Gas Reservoirs - SEC. 10 GAS RESERVOIRS 33 wells, vvhich in...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: SEC. 10 GAS RESERVOIRS 33 wells, vvhich in some cases vary from the well spacing because of the reservorr limlts. meg to the much smaller gradients, the three averages are much closer together than in the case of the Jones sand. Most engineers prefer to prepare an isobaric map as illustrated in Fig. 1.8 and to planimeter the areas between the isobaric lines and the isopach 2900 PSI ISOBARIC LINES 30 FT 20 FT l0 FT 0 ISOPACH LINES Fig. 1.8. Section of an isobaric and isopachous map. lines as shown in Fig. 1.8. Table 1.7, using data taken from Fig. 1.8, illustrates the method of obtaining the average volumetric pressure from this type of map. Table 1.7. VOLUMETRIC CALCULATIONS OF RESERVOIR PRESSURE (1) I (2) (3) (4) (5) (6) Pressure Area Acres* psia h ft A X h p X A X h A 25.5 2750 25 6375 175,313,000 B 15.1 2750 15 226.5 62,288,000 C 50.5 2850 25 1262.5 359,813,000 1) 30.2 2850 15 453.0 129,105,000 ‘ 2579.5 726,519,000 * Planimetered areas of Fig. 1.8 l b . fl 726,519,000 Average pressure on a v0 umc asxs — 25795 = 2817 psia ll. Calculation of Unit Recovery from Volumetric Gas Reser- voirs. In many gas reservoirs, particularly during the development period, the bulk volume is not known. In this case it is better to place the reservoir calculations on a unit basis, usually one acre—foot of bulk reservoir rock. 34 GAS RESERVOIRS CHAP. 1 Then one unit or one acre-foot of bulk reservoir rock contains Connate water: 43,560 Xfl) >< SW cubic feet Reservoir gas volume: 43,560 X 45 (1 — SW) cubic feet Reservoir pore volume: 43,560 X qb cubic feet The initial'standard cubic feet of gas in place in the unit is G = 43,560 X ¢ X (1 —— SW) X By SCF/ac-ft (1.19) G is in standard cubic feet Where the gas volume factor B gi is in standard cubic feet per cubic foot, SCF/cu ft, Eq. (1.7). The standard conditions are those used in the calculation of the gas volume factor, and they may be changed to any other standard by means of the ideal gas law. The porosity d) is expressed as a fraction of the bulk volume, and the connate water Sw as a fraction of the pore volume. For a reservoir under volumetric control there is no change in the interstitial water, so the reservoir gas volume re— mains the same. If B ga is the gas volume factor at the abandonment pres- sure, then the standard cubic feet of gas remaining at abandonment is i G, = 43,560 x a x (1 -— S,» x BE, SCF/ac-ft (1.20) Unit recovery is the difference between the initial gas in place and that re- maining at abandonment pressure, i.e., that produced at abandonment pressure, or ' Unit recovery = 43,560 X <23 X (1 —— SW) X (BS, - BM) SCF/ac-ft (1.21) The unit recovery is also called the initial um't reserve, which is generally lower than the initial unit in-place gas. The remaining reserve at any stage of depletion is the difference between this initial reserve and the unit production at that stage of depletion. The fractional recovery or recovery factor expressed in per cent of the initial in—place gas is "‘ Ga) = -' Bga) Recovery factor = G Bgi per cent (1.22) Experience with volumetric gas reservoirs indicates that the recoveries will range from 80 to 90 per cent. Some gas pipeline companies use an abandon— ment pressure of 100 psi per 1000 ft of depth. The gas volume factor in the Bell Gas Field at initial reservoir pressure is 188.0 SCF/cu ft and at 500 psia it is 27.6 SCF/cu ft. The initial unit reserve or unit recovery based on volumetric performance at an abandon- ment pressure of 500 psia is Unit recovery = 43,560 X 0.22 X (1 —- 0.23) X (188.0 — 27.6) = 1180M SCF/ac-ft 50 1.13. 1.14. GAS RESERVOIRS CHAP. 1 What volume will 100 lb of a gas of 0.75 specific gravity (air 1.000) occupy at 100°F and 100 psia? Ans: 276 cu ft. A 10 cubic-foot tank contains ethane at 25 psia surrounding a balloon 2 ft ' in diameter filled with methane at 35 psia. Neglecting the volume of the 1.15. 