Applied Reservoir Lectures

S.

akii

*

■

A u.

■

■

a

a

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

PPb - 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