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TAMU PETE 662 - Ch 5

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Chapter 5Production from Horizontal WellsPETE 662Production EngineeringVertical Well Flow PatternpepwfreTop ViewCross-Sectional ViewRadial drainage patternHorizontal Well Flow PatternsHeelToeCombination of radial, linear, and elliptical drainage patternsHorizontal WellsAdvantages Over Vertical Wells: Larger and more efficient drainage pattern leading to increased overall hydrocarbon recovery efficiency  Increased production rate due to greater wellbore length exposed to the pay zone  Reduced water and gas coning resulting from reduced drawdown in the reservoir for a given production rate Reduced pressure drop around the wellbore  Lower fluid velocities around the wellbore  A general reduction in sand production from reduced pressure drop around the wellbore and the resulting low fluid velocities around the wellboreHorizontal WellsImportant Factors Affecting Well Productivity: Vertical permeability – low vertical permeability or discontinuities detrimental to productivity  Permeability anisotropy in horizontal plane – higher well productivity from horizontal wells drilled normal to the large horizontal permeability direction Direction of maximum horizontal stress – direction of maximum horizontal permeability the same as that of maximum horizontal stressHorizontal WellsSteady-State Inflow Performance Joshi Modely-z planex-y planexzyqvqhHorizontal WellsSteady-State Inflow Performance Joshi ModelIntegrating flow long the x-y plane over the thickness h yields,  )15(2/2/ln222LLaaBpkqohAssumptions made in deriving Eq. 5-1:- elliptical drainage area with length of major axis = 2a- constant pressure at the drainage boundaryHorizontal WellsSteady-State Inflow Performance Joshi ModelIntegrating flow long the y-z plane over the well length L yields, )25(2ln2wovrhBpkqAssumptions made in deriving Eq. 5-2:- Radial flow from the vertical boundary located 2/h from the well- Pressure at the boundary = pressure at the elliptical horizontal boundaryHorizontal WellsSteady-State Inflow Performance Joshi Modelvqp /hqp /qp /=+From Eq. 5-1 and Eq. 5-2, for an isotropic formation, we obtain )35(2ln2/2/ln222woHrhLhLLaaBphkqHorizontal WellsSteady-State Inflow Performance For an anisotropic formation, Eq. 5-3 becomes (Economides, et al.)   )45(1ln2/2/ln2.14122aniwanianiowfeHIrhILhILLaaBpphkqwhere)55( VHanikkINote: Eq. 5-4 was derived assuming the horizontal well is centered in the drainage volumeHorizontal WellsSteady-State Inflow Performance Rearranging Eq. 5-4, we obtain the following IPR equation based on Joshi’s Model  )75(1ln2/2/ln2.14122aniwanianiHoewfIrhILhILLaahkqBppTo account for that the ends of the horizontal well are the foci of the ellipse in the horizontal plane, by equating the areas of the ellipse to that of a cylinder of radius, re, Joshi developed the following equation to relate ɑ with re.)65(2/25.05.025.05.04LrLaeHExample 5-1 Well Performance with Joshi ModelA 2000-ft horizontal lateral is producing from a 100-ft thick reservoir where kH= 10 md, kV= 1 md. The lateral has an rw= 3 in draining from a region 4000 ft in the direction of the well where pe= 4000 psi, µo= 5 cp, Bo= 1.1. Determine q when pwf= 2000 psi.Solution:  )75(1ln2/2/ln2.14122aniwanianiHoewfIrhILhILLaahkqBpp)55(3.162 110VHanikkIft 200024000aSubstituting in Eq. 5-7 and consolidate yields,)95(74.14000  qpwfFrom Eq.5-9, q = 1149 STB/d when pwf= 2000 psi.)95(74.14000  qpwfExample 5-1 Well Performance with Joshi ModelSolution (cont’d):Horizontal WellsSteady-State Inflow Performance Furui et al. Model In the cross-sectional area perpendicular to the wellbore, Furui Model assumes radial flow in the near wellbore region and linear flow beyond thatlprpSteady-State Inflow Performance Furui et al. ModelHorizontal Wells)105( lrppp )115(12ln2anianiwtrIIrrkLqp)135()()2/(hLkIyyqpanilbl)125( VHkkkwhere)145(222  hIyranittlprpFurui et al. ModelHorizontal WellsSteady-State Inflow Performance )105( lrpppCombining/consolidating Eqs. 5-11, 5-12, 5-13, 5-14 and substituting into Eq. 5-10 yields, )175()2//(12ln2anibanianiwaniIhyIIrhIkLqpFurui et al. ModelHorizontal WellsSteady-State Inflow Performance Defining skin factor as)185(2 skLqpskin )195()2//(12ln2 sIhyIIrhIkLqpanibanianiwaniWith skin included, Eq. 5-17 becomesSolving for q using field units, we obtainIncluding partial penetrating skin, sR, Eq. 5-20 becomes )215(224.11ln2.141)(RanibaniwaniowfesshIyIrhIBppkbq )205(224.11ln2.141)(shIyIrhIBppkLqaniLaniwaniowfewhere, b is the total length of the reservoir.Example 5-2 Well Performance with Furui et al. ModelA 2000-ft horizontal lateral is producing from a 100-ft thick (h = 100 ft) reservoir where kH= 10 md, kV= 1 md. The lateral has an rw= 3 in draining from a region 4000 ft in the direction of the well where pe= 4000 psi, µo= 5 cp, Bo= 1.1. Determine q when pwf= 2000 psi.Solution: )215(224.11ln2.141)(RanibaniwaniowfesshIyIrhIBppkbqmd 162.3110 VHkkk3.162 110VHanikkIft 17322ft 20002ft 400022b2222LybSTB/d 912 qBabu and Odeh ModelHorizontal WellsPseudosteady-State Inflow Performance  The


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TAMU PETE 662 - Ch 5

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