Applied reservoir eng

Applied Reservoir Engineering : Dr. Hamid Khattab
Reservoir DefinitionReservoir DefinitionReservoir DefinitionReservoir Definition
Cap rock
Res. Fluid
Reservoir rock
R iReservoir
Shallow Deep
offshare onshare offshare onshare

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 fluidsReservoir 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 βρg AMw γg Tc PC Z Cg βg µg βw rs µw SalinityCw
ρo γo APT rs βo βt µo Co

TR PR

Applied reservoir Engineering ContentsApplied reservoir Engineering Contents
1. Calculation of original hydrocarbon in place
i. Volumetric methodi. Volumetric method
ii. Material balance equation (MBE)
2 Determination of the reservoir drive mechanism2. 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

Calculation of original hydrocarbon in place byCalculation of original hydrocarbon in place byCalculation of original hydrocarbon in place byCalculation of original hydrocarbon in place by
volumetric methodvolumetric 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

Well Depth
1 h1
G
1 h1
2 h2
3 h3
4 h4
Goc
Gas
Oil4 h4
5 h5
6 h6
7 h
Woc
Oil
Water
7 h7
8 h8
9 h9
30
10
0
Isopach map

)1(. wiSBVN −= φ
)1()( wiSAh −= φ )()( wiφ
wiSAh
β
φ
6155
)1(43560 −
= SCFBblgβ
oiβ615.5
wiSAh
N
φ )1(7758 − STB
STBbbloβ
acresA:
oi
wi
N
β
φ )(
=
iSAhφ )1(7758 −
STB
SCF
fth :
fractionsSwi :,φ
gi
wiSAh
G
β
φ )1(7758
=
SCF fwi,φ

Calculation of (BV) using isopach mapCalculation of (BV) using isopach map
Area inch2C.L
( ) g p p( ) g p p
1. Trapozoidal method:
A110
Ao0 WOC
5.01 >−nn AA
A330
A220
[ ]AAAAA
h
BV nn +++++= − 22......22
2
1210
A550
A440
A’GOC
[ ]
[ ]AA
h
n
nn
′+
′
+
2
2
1210
A770
A660
O76
A770

Calculation of (BV) using isopach mapCalculation of (BV) using isopach map
Area inch2C.L
( ) g p p( ) g p p
2. Pyramid or cone method
5.01 ≤−nn AA A110
Ao0 WOC
[ ]AAAA
h
BV .
3
1010 ++=
A330
A220
[ ]AAAA
h
.
3
2121 +++
A550
A440
A’GOC
[ ] [ ]nnnn A
h
AAAnA
h
3
.
3
11 ++++ −−
A770
A660
O76
A770

Calculation of (BV) using isopach mapCalculation of (BV) using isopach map( ) g p p( ) g p p
3. Simpson method
Odd number of contour lines
[ ]nn AAAAAA
h
BV 24......424
3
13210 ++++++= −
[ ]nA
h
3
3
′
+
3

Converting map areas (inchConverting map areas (inch22) to acres) to acresConverting map areas (inchConverting map areas (inch22) to acres) to acres
Say : Scale 1 : 50000Say : Scale 1 : 50000
1 inch = 50,000 inch
acres56398
(50,000)
inch1
2
2
acres56.398
43560144
( )
inch1 =
×
=

Converting map areas (inchConverting map areas (inch22) to acres) to acres
Example 1 :
Gi th f ll i l i t d f it fGiven 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

Converting map areas (inchConverting map areas (inch22) to acres) to acresConverting map areas (inchConverting map areas (inch22) to acres) to acres
Solution :
[ ]
[ ] [ ] [ ]050050
6
512512
10
12261226
10
2640276298214022002250
2
10
+×+×+×+×+×+=BV
[ ] [ ] [ ]050050
3
6
512512
3
10
12261226
2
10
×+++×+++×+++
ftinch :7198 2
=
acres87.35
43560144
(15,000)
inch1
2
2
=
×
=
ftinch :7198
acresBV 39.25819387.357198 =×=∴
MMSTBN 38.250
4.1
)3.01(25.039.2581937758
=
−×××
=∴

