Reservoir Fluid Parameters Lab Report I have a petroleum engineering lab report that I would like to have some help with it. attached is the requirements a

Reservoir Fluid Parameters Lab Report I have a petroleum engineering lab report that I would like to have some help with it. attached is the requirements and an old lab. note use the same numbers. PNGE 332
Dr. Ebrahim Fathi
Experiment 3:
Determination of reservoir fluid parameters
Bader Bin Askar
800197943
Thursday, October 18, 2018
Cover letter:
Dear Dr.Fathi,
In this lab the students were meant to utilize three different factors, that are, differential
vaporization, flash vaporization and, separator test for hydrocarbon system including propane,
hexane, and methane. Using these three features in order to determine the surface volume of gas
and oil and make a relation between them. And, calculate Gas formation volume factor (Bg), oil
formation volume factor (Bo), and gas solubility (Rsolubility). Using the PVT simulator in order to
estimate the above unknowns, for one unit of volume. 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 =
???????????? ???????????????? ???? ?? ?????????? ???? ?????? ???? ?????????????????? ????????????????????
???????????? ???????????????? ???? ?? ?????????? ???? ?????? ???? ???????????????? ????????????????????
The oil formation volume factor is,
Bo =
???????????? ???? ?????? (?????????????????? ?????????????????? ??????)???????????????? ???? ?????????????????? ????????????????????
???????? ???????????? ???? ?????? ???????????????? ???? ?????????? ???????? ????????????????????
The gas solubility is,
Rs = volume of gas measured at standard conditions associated with a unit volume of
stack tank oil.
Theory, concept, and objective:
The main purpose of this experiment is to find the (Bg), oil formation volume factor (Bo),
and gas solubility (Rsolubility). Using the PVT simulator in order to estimate the above unknowns,
for one unit of volume. And relating them to make a connection to the volume of the surface.
And, plotting them versus the pressure.
Figure 1 flash liberation
Figure 2 differential liberation
The figures above clearly illustrate that there are 2 different ways to separate the gas from
the oil via the separator. The first figure (figure 1) shows how the flash liberation works, this
method works by making a sudden change in pressure to separate the oil from the gas, while
keeping the properties of the gas and oil the same. However, the other figure, explains how the
method of differential liberation. And this method works in a way that all of the oil in the system
is taken away and there is no lift of it in there. The only downfall of the first way is that it will
give a lower amount of volume. Using the flash vaporization test initially to determine the
bubble point. In addition, using the differential vaporization after that to reach to minimize the
pressure in order to get it lower than that of the bubble point. Eventually, the students will have
all they need to calculate the unknown parameters.
Experimental procedure:
PVT simulator
The students in the lab were given an amount of 0.4 mole, and 0.2 mole of hexane and
propane respectively, and a fixed pressure at 2000 psia and a temperature of 100 F. The liquid
and gas in the centrifuge and vessel needed to be collected through the different expansions of
the fluid under the cell conditions 71.6 F, and a pressure of 0 psia. As well as, the densities and
liquid masses, and chromatographic analyses needs to be collected, in the data sheets provided
by the professor. In this experiment, the steps are more difficult than the previous 2 labs, due to
some new features in the PVT simulator. Initially, using the function key F10, will verify the
composition of the initial conditions and record it on the data sheets. Then using what they
learned in experiment 1 and 2, the student need to find the bubble point pressure of the given
conditions. Recall, using F2 to open/close valves to let the displacement pump communicate.
More, F5 can add/remove a certain amount entered by the student of mercury to reach the bubble
point pressure. When the student reaches the bubble point pressure, using F6 to shake the fluid to
reach equilibrium. Worthy to be mentioned, that whenever the system undergoes any change the
F6 button must be used in order to mix the contents, and willingly not doing so, will result in
some inaccurate results.
Liquid expansion (flash), a sample at reservoir conditions will be send to the surface to
undergo the surface conditions. After reaching the bubble point pressure, the student needs to set
the back pressure regulator (BPR) to the next integer pressure above the current cell pressure.
