1678518855461.pdf

Applied Reservoir Engineering : Dr. Hamid Khattab
Reservoir Definition
Cap rock
Res. Fluid
Reservoir rock
R i
Reservoir
Shallow Deep
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Reservoir
rocks
Sedimentry Chemical
Sandstone Sand L.s Dolomit

Rock Properties
Porosity Saturation Permeability Capillary Wettability
Absolute Effective
So
Sw
Sg
Absolute
Eff ti
Relative
Primary Primary
Effective
Ratio
Secondary Seccondary

Reservoir fluids
Reservoir fluids
Water Oil Gas
Salt Fresh
Black
Volatile
Drey
Wet Condensate
Low volatile
High volatile Ideal
Real
(non ideal)

Fluid properties
Gas Oil Water
AM T P Z C β
ρg AMw γg Tc PC Z Cg βg µg βw rs µw Salinity
Cw
ρo γo APT rs βo βt µo Co

TR PR

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 recovery
Primary recovery
– Secoundry recovery by : Gas injection
Water injection

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

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

volumetric method
volumetric method
)
1
(
. wi
S
BV
N −
= φ
)
1
(
)
( wi
S
Ah −
= φ )
(
)
( wi
φ
wi
S
Ah
β
φ
615
5
)
1
(
43560 −
= SCF
Bbl
g
β
oi
β
615
.
5
wi
S
Ah
N
φ )
1
(
7758 − STB
STB
bbl
o
β
acres
A:
oi
wi
N
β
φ )
(
=
i
S
Ahφ )
1
(
7758 −
STB
SCF
ft
h :
fractions
Swi :
,
φ
gi
wi
S
Ah
G
β
φ )
1
(
7758
=
SCF f
wi
,
φ

Calculation of (BV) using isopach map
Area inch2
C.L
( ) g p p
( ) g p p
1. Trapozoidal method:
A1
10
Ao
0 WOC
5
.
0
1 >
−
n
n A
A
A3
30
A2
20
[ ]
A
A
A
A
A
h
BV n
n +
+
+
+
+
= − 2
2
......
2
2
2
1
2
1
0
A5
50
A4
40
A’
GOC
[ ]
[ ]
A
A
h
n
n
n
′
+
′
+
2
2
1
2
1
0
A7
70
A6
60
O
76
A7
70

Area inch2
C.L
( ) g p p
( ) g p p
2. Pyramid or cone method
5
.
0
1 ≤
−
n
n A
A A1
10
Ao
0 WOC
[ ]
A
A
A
A
h
BV .
3
1
0
1
0 +
+
=
A3
30
A2
20
[ ]
A
A
A
A
h
.
3
2
1
2
1 +
+
+
A5
50
A4
40
A’
GOC
[ ] [ ]
n
n
n
n A
h
A
A
An
A
h
3
.
3
1
1 +
+
+
+ −
−
A7
70
A6
60
O
76
A7
70

( ) g p p
( ) g p p
3. Simpson method
Odd number of contour lines
[ ]
n
n A
A
A
A
A
A
h
BV 2
4
......
4
2
4
3
1
3
2
1
0 +
+
+
+
+
+
= −
[ ]
n
A
h
3
3
′
+
3

Converting map areas (inch
Converting map areas (inch2
2) to acres
) to acres
2) 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
acres
56
.
398
43560
144
( )
inch
1 =
×
=

2) to acres
) to acres
Example 1 :
2) to acres
) to acres
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%,
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

2) to acres
) to acres
2) to acres
) to acres
Solution :
[ ]
[ ] [ ] [ ]
0
50
0
50
6
5
12
5
12
10
12
26
12
26
10
26
40
2
76
2
98
2
140
2
200
2
250
2
10
+
×
+
×
+
×
+
×
+
×
+
=
B
V
[ ] [ ] [ ]
0
50
0
50
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 39
.
258193
87
.
35
7198 =
×
=
∴
MMSTB
N 38
.
250
4
.
1
)
3
.
0
1
(
25
.
0
39
.
258193
7758
=
−
×
×
×
=
∴

