Energy Engineering - Fundamentals of Oil and Gas Engineering

Full exam

1 Solution set and marking schedule for Reservoir Engineering Exam 2021 Prepared by Martin Blunt (1) (i) Start from equation: g g gi p B P Wc B B G G            1 Then: gi g g gi p B B P Wc G B B G             1 Plot          g gi p B B G 1 on the y axis and gi g B B P   on the x axis. Slope = Wc and the y intercept when x=0 is the gas in place, G . Gp (million scf) P (Mpa) Bg (rb/scf) x=DP/(Bg-Bgi) y=Gp/(1-Bgi/Bg) 0 40 0.0018 58 38 0.0022 5000 319 113 36 0.0029 3636.363636 297.9090909 143 34 0.0035 3529.411765 294.4117647 167 32 0.0042 3333.333333 292.25 From the graph below, the intercept, G is approximately 2.4×10 8 scf (acceptable range 2-3) and the slope Wc = 0.016 rb/Pa. There evidence of an active aquifer albeit of modest size. (2 marks for method, 4 marks for table, 4 marks for graph and 8 marks for values, including correct units. Lose 2 marks for any values quoted to 3 or more significant figures.) 2 (ii) Recovery factor is 167/238 = 0.70 (2 marks). The gas saturation         G G B B S S p gi g gi g 1 = 0.51 (2 marks). From the relative magnitude of the two terms in the material balance equation this is 82% gas expansion and 18% water influx – so a fairly weak aquifer drive (8 marks). The field should be produced simply by dropping pressure leading to a recovery likely to be approximately 90% (2 marks), although there may be a problem of water production later in field life to be mitigated by slower production or only producing from near the top of the field (2 marks). (iii) With carbon dioxide injection, the reservoir volume injected is �������������� (4 marks). This is added to the water influx since it also contributes to production and pressure maintenance (4 marks) and converted back to a surface volume of gas to obtain (4 marks): ������� = ������ F1 − �������� ������� G + ������ � + �������������� ������� Carbon dioxide could be injected to improve recovery through maintaining pressure and by being trapped by the water influx rather than natural gas. Secondly it can be used for storage to prevent climate change (4 marks). (2) (i) Study the water pressure: the normal pressure ( Patm + wgz ) would be 34.61 MPa. Under-pressured since the measured water pressure at this depth is lower than the normal pressure (2 marks). (ii) Use g g D g D P P z o w o w owc ) ( 2 1 1 2          = 3309 m (8 marks); g g D g D P P z g o g o goc ) ( 3 2 2 3          = 3272 m. (8 marks). The oil column has a depth of 37 m to the nearest m (2 marks). Can use a graphical method if the depth is found to acceptable accuracy. (iii) The presence of oil below the free water level could be due to slow escape of oil from the cap rock over geological time, resulting in water movement and trapping of some oil in the pore space. Also regional water flows could y = 0.0161x + 238.37 R² = 0.9965 0 50 100 150 200 250 300 350 0 1000 2000 3000 4000 5000 6000 Gp/(1-Bgi/Bg) (million sm3) DP/(Bg-Bgi) (MPa) 3 have disturbed the oil/water contact over time. Finally some oil could be present from a previous oil accumulation or from primary migration – so- called paleo oil (2 marks for each explanation). (iv) The oil/water capillary pressure at the free oil level is simply  g h = 0.14 MPa (6 marks). For the gas/water capillary pressure at the top of the reservoir we need to include the effect of both the oil column and gas, or study the explicit difference in pressure between gas and water = 0.55 MPa (6 marks). (v) STOIIP, N = 51 million sm 3 (4 marks for the value in m 3). For the gas in place, we have NR si for solution gas and in the gas cap, the same equation but using Bg and the height of the gas column to find: G = 30×10 9 sm 3 (4 marks for the gas cap and 4 marks for including solution gas). 3. (i) Low salinity water injection makes the rock more water-wet. Hence it is likely to be ineffective for a rock that is already water-wet or mixed-wet where waterflood recovery is already favourable. Best used for more oil-wet formations. Polymer flooding works principally by improving sweep, but could boost local recovery in oil-wet and, possibly, mixed-wet rocks by supressing water flow: to see this consider what happens to the water fractional flow when the water viscosity is increased substantially. It is less effective in a water-wet rock which is all-shock in terms of solution already. Gas injection reduces the remaining oil saturation – this works well in a water-wet rock with a high residual oil saturation and may work for other wettabilities, but the oil relative permeability may still be low if the flood is not completely miscible (6 marks for each explanation – there is not a definitive answer, but sensible comments need to be made). (ii) You should obtain graphs as follows (5 marks for each graph). 4 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Relative permeability Water saturation Relative permeabilities 5 (iii) Perform tangent constructions on the fractional flow (5 marks) to obtain the saturation profile (5 marks) and recovery plots shown below (5 marks). Indicate the maximum possible recovery of N pD = 0.53. (iv) The discrepancy comes from assuming a one-dimensional displacement in the Buckley-Leverett analysis; in reality we have a heterogeneous three- dimensional reservoir with a less than 100% sweep efficiency, even though the local displacement efficiency is favourable (5 marks): the estimated sweep efficiency is simply the ratio of the Buckley-Leverett to actual recovery: 0.32/0.47 = 0.68 (4 marks). This can be improved through proper well placement or drilling additional wells, or through EOR methods that improve sweep efficiency, such as polymer flooding as mentioned above (4 marks). 