Reservoir engineering

1. General Material Balance Equation
Dr. Khaled Saeed Ba-Jaalah
Reservoir Engineering 1
Level-3 Semester-6
‫حضرمـــوت‬ ‫جــــامعة‬
HADRAMOUT UNIVERSITY
‫البترولية‬ ‫الهندسة‬ ‫قسم‬
DEPARTMENT OF PETROLEUM
ENGINEERING
‫البترول‬ ‫و‬ ‫الهندسـة‬ ‫كليـة‬
FACULTY OF ENGINEERING
& PETROLEUM
07/03/2021

2. General Material Balance Equation
The material balance equation is a simple reservoir
performance evaluation and prediction method.
 It treats the reservoir as a single tank or region
described by the average pressure and average
temperature at a given time.
2

3. General Material Balance Equation
2.1 Derivation of the general material balance
equation:
A) The following assumptions are made in deriving a
material balance equation:
1. The reservoir may have an initial vapor phase and
an initial liquid phase.
2. The gas is allowed to dissolve in the liquid phase.
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4. General Material Balance Equation
3. The oil can be volatile in the vapor
phase.
4. Water is allowed to invade the
reservoir from the aquifer during
production.
5. The water and rock are compressible
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5. General Material Balance Equation
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B) In making material balance calculations the
following production, reservoir, and laboratory data
are involved:
1. The initial reservoir pressure and the average
reservoir pressure at successive intervals after the
start of production.
2. The stock tank barrels of oil produced, measured
at 1 atm and 60oF, at any time or during any
production interval.

6. General Material Balance Equation
3. The total standard cubic feet of gas produced.
When gas is injected into the reservoir, this will
be the difference between the total gas produced
and that returned to the reservoir.
4. The ratio of the initial gas cap volume to the
initial oil volume, symbol m. The value of m is
determined from log and core data and from well
completion data.
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7. General Material Balance Equation
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5. The gas and oil volume factors and solutions gas-
oil ratios. These are obtained as functions of
pressure by laboratory measurements on bottom-
hole samples by differential and flash liberation
methods.
6. The quantity of water that has been produced.
7. The quantity of water that has been encroached
into the reservoir from the aquifer.

8. General Material Balance Equation
8
C) The following terms are used in the development
of the general material balance equation:
N: Initial (original) oil in place, STB
Boi : Initial oil formation volume factor, bbl/STB
Np: Cumulative oil produced, STB
Bo : Oil formation volume factor, bbl/STB
G: Initial reservoir gas, SCF
Bgi: Initial Gas formation volume factor, bbl/SCF
Gf : Amount of free gas in the reservoir, SCF
Gp: Cumulative gas produced, SCF
Rsi: Initial solution gas-oil ratio, SCF/STB

9. General Material Balance Equation
9
Rp: Cumulative produced gas-oil ratio, SCF/STB
Rs: Solution gas-oil ratio, SCF/STB
Bg: Gas formation volume factor, bbl/SCF
m: Ratio of initial volume of free gas to initial oil
volume, bbl/bbl
W: Initial reservoir water, bbl
Wp: Cumulative water produced, STB
Bw: Water formation volume factor, bbl/ STB
We: Water influx into reservoir, bbl
cw: Water isothermal compressibility, psi-1
∆P: Change in reservoir pressure (pi- p), psia

10. General Material Balance Equation
10
Swi: Initial water saturation
cf: Formation isothermal compressibility, psi-1
P.V: Total pore volume, bbl
Ginj : Cumulative gas injected, SCF
Bginj : Injected gas formation volume factor, bbl/SCF
Winj : Cumulative water injected, STB
Bwi: Injected water formation volume factor, bbl/
STB
HCPV: Hydrocarbon pore volume

11. General Material Balance Equation
11
Since the control volume we choose remains
constant then the sum of the volume changes of its
constituents must be zero.
Change in oil phase volume + Change in gas phase
volume + Change in water phase volume + Change in
solid phase volume = 0
A. Change in the oil volume:
Initial reservoir oil volume at

12. General Material Balance Equation
12
Oil volume at time t and pressure
Change in oil volume = decrease in oil volume
= initial oil volume – final oil volume

13. General Material Balance Equation
B. Change in gas volume:
Initial free gas volume
Reservoir free gas at pressure P and time t
Change in free gas volume = decrease in free gas volume
= initial gas volume – final gas volume
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14. General Material Balance Equation
14
The determine SCF free gas volume (Gf) from:
SCF free gas = SCF initial gas (free and dissolved )– SCF
gas produced– SCF gas remaining in solution

15. General Material Balance Equation
15
Substituting Gf from the previous relation:
C. Change in the water volume:
Initial reservoir water volume = W
Cumulative water produced at t
Reservoir volume of cumulative produced water

16. General Material Balance Equation
Volume of water encroached at t
Change in water volume = increase in water volume
= initial water volume – final water volume
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17. General Material Balance Equation
D. Change in rock volume:
Change in pore volume
Change in rock volume is negative of change in pore
volume.
Change in rock volume
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18. General Material Balance Equation
Combining the changes in water and rock volumes
results:
Change in water & rock volume
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19. General Material Balance Equation
The total volume of the hydrocarbon system
Initial oil volume + initial gas cap volume
19

