8

h8

9

h9

/

/

'E

30

Well

G Gas Goc

Oil Woc

Water

IfApplied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original hydrocarbon in place by volumetric method

N = BV .φ (1 − S wi ) = ( Ah)φ (1 − S wi ) 43560 Ahφ (1 − S wi ) = 5.615β oi

β g Bbl SCF

β o bbl STB

N=

7758 Ahφ (1 − S wi )

STB

G=

7758 Ahφ (1 − S wii )

SCF

β oi

β gi

A : acres h : ft

φ , S wi : ffractions

IfApplied Reservoir Engineering : Dr. Hamid Khattab

Calculation of ((BV)) using g isopach p map p 1. Trapozoidal method:

An An −1 > 0.5 BV =

h [A0 + 2 A1 + 2 A2 + ...... + 2 An−1 + 2 An ] 2

h′ + [An + A′] 2

C.L

Area inch2

0

Ao WOC

10

A1

20

A2

30

A3

40 50

A4 GOC A5

60

A6

70

A7

76

O

A’

IfApplied Reservoir Engineering : Dr. Hamid Khattab

Calculation of ((BV)) using g isopach p map p C.L

2. Pyramid or cone method

An An −1 ≤ 0.5 h BV = A0 + A1 + A0 . A1 3 h + A1 + A2 + A1 . A2 3 h h + An −1 + An + An −1 . An + [ An ] 3 3

[

[

[

]

]

]

Area inch2

0

Ao WOC

10

A1

20

A2

30

A3

40 50

A4 GOC A5

60

A6

70

A7

76

O

A’

IfApplied Reservoir Engineering : Dr. Hamid Khattab

Calculation of ((BV)) using g isopach p map p 3. Simpson method Odd number of contour lines

h BV = [A0 + 4 A1 + 2 A2 + 4 A3 + ...... + 4 An −1 + 2 An ] 3 h′ + [An ] 3

IfApplied Reservoir Engineering : Dr. Hamid Khattab

Converting map areas (inch2) to acres Say : Scale 1 : 50000

1 inch = 50,000 inch 2 ( (50,000) ) 1 inch 2 = = 398.56acres 144 × 43560

IfApplied Reservoir Engineering : Dr. Hamid Khattab

Converting map areas (inch2) to acres Example 1 :

Given the Gi th following f ll i planimetred l i t d areas of f an oit it of f reservoir. 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

86

Area inch2 : 250 200 140 98

76

40

26

12

5

0

IfApplied Reservoir Engineering : Dr. Hamid Khattab

Converting map areas (inch2) to acres Solution :

10 [250 + 2 × 200 + 2 × 140 + 2 × 98 + 2 × 76 + 2 × 40 + 26 ] 2 10 10 6 + 26 + 12 + 26 × 12 + 12 + 5 + 12 × 5 + 50 + 0 + 50 × 0 2 3 3

VB =

[

]

[

] [

= 7198inch 2 : ft 2 (15,000) 1 inch 2 = = 35.87 acres 144 × 43560

∴ BV = 7198 × 35.87 = 258193.39acres ∴N =

7758 × 258193.39 × 0.25 × (1 − 0.3) = 250.38MMSTB 1.4

]

IfApplied Reservoir Engineering : Dr. Hamid Khattab

Converting map areas (inch2) to acres By using Simpson method

BV = +

[

10 [250 + 4 × 200 + 2 ×140 + 4 × 98 + 2 × 76 + 4 × 40 + 2 × 26 + 4 ×12 + 2 × 5] 3

6 5+ 0+ 5 ×0 3

]

= 7156.6inch 2 . fft = 7156.6 × 35.87 = 256709.6acro. ft

7758 × 256709.6 × 0.25 × (1 − 0.3) ∴N = = 248.94 MMSTB 1 .4

∴ N av = (250.38 + 248.94) 2 = 249.66 MMSTB

IfApplied Reservoir Engineering : Dr. Hamid Khattab

Converting map areas (inch2) to acres Example 2 :

If the th reservoir i of f example l 1 is i a gas reservoir i and d βg=0.001 bbl/SCF. Calculate the original gas in place S l ti : Solution

G=

7758 × 258193.39 × 0.25 × (1 − 0.3) = 350.53MMSCF 0.001

IfApplied Reservoir Engineering : Dr. Hamid Khattab

Converting map areas (inch2) to acres Example 3 :

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

:

Area inch2 :

0(WOC) 10 350

20

30

310 270 220

33(GOC) 40 200

190

50

60

70

76

130 55

25

0

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

fM

Applied Reservoir Engineering : Dr. Hamid Khattab

Converting map areas (inch2) to acres Solution :

BVoil =

10 V ] = 9280inch 2 . ft [350 + 2 × 310 + 2 × 270 + 220] + 3 [220 + 200 2 2

[

V 7 10 10 BVgas = [200 + 190] + [190 + 130] + 130 + 55 + 130 × 55 32 2 2 10 6 + 55 + 25 + 55 × 25 + [25] = 4303.79inch 2 . ft 3 3

[

]

]

(20,000) 2 = 63.77acres 1 inch = 144 × 43560 2

7758 × 9280 × 63.77 × 0.25 × (1 − 0.3) = 618MMSTB 1.3 7758 × 4303.79 × 63.77 × 0.25 × (1 − 0.3) G= = 372.6MMSCF 0.001 N=

If-

aa

Applied Reservoir Engineering : Dr. Hamid Khattab

t

Material Balance ‫ ﻟﺗﺣدﯾد ﻧوع‬Reservoir

drive mechanism

‫ﻻﺑد ﻣن ﻣﻌرﻓﺔ‬

Water reservoir

P

Gas reservoir *

Bg

Gas

Gas Water

without bottom water drive

with bottom water drive Oil reservoir

If\

Applied Reservoir Engineering : Dr. Hamid Khattab

Oil reservoir I

Undersaturated

P>Pb

Oil

Oil

Water

without bottom i water drive

Oil

Depletion drive

Gas liberated res.

with bottom i water drive

Gas Oil

Gas cap drive

i

Saturated

1

P≤Pb

Oil

Gas

Gas Water

Oil W Water

Bottom water drive

Combination drive

If9E&

Applied Reservoir Engineering : Dr. Hamid Khattab

PVT data for gas and oil reservoirs Gas reservoirs

‫أول ﺣﺎﺟﺔ ﺑﺗﺟﯾﻠﻰ ھﯾﺎ دى وﻣن ﺧﻼﻟﮭﺎ‬ ‫ﺑﺣدد ﻧوع اﻟﺧزان اﻟﻠﻰ ﻋﻧدى‬ ‫وف ﺣﺎﻟﺔ اﻟﻐﺎز ھﻼﻗﻰ أﻛﯾد‬ Bg

A

Bg psia : Absolute psi : Absolute psig : Gauge

ZT Bg = 0.00504 P

>

P

P ----- Psia ‫ﻓﻠو ﻟﻘﯾت ﻣﺛﻼ‬ Psig 14.7 ‫ﻻزم ﺗﺟﻣﻊ‬

1W-. Applied Reservoir Engineering : Dr. Hamid Khattab

PVT data for gas and oil reservoirs Saturated oil reservoirs

i\

*ÿ..B t

Boi=Bti Bt = Bo+(rsi-rs)Bg

µo

ZT Bg = 0.00504 P

rsi Bo

/ /

Boi= Bti

rs 1 0

>

Bg

Pi

i

Applied Reservoir Engineering : Dr. Hamid Khattab

PVT data for gas and oil reservoirs A

Undersaturated oil reservoirs <

‫ﻻﺣظ ﻟو اﻟﺗﻐﯾر ف اﻟﺿﻐط طﻔﯾف ﻣﻊ‬ ‫اﻟوﻗت ﯾﺑﻘﻰ ﻛده ﻣﺷﻛﻠﺔ ف دﻗﺔ‬ > Material Balance ‫ﻋﺷﺎن ﻛل اﻟﺑﯾﺎﻧﺎت ﻣﻌﺗﻣدة ﻋل اﻟﺿﻐط‬

*

*

P1 > Pb

* +

+

#

saturated

undersat.

