Petroleum Engineering

Applied Reservoir Engineering

Applied Reservoir Engineering : Dr. Hamid Khattab

Reservoir Definition Cap rock Res. Fluid Reservoir rock

R Reservoir i

Shallow

offshare

Deep

onshare

offshare

onshare

Applied Reservoir Engineering : Dr. Hamid Khattab

Reservoir rocks

Sedimentry

Sandstone

Chemical

Sand

L.s

Dolomit

Applied Reservoir Engineering : Dr. Hamid Khattab

Rock Properties

Saturation

Porosity

Absolute

Effective

So

Absolute Sw

Capillary

Permeability

Relative

Sg Eff ti Effective

Primary

Secondary

Primary

Seccondary

Ratio

Wettability

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

Applied Reservoir Engineering : Dr. Hamid Khattab

Fluid properties

Oil

Gas

ρg AMw γg Tc PC Z

TR P R

Cg βg µg

ρo γo APT rs

Water

βw rs µw Cw Salinity

βo βt µo Co



Applied 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

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original hydrocarbon in place by volumetric method

Structural contour map

Well

Depth

1

D1

2

D2

3

D3

4

D4

5

D5

6

D6

7

D7

8

D8

9

D9

6

●

●7 ●4

3● 1●

●

●5

8

Scale:1:50000

Location map

●2 ●

9

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original hydrocarbon in place by volumetric method

30 10 0

Isopach map

Well

Depth

1

h1

2

h2

3

h3

4

h4

5

h5

6

h6

7

h7

8

h8

9

h9

G Gas Goc

Oil Woc

Water

Applied 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

Applied 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’

Applied 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’

Applied 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

Applied 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

Applied 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

Applied 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

]

Applied 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

Applied 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

Applied 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)

Applied Reservoir Engineering : Dr. Hamid Khattab

Converting map areas (inch2) to acres Solution :

BVoil =

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

[

7 10 10 BVgas = [200 + 190] + [190 + 130] + 130 + 55 + 130 × 55 2 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=

Applied Reservoir Engineering : Dr. Hamid Khattab

Reservoir drive mechanism Water reservoir

P

Gas reservoir

Bg

Gas

Gas Water

without bottom water drive

with bottom water drive Oil reservoir

Applied Reservoir Engineering : Dr. Hamid Khattab

Oil reservoir Undersaturated

P>Pb

Oil

Oil

Water

without bottom water drive

Oil

Depletion drive

with bottom water drive

Gas Oil

Gas cap drive

Saturated

P≤Pb

Oil

Gas

Gas

Oil W Water

Bottom water drive

Combination drive

Applied Reservoir Engineering : Dr. Hamid Khattab

PVT data for gas and oil reservoirs Gas reservoirs Bg

ZT Bg = 0.00504 P

P

Applied Reservoir Engineering : Dr. Hamid Khattab

PVT data for gas and oil reservoirs Saturated oil reservoirs

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

µo

ZT Bg = 0.00504 P

rsi Bo

Boi= Bti

rs 1 0

Bg

Pi

Applied Reservoir Engineering : Dr. Hamid Khattab

PVT data for gas and oil reservoirs Undersaturated oil reservoirs P1 > Pb undersat.

saturated

Bt

rsi=c

µo Bo rs

1 0

Bg

Applied Reservoir Engineering : Dr. Hamid Khattab

Laboratory measurment of PVT data

Gas SCF

Oil

Gas Oil

STB P = 14.7 psi T = 60o F

Oil

Pb saturated

Oil P > Pb undersaturated

Oil Pi

Applied Reservoir Engineering : Dr. Hamid Khattab

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

GBgi

(G − G )B

∆T

p

p∠pi

pi GBgi = (G − G p )Bg ∴G =

G p Bg Bg − Bgi

gi

1

Applied Reservoir Engineering : Dr. Hamid Khattab

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

Pppsi

G p SCF

Bg bbl SCF

Z

G

4000

0x10-6

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.

Applied Reservoir Engineering : Dr. Hamid Khattab

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 )

G p Bg

y1 G

2 x1

Another form:

Bg − Bgi

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

 Z Zi  Z ∴ G p = G  − p  p pi 

3

Gp

Z p

y2 G

x2 Z p−Zi Pi

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original gas in place by MBE Another form:

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

p Z

y3

G p  pi P  ∴ =  1 − Z G  Z i 

p p pi ∴ = i − Gp Z Z i GZ i at

p =0 Z

G = Gp

pi Zi

pi GZ i

G 0

x3

Gp

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original gas in place by MBE Example 5 :

Solution :

p

Solve the previous example using MBE as a straigh line

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

x1

y1

y2

x2

y3

x3

From Figgers

G = 200×106 STB

Applied Reservoir Engineering : Dr. Hamid Khattab

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

R

Gp

Wp

∆T

∴ Assuming

=0 causes an increase in G continuously

Applied Reservoir Engineering : Dr. Hamid Khattab

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

/ Assuming

is known

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

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

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

Oil

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

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

Applied Reservoir Engineering : Dr. Hamid Khattab

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

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

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

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-Ni)Bo

NBoi

P>Pb

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

∴N =

N p Bo Bo − Boi

(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

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 Considering Cw and Cf

o.b.p=1 psi/ftD

- 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

o.b.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

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

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

Pi

Voi

C

f

Co = −

Pi

Vo

= f (φ )

C w = f ( P , T , r s and salinity From the following charts

)

1 dV . V oi dP

=

1 V − Vi . V oi P − Pi

=

1 ( B o − B oi ) . B oi ∆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

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

∆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

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 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 • R p increases rapidly

• low R.F

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original oil in place by MBE Np

Gp

NBoi

(N − N )B p

∆T

Free gas 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 =

[

N p Bo + (R p − rs )Bg

]

Bo − Boi + (rsi − rs )Bg

o

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original oil in place by MBE Example 12 : 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

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

R . F = f (P & R P

B o + (R

p

− r s )B g

)

R.F ∝ 1 RP To increase R.F: • • • •

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

Applied Reservoir Engineering : Dr. Hamid Khattab

Calculation of original oil in place by MBE Example 13 : For example 12, 12 at P P=3400 3400 psi calculate: Sg and R.F R F without Gi and with Gi=60 Gp

Solution :

Sg

[

free gas = pore volume

]

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

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 =

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

m=

GB gi

Bgi 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 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

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  

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

Eo +

mB oi Eg B gi

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