Applied Reservoir Lectures

(1)

Applied Reservoir Engineering

S. ■ * A

a

u. ■

kii

■ a a 1 .

<®L.

■_J.—*:•ÿ._ .

(2)

Applied Reservoir Engineering : Dr. Hamid Khattab Reservoir Definition Reservoir Definition Reservoir Definition Reservoir Definition Cap rock Res. Fluid Reservoir rock R i Reservoir Shallow Deep

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h-Applied Reservoir Engineering : Dr. Hamid Khattab

Reservoir rocks

Sedimentry Chemical

Sandstone Sand L.s Dolomit

(4)

Applied Reservoir Engineering : Dr. Hamid Khattab

Rock Properties

Porosity Saturation Permeability Capillary Wettability

Absolute Effective So S_w S_g Absolute Eff ti Relative Primary Primary Effective Ratio Secondary Seccondary

Hi

I i

(5)

Reservoir fluids Reservoir fluids

Water Oil Gas

Salt Fresh

Black

Volatile Drey Wet Condensate

Low volatile

High volatile Ideal

Real (non ideal)

n

(6)

Fluid properties

Gas Oil Water

AM T P Z C β ρ_g AM_w γ_g T_c P_C Z C_g β_g µ_g _β w rs µw Cw Salinity ρ_o γ_o APT r_s β_o β_t µ_o C_o  T_R P_R

Hi

f r

(7)

Applied reservoir Engineering Contents Applied reservoir Engineering Contents

1. Calculation of original hydrocarbon in place

i. Volumetric method i. Volumetric method

ii. Material balance equation (MBE)

2 Determination of the reservoir drive mechanism 2. 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

5

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If-Applied Reservoir Engineering : Dr. Hamid Khattab

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 method volumetric method ●6 _●7 Well Depth 1 D₁ ● ● ●4 3 2 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 map Structural contour map

4i'\

--

N \ ✓ \ / / / /ÿ I y / _/ V ✓ _l / ✓ / _l \ / _I ✓ / ✓ -/ G I O s _I I I s I / / / / l / I _s / / \ _\ / / / V \ / sc*-: \ / \ / \ I \ *v _/ / \ _y

(9)

volumetric method volumetric method Well Depth 1 h₁ G 1 h₁ 2 h₂ 3 h₃ 4 h₄ Goc Gas Oil 4 h₄ 5 h₅ 6 h₆ 7 h Woc Oil Water 7 h₇ 8 h₈ 9 h₉ 30 10 0 Isopach map

1W-.

/ / o t i o _/ _I ✓ I ✓ _I Q / O _/ V Yo _/ \ / \ 40, / \ 'E

(10)

volumetric method volumetric method

)

1 (

.

S

_wi

BV

N

=

φ

−

)

1 (

)

(

Ah

−

S

_wi

=

(

)

φ

(

_wi

)

wi

S

Ah

β

φ

615

5 )

1 (

43560

−

=

β

_g

Bbl

SCF

oi

β

615 .

5

wi

S

Ah

N

7758

φ

(

1 −

)

STB

bbl

o

β

acres

A :

oi wi

N

β

φ

(

)

=

i

S

Ah

φ

(

1 )

7758

−

STB SCF

ft

h :

fractions

S

_wi

:

,

φ

gi wi

S

Ah

G

β

φ

(

1 )

7758

=

SCF

φ

,

wi

f

5

(11)

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

Area inch2 C.L ( ) g p p ( ) g p p 1. Trapozoidal method: A₁ 10 A_o 0 WOC

5 .

0

1

>

− n n

A

A₃ 30 A₂ 20

[

A A A A A

]

h BV = + 2 + 2 +...+ 2 _n₋ + 2 _n 2 0 1 2 1 A₅ 50 A₄ 40 A’ GOC

[

]

[

A A

]

h n n n ′ + ′ + 2 2 0 1 2 1 A₇ 70 A₆ 60 O 76 A₇ 70 5

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

Area inch2

C.L

( ) g p p

2. Pyramid or cone method

5 .

0

1

≤

− n n

A

10 A₁ A_o 0 WOC

[

A

]

h

BV

.

3

0

+

1

+

0 1

=

A₃ 30 A₂ 20

[

A

]

h

.

3

1

+

2

+

1 2

+

A₅ 50 A₄ 40 A’ GOC

[

A

_n

An

A

_n

A

_n

]

h

[ ]

A

_n

h

3 .

3

1

+

1

+

₋ ₋ A₇ 70 A₆ 60 O 76 A₇ 70 5

(13)

Calculation of (BV) using isopach map Calculation of (BV) using isopach map( )( ) gg pp pp

3. Simpson method

Odd number of contour lines

[

A

_n

A

_n

]

h

BV

4

2

4 ...

4

2

3

0

+

1

+

2

+

3

+

1

+

=

₋

[ ]

A

_n

h

3

3 ′

+

3

5

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Converting map areas (inch

Converting map areas (inch22_{) to acres}_{) to acres}

Say : Scale 1 : 50000 Say : Scale 1 : 50000 1 inch = 50,000 inch

acres

56

398 (50,000)

inch

1

2 2

₃₉₈

_.

₅₆

_acres

43560

144 (

)

inch

1 =

×

=

5

(15)

Example 1 :

Gi th f ll i l i t d f it f

Given the following planimetred areas of an oit of reservoir. Calculate the original oil place (N) if φ =25%, S_wi=30%, β_oi=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

5

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Solution :

[

]

[

26 12 26 12

] [

10 12 5 12 5

] [

6 50 0 50 0

]

10 26 40 2 76 2 98 2 140 2 200 2 250 2 10 + × + × + × + × + × + = B V

[

] [

50 0 50 0

]

3 6 5 12 5 12 3 10 12 26 12 26 2 10 × + + + × + + + × + + +

ft

inch :

7198

2

=

acres 87 . 35 43560 144 (15,000) inch 1 2 2 ₌ × =

ft

inch :

7198

acres BV = 7198×35.87 = 258193.39 ∴ MMSTB N 250.38 4 . 1 ) 3 . 0 1 ( 25 . 0 39 . 258193 7758× × × − ₌ = ∴ 5

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By using Simpson method

[

]

[

5 0 5 0

]

6 5 2 12 4 26 2 40 4 76 2 98 4 140 2 200 4 250 3 10 × + + + × + × + × + × + × + × + × + × + = BV

[

5 0 5 0

]

3 + + × + ft inch . 6 . 7156 2 = ft acro f . 6 . 256709 87 . 35 6 . 7156 × = =

)

3

0

1 (

25

0

6 256709

7758

MMSTB

N

248 .

94

4 .

1 )

3 .

0

1 (

25 .

0

6 .

256709

7758

×

−

₌

=

∴

MMSTB

N

_av

=

(

250 .

38 +

248 .

94 )

2 =

249 .