2) to acres
) to acres
By using Simpson method
2) to acres
) to acres
[ ]
[ ]
0
5
0
5
6
5
2
12
4
26
2
40
4
76
2
98
4
140
2
200
4
250
3
10
×
+
+
+
×
+
×
+
×
+
×
+
×
+
×
+
×
+
×
+
=
BV
[ ]
0
5
0
5
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 94
.
248
4
.
1
)
3
.
0
1
(
25
.
0
6
.
256709
7758
=
−
×
×
×
=
∴
MMSTB
Nav 66
.
249
2
)
94
.
248
38
.
250
( =
+
=
∴

2) to acres
) to acres
Example 2 :
If th i f l 1 i i d
2) to acres
) to acres
If the reservoir of example 1 is a gas reservoir and
βg=0.001 bbl/SCF. Calculate the original gas in place
S l ti
Solution :
)
3
0
1
(
25
0
39
258193
7758 −
×
×
×
MMSCF
G 53
.
350
001
.
0
)
3
.
0
1
(
25
.
0
39
.
258193
7758
=
×
×
×
=

2) to acres
) to acres
Example 3 :
2) to acres
) to acres
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 : 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)

2) to acres
) to acres
Solution :
2) to acres
) to acres
[ ] [ ]
3
10 2
[ ] [ ] ft
inch
BVoil .
9280
200
220
2
3
220
270
2
310
2
350
2
10 2
=
+
+
+
×
+
×
+
=
[ ] [ ] [ ]
BV 55
130
55
130
10
130
190
10
190
200
7
×
+
+
+
+
+
+
= [ ] [ ] [ ]
[ ] [ ] ft
inch
BVgas
.
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 6
.
372
001
.
0
)
3
.
0
1
(
25
.
0
77
.
63
79
.
4303
7758
=
−
×
×
×
×
=

Reservoir drive mechanism
Water reservoir
Water reservoir
P
Gas reservoir
Gas
Gas
Bg
Water
with bottom
water drive
without bottom
water drive
g
Oil reservoir

Oil reservoir
Undersaturated
P>Pb
Oil
Oil
Oil
Water
with bottom
without bottom Saturated
water drive
without bottom
water drive
Saturated
P≤Pb
Oil
Oil
Oil
Oil
W
Gas
Gas
Gas
Oil
Water
Gas
Combination
drive
Bottom water
drive
Gas cap
drive
Depletion
drive

PVT data for gas and oil reservoirs
Gas reservoirs
Bg
Gas reservoirs
P
ZT
Bg 00504
.
0
=
P

Saturated oil reservoirs
Boi=Bti
µo
B
rsi
Bt = Bo+(rsi-rs)Bg
P
ZT
Bg 00504
.
0
=
Bo
rs
Boi= Bti
P
Bg
0
1
P
0 Pi

Undersaturated oil reservoirs
saturated undersat.
P1 > Pb
Bt
µo
rsi=c
Bo
rs
Bg
1
0

Laboratory measurment of PVT data
Gas
Gas
SCF
Oil
P
Oil
Oil
P
Oil
Oil
SCF
STB
undersaturated
saturated
Pb P > Pb Pi
P = 14.7 psi
T = 60o F

Calculation of original gas in place by MBE
1. Gas reservoir without bottom water drive
p
G
T
∆ ( ) gi
p B
G
G −
gi
GB
( )
i
p
p∠
i
p
g
p B
G
G =
∴
( ) g
p
gi B
G
G
GB −
=
1
gi
g B
B
G
−
=
∴ 1

1. Gas reservoir without bottom water drive
Example 4 :
psi
P SCF
G p SCF
bbl
g
B 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 .
const
G ≠
Solution :
199.7
0.75
0.00107
58
3600
Using eq. (1)

MBE as an equation of a straight line
y1
( ) g
p
gi B
G
G
GB −
=
( )
B
B
G
B
G
∴ 2
g
p B
G
G
y1
( )
gi
g
g
p B
B
G
B
G −
=
∴
Another form:
2
gi
g B
B −
x1
( ) 






−
=






 i
p
p
T
Z
p
ZT
G
p
ZT
G 00504
.
0
00504
.
0 Z
Gp
y2





 i
p
p
p
p




−
=
∴ i
Z
Z
G
G
Z
3
p
p
G




=
∴
i
p
p
p
G
G
p 3
i
i P
Z
p
Z −
x2

Another form:
( ) g
p
gi B
G
G
GB −
=
( ) 
 Z
Z
p
i
i
Z
p
( ) 