After that, the gas collected in the vessel needs to be evacuated by using the key function F8,
hence, the pressure will drop from 14.7 psia to 0 psia. Close valve 9, and then the liquid will
expand from the cell to the surface along the path through valves 14,16,18, and 19. Then, any
liquid passes by the BPR encounters a pressure of 0 psia, and flashes to both liquid and gas.
Therefore, open the path mentioned above, valves 14, 16 but before opening valves 18, and 19
the density needed to be recorded and can be found in the densimeter, D-2. After doing so, open
valves 18 and 19 in order to make the path. However, instead of using the hand pump the
students were meant to use the servo pump, that is located on the lower left of the screen.
Because, the servo pump can be operated at a given rate. Using the key function F3 to enter the
rate into the servo pump control panel. The maximum delivery is 6 and a rate of 10. The 6 cc, is
at the cell conditions and it will expand to reach the gas and liquid at the surface conditions.
Before activating the servo pump, the students needed to observe how the volume in the cell is
decreasing. A maximum pressure is reached in the cell at a pressure of 1769 psia. More, when
the pump stops, the liquid will drain to the surface at BPR and, on the other hand, the gas left
will be still in the cell at a pressure of 1391 psia. So, the gas will be in the gas vessel and the
liquid will be in the centrifuge tube, record the pressure of the gas and the liquid masses in the
data sheet. To record the needed data first close valves 18 and 19, then open valves 20, 21 to
make a connection between the helium supply and the chromatograph. Using F9 will charge the
contents into the chromatograph, after waiting a minute for the charging to complete. An analysis
of the liquid and gas can now be obtained from the data given in the simulator by pressing F10.
Differential Liberations (gas expansions), the gas is released from an oil sample by
decreasing the pressure and, before reaching an equilibrium with the liquid phase. In this case,
the initial conditions are the same for the previous part, however, the volume in this case is 10cc
of liquid in the cell. This means, that there is less liquid volume, meaning there is a lot of room
for the gas. Likewise, the expansion for the gas is similar to that in the liquid, the only different
is the path of the gas through the valves. The path is connected by valves, 13, 15, 17, and 19.
Opening valve 9 will make some room for the evolved gas that will be produced. But first, the
bubble point is needed to be found. Observe and recorded the required data. For each liberation a
reduction of pressure is required by an amount of 150 psia, from the last value. On table 2 in
page 8, a list of the amount that needs to be withdrawn for each liberation.
Differential Liberation.
Set the BPR using F7 to the next higher integer, and then, F8 to evacuate the gas in the
vessel. Next, make the path mentioned above by opening valves 13, and 15. Now the
density needs to be recorded, but this time from D-1, instead of D-1, because this
density is for the gas. Now continue the path by opening valves, 17 and 19. Setting the
servo pump to 10cc/min and the maximum delivery can be found in Table 2, close
valves 9, 7 and then run the servo pump. Now, close valve 13 and then mix the
contents to reach equilibrium by using F6. At the end, close valves 15,17, and 19, but
open 20 and 21, charge the chromatograph by using the same function key F9 and then
obtain the composition using F10. Close all valves expect 8 and then open 9, then
repeat the procedures for the different liberations.