%

♦

*

+

Bt ■

*

*

#

ro'P t si=c

■

µo

"I

*

Bo

♦

♦

♦

/

1 0

/

M-Q

rs Bg

►

IfZ:-

Applied Reservoir Engineering : Dr. Hamid Khattab

Laboratory measurment of PVT data V

Rs=0

V

V V V

Gas SCF

Oil

Gas Oil

STB P = 14.7 psi T = 60o F

Oil

Pb saturated

Oil P > Pb undersaturated

Oil Pi

>

IfZ:-

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original gas in place by MBE A

1. Gas reservoir without bottom water drive

‫أﺑﺳط اﻷﻧواع‬

> Gp

GBgi

(G − G )B

∆T

p

p∠pi

pi GBgi = (G − G p )Bg ∴G =

G p Bg Bg − Bgi

gi

‫ھﻧﺎ ﺑﮭﻣل ﺗﺄﺛﯾر ﺗﻣدد ﻛل ﻣن‬ Connate water Rock

o 1

IfZ:-

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original gas in place by MBE 1. Gas reservoir without bottom water drive Example 4 :

Initial ‫ده ﻛده‬ 0 ‫ﻋﺷﺎن‬Gp

*10^6

Pppsi

G p SCF

Bg bbl SCF

Z

G

4000

0x10 0 -6

0.00077

0.83

0x106

3900

12

0.00084

0.81

201.6

3800

27

0.00089

0.79

200.2

3700

37

0.00095

0.77

195.2

3600

58

0.00107

0.75

199.7

Solution : Using eq. (1)

G ≠ const.

If-

aa

Applied Reservoir Engineering : Dr. Hamid Khattab

A

Calculation of original gas in place by MBE MBE as an equation of a straight line

GBgi = (G − G p )Bg

∴ G p Bg = G (Bg − Bgi ) Another form:

o

G p Bg

G

> 2 x1

A

‫ ھذا اﻟﺧط ﻻﺑد‬: ‫ﻻﺣظ‬ ‫وﻻﻣﻔر ﻣن ﻣروره ﺑﻧﻘطﺔ‬ ‫اﻷﺻل‬

Bg − Bgi

 ZT Z iT   ZT    = G (0.00504 ) G p  0.00504 − p  pi    p

o

 Z Zi  Z ∴ G p = G  − p  p pi 

y1

3

Gp

Z p

y2 G

> x2 Z p−Zi Pi

If-

aa

‫دم أﻛﺛر ف اﻟﺷرﻛﺎت ﻋﻧﮭﺎ وذﻟك ﻟﻛﺛرة اﻟﻧﻘﺎط اﻟﻣﺗﺎﺣﺔ‬Applied ‫ﻘﺔ ﺗﺳﺗﺧ‬Reservoir ‫ذا اﻟطرﯾ‬Engineering ‫ھ‬ : Dr. Hamid Khattab ‫وﻟﻛن ف اﻻﻣﺗﺣﺎن ﺣل ﺑﺎﻟطرﯾﻘﺔ اﻟﺳﺎﺑﻘﺔ‬

Calculation of original gas in place by MBE A

Another form:

GBgi = (G − G p )Bg  Z Z 0 . 00504 i G = (G − G p ) 0 . 00504 pi p  

p Z

at

p =0 Z

G = Gp

\

\

\ \

pi GZ i \

\

y3

G p  pi P  ∴ =  1 − Z G  Z i 

p p pi ∴ = i − Gp Z Z i GZ i

pi Zi

\ \

\

S \

\ \

\ \

\

G 0

x3

Gp

Ifh-

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original gas in place by MBE Example 5 :

Solution :

p

/ using MBE as a straigh line Solve the previous example Bg − Bgi

G pBg

Z P

Z p − Zi Pi

P Z

Gp

4000

―

―x104

2.075x10-4

― x10-5

4819

0x10-6

3900

0.00005

1.068

2.25

1.75

4441

12

3800

0.00012

2.403

2.39

3.15

4177

27

3700

0.00018

3.515

2.66

5.91

3896

37

3600

0.00030

5.990

2.88

8.09

3421

56

y2

x2

y3

x3

x1 From Figgers

y1

>

SCF G = 200×106 STB

‫‪Gas Res. with bottom water drive‬‬ ‫ﺧﻼص ﻛده ھوا ﺣددﻟك ﻣﺗﻌﻣﻠش اﻟﺣرﻛﺔ‬ ‫اﻟﺗﺄﻛﯾدﯾﺔ دى‬

‫ﻻﺣظ ﻟو ﻣداﻟﻛش اﻟﺧزان اﻟﻠﻰ ﻓﯾﮫ ﻏﺎز ده‬ ‫‪With water drive or not‬‬ ‫‪== You should firstly check.‬‬ ‫ﺑﺗﻔﺗرض اﻻول اﻧﮫ ﻣﻔﮭوش ﻣﯾﺎه وﺗﺣط اﻟﻧﻘط وﺗرﺳم‬ ‫طﻠﻌت ﺧط ﻣﺳﺗﻘﯾم ﯾﺑﻘﻰ ﺗﻣﺎم‬ ‫ﻣطﻠﻌﺗش ﺧط ﻣﺳﺗﻘﯾم ﯾﺑﻘﻰ ﻛده ﻋﻧدك ﻣﯾﺎه ﺗﺑدأ ﺗﺣﺳب ﺗﺎﻧﻰ ﺣﺳﺎﺑﺎﺗك‬

‫اﻟﻣﺳﺗوى اﻟﻠﻰ ﺗﺣت اﻟزﯾت ھوا‬ ‫‪Oil Water Contact‬‬ ‫ﺳواء ﺑﻘﮫ ﻛﺎن ﺗﺣت اﻟزﯾت ﻣﯾﺎه وﻻ ﻷ‬ ‫ﻻﺣظ ‪ :‬اﺣﻧﺎ ﻛﺗﺑﻧﺎ‬ ‫‪We : for encroachment‬‬ ‫وﺑﻧﻧطﻘﮭﺎ‬ ‫‪influx or encroachment‬‬ ‫‪Not Wi to be not conflicted with injection‬‬

- .

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original gas inAplace by MBE 2.Gas reservoir with bottom water drive

R

GBg*

Pi

GBgr(ÿGp)pg+(We-WpBw) GjPs-fye-WpBw) G=

>

(G-G.)B

Gp

Wp

(wg-WpBw)

∆T Influx encroachment

Bg-Bgi

we

∴ Assuming

P
=0 causes an increase in G continuously

Bw ‫* ﻣﺑﻧﺿرﺑش ھﻧﺎ‬ ‫ﻻن دى ﻣﯾﺎه ﺗﺣت ﻣطﻠﻌﺗش‬ ‫ﻓوق ﺧﺎﻟص‬

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original gas in place by MBE MBE as a straight line F/ 45o ∴

∴

Assuming

NG

Flowing ‫ودى ﻣش ﺛﺎﺑﺗﺔ ﯾﻌﻧﻰ ھﺗﺧﺗﻠف ﻣن ﻧوع ﻵﺧر‬

is known

Expansion

/

33

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original gas in place by MBE Example 6 : A gas reservoir i with i h a kknown b bottom water d drive i h has the h ffollowing ll i data: =0 and

Ppsi

G p SCF

Bg bbl SCF

We bbl

0

4000

0x109

0.00093

0x10-6

1

3900

27.85

0.00098

2.297

2

3800

72.33

0.00107

7.490

3

3700

113.85

0.00117

13.308

4

3600

151.48

0.00125

18.486

T years

‫ھﻧﺎ ﻣﻠوش ﻻزﻣﺔ ﻋﺷﺎن‬ ‫أﻧت واﺧد ﻗﯾﻣﺔ اﻟﻣﯾﺎه‬ ‫ﻟﻛن ﻟو ﻣش ﻣﻌﺎك‬ ‫ھﺗﻛون ﻟﯾﮭﺎ دور طﺑﻌﺎ‬