−
=
p
Z
G
G
G
p
Z
p
i
i
00504
.
0
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 p
G
x3
G
0
0
=
Z
p
p
G
G =
p

Example 5 :
Solve the previous example using MBE as a straigh line
Solution :
p P
Z i
i P
Z
p
Z −
g
p B
G
gi
g B
B − Z
P p
G
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 STB
G 6
10
200×
=

2.Gas reservoir with bottom water drive
R
p
G
p
W
T
∆
∴
Assuming =0 causes an increase in G continuously

MBE as a straight line
F/
45o
∴
N
∴
/
Assuming is known
33

A i i h k b d i h h f ll i
Example 6 :
A gas reservoir with a known bottom water drive has the following
data: =0 and
B
0x10-6
0.00093
0x109
4000
0
We bbl
T years psi
P SCF
G p SCF
bbl
g
B
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

Tyear F Eg F/Eg x109 We/Eg x109
Solution
Tyear F Eg F/Eg x10 We/Eg x10
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
We/Eg

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

Example: 7
Calculate the gas cap volume change if G=40x109 SCF
P Gpc x109 Bg
4000 0 0.0020
3900 4 0 0022
3900 4 0.0022
3800 7 0.0025
3700 10 0.0028
3600 13 0.0031
3500 17 0.0035

Solution
Assuming gas cap expansion = (G-Gpc).Bg-Ggi
Pressure Gas cap change x103 type
4000 - -
3900 -800 shrinkage
3800 +2500 expansion
3600 +3700 shrinkage
Shrinkage at P=3600 may be due PVT or Gp data
Shrinkage at P=3600 may be due PVT or Gpc data

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

1- Under-saturated oil reservoirs without bottom water
Np
NBoi
P P
(N-Ni)Bo
P>P
Pi>Pb
P>Pb
neglecing Cw and Cf
NBoi=(N-Np)Bo
o
p B
N
N
∴ (1)
oi
o
p
B
B
N
−
=
∴ (1)

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 Cw
and Cf using the following data
P Np x106 Bo
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

Solution
Pressure NpBo x106 Bo-Boi N x106
4000 - - -
3800 2.179 0.02 108.95
3600 5 539 0 05 110 78 const
N ≠
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
NBoi = (N-Np)Bo
F
F = NEo
From Fig:
6
N
STB
x
N 6
10
110
≠
Eo

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

NBoi
Vp
B
N
N
NB f
w
o
p
oi ∆
+
−
= )
( ,
Pi>Pb
dp
V
C
dVp
dVp
C
Vp
Vp
Vp
f
f
w
f
w
→
∆
+
∆
=
∆
1
,
(N-Np)Bo
dVp
dp
V
C
dVp
dp
V
C p
f
f
f
p
f =
→
=
1
.
P>Pb
∆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
V
V
S p
w
w
w
p
w
w
p
w
w =
→
=
→
=

dp
V
C
S
C
dVp p
f
w
w
w
f +
=
∴ )
(
,
S
NB
Vp
S
Vp
NB
w
oi
w
oi
−
=
→
−
=
)
1
(
)
1
(
dp
NB
S
C
S
C
dVp oi
f
w
w
w
f
−
+
=
∴ )
1
(
,
p
NB
S
C
S
C
B
N
N
NB
S
oi
f
w
w
o
p
oi
w
∆
+
+
−
=
∴ )
1
(
)
(
1
S
oi
w
o
p
oi
−
1

B
N
N
o
p
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)

P
P
P i −
=
∆
Pi
Pi
V
V
dP
dV
V
C
oi
o −
=
1
.
1
P
B
B
B
C
oi
oi
o
o
∆
−
=
Pi
B
B
P
P
V
V
V
oi
o
i
i
oi
−
−
−
=
)
(
1
.
1
Voi
Vo
P
B
oi
o
oi ∆
=
)
(
.
)
salinity
and
,
,
(
)
(
s
w
f
r
T
P
f
C
f
C
=
= φ
From the following charts

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
B
B
C oi
o −
=
18
2.9x10-6
―
―
4000
R1(Fig.2)
rsf(Fig.1)
Cwp(Fig.4)
∆P=(Pi-P)
P
p
B
C
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

