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 = 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 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠 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. Lab Report #3: Determination of Reservoir Fluid parameters Mohammed Alsemail 800199817 4/6/2018 Cover letter: Dear Dr. Fathi, The Third lab was done in the ESB the goal is to find the surface volume of oil and gas, then finding the relation to reservoir volume. Another goal here was to find the three unknowns(Bg,Bo,Rs) using three methods which are: flash vaporization, differential vaporization and separator test on methane, propane and hexane. Figure 1: PVT simulator After that, the three variables are found using the PVT simulator, then the quantities of gas and oil that will result in surface conditions are predicted. Theory, Concepts, and Objective of the experiment: First, the objective for this experiment was to find three variables which are, (Bo, Bg, and Rs) Following are the two methods to reach oil-gas separation which are the following: - flash liberation In this method the gas keeps its contact with the oil so the system keeps the same composition - the differential liberation The gas leaves the system. This means that every method will give a different result. The oil volume in the flash liberation is expected to be higher, the reason for that is the ejected light hydrocarbons which is going to cause a change in the heavylight ratio components. The steps which will be done on the pvt are the following. First, doint the flash liberation: 1- find the bubble point 2- the pressure must be higher than that for the differential liberation: 1- get the pressure to lower than the bubble point 2- measure the gas at the standard conditions after that claculate the three variables. Experimental procedure: In this experiment PVT simulator was used in the computer lab to avoid the dangerous of mercury. Figure 2: PVT simulator - 0.4 moles of hexane and 0.2 of propane were used respectively at 100 F and 1000 psia - After every step the data should be recorded by F10 - The gas is transferred to another cell, p=0, T=71.6 F using the fluid expansion - The bubble point is reached after withdrawing 1572 cc of mercury - The cell must be shaken everytime for the accurate data to show up - The back pressure regulator was used to hold the pressure at 1391 psia (F7) - The other cell is vaccumed completely (F8) - The servo pump is used at a rate of 10 cc/min and a 6 cc maximum (F3 to set the data, F4 to transfer) - The pressure was changed to 1769 psia and the data was recorded - The gas and liquid are obtained - Rsj and Cbf can be calculated using the data - 0.0262 of mercury is needed to find the bubble point - The differential liberation was done during 8 steps with reducing the pressure by 150 every time - 1.22 cc Hg withdrawn, pressure drop from 1390.2 to 1256 and the data was recorded. - The back pressure regulator is set back to 1391 psia - The densitometer is used to find the gas density - The servo pump is used at a rate of 10 cc/min with a maximum as indicated in the table 2 in the lab Figure 3: differential liberation - Valve 13 is closed and the cell is shaken and the data is recorded - The last 5 steps are repeated for the 8 liberations. Results and Calculations: 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 stage n V F1 F E 0 1 0 0.00717243 0 169.9260186 0 16.5619901 16.56 101.14644 Ma 18.226015 19.67533 20.095565 20.03297 20.34567 20.550835 21.43597 22.795635 26.522745 gas density 0.084 2 3 4 5 6 7 8 0.00666416 0.00676884 0.00630994 0.00589854 0.00529017 0.00484303 0.00443845 157.8841593 160.3642983 149.4922624 139.7456013 125.3323703 114.7389096 105.1538366 15.3883196 15.6300486 14.5703959 13.620429 12.2156306 11.1831296 10.248912 32.29 48.18 63.07 76.84 89.22 100.4 110.57 84.8839566 70.9576541 57.0581154 43.8073985 30.9462643 20.7860343 10.7190455 0.072 0.06 0.049 0.038 0.028 0.02 0.012 P 2000 1391 1256 1113 Vb 10.26 10.26 10.26 10.26 Vo 10 10.26 9.78 9.32 Vg 0 0 1.69 1.86 Vg+Vo 10 10.26 11.47 11.18 Total Gas 0 0 1.69 3.55 Vo/Vb 0.97465887 1 0.95321637 0.90838207 Bo 2.44889163 2.51256281 2.39501602 2.282367 Rs 450.67 446.6 404.96 365.38 0.00988666 0.01178079 952 785 615 450 10.26 10.26 10.26 10.26 8.86 8.44 8.06 7.71 2.6 2.62 3.19 4.05 11.46 11.06 11.25 11.76 6.15 7.09 8.41 9.86 0.86354776 0.82261209 0.78557505 0.75146199 2.16971799 2.06686454 1.97380666 1.88809545 325.43 287.97 253.33 222.19 0.01409291 0.01752599 0.02282719 0.03231408 299 165 10.26 10.26 7.39 7.08 5.52 9.8 12.91 16.88 12.76 19.39 0.7202729 0.69005848 1.80973092 1.73381528 194.09 168.49 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 Bg P Vs. Bo 3 2.5 Bo 2 1.5 1 0.5 0 0 500 1000 1500 2000 1500 2000 Pressure Figure 4: P vs. Bo P Vs. Rs 500 450 400 350 Rs 300 250 200 150 100 50 0 0 500 1000 Pressure Figure 5: P 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 6: P vs Bg From the three graphs: - The oil causes the pressure to decrease - The gas volume increases before the bubble point and decreases after it - When the gas volume increases the gas volume factor becomes less than 1 at surface conditions - When the pressure of the gas increase the gas volume factor becomes more than 1. Analysis and Discussion: The fact that PVT simulator which used Peng Robinson equation was used means that there are many sources of error one of these is that the following assumptions were made: - Mercury is incompressible - The tube has no volume - The thermodynamic equilibrium happens instantaneously - Isothermal pressure is neglected - Joule-Thompson effect is ignored Also, the program has a (+/-)0.001 error percentage. However, the effect of these error sources will keep the result acceptable. Conclusion In conclusion, the objective which is to obtain the three variables ( B g, Bo, and Rs) was done successfully. Also, it is approached that the variables change with the change of the conditions References - Lab manual. - The book: William, D. McCain, Jr. The properties of petroleum fluids, 1990, 2.
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