66 ∴

5

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Example 2 :

If th i f l 1 i i d

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

S l ti Solution : ) 3 0 1 ( 25 0 39 258193 7758× × × − MMSCF G 350.53 001 . 0 ) 3 . 0 1 ( 25 . 0 39 . 258193 7758 = × × × = 5

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Example 3 :

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

C.L : 0(WOC) 10 20 30 33(GOC) 40 50 60 70 76

Area inch2 _: ₃₅₀ _{310 270 220} ₂₀₀ _{190 130 55} _{25 0}

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

5

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Solution :

[

] [

3

]

10

[

] [

]

_inch2 _ft BV_oil 220 200 9280 . 2 3 220 270 2 310 2 350 2 10 ₊ _× ₊ _× ₊ ₊ ₊ ₌ 2 =

[

]

[

]

[

]

BV = 7

[

200+190

]

+10

[

190+130

]

+10

[

130+55+ 130×55

]

[

]

[ ]

inch ft BV_gas . 79 . 4303 25 3 6 25 55 25 55 3 10 55 130 55 130 2 130 190 2 190 200 2 2 = + × + + + × + + + + + + = acres 77 . 63 43560 144 (20,000) inch 1 2 2 ₌ × = MMSTB N 618 3 . 1 ) 3 . 0 1 ( 25 . 0 77 . 63 9280 7758 = − × × × × = MMSCF G 372.6 001 . 0 ) 3 . 0 1 ( 25 . 0 77 . 63 79 . 4303 7758× × × × − ₌ =

mmm

fM

V

3

(21)

Reservoir drive mechanism Reservoir drive mechanism Reservoir drive mechanism Reservoir drive mechanism

Water reservoir Water reservoir P Gas reservoir Gas Gas B_g Water with bottom water drive without bottom water drive g Oil reservoir

If-aa

t

* ﻻ ﺑ د ﻣ ن ﻣ ﻌ ر ﻓ ﺔ Material Balance عوﻧ دﯾدﺣ ﻟﺗ

(22)

Applied Reservoir Engineering : Dr. Hamid Khattab Oil reservoir Undersaturated P>P_b Oil Oil Oil Water with bottom

without bottom _Saturated

water drive without bottom

water drive Saturated

P≤P_b Oil Oil Oil Oil W Gas Gas Gas

Oil Gas _Water

Combination drive Bottom water drive Gas cap drive Depletion drive TT

If-\ I i i i

1

Water

(23)

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

Gas reservoirs B_g Gas reservoirs P ZT B_g = 0.00504 P

If-9E& A > أ و ل ﺣ ﺎ ﺟ ﺔ ﺑﺗ ﺟ ﯾﻠ ﻰ ھ ﯾﺎ د ى و ﻣ ن ﺧ ﻼ ﻟ ﮭ ﺎ ﺑ ﺣ د د ﻧ و ع اﻟ ﺧ ز ا ن اﻟ ﻠ ﻰ ﻋ ﻧ د ى و ف ﺣ ﺎﻟ ﺔ اﻟ ﻐ ﺎ ز ھ ﻼ ﻗ ﻰ أ ﻛ ﯾ د Bg P --- Psia ﻓﻠ و ﻟﻘ ﯾ ت ﻣ ﺛ ﻼ Psig ﻻ ز م ﺗ ﺟ ﻣ ﻊ 14.7 psia : Absolute psi : Absolute psig : Gauge

(24)

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

Saturated oil reservoirs

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

B_oi=B_ti µ_o B r_si B_t= B_o+(r_si-r_s)B_g P ZT B_g = 0.00504 B_o r_s B_oi= B_ti P B_g 0 1 P 0 P_i

1W-.

i\ *ÿ..B_t / / >

(25)

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

Undersaturated oil reservoirs

saturated undersat. P₁ > P_b B_t µ_o rsi=c B_o r_s B_g 1 0 IT_{• •} • i A > < * * * + + # % ♦ *+ * *# o'P ■ ■ t "I * ♦ ♦ / M-Q ♦ / ► ﻻ ﺣ ظ ﻟ و اﻟ ﺗ ﻐ ﯾ ر ف اﻟ ﺿ ﻐ ط ط ﻔﯾ ف ﻣ ﻊ اﻟ و ﻗ ت ﯾﺑ ﻘ ﻰ ﻛ د ه ﻣ ﺷ ﻛ ﻠ ﺔ ف د ﻗ ﺔ Material Balance ﻋ ﺷ ﺎ ن ﻛ ل اﻟ ﺑﯾ ﺎﻧ ﺎ ت ﻣ ﻌ ﺗ ﻣ د ة ﻋ ل اﻟ ﺿ ﻐ ط

(26)

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

Gas Gas SCF Oil P Oil Oil P Oil Oil SCF STB undersaturated saturated P_b P > P_b P_i P = 14.7 p_si T = 60o _F

If- Z:-V V V V V > Rs=0

(27)

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

1. Gas reservoir without bottom water drive

p G T ∆

(

G−Gp

)

Bgi gi GB

(

)

i p p∠ i p g pB G G = ∴

(

p

)

g gi G G B GB = − 1 gi g B B G − = ∴ 1

If- Z:-A >

o

أﺑ ﺳ ط ا ﻷ ﻧ و ا ع ھ ﻧﺎ ﺑ ﮭ ﻣ ل ﺗﺄ ﺛﯾ ر ﺗ ﻣ د د ﻛ ل ﻣ ن Connate water Rock

(28)

1. Gas reservoir without bottom water drive

Example 4 : psi P GpSCF Bg_bbl _SCF Z

G

201.6 0.81 0.00084 12 3900 0x106 0.83 0.00077 0x10-6 4000 p G 195.2 0.77 0.00095 37 3700 200.2 0.79 0.00089 27 3800 G ≠ const. Solution : 199.7 0.75 0.00107 58 3600 Using eq. (1)

If- Z:-Initial هدﻛ هد ﻋ ﺷ ﺎ ن 0 Gp 0 *10^6

(29)

MBE as an equation of a straight line

y₁

(

p

)

g gi

G

B

GB

=

−

(

B

)

G

B

G

∴

2 g pB G G y₁

(

g gi

)

g p

B

G

B

G

=

−

∴

Another form: 2 gi g B B − x₁

(

)

      − =       _i p p T Z p ZT G p ZT G 0.00504 0.00504 Z G_p y2       i p p p p    − = ∴ Z _G _G Z Zi 3 p p G    = ∴ i p p p G G p 3 i i P Z p Z − x₂

If-aa

A

o

> A

o

> ﻻ ﺣ ظ : ھ ذا اﻟ ﺧ ط ﻻ ﺑ د و ﻻ ﻣ ﻔ ر ﻣ ن ﻣ ر و ر ه ﺑﻧ ﻘ ط ﺔ ا ﻷ ﺻ ل

(30)

Another form:

(

p

)

g gi G G B GB = −

(

)

 Z  Z p i i Z p

(

)

_      − = p Z G G G p Z p i i ₀_.₀₀₅₀₄ 00504 . 0 p G P _  Z y₃ i i GZ p p p p i i p Z p G G Z P       − = ∴ 1 G p i i i i _G GZ p Z p Z p ₌ ₋ ∴ at Gp x₃ G 0 0 = Z p p G G = p

If-aa

A \ \ \ \ \ \ \ \ \ S \ \ \ \ \ \ ھ ذا اﻟ ط ر ﯾﻘ ﺔ ﺗ ﺳ ﺗ ﺧ د م أ ﻛ ﺛ ر ف اﻟ ﺷ ر ﻛ ﺎ ت ﻋ ﻧ ﮭ ﺎ و ذﻟ ك ﻟ ﻛ ﺛ ر ة اﻟ ﻧﻘ ﺎ ط اﻟ ﻣ ﺗﺎ ﺣ ﺔ و ﻟ ﻛ ن ف ا ﻻ ﻣ ﺗ ﺣ ﺎ ن ﺣ ل ﺑﺎ ﻟ ط ر ﯾﻘ ﺔ اﻟ ﺳ ﺎﺑ ﻘ ﺔ

