# Productivity Formulas for a Partially Penetrating Vertical Well in a Circular Cylinder Drainage Volume

Authors: Jing Lu; Tao Zhu; Djebbar Tiab; Jalal Owayed
Journal: Mathematical Problems in Engineering
 (2009)

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## Abstract

Taking a partially penetrating vertical well as a uniform line sink in three-dimensional space, by developing necessary mathematical analysis, this paper presents steady state productivity formulas for an off-center partially penetrating vertical well in a circular cylinder drainage volume with constant pressure at outer boundary. This paper also gives formulas for calculating the pseudo-skin factor due to partial penetration. If top and bottom reservoir boundaries are impermeable, the radius of the cylindrical system and off-center distance appears in the productivity formulas. If the reservoir has a gas cap or bottom water, the effects of the radius and off-center distance on productivity can be ignored. It is concluded that, for a partially penetrating vertical well, different productivity equations should be used under different reservoir boundary conditions.


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## Body

## 1. Introduction

Well productivity is one of primary concerns in oil field development and provides the basis for oil field development strategy. To determine the economical feasibility of drilling a well, the engineers need reliable methods to estimate its expected productivity. Well productivity is often evaluated using the productivity index, which is defined as the production rate per unit pressure drawdown. Petroleum engineers often relate the productivity evaluation to the long time performance behavior of a well, that is, the behavior during pseudo-steady-state or steady-state flow.The productivity index expresses an intuitive feeling, that is, once the well production is stabilized, the ratio of production rate to some pressure difference between the reservoir and the well must depend on the geometry of the reservoir/well system only. Indeed, a long time ago, petroleum engineers observed that in a bounded reservoir or a reservoir with strong water drive, the productivity index of a well stabilizes in a long time asymptote.When an oil reservoir is bounded with a constant pressure boundary (such as a gas cap or an aquifer), flow reaches the steady-state regime after the pressure transient reaches the constant pressure boundary. Rate and pressure become constant with time at all points in the reservoir and wellbore once steady-state flow is established. Therefore, the productivity index during steady-state flow is a constant.Strictly speaking, steady-state flow can occur only if the flow across the drainage boundary is equal to the flow across the wellbore wall at well radius, and the fluid properties remain constant throughout the oil reservoir. These conditions may never be met in an oil reservoir; however, in oil reservoirs produced by a strong water drive, whereby the water influx rate at reservoir outer boundary equals the well producing rate, the pressure change with time is so slight that it is practically undetectable. In such cases, the assumption of steady-state is acceptable.In many oil reservoirs the producing wells are completed as partially penetrating wells; that is, only a portion of the pay zone is perforated. This may be done for a variety of reasons, but the most common one is to prevent or delay the unwanted fluids into the wellbore. If a vertical well partially penetrates the formation, there is an added resistance to flow in the vicinity of the wellbore. The streamlines converge and the area for flow decreases, which results in added resistance.The problem of fluid flow into wells with partial penetration has received much attention in the past, the exact solution of the partial penetration problem presents great analytical problems. Brons and Marting [4], Papatzacos [5], Basinev [3] developed solutions to the two dimensional diffusivity equation, which included flow of fluid in the vertical direction. They only obtained semianalytical and semiempirical expressions to calculate the added resistance due to partial penetration.The primary goal of this study is to present new steady-state productivity formulas for a partially penetrating vertical well in a circular cylinder drainage reservoir with constant pressure at outer boundary. Analytical solutions are derived by making the assumption of uniform fluid withdrawal along the portion of the wellbore open to flow. The producing portion of a partially penetrating vertical well is modeled as a uniform line sink. This paper also gives new expressions for calculating the added resistance due to partial penetration, by solving the three-dimensional Laplace equation.