Continue
Cw=CwpxR2
R2 (Fig.3)
rs= rsf x 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

Continue
B
N
p
C
B
B
N
N
e
oi
o
p
∆
=
P NpBo NBoiCe∆P N
4000 ― ― ―
3800 2.179x016 0.0218 108.2x106
3600 5.359 0.0536 107.9
N ≠ C
3400 11.389 0.1131 106.5
3200 14 699 0 1470 105 1
3200 14.699 0.1470 105.1

Use MBE as a straight line as follows:
P
C
NB
B
N e
oi
o
p ∆
=
B
N
F
Plot the fig
o
NE
F = o
p B
N
F =
6
10
100×
=
N
Plot the fig. 10
100×
=
N
STB
N 6
10
100×
=
P
C
B
E ∆P
C
B
E e
oi
o ∆
=

U d t t d il i ith b tt t
Undersaturated oil reservoir with bottom water
p
N
p
W p
W
(N-Np)Bo
NBoi
P>Pb
Pi>Pb
Assuming (We) is known and neglect Cw+Cf
( ) ( )
B
w
W
B
N
N
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
Assuming We=0 will cuse an increase in (N)

Example 11 :
Using the following data in the undersaturated oil reservoir with a
known (We), neglecting Cw & Cf calculate (N): wp= 0
P Np Bo We
4000 ―x106 1.40 ―x106
3800 2.334 1.45 1.135
3600 5 362 1 42 2 416
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

Solution :
Solution :
( )
oi
o
w
p
e
o
p
B
B
B
w
W
B
N
N
−
−
−
=
P NpBo Bo-Boi 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

E
F
Rearrange MBE as a straight line
o
E
o
45
[ ] e
i
o
w
p
o
p W
B
B
N
B
W
B
N +
−
=
+ 0
W
E
N
F +
=
110
=
N
e
o W
E
N
F +
=
o
e
o E
W
N
E
F +
=
∴
o
e E
W
[ ]
i
o
o B
B
E 0
−
= o
e E
W
p o
E
F
o
p B
N
F =
[ ]
i
o
o 0
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

Example 11 :
Solve examole (10) considering Cw and Cf effect
Solution :
So ut on
Cw, Co, Cf and Ce 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 C
B
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 C
B
E ∆
P
e
oi
o C
B
E ∆
As in Fig. 6
10
100×
=
N

B S t t d il i
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
=
• p
W
rapidly
increases
Rp
•
F
R
low .
•

p
G
p
N
T
∆
oi
NB ( ) o
p B
N
N −
p
Free gas
( )
b
i P
P≤
Free gas
p
( ) gas
free
B
N
N
NB o
p
oi +
−
=
( ) SCF
R
N
r
N
N
Nr
gas
free p
p
s
p
si −
−
−
=
( ) ( )
[ ] g
p
p
s
p
si
o
p
oi B
R
N
r
N
N
Nr
B
N
N
NB −
−
−
+
−
=
∴
( )
[ ]
( )
[ ]
( ) g
s
si
oi
o
g
s
p
o
p
B
r
r
B
B
B
r
R
B
N
N
−
+
−
−
+
=
∴

Example 12 :
Calucaltion (N) for a depletion drive reservoir has the following data : Swi=30%
91 50
614
0 001273
1 423
674
3 87
3800
― x106
718
0.001041
1.492
718
― x106
4000
N
rs
Bg
Bo
RP
NP
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

( )
[ ] ( )
[ ]
g
s
si
oi
o
g
s
p
o
p B
r
r
B
B
N
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
o
E
STB
N
Fig
From 6
10
96
: ×
=
o

( ) g
s
si
oi
o
p B
r
r
B
B
N
F
R
−
+
−
( )
( ) g
s
p
o
g
s
si
oi
o
p
B
r
R
B
N
F
R
−
+
=
=
.
( )
P
R
P
f
F
R &
. = ( )
P
R
P
f
F
R &
.
P
R
F
R 1
. ∝
To increase R.F:
• Working over high producing GOR wells
Working over high producing GOR wells
• Shut-in ,, ,, ,, ,, ,,
• Reduce (q) of ,, ,, ,, ,,
R i j t f d d
• Reinject some of gas produced

