Unformatted text preview: PGE331 Exam 2 Key Open book, notes. Show all work and read each problem carefully. 1. Material Balance Response. Ignore the connate water expansion and hydrocarbon pore volume expansion in this problem. The initial hydrocarbon is an undersaturated oil. a. (15 points) The figure below shows the cumulative underground withdrawal F plotted against the oil expansion factor Eo. 6 ) 4 MMrb F2 ( 0.05 0.10 0.15 Eo (rb/STB) What do you think the drive mechanism is? Why do you say this? What is the original hydrocarbon (both gas and oil) in place? The straight line indicates that the drive mechanism is some sort of expansion drive, probably solution gas drive. Can't say more than this with the information given. The original oil in place is the slope of the line = 4.5 = 30 0.15 0 0.00 S Can't say what the original gas in place is because the solution gasoil ratio is not given. Taking a typical value from Table 13 = (30 ) 500 = 15000 = 15 Grading: Drive mechanism HCs in place S 5 10 The following problems are sketches (no calculation required). The figure is for reservoir A and you are to supply the points or curve for reservoir B. Reservoir B is identical to reservoir A in every respect except for the item noted. 1 b. (10 points) Reservoir B has a substantial water drive. 6 ) 4 MMrb F2 ( 0 0.00 0.05 0.10 Eo (rb/STB) 0.15 The curve will no longer be a straight line because the water influx will support the pressure and thereby cause the expansion factor to be less at a given F than in reservoir A. c. (10 points) Reservoir B has onehalf the original hydrocarbon in place. 6 ) 4 MMrb F2 ( 0 0.00 0.05 0.10 Eo (rb/STB) 0.15 Reservoir B is exactly the same as reservoir A with onehalf the slope. d. (10 points) Reservoir B has a substantial gas cap. 6 ) 4 MMrb F2 ( 0 0.00 0.05 0.10 Eo (rb/STB) 0.15 This is the same explanation as part b except the pressure support is caused by the gas cap expansion. 2 2. Material Balance Calculation. (30 points) Use the PVT data in Table 132 and enclosed in this problem. This reservoir is being depleted by a strong water drive. In fact, the water drive is so strong that the average pressure has not declined from its original value of 4658 psia even though 25,000 STB of oil, 0.1 Bscf of gas and 2,000 STB of water have been produced. Estimate the amount of water influx (in RB or. m3) needed for this to occur. You can take the formation volume factor of water to be 1 rb/STB. Neither gas nor water has been injected. The material balance equation is = + + where the terms are derived as in the text. But if there is no pressure decline all of the expansion factors are zero and the equation becomes = so that the problem becomes on of evaluating the withdrawal. We have S   = = + + (  ) 1 (  ) 1 S S Since there has been no injection   = + + 1 1 There are no simplifications possible from here since the hydrocarbon is a volatile oil. We will evaluate it term by term S  = 1 1 2.707  0.83 2845 3 10 = (25,000 ) 12899 1 1 116 2845 6 10 S 3  = 1 106 103 0.83  116 2.707 1 1 = (0.1 ) 1 1 116 2845 106 77015 Finally (remembering that the formation volume factor for water is unity) S = 12899 + 77015 + 2500 = 91913 = 91913 Grading: Correct starting equation Correct working equation Units Answer 5 10 10 5 S 3. Productivity Index Calculations. (30 points) You are to decide how best to drain the sealed boxshaped reservoir given below. The three options available are given below. Vertical well h Horizontal well h Vertical well h L Fracture L L W W W The development decision will be based entirely on which option has the largest productivity index. From the properties given below, recommend the best strategy. Recall that the productivity index (based on average pressure) for a vertically fractured well is = (1 ) 2  (1+ ) 3 2 where the symbols have meaning consistent with HW7. The data is S = = 0.1 , = 0.01 , = 0.1 , = 3, = 500 , = 100 , = 250 , = 8 . Since they are all sealed f=0 in all cases. The viscosity suggests that this is mostly gas production or of a very gassy oil. Will do the fracture case first. 4 S = 2(125 )(100 )(0.1 )  = = 15 0 (1 ) 1 2  (1+ ) (0.1 )(250 ) 2(3))  (1+ 3 3 2 S Recall from the HW that the area is only 1/2 of the fracture. Recall also that the above is based on the production rate into 1/4 of the fracture. The well productivity index would be = 4 60 =  S Since the problem did not ask for productivity index, the units are not important. All that is important is that they be consistent for comparison. The PI for the other two cases is... = 2 1 4 + 2 2 For the vertical well.... S 4 4(500 )(250 ) = = 12.61 2 2 2 0.305 8 (22.6) (1.781) 12 1 where the shape factor is from Table 72. S = 2(0.1 )(100 )  = 67.51 1 (0.1 ) (12.61) + 3 2 And for the horizontal well.... S 4 4(100 )(250 ) = = 13.26 2 2 2 0.305 8 (2.36) (1.781) 12 1 S 5 = 2 (0.1 )(0.01 ) (500 )  = 103.1 1 (0.1 ) (13.26) + 3 2 Based solely on PI, the horizontal well is better. S Grading.... Units Correct shape factor Hor. Well perm Knowledge that f = 0 Use of PIs 10 5 5 5 5 6 ...
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