Expert Answer:Make a lab report, using both instructor material

Answer & Explanation:please use instructor material to make a lab report with the help of these old lab reports into one new lab report.book name:William D. McCain, Jr., The Properties of Petroleum Fluids, 2nd edition (PennWell Publishing Company, 1990). “or the third edition”use the numbers given in (Expansions numbers used in the report)I have added a power point explaining the experiment. and some other materials.
adiq_lab3_5452_.docx

copy_of_lab.xlsx

expansions_numbers_used_in_the_report.pdf

experiment_3_instruction_.pdf

lab_3__5437_.docx

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1
adiq
W01 – W02
Experiment 3: Multi-Component Systems
Determination of Reservoir Fluid Parameters
Cover Letter
2
Dear Dr.
On Friday March, the third experiment of the course took place at one of the Engineering
computer labs. This lab main idea was the determination of reservoir fluid parameters which
leads to its purpose which was aiming to determine and relate surface volumes of oil and gas to
reservoir volumes and the opposite. To be more specific, there are three parameters have to be
determined to get some volume data which are the oil formation volume factor Bo, the gas
formation volume factor Bg, and the gas solubility factor Rs. Moreover, the PVT Simulator was
the method used obtaining all the data required for this lab. However, this will result in a
constant temperature diagram of pressure composition which uses the carbon dioxide in a
constant temperature of 71.6 OF for the assigned group.
In this particular lab, there were three factors that leads to obtain the surface volume and
the three parameters using the formulas. These factors are flash vaporization, differential
vaporization, and a separator test for hydrocarbon system which includes methane, propane, and
hexane. The procedure of doing this lab was to find the bubble point and the dew point by adding
some carbon dioxide to the cell. The initial pressure of the cell was 2000.0psi and 10.0 n-butane
in the beginning. Getting the bubble point can be done with removing mercury from the cell.
Then, mercury have to be injected to the carbon dioxide charging vessel to be added to the cell.
However, after injecting the carbon dioxide gradually until reaching the bubble point where the
gas volume was 0.002cc ±0.00, carbon dioxide in the cell have to be collected as a mole fraction
for the ten different bubble points. Those mole fractions will be from 0.1 to 0.9. After reaching
the 10 different bubble points, some amounts of carbon dioxide have to be added to the cell to
get the dew points pressure, but the volume of the cell is too small to allow all the liquid to be
expanded to the gas phase. Moreover, the other system cell that can be used to get the dew points
3
pressure has only 1.0 cc of n-butane which makes it possible for all the liquid to turn into gas.
Finally, after collecting all the data regarding the several bubble points and dew points, the
resulted mole fraction increases as the bubble point pressure increases. Nevertheless, different
cell pressures derived to the direct correlation between pressure and mole fraction of the carbon
dioxide as the highest-pressure bubble point has the highest mole fraction of CO2. For instance,
the bubble point pressure before adding carbon dioxide was 32.562 psia. After that, the pressure
increases respectively as the mole fraction of carbon dioxide increases as the pressure reached
784.31 psia while the CO2 was 0.9 which illustrates the relation between the pressure, and the
molar fraction.
4
Theory, concepts and objective of the experiment
This experiment has an objective that aims to determine several bubble points and the
dew points pressure with a constant temperature by changing the molar fraction, also to obtain
the relation between molar fraction and presser. The PVT lab simulator had to be used
determining those points which is the available method for us. In the PVT simulator, certain
steps have to be followed to collect the data regarding the bubble and dew points. Firstly, the
valves between the hand pump, the cell, and the CO2 vessel have to be opened. Then, CO2
should be injected gradually to the cell using the mercury until reaching the bubble point where
the fist gas bubble appears. On the other hand, the dew point will be found when the last drop of
liquid remains in the cell, but it is impossible to get it with the 10cc n-butane cell so a 1cc nbutane cell was used.
From fig.2-2, the relation between pressure and molar fraction of CO2 is shown as
pressure increases if the molar fraction increases. This diagram has two lines which illustrates
the relation between for the previous pressure CO2 molar fraction assumption (fig.2-2). In
addition, the line with a linear slope illustrates the pressure while the other line corresponds to
the molar fraction of CO2. The equilibrium line is the line that connects both of those lines.
