initial reservoir pressure and temperature in the Midale Marly formation in Sout
ID: 230997 • Letter: I
Question
initial reservoir pressure and temperature in the Midale Marly formation in Southeast Saskatchewan are about 21.0 MPa and 58°C. Density and viscosity of the Midale crude are measured to be 888.2 kg/m3 and 15.2 cP at 20oC and atmospheric pressure, respectively. The bubble-point pressure of the Marly oil is 14.5 MPa. The true vertical depth of a well is 1463 m, while the horizontal section extends to 600 m in the payzone. The current oil production is 250 STB/d at the reservoir pressure of 17.0 MPa. As for a sample of Midale crude with volume of 330 cc at reservoir conditions, it passes through a separator and into a stock tank at atmospheric pressure and 60°F. The liquid volume in the stock tank is measured to be 285 cc, while the separator produced 0.648 scf of gas, and the stock tank produced 0.089 scf of gas. 1). Calculate the oil specific gravity and API gravity. 2). Calculate the oil formation volume factor and solution gas-oil ratio. 3). What type of reservoir fluid is in the Midale formation? Is the reservoir oil saturated or undersaturated? Provide your justifications. 4). Describe the behaviour with your justification if the reservoir pressure is reduced to 12.0 MPa. 2. The initial reservoir pressure and temperature in the Midale Marly formation in Southeast Saskatchewan are about 21.0 MPa and 58°C. Density and viscosity of the Midale crude are measured to be 888.2 kg/m and 15.2 cP at 20°C and atmospheric pressure, respectively. The bubble-point pressure of the Marly oil is 14.5 MPa. The true vertical depth of a well is 1463 m, while the horizontal section extends to 600 m in the payzone. The current oil production is 250 STB/d at the reservoir pressure of 17.0 MPa. As for a sample of Midale crude with volume of 330 cc at reservoir conditions, it passes through a separator and into a stock tank at atmospheric pressure and 60°F. The liquid volume in the stock tank is measured to be 285 cc, while the separator produced 0.648 sef of gas, and the stock tank produced 0.089 scf of gas 1). Calculate the oil specific gravity and API gravity 2). Calculate the oil formation volume factor and solution gas-oil ratio. 3). What type of reservoir fluid is in the Midale formation? Is the reservoir oil saturated or undersaturated? Provide your justifications. 4). Describe the behaviour with your justification if the reservoir pressure is reduced to 12.0 MPa.Explanation / Answer
Pi = 21 MPa = 3045 psia
reservoir temp = 58 C = 136.4 F
oil density = 888.2 kg/m^3
viscosity = 15.2 cP @ 20 C
Pb= 14.5 MPa = 2102.5 psia
TVD = 1463 m
Pe = 17 MPa
q= 250 STB/d
1) oil specific gravity = 888.2 / 1000 = 0.8882
specific gravity = 141.5 / (131.5 + API)
API = 27.8
2) oil formation volume factor = vol of oil at reservoir condition(plus gas) / vol of oil at std condition
Bo= 330/285 = 1.158 cc/cc
solution gas oil ration Rs at std condition = volume of gas / volume of oil at std condition
Rs = (0.648 + 0.089) / 285 = 0.0026 scf/cc = 413.4 scf/STB
3) since oil formation vol factor is below 2, GOR is between 200-700 scf/STB and API oil gravity between 15-40 degree, hence it is an ordinary black oil reservoir. the reseroir pressure (17 MPa) is greater than bubble point pressure (14.5 MPa) hence it is undersaturated oil reservoir.
4) if the reservoir pressure is reduced to 12 MPa then the reservoir pressure would be below bubble point pressure and solution gas start evolving from the oil. it wud support in making the secondary gas cap. and as soon as the gas concentration crosses critical saturation value, then the gas will start moving in the formation and resulting in two-phase flow in the reservoir. it would result in the increased GOR if the depletion drive is solution deplition drive. GOR would increase to a peak value and then start decresing.
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