Calculate the original gas in place 34

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original gas in place by MBE Solution Tyear

F

Eg

F/Eg x109

We/Eg x109

1

27.2x106

0.00005

546

45.93

2

77.39

0.00014

553

53.04

3

133.20

0.00024

555

55.44

4

189.35

0.00032

554

54.25

F/E g

45

From Fig:

G=500x109

SCF

G=500x109 We/Eg

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original gas in place by MBE Gas Cap Expansion an Shrinkage G c Gp

Gpc gas

expansion GOC

GOC

GOC shrinkage

Oil

Oil

‫ده اﻟﻣﺳﺗوى اﻻﺻﻠﻰ‬

Shrinkage g due to: poor p planning p g or accident and corrosion - Assume gas cap expansion = (G-Gpc).Bg-GBgi - Assume gas cap shrinkage = GBgi - (G-Gp (G Gpc)Bg Gpc: gas produced from the gas cap and my be = zero

Gas Cap ‫اﻟﻐﺎز اﻟﻣﻧﺗﺞ‬ ‫ﻣن‬

‫ھﻌرف ﻣﻧﯾن اﻟﻐﺎز اﻟﻠﻰ طﺎﻟﻊ ده طﺎﻟﻌﻠﻰ ﻣن‬ ‫!! ‪Gas Cap or oil zone‬‬ ‫ﻣﻊ اﻟﻌﻠم ان ﻛده ﻛده اﻟزﯾت ﺑﯾﻛون ﻓﯾﮫ ﻏﺎز داﯾب‬ ‫دى ﺑﻘﮫ ﺑﺗﯾﺟﻰ ﻣن اﻟﺧﺑرة ﯾﻌﻧﻰ أﻧت ﻋﺎﻣل‬ ‫‪ PVT‬وﻋﺎرف ﺗﻛوﯾن اﻟﻐﺎز اﯾﮫ اﻟﻠﻰ داﯾب ف اﻟزﯾت‬ ‫وﻏﺎﻟﺑﺎ ﺑﯾﻛون اﻟﻐﺎز اﻟﻠﻰ داﯾب ف اﻟزﯾت ده ﻣﻛوﻧﺎﺗﮫ أﺗﻘل ﻣن اﻟﻠﻰ ف طﺑﻘﺔ اﻟﻐﺎز‬

‫ﻣﻣﻛن ﺗﻼﻗﻰ اﻟﻐﺎز ﻋﻠﻰ اﻟﺳطﺢ ﻧﺗﯾﺟﺔ‬ ‫إﻧﮫ ﻓﯾﮫ ﻣﺛﻼ‬ ‫‪- failure in the Csg.‬‬ ‫ودى ھﺗﻛون ﻣﺻﺎﺣﺑﺔ ‪Shrinkage‬‬ ‫اﻟﻐﺎز ﺗﻣدد أﺻﻼ ووﺻل ﻷول ‪-‬‬ ‫‪Perforation‬‬ ‫وده ﻣﺻﺎﺣب ‪Expansion‬‬

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original gas in place by MBE Example: 7 Calculate the gas cap volume change if G=40x109 SCF

Cumulative

P

Gpc x109

Bg

4000

0

0.0020

3900

4

0 0022 0.0022

3800

7

0.0025

3700

10

0.0028

3600

13

0.0031

3500

17

0.0035

‫ ﻣش ﻣﮭم‬... ‫ﻟو ﻣداﻟﻛش ﻗﯾﻣﺗﮫ‬ ‫أﺻﻼ ﻹن أﻧت ھدﻓك ﺗﻌرف‬ ‫ھﯾﺗﻣدد وﻻ ھﯾﻧﻛﻣش وﺑﻣﻘدار‬ ‫أد اﯾﮫ‬

‫أى ﺗﻐﯾر طﻔﯾف ﻧﺗﯾﺟﺔ اﻟﺧطﺄ‬ ‫ ف ﻗﯾم‬Bg ‫ھﯾﻛون اﻟﺧطﺄ واﺿﺢ ف‬ ‫ﺣﺳﺎﺑﺎﺗك‬

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original gas in place by MBE Solution Assuming gas cap expansion = (G-Gpc).Bg-Ggi

‫أول ﻗﯾﻣﺔ ﺳﺎﻟب‬ ‫وﺑﻌد ﻛده ﻣوﺣب‬ ‫ﯾﺑﻘﻰ اﻟﻧﻘط دى ﻓﯾﮭﺎ‬ ‫ﻣﺷﻛﻠﺔ اﻣﺎ‬ production data PVT data 4000 ‫اﻟﻘﯾم وﺻﻠت ل‬ ‫وﺑﻌد ﻛده ﻗﻠت ﯾﺑﻘﻰ ﯾﺎ‬ ‫ﻣﯾرو ﻧﻔس اﻟﻣﺷﺎﻛل اﻟﻠﻰ‬ ‫ﻓوق‬

Pressure

Gas cap change x103

type

4000

-

-

3900

-800

shrinkage

3800

+2500

expansion

3700

+4000

expansion

3600

+3700

shrinkage

3500

+5000

expansion

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

‫ﻻﺣظ ‪ :‬ﻟو اﻟﺿﻐط اﻻوﻟﻰ ﻋﻧدك ﻛﺎن ‪ 5000‬وادﯾك اﻟﺑﯾﻧﺎت ﺑﺗﺎﻋﺔ اﻻﻧﺗﺎج وأﻧت ﻟﻘﯾت إن ﻗﯾﻣﺔ‬ ‫‪ Pb =4700‬اﻟﻠﻰ ﻋﻧدھﺎ ﻛﺎﻧت ‪ B‬أﻛﺑر ﻗﯾﻣﺔ ﻓﻠﻣﺎ ﺗﺣل ﻣش ھﺗﻌرف ﺗﺣﺳب ف اﻟﻔﺗرة اﻟﺻﻐﯾرة دى‬ ‫ﻓﺗﻔﺗرض ان اﻟﺿﻐط اﻷوﻟﻰ ﻟﻠﺧزان ‪ 4700‬وﺗﺷﺗﻐل ع ھذا اﻻﺳﺎس وﺗﺻﻔر اﻻرﻗﺎم ﺑﺗﺎﻋﺔ اﻻﻧﺗﺎج ﯾﻌﻧﻰ ﺗطرح ﻣن‬ ‫‪cumulative - production at Pb‬‬ ‫‪Boi : Bo but at Pb‬‬ ‫وﻟﻣﺎ ﺗطﻠﻊ ﻗﯾﻣﺔ اﻟزﯾت ﻣﺗﻧﺳﺎش ﺗﺑﻘﻰ ﺗﺟﻣﻊ ﻋﻠﯾﮫ ﻛﻣﯾﺔ اﻻﻧﺗﺎج اﻟﻠﻰ طﻠﻌﻠك ﻗﺑل ﻣﺎ ﺗوﺻل ﻟل ‪4700‬‬ ‫طب ﻟو اداﻟك ‪ 5‬ﻗﯾم ﻓوق ‪ Pb‬وﻧﻘطﺗﯾن ﻣﺛﻼ ﺑﻌدھﺎ ﺧﻼص اﺷﺗﻐل ع اﻟﺧﻣس ﻧﻘط وﺧﻼص‬ ‫وﺗﻌﺎﻟﻰ ﻋﻧد آﺧر ﻗﯾﻣﺔ ﻟﻼﻧﺗﺎج واﺟﻣﻌﮭﺎ ﻋل اﻟﻧﺎﺗﺞ اﻟﻧﮭﺎﺋﻰ‬ ‫طب ﻟو ادﯾﺗﻠك ‪ 3‬ﻧﻘط ﻓوق و ‪ 3‬ﺗﺣت ﯾﺑﻘﻰ ﺑﺗﺷﺗﻐل ﻋل ‪ 3‬دول ﻟوﺣدھم و ‪ 3‬دول ﻟوﺣدھم‬ ‫واﻟﻔرق ﺑﯾن اﻻﺗﻧﯾن ھوا ﻗﯾﻣﺔ اﻻﻧﺗﺎج ﻋﻧد‬ ‫‪Pb‬‬