Critical temperature of the substance is the temperature where the vapor of the substance cannot
be liquefied despite of the applied pressure. Nevertheless, the area above the linear line belongs
to the liquid phase of substance while the area below the curved line belongs to the gas phase.
5
Fig.1. PVT Lab Simulator (Lab).
6
Fig. 2-1. Typical phase diagram of pure substance.
Fig. 2-2. Molar Fraction VS Pressure
7
Experimental procedure
The experiment with inserting the code with the assigned group number and starting the
PVT simulator. That started with a pressure of 2000psi with a 10.0cc of n-butane. The first step
began by releasing four of the valves 08, 09 to be able to remove the mercury from the cell, and
also valves 04 and 11 have to be opened to let the carbon dioxide move to the cell. To open each
valve using the PVT simulator it is simple as pressing the f2 button then type the number of the
valve. The following step is to inject a specific amount of mercury by pressing f5 and entering
the required amount will be shown in the hand pomp box, then close the three valves 01, 04, 11,
and open valve 08. This process of removing mercury and controlling those certain valves will
allow the CO2 to enter the cell gradually until reaching the bubble and the dew points. To collect
the required data for this experiment, there are two boxes to get results and data from which are
the pressure box and the box that can be shown on the upper right hand of the PVT simulator
after pressing f10. In the other hand, the pressure box is next to the valve number 08 and below
the cell. Moreover, the LC pump have to be used after any addition or removal from the cell to
get some accurate readings and maintain the cell. Finally, after reaching the bubble points where
the first bubble of gas appears with an approximate 0.002cc ±0.001 gas volume, and while
continuing the same process to reach the dew point where the volume of the liquid will be
0.002cc ±0.001, information and data regarding the pressure and upper left box have to be
recorded.
8
Results and calculations
For the results, G1 data and information was collected from the experiment at the bubble
points and dew points. In addition, the volume of carbon dioxide is calculated to be able to
calculate its density. As sown in the following figures, the bubble point can be reached when the
first bubble of gas appears where the gas volume start increasing from 0.000cc. In the other
hand, the dew point was reached when the last drop of the liquid changes to the gas phase.
CO2
Pressure
XCO2
n-butane
YCO2
Xn-butane
Yn-butane
0.0
32.56
0.000
0.000
1.000
1.000
0.1
140.90
0.094
0.750
0.906
0.250
0.2
220.70
0.190
0.848
0.810
0.152
0.3
343.30
0.286
0.886
0.714
0.114
0.4
432.63
0.384
0.906
0.616
0.095
0.5
512.32
0.483
0.918
0.517
0.082
0.6
523.02
0.584
0.926
0.416
0.074
0.7
614.21
0.686
0.934
0.314
0.066
0.8
700.41
0.789
0.942
0.211
0.058
0.9
784.31
0.893
0.951
0.107
0.049
Collected lab data
22−16.9
10 MPa Density = 878.2 + (26.9−16.9) ∗ (802.1 − 878.2) = 839.38 kg/m3
22−16.9
15 MPa Density= 920.7 + (26.9−16.9) ∗ (866.4 − 920.7) = 893.05 kg/m3
9
𝑍co2 × 𝑛𝐶4
vco2 = vco2 (
1− 𝑍co2
)
CO2 Density= 880.17
CO2 Molecular Weight = 44.010
vco2 =49.99
VCO2 = 49.99
𝑐𝑐
0.5 × 0.1025 𝑚𝑜𝑙𝑒
) = 5.12 𝑐𝑐
×(
𝑚𝑜𝑙
1 − 0.5
The previous equations aims to get the volume required of the carbon dioxide which is 5.12 cc.
Analysis and discussion
In this experiment, students were assigned to determine the reservoir fluid parameters.
