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Material balance Equation
1. Chapter 3
Material Balance Applied to Oil Reservoirs
§ 3.1 Introduction
-The Schilthuis material balance equation
- Basic tools of reservoir engineering
=> Interpreting and predicting reservoir performance.
-Material balance
1. zero dimension – this chapter
2. multi-dimension (multi-phase) – reservoir simulation
2. § 3.2 General form of the material balance equation for a
hydrocarbon reservoir
Underground withdrawal (RB)
= Expansion of oil and original dissolved gas (RB)………(A)
+ Expansion of gascap gas (RB) ……………… ………(B)
+ Reduction in HCPV due to connate water expansion and decrease in
the pore volume (RB)……………………… …….……(C)
3. 1
N = initial oil in place ( STB ) = Vφ (1 − S wc ) ( STB )
Boi
initial H .C. Vol. of the gascap SCF ft 3 bbl
m= (= const.) [=] or or
initial H .C. Vol. of the oil SCF ft 3 bbl
N p = cumulative oil production (STB )
cum. gas production ( SCF )
R p = cumulative gas − oil rato =
cum. oil production ( STB )
4. Expansion of oil & originally dissolved gas
Liquid exp ansion (oil exp ansion) = liq. (at p ) − liq. (at pi )
RB
= NBo − NBoi = N ( Bo − Boi )[=]STB [=]RB (3.1)
STB
Liberated gas exp ansion = − [ solution gas (at p) − solution gas (at pi )]
SCF RB
= NRsi B g − NRs B g = N ( Rsi − Rs ) B g [=]STB [=]RB (3.2)
STB SCF
5. Expansion of the gascap gas
Expansion of the gascap gas =gascap gas (at p) –gascap (at pi)
SCF RB
The total volume of gascap gas = mNBoi [=] STB [=]RB
SCF SCF
1 1
or G = mNBoi [=]RB [=]SCF
B gi RB
SCF
1 RB
Amount of gas (at p ) = mNBoi B g [=]SCF [=]RB
B gi SCF
Bg
Expansion of the gascap gas = mNBoi − mNBoi [ =]RB
B gi
Bg
= mNBoi ( − 1) ( RB) (3.3)
B gi
6. Change in the HCPV due to the connate water expansion &
pore volume reduction
d ( HCPV ) = − dVw + dV f V f = the total pore vol. = HCPV /(1 − S w )
= −(c wVw + c f V f )∆p Vw = the connate water vol. = V f ⋅ S wc
HCPV
= −(c wV f ⋅ S wc + c f V f ) ∆p = −V f (c w S wc + c f )∆p = − (c w S wc + c f ) ∆p
(1 − S w )
c w S wc + c f
= −(1 + m) NBoi ( ) ∆p
1 − S wc
7. Underground withdrawal
Pr oduction at surface N p ( STB ) − oil N p R p ( SCF ) − gas
Underground withdrawal N p Bo ( RB) − oil ( N p R p − N p Rs ) B g ( RB) − gas
⇒ Underground withdrawal = N p Bo + N p ( R p − Rs ) B g = N p [ Bo + ( R p − Rs ) B g ]
8. The general expression for the material balance as
Bg
N p [ Bo + ( R p − Rs ) B g ] = N ( Bo − Boi ) + N ( Rsi − Rs ) B g + mNBoi ( − 1)
B gi
c w S wc + c f
+ (1 + m) NBoi ( )∆p + (We − W p ) Bw
1 − S wc
( Bo − Boi ) + ( Rsi − Rs ) B g Bg c w S wc + c f
⇒ N p [ Bo + ( R p − Rs ) B g ] = NBoi + m − 1 + (1 + m)
1− S ∆ p
Boi B
gi wc
+ (We − W p ) Bw (3.7)
Note : Bo , Rs , B g = f ( p)
We = f ( p, t )
Simple form : Pr oduction = Expansion of reservoir fluids
dV = c ⋅ V ⋅ ∆p
Main difficulty : measuring p
9. F = N ( Eo + mE g + mE f ,w ) + We Bw (3.12)
where
F = N p [ Bo + ( R p − R s ) B g ] + W p B w [=]RB
E o = ( Bo − Boi ) + ( Rsi − Rs ) B g [=] RB
STB
Bg
E g = Boi ( − 1) [=] RB
B gi STB
c w S wc + c f
E f , w = (1 + m) Boi ( )∆p RB
1 − S wc STB
10. F = N ( E o + mE g + mE f , w ) + We Bw (3.12)
No initial gascap, negligible water influx c f & cw → 0
Eq.(3.12) ⇒ F = NE o (3.13)
With water influx eq(3.12) becomes
F We
= N + (3.14)
Eo Eo
Eq.(3.12) having a combination drive-all possible sources of energy.
