Page 51 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 41 3.1.2007 8:30pm Compositor Name: SJoearun

RESERVOIR DELIVERABILITY 3/41

0:007082kh k ro 1 2 2
Reservoir Reservoir q o ¼ (p p ) (3:55)
wf
i
pressure ¼ 2,250 psig pressure ¼ 1,800 psig ln r e m o B o i 2p i
r w
p wf (psig) q (stb/day) p wf (psig) q (stb/day)
or
2,250 0 1,800 0 q o ¼ J (p p ), (3:56)
0
2
2
2,025 217 1,620 129 i i wf
1,800 414 1,440 246 where
1,575 591 1,260 351
1,350 747 1,080 444 0 0:007082kh k ro 1
1,125 884 900 525 J ¼ m o B o 2p i : (3:57)
i
r e
900 1000 720 594 ln i
r w
675 1096 540 651
450 1172 360 696
The derivative of Eq. (3.45) with respect to the flowing
225 1227 180 729
bottom-hole pressure is
0 1263 0 750
dq o 0
¼ 2J p wf : (3:58)
i
dp wf
3.6.2 Fetkovich’s Method
The integral form of reservoir inflow relationship for This implies that the rate of change of q with respect to p wf
multiphase flow is expressed as is lower at the lower values of the inflow pressure.
Next, we can modify Eq. (3.58) to take into account that
p ð e
0:007082kh in practice p e is not constant but decreases as cumulative
0
q ¼ f (p)dp, (3:53) production increases. The assumption made is that J will
i
ln r e decrease in proportion to the decrease in average reservoir
r w p wf
(drainage area) pressure. Thus, when the static pressure is
where f(p) is a pressure function. The simplest two-phase p e ( < p i ), the IPR equation is
flow case is that of constant pressure p e at the outer
boundary (r e ), with p e less than the bubble-point pressure 0 p e 2 2
e
so that there is two-phase flow throughout the reservoir. q o ¼ J i (p p ) (3:59)
wf
p i
k ro
Under these circumstances, f(p) takes on the value , or, alternatively,
m o B o
where k ro is the relative permeability to oil at the satura- 0 2 2
tion conditions in the formation corresponding to the q o ¼ J (p p ), (3:60)
wf
e
pressure p. In this method, Fetkovich makes the key as- where
sumption that to a good degree of approximation, the
k ro
0
expression is a linear function of p, and is a straight J ¼ J 0 p e : (3:61)
m o B o i
p i
line passing through the origin. If p i is the initial formation
pressure (i.e., p e ), then the straight-line assumption is These equations may be used to extrapolate into the
future.
k ro k ro p
¼ : (3:54)
m o B o m o B o p i Example Problem 3.7 Using Fetkovich’s method plot the
IPR curves for a well in which p i is 2,000 psia and
0
2
Substituting Eq. (3.54) into Eq. (3.53) and integrating the J ¼ 5 10 4 stb=day-psia . Predict the IPRs of the well
i
latter gives at well shut-in static pressures of 1,500 and 1,000 psia.
2,500
Reservoir pressure = 2,250 psig
Reservoir pressure = 1,800 psig
2,000
p wf (psig) 1,500
1,000
500
0
0 200 400 600 800 1,000 1,200 1,400
q (Stb/Day)
Figure 3.19 IPR curves for Example Problem 3.6.