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: Sg and R.F without Gi and
with Gi=60 Gp
gas
free
S =
( )
[ ]
i B
R
N
r
N
N
Nr
gas
free −
−
−
=
volume
pore
S g =
( )
[ ] g
p
p
s
p
si B
R
N
r
N
N
Nr
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
6
10
62
.
204
3
.
0
1
492
.
1
10
96
)
1
(
×
=
−
×
×
=
−
=
( )
w )
(
%
7
.
13
137
.
0
10
62
.
204
10
05
.
28
6
6
=
=
×
×
=
∴ g
S

( ) g
s
si
oi
o B
r
r
B
B
F
R
−
+
−
( )
( ) g
s
p
o
g
s
si
oi
o
G
without
B
r
R
B
F
R i
−
+
=
.
( ) 0022
.
0
406
718
492
.
1
286
.
1 ×
−
+
− ( )
( )
%
7
.
6
067
.
0
0022
.
0
406
3077
286
.
1
=
=
×
−
+
=
( )
( )
g
s
si
oi
o
G
with
B
r
R
B
B
r
r
B
B
F
R i
−
+
−
+
−
=
%
60
.
( )
( ) 0022
0
406
3077
4
0
286
1
0022
.
0
406
718
492
.
1
286
.
1
×
−
×
+
×
−
+
−
=
( ) g
s
p
o B
r
R
B +
( )
%
49
.
15
1549
.
0
0022
.
0
406
3077
4
.
0
286
.
1
=
=
×
×
+

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

Calculation of OOIP by MBE
y
y
p
N
p
G
gi
GB
(N-Np)Bo
free gas
oi
NB
gi
oi
gi
NB
B
m =
( ) gas
free
B
N
N
NB
GB o
p
oi
gi +
−
=
+
P>Pb
i
P
[ ] ( ) p
p
s
p
si R
N
r
N
N
G
Nr
gas
free −
−
−
+
=
( )
[ ]
g
s
p
o
p B
r
R
B
N
N
−
+
=
∴
( ) ( )
gi
g
gi
oi
g
s
si
oi
o B
B
B
B
m
B
r
r
B
B
N
−
+
−
+
−
∴
( ) ( )
oi
B
R
N
N
N
mNB
N
B
N
N
NB
NB 



+
+
+
∴ ( ) ( ) g
p
p
s
p
gi
oi
si
o
p
oi
oi B
R
N
r
N
N
B
Nr
p
B
N
N
NB
mNB






−
−
−
+
+
−
=
+
∴
This equation contains two unknown (m and N)

y
y
Rearrange MBE to give a straight line equation
( )
[ ] ( )
[ ] ( )
gi
g
oi
g
s
si
oi
o
g
s
p
o
p B
B
B
mNB
B
r
r
B
B
N
B
r
R
B
N −
+
−
+
−
=
−
+ g
g
gi
g
g
p
p
B
E
F
g
o GE
NE
F +
=
o
E
G
o
g
o E
E
G
N
E
F
+
=
∴
N
N
o
g E
E

y
y
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 1060 1 2222 450 0 00096
3800 5.905 1060 1.2222 450 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

y
y
Solution :



+
=
∴ g
E
E
G
N
E
F
P F Eo Eg F/Eo Eg/Eo



+
∴
o
o E
G
N
E
4000 ―x106 0 0 ―x106 ―
3900 5.807 0.0145 0.00005 398.8 0.0034
3800 10.671 0.0287 0.00009 371.8 0.0031
3700 17.302 0.0469 0.00014 368.5 0.0029
3600 24.094 0.0677 0.00020 355.7 0.0028
3500 31.898 0.09268 0.00026 340.6 0.0027
3400 41 130 0 1207 0 00033 340 7 0 0027
3400 41.130 0.1207 0.00033 340.7 0.0027

y
y
F
o
E
9
10
826 ×
=
G
From Fig.
N = 115 x 106 STB
6
10
115 ×
=
N 10
115 ×
N
o
g E
E
2511
1
10
115 6
×
×
×
m
mNB
00087
.
0
2511
.
1
10
115
10
826 9 ×
×
×
=
=
×
=
m
B
mNB
G
gi
oi
5
0
∴ 5
.
0
=
∴m

y
y
Another solution
Assume several values of (m) until the straight line going
through the origin as follows:
g
o GE
NE
F +
=
mNB
g
gi
oi
o E
B
mNB
NE +
=