Results and calculations:
Table 1 the data calculated
stage
0
1
2
3
4
5
6
7
8
stage
0
1
2
3
4
5
6
7
8
Gas
Vap
0
1.68
1.86
2.26
2.62
3.19
4.05
5.52
9.81
m
R
T
P
0
0.14112
0.13392
0.1356
0.12838
0.12122
0.1134
0.1104
0.11772
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
n
0
0.0071724
3
0.0066641
6
0.0067688
4
0.0063099
4
0.0058985
4
0.0052901
7
0.0048430
3
0.0044384
5
V
F1
0
0
16.561990
1
15.388319
6
15.630048
6
14.570395
9
169.9260186
157.8841593
160.3642983
149.4922624
139.7456013
125.3323703
114.7389096
105.1538366
13.620429
12.215630
6
11.183129
6
10.248912
C1
C3
(16.05) (44.11)
0.8083
0.08
0.8986
0.08
0.8973
0.09
0.8934
0.09
0.8854
0.1
0.8707
0.11
0.8434
0.14
0.7907
0.19
0.6669
0.3
C6
(86.2)
0.02
0.02
0.02
0.02
0.02
0.02
0.02
0.02
0.03
F
E
16.56
101.14644
32.29
48.18
63.07
76.84
89.22
100.4
110.57
84.883956
6
70.957654
1
57.058115
4
43.807398
5
30.946264
3
20.786034
3
10.719045
5
Ma
18.226015
19.67533
20.095565
20.03297
20.34567
20.550835
21.43597
22.795635
26.522745
gas
density
0.084
0.072
0.06
0.049
0.038
0.028
0.02
0.012
Table 2 the dataa versus the pressure
P
Vb
Vo
Vg
Vg+Vo
2000
1391
1256
1113
952
785
615
450
299
165
10.26
10.26
10.26
10.26
10.26
10.26
10.26
10.26
10.26
10.26
10
10.26
9.78
9.32
8.86
8.44
8.06
7.71
7.39
7.08
0
0
1.69
1.86
2.6
2.62
3.19
4.05
5.52
9.8
10
10.26
11.47
11.18
11.46
11.06
11.25
11.76
12.91
16.88
Gas
Molar
Mass
35.33
Total
Gas
0
0
1.69
3.55
6.15
7.09
8.41
9.86
12.76
19.39
Vo/Vb
Bg
Bo
Rs
0.97465887
1
0.95321637
0.90838207
0.86354776
0.82261209
0.78557505
0.75146199
0.7202729
0.69005848
–0.00988666
0.01178079
0.01409291
0.01752599
0.02282719
0.03231408
0.04810922
0.09329189
2.44889163
2.51256281
2.39501602
2.282367
2.16971799
2.06686454
1.97380666
1.88809545
1.80973092
1.73381528
450.67
446.6
404.96
365.38
325.43
287.97
253.33
222.19
194.09
168.49
Gas Mass
Gas
moles
Vgas -sc
B oil-sc
Rsi
1.568
0.045
1065.39
2.386
446.596
P Vs. Bo
3
2.5
Bo
2
1.5
1
0.5
0
0
500
1000
Pressure
Figure 3 pressure vs. Bo
1500
2000
P Vs. Rs
500
450
400
350
Rs
300
250
200
150
100
50
0
0
500
1000
1500
2000
1500
2000
Pressure
Figure 4 pressure vs. Rs
P Vs. Bg
0.1
0.09
0.08
0.07
Bg
0.06
0.05
0.04
0.03
0.02
0.01
0
0
500
1000
Pressure
Figure 5 pressure vs Bg
From the above graphs:
•
The pressured decreased until reaching the bubble point due to it being taken out from the
reservoir
•
After reaching the bubble point the Vgas increased but, declined after the bubble point
•
While the gas is being developed Vgas increased, and since it is less than one, at the surface
conditions while the pressure is big, however, the Voil factor is larger than 1 at the point where
the pressure started to decrease
Analysis and discussion:
Utilizing the PVT simulator in this experiment in order to complete it. However, the
problem is that the PVT simulator can draw many errors that can cause some problems in the
calculation. Due to, the simulator using the Peng-Robinson equation, as well as, the assumptions
that was made before. Assumptions such as, the mercury is incompressible, and that meant that
the volume was taken is the same. More, the liquid is never lost in between the tubes because it
is a system in the end and it has to have some drawbacks. The visual cell has a constant
temperature due to the change that occur on the pressure. Worthy to be mentioned, the
experiment has the assumption that there is an equilibrium in thermodynamics. Errors can occur
everywhere, while the system has its error as well as the students did some mistakes that can give
not perfect values. Although, there are so many errors, it is not going to affect the values of this
experiment by a great deal.