By using Simpson method
[ ]
[ ]0505
6
5212426240476298414022004250
3
10
×+++
×+×+×+×+×+×+×+×+=BV
[ ]0505
3
×+++
ftinch .6.7156 2
=
ftacro
f
.6.25670987.356.7156 =×=
)301(25062567097758
MMSTBN 94.248
4.1
)3.01(25.06.2567097758
=
−×××
=∴
MMSTBNav 66.2492)94.24838.250( =+=∴

Example 2 :
If th i f l 1 i i d
If the reservoir of example 1 is a gas reservoir and
βg=0.001 bbl/SCF. Calculate the original gas in place
S l tiSolution :
)301(250392581937758 −×××
MMSCFG 53.350
001.0
)3.01(25.039.2581937758
=
×××
=

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 : 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)

Solution :
[ ] [ ]310 2
[ ] [ ] ftinchBVoil .9280200220
2
3
22027023102350
2
10 2
=+++×+×+=
[ ] [ ] [ ]BV 5513055130
10
130190
10
190200
7
×++++++= [ ] [ ] [ ]
[ ] [ ] ftinch
BVgas
.79.430325
3
6
25552555
3
10
5513055130
2
130190
2
190200
2
2
=+×+++
×++++++=
acres77.63
43560144
(20,000)
inch1
2
2
=
×
=
MMSTBN 618
3.1
)3.01(25.077.6392807758
=
−××××
=
MMSCFG 6.372
001.0
)3.01(25.077.6379.43037758
=
−××××
=

Reservoir drive mechanismReservoir drive mechanismReservoir drive mechanismReservoir drive mechanism
Water reservoirWater reservoir
P
Gas reservoir
Gas
GasBg
Water
with bottom
water drive
without bottom
water drive
g
Oil reservoir

Oil reservoir
Undersaturated
P>Pb
OilOil
Oil
Water
with bottomwithout 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 reservoirsPVT data for gas and oil reservoirsPVT data for gas and oil reservoirsPVT data for gas and oil reservoirs
Gas reservoirs
Bg
Gas reservoirs
P
ZT
Bg 00504.0=
P

PVT data for gas and oil reservoirsPVT data for gas and oil reservoirs
Saturated oil reservoirs
PVT data for gas and oil reservoirsPVT data for gas and 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
P0 Pi

PVT data for gas and oil reservoirsPVT data for gas and oil reservoirsPVT data for gas and oil reservoirsPVT data for gas and oil reservoirs
Undersaturated oil reservoirs
saturated undersat.
P1 > Pb
Bt
µo
rsi=c
Bo
rs
Bg
1
0

Laboratory measurment of PVT dataLaboratory measurment of PVT dataLaboratory measurment of PVT dataLaboratory measurment of PVT data
GasGas
SCF
Oil
P
Oil
Oil
P
OilOil
SCF
STB
undersaturatedsaturated
Pb P > Pb Pi
P = 14.7 psi
T = 60o F

Calculation of original gas in place by MBECalculation of original gas in place by MBECalculation of original gas in place by MBECalculation of original gas in place by MBE
1. Gas reservoir without bottom water drive
pG
T∆ ( ) gip BGG −giGB
( )
ipp∠ip
gp BG
G =∴
( ) gpgi BGGGB −=
1
gig BB
G
−
=∴ 1

1. Gas reservoir without bottom water drive
Example 4 :
psiP SCFG p SCFbblgB Z G
201.60.810.00084123900
0x1060.830.000770x10-64000
p
G
195.20.770.00095373700
200.20.790.00089273800 .constG ≠
Solution :
199.70.750.00107583600
Using eq. (1)

MBE as an equation of a straight line
y1
( ) gpgi BGGGB −=
( )BBGBG∴ 2
gp BG
G
y1
( )giggp BBGBG −=∴
Another form:
2
gig BB −
x1
( ) 





−=




 i
p
p
TZ
p
ZT
G
p
ZT
G 00504.000504.0 Z
Gp
y2




 i
p
ppp



−=∴ iZZ
GG
Z
3
p
p
G



=∴
i
p
pp
GG
p 3
ii PZpZ −
x2

Another form:
( ) gpgi BGGGB −=
( )  ZZ
p
i
i
Z
p
( ) 