(31)

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

Example 5 :

Solve the previous example using MBE as a straigh line

Solution : p _Z _P i i P Z p Z − g p B G gi g B B − _P _Z G_p 12 4441 1.75 2.25 1.068 0.00005 3900 0x10-6 4819 ― x10-5 2.075x10-4 ―x104 ― 4000 37 3896 5.91 2.66 3.515 0.00018 3700 27 4177 3.15 2.39 2.403 0.00012 3800 x₃ y₃ x₂ y₂ y₁ x₁ 56 3421 8.09 2.88 5.990 0.00030 3600 From Figgers

_G

=

₂₀₀

×

₁₀

6

_STB

If-

h-/

> SCF

(32)

ﻻ ﺣ ظ ﻟ و ﻣ دا ﻟ ﻛ ش اﻟ ﺧ ز ا ن اﻟ ﻠ ﻰ ﻓﯾ ﮫ ﻏ ﺎ ز د ه

With water drive or not

== You should firstly check.

ﺑﺗ ﻔﺗ ر ض ا ﻻ و ل اﻧ ﮫ ﻣ ﻔ ﮭ و ش ﻣ ﯾﺎ ه و ﺗ ﺣ ط اﻟ ﻧﻘ ط و ﺗ ر ﺳ م ط ﻠ ﻌ ت ﺧ ط ﻣ ﺳ ﺗﻘ ﯾ م ﯾﺑ ﻘ ﻰ ﺗ ﻣ ﺎ م ﻣ ط ﻠ ﻌ ﺗ ش ﺧ ط ﻣ ﺳ ﺗﻘ ﯾ م ﯾﺑ ﻘ ﻰ ﻛ د ه ﻋ ﻧ د ك ﻣ ﯾﺎ ه ﺗﺑ دأ ﺗ ﺣ ﺳ ب ﺗﺎ ﻧ ﻰ ﺣ ﺳ ﺎﺑ ﺎﺗ ك اﻟ ﻣ ﺳ ﺗ و ى اﻟ ﻠ ﻰ ﺗ ﺣ ت اﻟ ز ﯾ ت ھ و ا

Oil Water Contact

ﺳ و ا ء ﺑﻘ ﮫ ﻛ ﺎ ن ﺗ ﺣ ت اﻟ ز ﯾ ت ﻣ ﯾﺎ ه و ﻻ ﻷ ﻻ ﺣ ظ : ا ﺣ ﻧﺎ ﻛ ﺗﺑ ﻧﺎ We : for encroachment و ﺑﻧ ﻧ ط ﻘ ﮭ ﺎ influx or encroachment

Not Wi to be not conflicted with injection

Gas Res. with bottom water drive ﺧ ﻼ ص ﻛ د ه ھ و ا ﺣ د دﻟ ك ﻣ ﺗ ﻌ ﻣ ﻠ ش اﻟ ﺣ ر ﻛ ﺔ اﻟ ﺗﺄ ﻛ ﯾ د ﯾ ﺔ د ى

(33)

2.Gas reservoir with bottom water drive

R p G p W T ∆ ∴

Assuming =0 causes an increase in G continuously

iksL

- . A

(G-G.)B

GB_g* _>

(wg-WpBw)

P<Pi Pi

GBgr(ÿGp)pg+(We-WpBw)

GjPs-fye-WpBw)

G= Bg-Bgi

we

Influx encroachment Bw ﻧﺎھ شﺑرﺿ ﺑﻧﻣ * ﻻ ن د ى ﻣ ﯾﺎ ه ﺗ ﺣ ت ﻣ ط ﻠ ﻌ ﺗ ش ﻓ و ق ﺧ ﺎﻟ ص

(34)

MBE as a straight line

F/ 45o ∴ N ∴ / Assuming is known 33 Flowing و د ى ﻣ ش ﺛﺎ ﺑﺗ ﺔ ﯾ ﻌ ﻧ ﻰ ھ ﺗ ﺧ ﺗﻠ ف ﻣ ن ﻧ و ع ﻵ ﺧ ر Expansion G

(35)

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

Calculation of original gas in place by MBE Calculation of original gas in place by MBE Example 6 :

A gas reservoir with a known bottom water drive has the following

data: _{=0 and} B 0x10-6 0.00093 0x109 4000 0 W_{e bbl} T _years psi P G_pSCF Bg_bbl _SCF 7.490 0.00107 72.33 3800 2 2.297 0.00098 27.85 3900 1 13.308 0.00117 113.85 3700 3 18.486 0.00125 151.48 3600 4 34

Calculate the original gas in place

ھ ﻧﺎ ﻣ ﻠ و ش ﻻ ز ﻣ ﺔ ﻋ ﺷ ﺎ ن أﻧ ت و ا ﺧ د ﻗﯾ ﻣ ﺔ اﻟ ﻣ ﯾﺎ ه ﻟ ﻛ ن ﻟ و ﻣ ش ﻣ ﻌ ﺎ ك ھ ﺗ ﻛ و ن ﻟﯾ ﮭ ﺎ د و ر ط ﺑ ﻌ ﺎ

(36)

T_year F E_g F/E_g x109 _W

e/Egx109

Solution

T_year F E_g F/E_gx10 W_e/E_gx10 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 F/E g 45 From Fig: G=500x109 _SCF G=500x109 W_e/E_g

(37)

Gas Cap Expansion an Shrinkage

G

Gp_c gas GOC Gp_c expansion GOC GOC Oil shrinkage Oil

Shrinkage due to: poor planning or accident and corrosiong p p g - Assume gas cap expansion = (G-Gp_c).B_g-GB_gi

Assume gas cap shrinkage = GB (G Gp )B - Assume gas cap shrinkage = GB_gi- (G-Gp_c)B_g

Gp_c: gas produced from the gas cap and my be = zero

د ه اﻟ ﻣ ﺳ ﺗ و ى ا ﻻ ﺻ ﻠ ﻰ Gas Cap ﺞﻧﺗﻣ اﻟ زﺎﻐ اﻟ ﻣ ن

(38)

ھ ﻌ ر ف ﻣ ﻧﯾ ن اﻟ ﻐ ﺎ ز اﻟ ﻠ ﻰ ط ﺎﻟ ﻊ د ه ط ﺎﻟ ﻌ ﻠ ﻰ ﻣ ن

Gas Cap or oil zone !!