The mole fraction pressure figure above illustrates data collected for G1 in the experiment. This
particular group produced a graph that shows the relation between the molar fraction of the CO2
and the pressure where the temperature is constant. The pressure increases gradually as the molar
fraction increases by adding CO2 to the cell until reaching the bubble point. After reaching the
bubble point by adding carbon dioxide to the cell, a different cell has to be used to determine the
10
dew points and the previous cell is small to turn all the liquid into gas and reach. All in all, the
data that was collected in this lab was from the PVT lab simulator which uses the Robinson’s
equation𝑃 = 𝑉
𝑅𝑇
𝑚
𝑎
− 𝑣2 +2𝑣
−𝑏
𝑚
𝑚 𝑏−𝑏
2
), and this may result in some errors. That is because as it is a
simulator application or program that may include some technical issues such getting wrong
readings as a result of not using the LC pump or other errors, and also Joule-Thomson impacts
are not considered in the PVT simulator.
Conclusions
All in all, determining the several parameters in this experiment resulted in obtaining the
oil formation volume factor Bo, the gas formation volume factor Bg, and the gas solubility factor
Rs which illustrates the relation of reservoir as well as surface in a certain condition. and the dew
points pressure of the binary system that has a constant temperature illustrates the relation in the
PVT simulator as pressure and mole fraction of carbon dioxide have a direct correlation when as
both increases together. Also, determining the dew point was impossible until having another
11
simulator system cell with 1 cc of n-butane to 1 cc. The collected data resulted in a pressure mole
fraction graph of CO2 that helps obtaining more information as the critical point.
References
William D. McCain, Jr., The Properties of Petroleum Fluids, 2nd edition (PennWell
Publishing Company, 1990).
Lab Manual.
12
R
T
P
C1
C3
(16.05) (44.11)
C6
(86.2)
Ma
n
V
F1
F
82,06
82,06
82,06
82,06
82,06
82,06
82,06
82,06
82,06
288,71
288,71
288,71
288,71
288,71
288,71
288,71
288,71
288,71
1
1
1
1
1
1
1
1
1
0,8083
0,8986
0,8973
0,8934
0,8854
0,8707
0,8434
0,7907
0,6669
0,08
0,08
0,09
0,09
0,1
0,11
0,14
0,19
0,3
0,02
0,02
0,02
0,02
0,02
0,02
0,02
0,02
0,03
18,23
19,68
20,1
20,03
20,35
20,55
21,44
22,8
26,52
0
0,0072
0,0067
0,0068
0,0063
0,0059
0,0053
0,0048
0,0044
0
169,9
157,9
160,4
149,5
139,7
125,3
114,7
105,2
0
16,562
15,388
15,63
14,57
13,62
12,216
11,183
10,249
16,6
32,3
48,2
63,1
76,8
89,2
100
111
Gas Mass1,568
C1
16,1
Gas moles0,045
Vgas -sc1065,4
B oil-sc 2,386
Rsi 446,6
C3
C6
44,1
86,2
Vo/Vb
0,9747
1
0,9532
0,9084
0,8635
0,8226
0,7856
0,7515
0,7203
0,6901
Rs
450,67
446,6
404,96
365,38
325,43
287,97
253,33
222,19
194,09
168,49
Bg
P
2000
1391
1256
1113
952
785
615
450
299
165
Vb
10,26
10,26
10,26