1.16. 1.17. 1.18. 1.i9. 1.20. 1.21. rubber in the balloon and assuming isothermal conditions, calculate the final pressure when the balloon is burst. Ans: 29.19 psia. (a) What per cent methane by weight does a gas of 0.65 specific gravity contain which is composed only of methane and ethane? What per cent by volume? Ans: 67.8 per cent, 79.6 per cent. (b) Explain why the per cent by volume is greater than the per cent by weight. ‘ (a) A 1500 cubic—foot tank contains methane at 30 psia and 80°F. To it are added: 1.80 moles of ethane at 14.4 psia and 60°F, 25 1b of butane at 75°F, 4.65 X 1026 molecules of propane at 30°F, and 500 SCF (14.7 psia and 60°F) of nitrogen. If the final temperature of the mixture is 60°F, what is the final pressure of the tank? Ans: 48.43 psia. ' (b) Of What significance are the temperatures given in part (a), with the moles of ethane, the pounds of butane, and the molecules of propane‘.) Explain. A 50 cubic-foot tank contains gas at 50 psia and 50°F. It is connected to another tank which contains gas at 25 psia and 50°F. When the valve between the two is opened, the pressure equalizes at 35 psia at 50°F. What is the volume of the other tank? Ans: 75 cu ft. ' , What is the weight of one molecule of pentane? Ans: 26.3 X 10‘“. Gas was contracted at 5.5 cents per M CF at contract conditions of 14.4 psia and 80°F. What is the equivalent price at a legal temperature of 60°F and pressure of 15.025 psia? Ans: 5.96 cents. What is the approximate weight, in tens, of one MM CF of natural gas? Ans: 25 tons for 0.65 specific gravity. A cylinder is fitted with a leak—proof piston and calibrated so that the volume within the cylinder can be read from a scale for any position of the piston. The cylinder is immersed in a constant temperature bath, maintained at 160°F, which is the reservoir temperature of the Sabine Gas Field. Forty- five thousand cu cm of the ‘gas, measured at 14.7 psia and 60°F, is charged into the cylinder. The volume is decreased in the steps indicated below, and the corresponding pressures are read with a dead weight tester after tem— perature equilibrium is reached. 453 1500 142.2 6000 156.5 5000 180 4000 265 2500 964 750 V, cu cm —— 2529 p, psia —— 300 (a) Calculate and place in tabular form the ideal volumes for the 45,000 cu cm, at 160°F at each pressure, and the gas deviation factors. CHAP. 1 1.22. 1.24. 1.25. AGAS RESERVOIRS 51 (b) Calculate the gas volume factors at each pressure, in units of cubic feet of reservon space per standard cubic foot of gas and also in units of standard cublc feet per cubic foot of reservoir space. (0) Plot the deviation factor and the gas volume factors calculated in part (b) versus pressure on the same graph. ((1) Express the gas volume factor at 2500 psia and 160°F in units of cu ft/SCF,SCF/cu ft, bbl/SCF, and SCF/bbl. Ans: 0.00590, 169.5, 0.00105, 952. (a) If the Sabine Field gas gravity is 0.65, calculate the deviation factors from zero to 6000 psia at 160°F, in 1000-pound increments, using the gas gravity correlation of Fig. 1.2. (b) Using the critical pressures and temperatures in Table 1.5, calculate and plot the deviation factors for the Sabine gas at several pressures and 160°F. The gas analysis is as follows: Component C, C2 Ca fC. nC. iCs nC. Mole Fraction 0.875 0.083, 0.02l 0.006 0.008 0.003 0.002 Component 0+7 Mole Fraction 0.001 0.001 1 Use the molecular weight and critical temperature and pressure of octane for the heptanes—plus. Plot the data of Prob. 21(a) and Prob. 22(a) on the same graph for comparison. (c) Below what pressure at 160°F may the ideal gas law be used for the gas of the Sabine Field if errors are to be kept within two per cent? Ans: 180 psia. (d) Will a reservoir contain more SCF of a real or of an ideal gas at similar ’ conditions? Explain. 