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation C l l ti of f original i i l oil il iin place l b by MBE a) Under-saturated oil reservoirs Characteristics - P>Pb - No free gas, no Wp - Large volume - Limited K - Low flow rate - Produce by Cw and Cf

‫ أﻧﺎ ﻣش ﻋﺎﯾزه ﯾوﺻل ﺑﺳرﻋﺔ ﻟﻠﺿﻐط‬Pb ‫ﻋﻠﺷﺎن ﺗﻌرف ﺗﺣﺳب ﺑراﺣﺗك وده ﺑﻘﮫ ﯾﺣﻘق‬ ‫اﻟﺷروط دى‬

compressibility of connate water ‫ﻣﯾﻧﻔﻌش ﺧﺎﻟص ﺗﮭﻣل ھذه‬ ‫اﻟﺗﺄﺛﯾرات‬

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation C l l ti of f original i i l oil il iin place l b by MBE 1- Under-saturated oil reservoirs without bottom water Np (N-Nip)Bo

NBoi

P>Pb

Pi>P Pb neglecing Cw and Cf NBoi=(N-Np)Bo

∴N =

N p Bo Bo − Boi

‫ده ھﻧﺎ ﺑس ﻟﻛن ﻣش ھﻧﮭﻣﻠﮭﺎ ﺧﺎﻟص ﺑﻌد ﻛده‬ ‫واﻟﻣﺛﺎل ده ﺗوﺿﯾﺣﻰ ﻓﻘط ﻟﻛن ﻣش ده اﻟﻣظﺑوط‬‫وھذه اﻟﻣﻌﺎدﻟﺔ ﻻ ﺗﺳﺗﺧدﻣﮭﺎ ﺗﺎﻧﻰ ﺧﺎﻟص‬

(1)

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation C l l ti of f original i i l oil il iin place l b by MBE Example: 8 Calculate C l l t the th original i i l oil il in i place l assuming i no water t drive d i and d neglecting l ti Cw and Cf using the following data

P

Np x106

Bo

4000

0

1.40

3800

1 535 1.535

1 42 1.42

3600

3.696

1.45

3400

7.644

1.49

3200

9.545

1.54

‫ﻣن ھذه اﻟﻘﯾم ﺗﻌرف ﻧوع اﻟﺧزان‬ ‫اﻟﻘﯾم ﺗﺗزاﯾد ﯾﺑﻘﻰ ﺗﻣﺎم‬

Applied Reservoir Engineering : Dr. Hamid Khattab

Solution

Calculation C l l ti of f original i i l oil il iin place l b by MBE

Pressure

NpBo x106

Bo-Boi

N x106

4000

-

-

-

3800

2.179

0.02

108.95

3600

5 539 5.539

0 05 0.05

110 78 110.78

3400

11.389

0.09

126.64

3200

14.699

0.14

104.99

rearrange MBE as a straight line NBoi = (N-Np)Bo

N ≠ const

F

F = NEo

From Fig: N =≠ 110 x10 STB 6

N Eo

‫‪Applied Reservoir Engineering : Dr. Hamid Khattab‬‬

‫‪Calculation‬‬ ‫‪C l l ti of‬‬ ‫‪f original‬‬ ‫‪i i l oil‬‬ ‫‪il iin place‬‬ ‫‪l‬‬ ‫‪b‬‬ ‫‪by MBE‬‬ ‫‪o.b.p=1 psi/ftD‬‬

‫‪Considering Cw and Cf‬‬ ‫‪- overburden pressure = 1 psi/ftD‬‬ ‫‪- rock strength = 0.5 psi/ftD‬‬ ‫‪- reservoir‬‬ ‫‪r s rv ir pressure‬‬ ‫‪pr ssur = 0.5‬‬ ‫‪0 5 psi/ftD‬‬

‫اﻟزﯾت ﻛده ﺑﯾﺣﺑس ﻧص اﻟظﻐط‬ ‫اﻟﻠﻰ ﻧﺎﺗﺞ ﻣن اﻟطﺑﻘﺎت اﻟﻌﻠﯾﺎ‬

‫اﻟﺻﺧر ﺑﯾﺷﯾل ﻛده ﺗﻘرﯾﺑﺎ ﻧص‬ ‫وزن اﻟطﺑﻘﺎت اﻟﻌﻠﯾﺎ‬

‫ﻣﻊ اﻻﻧﺗﺎج اﻟﺿﻐط ﺑﺗﺎع اﻟزﯾت ﺑﯾﻘل ﻓﺑﺎﻟﺗﺎﻟﻰ اﻟﻣﺟﻣوع ﻣش ﺑﯾﺳﺎوى ‪ 1‬ﺑﺗﺎع ﺿﻐط اﻟزﯾت وﺿﻐط اﻟﺻﺧر‬ ‫‪ :‬ﻓﺗﺣﺻل اﻣﺎ‬ ‫‪o.b.p‬‬ ‫ﻗﺷرة اﻟﻣﯾﺎه اﻟﻠﻰ ﺣوﻟﯾن اﻟﺻﺧر ﺗﺗﻣدد ‪ ---‬ﺣﺑﯾﺑﺎت اﻟﺻﺧر ﻧﻔﺳﮭﺎ ﺗﺗﻣدد ‪ ---‬اﻻﺗﻧﺗﯾن ﯾﺗﻣددوا ﻣﻊ ﺑﻌض‬ ‫ﺑﯾﻌوﺿوا ف اﻟﺑداﯾﺔ اﻟﺗﺄﺛﯾر ﺑﺗﺎع اﻟﺳﺣب ﺑس ﺑد ﻛده ﻣﻣﻛن ﻻ‬ ‫ﻟذﻟك ﻧﺗﯾﺟﺔ اﻟﺗﻣدد ده ﺑﻌد ﻛده ﺑﺗﻼﻗﻰ ف ﺣﯾﺎة اﻟﺧزان ف اﻵﺧر ﺷروخ‬ ‫‪ crushes‬وﺑدأت ﻓواﻟق ﺗظﮭر ﻣﻛﺎﻧﺗش ﻣوﺟودة ﻗﺑل ﻛده‬ ‫وﻻﺣظ ده ﻻزم ﯾﺗم ﻗﺑل اﻟﺧزان ﻣﺎ ﯾوﺻل ﻟﻠﺿﻐط اﻟﺑﺧﺎرى ﯾﻌﻧﻰ ﻣﻔﯾش ﻏﺎز اﺗﻛون ﻋﺷﺎن ﻣﯾﺎﺧدش ھوا اﻟﺣﺟم‬ ‫ده‬

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation C l l ti of f original i i l oil il iin place l b by MBE Considering Cw and Cf

pore volume

NB oi = ( N − N p ) Bo + ∆ Vp w , f ∆ Vp w , f = ∆ Vp w + ∆ Vp f

‫ اﻟﻣﻔروض ﻓﯾﮫ اﺷﺎرة‬‫ ﺑس اﻧﺎ ﺑدﻟﻠت ﻓرق‬V

1 dVp f Cf= . → dVp f = C f V p dp V p dp 1 dVp w C w= . → dVp w = C wVw dp V w dp Vw Sw = → Vw = S wV p → dVp w = C w S wV p dp Vp

NBoi Pi>Pb

(N-Np)Bo ∆Vp,,w P>Pb

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation C l l ti of f original i i l oil il iin place l b by MBE Considering Cw and Cf

∴ dVp f , w = (Cw S w +C f )V p dp NBoi NBoi = Vp (1 − S w ) → Vp = (1 − S w ) ∴ dVp f , w = (

C w S w +C f 1 − Sw

) NBoi dp

∴ NBoi = ( N − N p ) Bo + (

C w S w +C f 1 − Sw

) NBoi ∆p

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation C l l ti of f original i i l oil il iin place l b by MBE Considering Cw and Cf