+
= g
gi
oi
o E
B
mB
E
N
F

y
y
oi
E
mB
E +
F
P
m = 0.6
m = 0.5
m = 0.4
g
gi
oi
o E
B
E +
0 106
0 093
0 081
10 671
3800
0.057
0.051
0.043
5.807
3900
0
0
0
0x106
4000
0 240
0 211
0 183
24 094
3600
0.167
0.147
0.127
17.302
3700
0.106
0.093
0.081
10.671
3800
0.405
0.358
0.311
41.130
3400
0.318
0.244
0.243
31.898
3500
0.240
0.211
0.183
24.094
3600

y
y
F
From Fig.
m = 0 5
m = 0.5
N = 115 x 106 STB
oi
E
mB
E + g
gi
oi
o E
B
E +

3. Water drive reservoirs
Edge water Bottom water
Finite Infinite Finite Infinite
Oil Oil
Water
W W
Water

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

y
y
(N-Np)Bo
NBoi free gas
( ) ( ) gas
free
B
w
W
B
N
N
NB w
p
e
o
p
oi +
−
+
−
=
( ) p
p
s
p
si R
N
r
N
N
Nr
gas
free −
−
−
=
( ) ( ) ( )
[ ]
( )
[ ] ( )
B
w
W
B
r
R
B
N +
( ) ( ) ( )
[ ] g
p
p
s
p
si
w
p
e
o
p
oi B
R
N
r
N
N
Nr
B
w
W
B
N
N
NB −
−
−
+
−
+
−
=
∴
( )
[ ] ( )
( ) g
s
si
oi
o
w
p
e
g
s
p
o
p
B
r
r
B
B
B
w
W
B
r
R
B
N
N
−
+
−
−
−
−
+
=
∴

y
y
Rearrange MBE as an equation of a straight line:
( )
[ ] ( )
[ ] e
g
s
si
oi
o
w
p
g
s
p
o
p W
B
r
r
B
B
N
B
w
B
r
R
B
N +
−
+
−
=
+
−
+
∴
e
o W
E
N
F +
=
o
e
o E
W
N
E
F
+
=
∴
o
o

y
y
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

y
y
Solution :
o
e
o E
W
N
E
F +
=
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
3700 41.862 0.073 573.45 373.5
3600 72.042 0.140 514.38 314.6
o
E
F
o
o
45
From Fig.
6
10
200 ×
=
N
o
e E
W
N = 200 x 106

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

y
y
free gas
i
GB gi
GB
(N-Np)Bo
oi
gi
NB
GB
m =
oi
NB
gi
Pi
( ) ( ) gas
free
B
w
W
B
N
N
GB
NB w
p
e
o
p
gi
oi +
−
+
−
=
+
( ) R
N
N
N
N
G
f
P<Pi
Pi
( ) p
p
s
p
si R
N
r
N
N
Nr
G
gas
free −
−
−
+
=
( ) ( )
( )
[ ]
w
p
e
o
p
oi
oi B
w
W
B
N
N
mNB
NB −
+
−
=
+
∴
( )
[ ] ( )
B
w
W
B
r
R
B
N −
−
−
+
( )
[ ] g
p
p
s
p
si B
R
N
r
N
N
Nr −
−
−
+
( )
[ ] ( )
( ) ( )
gi
g
gi
oi
g
s
si
oi
o
w
p
e
g
s
p
o
p
B
B
B
mB
B
r
r
B
B
B
w
W
B
r
R
B
N
N
−
+
−
+
−
+
=
∴

y
y
This equation includes 3 unknown (We, m & N)
Rearange this equation as a straight line equation
( )
[ ] ( )
[ ] ( ) e
gi
g
oi
g
s
si
oi
o
w
p
g
s
p
o
p W
B
B
B
mB
N
B
r
r
B
B
N
B
w
B
r
R
B
N +








−
+
−
+
−
=
+
−
+
∴
g q g q
( )
[ ] [ ] ( )
g
g
gi
g
p
g
p
p
B 



e
g
oi
o W
E
B
mB
E
N
F +








+
= g
gi
B 



e
mB
W
N
mB
F
+
=
∴
g
gi
oi
o E
B
mB
E
F
+
o
45
g
gi
oi
o
g
gi
oi
o E
B
mB
E
E
B
mB
E +
+
45
N
If We is assumed to be known and m is calculated
by geological dat. N can be obtained
g
gi
oi
o
e
E
B
mB
E
W
+
N