1.23. A high-pressure cell has a volume of 0.330 cu ft and contains gas at 2500 psia and 130°F, at which conditions its deviation factor is 0.75. When 43.6 SCF measured at 14.7 psia and 60°F are bled from the cell through a wet test meter, the pressure dropped to 1000 psia, the temperature remaining at 130°F. What is the gas deviation factor at 1000 psia and 130°F? Ans: 0.885. (a) Calculate the bulk volume of the gas cap of a reservoir whose areal extent is 940 acres, i.e., the area enclosed by the zero thickness contour. The areas enclosed by the 4, 8, 12, 16, and 20-ft isopach lines are 752, 526, 316, 142, and 57 acres, respectively. The greatest thickness within the 20—ft isopach line is 23 ft. Ans: 8959 ac—ft. (b) Show that when the ratio of the areas enclosed by two successive contours is 0.50, the error introduced by using the trapezoidal formula is 2 per cent greater than the pyramidal formula. (c) What error is introduced by using the trapezoidal formula instead of the pyramidal formula when the ratio of the areas is 0.333? Ans: 4.7 per cent. A volumetric gas field has an initial pressure of 4200 psia, a porosity of 17.2 per cent, and connate water of 23 per cent. The gas volume factor at 4200 psia is 292 SCF/cu ft and at 750 psia is 54 SCF/cu ft. (a) Calculate the initial in—placc gas in standard cubic feet on a unit basis. Ans: 1.68MM SCF/ac-ft. 52 1.26. GAS RESERVOIRS Cup. I (b) Calculate the initial gas reserve in standard cubic feet on a unit basis, assuming an abandonment pressure of 750 psia. Ans: 1.37MM SCF/ac-ft. (c) Explain why the calculated initial reserve depends upon the abandonment pressure selected. (d) Calculate the initial reserve of a 640-acre unit whose average net pro- ductive formation thickness is 34 ft, assuming an abandonment pressure of 750 psia. Ans: 29.8MMM SCF. (8) Calculate the recovery factor based on an abandonment pressure of 750 psia. Ans: 81.5 per cent. The discovery well N o. 1 and wells No. 2 and No. 4 produce gas in the 7500-ft reservoir of the Echo Lake Field, Fig. 1.10. Wells Nos. 8 and 7 were dry in the 7500-ft reservoir; however, together with their electric logs and the one from well No. 1, the fault which seals the northeast side of the reservoir was established. The logs of wells Nos. 1, 2, 4, 5, and 6 were used to construct the map of Fig. 1.10, to locate the gas-water contact and to determine the average net sand thickness. The reservoir had been producing for 18 months when well No. 6 was drilled at the gas-water contact. The static wellhead pressures of the producing wells showed virtually no decline during the 18 month period prior to drilling well No. 6, and averaged near 3400 psia. The following data were available from electric logs, core analysis, etc. Average well depth = 7500 ft Average static wellhead pressure = 3400 psia Reservoir temperature = 175°F Gas specific gravity = 0.700 (air = Average porosity = 27 per cent Average connate water = 22 per cent Standard conditions = 14.7 psia and 60°F Bulk volume of productive reservoir rock at the time No. 6 was drilled = 22,500 ac-ft. 1.00) Fig. 1.10. Echo Lake Field, subsurface map, 7500 ft reservoir. CHAP. 1A8 RESERVOIRS 53' (a) Calculate the reservoir pressure using Eq. (1.12). Ans: 4038 psia. (b) Estimate the gas deviation factor, and the gas volume factor in standard cubic feet'per cubic foot. Ans: 0.90;249. (0) Calculate the reserve at the time well No. 6 was drilled assuming a residual gas saturation of 30 per cent. Ans: 33.8MMM SCF. (d) Discuss the location of well No. 1 with regard to the over-all gas recovery. (6) Discuss the effect of sand uniformity on over-all recovery, e.g., a uniformly permeable'sand versus a sand in two beds of equal thickness, one of which 4» has a permeability of 500 millidarcys, and the other, 100 millidarcys. 