∴N = B o − B oi + ( Q Co =

1− Sw

f

) B oi ∆ p

B o − B oi → B o − B oi = C o B oi ∆ p B oi ∆ p

∴N = [C o + where

N p Bo CwSw + C

N p Bo CwSw + C 1− Sw

= f

] B oi ∆ p

[

So = 1− Sw

N p Bo ∴N = C S + CwSw + C f [ o o ] B oi ∆ p 1− Sw N =

N p Bo B oi C e ∆ p

(2)

CoSo + 1− Sw

N p Bo CwSw + C 1− Sw

f

] B oi ∆ p

Effective compressibility

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation C l l ti of f original i i l oil il iin place l b by MBE Considering Cw and Cf

∆ P = Pi − P B − Boi Co = o Boi ∆ P ‫ﺗﺣﺳب ﻋﻧد ﻛل‬ ‫ﻗﯾﻣﺔ ﺿﻐط‬

C

f

Co = −

Pi Pi

Voi

Vo

=

1 V − Vi . V oi P − Pi

=

1 ( B o − B oi ) . B oi ∆P ‫ھﻧﺎ‬ ‫ ﺛﺎﺑﺗﺔ‬Rs

= f (φ )

C w = f ( P , T , r s and salinity

1 dV . V oi dP

)

From the following charts ‫ھﻧﺎ ﺑﻌﺗﺑر اﻟﺣرارة ﺛﺎﺑﺗﺔ‬

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation C l l ti of f original i i l oil il iin place l b by MBE

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation C l l ti of f original i i l oil il iin place l b by MBE Example: 9 Solution P

Solve l example l (8) ( ) considering d the h effect ff of f Cw and d Cf fresh water

∆P=(Pi-P) ∆P=(Pi P) C o =

B o − B oi B oi ∆ p

Cwp(Fig.4) (Fig 4)

rsf(Fig.1) (Fig 1)

4000

―

―

2.9x10-6

18

3800

200

7.143x10-5

2.93

17.2

3600

400

8.928

2.95

16.8

3400

600

10.714

2.98

16

3200

800

12.500

3.00

15.2

R1(Fig 2) R1(Fig.2)

0.85

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation C l l ti of f original i i l oil il iin place l b by MBE Continue correction for Rs

Correction for comp.

fresh without salt and gas

P

rs= rsf x R1

R2 (Fig.3)

Cw=CwpxR2

Co So + Cw S w + C f 1− Sw

4000

15.3

1.4

3.30x10-6

―

3800

14.62

1.13

3.311

7.725x10-5

3600

14.28

1.11

3.247

9.570

2400

13.60

1.104

3.289

13.569

3200

12.92

1.09

3.17

13.143

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation C l l ti of f original i i l oil il iin place l b by MBE Continue

N=

N p Bo BoiCe ∆p

P

NpBo

NBoiCe∆P

N

4000

―

―

―

3800

2.179x016

0.0218

108.2x106

3600

5.359

0.0536

107.9

3400

11.389

0.1131

106.5

3200

14 699 14.699

0 1470 0.1470

105 1 105.1

N

≠ C

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation C l l ti of f original i i l oil il iin place l b by MBE Use MBE as a straight line as follows:

N p Bo = NBoi Ce ∆P

F = NEo Plot the fig. fig

F = N p Bo N = 100 × 10 6

N = 100× 106 STB ‫ﻻﺣظ ﻟﻣﺎ أھﻣﻠت ﺗﺄﺛﯾر ﺗﻣدد اﻟﻣﯾﺎه‬ ‫و اﻟﺻﺧر ﻛﺎﻧت اﻟﻘﯾﻣﺔ‬ 110*10^6

Eo = Boi Ce ∆ ∆P P

Applied Reservoir Engineering : Dr. Hamid Khattab

U d Undersaturated t t d oil il reservoir i with ith b bottom tt water t Np

Wp NBoi

Pi>Pb Assuming (We) is known and neglect Cw+Cf

NBoi = (N − N p )Bo + (We − w p Bw ) ∴N =

N p Bo − (We − w p Bw ) Bo − Boii

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

(N-Np)Bo

P>Pb

Applied Reservoir Engineering : Dr. Hamid Khattab

U d Undersaturated t t d oil il reservoir i with ith b bottom tt water t 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 5.362

1 42 1.42

2 416 2.416

3400

10.033

1.49

3.561

3200

12 682 12.682

1 54 1.54

4 832 4.832

Applied Reservoir Engineering : Dr. Hamid Khattab

U d Undersaturated t t d oil il reservoir i with ith b bottom tt water t Solution :

N=

N p Bo − (We − w p Bw ) Bo − Boi

P

NpBo

Bo-Boi

N

4000

―x10 106

―

―x10 106

3800

3.314

0.02

108.5

3600

7 775 7.775

0 05 0.05

107 1 107.1

3400

14.950

0.09

126.5

3200

19 531 19.531

0 14 0.14

105 0 105.0

N≠C

Applied Reservoir Engineering : Dr. Hamid Khattab

U d Undersaturated t t d oil il reservoir i with ith b bottom tt water t Rearrange MBE as a straight line

F

N p Bo + W p Bw = N [Bo − B0i ] + We F = N Eo

Eo 45 o

+ We

∴ F Eo = N + We Eo

N = 110

We Eo

p

E o = [B o − B 0 i ]

F = N p Bo

F Eo

4000 3800 3600 3400 3200

― 0.02 0 05 0.05 0.09 0.14

― x10-6 3.314 7 775 7.775 14.980 19.531

― 165.7 155 5 155.5 166.4 139.5

We Eo

― x10-6 56.75 48 32 48.32 39.56 34.51

Applied Reservoir Engineering : Dr. Hamid Khattab

Undersaturated oil reservoir with bottom water Example 11 : Solution So ut on :

P

Solve examole (10) considering Cw and Cf effect

Cw, Co, Cf and Ce are the same as example (9)

∆P

Ce

Boi C e∆ P

F N P Bo = Eo B oi C e ∆ P

We We = Eo B oi C e ∆ P

4000

― x10-5

―

―

―

―

3800

7.785

200

0.0218

152.01 x106

52.06 x106

3600

9.570

400

0.0536

145.06

45.07

3400

13.568

600

0.1139

131.25

31.26

3200

13.143

800

0.1470

132.86

32.87 F

Plot

N P Bo F = E o Boii C e∆P

As in Fig.

vs

N = 100×106

We We = E o Boii C e∆P

Eo

45 o

N = 100 × 10 6

We E o

Applied Reservoir Engineering : Dr. Hamid Khattab

B S B. Saturated t t d oil il reservoirs i 1 Depletion 1. D l ti d drive i reservoirs i Characteristics

• P ≤ Pb • Wp = 0

‫ف اﻟﻐﺎﻟب ﻣﻔﯾش ﻣﯾﺎه ﺑﺗﻛون ﻣوﺟودة‬

• R p increases rapidly

producing gas oil ratio

• low R.F Rp : producing GOR ( Rp ) instantaneous Rs

‫ ﻛل اﻟزﯾت اﻟﻠﻰ أﻧﺗﺞ‬/ ‫ﻛل اﻟﻐﺎز اﻟﻠﻰ أﻧﺗﺞ ﻟﺣد دﻟوﻗﺗﻰ‬ ‫ف اﻟﻠﺣظﺔ اﻟﻠﻰ أﻧﺎ ﺑﺗﻛﻠم ﻓﯾﮭﺎ طﺎﻟﻊ ﻛﻣﯾﺔ ﻏﺎز أد إﯾﮫ وﻛﻣﯾﺔ زﯾت أد إﯾﮫ‬ ‫اﻟذوﺑﺎﻧﯾﺔ ﺑﺗﺎﻋﺔ اﻟﻐﺎز ف اﻟزﯾت‬

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original oil in place by MBE ‫ﺟزء ﺑﯾﺧرج ﻣﻊ اﻻﻧﺗﺎج زى اﻟرﻏﺎوى‬

Np

Gp

NBoi

(N − N )B p

∆T

Free gas

SCF/ STB

p

P≤ i P b

NBoi = (N − N p )Bo + free gas free gas = Nr si − (N − N

[

p

)r

s

− N pRp

SCF

]