Example 16 :
y
y
Calculate the original oil in place (N)for the following combination drive
reservoir assuming that m=0.5 and values of (We) are given:
P Np Bo rs Rp Bg We
4000 0x106 1 351 600 600 0 00100 0x106
4000 0x106 1.351 600 600 0.00100 0x106
3800 4.942 1.336 567 1140 0.00105 0.515
3600 8.869 1.322 540 1150 0.00109 1.097
3400 17.154 1.301 491 1325 0.00120 3.011

Solution :
y
y
―
―
―
―
0x106
4000
Eo
F
P g
gi
oi
o E
B
mB
E
F
+ g
gi
oi
o
e
E
B
mB
E
W
+
g
gi
oi
o E
B
mB
E +
13 95
183 95
0 2159
0 0808
39 715
3400
11.29
181.29
0.0972
0.0364
17.622
3600
9.66x106
179.66x106
0.0533
0.0196
9.576
3800
0x10
4000
13.95
183.95
0.2159
0.0808
39.715
3400
g
i
oi
o E
B
mB
E
F
+
F F gi
B
o
45
From Fig.
STB
N 6
10
170×
=
g
gi
oi
o
e
E
B
mB
E
W
+
6
10
170 ×
=
N

Uses of MBE
¾ Calculation of (N), (G) and (We)
¾ Prediction of future performance
Difficulties of its application
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
¾High permeability
¾ Homogeneous formation
¾ Low oil viscosity
¾N ti t d i
¾No active water drive
¾ No large gas cap

S l ti f PVT d t f MBE ppli ti s
Selection of PVT data for MBE applications
Depletion drive flash
Gas cap drive differential
C bi ti (fl h diff )
rs
Combination (flash + diff.)
Water drive flash
Low volatile oil differential
rs
High volatile oil flash
Moderate volatile (flash + diff.)
p
p
dif
flash f
f

Water in flux
Due to: Cw, Cf and artesian flow
We
Oil
Bottom water Edge water Linear flux
Oil
Oil
water
W
W
water

Flow regimes
Steady state semi-steady state Unsteady state
Outer boundary condition
Infinite Limited
Infinite Limited

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

Hydraulic analog
Hydraulic analog
( )
∝
∆
∝
P
P
dt
dW
P
q Pi
( )
( )
( )
∑ ∆
−
=
−
∝
P
P
k
W
P
P
k
dt
dW
P
P
dt
dW
i
e
i
e
Pw
( )
∑ ∆
−
= t
P
P
k
W i
e
x
screen sand
q
constant
influx
water
:
k
( ) curve
(Pust)
under
area
:
∑ ∆
− t
P
Pi
C l l f K
Calculation of K:
Water influx rate = oil rate + gas rate + water prod. rate
dW
dN
dN
dW
)
(
)
( P
P
k
B
dt
dW
B
r
R
dt
dN
B
dt
dN
dt
dW
i
w
P
g
s
p
P
o
P
e
−
=
+
−
+
=

Example :
C l l t K i th f ll i d t P 3500 i P 3340
Calculate K using the following data: Pi=3500 psi, P=3340
(Bo=0.00082 bbl/SCF), Qw=0, Bw=1.1 bbl/STB and Qo=13300 STB/day
Solution :
psi
day
bbl
k
day
bbl
dt
dWe
/
/
130
)
3340
3500
(
20800
/
20800
0
00082
.
0
)
700
900
(
13300
4
.
1
13300
=
=
∴
=
+
×
−
×
+
×
=
)
3340
3500
( −
Pi
t1 t2 t3 t4
∆t1 ∆t2 ∆t3 ∆t4
A1 A2 A3 A4
Calculation of ( )
∑ ∆
− t
P
Pi
P1
A1 A2 A3 A4
( )
( )
1
1
4
3
2
1
2
t
P
P
A
A
A
A
t
P
P
i
i
∆
−
=
+
+
+
=
∆
−
∑
P2
P3
( ) ( )
( ) ( )
3
2
2
2
1
2
t
P
P
P
P
t
P
P
P
P
i
i
i
i
∆
−
+
−
+
∆
−
+
−
+
P3
P4
( ) ( )
4
4
3
3
2
2
t
P
P
P
P
t
i
i
∆
−
+
−
+
∆
+