1.27. 1.28. The “M” Sand is a small gas reservoir with an initial bottom-holepressure of 3200 psia and bottom-hole temperature of 220°F. It is desired to inventory the gas in place at three production intervals. The pressure-production history and gas volume factors in cubic feet per standard cubic foot at standard conditionsof 14.7 psia and 60°F are as follows: Cumulative Gas Gas Volume Pressure Production Factor psia MM CF cu ft/SCF 3200 -. 0 ’ 0.0052622 2925 79 0.0057004 2525 221 0.0065311 2125 452 0.0077360 (a) Calculate the initial gas in place using production data at the end of each of the production intervals, assuming volumetric behavior. Ans: 1028, 1138, and 1414MM SCF. (b) Explain why the calculations of part (a) indicate a water drive. (0) Show that a water drive exists by plotting the cumulative production versus p/z. .4 (d) Based on electric log and core data, volumetric calculations on the “M” Sand showed that the initial volume of gas in place is 1018MM SCF. If the sand is under a partial water drive, What is the volume of water encroached at the end of each of the periods? There was no appreciable water production. Ans: 756, 27,000, and 174,200 bbl. When theSabine Gas Field was brought in, it had a reservoir pressure of 1700 psia, and a temperature of 160°F. After 5.00MMM SCF (14.7 psia plus 4 oz and 80°F) was produced, the pressure had fallen to 1550 psia. If the reservoir is assumed to be under volumetric control, using the deviation factors of Prob. 21, calculate the following: \a) The hydrocarbon pore volume of the reservoir. Ans: 433 X 106 cu ft. (b) The pounds of gas initially in place if the gas gravity is 0.65. Ans: 2.43MMM lb. (0) The SCF (14.7 psia and 60°F) of gas initially in place. Ans: 48.9MMM SCF. (d) The SCF produced when the pressure falls to 1550, 1400, 1100, 500, and 200 psia. Plot cumulative recovery in SCF versus p /2. Ans: 4.9, 9.6, 18.7, 35.9, 43.9MMM SCF. 54 1.29. 1.30. 1.31. 1.32. GAS RESERVOIRS CHAP. 1 (c) From your graph find how much gas can be obtained without the use of compressors for delivery into a pipeline operating at 750 psia. Ans: 29.3MMM SCF. ~’(f) What is the approximate pressure drop per MMM SCF of production? Ans: 32 psi. (g) Calculate the minimum value of the initial reserve if the produced gas measurement is accurate to :t 5 per cent and if the average pressures are accurate to :1: 12 psi when 5.00MMM SCF (14.7 psia plus 4 oz and 80°F) have been produced and the reservoir pressure has dropped to 1550 psia. Ans: 24.8MMM SCF at 750 psia abandonment pressure. If, however, during the production of 5.00MMM SCF of gas in the pre- ceding problem, 4.00MM bbl of water had encroached into the reservoir and still the pressure had dropped to 1550 psia, calculate the initial in—place gas. Compare with Prob. 28 (c). Ans: 26.1MMM SCF. (a) The gas cap of the St. John Oil Field had a bulk volume of 17,000 acre- feet when the reservoir pressure had declined to 634 psig. Core analysis shows an average porosity of 18 per cent, and an average interstitial water of 24 per cent. It is desired to increase the recovery of oil from the field by repressuring the gas cap to 1100 psig. Assuming that no additional gas dis— solves in the oil during repressuring, calculate the SCF (14.7 psia and 60°F) required. The deviation factors for both the reservoir gas and the injected gas are 0.86 at 634 psig and 0.78 at 1100 psig, both at 130°F. Ans: 4.1MMM SCF. (b) If the injected gas has a deviation factor 0.94 at 634 psig and 0.88 at 1100 psig, and the reservoir gas deviation factors are as above, recalculate the injected gas required. Ans: 3.6MMM SCF. (C) Is the assumption that no additional solution gas enters the reservoir oil a valid one? ((1) Considering the possibility of some additional solution gas and the production of oil during the time of injection, will the figure of (a) be max- imum or minimum? Explain. (e) Explain why the gas deviation factors are higher (deviation less) for the injected gas in part (1)) than for the. reservoir gas. (a) A well drilled into. a gas cap for gas recycling purpOses is found to be in an isolated fault block. After injecting 50MM SCF (14.7 psia and 60°F), the pressure increased from 2500 to 3500 psia. Deviation factors for the gas are 0.90 at 3500 and 0.80 at 2500 psia and the bottom—hole temperature is 160°F. What is the cubic feet of gas storage space in the fault block? Ans: 1.15l\*l..