Conclusion:
To conclude with, in this experiment the students had to achieve various objectives. The
main reason of doing this experiment however, is to get certain parameters to calculate Bgas, Boil,
and Rsolubility as well as, make a connection between the reservoir condition and the surface
conditions. There are two different methods used in this lab to get the separation of oil-gas the
flash liberation and the differential liberation. However, as the conditions is changing, the
parameters are changing. In the end, the gas and oil mainly rely on the environments of the
reservoir.
References:
1. Fathi, Ebrahim. “Lab manual”. West Virginia University Web. 18 Oct. 2018.
https://ecampus.wvu.edu/bbcswebdav/pid-5123993-dt-content-rid44316507_1/courses/star82881.201808/LAb%20manual.pdf
2. William, D. McCain, Jr. The properties of petroleum fluids, 1990, 2.
Determination of Reservoir Fluid parameters
HAMAD ALSUBAIE
PNGE332-W01
April 2, 2018
Dr. Fathi
The object of this experiment is to determine the three different parameters using the flash
vaporization and differential vaporization in addition to the separator tests. Also, find the relation
between the volume of the gas and the oil at surface to the volume in the reservoir. These three
parameters include, Gas formation factor, Bg, gas solubility, Rs, and the oil formation factor. The
PVT simulator was used to help finding the values of these parameters. Also, helps finding
equation of the gas and oil at the surface in a single unite volume of reservoir oil at its condition.
1. Theory, concepts and objective of the experiment
The goal of this experiment is run different tests to find different parameters and plot them
using Excel with respect to the pressure. The different parameters include: Bo: which is the volume of the oil including dissolved gas measured at reservoir condition
over the unit volume of oil measured at stock tank conditions.
Bg: is the volume occupied by number of moles of gas at reservoir condition over the one at
the standard conditions.
Rs: is the volume of gas measured at standard condition associated with a unit volume of
stock tank
Figure 1 flash liberation
Figure 2 differential liberation
As shown above there is two ways to separate the gas from the oil using the separator. Figure
1 show the flash liberation method where both the composition of the system and the gas
developed from the system kept constant with the oil. Figure 2 shows the differential liberation
method where the gas was taken from the system. The first method will give a less oil volume
due to ejected of the light hydrocarbons which changes in more heaver components.
In the begging the bubble point was found using the flash vaporization test. Than the
differential vaporization was used by reducing the pressure to be less than the bubble point than
moving it to the standard conditions. Finally calculating the needed parameters.
2. Experimental procedure
A 0.4 mole of the hexane and the methane in addition to 0.2 mole of the propane at fixed
pressure and temperature of 100oF and 1000 Psia respectively were used. The gas will be
transferred to another cell with a pressure of 0 Psia and 71.6 oF temperature. The transferring will
be done using the expansion of the fluid. Before that, the initial condition and component should
be recorded using key F10. First, the bubble point should be obtained after 1.572 cc of Hg was
drown. Because this experiment on a mixture not a single component the cell must be circulate to
make sure the results is accurate, this can be done with the key F6 after withdrawing the Hg.
The pressure inside the cell was hold on 1391 Psia using the back pressure regulator on the
key F7. The other cell should be vacuumed completely using the key F8. The gas now can be
transferred to the other cell in the surface condition. To do that the servo pump was used instead
of the hand pump to regulate the rate at 10 cc/min and max of 6 cc using F3 to set that and F4 to
start the transfer. The pressure was adjusted to 1769 Psia, when the pump stopped the pressure
should be recorded using F10. Than He supply was fixed to the chromatograph, and the gas and
liquid were obtained. After that, Rsj and Cbf can be calculated using the data above.
Second, the differential liberation at the begging condition. To a 0.0262 cc of Hg should be
adjusted to find the bubble point, it is important to circulate the cell after each step to get the
optimum results.
Third, the differential liberation was done using 8 steps dropping 150 Psia of the pressure in
each step. The first step will cause the pressure to drop from 1390.2 Psia to 1256 Psia after 1.22
cc of the Hg was withdrawn. All the needed information should be recorded at this step. Than the
back pressure regulator was set to 1391 Psia. Than the densitometer was used to find the density
of the gas. Again, the servo pump was used at 10 cc/min and a maximum of the data in table 2 in
the lab (figure 3 below). Than valve 13 was closed and the cell was circulated to reach
equilibrium. Than the steps were repeated to the other seven trials.