ﻣ ﻊ اﻟ ﻌ ﻠم ا ن ﻛ د ه ﻛ د ه اﻟ ز ﯾ ت ﺑﯾ ﻛ و ن ﻓﯾ ﮫ ﻏ ﺎ ز دا ﯾ ب د ى ﺑﻘ ﮫ ﺑﺗ ﯾ ﺟ ﻰ ﻣ ن اﻟ ﺧ ﺑ ر ة ﯾ ﻌ ﻧ ﻰ أﻧ ت ﻋ ﺎ ﻣ ل و ﻋ ﺎ ر ف ﺗ ﻛ و ﯾ ن اﻟ ﻐ ﺎ ز اﯾ ﮫ اﻟ ﻠ ﻰ دا ﯾ ب ف اﻟ ز ﯾ ت PVT و ﻏ ﺎﻟ ﺑﺎ ﺑﯾ ﻛ و ن اﻟ ﻐ ﺎ ز اﻟ ﻠ ﻰ دا ﯾ ب ف اﻟ ز ﯾ ت د ه ﻣ ﻛ و ﻧﺎ ﺗ ﮫ أﺗ ﻘ ل ﻣ ن اﻟ ﻠ ﻰ ف ط ﺑﻘ ﺔ اﻟ ﻐ ﺎ ز ﻣ ﻣ ﻛ ن ﺗ ﻼ ﻗ ﻰ اﻟ ﻐ ﺎ ز ﻋ ﻠ ﻰ اﻟ ﺳ ط ﺢ ﻧﺗ ﯾ ﺟ ﺔ إﻧ ﮫ ﻓﯾ ﮫ ﻣ ﺛ ﻼ - failure in the Csg. Shrinkage ﺔﺑﺣﺎﺻﻣ نوﻛﺗھ ىدو - لوﻷ لﺻوو ﻼﺻأ ددﻣﺗ زﺎﻐاﻟ Perforation Expansion بﺣﺎﺻﻣ هدو

(39)

Calculation of original gas in place by MBE Calculation of original gas in place by MBE Example: 7

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

P Gp_cx109 _B g 4000 0 0.0020 3900 4 0 0022 3900 4 0.0022 3800 7 0.0025 3700 10 0.0028 3600 13 0.0031 3500 17 0.0035 أ ى ﺗ ﻐ ﯾ ر ط ﻔﯾ ف ﻧﺗ ﯾ ﺟ ﺔ اﻟ ﺧ ط ﺄ ف ﻗﯾ م Bg ھ ﯾ ﻛ و ن اﻟ ﺧ ط ﺄ و ا ﺿ ﺢ ف ﺣ ﺳ ﺎﺑ ﺎﺗ ك Cumulative ﻟ و ﻣ دا ﻟ ﻛ ش ﻗﯾ ﻣ ﺗ ﮫ . . . ﻣ ش ﻣ ﮭ م أ ﺻ ﻼ ﻹ ن أﻧ ت ھ د ﻓ ك ﺗ ﻌ ر ف ھ ﯾﺗ ﻣ د د و ﻻ ھ ﯾﻧ ﻛ ﻣ ش و ﺑ ﻣ ﻘ دا ر أد اﯾ ﮫ

(40)

Solution

Assuming gas cap expansion = (G-Gp_c).B_g-G_gi

Pressure Gas cap change x103 _type

4000 - -3900 -800 shrinkage 3800 +2500 expansion 3800 +2500 expansion 3700 +4000 expansion 3600 +3700 shrinkage 3500 +5000 expansion

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

أ و ل ﻗﯾ ﻣ ﺔ ﺳ ﺎﻟ ب و ﺑ ﻌ د ﻛ د ه ﻣ و ﺣ ب ﯾﺑ ﻘ ﻰ اﻟ ﻧﻘ ط د ى ﻓﯾ ﮭ ﺎ ﻣ ﺷ ﻛ ﻠ ﺔ ا ﻣ ﺎ production data PVT data اﻟ ﻘﯾ م و ﺻ ﻠ ت ل 4000 و ﺑ ﻌ د ﻛ د ه ﻗﻠ ت ﯾﺑ ﻘ ﻰ ﯾﺎ ﻣ ﯾ ر و ﻧﻔ س اﻟ ﻣ ﺷ ﺎ ﻛ ل اﻟ ﻠ ﻰ ﻓ و ق

(41)

ﻻ ﺣ ظ : ﻟ و اﻟ ﺿ ﻐ ط ا ﻻ و ﻟ ﻰ ﻋ ﻧ د ك ﻛ ﺎ ن 5000 و اد ﯾ ك اﻟ ﺑﯾ ﻧﺎ ت ﺑﺗ ﺎ ﻋ ﺔ ا ﻻ ﻧﺗ ﺎ ج و أﻧ ت ﻟﻘ ﯾ ت إ ن ﻗﯾ ﻣ ﺔ ىد ةرﯾﻐﺻ اﻟ ةر ﻔﺗ اﻟ ف بﺳﺣﺗ فرﻌﺗھ شﻣ لﺣﺗ ﺎﻣ ﻓﻠ ﺔﻣ ﻗﯾ رﺑﻛأ B ت ﺎﻧﻛ ﺎھدﻧﻋ ﻰﻠ اﻟ Pb =4700 ﻓﺗ ﻔﺗ ر ض ا ن اﻟ ﺿ ﻐ ط ا ﻷ و ﻟ ﻰ ﻟﻠ ﺧ ز ا ن 4700 و ﺗ ﺷ ﺗ ﻐ ل ع ھ ذا ا ﻻ ﺳ ﺎ س و ﺗ ﺻ ﻔ ر ا ﻻ ر ﻗﺎ م ﺑﺗ ﺎ ﻋ ﺔ ا ﻻ ﻧﺗ ﺎ ج ﯾ ﻌ ﻧ ﻰ ﺗ ط ر ح ﻣ ن cumulative - production at Pb Boi : Bo but at Pb و ﻟ ﻣ ﺎ ﺗ ط ﻠ ﻊ ﻗﯾ ﻣ ﺔ اﻟ ز ﯾ ت ﻣ ﺗﻧ ﺳ ﺎ ش ﺗﺑ ﻘ ﻰ ﺗ ﺟ ﻣ ﻊ ﻋ ﻠﯾ ﮫ ﻛ ﻣ ﯾ ﺔ ا ﻻ ﻧﺗ ﺎ ج اﻟ ﻠ ﻰ ط ﻠ ﻌ ﻠ ك ﻗﺑ ل ﻣ ﺎ ﺗ و ﺻ ل ﻟ ل 4700 و ﻧﻘ ط ﺗﯾ ن ﻣ ﺛ ﻼ ﺑ ﻌ د ھ ﺎ ﺧ ﻼ ص ا ﺷ ﺗ ﻐ ل ع اﻟ ﺧ ﻣ س ﻧﻘ ط و ﺧ ﻼ ص Pb قوﻓ م ﻗﯾ 5 ك اﻟ اد وﻟ بط و ﺗ ﻌ ﺎﻟ ﻰ ﻋ ﻧ د آ ﺧ ر ﻗﯾ ﻣ ﺔ ﻟ ﻼ ﻧﺗ ﺎ ج و ا ﺟ ﻣ ﻌ ﮭ ﺎ ﻋ ل اﻟ ﻧﺎ ﺗ ﺞ اﻟ ﻧ ﮭ ﺎﺋ ﻰ ط ب ﻟ و اد ﯾﺗ ﻠ ك 3 ﻧﻘ ط ﻓ و ق و 3 ﺗ ﺣ ت ﯾﺑ ﻘ ﻰ ﺑﺗ ﺷ ﺗ ﻐ ل ﻋ ل 3 د و ل ﻟ و ﺣ د ھ م و 3 د و ل ﻟ و ﺣ د ھ م و اﻟ ﻔ ر ق ﺑﯾ ن ا ﻻ ﺗﻧ ﯾ ن ھ و ا ﻗﯾ ﻣ ﺔ ا ﻻ ﻧﺗ ﺎ ج ﻋ ﻧ د Pb