−=
p
Z
GGG
p
Z
p
i
i
00504.000504.0
pGP 


Z
y3 i
i
GZ
p
ppp
i
ip
Z
p
G
G
Z
P






−=∴ 1
G
p
i
i
i
i
G
GZ
p
Z
p
Z
p
−=∴
at pG
x3
G
0
0=
Z
p
pGG =
p

Calculation of original gas in place by MBECalculation of original gas in place by MBE
Example 5 :
Solve the previous example using MBE as a straigh line
Solution :
p PZ ii PZpZ −gp BGgig BB − ZP pG
1244411.752.251.0680.000053900
0x10-64819― x10-52.075x10-4―x104―4000
3738965.912.663.5150.000183700
2741773.152.392.4030.000123800
x3y3x2y2y1x1
5634218.092.885.9900.000303600
From Figgers STBG 6
10200×=

2.Gas reservoir with bottom water drive
R
pG
pW
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
0x10-60.000930x10940000
We bblT years psiP SCFG p SCFbblgB
7.4900.0010772.3338002
2.2970.0009827.8539001
13.3080.00117113.8537003
18.4860.00125151.4836004
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/EF/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 corrosiong 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 00223900 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 expansion3800 +2500 expansion
3700 +4000 expansion
3600 +3700 shrinkage
3500 +5000 expansion
Shrinkage at P=3600 may be due PVT or Gp dataShrinkage at P=3600 may be due PVT or Gpc data

C l l ti f i i l il i l b MBEC l l ti f i i l il i l b MBECalculation of original oil in place by MBECalculation 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

C l l ti f i i l il i l b MBEC l l ti f i i l il i l b MBE
1- Under-saturated oil reservoirs without bottom water
Calculation of original oil in place by MBECalculation of original oil in place by MBE
Np
NBoi
P P
(N-Ni)Bo
P>PPi>Pb
P>Pb
neglecing Cw and Cf
NBoi=(N-Np)Bo
op BN
N∴ (1)
oio
p
BB
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 423800 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 constN ≠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
STBxN 6
10110≠
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
VpBNNNB fwopoi ∆+−= )( ,
Pi>Pb
dpVCdVp
dVp
C
VpVpVp
f
fwfw
→
∆+∆=∆
1
,
(N-Np)Bo
dVp
dpVCdVp
dpV
C pff
f
p
f =→=
1
.
P>Pb
∆Vp,,w
V
dpVCdVp
dp
dVp
V
C www
w
w
w =→= .
1
dpVSCdVpVSV
V
V
S pwwwpww
p
w
w =→=→=

dpVCSCdVp pfwwwf +=∴ )(,
S
NB
VpSVpNB
w
oi
woi
−
=→−=
)1(
)1(
dpNB
S
CSC
dVp oi
fww
wf
−
+
=∴ )
1
(,
pNB
S
CSC
BNNNB
S
oi
fww
opoi
w
∆
+
+−=∴ )
1
()(
1
S
oi
w
opoi
−1

BN
N
op
BB
pB
S
CSC
BB
N
oi
w
fww
oio
op
∆
−
+
+−
=∴
)
1
(
BNBN
N
pBCBB
pB
BB
C
opop
oiooio
oi
oio
o
==∴
∆=−→
∆
−
=Q
SSwhere
pB
S
CSC
S
SC
pB
S
CSC
C
N
wo
oi
w
fww
w
oo
oi
w
fww
o
−=
∆
−
+
+
−
∆
−
+
+
∴
1
]
11
[]
1
[
pB
S
CSCSC
BN
N
oi
fwwoo
op
wo
∆
++
=∴
]
1
[
pCB
BN
N
S
eoi
op
w
∆
=
−1
(2)

PPP i −=∆
Pi
Pi
VV
dP
dV
V
C
oi
o −=
1
.
1
PB
BB
C
oi
oio
o
∆
−
=
Pi
BB
PP
VV
V
oio
i
i
oi
−
−
−
=
)(1
.
1
Voi
Vo
PB
oio
oi ∆
=
)(
.
)salinityand,,(
)(
sw
f
rTPfC
fC
=
= 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
BB
C oio −
=
182.9x10-6――4000
R1(Fig.2)rsf(Fig.1)Cwp(Fig.4)∆P=(Pi-P)P
pB
C
oi
o
∆
=
0.8516.82.958.9284003600
17.22.937.143x10-52003800
15.23.0012.5008003200
162.9810.7146003400

