Reservoir Engineering Module PDF

Reservoir Engineering Module Reservoir Pressure and Temperature Dr. Mohammed Abdalla Ayoub Dr. Syed Mohammad Mahmood  Lesson Outline To understand the reservoir pressure To describe the hydrocarbon reservoirs based on the fluid content. To calculate the fluid...

Reservoir Engineering Module Reservoir Pressure and Temperature Dr. Mohammed Abdalla Ayoub Dr. Syed Mohammad Mahmood  Lesson Outline To understand the reservoir pressure To describe the hydrocarbon reservoirs based on the fluid content. To calculate the fluid contacts in the reservoir. 2  Reservoir Pressures Magnitude and variation of pressures in a reservoir are an important aspect of reservoir understanding during exploration and production phase. Oil and gas occur at a range of sub- surface depths. At these depths, pressure exists as a result of: ❑ The weight of depositional process ❑ The fluids contained in porous media. 3  Reservoir Pressures The total reservoir pressure at any depth resulting from the combined weight of the formation rock and fluids. This pressure is known as the overburden pressure. At any depth, the overburden pressure or the total pressure is the summation of rock or grain pressure and fluids pressure OP = FP + GP Function of depth, density Pressure gradient 1 psi./ ft OP at depth D = 1.0 x D A reduction in fluid pressure will lead to a corresponding increase in the grain pressure. 4  Reservoir Pressures Overburden pressure is balanced in part by the pressure of the grains of rock under compaction and the fluids pressure within pores (pore pressure). In deposited rocks, like reservoirs, fluid pressure Path of well arises from the continuity of the aqueous phase from surface to the depth. This pressure is called hydrostatic pressure. Value depends on the density of fluid. Water - salinity 0.433 psi/ft - fresh water 0.45 psi/ft for saline water 55,000ppm. 0.465 psi for 88,000ppm 5  Fluid Pressure-normal Pressure Fluid pressure regimes in hydrocarbon columns are dictated by prevailing water pressure in the vicinity of reservoir. In the normal case, water pressure at any depth can be calculated as Atmos. Pressure 0 psig. Or 14.7psia. dP/dD is the water pressure gradient and dependent on the salinity and for pure water has the value of 0.4335 psi/ft. Assumes continuity of water pressure from surface and constant salinity 6  Fluid Pressure-abnormal Pressure Under certain conditions, fluid pressure is not normally pressured. Hence it is called abnormal pressure. Causes of Abnormal Pressure ❑ Thermal effects-expansion or contraction of water ❑ Rapid burial of sediments ❑ Geological changes. ❑ Osmotic effects via salinity differences Overpressured reservoirs. Hydrostatic pressure greater than normal pressure Underpressured reservoirs Hydrostatic pressure below normal pressure 7  Fluid Pressure-abnormal Pressure In the case of abnormal pressure, the water pressure at any depth can be calculated as  dP   Pw =   * D  + 14.7 + C  dD  water  C - constant positive - overpressured C - constant negative - underpressured For the fluid in any sand to be abnormally pressured, the sand must be effectively sealed off from the surrounding strata so that hydrostatic pressure continuity to the surface can not be established. 8  Fluid Pressures-hydrocarbon Systems Hydrocarbon pressure regimes different since densities of oil and gas are less than water. Pressure 0 Depth 9  Pressure Distribution for an Oil Reservoir with a Gas-cap and Oil Water Contact The pressure in the water at the OWC can be calculated as: Pw = 0.45*D +14.7 psi Since the Po is equal to Pw at the OWC, so Po = Pw = 0.45 *5500 + 14.7 Po = Pw = 2490 psi Pressure, psi Exploration well The linear equation for Po above the OWC is 2250 2375 2500 then Po = 0.35 D + constant 2369 Since Po at 5500 = 2490, the constant can be 2265 evaluated to give Gas Po = 0.35 D + 565 GOC: This equation can be used to evaluate the Po at 5200 Po=Pg = 2385 any depth. Depth, ft At the GOC at 5200ft, the Po and Pg is equal and the Po can be calculated by the oil equation to be At 5250 ft Oil 2385 psi. Po =2402 5500 The linear equation for the gas can then be determined as OWC: Po=Pg = 2490 Po = Pg = 0.08 D + constant water Since Po at 5200= 2385 psi, the constant can be evaluated to give Pg = 0.08 D + 1969 This equation can be used to evaluate the Pg at any depth. 10 Pressure Distribution for an Oil Reservoir with a Gas-cap and Oil Water Contact 11  Reservoir Temperature Earth temperature increases from surface to center. Heat flow outwards generates a geothermal gradient. Conforms to local and regional gradients as influenced by lithology. Obtained from wellbore temperature surveys. Reservoir geothermal gradients around 1.6oF/100ft ( 0.029K/m). Because of large thermal capacity and surface area of porous reservoir, flow processes in a reservoir occur at constant temperature. Local conditions , eg around the well can be influenced by transient cooling or heating effects of injected fluids. 12 QUESTIONS? 13  Exercise 1 If the average pressure gradient in a region is 0.47 psi/ft, calculate the pore pressure in a normally pressurised formation at 7400ft. Convert the pressure from psi to KPa, then express the pressure in MPa. What is the pressure gradient in KPa/m? Multiply KPa by 0.145 to get psi. 1 US foot = 0.3048m Pressure in formation = 0.47 * 7400 = 3478 psi Converting to KPa = 3478 / 0.145 = 23986 KPa Converting to MPa = 23986 / 1000 = 23.99 MPa SOLUTION Pressure gradient = 0.47 psi/ft = (0.47 / 0.145) KPa/ft = 3.2414 KPa/ft = (3.2414 /0.3048) KPa/m = 10.63 KPa/M Internal 14  Exercise 2 If the pressure in a reservoir at the OWC is 3625 psi, calculate the pressure at the top if there is a 600ft continuous oil column. If a normal pressure gradient exists outwith the reservoir, calculate the pressure differential at the top of the reservoir. Redo the calculations for a similar field, but this time containing gas. Pressure at seal = 3625 - (600*0.35) = 3415 psi To calculate the pressure differential across seal, look at fluid gradient differential from OWC to seal 600ft above… SOLUTION Differential = (0.45-0.35) * 600 = 60 psi If the reservoir is gas then the differential becomes… (0.45 – 0.08) * 600 = 222 psi higher in the reservoir than surrounding area 15 Internal