LECTURE – 8
LAB MEASUREMENT OF CRUDE OIL PROPERTIES
PVT Cell
Mercury reservoir
PVT cell
Pressure Gauge
Thermal
jacket
Mercury pump
 PVT cell maintained at constant temperature
 Pressure controlled by positive displacement Hg-pump
 Plunger movement calibrated for volume of Hg injected or
withdrawn
 PVT cell is opaque and so volume of individual fractions
not known. Total fluid volume and change with pressure
 Bubble point established by “feel” – small change in
pressure results in large change in volume
 Two types of experiments performed:
- Flash liberation
- Differential liberation
Flash liberation
oil
v
Vt=v/vb
v =vb
Vt=1
oil
Hg
P > Pb
gas
Vt=v/vb
oil
Hg
P = Pb
Hg
P < Pb
 No gas removed from the PVT cell – gas remains in
equilibrium with oil
 Overall hydrocarbon composition remains same
(components are re-distributed between the gas and liquid
phase)
 vb is known (established by “feel”)
 Vt expressed with respect to the volume at bubble point
pressure. Hence at bp pressure Vt=1 (reservoir barrel)
 Experiment starts with pressure far above bp pressure and
then is gradually reduced.
 Typical data:
Pressure
Relative total vol.
(psia)
Vt = v/vb (rb/rbb)
5000
0.9810
Initial res.
4000
0.9925
pressure
3500
0.9975
Bubble point
3330
1.0000
3000
1.0270
2400
1.1060
Differential Liberation
Vto = 1
oil
gas
vg
gas Vto=vo/vb
vo
oil
Hg
Hg
P = Pb
oil
Hg
P < Pb
 At each stage of depletion gas is physically removed
from contact with oil by injecting Hg at constant
pressure (by keeping the valve downstream of cell
open)
 There is continual change of fluid composition in cell
 Remaining hydrocarbons in cell progressively become
heavier (mol. wt. continually increases)
 Experiment starts at bubble point (since for P>Pb flash
and differential liberation are the same)
 Vtg=vg/vb typically measured relative to the unit volume
of bubble point oil (bbl/bblb). Typical data:
Pressure
Vtg at p and T Vtsc, at sc Vto at P and T
3330
0.0460
8.5211
1.0000
2400
0.0535
6.9457
0.9449
1500
0.0687
6.2333
0.9022
300
0.3728
6.2297
0.8459
14.7 (200oF)
0.8296
14.7 (60oF)
0.7794
Calculation using differential liberation data
 Gas formation volume factor Bg:
Bg is the ratio of gas volume at P and T to the
equivalent volume at standard conditions
At 2400 psia: Bg 
Vt g
Vt sc

0.0535
 0.0077 bbl./STB or
6.9457
0.0077 cuft./SCF
 Gas deviation factor z:
zT
0.0077  2400
Bg  0.02829
z
 0.99
0.02829  660
p
 Oil formation volume factor Bo:
Bo is the ratio of oil volume at reservoir P and T to the
volume at stock tank conditions
From the table, at stock tank conditions Vt o  0.7794
At 2400 psia: Bo 
Vto press
Vto ST

0.9449
1.212 bbl/STB
0.7794
 Solution gas ratio Rso:
Cummulative gas liberated at the end of the experiment:
Vtcum g = 8.5211 + 6.9457 + 6.2333 + 6.2297 = 27.9298
Cummulative gas produced to 2400 psia:
Vt2400 g = 8.5211 + 6.9457 = 15.4668
Therefore: Rso 
(Vtcum g Vt 2400 g )
VtoST
(27.9298  15.4668)

15.99 stb/stb
0.7794
or 89.78 SCF/STB
From laboratory data to field conditions
 From above, computation of fluid properties hinges on
oil at stock tank conditions being invariant no matter
what the liberation process.
 For volatile oils that contain higher fraction of
intermediate hydrocarbons (butane + pentane),
components partition between gas and liquid differently
depending on separation process.
 In general, the volume of oil remaining at the lowest
pressure will be smaller for flash process.
 So which type of lab. data to use for calculating field
values – a combination of both type of data required.
 In the reservoir, close to the well, process similar to
differential liberation, since liberated gas is more
mobile and quickly moves away from contact with oil
 At locations away from well, process similar to flash
liberation
 In surface separator, however, there is flash expansion
of oil i.e. gas stays in equilibrium with oil.
 If multi-stage separation, then gas is physically
removed from the first stage separator, before the oil
enters the second stage separator at different P and T
conditions non-isothermal differential liberation
Combining lab. data with separator tests
 Robust oil properties obtained by combining flash and
differential liberation data
 Oil at bubble point pressure and reservoir T introduced into
separators at different fixed pressure and temperature.
Separator
P
200
150
100
50
Formation Vol.
factor, Bofb
(bblb/stb)
1.253
1.251
1.261
1.276
T
80
80
80
80
Separator GOR
Rsif
(scf/stb)
512
510
515
526
 Optimum separator pressure gives the smallest formation
volume factor (since then you lose the least due to shrinkage)
and correspondingly one with smallest GOR (maximizing
volume of oil)
 Earlier Vto was the relative volume at P and T w.r.t to
bubble point volume (bbl/bblb). Therefore field Bo
equals:
Bo 
voil
voil ST

voil
v
 b  Vto  Bof b bbl/STB
vb voil ST
 Previously we calculated cumulative gas liberated at
any pressure: Vt press g (STB/bblb). Therefore:
( Rsi f  Rso ) SCF/STB 
 Vt press g 
Volume of gas evolved
v oil ST
vb
voil ST
 5.615 Vt press g  Bofb  5.615 scf/stb
Oil PVT Calculations
Given the following differential data:
Pressure
Vtg at P and T Vtsc, at sc Vto at P and T
3330
0.0460
8.5211
1.0000
2400
0.0535
6.9457
0.9449
1500
0.0687
6.2333
0.9022
300
0.3728
6.2297
0.8459
14.7 (200oF)
0.8296
o
14.7 (60 F)
0.7794
 Oil formation volume factor Bo:
0.9449
1.212 bbl/STB
At 2400 psia: Bo 
(1)
0.7794
Designate this oil FVF as Bod – from differential test
Given the following separator data:
Separator
Formation Vol. Separator GOR
factor, Bofb
Rsif
P
T
(bblb/stb)
(scf/stb)
150
80
1.251
510
We calculated: At 2400 psia:
Bo  Bof b Vt o = 1.251*0.9449 = 1.182 bbl/STB
Vto =
voil d
vb d
where voil is the volume at P-T and subscript d
indicates differential test data
voil d
Therefore,
Bo  Bof b 
v ST d
vb
d
v ST d
 Bof b 
Bod
Bbd
bbl/STB
Oil PVT Calculations
From the differential liberation data:
Solution gas ratio Rso:
Vt cum g = 8.5211 + 6.9457 + 6.2333 + 6.2297 = 27.9298
Vt 2400 g = 8.5211 + 6.9457 = 15.4668
Therefore:
(27.9298  15.4668)  5.615
 89.78 STB/STB
0.7794
vcum g
Remember Vt cum g 
, therefore
vb
vcum g
vb
V
Rso d 
 5.615  2400  5.615  Rsid V2400  Bbd  5.615
v ST d
v ST d
vb
vb
Rso d 
vcum
 5.615
vr
From the separator data , we calculated
( Rsi f  Rso ) SCF/STB  V2400 5.615 Bofb SCF/STB
where Rsid 
Therefore
Rso  Rsi f  V2400  5.615 Bofb  Rsi f  ( Rsid  Rsod ) 
Bofb
Bbd