11. § 3.4 Reservoir Drive Mechanisms
Reservoir drive mechanism
-reducing the M.B to a compact form to
quantify reservoir performance
- Solution gas drive -determining the main producing
- Gascap drive characteristics,
-Natural water
drive
In terms of for example, GOR; water cut
- Compaction drive -determining the pressure decline in the
reservoir
- estimating the primary recovery factor
12. § 3.5 Solution gas drive
(a) above the B.P. pressure (b) below the B.P. pressure
13. Above the B.P. pressure
- no initial gascap, m=0
- no water flux, We=0 ; no water production, Wp=0
- Rs=Rsi=Rp
from eq.(3.7)
( Bo − Boi ) + ( R si − R s ) B g Bg c w S wc + c f
N p [ Bo + ( R p − R s ) B g ] = NBoi + m − 1 + (1 + m)
1− S
∆p
Boi B
gi wc
+ (We − W p ) B w (3.7)
Note : ( R p − Rs ) = 0 ; ( Rsi − Rs ) = 0 ; m = 0 ;We = 0 ;W p = 0
( Bo − Boi ) (c w s wc + c f )
⇒ N p Bo = NBoi [ + ∆p ]
Boi 1 − S wc
cw S w + c f 1 dVo 1 dBo
⇒ N p Bo = NBoi [co + ( )]∆p co = − =−
1 − S wc Vo dp Bo dp
co S o + c w S w + c f − ( Boi − Bo ) ( Bo − Boi )
⇒ N p Bo = NBoi ( )∆p (3.17) ≈ =
1 − S wc Boi ∆p Boi ∆p
or N p Bo = NBoi ce ∆p (3.18) S o + S wc = 1
co S o + c w S w + c f
where ce = = the effective, saturation − weighted compressibility
1 − S wc
14. Exercise3.1 Solution gas drive, undersaturated oil reservoir
if p − pi → pb
Determine R.F. PVT − table 2.4 ( p.65)
c w = 3 × 10 −6 psi −1 c f = 8.6 × 10 −6 psi −1 S w = 0.2
Solution:
1 dVo 1 dBo
FromTable2.4(p.65) co = − =−
Vo dp Bo dp
Bob − Boi
pi = 4000 psi, Boi = 1.2417 RB co =
STB Boi ∆p
1.2511 − 1.2417
pb = 3330 psi, Bob = 12511 RB = = 11.3 ×10 −6 psi −1
STB 1.2417(4000 − 3330)
Eq(3.18)
N p Bo = NBoi c e ∆p
Np Boi
R.F . = = ce ∆p
N Pb
Bob
1.2417
= ×22.8 ×10 −6 ×( 4000 −3330)
1.2511
= 0.015 =1.5%
22. Exercise3.2 Solution gas drive; below bubble point pressure
Reservoir-described in exercise 3.1
Pabandon = 900psia
(1) R.F = f(Rp)? Conclusion?
(2) Sg(free gas) = F(Pabandon)?
Solution:
(1) From eq(3.7)
( Bo − Boi ) + ( R si − R s ) B g Bg c w S wc + c f
N p [ Bo + ( R p − Rs ) B g ] = NBoi + m − 1 + (1 + m)
1− S
∆ p
Boi B
gi wc
+ (We − W p ) B w (3.7)
for solution gas below B.P.
−m =0 no initial gas cap
− We = 0 ; W p = 0
c w S wc + c f
− NBoi ( )∆p is negligible if S g is developed
1 − S wc
23. Eq(3.7) becomes
N p [ Bo + ( R p − Rs ) B g ] = N [( Bo − Boi ) + ( Rsi − Rs ) B g ] (3.20)
Np ( Bo − Boi ) + ( Rsi − Rs ) B g
R.F . = =
N Bo + ( R p − R s ) B g
Np (1.0940 − 1.2417) + (510 − 122) × 0.00339 344
R.F . p =900 = = =
N p =900
1.0940 + ( R p − 122) × 0.00339 R p + 201
1
Conclusion: RF ≈
Rp
From Fig .3.3 ( p.55)
Np
R p = 500 SCF ⇒ = 49% = 0.49
STB N 900
24.
25. (2) the overall gas balance
liberated total gas gas still
gas in the = amount −
− produced dissolved
reservoir of gas at surface in the oil
NBoi HCPV
= pore volume = for p > p b
1 − S wc (1 − S wc )
NBoi S g
⇒ = NR si B g − N p R p B g − ( N − N p ) Rs B g
(1 − S wc )
[ N ( Rsi − Rs ) − N p ( R p − Rs )]B g (1 − S wc )
⇒ Sg = (3.21)
NBoi
Np
[ N ( Rsi − Rs ) − Np( R p − Rs )]Bg (1 − S wc ) [( Rsi − Rs ) − ( R p − Rs )]
Sg = = N Bg (1 − S wc )
NBoi Boi
[(510 − 122) − 0.49(500 − 122)]
= × 0.00339 × 0.8 = 0.4428
1.2417