METODO DE BUCKLEY ‐ LEVERETT L= a= h= Ø= Swi = Kabs = = µw =
58 5850 650 30 25 20 120 . 0.85
pies pies pies % % md
qi/pz = α= βo = K= WOR =
225 26 1.25 0.12 37
µw/µo =
0.425
ρo =
0.82 gr/cc
qt = Δρ =
ρw =
1.0 gr/cc
qd =
N=
cps
3 pozos inyectores
qo =
Sw
Kr w
Kr o
K r o /K r w
0.20
0.000
0.800
infinito
0.25
0.002
0.610
0.30
0.009
0.35
BPD grados bbl/bbf d
0.035 BPD/pie2 0.18 675 BPD 3790. 379 0.12 125 5 ft ft3/ 3/di diaa
Sw1 = Sw2 = ko/kw1 = ko/kw2 = ΔSw = a=
0.385 0.57 10 1 0.185 1205 .
T1 =
100
T2 =
125
T3 =
150
fw
d f w /d s w
x 1 (100)
x 2 (125)
x 3 (150)
305.0000
0. 0.00734
0.09455
7.351
9.189
11.027
0.470
52.22 .2222
0.04 .04176
0.51 .51348
39.92 .921
49.901
59.88 .881
0.020
0.370
18.50 .5000
0.11 .11005
1.24 .24593
96.86 .866
121.08 .082
145.29 .299
0.40
0.033
0.285
8.63 .6364
0.21 .21004
2.09 .09433
162.82 .826
203.53 .533
244.23 .239
0.45
0.051
0.220
4.31 .3137
0.34 .34776
2.84 .84243
220.98 .988
276.23 .235
331.48 .482
0.50
0.075
0.163
2.17 .1733
0.51 .51419
3.10 .10670
241.53 .534
301.91 .918
362.30 .301
0.55
0.100
0.120
1.20 .2000
0.65 .65695
2.78 .78394
216.44 .441
270.55 .551
324.66 .661
0.60
0.132
0.081
0.61 .6136
0.78 .78886
2.04 .04200
158.75 .757
198.44 .447
238.13 .136
0.65
0.170
0.050
0.29 .2941
0.88 .88592
1.22 .22927
95.57 .571
119.46 .464
143.35 .357
0.70
0.208
0.027
0.12 .1298
0.94 .94601
0.61 .61672
47.94 .948
59.935
71.92 .922
A.‐ CALCULO DEL FLUJO FRACCIONAL (fw) Ej. Sw= 0,25 1 ‐ (0.488 *0,12*0,610*(1,0 ‐ 0,82) Sen 26°) fw = 2*0,035 1+ 305*0,85 2.0
= 0,00734
B.‐ RECUPERACION AL ROMPIMIENTO O IRRUPCION DE AGUA
(Swf ‐ Swi) * 100 (1 ‐ Swi)
R = Np/N
R =
(0,62 ‐ 0,20) * 100 (1 ‐ 0,20)
Swf = Swi =
0.62 (De Gráfica) 0.20
52.50 %
C.‐ TIEMPO QUE SE REQUIERE PARA ALCANZAR LA IRRUPCION DE AGUA
POR DEFINICION :
ts =
δfw
=
δSw
Sŵf
1
1
(Swf ‐ Swi)
(0,62 ‐ 0,20)
L *Ø*A qo* δfw δSw
2.38
De gráfico Swf =
5850*0,25*19500 3790 * 2,38
ts =
0.62
3160 DIAS 8.7 AÑOS
Swf
D.‐ VOLUMEN DE PETROLEO RECUPERADO AL ROMPIMIENTO O IRRUPCION DE AGUA
Np = Ø*A*L (Swf ‐ Swi) βo
0,25*19500*5850* (0,62 ‐ 0,20) 1.25
9.58 MM PC
G.‐ EL VOLUMEN DE AGUA QUE SE REQUIERE INYECTAR PARA ALCANZAR UN WOR = 37
w=
ny* p
w ‐ w (1 ‐ fw2)
ny =
Vw = (Sw2 ‐ Swf) * Vp
(0,73 ‐ 0,62) * 2,15 1 ‐ 0,974
(1 ‐ fw2)
9.00 MM BBL
Vw =
H.‐ EL TIEMPO NECESARIO PARA ALCANZAR UN WOR = 37
t = fw2* Vp Qd
0,974*2150000 675
3106 DIAS
t=
9 AÑOS LA DERIVADA (dfw/dSw) : = δSw
* *
=
= (Sw2 ‐ Sw1)
(1+(µw/µo)(Kro/Krw))^2
Ej. Sw = 0,25 δfw δSw
= (0,85/2,0)*12,45*305 (1 + (0,85/2,0)*305)^2
= 0,09455
= Xf =
Wi Ø*A
t
*
δfw
qt * t
δfw
δSw Swf
Ø*A
δSw Swf
b = Ln (10/1) 0,57 ‐ 0,385 .
,
= 12,45
= ,
5,615*675*100*0,09455 0,25 * 19500
= 7,351
10.0
1.00 0.95
9.0
0.90 0.85
8.0
0.80 0.75
7.0
0.70
w0.65 F l a 0.55 n o i c 0.50 c a r 0.45 F o 0.40 j u 0.35 l F
6.0
0.30
3.0
0.60
5.0
4.0
0.25
2.0
0.20 0.15
1.0
0.10 0.05
0.0
0.00
0.0
0.1
0.2
0.3
0.4
0.5
0.6
Saturación de agua (Sw)
0.7
0.8
0.9
1.0
w s d / w f d
100.000
w K / o K s a v i t a l e r s e d a d i l i b a e m r e p e d n ó z a R
10.000
1.000
0.100
0.010
0.001
0.0
0.1
0.2
0.3
0.4
0.5
0.6
Saturación de agua (Sw)
0.7
0.8
0.9
1.0