∴ NBoi = (N − N p )Bo + Nrsi − (N − N p )rs − N p R p Bg ∴N =

o

[

N p Bo + (R p − rs )Bg

]

Bo − Boi + (rsi − rs )Bg

‫ ﺣﺎطط‬: ‫ﻻﺣظ‬ ‫اﻟﻐﺎز ﺗﺣت اﻟزﯾت‬ ‫ﻋﻠﺷﺎن ﻣﺗﻔﺗﻛرھﺎش‬ ‫إﻧﮭﺎ‬ gas cap ‫ده ﻓﻘﺎﻋﺎت ﻏﺎز‬ ‫ﻣﻧﻔﺻﻠﺔ ﻋن‬ ‫ﺑﻌﺿﮭﺎ اﻟﺑﻌض ﻟو‬ ‫اﺗﺟﻣﻌوا ﻣﻊ ﺑﻌض‬ ‫ﯾﻌﻣﻠوا اﻟﺣﺟم ده‬ ‫ﻻﻧﮭم ﻟﺳﮫ‬ ‫ﻣوﺻﻠوش ﻟل‬ critical saturation

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original oil in place by MBE Example 12 :

‫ﺑﺗزﯾد ﻛﻠﻣﺎ ﻗل اﻟﺿﻐط وذﻟك ﻟزﯾﺎدة ﻛﻣﯾﺔ‬ ‫اﻟﻐﺎز اﻟﻣﻧﻔﺻﻠﺔ‬

‫ﻛل ﻣﺎ اﻟﺿﻐط ﯾﻘل ﻛل ﻣﺎ ﺗﻘل اﻟذوﺑﺎﻧﯾﺔ‬

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

NP

RP

Bo

Bg

rs

N

4000

― x106

718

1.492

0.001041

718

― x106

3800

3 87 3.87

674

1 423 1.423

0 001273 0.001273

614

3600

5.26

1937

1.355

0.001627

510

3400

6.44

3077

1.286

0.002200

400

Solution

P

91 50 91.50 96.02 96.01

As shown N ≠ const., so rearrange MBE as a straight line ‫اﻟﻘﯾم ﺑﺗﻘل ﯾﺑﻘﻰ ع طول‬ saturated

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original oil in place by MBE

[

]

[

N p B o + (R p − rs )B g = N B o − B oi + (rsi − rs )B g F

=N

]

Eo

Solution : P

F

Eo

4000

00x10 106

0

3800

5.802

0.0634

3600

19 339 19.339

0 2014 0.2014

3400

46.124

0.4804

From Fig : N = 96 × 10 STB 6

F N = 96 × 10 6

Eo

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original oil in place by MBE

R .F =

N

p

N

=

B o − B oi + (r si − r s )B g B o + (R

R . F = f (P & R P

R.F ∝ 1 RP To increase R.F: • • • •

p

− r s )B g

) production data

‫اﻟﺗﺟﻛم ف اﻟﺿﻐط ﻛش‬ ‫ھﻌرف اﻻ ﻟو ﻋﻣﻠت ﺑﻘﮫ‬ secondary recovery

Working over high producing GOR wells Shut-in ,, ,, ,, ,, ,, Reduce (q) of ,, ,, ,, ,, R i j t some of Reinject f gas produced d d

‫ﻣﺗﻧﻔﻌش ھذه اﻟطرﯾﻘﺔ ف ﺣﺎﻟﺔ‬ ‫‪depletion drive‬‬ ‫ﻟذﻟك ﻻ ﯾﻧﺻﺢ ﺑﻔﻌل ھذا ف ھذه اﻟﺣﺎﻟﺔ‬

‫ﻓﯾﮫ ﻧﺎس ﺑﻘﮫ ﺑﯾﻘﻠك ﻻ اﺣﻧﺎ ﻣﻣﻛن ﻧﺿﺦ ﻏﺎز ف اﻣﺎﻛن ﻗرﯾﺑﺔ ﻣن ﺑﻌﺿﮭﺎ‬ ‫ﺑﺣﯾث ﻧﻌﻣل‬ ‫‪artificial gas cap‬‬ ‫وﻣﻣﻛن اﻟﻐﺎز اﻟﻠﻰ ﺑﯾﻧﻔﺻل اﺛﻧﺎء اﻻﻧﺗﺎج ﯾﺗﺟﻣﻊ وﯾﺗﺣد ف اﻋﻠﻰ اﻟطﺑﻘﺔ‬ ‫‪ secondary gas cap‬وﯾﻛون‬ ‫ﻟو ﻋﻧدك أﺻﻼ‬ ‫‪gas cap‬‬ ‫وﻋﻧدك طﺑﻌﺎ اﻟﻐﺎز ﻣﻊ اﻻﻧﺗﺎج ﺑﯾﻧﻔﺻل وﯾورح ﻟﻣﻧطﻘﺔ اﻟﻐﺎز ﻓﻼزم ﺗﺗﺎﺑﻊ ﻛوﯾس ﺟدا ﺟدا‬ ‫ان اﻟﻐﺎز ﻣوﺻﻠش‬ ‫‪critical saturation‬‬ ‫وده ﻋﺷﺎن ﻣﺗﺧﺳرش طﺎﻗﺔ اﻟﺧزان ‪ pvt‬واﻧت ﺑﺗﺑﻘﻰ ﻋﺎرﻓﮭﺎ ﻣن‬

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original oil in place by MBE Example 13 :

Solution :

gas injection

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

Sg

[

free gas = pore volume

]

free gas = Nrsii − (N − N p )rs − N p R p B g

‫ﻟو ﺳﺄﻟك ھل ﯾﻧﻔﻊ ﺗﺿﺦ ﻛل ھذه اﻟﻛﻣﯾﺔ ؟‬ ‫ ﯾﺑﻘﻰ ﻻزم ﺗﺣﺳب‬critical gas saturation ‫وﺗﻘﺎرن‬

96 × 10 6 × 718 − (96 − 6.44 )× 10 6 × 406 −  6 = × 0 . 0022 = 28 . 05 × 10 bbls  6 6.44 × 10 × 3077  

NB oi 96 × 10 6 × 1.492 pore volume = = = 204 .62 × 10 6 bbls (1 − 0.3) (1 − S w ) ∴ Sg

28 . 05 × 10 6 = = 0 . 137 = 13 . 7 % 204 . 62 × 10 6

‫ﯾﺑﻘﻰ ﻟو ﻋرﻓت إن‬ ‫‪critical gas saturation = ...‬‬ ‫ﯾﺑﻘﻰ اﻟﻣﻘﺎم اﻟﻠﻰ ھوا ﺣﺟم اﻟﻔراﻏﺎت ﺛﺎﺑت ﻣش ھﻌرف اﻏﯾره‬ ‫اﻟﺑﺳط اﻟﻠﻰ ھﺑدأ اﻏﯾره ﺑﺣﯾث اﻧﮫ ﻣﯾوﺻﻠش ﻟﻠﻘﯾﻣﺔ اﻟﻛرﯾﺗﯾﻛﺎل‬ ‫ھﻼﺛﻰ إﻧﻰ ﻣش ھﻌرف اﻏر أى ﺣﺎﺟﺔ ﻏﯾر‬ ‫‪Rp‬‬ ‫اﻟﻠﻰ ﺑﺗﺣﻛم ﻓﯾﮭﺎ ﺑﻛﻣﯾﺔ اﻟﻐﺎز اﻟﻠﻰ ﺑرﺟﻌﮫ ﺗﺎﻧﻰ ﻟﻠﺧزان وﺑﻛده ﻻزم‬ ‫ﻣﺗﻌدﯾش اﻟﻘﯾﻣﺔ دى ﯾﺎ ﺣﺞ أﻣﯾر ﻋﻠﺷﺎن ﻣﺗوﺻﻠش ان اﻟﻠﻰ ﺑﺗﺿﺧﮫ‬ ‫ﺑﯾطﻠﻌﻠك ﺗﺎﻧﻰ ع اﻟﺳطﺢ وﻛﺄﻧﮭﺎ ارﺑﺔ ﻣﺧروﻣﺔ‬

