MAB - Pete 310
Fall, 2002
page 1
PETE 310
Application of the Crude Oil PVT Properties in
Reservoir Engineering Problems
Radial Flow of Crude Oils
Using Darcy's law in differential form to a crude oil
flowing in a radial system gives:
qo =
1.1271 (2rh)kh dP
1000o
dr
= res bbl
day
(1)
To express the oil flow rate in STB/day we need to
include the oil formation volume factor.
Qo =
1.1271 (2rh)kh dP
1000oB o
dr
= STB
day
(2)
Separate variables and integrate by imposing the
inner (wellbore) and outer (drainage area) boundaries.
re
Qo
rw
dr= 1.1271 (2rh)kh
r
1000o B o
Pe
dP
(3)
P wf
Notice that we have assumed constant (or average)
properties to integrate equation (3)!
MAB - Pete 310
Fall, 2002
page 2
Qo = 0.00708 k h  P
o B o ln re/rw
Where:
Qo
h
k
∆P
=
=
=
=
=
Bo =
µo =
rw =
re =
oil flow rate , STB/day
pay zone thickness, ft
permeability , md (milidarcies)
difference between reservoir pressure and
bottom-hole flowing pressure
(P - Pwf), psia
formation volume factor of oil at reservoir
pressure, bbl/STB
oil viscosity, cp
wellbore radius, ft
drainage radius, ft
The following sketch helps to identify the process
(4)
MAB - Pete 310
Fall, 2002
page 3
Qo
Pe
h
P wf
re
magnified
wellbore
rw
The application of this equation is limited to
steady-state and laminar flow.
The fluid and the reservoir are homogeneous.
MAB - Pete 310
Fall, 2002
page 4
Oil production example:
Estimate the daily oil production of WELL77J in
STB/day.
The following rock and fluid propertieshave been
given to you.
k =
h =
re =
rw =
∆P =
Pe =
Pwf =
Rs =
time)
gg =
˚API =
T =
1000 md
40 ft
700 ft
0.5 ft
300 psia.
3300 psia
3000 psia
800 SCF/STB ( has remained constant over
0.9 (specific gravity of separator gas)
40
200 ˚F (reservoir temperature)
The PVT lab could not finish a complete Differential
Depletion Test at Reservoir Temperature T=200˚F,
but they gave to you the following data.
MAB - Pete 310
Fall, 2002
page 5
Pressure (psia) Oil Density (gm / cc)
5000
4500
4100
3500
0.681
0.676
0.673
0.666
3000
2500
2350
2100
1850
1600
0.660
0.652
0.665
0.673
0.689
0.697
You have reasons to believe that the reservoir
pressure is above the bubble point pressure of the
reservoir oil because the producing gas-oil ratio has
remained constant. In addition you can (and must )
estimate the bubble point pressure of this reservoir
fluid by using the data of the PVT report.
A plot of pressure vs density ( or pressure vs volume)
at fixed temperature should give you the bubble point
pressure of this oil.
MAB - Pete 310
Fall, 2002
page 6
Experimental Determination of Bubble Point Pressure
0.70
T = 200 ÞF
Density (gr/cc)
0.69
0.68
0.67
0.66
0.65
1000
2000
3000
4000
5000
Pb
Pressure (psia)
We will take the bubble point pressure as Pb = 2,500
psia.
Or alternatively we can obtain this pressure from the
type of plot we are more familiar with
6000
MAB - Pete 310
Fall, 2002
page 7
1.54
T = 200ÞF
Specific Volume (cc/gr)
1.52
1.50
1.48
1.46
1.44
1.42
1000
2000
Pb
3000
4000
5000
6000
Pressure (psia)
Now we are sure that our reservoir is above the
bubble point pressure.
To evaluate the formation volume factor to be used in
equation (4) we first require its value at the bubble
point pressure and then we correct it for a higher
pressure using the oil compressibility.
Bob can be calculated from the following formula:
STO + 0.0135R s g
Bo b =
OR
(5)
MAB - Pete 310
Fall, 2002
page 8
Where:
 STO is obtained from the API gravity of the stock
tank oil:
STO =
141.5
w = 141.5 62.37 = 51.46 lbm /cu ft (6)
API + 131.5
40 + 131.5
The density at reservoir conditions is obtained from
the PVT report. (We interpolate between the two
pressures closest to the reservoir pressure).
We can use a linear interpolation
Pressure (psia) Oil Density
(gm/cc)
3500
0.666
3300
0.664
3000
0.660
thus,
OR = 0. 666 - 0.666 - 0.660 200 = 0.664 gm /cc
(7)
OR = (0.664)(62.37) = 41.414 lbm / cu ft
(8)
3500-3000
MAB - Pete 310
Fall, 2002
page 9
The formation volume factor at the bubble point is
then:
B o b=
51.56 + 0.01358000.9
= 1.48 bbl /STB
41.414
(9)
....but we need the formation volume factor above the
bubble-point pressure
Therefore we need to correct for the compressibility of
the oil above the Pb.
The following equation is to be used:
Bo = B obexp co Pb - P
(10)
We need to find out a value for the isothermal
compressibility.
We must use the data provided in the PVT report :

co = 1
 P
T
= 1 0.666-0.660 = 18.10-6 psi -1
0.664 3500- 3000
(11)
MAB - Pete 310
Fall, 2002
page 10
Therefore the formation volume factor at 3300 psia is:
Bo = 1.48exp 18.10-6 2500 - 3300 = 1.46 bbl / STB (12)
To estimate the viscosity of the oil at reservoir
conditions we will make use of the equations listed in
Appendix B of your text book. The charts from
Chapter 11 in your text book could also be used.
To evaluate the viscosity we follow a three step
procedure:
(1) Evaluate Dead oil viscosity ( µoD ):
log (log (µoD + 1))
0.5644 log T
= 1.8653 - 0.025086API (13)
= 1.8653 - 0.025086 ( 40 ) - 0.5644 log
200
= -0.437
log (µoD + 1)
= 10 - 0.437 = 0.366
µoD
= 10 - 0.366 - 1 = 1.321 cp
MAB - Pete 310
Fall, 2002
page 11
(2) Evaluate Saturated Oil Viscosity ( µob ) :
µob = A µoD B
(14)
A
B
µob
= 10.715 (Rs + 100)- 0515
= 5.44(Rs + 150)- 0338
= 0.374 cp
(3) Evaluate Oil viscosity above Pb
µo = µob ( Pb / P ) B
(15)
B
= C1P C2 exp (C3 + C4P )
= 2.6 P 1.187 exp ( -11.513 - 8.98x10-5 P )
= 0.29
µo = 0.374 ( 3300 / 2500 ) 0.29 = 0.405 cp
Finally the daily oil flow rate is:
Qo = 0.00708 k h  P
o B o ln re/rw
MAB - Pete 310
Fall, 2002
Qo =
0.007081000 40 300
0.405 1.46 ln 700/0.5
Pretty good well !
page 12
= 19,834.193 STB /day