10,26
10,26
10,26
10,26
10,26
10,26
10,26
Vo
10
10,3
9,78
9,32
8,86
8,44
8,06
7,71
7,39
7,08
stage Gas Vap m
0
1
2
3
4
5
6
7
8
0
1,68
1,86
2,26
2,62
3,19
4,05
5,52
9,81
0
0,141
0,134
0,136
0,128
0,121
0,113
0,11
0,118
35,33
Gas Molar Mass
Cbf
P
2000
1391
1256
1113
952
785
615
450
299
165
0,398
Vb
10,26
10,26
10,26
10,26
10,26
10,26
10,26
10,26
10,26
10,26
Vg+Vo
10
10,26
11,47
11,18
11,46
11,06
11,25
11,76
12,91
16,88
Vo
10
10,26
9,78
9,32
8,86
8,44
8,06
7,71
7,39
7,08
Vg Total Gas Bo
0
0 2,44889
0
0 2,51256
1,69 1,69 2,39502
1,86 3,55 2,28237
2,6
6,15 2,16972
2,62 7,09 2,06686
3,19 8,41 1,97381
4,05 9,86 1,8881
5,52 12,8 1,80973
9,8
19,4 1,73382
P Vs. Bg
0,01
0,012
0,014
0,018
0,023
0,032
0,048
0,093
P Vs. Bo
P Vs. Bo
P Vs. Bg
3
0,1
0,09
2,5
0,08
0,07
2
Bo
Bg
0,06
0,05
0,04
1,5
1
0,03
0,02
0,5
0,01
0
0
0
500
1000
1500
Pressure
2000
2500
0
500
1000
1500
Pressure
E
101
84,9
71
57,1
43,8
30,9
20,8
10,7
gas
density
0,084
0,072
0,06
0,049
0,038
0,028
0,02
0,012
Vg Vo+Vg Total Gas
0
10
0
0
10,26
0
1,69 11,47
1,69
1,86 11,18
3,55
2,6 11,46
6,15
2,62 11,06
7,09
3,19 11,25
8,41
4,05 11,76
9,86
5,52 12,91
12,76
9,8 16,88
19,39
P Vs. Bo
Stage Gas Vap
Gas
Density
m
C1
C3
C6
Ma
n
V
0,1411
0,1339
0,1356
0,1284
0,1212
0,1134
0,1104
0,1177
0,8083
0,8986
0,8973
0,8934
0,8854
0,8707
0,8434
0,7907
0,6669
0,08
0,08
0,09
0,09
0,1
0,11
0,14
0,19
0,3
0,02
0,02
0,02
0,02
0,02
0,02
0,02
0,02
0,03
18,23
19,68
20,10
20,03
20,35
20,55
21,44
22,80
26,52
0
0,00717
0,00666
0,00677
0,00631
0,00590
0,00529
0,00484
0,00444
0
169,93
157,88
160,36
149,49
139,75
125,33
114,74
105,15
0
1
2
3
4
5
6
7
8
0
1,68
1,86
2,26
2,62
3,19
4,05
5,52
9,81
0,084
0,072
0,06
0,049
0,038
0,028
0,02
0,012
P
R
T
1
82,06
288,71
Vo/Vb
#####
#####
#####
#####
#####
#####
#####
#####
#####
#####
Bo
2,4489
2,5126
2,3950
2,2824
2,1697
2,0669
1,9738
1,8881
1,8097
1,7338
Cbf
0,398
Bg
0,010
0,012
0,014
0,018
0,023
0,032
0,048
0,093
C1
C3
C6
(16.05) (44.11) (86.2)
Rs
450,67
446,6
404,96
365,38
325,43
287,97
253,33
222,19
194,09
168,49
P Vs. Rs
P Vs. Rs
P Vs. Bo
500
450
400
350
Rs
300
250
200
150
100
50
0
1500
Pressure
2000
2500
0
500
1000
1500
Pressure
2000
2500
F1
F
0
16,56 16,56
15,39 32,29
15,63 48,18
14,57 63,07
13,62 76,84
12,22 89,22
11,18 100,4
10,25 110,57
E
101,15
84,88
70,96
57,06
43,81
30,95
20,79
10,72
1st Expansion:
Hand Pump =
Vol w/drawn =
Cell P =
Hand pump =
Cell Contents:
BPR =
Servo Pump:
Gas Density =
Cyl. P =
Cell Contents:
2nd Expansion:
Hand Pump =
Vol w/drawn =
Cell P =
Hand pump =
Cell Contents:
BPR =
Servo Pump:
Gas Density =
Cyl. P =
Cell Contents:
0.262cc
1.2cc
1256psia
1.462cc
Liq =
CH4
C 3H 8
C6H14
9.775cc
X(i)
0.3647
0.2083
0.427
Vap =
Y(i)
0.8986
0.0828
0.0186
1.688cc
1256psia
1.68cc
0.084g/cc
2.515psia
Liq =
CH4
C 3H 8
C6H14
9.775cc
X(i)
0
0
0
Vap =
Y(i)
0.8986
0.0828
0.0186
0.007cc
1.462cc
1.4cc
1113psia
2.862cc