Figure 3 differential liberations
Figure 4 PVT simulator
3. Results and calculations
Table 1 the data calculated
stage
0
1
2
3
4
5
6
7
8
Gas
Vap
0
1.68
1.86
2.26
2.62
3.19
4.05
5.52
9.81
m
R
T
P
0
0.14112
0.13392
0.1356
0.12838
0.12122
0.1134
0.1104
0.11772
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
C1
C3
(16.05) (44.11)
0.8083
0.08
0.8986
0.08
0.8973
0.09
0.8934
0.09
0.8854
0.1
0.8707
0.11
0.8434
0.14
0.7907
0.19
0.6669
0.3
C6
(86.2)
0.02
0.02
0.02
0.02
0.02
0.02
0.02
0.02
0.03
Ma
18.226015
19.67533
20.095565
20.03297
20.34567
20.550835
21.43597
22.795635
26.522745
gas
density
stage
n
V
F1
F
E
0
1
2
3
4
5
6
7
8
0
0.00717243
0.00666416
0.00676884
0.00630994
0.00589854
0.00529017
0.00484303
0.00443845
0
169.9260186
157.8841593
160.3642983
149.4922624
139.7456013
125.3323703
114.7389096
105.1538366
0
16.5619901
15.3883196
15.6300486
14.5703959
13.620429
12.2156306
11.1831296
10.248912
16.56
32.29
48.18
63.07
76.84
89.22
100.4
110.57
101.14644
84.8839566
70.9576541
57.0581154
43.8073985
30.9462643
20.7860343
10.7190455
0.084
0.072
0.06
0.049
0.038
0.028
0.02
0.012
Table 2 the dataa versus the pressure
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
10.26
9.78
9.32
8.86
8.44
8.06
7.71
7.39
7.08
Vg
0
0
1.69
1.86
2.6
2.62
3.19
4.05
5.52
9.8
Vg+Vo
10
10.26
11.47
11.18
11.46
11.06
11.25
11.76
12.91
16.88
Total Gas
0
0
1.69
3.55
6.15
7.09
8.41
9.86
12.76
19.39
Vo/Vb
0.97465887
1
0.95321637
0.90838207
0.86354776
0.82261209
0.78557505
0.75146199
0.7202729
0.69005848
Bo
2.44889163
2.51256281
2.39501602
2.282367
2.16971799
2.06686454
1.97380666
1.88809545
1.80973092
1.73381528
Rs
450.67
446.6
404.96
365.38
325.43
287.97
253.33
222.19
194.09
168.49
Bg
0.00988666
0.01178079
0.01409291
0.01752599
0.02282719
0.03231408
0.04810922
0.09329189
Gas Molar
Mass
35.33
Gas Mass
Gas moles
Vgas -sc
B oil-sc
Rsi
1.568
0.045
1065.39
2.386
446.596
P Vs. Bo
3
2.5
Bo
2
1.5
1
0.5
0
0
500
1000
1500
2000
1500
2000
Pressure
Figure 5 pressure vs. Bo
P Vs. Rs
500
450
400
350
Rs
300
250
200
150
100
50
0
0
500
1000
Pressure
Figure 6 pressure vs. Rs
P Vs. Bg
0.1
0.09
0.08
0.07
Bg
0.06
0.05
0.04
0.03
0.02
0.01
0
0
500
1000
1500
2000
Pressure
Figure 7 pressure vs Bg
We notice from the three figures above that the oil cause the pr…
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Every paper is subject to a strict editorial and revision process. This is to ensure that your document is complete and accurate and that all of your instructions have been followed carefully including creating reference lists in the formats APA, Harvard, MLA, Chicago / Turabian.

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Prices and Discounts

We are happy to say that we offer some of the most competitive prices in this industry. Since many of our customers are students, job seekers and small entrepreneurs, we know that money is a problem. Therefore, you will find better prices with us compared to writing services of this calibre.