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Abnormal Pressure from Anticline Gas Cap.docx

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Abnormal Pressure from Anticline Gas Cap I got a question about how an anticline gas cap can create the abnormal pressure. The anticline with gas cap can be the potential high pressurized zone. Because of reservoir connectivity between fluid underneath gas and gas reservoir, it can generate abnormal pressure. This example below demonstrates you how this situation could happen. Well#1 was already drilled into oil reservoir at 6500’MD/6000’ TVD and its formation pressure is 0.52 psi/ft. The seco
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  Abnormal Pressure from Anticline Gas Cap I got a question about how an anticline gas cap can create the abnormal pressure. The anticline with gas cap can be the potential high pressurized zone. Because of reservoir connectivity between fluid underneath gas and gas reservoir, it can generate abnormal pressure. This example below demonstrates you how this situation could happen. Well#1   was already drilled into oil reservoir at 6500’MD/6000’ TVD and its formation pressure is 0.52 psi/ft. The second well, well#2 , is planned to drill into the gas cap, which has reservoir connectivity to the well#1, at 5400’MD/4800’ TVD and a geologist estimates gas cap thickness of 950 ft and pressure gradient is 0.1 psi/ft. What is minimum mud weight to be able to successfully drill well#2?  Firstly, I would like to share which basic concepts I will use for this case . Hydrostatic Pressure Pressure gradient Convert pressure to equivalent mud weight Equivalent Mud Weight  You need to determine formation pressure of well#1.    Well#1: Well depth: 6,500’MD/6,000’TVD and  Pressure gradient: 0.52 psi/ft Formation pressure of well#1 = 0.52 x 6000 = 3120 psi Convert pressure to equivalent mud weight = 3120 ÷ (0.052 x 6000) = 10.0 ppg You may think that 10.0 ppg should be good mud weight to drill well#2.  – >  Wrong answer.   Let’s take a look at the well#2.   Pressure at the bottom of gas cap = formation pressure of well 1 @ 6000’TVD –  hydrostatic pressure of reservoir fluid Pressure at the bottom of gas cap = 3120 psi  –  0.52 x (6000  –  5750) = 2990 psi Determine the reservoir pressure at the top of gas cap:  Pressure at the top of gas cap = Pressure at the bottom of gas cap  –  hydrostatic of gas gradient Pressure at the top of gas cap = 2990  –  (0.1×950) = 2895 psi Since we know the formation pressure, we can calculate to equivalent mud weight (EMW). EMW of well#2 = 2895 ÷ (0.052 x 4800) = 11.6 ppg You need mud weight at least 11.6 ppg  to drill well#2 successfully.   Conclusion:  This case clearly shows you that the anticline gas cap can be high pressurized zone and you may not be able to drill the well with the same mud weight as the adjacent drilled into the same reservoir.
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