## 2. Literature Review

Putting Darcy's equation into the equation of continuity, the productivity formula of a fully penetrating vertical well in a homogeneous, isotropic permeability reservoir is obtained [1, page 52]:
(2.1)Qw=FD2πKH(Pe-Pw)/(μB)ln(Re/Rw),
where Pe is outer boundary pressure, Pw is flowing wellbore pressure, K is permeability, H is payzone thickness, μ is oil viscosity, B is oil formation volume factor, Re is drainage radius, Rw is wellbore radius, and FD is the factor which allows the use of field units, and it can be found in a Table 1 at page 52 of [1].Formula (2.1) is applicable for a fully penetrating vertical well in a circular drainage area with constant pressure outer boundary.If a vertical well partially penetrates the formation, there is an added resistance to flow which is limited to the region around the wellbore, this added resistance is included by introducing the pseudo-skin factor,Sps. Thus, Formula (2.1) may be rewritten to include the pseudo-skin factor due to partial penetration as [2, page 92]
(2.2)Qw=FD2πKH(Pe-Pw)/(μB)ln(Re/Rw)+Sps.Define partial penetration factorη:
(2.3)η=LpH,
where Lp is the producing well length, that is, perforated interval.Several authors obtained semianalytical and semiempirical expressions for evaluating pseudo-skin factor due to partial penetration.Bervaldier's pseudo-skin factor formula [3]:
(2.4)Sps=(1η-1)[ln(Lp/Rw)(1-Rw/Lp)-1].Brons and Marting's pseudo-skin factor formula [4] is as follows:
(2.5)Sps=(1η-1)[ln(hD)-G(η)],
where
(2.6)hD=(H2Rw)(KhKv)1/2,G(η)=2.948-7.363η+11.45η2-4.675η3.Papatzacos's pseudo-skin factor formula [5] is as follows:
(2.7)Sps=(1η-1)ln(πhD2)+(1η)ln[(η2+η)(I1-1I2-1)1/2],
where
(2.8)I1=Hh1+0.25Lp,I2=Hh1+0.75Lp,
and h1 is the distance from the top of the reservoir to the top of the open interval.It must be pointed out that the aforementioned formulas are only applicable to a reservoir with both impermeable top and bottom boundaries.