(42)

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

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

a) Under-saturated oil reservoirs

Characteristics P>P - P>P_b - No free gas, no W_p - Large volume Limited K - Limited K

- Low flow rate

- Produce by C_w and C_f أﻧ ﺎ ﻣ ش ﻋ ﺎﯾ ز ه ﯾ و ﺻ ل ﺑ ﺳ ر ﻋ ﺔ ﻟﻠ ﺿ ﻐ ط Pb ﻋ ﻠ ﺷ ﺎ ن ﺗ ﻌ ر ف ﺗ ﺣ ﺳ ب ﺑ ر ا ﺣ ﺗ ك و د ه ﺑﻘ ﮫ ﯾ ﺣ ﻘ ق اﻟ ﺷ ر و ط د ى compressibility of connate water ﻣ ﯾﻧ ﻔ ﻌ ش ﺧ ﺎﻟ ص ﺗ ﮭ ﻣ ل ھ ذ ه اﻟ ﺗﺄ ﺛﯾ ر ا ت

(43)

1- Under-saturated oil reservoirs without bottom water

N_p NB_oi P P (N-N_i)B_o P>P P_i>P_b P>Pb neglecing C_w and C_f NB_oi=(N-N_p)B_o o pB N N ∴ (1) oi o p B B N − = ∴ (1) p د ه ھ ﻧﺎ ﺑ س ﻟ ﻛ ن ﻣ ش ھ ﻧ ﮭ ﻣ ﻠ ﮭ ﺎ ﺧ ﺎﻟ ص ﺑ ﻌ د ﻛ د ه و اﻟ ﻣ ﺛﺎ ل د ه ﺗ و ﺿ ﯾ ﺣ ﻰ ﻓﻘ ط ﻟ ﻛ ن ﻣ ش د ه اﻟ ﻣ ظ ﺑ و ط -و ھ ذ ه اﻟ ﻣ ﻌ ﺎ دﻟ ﺔ ﻻ ﺗ ﺳ ﺗ ﺧ د ﻣ ﮭ ﺎ ﺗﺎ ﻧ ﻰ ﺧ ﺎﻟ ص

(44)

Example: 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 C_w and C_f using the following data

P N_p x106 _B o 4000 0 1.40 3800 1 535 1 42 3800 1.535 1.42 3600 3.696 1.45 3400 7.644 1.49 3200 9.545 1.54 ﻣ ن ھ ذ ه اﻟ ﻘﯾ م ﺗ ﻌ ر ف ﻧ و ع اﻟ ﺧ ز ا ن اﻟ ﻘﯾ م ﺗﺗ ز اﯾ د ﯾﺑ ﻘ ﻰ ﺗ ﻣ ﺎ م

(45)

Applied Reservoir Engineering : Dr. Hamid Khattab C l l ti f i i l il i l b MBE C l l ti f i i l il i l b MBE Solution Pressure N_pB_o x106 _B o-Boi N x106

4000 - - -3800 2.179 0.02 108.95 3600 5 539 0 05 110 78 N ≠ const 3600 5.539 0.05 110.78 3400 11.389 0.09 126.64 3200 14.699 0.14 104.99

rearrange MBE as a straight line

NB_oi= (N-N_p)B_o F F = NE_o From Fig: 6_STB N x N _≠₁₁₀ ₁₀6 E_o =

(46)

o.b.p=1 psi/ftD Considering C_w and C_f

- 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 اﻟ ﺻ ﺧ ر ﺑﯾ ﺷ ﯾ ل ﻛ د ه ﺗﻘ ر ﯾﺑ ﺎ ﻧ ص و ز ن اﻟ ط ﺑﻘ ﺎ ت اﻟ ﻌ ﻠﯾ ﺎ اﻟ ز ﯾ ت ﻛ د ه ﺑﯾ ﺣ ﺑ س ﻧ ص اﻟ ظ ﻐ ط اﻟ ﻠ ﻰ ﻧﺎ ﺗ ﺞ ﻣ ن اﻟ ط ﺑﻘ ﺎ ت اﻟ ﻌ ﻠﯾ ﺎ ﻣ ﻊ ا ﻻ ﻧﺗ ﺎ ج اﻟ ﺿ ﻐ ط ﺑﺗ ﺎ ع اﻟ ز ﯾ ت ﺑﯾ ﻘ ل ﻓﺑ ﺎﻟ ﺗﺎ ﻟ ﻰ اﻟ ﻣ ﺟ ﻣ و ع ﻣ ش ﺑﯾ ﺳ ﺎ و ى 1 ﺑﺗ ﺎ ع ﺿ ﻐ ط اﻟ ز ﯾ ت و ﺿ ﻐ ط اﻟ ﺻ ﺧ ر ﻓﺗ ﺣ ﺻ ل ا ﻣ ﺎ : ﻗ ﺷ ر ة اﻟ ﻣ ﯾﺎ ه اﻟ ﻠ ﻰ ﺣ و ﻟﯾ ن اﻟ ﺻ ﺧ ر ﺗﺗ ﻣ د د -ﺣ ﺑﯾ ﺑﺎ ت اﻟ ﺻ ﺧ ر ﻧﻔ ﺳ ﮭ ﺎ ﺗﺗ ﻣ د د -ا ﻻ ﺗﻧ ﺗﯾ ن ﯾﺗ ﻣ د د و ا ﻣ ﻊ ﺑ ﻌ ض ﺑﯾ ﻌ و ﺿ و ا ف اﻟ ﺑ دا ﯾ ﺔ اﻟ ﺗﺄ ﺛﯾ ر ﺑﺗ ﺎ ع اﻟ ﺳ ﺣ ب ﺑ س ﺑ د ﻛ د ه ﻣ ﻣ ﻛ ن ﻻ ﻟذ ﻟ ك ﻧﺗ ﯾ ﺟ ﺔ اﻟ ﺗ ﻣ د د د ه ﺑ ﻌ د ﻛ د ه ﺑﺗ ﻼ ﻗ ﻰ ف ﺣ ﯾﺎ ة اﻟ ﺧ ز ا ن ف ا ﻵ ﺧ ر ﺷ ر و خ و ﺑ دأ ت ﻓ و اﻟ ق ﺗ ظ ﮭ ر ﻣ ﻛ ﺎﻧ ﺗ ش ﻣ و ﺟ و د ة ﻗﺑ ل ﻛ د ه crushes و ﻻ ﺣ ظ د ه ﻻ ز م ﯾﺗ م ﻗﺑ ل اﻟ ﺧ ز ا ن ﻣ ﺎ ﯾ و ﺻ ل ﻟﻠ ﺿ ﻐ ط اﻟ ﺑ ﺧ ﺎ ر ى ﯾ ﻌ ﻧ ﻰ ﻣ ﻔﯾ ش ﻏ ﺎ ز اﺗ ﻛ و ن ﻋ ﺷ ﺎ ن ﻣ ﯾﺎ ﺧ د ش ھ و ا اﻟ ﺣ ﺟ م د ه

(47)

Considering C_w and C_f NB_oi

Vp

B

N

NB

_oi

=

(

−

_p

)

_o

+

∆

_w_, _f P_i>P_b

dp

V

C

dVp

C

Vp

f f w f w

→

∆

+

∆

=

∆

1

, (N-N_p)B_o

dVp

dp

V

C

dVp

dp

V

C

f _f _f _p p f

=

→

=

1 .

