initial reservoir pressure and temperature in the Midale Marly formation in Sout
ID: 1718243 • 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 20C 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 ck 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
oil specific gravity = density of object/density of water
=888.2 / 1000
specific gravity of oil is =0.8882
API gravity =( 141.5/specific gravity) -131.5
= 27.8
c) mirly crude fluid is in the midale formation . the oil is undersaturated condition because initially there will be crude formation will be done . by increasing the temperature and the presurre of the fluid the oil formation wll be done.if the oil is saturated then it will get evaporates.
d) if the pressure is reduced to 12MPa then the oil is in undersaturated condition which will turn to crude formation
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