‘\l cu it. (b) ‘lfj-the average porosity is 16 per cent, average connatc water is 24 per eentfiand average sand thickness is 12 feet, what is the areal extent of the fault block? .1718: 18 acres. The initial volume of gas in place in the “P” Sand reservoir of the Holden Field'.-is‘ ealculated from electric log and core data to be QOOMMM SCF CRAP. l GAS RESERW )1 RS 55 (14.7 psia and 60°F) underlying 2250 productive acres, at an initial pressure of 3500 pane and 140°F. The pressure-production history is ‘ Production Gas Deviation Pressure, psra MMM SCF Factor at 140°!“ 3500 (initial) 0.0 0.85 2500 75.0 0.82 (a) What is the initial volume of gas in place as calculated from the pressure— production history assuming no water influx? Ans: 289MMM SCF. (b) Assuming uniform sand thickness, porosity, and connate water, if the volume of gas in place from pressure-production data is believed to be correct, how many acres of extension to the present limits of the “P” Sand are predicted? Am: 1000 acres. (0) If, on the other hand, the gas in place calculated from the log and core data is believed to be correct, how much water influx must have occurred during the 75MMM SCF of production to make the two figures agree? Ans: 22.8MM bbl. 1.33. Explain why initial calculations of gas in place are likely to be in greater error during the early life of depletion type reservoirs. Will these factors make the predictions high or low? Explain. L34. A gas reservoir under partial water drive produced 12.0MMM SCF (14.7 psia and 60°F) when the average reservoir pressure had dropped from 3000 psia to 2200 psia. During the same interval an estimated 5.20MM bbl of water entered the reservoir based on the volume of the invaded area. If the gas deviation factor at 3000 psia and bottom-hole temperature of 170°F is 0.88 and at 2200 psia is 0.78, what is the initial volume of gas in place measured at 14.7 psia and 60°F? Ans: 42.9MMM SCF. 1.35. A gas-producing formation has a uniform thickness of 32 ft, a porosity of 19 per cent, and connate water saturation of 26 per cent. The gas deviation factor is 0.83 at the initial reservoir pressure of 4450 psia and reservoir temperature of 175°F. (8.0. 14.7 psia and 60°F) (a) Calculate the initial in—place gas per acre-foot of bulk reservoir rock. Ans: 1.83MM SCF. (b) How many years will it take a well to deplete by 50 per cent a 640—acre unit at the rate of 3MM SCF/day? Ans: 17.1 years. (c) If the reservoir is under an active water drive so that the decline in reservoir pressure is negligible, and during the production of 50.4MMM SCF of gas, measured at 14.7 psia and 60°F, water invades 1280 acres, what is the per cent recovery by water drive? Ans: 67.24 per cent. (d) What is the gas saturation as per cent of total pore space in the water invaded portion of the reservoir? Ans: 24.24 per cent. pm. 36. Calculate the daily gas production including the condensate and water gas equivalents for a reservoir with the following daily production. ...
View Full Document

This note was uploaded on 02/02/2011 for the course PETE 100 taught by Professor Eric during the Spring '11 term at University of Louisiana at Lafayette.

Page1 / 8

Gas Reservoirs - SEC. 10 GAS RESERVOIRS 33 wells, vvhich in...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online