P>P_b ∆Vp_,,w

V

dp

V

C

dVp

dp

dVp

V

C

w _w _w _w w w

=

.

→

=

1 dp

V

S

C

dVp

V

S

V

S

_w _w _p _w _w _w _p p w w

=

→

=

→

=

pore volume اﻟ ﻣ ﻔ ر و ض ﻓﯾ ﮫ ا ﺷ ﺎ ر ة - ﺑ س اﻧ ﺎ ﺑ دﻟ ﻠ ت ﻓ ر ق V

(48)

Considering C_w and C_f

dp

V

C

S

C

dVp

_f _w

=

_w _w

+

_f _p

∴

_,

(

)

S

NB

Vp

S

Vp

NB

w oi w oi

=

−

→

=

₋

)

1 (

)

1 (

dp

NB

S

C

S

C

dVp

_f _w w w f _oi

−

+

=

∴

)

1 (

,

p

NB

S

C

S

C

B

N

NB

S

oi f w w o p oi w

∆

+

−

=

∴

)

1 (

)

(

1 S

_w oi o p oi

₋

1

(49)

Applied Reservoir Engineering : Dr. Hamid Khattab C l l ti f i i l il i l b MBE C l l ti f i i l il i l b MBE B N N p o

Considering C_w and C_f B B p B S C S C B B N oi w f w w oi o o p ∆ − + + − = ∴ ) 1 ( B N B N N p B C B B p B B B C o p o p oi o oi o oi oi o o = = ∴ ∆ = − → ∆ − = Q S S where p B S C S C S S C p B S C S C C N w o oi w f w w w o o oi w f w w o − = ∆ − + + − ∆ − + + ∴ 1 ] 1 1 [ ] 1 [ p B S C S C S C B N N oi f w w o o o p w o ∆ + + = ∴ ] 1 [ p C B B N N S e oi o p w ∆ = − 1 (2) Effective compressibility

(50)

Considering C_w and C_f P P P = _i − ∆ P_i P_i V V dP dV V C oi o = − 1 . 1 P B B B C oi oi o o _∆ − = P_i B B P P V V V oi o i i oi − − − = ) ( 1 . 1 V_oi V_o P B oi o oi ∆ = . ( ) ) salinity and , , ( ) ( s w f r T P f C f C = =

φ

From the following charts

ﺗ ﺣ ﺳ ب ﻋ ﻧ د ﻛ ل ﻗﯾ ﻣ ﺔ ﺿ ﻐ ط ھ ﻧﺎ ﺛﺎ ﺑﺗ ﺔ Rs ھ ﻧﺎ ﺑ ﻌ ﺗﺑ ر اﻟ ﺣ ر ا ر ة ﺛﺎ ﺑﺗ ﺔ

(51)

(52)

Example: 9

l l ( ) d h ff f d

Solution Solve example (8) considering the effect of Cw and Cf

R1(Fig 2) r (Fig 1) C (Fig 4) ∆P=(Pi P) P C = Bo − Boi 18 2.9x10-6 ― ― 4000 R1(Fig.2) r_sf(Fig.1) C_wp(Fig.4) ∆P=(Pi-P) P C _B _p oi o = _∆ 0.85 16.8 2.95 8.928 400 3600 17.2 2.93 7.143x10-5 200 3800 15.2 3.00 12.500 800 3200 16 2.98 10.714 600 3400 fresh water

(53)

Continue

C_w=C_wpxR₂ R₂(Fig.3) r_s= r_sfx R₁ P w f w w o o S C S C S C − + + 1 7.725x10-5 3.311 1.13 14.62 3800 ― 3.30x10-6 1.4 15.3 4000 13.569 3.289 1.104 13.60 2400 9.570 3.247 1.11 14.28 3600 13.143 3.17 1.09 12.92 3200

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Applied Reservoir Engineering : Dr. Hamid Khattab C l l ti f i i l il i l b MBE C l l ti f i i l il i l b MBE Continue B N

p C B B N N e oi o p ∆ = P N_pB_o NB_oiC_e∆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 1 3200 14.699 0.1470 105.1

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Use MBE as a straight line as follows:

P C NB B N_p _o = _oi _e∆ B N F

Plot the fig

o

NE

F

=

F = NpBo 6 10 100× = N

Plot the fig. N =100×10

STB N ₌₁₀₀_×₁₀6 P C B E B C ∆P E_o = _oi _e∆ ﻻ ﺣ ظ ﻟ ﻣ ﺎ أ ھ ﻣ ﻠ ت ﺗﺄ ﺛﯾ ر ﺗ ﻣ د د اﻟ ﻣ ﯾﺎ ه و اﻟ ﺻ ﺧ ر ﻛ ﺎﻧ ت اﻟ ﻘﯾ ﻣ ﺔ 110*10^6

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U d t t d il i ith b tt t

Undersaturated oil reservoir with bottom water Undersaturated oil reservoir with bottom water

p N p W_p W (N-N_p)B_o NB_oi P>P_b P_i>P_b

Assuming (W_e) is known and neglect C_w+C_f

(

N N

) (

B W w B

)

NB =

(

−

) (

+ −

)

(

)

i w p e o p w p e o p oi B B B w W B N N B w W B N N NB − − − = ∴ + = oi o B B

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Undersaturated oil reservoir with bottom water Undersaturated oil reservoir with bottom water Example 11 :

Using the following data in the undersaturated oil reservoir with a known (W_e), neglecting C_w & C_f calculate (N): w_p= 0

P N_p B_o W_e 4000 ―x106 _1.40 _―x106 3800 2.334 1.45 1.135 3600 5 362 1 42 2 416 3600 5.362 1.42 2.416 3400 10.033 1.49 3.561 3200 12 682 1 54 4 832 3200 12.682 1.54 4.832

(58)

Solution : Solution :

(

)

oi o w p e o p B B B w W B N N − − − = P N_pB_o B_o-B_oi N 4000 106 ₁₀6 4000 ―x106 _― _―x106 3800 3.314 0.02 108.5 3600 7 775 0 05 107 1 N ≠ C 3600 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

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E F

Rearrange MBE as a straight line

o E o 45

[

_o _i

]

_e w p o pB W B N B B W N + = − ₀ + W E N F = + 110 = N e o W E N F = + o e o N W E E F = + ∴ o e E W

[

o i

]

o B B E = − ₀ We Eo p o

[

o 0i

]

F = N p Bo F E o 48 32 155 5 7 775 0 05 3600 56.75 165.7 3.314 0.02 3800 ― x10-6 ― ― x10-6 ― 4000 p p 34.51 139.5 19.531 0.14 3200 39.56 166.4 14.980 0.09 3400 48.32 155.5 7.775 0.05 3600

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Undersaturated oil reservoir with bottom water Undersaturated oil reservoir with bottom water Example 11 :