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original oil in place by MBE

R . F without

Gi

B o − B oi + (rsi − rs )B g

=

B o + (R p − rs )B g

1 . 286 − 1 . 492 + (718 − 406 ) × 0 . 0022 1 . 286 + (3077 − 406 ) × 0 . 0022 = 0 . 067 = 6 . 7 % =

R.Fwith 60% Gi = 60 ‫ھﺿﺦ‬% ‫ﯾﺑﻘﻰ اﻟﻠﻰ ﺑﯾطﻠﻊ ﻓوق‬ 40%

Bo − Boi + (rsi − rs )Bg Bo + (R p − rs )Bg

1.286 − 1.492 + (718 − 406 )× 0.0022 = 1.286 + (0.4 × 3077 − 406 )× 0.0022 = 0.1549 = 15.49%

Applied Reservoir Engineering : Dr. Hamid Khattab

2 Gas Cap reservoir 2. 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/qo

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE Np

Gp

GB gi

G B gi m= NBoi

free gas (N-Np)Bo

NBoi Pi

P>Pb

GBgi + NBoi = (N − N p ) Bo + free gas free gas = [Nr

si

[

+ G ] − (N − N

p

)r

s

− N

p

R

p

]

N p Bo + (R p − rs )Bg ∴N = B Bo − Boi + (rsi − rs )Bg + m oi (Bg − Bgi ) Bgi   mNBoi ∴ mNB NBoi + NBoi = (N − N p ) Bo + p  Nr N si + − (N − N p )rs − N p R p  Bg Bgi   This equation contains two unknown (m and N)

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE ‫ﻧﻔس اﻟﻣﻌﺎدﻟﺔ اﻟﺳﺎﺑﻘﺔ وﻟﻛن‬ ‫زاد ھذا ﻟﺟزء اﻟﻣظﻠل‬

Rearrange MBE to give a straight line equation

[

]

[

]

N p Bo + (R p − rs )Bg = N Bo − Boi + (rsi − rs )Bg + F

F = NEo + GE g

mNBoi (Bg − Bggi ) Bgi Eo

G

Eg F ∴ = N +G Eo Eo

2 ‫ھذه اﻟﻣﻌﺎدﻟﺔ ﻓﯾﮭﺎ‬ ‫ﻣﺟﮭوﻟﯾن ﻟذﻟك ﻻ‬ ‫ﯾﺳﺗﺧدم ھذه اﻟطرﯾﻘﺔ اﻻ‬ ‫اذا ﻛﺎﻧت ﻣﺳﺄﻟﺔ اﻣﺗﺣﺎن‬

N

Eg Eo

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE Example 14 : Calculate (N) and (m) for the following gas cap reservoir

P

Np

Rp

Bo

rs

Bg

4000

―x106

510

1.2511

510

0.00087

3900

3.295

1050

1.2353

477

0.00092

3800

5 905 5.905

1060

1 2222 1.2222

450

0 00096 0.00096

3700

8.852

1160

1.2122

425

0.00101

3600

11.503

1235

1.2022

401

0.00107

3500

14.513

1265

1.1922

375

0.00113

3400

17.730

1300

1.1822

352

0.00120

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE Solution :

∴F

E  = N + G g Eo Eo  

P

F

Eo

Eg

F/Eo

Eg/Eo

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 41.130

0 1207 0.1207

0 00033 0.00033

340 7 340.7

0 0027 0.0027

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE F

From Fig. N = 115 x

106

Eo

G = 826 × 10 9

STB

N = 115 × 10 6

6 mNB m × 115 × 10 ×1.2511 9 oi G = 826 × 10 = = Bgi 0.00087

∴ = 0.5 ∴m

E g Eo

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE Another solution Assume several values of (m) until the straight line going through the origin as follows:

F = NEo + GE g

mNBoi = NEo + Eg Bgi

 mBoi  F = N  Eo + Eg  Bgi  

‫ده اﻟﻣﺟﮭول اﻟوﺣﯾد‬ m

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE

P

Eo +

F

mB oi Eg B gi

m = 0.4

m = 0.5

m = 0.6

4000

0x106

0

0

0

3900

5.807

0.043

0.051

0.057

3800

10 671 10.671

0 081 0.081

0 093 0.093

0 106 0.106

3700

17.302

0.127

0.147

0.167

3600

24 094 24.094

0 183 0.183

0 211 0.211

0 240 0.240

3500

31.898

0.243

0.244

0.318

3400

41.130

0.311

0.358

0.405

‫‪Applied Reservoir Engineering : Dr. Hamid Khattab‬‬

‫‪Calculation of OOIP by‬‬ ‫‪y MBE‬‬ ‫‪F‬‬ ‫‪From Fig.‬‬ ‫‪m = 0.5‬‬ ‫‪05‬‬

‫‪N = 115 x 106 STB‬‬

‫اﻟﻘﯾﻣﺔ اﻟﻛﺑﯾرة‬ ‫ﺑﺗﻐرق‬

‫‪mB oi‬‬ ‫‪Eg‬‬ ‫‪B gi‬‬

‫‪Eo +‬‬

‫ﺗﺳﺗﺧدم ھذه اﻟطرﯾﻘﺔ ف اﻟﺷرﻛﺎت‬ ‫او ھوا ﻟو ﻗﺎﻟك اﺷﺗﻐل ع اﻟﻘﯾﻣﺔ دى او دى‬ ‫ﺑس ﻋﯾب اﻟﺷﻐل ﺑﺗﺎﻋﺗﻧﺎ ان اﻟﺑﯾﻧﺎت ﺑﺗﺎﻋﺗﻧﺎ ﻛﻠﮭﺎ‬ ‫ﺑﺗﯾﺟﻰ ف اﻟﺣﺗﺔ ااﻟﻠﻰ‬

Applied Reservoir Engineering : Dr. Hamid Khattab

3. Water drive reservoirs

Edge water

Finite

Bottom water

Infinite

Oil

W

Finite

Infinite

Oil

W

Water

Applied Reservoir Engineering : Dr. Hamid Khattab

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

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE (N-Np)Bo NBoi

free gas

NBoi = (N − N p )Bo + (We − w p Bw ) + free gas free gas = Nrsi − (N − N p )rs − N p R p

[

]

∴ NBoi = (N − N p )Bo + (We − w p Bw ) + Nrsi − (N − N p )rs − N p R p Bg

∴N =

[

]

N p Bo + (R p − rs )Bg − (We − w p Bw ) Bo − Boi + (rsi − rs )Bg

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE Rearrange MBE as an equation of a straight line:

[

]

[

]

∴ N p Bo + (R p − rs )B g + w p Bw = N Bo − Boi + (rsi − rs )B g + We F = N

We F =N+ ∴ Eo Eo

Eo

+We

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE 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

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE Solution :

F

Eo

=N+

We

Eo

P

F

Eo

F/Eo

We/Eo

4000

―x106

―

―x106

―x106

3900

5.111

0.006

851.89

652

3800

18.420

0.024

767.52

568

3700

41.862

0.073

573.45

373.5

3600

72.042

0.140

514.38

314.6

F

From Fig.