Continue
Cw=CwpxR2R2 (Fig.3)rs= rsf x R1P
w
fwwoo
S
CSCSC
−
++
1
7.725x10-53.3111.1314.623800
―3.30x10-61.415.34000
13.5693.2891.10413.602400
9.5703.2471.1114.283600
13.1433.171.0912.923200

Continue
BN
pCB
BN
N
eoi
op
∆
=
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 13200 14.699 0.1470 105.1

Use MBE as a straight line as follows:
PCNBBN eoiop ∆=
BNF
Plot the fig
oNEF = op BNF =
6
10100×=NPlot the fig. 10100×=N
STBN 6
10100×=
PCBE ∆PCBE eoio ∆=

U d t t d il i ith b tt tU d t t d il i ith b tt tUndersaturated oil reservoir with bottom waterUndersaturated oil reservoir with bottom water
pN
pW pW
(N-Np)Bo
NBoi
P>PbPi>Pb
Assuming (We) is known and neglect Cw+Cf
( ) ( )BwWBNNNB −+−= ( ) ( )
( )
i
wpeop
wpeopoi
BB
BwWBN
N
BwWBNNNB
−
−−
=∴
+=
oio BB
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 4163600 5.362 1.42 2.416
3400 10.033 1.49 3.561
3200 12 682 1 54 4 8323200 12.682 1.54 4.832

Solution :Solution :
( )
oio
wpeop
BB
BwWBN
N
−
−−
=
P NpBo Bo-Boi N
4000 106 1064000 ―x106 ― ―x106
3800 3.314 0.02 108.5
3600 7 775 0 05 107 1 N ≠ C3600 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

U d t t d il i ith b tt tU d t t d il i ith b tt t
E
FRearrange MBE as a straight line
Undersaturated oil reservoir with bottom waterUndersaturated oil reservoir with bottom water
oE
o
45
[ ] eiowpop WBBNBWBN +−=+ 0
WENF +=
110=N
eo WENF +=
oeo EWNEF +=∴
oe EW
[ ]ioo BBE 0−= oe EWp oEFop BNF =[ ]ioo 0
48 32155 57 7750 053600
56.75165.73.3140.023800
― x10-6―― x10-6―4000
p p
34.51139.519.5310.143200
39.56166.414.9800.093400
48.32155.57.7750.053600

Undersaturated oil reservoir with bottom waterUndersaturated oil reservoir with bottom water
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)
Peoi
e
o
e
CB
W
E
W
∆
=PeoiCB ∆
Peoi
oP
o CB
BN
E
F
∆
=P∆eCP
45.07145.060.05364009.5703600
52.06 x106152.01 x1060.02182007.7853800
――――― x10-54000
32.87132.860.147080013.1433200
31.26131.250.113960013.5683400
45.07145.060.05364009.5703600
F
oE
F
o
45
Pi
ee
CB
W
E
W
∆
=
Pi
oP
CB
BN
E
F
∆
=Plot vs
6
10100 ×=N
oe EW
Peoio CBE ∆Peoio CBE ∆
As in Fig. 6
10100×=N

B S t t d il iB S t t d il iB. Saturated oil reservoirsB. Saturated oil reservoirs
1 D l ti d i i1. Depletion drive reservoirs
Characteristics
bPP ≤• b
0=• pW
rapidlyincreasesRp•
FRlow .•

pG
pN
T∆oiNB ( ) op BNN −
p
Free gas
( )
bi PP≤
Free gas
p
( ) gasfreeBNNNB opoi +−=
( ) SCFRNrNNNrgasfree ppspsi −−−=
( ) ( )[ ] gppspsiopoi BRNrNNNrBNNNB −−−+−=∴
( )[ ]( )[ ]
( ) gssioio
gspop
BrrBB
BrRBN
N
−+−
−+
=∴