Example :
The pressure history of a steady-state water drive reservoir is
given as follows:
Tdays : 0 100 200 300 400
Ppsi : 3500 3450 3410 3380 3340
If k=130 bbl/day/psi,
calculate We at 100, 200,300 & 400 days
e , , y

100 100 100 100 t
Solution : 50
90
P
t
( ) bbls
We100 000
,
325
0
100
3450
3500
130 





−
−
=
120
160
P
( )
We
e
200
100
1235000
100
2
90
50
100
2
50
130
,
2
=





 +
+
×
=




We
3
200
160
120
120
90
90
50
50
10
2606
100
2
120
90
100
2
90
50
100
2
50
130

 +
+
+
×
=





 +
+
+
+
×
=
bbls
We
3
200 10
4420
100
2
160
120
100
2
120
90
100
2
90
50
100
2
50
130 ×
=





 +
+
+
+
+
+
×
=

Semi
Semi-
-steady
steady-
-state water influx
state water influx
Semi
Semi steady
steady state water influx
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
increases with time, there for (re/rw) is replaced by a time
dependent function (re/rw)→at
P
P
C
P
P
C
P
P
kh
dW × −
)
(
)
(
)
(
10
08
7 3
P
P
C
dW
at
n
P
P
C
r
r
n
P
P
C
r
r
n
P
P
kh
dt
dW i
w
e
w
e
w
e
w
e
e −
→
−
=
−
×
=
∴
)
(
)
(
)
(
)
(
)
(
)
(
)
(
10
08
.
7
l
l
l
µ
P
P
at
n
P
P
C
dt
dW i
e −
=
∴
)
(
)
(
)
(
l
t
at
n
P
P
C
W i
e ∆
−
=
∴ ∑ )
(
)
(
l

The two unknown constants (a and C) are determined as:
(at)
ln
C
dt
dW
P
P
e
i 1
)
(
)
(
=
−
)
(
)
(
dt
dW
P
P
e
i −
t
a
n ln
l
C
C
dt
dW
P
P
e
i 1
1
)
(
)
(
+
=
−
∴
)
( We
1
C
1
Plott this equation as a straight line:
t
ln
a
n
l
C
1
Gives slop = and intercept =
C
1
C
C

Example 18:
Example 18:
Using the following data calculate (a) and (c)
Tmonth P We MBE ∆We (Wen+1-Wen-1) ∆We/ ∆t (Pi-P)
0 3793 0x103 0 0 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
tion
15 3680 309 154 1687 113
18 3643 480 197 2158 150
21 3595 703 249 2727 198
Solut
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

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
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 3844 308 6 816 0 081
30 912.5 3844 308 6.816 0.081
)
(
)
(
dt
dW
P
P
i −
002
.
0
1
=
C
From Fig.
)
( dt
dWe
∴C = 50
Using any point in the straight line
C
002
.
0
1
=
C
Using any point in the straight line
a = 0.064
∑
− P
P
W i
50
t
ln
∑
=
∴
ln(0.064t)
W i
e 50

Example 18:
Example 18:
Using data of example (18) calculate the cumulative water influx (We)
after 39 months (1186.25 days) where the pressure equals 3379 psi
Solution :
P
P
P
P 

dt
t
a
P
P
t
a
P
P
W
W i
i
e
e 




 −
+
−
×
+
= 2
50
2
39
1
36
36
39
ln
ln
[ ]
3416
3793
3379
3793
3 
 −
−
[ ]
1095
25
.
1186
2
)
1095
064
.
0
(
3416
3793
)
25
.
1186
064
.
0
(
3379
3793
50
10
2388 3
×
×






×
+
×
×
+
×
=
ln
ln
3
3 3
3
10
508
.
420
10
2388 ×
+
×
=
bbls
3
10
2809×
=

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

Hydraulic analog
Hydraulic analog
Hydraulic analog
Hydraulic analog
Pi
P2
P1
Pw
q
x q
screen
sand sand sand

Physical analog
Physical analog
Physical analog
Physical analog

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