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Fractional flow Water saturation 6 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0 0.5 1 1.5 2 2.5 3 Water saturation Dimensionless velocity Buckley-Leverett solution 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 0.5 1 1.5 2 Pore volumes of oil produced Pore volumes of water injected Dimensionless recovery 7 4. (i) In this case the initial pressure is likely to be at the bubble point, indicating the possible presence of a gas cap (4 marks). This is evident from the decreasing values of Bo the falling Rs and increasing Rp initially close to Rs indicative of preferential gas production (2 marks for each comment up to 6 marks). You cannot have gas and oil initially in equilibrium with a gas cap with the initial pressure below the bubble point. (ii) From the simplification of the oil material balance equation below (3 marks), provide the table below (3 marks) and the graph (3 marks). The size of the oil field is 36 MMstb and m =2.8 (6 marks for values and remove two marks for any values quoted to more than 2 significant figures). The total gas in place, including gas in the gas cap and solution gas is 56 Gscf (5 marks). The approach here is to reduce this to an equation of a straight line: ������ /������� = ������ + ������������������ �/������� Np (MMstb) Rp (scf/stb) P (Mpa) Rs (scf/stb) Bo Bg (rb/scf) F Eo Eg F/Eo 0 0 48 500 1.231 0.00314 13 615 47 440 1.219 0.00356 23.946 0.2016 0.164656 118.7798 19 786 46 390 1.211 0.00412 54.00788 0.4332 0.384197 124.6719 26 913 45 340 1.205 0.00512 107.6078 0.7932 0.776236 135.6628 28 1023 44 310 1.197 0.00603 153.8989 1.1117 1.13299 138.4357 (iii) The recovery mechanism can be quantified from the relative contribution of NE o and NmE g in the table above are approximately 74% gas cap expansion and 26% oil expansion and solution gas drive. The production comes principally from the large gas cap (5 marks). This is also seen in the growing and large gas production from the field. The recovery factor is 78% for a less than 10% drop in absolute pressure. This very favourable and – for this small field – continued primary production is likely to the preferable option y = 101.02x + 35.909 R² = 0.993 0 20 40 60 80 100 120 140 160 0 0.2 0.4 0.6 0.8 1 1.2 Volume of fluid produced/epxansion of oil, F/E o (stb) Expansion of gas/expansion of oil, E g /E o 8 (5 marks). Pressure maintenance may be required to prevent excessive gas production, but this depends on whether or not there is a market for the gas. This could be gas re-injection into the gas cap if there is no gas market, or water injection into the base of the oil column (5 marks). Additional data on permeability, reservoir structure and heterogeneity would be needed to design water injection; an economic analysis is required in any event, and in particular to assess the trade-off between gas injection and gas sales, but in this case with a very high recovery factor there is unlikely to be much additional recovery (5 marks). 5. (i) Here the discussion can follow the notes, but the key things I am looking for – in the student’s own words – are the initial conservation equation for capillary-controlled flow (2 marks): ������ ������������ � ������������ + ������������ � ������������ = 0 ������� = ������ ������� ������� ������� ������������ � ������������ a definition of the capillary dispersion D (3 marks): ������ (������� ) = − ������������ � ������� ������������ � ������������ � ������������ � and the final equation (2 marks): ������ ������6������ ������������ �6 = −2 ������ T There then needs a clear description of approaches (integral or finite difference) to solve this equation iteratively (5 marks). 6 marks for drawing an explanatory diagram. (ii) D governs how rapidly imbibition occurs or – more specifically – the diffusive spreading of the water (3 marks). Large D results in a spread-out imbibition front, with low saturation gradients and fast imbibition (3 marks). D=0 is the limit of no dispersion – seen at the leading edge of the imbibition front into an irreducible water saturation – where we have an incipient shock, or an infinite saturation gradient (6 marks). (iii) Time to diffuse scales as length squared. So we expect first an increase in time by a factor of 4 for a layer twice as thick (3 marks). But the pore space is shrunk by a factor of 2. This is subtle: the permeability decreases by a factor of 4 (proportional to pore area) but capillary pressure increases by a factor of 2 (proportional to 1/pore size) (6 marks): this results in an intrinsic halving of imbibition rate (3 marks). Overall then the time increases by a 9 factor of eight to 120 s (2 marks). Only credit the final answer with the proper explanation – no marks for a lucky guess! (iv) In a hydrophobic material there is no spontaneous imbibition (4 marks). If there is some imbibition it will be much slower (2 marks).