Liq =
CH4
C 3H 8
C6H14
9.318cc
X(i)
0.3263
0.2171
0.4566
Vap =
Y(i)
0.8973
0.0857
0.0170
1.864cc
1113psia
1.86cc
0.072g/cc
2.41psia
Liq =
CH4
C 3H 8
C6H14
9.318cc
X(i)
0
0
0
Vap =
Y(i)
0.8973
0.0857
0.0170
0.003cc
3rd Expansion:
Hand Pump =
2.862cc
Vol w/drawn =
1.8cc
Cell P =
Hand pump =
X(i)
Y(i)
CH4
0.2816
0.8934
951.5psia
C3H8
0.2271
0.0908
4.662cc
C6H14
0.4913
0.0158
Cell Contents:
Liq =
BPR =
952psia
Servo Pump:
2.26cc
Gas Density =
Cyl. P =
8.862cc
Vap =
2.6cc
X(i)
Y(i)
CH4
0
0.8934
0.060g/cc
C3H8
0
0.0908
2.437psia
C6H14
0
0.0158
Cell Contents:
Liq =
8.862cc
Vap =
0.002cc
4thst Expansion:
Hand Pump =
4.662cc
Vol w/drawn =
2.2cc
Cell P =
Hand pump =
X(i)
Y(i)
CH4
0.2337
0.8854
784.35psia
C3H8
0.2372
0.0995
6.862cc
C6H14
0.5291
0.0151
Cell Contents:
Liq =
BPR =
785psia
8.442cc
Vap =
X(i)
2.622cc
Y(i)
Servo Pump:
2.62cc
CH4
0
0.8854
Gas Density =
0.049cc
C3H8
0
0.0995
C6H14
0
Cyl. P =
Cell Contents:
2.266psia
Liq =
8.442cc
0.0151
Vap =
0.004cc
5th Expansion:
Hand Pump =
6.862cc
Vol w/drawn =
2.8cc
Cell P =
Hand pump =
Cell Contents:
X(i)
Y(i)
CH4
0.1832
0.8707
614.55psia
C3H8
0.2470
0.1141
9.662cc
C6H14
0.5698
0.0152
Liq =
BPR =
615psia
Servo Pump:
3.19cc
Gas Density =
Cyl. P =
Cell Contents:
6 th Expansion:
Hand Pump =
Vol w/drawn =
Cell P =
Hand pump =
Cell Contents:
8.056cc
Vap =
3.19cc
X(i)
Y(i)
CH4
0
0.8707
0.038g/cc
C3H8
0
0.1141
2.103psia
C6H14
0
0.0152
Liq =
9.662cc
3.7cc
449.49psia
13.362cc
Liq =
BPR =
450psia
Servo Pump:
4.05g/cc
Gas Density =
0.028g/cc
Cyl. P =
1.901psia
Cell Contents:
Liq =
7th Expansion:
Hand Pump =
13.362cc
Vol w/drawn =
5.2cc
Cell P = 298.95psia
Hand pump =
18.562cc
Cell Contents:
Liq =
8.056cc
Vap =
0.003cc
11.246
0.12122
CH4
C 3H 8
C6H14
7.707cc
X(i)
0.1322
0.2552
0.0126
Vap =
Y(i)
0.8434
0.1401
0.0166
4.052cc
11.759
CH4
C 3H 8
C6H14
7.707cc
X(i)
0
0
0
Vap =
Y(i)
0.8434
0.1401
0.0166
0.008cc
0.1134
CH4
C 3H 8
C6H14
7.394cc
X(i)
0.0839
0.26
0.6561
Vap =
Y(i)
0.7907
0.1889
0.0203
5.519cc
12.913
BPR =
Servo Pump:
Gas Density =
Cyl. P =
Cell Contents:
psia
5.51cc
0.02g/cc
1.685psia
Liq =
CH4
C 3H 8
C6H14
7.394cc
X(i)
0
0
0
Vap =
Y(i)
0.7907
0.1889
0.0203
0.001cc
0.1102
8th Expansion:
Hand Pump =
Vol w/drawn =
Cell P =
Hand pump =
Cell Contents:
18.562cc
9.5cc
164.67psia
28.062cc
Liq =
CH4
C 3H 8
C6H14
7.079cc
X(i)
0.0396
0.2569
0.7035
Vap =
Y(i)
0.6667
0.3022
0.0311
9.816cc
16.895
BPR =
Servo Pump:
Gas Density =
Cyl. P =
Cell Contents:
165psia
9.81cc
0.012g/cc
1.613psia
Liq =
CH4
C 3H 8
C6H14
7.078cc
X(i)
0
0
0
Vap =
Y(i)
0.6667
0.3022
0.0311
0.027cc
0.11772
I ) find your initial condition
Initial Condidtions:
Cell T =
Amb T =
P=
V=
Chromatographic Analysis
of Cell Contents:
X(i)
Y(i)
CH4
C 3H 8
C6H14
II) Find your bubble point pressure (Pb) and Volume (Vb)
III) For each gas expansion stages find the information required in the following table.