Solve examole (10) considering C_w and C_f effect

Solution : So ut on

C_w, C_o, C_f and C_e are the same as example (9)

P e oi e o e C B W E W ∆ = P e oiC B _∆ P e oi o P o B C B N E F ∆ = P ∆ e C P 45.07 145.06 0.0536 400 9.570 3600 52.06 x106 152.01 x106 0.0218 200 7.785 3800 ― ― ― ― ― x10-5 4000 32.87 132.86 0.1470 800 13.143 3200 31.26 131.25 0.1139 600 13.568 3400 45.07 145.06 0.0536 400 9.570 3600 F o E F o 45 P i e e C B W E W ∆ = P i o P C B B N E F ∆ = Plot vs 6 10 100× = N o e E W P e oi o B C E _∆ P e oi o B C E _∆ As in Fig. _N ₌₁₀₀_×₁₀6

(61)

B S t t d il i

B. Saturated oil reservoirs B. Saturated oil reservoirs

1 D l ti d i i

1. Depletion drive reservoirs Characteristics b P P ≤ • _b 0 = • W_p rapidly increases R_p • F R low . •

producing gas oil ratio ف اﻟ ﻐ ﺎﻟ ب ﻣ ﻔﯾ ش ﻣ ﯾﺎ ه ﺑﺗ ﻛ و ن ﻣ و ﺟ و د ة ﻛ ل اﻟ ﻐ ﺎ ز اﻟ ﻠ ﻰ أﻧ ﺗ ﺞ ﻟ ﺣ د دﻟ و ﻗﺗ ﻰ / ﻛ ل اﻟ ز ﯾ ت اﻟ ﻠ ﻰ أﻧ ﺗ ﺞ ف اﻟ ﻠ ﺣ ظ ﺔ اﻟ ﻠ ﻰ أﻧ ﺎ ﺑﺗ ﻛ ﻠم ﻓﯾ ﮭ ﺎ ط ﺎﻟ ﻊ ﻛ ﻣ ﯾ ﺔ ﻏ ﺎ ز أد إﯾ ﮫ و ﻛ ﻣ ﯾ ﺔ ز ﯾ ت أد إﯾ ﮫ اﻟ ذ و ﺑﺎ ﻧﯾ ﺔ ﺑﺗ ﺎ ﻋ ﺔ اﻟ ﻐ ﺎ ز ف اﻟ ز ﯾ ت Rp : producing GOR ( Rp ) instantaneous Rs

(62)

p

G

p

N

T ∆ oi NB

(

N − Np

)

Bo p Free gas

(

)

b i P P≤ Free gas p

(

N N

)

B free gas NB_oi = − _p _o +

(

N N

)

r N R SCF Nr gas free = _si − − _p _s − _p _p

(

p

)

o

[

si

(

p

)

s p p

]

g oi N N B Nr N N r N R B NB = − + − − − ∴

(

)

[

(

)

]

[

]

(

_si _s

)

_g oi o g s p o p B r r B B B r R B N N − + − − + = ∴ ﺟ ز ء ﺑﯾ ﺧ ر ج ﻣ ﻊ ا ﻻ ﻧﺗ ﺎ ج ز ى اﻟ ر ﻏ ﺎ و ى ﻻ ﺣ ظ : ﺣ ﺎ ط ط اﻟ ﻐ ﺎ ز ﺗ ﺣ ت اﻟ ز ﯾ ت ﻋ ﻠ ﺷ ﺎ ن ﻣ ﺗﻔ ﺗ ﻛ ر ھ ﺎ ش إﻧ ﮭ ﺎ gas cap د ه ﻓﻘ ﺎ ﻋ ﺎ ت ﻏ ﺎ ز ﻣ ﻧﻔ ﺻ ﻠ ﺔ ﻋ ن ﺑ ﻌ ﺿ ﮭ ﺎ اﻟ ﺑ ﻌ ض ﻟ و اﺗ ﺟ ﻣ ﻌ و ا ﻣ ﻊ ﺑ ﻌ ض ﯾ ﻌ ﻣ ﻠ و ا اﻟ ﺣ ﺟ م د ه ﻻ ﻧ ﮭ م ﻟ ﺳ ﮫ ﻣ و ﺻ ﻠ و ش ﻟ ل critical saturation SCF/ STB

(63)

Calculation of original oil in place by MBE Calculation of original oil in place by MBE Example 12 :

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

91 50 614 0 001273 1 423 674 3 87 3800 ― x106 718 0.001041 1.492 718 ― x106 4000 N r_s B_g B_o R_P N_P P ion 96.01 400 0.002200 1.286 3077 6.44 3400 96.02 510 0.001627 1.355 1937 5.26 3600 91.50 614 0.001273 1.423 674 3.87 3800 Solut 96.01 400 0.002200 1.286 3077 6.44 3400

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

ﻛ ل ﻣ ﺎ اﻟ ﺿ ﻐ ط ﯾﻘ ل ﻛ ل ﻣ ﺎ ﺗﻘ ل اﻟ ذ و ﺑﺎ ﻧﯾ ﺔ ﺑﺗ ز ﯾ د ﻛ ﻠ ﻣ ﺎ ﻗ ل اﻟ ﺿ ﻐ ط و ذﻟ ك ﻟ ز ﯾﺎ د ة ﻛ ﻣ ﯾ ﺔ اﻟ ﻐ ﺎ ز اﻟ ﻣ ﻧﻔ ﺻ ﻠ ﺔ اﻟ ﻘﯾ م ﺑﺗ ﻘ ل ﯾﺑ ﻘ ﻰ ع ط و ل saturated

(64)

(

)

[

_o _p _s _g

]

[

_o _oi

(

_si _s

)

_g

]

p B R r B N B B r r B

N + − = − + −

o E N F = Solution : P F Eo 4000 0 106 ₀ Solution F 4000 0x106 ₀ 3800 5.802 0.0634 3600 19 339 0 2014 6 10 96× = N 3600 19.339 0.2014 3400 46.124 0.4804 6 _STB Eo N Fig From _: ₌ ₉₆_×₁₀6 o

(65)

(

_si _s

)

_g oi o p B B r r B N F R − + −

(

)

(

_p _s

)

_g o g s si oi o p B r R B N F R − + = = .

(

P R _P

)

f F R .F = f

(

P & R _P

)

R . & P

R

F

R

.

