Chapter 2
Pressure-Volume-Temperature for Oil
PVT analysis – PVT relationship

Gas
scf
Oil
simple
condition  underground condition
 complex (because of bubble po int)
Three main oil PVT parameters
 The
three main parameters required to relate
surface to reservoir volumes for an oil
reservoir:
- Rs: The solution (or dissolved) gas-oil ratio
- Bo: The oil formation volume factor
- Bg: The gas formation volume factor
Definition of the basic oil PVT parameters
- Rs: The solution (or dissolved) gas-oil ratio
[]
SCF gas
STB oil
at reservoir P&T
- Bo: The oil formation volume factor
[ ]
RB (oil  dissolved gas )
STB oil
at reservoir P&T
- Bg: The gas formation volume factor
[]
Note :
RB free gas
SCF gas
T  60 F
Standard Condition
p  14.7 psia
R s , Bo , B g  f ( P , T )
For T = const.
R s , Bo , B g  f ( P )
Determination and Conversion of PVT
Data

Determination of three main oil PVT parameters(RS, Bo, Bg)
-> PVT = f (P only) by routine laboratory analysis

Conversion of PVT data, as presented by the laboratory, to
the form required in the field,
Laboratory - an absolute set of measurements
Field - depend up as the manner of surface
separation of the gas and oil
The complexity of Oil PVT

For gas, PVT relation
pV  nzRT
E  35.37
p
[] scf
rcf
zT
-- Simple relation

For oil, PVT relation > Complex;
PVT parameters must be measured by
laboratory analysis of crude oil samples.
>Relationship between surface and
reservoir hydrocarbon volumes.

The complexity of Oil PVT
The complexity in relating surface volumes of hydrocarbon
production to their equivalent volumes in the reservoir can
be appreciated by considering the following figures:
Undersaturated oil
saturated oil
gas saturated oil
+ free gas (or liberated solution gas)
They are traveling in reservoir at different velocity
How to divide the observed surface gas
production into liberated and dissolved gas
volumes in the reservoir?
 Control
in relating surface volumes of
production to underground withdrawal is
gained by knowing the three oil PVT
parameters which can be measured by
laboratory experiments performed on
samples of the reservoir oil, plus its
originally dissolved gas.
Oil Reservoir & Surface Volume
– Above Bubble Point
Undersaturated oil
Oil Reservoir & Surface Volume
– Below Bubble Point
Saturated oil
gas saturated oil
+ free gas (or liberated
solution gas)
They are traveling in
reservoir at different
velocity
The instantaneous gas-oil ratio or
producing gas-oil ratio
R( SCF
STB
)
the instantaneous gas-oil
ratio or producing
gas-oil ratio
(Underground withdrawal )
STB
 Bo  ( R  Rs )  Bg
( RB
STB
)
Bo as Function of Pressure
Rs as Function of Pressure
Bg and E as Function of Pressure
Producing Gas-oil Ratio (R) as Function of Pressure
Exercise 2.1 - Underground withdrawal
Given:
q o  x ( STB )
D
q g  y ( SCF )
D
measured at t during the
producing life
Calculate:
[in _( RB )]  ?
D
(1) Underground withdrawal rate
expressed in
terms of x & y
(2) Underground withdrawal rate [in _( RB D)]  ?
if pres = 2400 psia , qo =2500 STB/D , qg=2.125 MMscf/D
and
PVT data in fig. 2.5(a)~(c) (p.51) or table 2.4 (P.65)
(3) Pressure gradient of oil =?
if
 o  52.8 lbm / ft3(at s.c),
r
g
 0.67 (air  1), p  2400 psia
(1) R = y/x = qg/qo [=] SCF/STB
p is known  Bo ; Rs ; Bg
Underground withdrawal
qo  x( STB )  Bo ( RB
)  x  Bo ( RB )
D
STB
D
y
q g  x( STB )  (  Rs )( SCF
)  Bg ( RB
)
D x
STB
SCF
y
 x(  Rs ) B g
( RB )
D
x
y
q t  q o  q g  x  Bo  x  (  R s )  B g
x

P = 2400 psia
from table 2.4  Bo =1.1822RB/STB
Rs = 352 SCF/STB
Bg = 0.0012 RB/SCF
x = qo = 2500 STB/D;
y = qg = 2.125 MMSCF/D
 qo  2500  1.1822  2955.5 RB
D
2.125  10 6
q g  2500  (
 352)  0.0012  1494 RB
D
2500
qt  qo  q g  2955.5  1494  4449.5 RB
D
(3)
dP
 o g
dD
To find  o
, applying mass balance, such as
Mass of 1 STB of oil
+
=
Rs scf dissolved gas at
standard condition
Mass of Bo RB of oil
+
dissolved gas in the
reservoir condition
[ osc
lbm
5.615 SCF
lbm
SCF
(
) 1 ( STB)
]   gsc (
)[ Rs
1 ( STB)]
SCF
1 STB
SCF
STB
lbm
RB
5.61458 ft 3
 or ( 3 ) 1 ( STB)  Bo (
)
ft
STB
1 RB
 5.615  osc  Rs   gsc  5.615Bo or
 or 
5.615osc  Rs  gsc
5.615Bo

5.615osc  Rs [rg  0.0763]
5.615Bo
5.615  52.8  352[0.67  0.0763]

 47.37 lbm / ft 3
5.615 1.1822

dP
lbm 1 slug
slug ft
ft
 or g  (47.37 3 
)  (32.2 2 )  47.37  3  2
s
dD
ft 32.2 lbm
ft s
1 ft 2
47.37 lb f 1
 47.37 3 

2
ft 144 in
144 in 2 ft
lb f
 0.329 psi / ft