Eo

45 o

N = 200 x 106 N = 200 × 10 6

We E o

Applied Reservoir Engineering : Dr. Hamid Khattab

4. Combination drive reservoir Characteristics: -Increase I Wp from f llow structure t t wells ll -Increase GOR from high structure wells y rapid p decline of fP -Relativity -R.F > R.Fwater influx

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE

m=

GBgii

GBgi

NBoi

NBoi

free gas (N-Np)Bo

Pi

P
NBoi + GBgi = (N − N p )Bo + (We − w p Bw ) + free gas

f gas = G + Nr free N si − (N − N p )rs − N p R p

∴ NBoi + mNBoi = (N − N p )Bo + (We − w p Bw )

[

]

+ Nrsi − (N − N p )rs − N p R p Bg

[

]

N p Bo + (R p − rs )Bg − (We − w p Bw ) ∴N = mBoi ( Bo − Boi + (rsi − rs )Bg + Bg − Bgi ) Bgi

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE This equation includes 3 unknown (We, m & N) Rearange g this equation q as a straight g line equation q

 mBoi  (Bg − Bggi ) + We ∴ N p Bo + (R p − rs )Bg + w p Bw = N Bo − Boi + (rsi − rs )Bg + N   Bgi 

[

]

[

 mBoi  E g  + We F = N  Eo + Bgi  

We F ∴ =N+ mBoi mBoi Eo + Eg Eo + Eg Bgi Bgi

]

F mBoi Eg Eo + Bgi

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

45 o

N

We mB oi Eo + Eg B gi

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE Example 16 : 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 1.351

600

600

0 00100 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

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of OOIP by y MBE Solution :

P

F

Eo

4000 3800 3600 3400

0x106 9.576 17.622 39 715 39.715

― 0.0196 0.0364 0 0808 0.0808

― 0.0533 0.0972 0 2159 0.2159

We F mB oi mB oi Eg + Eg Eo + B gi B gi

― 179.66x106 181.29 183 95 183.95

― 9.66x106 11.29 13 95 13.95

F mBoi Eg Eo + Bgii

F From Fig. F

N = 170×106

mB oi Eo + Eg E o B gi

45

o

STB N = 170 × 10 6

We mB oi Eo + Eg B gi

Applied Reservoir Engineering : Dr. Hamid Khattab

Uses of MBE ¾ Calculation of (N), (G) and (We) ¾ Prediction of future performance Difficulties of its application ¾ Lackof PVT data ¾ Assume constant gas composition ¾ Production data (NP, GP and WP) ¾ Pi and We calculations Limitation of MBE application ¾ Thick formation ¾High permeability ¾ Homogeneous formation ¾ Low oil viscosity ¾N active ¾No ti water t d drive i ¾ No large gas cap

Applied Reservoir Engineering : Dr. Hamid Khattab

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

Depletion drive

flash

Gas cap drive

differential

C bi ti Combination

(fl (flash h + diff.) diff )

Water drive

flash

Low volatile oil

differential

High volatile oil

flash

Moderate volatile

(flash + diff.)

rs

p

flash f

dif f

Applied Reservoir Engineering : Dr. Hamid Khattab

Water in flux Due to: Cw, Cf and artesian flow We Bottom water

Oil

Oil water

Linear flux

Edge water

W

W

Applied Reservoir Engineering : Dr. Hamid Khattab

Flow regimes

Steady state

semi-steady state

Unsteady state

Outer boundary condition Infinite

Limited

Applied Reservoir Engineering : Dr. Hamid Khattab

Steady state water influx - Open external boundary - ∆P/∆r = C with time - qe=qw=C with time - Strong We - Steady y state equation q (Darcy ( y law)) pe

qw

qe

pw rw

r

re

Applied Reservoir Engineering : Dr. Hamid Khattab

Hydraulic analog

q ∝ ∆P

Pi

dWe dt ∝ (Pi − P )

Pw

dWe dt = k (Pi − P )

We = k ∑ (Pi − P )∆t k : water influx constant

x

screen

sand

∑ (P − P )∆t : area under (Pust) curve i

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

dWe dN P dN P dWP = Bo + ( R p − rs ) Bg + Bw = k ( Pi − P ) dt dt dt dt

q

Applied Reservoir Engineering : Dr. Hamid Khattab

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

Solution : dWe dt = 13300 × 1.4 + 13300 × (900 − 700) × 0.00082 + 0 = 20800 bbl / day 20800 = 130 bbl / day / psi ∴k = (3500 − 3340) Calculation of

∑ (P − P )∆t i

∑ (P − P )∆t = A i

1

=

Pi

1

2 (P − P1 ) + (Pi − P2 ) ∆ t + i 2 2 (P − P2 ) + (Pi − P3 ) ∆ t + i 3 2 (P − P3 ) + (Pi − P4 ) ∆ t + i 4 2

t1

A1

+ A2 + A3 + A4

(Pi − P1 ) ∆ t

∆t1

P1 P2 P3 P4

∆t2

A2

t2

∆t3

A3

t3

∆t4

A4

t4

Applied Reservoir Engineering : Dr. Hamid Khattab

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

0

100

200

300

400

3500

3450

3410

3380

3340

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

Applied Reservoir Engineering : Dr. Hamid Khattab 100

100

t

100

50

Solution : P

 3500 − 3450  ( ) Wee100 130 100 − 0 325,000 bbls = 100   2  50 + 90  50  We 200 = 130  × 100 + 100  = 1235000 2 2 

100

90 120 160

50 + 90 90 + 120   50 We 200 = 130  × 100 + 100 + 100  = 2606 × 10 3 2 2 2  50 + 90 90 + 120 120 + 160  50  We 200 = 130  × 100 + 100 + 100 + 100  = 4420 × 10 3 bbls 2 2 2 2 

Applied Reservoir Engineering : Dr. Hamid Khattab

Semi--steady Semi steady--state 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 dependent function (re/rw)→at −3

dWe 7.08×10 kh(Pe − Pw ) C(Pe − Pw ) C(Pi − P) ∴ = = → dt µ ln(re rw ) ln(re rw ) ln(at) dWe C(Pi − P) ∴ = dt ln(at) (Pi − P) ∴We = C∑ ∆t ln(at)

Applied Reservoir Engineering : Dr. Hamid Khattab

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

( Pi − P ) 1 = ln (at) (dWe dt ) C

( Pi − P) 1 1 = ln a + ln t ∴ C (dWe dt ) C

(Pi − P) (dW We dt)

1 C

Plott this equation as a straight line:

1 1 Gives slop = and intercept = ln a C C

ln t

Applied Reservoir Engineering : Dr. Hamid Khattab

Example 18:

Using the following data calculate (a) and (c) P

We MBE

∆We (Wen+1-Wen-1)

∆We/ ∆t

(Pi-P)

0

3793

0x103

0

0

0

3

3788

4.0

12.4

136

5

6

3774

24.8

35.5

389

19

9

3748

75.5

73.6

806

45

12

3709

172

116.8

1279

84

15

3680

309

154

1687

113

18

3643

480

197

2158

150

21

3595

703

249

2727

198

24

3547

978

291

3187

246

27

3518

1286

319

3494

275

30

3485

1616

351

3844

308

33

3437

1987

386

4228

356

36

3416

2388

407

4458

377

Soluttion

Tmonth

Applied Reservoir Engineering : Dr. Hamid Khattab

tmonth

tdays

∆We/ ∆t

(Pi-P)

Ln t

(Pi-P)/ dWe/ dt

0

0

0

0

―

―

6

182.5

389

19

5.207

0.049

12

365

1279

84

5.900

0.066

18

547.5

2158

150

6.305

0.070

24

780

3187

246

6.593

0.077

30

912 5 912.5

3844

308

6 816 6.816

0 081 0.081

From Fig.

1 = 0 . 002 C ∴C = 50

Using any point in the straight line a = 0.064

∴We = 50∑

(Pi − P) (dWe dt)

1 C

= 0 . 002

Pi − P ln(0.064t) ln t

Applied Reservoir Engineering : Dr. Hamid Khattab

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 : P − P  P −P We39 = We36 + 50 ×  i 36 + i 39 2 dt ln a t 2  ln a t1 

 3793 − 3379  3793 − 3416 = 2388 ×103 + 50 ×  + 2 × [1186.25 × - 1095] ln (0.064 ×1186.25) ln (0.064 ×1095) 

= 2388 × 103 + 420.508 × 103

= 2809×103 bbls

Applied Reservoir Engineering : Dr. Hamid Khattab

Unsteady--state water influx Unsteady - P and q = C with time - q = 0 at re, q=qmax at rw - Closed extended boundry - We due to Cw and Cf

rw

Applied Reservoir Engineering : Dr. Hamid Khattab

Hydraulic analog

Pi P2 P1 Pw x

screen

sand

sand

sand

q

Applied Reservoir Engineering : Dr. Hamid Khattab

Physical analog