20. General Material Balance Equation
Recognizing that
Therefore change in water & rock volume
Summing all changes and equating them to zero
gives:
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21. General Material Balance Equation
Opening the terms and adding and subtracting
Grouping the terms:
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22. General Material Balance Equation
Substituting the following definition:
Rearranging to obtain final form:
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23. General Material Balance Equation
Remembering that:
The MBE for gas cap drive or combination drive
reservoirs may be written as:
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24. General Material Balance Equation
Example (2-1):
Given:
The ratio of the original gas-cap volume to the original oil
volume = 0.175
Initial reservoir pressure = 2700 psia
Initial oil FVF = 1.340 bbl/STB
Initial gas volume factor = 0.001116 bbl/SCF
Initial dissolved GOR = 562 SCF/STB
Reservoir pressure at end of the interval = 2300 psia
Average produced GOR = 700 SCF/STB
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25. General Material Balance Equation
Oil produced during the interval = 20MM STB
Vol. of water produced = 1.05MM STB
Two-phase FVF at 2300 psia = 1.4954 bbl/STB
Vol. of water encroached = 11.58MM bbl
FVF of the water = 1.028 bbl/STB
Gas volume factor at 2300 psia = 0.001510 bbl/ SCF
Determine OOIP and OGIP, , neglect connate water
and rock expansion?
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26. General Material Balance Equation
Solution:
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𝑚𝐵𝑡𝑖
𝐵𝑔𝑖
𝐵𝑔 − 𝐵𝑔𝑖 =
0.175 × 1.34
0.001116
0.001510 − 0.001116 = 0.0828
𝑏𝑏𝑙
𝑆𝑇𝐵
𝑁𝑃 𝐵𝑡 + 𝑅𝑃 − 𝑅𝑠𝑜𝑖 𝐵𝑔 = 20 × 106
1.4954 + 700 − 562 × 0.001116 = 34.08 𝑀𝑀𝑏𝑏𝑙

27. General Material Balance Equation
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28. General Material Balance Equation
Example (2-2):
A combination drive reservoirs has the following
production data and PVT information:
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Pi = 5070 psia N = 10.992 MMSTB G = 31.976 MMMSCF
P (psia) NP
(MMSTB)
GP
(MMMSCF)
Bo
(bbl/STB)
Bg
(bbl/SCF)
Rs
(SCF/STB)
5070 0 0 2.695 0.000926 2909
4000 0.85 2.61 2.093 0.000933 1826

29. General Material Balance Equation
1. Calculate cumulative production gas oil ratio?
2. Calculate size of gas cap?
3. Calculate water influx?
Solution:-
1. Calculate RP:-
29
𝑅𝑃 =
𝐺𝑃
𝑁𝑃
=
2.61 × 109
0.85 × 106
= 3070.6
𝑆𝐶𝐹
𝑆𝑇𝐵

30. General Material Balance Equation
2. Calculate m:-
3. Calculate We:-
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𝑚 =
𝐺𝐵𝑔𝑖
𝑁𝐵𝑜𝑖
=
31.976 × 109
× 0.000926
10.992 × 106 × 2.695
= 0.9995
𝑁 𝐵𝑡 − 𝐵𝑡𝑖 +
𝑚𝑁𝐵𝑡𝑖
𝐵𝑔𝑖
𝐵𝑔 − 𝐵𝑔𝑖 + 1 + 𝑚 𝑁𝐵𝑡𝑖
𝑐𝑤𝑠𝑤𝑖 + 𝑐𝑓
1 − 𝑠𝑤𝑖
∆𝑃
+ 𝑊
𝑒 = 𝑁𝑃 𝐵𝑡 + 𝑅𝑃 − 𝑅𝑠𝑜𝑖 𝐵𝑔 + 𝑊𝑃𝐵𝑤

31. General Material Balance Equation
Calculate Bt:
Calculate N (Bt – Bti):-
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𝐵𝑡𝑖 = 𝐵𝑜𝑖 = 2.695
𝑏𝑏𝑙
𝑆𝑇𝐵
𝐵𝑡 = 𝐵𝑜 + 𝑅𝑠𝑖 − 𝑅𝑠 𝐵𝑔 = 2.093 + 2909 − 1826 × 0.000926
= 3.096
𝑏𝑏𝑙
𝑆𝑇𝐵
𝑁 𝐵𝑡 − 𝐵𝑡𝑖 = 0.85 × 106 × 3.096 − 2.695
= 0.3409 𝑀𝑀𝑏𝑏𝑙

32. General Material Balance Equation
Calculate
𝑵𝒎𝑩𝒕𝒊
𝑩𝒈𝒊
𝑩𝒈 − 𝑩𝒈𝒊 :
Calculate 𝑵𝑷 𝑩𝒕 + 𝑹𝑷 − 𝑹𝒔𝒐𝒊 𝑩𝒈 :
32
𝑁𝑚𝐵𝑡𝑖
𝐵𝑔𝑖
𝐵𝑔 − 𝐵𝑔𝑖 =
0.85 × 106
× 0.9995 × 2.695
0.000926
0.000933 − 0.000926
= 0.0173 𝑀𝑀𝑏𝑏𝑙
𝑁𝑃 𝐵𝑡 + 𝑅𝑃 − 𝑅𝑠𝑜𝑖 𝐵𝑔
= 0.85 × 106
3.096 + 3070.6 − 2909 × 0.000933 = 2.76 𝑀𝑀𝑏𝑏𝑙

33. General Material Balance Equation
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0.3409 × 106
+ 0.0173 × 106
+ 0 + 𝑊
𝑒
= 2.76 × 106
+ 0
∴ 𝑊
𝑒 = 2.4018 𝑀𝑀𝑏𝑏𝑙

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