∝

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

production data اﻟ ﺗ ﺟ ﻛ م ف اﻟ ﺿ ﻐ ط ﻛ ش ھ ﻌ ر ف ا ﻻ ﻟ و ﻋ ﻣ ﻠ ت ﺑﻘ ﮫ secondary recovery

(66)

ﻓﯾ ﮫ ﻧﺎ س ﺑﻘ ﮫ ﺑﯾ ﻘﻠ ك ﻻ ا ﺣ ﻧﺎ ﻣ ﻣ ﻛ ن ﻧ ﺿ ﺦ ﻏ ﺎ ز ف ا ﻣ ﺎ ﻛ ن ﻗ ر ﯾﺑ ﺔ ﻣ ن ﺑ ﻌ ﺿ ﮭ ﺎ ﺑ ﺣ ﯾ ث ﻧ ﻌ ﻣ ل

artificial gas cap

و ﻣ ﻣ ﻛ ن اﻟ ﻐ ﺎ ز اﻟ ﻠ ﻰ ﺑﯾ ﻧﻔ ﺻ ل اﺛ ﻧﺎ ء ا ﻻ ﻧﺗ ﺎ ج ﯾﺗ ﺟ ﻣ ﻊ و ﯾﺗ ﺣ د ف ا ﻋ ﻠ ﻰ اﻟ ط ﺑﻘ ﺔ و ﯾ ﻛ و

ن secondary gas cap

ﻣ ﺗﻧ ﻔ ﻌ ش ھ ذ ه اﻟ ط ر ﯾﻘ ﺔ ف ﺣ ﺎﻟ ﺔ depletion drive ﻟذ ﻟ ك ﻻ ﯾﻧ ﺻ ﺢ ﺑﻔ ﻌ ل ھ ذا ف ھ ذ ه اﻟ ﺣ ﺎﻟ ﺔ ﻟ و ﻋ ﻧ د ك أ ﺻ ﻼ gas cap و ﻋ ﻧ د ك ط ﺑ ﻌ ﺎ اﻟ ﻐ ﺎ ز ﻣ ﻊ ا ﻻ ﻧﺗ ﺎ ج ﺑﯾ ﻧﻔ ﺻ ل و ﯾ و ر ح ﻟ ﻣ ﻧ ط ﻘ ﺔ اﻟ ﻐ ﺎ ز ﻓ ﻼ ز م ﺗﺗ ﺎﺑ ﻊ ﻛ و ﯾ س ﺟ دا ﺟ دا ا ن اﻟ ﻐ ﺎ ز ﻣ و ﺻ ﻠ ش critical saturation و اﻧ ت ﺑﺗ ﺑﻘ ﻰ ﻋ ﺎ ر ﻓ ﮭ ﺎ ﻣ ن pvt نازﺧ اﻟ ﺔ ﺎﻗط شرﺳﺧﺗﻣ نﺎﺷﻋ هدو

(67)

Calculation of original oil in place by MBE Calculation of original oil in place by MBE Example 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: S_g and R.F without Gi and with G_i=60 G_p gas free S =

(

)

[

Nr i N N r N R

]

B gas free = − − − volume pore S _g =

(

)

[

Nrsi N N p rs N pRp

]

Bg gas free ( ) bbls 6 6 6 6 10 05 . 28 0022 . 0 3077 10 44 . 6 406 10 44 . 6 96 718 10 96 × = ×         × × − × × − − × × = 3077 10 44 . 6 _ 

(

)

bbls S NB volume pore w oi 6 ₂₀₄_.₆₂ ₁₀6 3 . 0 1 492 . 1 10 96 ) 1 ( − = × × × = − =

(

)

w) ( % 7 . 13 137 . 0 10 62 . 204 10 05 . 28 6 6 = = × × = ∴ S _g gas injection % ﻟ و ﺳ ﺄﻟ ك ھ ل ﯾﻧ ﻔ ﻊ ﺗ ﺿ ﺦ ﻛ ل ھ ذ ه اﻟ ﻛ ﻣ ﯾ ﺔ ؟ ﯾﺑ ﻘ ﻰ ﻻ ز م ﺗ ﺣ ﺳ ب critical gas saturation و ﺗﻘ ﺎ ر ن

(68)

ﯾﺑ

ﻘ

ﻰ

ﻟ

و

ﻋ

ر

ﻓ

ت

إ

ن

critical gas saturation = ...

ﯾﺑ

ﻘ

ﻰ

اﻟ

ﻣ

ﻘﺎ

م

اﻟ

ﻠ

ﻰ

ھ

و

ا

ﺣ

ﺟ

م

اﻟ

ﻔ

ر

ا

ﻏ

ﺎ

ت

ﺛﺎ

ﺑ

ت

ﻣ

ش

ھ

ﻌ

ر

ف

ا

ﻏ

ﯾ

ر

ه

اﻟ

ﺑ

ﺳ

ط

اﻟ

ﻠ

ﻰ

ھ

ﺑ

دأ

ا

ﻏ

ﯾ

ر

ه

ﺑ

ﺣ

ﯾ

ث

اﻧ

ﮫ

ﻣ

ﯾ

و

ﺻ

ﻠ

ش

ﻟﻠ

ﻘﯾ

ﻣ

ﺔ

اﻟ

ﻛ

ر

ﯾﺗ

ﯾ

ﻛ

ﺎ

ل

ھ

ﻼ

ﺛ

ﻰ

إﻧ

ﻰ

ﻣ

ش

ھ

ﻌ

ر

ف

ا

ﻏ

ر

أ

ى

ﺣ

ﺎ

ﺟ

ﺔ

ﻏ

ﯾ

ر

Rp

اﻟ

ﻠ

ﻰ

ﺑﺗ

ﺣ

ﻛ

م

ﻓﯾ

ﮭ

ﺎ

ﺑ

ﻛ

ﻣ

ﯾ

ﺔ

اﻟ

ﻐ

ﺎ

ز

اﻟ

ﻠ

ﻰ

ﺑ

ر

ﺟ

ﻌ

ﮫ

ﺗﺎ

ﻧ

ﻰ

ﻟﻠ

ﺧ

ز

ا

ن

و

ﺑ

ﻛ

د

ه

ﻻ

ز

م

ﻣ

ﺗ

ﻌ

د

ﯾ

ش

اﻟ

ﻘﯾ

ﻣ

ﺔ

د

ى

ﯾﺎ

ﺣ

ﺞ

أ

ﻣ

ﯾ

ر

ﻋ

ﻠ

ﺷ

ﺎ

ن

ﻣ

ﺗ

و

ﺻ

ﻠ

ش

ا

ن

اﻟ

ﻠ

ﻰ

ﺑﺗ

ﺿ

ﺧ

ﮫ

ﺑﯾ

ط

ﻠ

ﻌ

ﻠ

ك

ﺗﺎ

ﻧ

ﻰ

ع

اﻟ

ﺳ

ط

ﺢ

و

ﻛ

ﺄﻧ

ﮭ

ﺎ

ا

ر

ﺑ

ﺔ

ﻣ

ﺧ

ر

و

ﻣ

ﺔ

(69)

(

_si _s

)

_g oi o B r r B B F R − + −

(

)

(

p s

)

g o g s si oi o G without B r R B F R i = + − .

(

718 406

)

0.0022 492 . 1 286 . 1 − +

(

−

)

×

(

)

% 7 . 6 067 . 0 0022 . 0 406 3077 286 . 1 = = × − + =

(

)

(

si

)

s g oi o G with

B

r

R

B

r

B

F

R

i

+

−

+

−

=

% 60

.

(

)

(

0 4 3077 406

)

0 0022 286 1 0022 . 0 406 718 492 . 1 286 . 1 × − × + × − + − =

(

_p _s

)

_g o

R

r

B

+

(

)

% 49 . 15 1549 . 0 0022 . 0 406 3077 4 . 0 286 . 1 = = × × + ھ ﺿ ﺦ 60 % ﯾﺑ ﻘ ﻰ اﻟ ﻠ ﻰ ﺑﯾ ط ﻠ ﻊ ﻓ و ق 40%

(70)

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

Characteristics • P falls slowly • No Wp

• High GOR for high structure wells • R.F > R.F_depletion