Remember you need to find the gas and liquid properties at standard conditions. The gas vessel
is your separator and centrifuge works as your Stock Tank. Therefore for all the stages the
pressure and temperature for separator and Stock Tank is standard conditions.
stage
Gas vaporization
m
R
T
P
C1(16.05)
C3
44.11
C6
86.2
Ma
n
V
F1
F
E
0
1
2
3
4
5
6
7
8
= Gas Actually Expanded (cc)= gas cell content before expansion – gas cell content
after expansion
Gas vaporization
mass of gas = m = gas density gr/cc * Gas Actually Expanded cc = gr
R= Universal gas constant (make sure you use the right dimension i.e. cm3 , atm, K, mol)
T = Temperature standard
P= pressure standard
C1, C3, C6 = since you want to find the gas volume that would be gas mole fractions in gas
chromatography
Ma= apparent molecular weight (exactly the same way you were doing in homeworks)
n = m/Ma
V= gas volume you have all the information just use ideal equation of state for each row
F1= relative gas volume at standard condition = V/Vb
F= Cumulative relative gas volume at standard condition (just do the cumulative of F1)
E= gas expansion factor = gas volume (V) / Gas Actually Expanded
III)
Determination of Shrinkage Factor, cbf, from Liquid Expansion:
Mass of liquid after expansion =
Oil collected at standard conditions is almost entirely composed of hexane (see gas
chromatography)
Density of liquid hexane =
Density of liquid methane =
Density of liquid propane =
0.664 g/cc
0.300 g/cc
0.507 g/cc
Density of liquid= Sum( liquid mole fraction of component * Density of component)
Volume of oil at surface = Mass of liquid after expansion/ Density of liquid
Volume Displaced = maximum delivery by servo pump=
Cbf= Volume of oil at surface/ Volume Displaced
Determination of Initial Solution Gas Ratio, Rsi:
Collected Gas Composition, y(i)
Methane
Propane
Hexane
Apparent Molar Mass of Gas :
Mass of Gass:
Moles of Gas:
Volume of collected gas at
s.c.:
Amount of oil collected at
s.c.(From cbf calculation you
have it):
Initial Solution Gas Ratio:
IV. Find the Bg Bo and Rs
g/mol
g
moles
Cc
cc
However, in order to calculate the needed parameters students were meant to use these
three formulas in order to get them:
The gas formation volume factor is,
Bg =
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑐𝑐𝑢𝑝𝑖𝑒𝑑 𝑏𝑦 𝑛 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑔𝑎𝑠 𝑎𝑡 𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑐𝑐𝑢𝑝𝑖𝑒𝑑 𝑏𝑦 𝑛 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑔𝑎𝑠 𝑎𝑡 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠
The oil formation volume factor is,
Bo =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑜𝑖𝑙 (𝑖𝑛𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑑𝑖𝑠𝑠𝑜𝑙𝑣𝑒𝑑 𝑔𝑎𝑠)𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑎𝑡 𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠
𝑢𝑛𝑖𝑡 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑜𝑖𝑙 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑎𝑡 𝑠𝑡𝑜𝑐𝑘 𝑡𝑎𝑛𝑘 𝑐𝑜𝑛𝑑� …
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