## 3. Partially Penetrating Vertical Well Model

Figure1 is a schematic of an off-center partially penetrating vertical well. A partially penetrating well of drilled length L drains a circular cylinder porous volume with height H and radius Re.Figure 1
Partially penetrating vertical well model.The following assumptions are made.(1)
The reservoir has constantKx,Ky,Kz permeabilities, thickness H, porosity ϕ. During production, the partially penetrating vertical well has a circular cylinder drainage volume with height H and radius Re. The well is located at R0 away from the axis of symmetry of the cylindrical body.(2)
At timet=0, pressure is uniformly distributed in the reservoir, equal to the initial pressure Pi. If the reservoir has constant pressure boundaries (edge water, gas cap, bottom water), the pressure is equal to the initial value at such boundaries during production.(3)
The production occurs through a partially penetrating vertical well of radiusRw, represented in the model by a uniform line sink, the drilled well length is L, the producing well length is Lp.(4)
A single phase fluid, of small and constant compressibilityCf, constant viscosity μ, and formation volume factor B, flows from the reservoir to the well. Fluid properties are independent of pressure. Gravity forces are neglected.(5)
There is no water encroachment and no water/gas coning. Edge water, gas cap, and bottom water are taken as constant pressure boundaries, multiphase flow effects are ignored.(6)
Any additional pressure drops caused by formation damage, stimulation, or perforation are ignored, we only consider pseudo-skin factor due to partial penetration.The porous media domain is:(3.1)Ω={(x,y,z)∣x2+y2<Re2,0<z<H},
where Re is cylinder radius, Ω is the cylindrical body.Located atR0 away from the center of the cylindrical body, the coordinates of the top and bottom points of the well line are (R0,0,0) and (R0,0,L), respectively, while point (R0,0,L1) and point (R0,0,L2) are the beginning point and end point of the producing portion of the well, respectively. The well is a uniform line sink between (R0,0,L1) and (R0,0,L2), and there hold
(3.2)Lp=L2-L1,Lp≤L≤H.We assume(3.3)Kx=Ky=Kh,Kz=Kv,
and define average permeability:
(3.4)Ka=(KxKyKz)1/3=Kh2/3Kv1/3.The reservoir initial pressure is a constant:(3.5)P|t=0=Pi.The pressure at constant pressure boundaries (edge water, gas cap, bottom water) is assumed to be equal to the reservoir initial pressure during production:(3.6)Pe=Pi.Suppose point(R0,0,z′) is in the producing portion, and its point convergence intensity is q, in order to obtain the pressure at point (x,y,z) caused by the point (R0,0,z′), according to mass conservation law and Darcy's law, we have to obtain the basic solution of the diffusivity equation in Ω [6, 7]:
(3.7)Kh∂2P∂x2+Kh∂2P∂y2+Kv∂2P∂z2=μqBδ(x-R0)δ(y)δ(z-z′),inΩ,
where Ct is total compressibility coefficient of porous media, δ(x-R0),δ(y),δ(z-z′) are Dirac functions.In order to simplify the equations, we take the following dimensionless transforms:(3.8)xD=(xL)(KaKh)1/2,yD=(yL)(KaKh)1/2,zD=(zL)(KaKv)1/2,(3.9)LD=(KaKv)1/2,HD=(HL)(KaKv)1/2,(3.10)L1D=(L1L)(KaKv)1/2,L2D=(L2L)(KaKv)1/2,(3.11)LpD=L2D-L1D=(LpL)(KaKv)1/2,(3.12)ReD=(ReL)(KaKh)1/2,R0D=(R0L)(KaKh)1/2.The dimensionless wellbore radius is [8]
(3.13)RwD≈[(KhKv)1/4+(KhKv)-1/4](Rw2L).Assumeq is the point convergence intensity at the point sink (R0,0,z′), the partially penetrating well is a uniform line sink, the total productivity of the well is Q, and there holds
(3.14)q=QLpD.Define the dimensionless pressures:(3.15)PD=KaL(Pe-P)μqB,PwD=KaL(Pe-Pw)μqB.Then (3.7) becomes [6, 7]
(3.16)∂2PD∂xD2+∂2PD∂yD2+∂2PD∂zD2=-δ(xD-R0D)δ(yD)δ(zD-zD′),inΩD,
where
(3.17)ΩD={(xD,yD,zD)∣xD2+yD2<ReD2,0<zD<HD}.If pointr0 and point r are with distances ρ0 and ρ, respectively, from the axis of symmetry of the cylindrical body, then the dimensionless off-center distances are
(3.18)ρ0D=(ρ0L)(KaKh)1/2,ρD=(ρL)(KaKh)1/2.There holds(3.19)(πHD)(2ReD-ρ0D-ρD-ρ0DρD)=(KvKh)1/2(πL2H)(4ReL-2ρ0L-2ρL-2ρ0ρL)=(KvKh)1/2(πReH)(2-ρ0Re-ρRe-ρ0ρRe)=(KvKh)1/2(πReH)(2-ϑ0-ϑ-ϑ0ϑ),
where
(3.20)ϑ0=ρ0Re,ϑ=ρRe.Since the reservoir is with constant pressure outer boundary (edge water), in order to delay water encroachment, a producing well must keep a sufficient distance from the outer boundary. Thus in this paper, it is reasonable to assume that(3.21)ϑ0≤0.6,ϑ≤0.6If(3.22)ϑ0=ϑ=0.6,KvKh=0.25,ReH=15,
then
(3.23)(KvKh)1/2(πReH)(2-ϑ0-ϑ-ϑ0ϑ)=0.251/2×(π×15)×(2.0-0.6-0.6-0.6×0.6)=4.7124,exp(-4.7124)=8.983×10-3.Moreover  if(3.24)ϑ0=ϑ=0.5,KvKh=0.5,ReH=10,
then
(3.25)(KvKh)1/2(πReH)(2-ϑ0-ϑ-ϑ0ϑ)=0.51/2×(π×10)×(2.0-0.5-0.5-0.5×0.5)=11.107,exp(-11.107)=1.501×10-5.Recall (3.19), there holds
(3.26)exp[-(πHD)(2ReD-ρ0D-ρD-ρ0DρD)]=(KvKh)1/2(πReH)(2-ϑ0-ϑ-ϑ0ϑ),
since there holds (3.21), and according to the aforementioned calculations in (3.22), (3.23), (3.24), and (3.25), we obtain
(3.27)exp[-(πHD)(2ReD-ρ0D-ρD-ρ0DρD)]≈0.Because(3.28)0<(πHD)(2ReD-ρ0D-ρD-ρ0DρD)<(πHD)(2ReD-ρ0D-ρD),
thus
(3.29)exp[-(πHD)(2ReD-ρ0D-ρD-ρ0DρD)]>exp[-(πHD)(2ReD-ρ0D-ρD)].Combining (3.27) and (3.29), we have
(3.30)exp[-(πHD)(2ReD-ρ0D-ρD)]≈0.