Example 12 :
Calucaltion (N) for a depletion drive reservoir has the following data : Swi=30%
91 506140 0012731 4236743 873800
― x1067180.0010411.492718― x1064000
NrsBgBoRPNPP
ion
96.014000.0022001.28630776.443400
96.025100.0016271.35519375.263600
91.506140.0012731.4236743.873800
Solut
96.014000.0022001.28630776.443400
As shown N ≠ const., so rearrange MBE as a straight line

( )[ ] ( )[ ]gssioiogspop BrrBBNBrRBN −+−=−+
oENF =
Solution :
P F Eo
4000 0 106 0
Solution
F
4000 0x106 0
3800 5.802 0.0634
3600 19 339 0 2014
6
1096×=N
3600 19.339 0.2014
3400 46.124 0.4804
6
oE
STBNFigFrom 6
1096: ×=
o

( ) gssioiop BrrBBN
FR
−+−
( )
( ) gspo
gssioiop
BrRBN
FR
−+
==.
( )PRPfFR &. = ( )PRPfFR &.
PRFR 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

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
gasfree
S =
( )[ ]i BRNrNNNrgasfree −−−=
volumepore
S g =
( )[ ] gppspsi BRNrNNNrgasfree
( )
bbls6
6
66
1005.280022.0
30771044.6
4061044.6967181096
×=×








××
−××−−××
=
30771044.6 
( )
bbls
S
NB
volumepore
w
oi 6
6
1062.204
3.01
492.11096
)1(
×=
−
××
=
−
=
( )w )(
%7.13137.0
1062.204
1005.28
6
6
==
×
×
=∴ gS

( ) gssioio BrrBB
FR
−+−
( )
( ) gspo
gssioio
Gwithout
BrRB
FR i
−+
=.
( ) 0022.0406718492.1286.1 ×−+− ( )
( )
%7.6067.0
0022.04063077286.1
==
×−+
=
( )
( )
gssioio
Gwith
BrRB
BrrBB
FR i
−+
−+−
=%60.
( )
( ) 002204063077402861
0022.0406718492.1286.1
×−×+
×−+−
=
( ) gspo BrRB +
( )
%49.151549.0
0022.040630774.0286.1
==
××+

2 Gas Cap reservoir2 Gas Cap reservoir2. Gas Cap reservoir2. 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/qoUltimate R.F ∝ Kv, gas cap size, 1/µo, 1/qo

Calculation of OOIP by MBECalculation of OOIP by MBEyy
pN
pG
giGB
(N-Np)Bo
free gas
oiNB
gi
oi
gi
NB
B
m =
( ) gasfreeBNNNBGB opoigi +−=+
P>PbiP
[ ] ( ) ppspsi RNrNNGNrgasfree −−−+=
( )[ ]gspop BrRBN
N
−+
=∴
( ) ( )gig
gi
oi
gssioio BB
B
B
mBrrBB
N
−+−+−
∴
( ) ( )oi
BRNNN
mNB
NBNNNBNB 



+++∴ ( ) ( ) gppsp
gi
oi
siopoioi BRNrNN
B
NrpBNNNBmNB





−−−++−=+∴
This equation contains two unknown (m and N)

Rearrange MBE to give a straight line equation
( )[ ] ( )[ ] ( )gig
oi
gssioiogspop BB
B
mNB
BrrBBNBrRBN −+−+−=−+ gg
gi
ggpp
B
E
F
go GENEF +=
oE
G
o
g
o E
E
GN
E
F
+=∴
NN
og EE

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 000963800 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

Solution :



+=∴ g
E
E
GN
E
F
P F Eo Eg F/Eo Eg/Eo



+∴
oo E
GN
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 00273400 41.130 0.1207 0.00033 340.7 0.0027

F
oE
9
10826 ×=G
From Fig.
N = 115 x 106 STB
6
10115 ×=N 10115 ×N
og EE
2511110115 6
×××mmNB
00087.0
2511.110115
10826 9 ×××
==×=
m
B
mNB
G
gi
oi
50∴ 5.0=∴m

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

oi
E
mB
E +
FP
m = 0.6m = 0.5m = 0.4
g
gi
oi
o E
B
E +
0 1060 0930 08110 6713800
0.0570.0510.0435.8073900
0000x1064000
0 2400 2110 18324 0943600
0.1670.1470.12717.3023700
0.1060.0930.08110.6713800
0.4050.3580.31141.1303400
0.3180.2440.24331.8983500
0.2400.2110.18324.0943600

