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SOLUCIONARIO DE LOS PROBLEMAS PRODUCCION - I 16/10/2008 ALUMNO: HECTOR ENRIQUE LEIVA ESTEBAN PROFESOR: ING. LUIS DEL CASTILLO RODRIGUEZ PERIODO ACADEMICO: 2008-I

INDICE

I.

OBJETIVO

II.

CONTENIDO TEORICO

III.

ENUNCIADOS

IV.

SOLUCIONES

V.

BIBLIOGRAFIA

I.

II.

OBJETIVO

•

Desarrollar la ecuación de Vogel para Reservorios Saturados con un flujo Pseudoestable.

•

Determinar la curva IPR teniendo en cuenta de que trabajamos con un Reservorio Saturado con EF≠1.

•

Compararemos como es la producción de este tipo de Reservorios teniendo en cuenta para EF=1, EF1 a través de sus respectivas curvas IPR.

CONTENIDO TEORICO OIL INFLOW PERFORMANCE FOR TWO-PHASE RESERVOIR Vogel (1968) introduced an empirical relationship for qo based on a number of history- matching simulations. The relationship, normalized for the absolute open flow potential, qomax is

where, for pseudo-steady state,

and therefore

The convenience of the Vogel correlation is that it allows the use of the properties of only oil in a two-phase system. The viscosity and the formation volume factor must be taken at .

GENERALIZED VOGEL INFLOW PERFORMANCE If the reservoir pressure is above the bubble point and yet the flowing bottomhole pressure is below, a generalized inflow performance can be written. This can be done for transient, steady state, and pseudo-steady state. At first qb, the flow rate, where Pwf=Pb, can be written as

Where PD is the transient dimensionless pressure drop or is equal to for steady state or

for pseudo-steady state.

The productivity index above the bubble point is simply

And is related to qV (denoted here as “Vogel” flow) by

Finally,

III.

ENUNCIADOS PROBLEMA 1: Datos: Pr=2400 psi < Pb De la prueba q=200 b/d para Pwf=1800 psi. Del buildup la EF=0.6. Calcular el IPR del pozo para EF=0.8, EF=1 y EF=1.2.

PROBLEMA 2: Datos: Pr=2500 psi Pb=2000 psi qo=150 b/d Pwf=1500 psi EF=0.8 Obtener las curvas IPR para EF=0.8,1 y 1.2 respectivamente.

IV.

SOLUCION

PROBLEMA 1: Pr= EF= qo= Pwf= 1-(Pwf/Pr)= qomaxEF=1= Restricció n=

2400 0.8 200 1800 0.25 609.7560 98 400

qo para EF = Pwf 2400 2000 1800 1600 1300 1200 700 300 0

(1Pwf/Pr) 0.8 1 0 0 0 0.166666 137.6693 169.3766 67 77 94 243.9024 0.25 200 39 0.333333 257.9945 311.6531 33 8 17 0.458333 336.8563 400.5758 33 69 81 360.9756 426.8292 0.5 1 68 0.708333 465.3116 532.6897 33 53 02 529.2682 586.8902 0.875 93 44 565.8536 609.7560 1 59 98

1.2 0 200 285.3658 54 360.9756 1 456.0975 61 482.9268 29 580.4878 05 655.6097 56

PROBLEMA 2:

Procedimiento: Caso II – Prueba Pwf < Pb 1) Calculamos J usando datos de la prueba en la ecuación:

2) El IPR para Pwf >= Pb es lineal. 3) q0 para valores de Pwf < Pb será:

Teniendo en cuenta que:

La ecuación para qo queda:

Pr= EF=

2500 psi 0.8

qo=

150 b/d

Pwf=

1500 psi

Pb=

2000 psi

(1-Pwf/Pb)

0.25

JEF=0.8= 0.15697674 JEF=1.2= 0.23546512 qomaxEF=1Vogel= 218.023256 b/d qomaxEF=1.2Vogel= 234.418605 b/d Restricción = 571.428571 psi

qo para EF= Presión

(1Pwf/Pb)

0.8 1 1.2 0.156976 0.196220 0.235465 74 93 12

J*= 2500 2200 2000 1500

0

0 0 0 47.09302 58.86627 70.63953 300 2 9 6 98.11046 117.7325 500 78.48837 5 6 0.25 149.9999 185.3197 219.7674

1200

0.4

1000

0.5

750

0.625

500

0.75

250

0.875

100

0.95

0

1

96 186.2092 97 207.5581 34 231.1046 45 251.1627 84 267.7325 51 275.9999 93 280.8139 46

67 227.1802 32 250.7267 44 275.2543 6 294.3313 95 307.9578 48 313.5174 41 316.1337 21

45 265.9186 09 290.4069 81 313.9534 93 329.6511 68 337.5000 05 338.4418 66 352.1511 65

PROBLEMA 3:

•

El corte de agua que se produce al final del reservorio.

•

Un cuadro donde se representa todas las producciones de agua y petróleo y el total también.

Datos: Arena 1

Arena 2

Pr1 =

2500

Pwf =

1500

Pwf/Pr1 =

0.6

qo =

60

P*wf/Pr1 =

0.7592

Pr2 =

1800

Pwf =

1500

qw/(Pr2-Pwf) =

0.4

qw =

120

Pr2/P*wf =

0.948366 7

Solución:

q2max =

101.3513 51 720

J*2 =

P*wf =

1898

q1 =

q1max =

J*1 =

q2 =

1. Hallando qmax1:

2. Hallando qmax2:

0.072972 97 0.4 39.22827 24 65.38113 99

3. Necesitamos hallar el Pwf*=

p=1898

Pwf

Arena 1

Arena 2

Total

2500

0

-280

2400

7.167567 543

-240

2200

20.72432 425

-160

2000

33.24324 313

-80

-280 232.8324 32 139.2756 76 46.75675 69 0.028272 3 44.72432 42 135.1675 67 224.5729 73 312.9405 4 400.2702 7 486.5621 62 571.8162

1898 1800 1600 1400 1200 1000 800 600

39.22827 23 44.72432 417 55.16756 738 64.57297 275 72.94054 029 80.27026 999 86.56216 186 91.81621

-39.2 0 80 160 240 320 400 480

Corte de Agua

0.5918579 55 0.7124632 94 0.7669188 52 0.7994598 25 0.8220943 41 0.8394305

400 200 0

59 96.03243 21 99.21081 047 101.3513 51

560 640 720

16 49 656.0324 0.8536163 32 34 739.2108 0.8657882 1 04 821.3513 0.8766041 51 46

RESERVORIOS ESTRATIFICADOS

V.

BIBLIOGRAFIA • • •

Reservoir Engineering Handbook – Tarek Ahmed (pgs. 321 – 330) Petroleum Production Systems – Michael Economides (pgs. 17-31) Fundamentals of Reservoir Engineering – LP. Dake (pgs. 148-161)