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

3. Water drive reservoirs3. Water drive reservoirs
Edge water Bottom water
Finite Infinite Finite Infinite
Oil Oil
Water
W W
Water

3. Water drive reservoirs3. Water drive reservoirs
Characteristics
-P decline very gradually
-Wp high for lower structure wells
-Low GOR
-R.F > R.Fgac cap > R.Fdepletion

(N-Np)Bo
NBoi free gas
( ) ( ) gasfreeBwWBNNNB wpeopoi +−+−=
( ) ppspsi RNrNNNrgasfree −−−=
( ) ( ) ( )[ ]
( )[ ] ( )BwWBrRBN +
( ) ( ) ( )[ ] gppspsiwpeopoi BRNrNNNrBwWBNNNB −−−+−+−=∴
( )[ ] ( )
( ) gssioio
wpegspop
BrrBB
BwWBrRBN
N
−+−
−−−+
=∴

Rearrange MBE as an equation of a straight line:
( )[ ] ( )[ ] egssioiowpgspop WBrrBBNBwBrRBN +−+−=+−+∴
eo WENF +=
o
e
o E
W
N
E
F
+=∴
oo

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

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 0.073 573.45 373.53700 41.862 0.073 573.45 373.5
3600 72.042 0.140 514.38 314.6
oE
F
o
o
45From Fig.
6
10200 ×=N
oe EW
N = 200 x 106

4. Combination drive reservoir4. 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 Py p f
-R.F > R.Fwater influx

free gas
iGB giGB
(N-Np)Bo
oi
gi
NB
GB
m =
oiNB
gi
Pi
( ) ( ) gasfreeBwWBNNGBNB wpeopgioi +−+−=+
( ) RNNNNGf
P<Pi
Pi
( ) ppspsi RNrNNNrGgasfree −−−+=
( ) ( )
( )[ ]
wpeopoioi BwWBNNmNBNB −+−=+∴
( )[ ] ( )BwWBrRBN −−−+
( )[ ] gppspsi BRNrNNNr −−−+
( )[ ] ( )
( ) ( )gig
gi
oi
gssioio
wpegspop
BB
B
mB
BrrBB
BwWBrRBN
N
−+−+−
+
=∴

This equation includes 3 unknown (We, m & N)
Rearange this equation as a straight line equation
( )[ ] ( )[ ] ( ) egig
oi
gssioiowpgspop WBB
B
mB
NBrrBBNBwBrRBN +








−+−+−=+−+∴
g q g q
( )[ ] [ ] ( )gg
gi
gpgpp
B 
eg
oi
o WE
B
mB
ENF +








+= g
giB 
e
mB
W
N
mB
F
+=∴
g
gi
oi
o E
B
mB
E
F
+
o
45
g
gi
oi
og
gi
oi
o E
B
mB
EE
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

Calculation of OOIP by MBECalculation of OOIP by MBE
Example 16 :
yy
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 0x1064000 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

Calculation of OOIP by MBECalculation of OOIP by MBE
Solution :
yy
――――0x1064000
EoFP 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 95183 950 21590 080839 7153400
11.29181.290.09720.036417.6223600
9.66x106179.66x1060.05330.01969.5763800
0x104000
13.95183.950.21590.080839.7153400
g
i
oi
o E
B
mB
E
F
+
F F giB
o
45From Fig.
STBN 6
10170×=
g
gi
oi
o
e
E
B
mB
E
W
+
6
10170 ×=N

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

S l ti f PVT d t f MBE ppli ti sSelection of PVT data for MBE applications
Depletion drive flash
Gas cap drive differential
C bi ti (fl h diff )
rsCombination (flash + diff.)
Water drive flash
Low volatile oil differential
rs
High volatile oil flash
Moderate volatile (flash + diff.)
pp
difflash ff

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

Flow regimes
Steady state semi-steady state Unsteady state
Outer boundary condition
Infinite LimitedInfinite 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
rrw
rer

Hydraulic analogHydraulic analog
( )∝
∆∝
PPdtdW
Pq Pi
( )
( )
( )∑ ∆
−=
−∝
PPkW
PPkdtdW
PPdtdW
ie
ie
Pw
( )∑ ∆−= tPPkW ie
x
screen sand
q
constantinfluxwater:k
( ) curve(Pust)underarea:∑ ∆− tPPi
C l l f KCalculation of K:
Water influx rate = oil rate + gas rate + water prod. rate
dWdNdNdW
)()( PPkB
dt
dW
BrR
dt
dN
B
dt
dN
dt
dW
iw
P
gsp
P
o
Pe
−=+−+=

Example :
C l l t K i th f ll i d t P 3500 i P 3340Calculate 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 :
psidaybblk
daybbldtdWe
//130
)33403500(
20800
/20800000082.0)700900(133004.113300
==∴
=+×−×+×=
)33403500( −
Pi
t1 t2 t3 t4∆t1 ∆t2 ∆t3 ∆t4
A1 A2 A3 A4
Calculation of ( )∑ ∆− tPPi
P1
A1 A2 A3 A4
( )
( )
1
1
4321
2
t
PP
AAAAtPP
i
i
∆
−
=
+++=∆−∑
P2
P3
( ) ( )
( ) ( )32
2
21
2
t
PPPP
t
PPPP
ii
ii
∆
−+−
+
∆
−+−
+
P3
P4
( ) ( )
4
43
3
2
2
t
PPPP
t
ii
∆
−+−
+
∆+

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 dayse , , y

100 100 100 100 t
Solution : 50
90
P
t
( ) bblsWe100 000,3250100
34503500
130 



−
−
=
120
160
P
( )
We
e
200
100
1235000100
2
9050
100
2
50
130
,
2
=


 +
+×=

We
3
200
16012012090905050
102606100
2
12090
100
2
9050
100
2
50
130
 +++
×=


 +
+
+
+×=
bblsWe
3
200 104420100
2
160120
100
2
12090
100
2
9050
100
2
50
130 ×=




 +
+
+
+
+
+×=

SemiSemi--steadysteady--state water influxstate water influxSemiSemi steadysteady state water influxstate water influx
As the water drains from the aquifer, the aquifer radius (re)
increases with time, there for (re/rw) is replaced by a timeincreases with time, there for (re/rw) is replaced by a time
dependent function (re/rw)→at
PPCPPCPPkhdW × −
)()()(10087 3
PPCdW
atn
PPC
rrn
PPC
rrn
PPkh
dt
dW i
we
we
we
wee −
→
−
=
−×
=∴
)(
)(
)(
)(
)(
)(
)(1008.7
lllµ
PP
atn
PPC
dt
dW ie −
=∴
)(
)(
)(
l
t
atn
PP
CW i
e ∆
−
=∴ ∑ )(
)(
l

The two unknown constants (a and C) are determined as:
(at)ln
CdtdW
PP
e
i 1
)(
)(
=
−
)(
)(
dtdW
PP
e
i −
tan lnl
CCdtdW
PP
e
i 11
)(
)(
+=
−
∴
)( We
1
C
1
Plott this equation as a straight line:
tln
anl
C
1
Gives slop = and intercept =
C
1
CC

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 0.06612 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 08130 912.5 3844 308 6.816 0.081
)(
)(
dtdW
PPi −002.0
1
=
C
From Fig.
)( dtdWe
∴C = 50
Using any point in the straight line
C
002.0
1
=
C
Using any point in the straight line
a = 0.064
∑
− PP
W i
50
tln
∑=∴
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 :
PPPP 
dt
ta
PP
ta
PP
WW ii
ee 




 −
+
−
×+= 250
2
39
1
36
3639
lnln
[ ]34163793337937933  −−
[ ]109525.11862
)1095064.0(
34163793
)25.1186064.0(
33793793
50102388 3
××





×
+
×
×+×=
lnln
33 33
10508.420102388 ×+×=
bbls3
102809×=

UnsteadyUnsteady--state water influxstate water influxUnsteadyUnsteady state water influxstate 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 analogHydraulic analogHydraulic analogHydraulic analog
Pi
P2
P1
Pw
qx q
screen
sand sand sand

Physical analogPhysical analogPhysical analogPhysical analog

Applied reservoir eng

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