## 4. Boundary Conditions

In this paper, we always assume constant pressure lateral boundary:(4.1)P(x,y,z)=Pe=Pi,
on cylindrical lateral surface:
(4.2)Γ={(x,y,z)∣x2+y2=Re2,0<z<H}.Recall (3.15), the dimensionless form of constant pressure lateral boundary condition is
(4.3)PD(xD,yD,zD)=0,
on
(4.4)ΓD={(xD,yD,zD)∣xD2+yD2=ReD2,0<zD<HD}.Also we have the following dimensionless equations for top and bottom boundary conditions:(i)
If the circular cylinder drainage volume is with top and bottom impermeable boundaries, that is, the boundaries atz=0 and z=H are both impermeable (e.g., the reservoir does not have gas cap drive or bottom water drive), then
(4.5)∂PD∂zD|zD=0=0;∂PD∂zD|zD=HD=0.(ii)
If the circular cylinder drainage volume is with impermeable boundary atz=H, constant pressure boundary at z=0, (e.g., the reservoir has gas cap drive), then
(4.6)PD|zD=0=0;∂PD∂zD|zD=HD=0.(iii)
If the circular cylinder drainage volume is with impermeable boundary atz=0, constant pressure boundary at z=H (e.g., the reservoir has bottom water drive), then
(4.7)PD|zD=HD=0;∂PD∂zD|zD=0=0.(iv)
If the circular cylinder drainage volume is with top and bottom constant pressure boundaries, that is, the boundaries atz=0 and z=H are both constant pressure boundaries (e.g., the reservoir has both gas cap drive and bottom water drive), then
(4.8)PD|zD=0=0;PD|zD=HD=0.


## 5. Point Sink Solutions

For convenience, we use dimensionless variables given by (3.8) through (3.13), but we drop the subscript D. In order to obtain the dimensionless pressure of a point sink in a circular cylinder reservoir, we need to solve a dimensionless Laplace equation in dimensionless space:
(5.1)∂2P∂x2+∂2P∂y2+∂2P∂z2=-δ(x-x′)δ(y)δ(z-z′),inΩ,
where
(5.2)Ω={(x,y,z)∣x2+y2<Re2,0<z<H}.The following initial reservoir condition and lateral reservoir boundary condition will be used to obtain point sink pressure in a circular cylinder reservoir with constant pressure outer boundary:(5.3)P(t,x,y,z)|t=0=0,inΩ,P(t,x,y,z)=0,onΓ,
where Γ={(x,y,z)∣x2+y2=Re2,0<z<H}.The problem under consideration is that of fluid flow toward a point sink from an off-center position within a circular of radiusRe. We want to determine the pressure change at an observation point with a distance ρ from the center of circle.Figure2 is a geometric representation of the system. In Figure 2, the point sink r0 and the observation point r are with distances ρ0 and ρ, respectively, from the circular center; and the two points are separated at the center by an angle θ. The inverse point of the point sink r0 with respect to the circle is point r*. Point r* is with a distance ρ* from the center, and ρ1 from the observation point. The inverse point is the point outside the circle, on the extension of the line connecting the center and the point sink, and such that
(5.4)ρ*=Re2ρ0.Figure 2
Geometric representation of a circular system.AssumeR′ is the distance between point r and point r0, then
(5.5)R′=ρ2+ρ02-2ρρ0cosθ.If the observation pointr is on the drainage circle, ρ=Re, then
(5.6)R′=Re2+ρ02-2Reρ0cosθ,Re>ρ0>0.If the observation pointr is on the wellbore, then
(5.7)R′=Rw.Define(5.8)λn=nπH.


### 5.1. Impermeable Upper and Lower Boundaries

If upper and lower boundaries are impermeable,(5.9)∂P∂z|z=0=0,∂P∂z|z=H=0,
obviously for such impermeable boundary conditions, we have
(5.10)δ(z-z′)=∑n=0∞‍cos(λnz′)cos(λnz)(Hdn),
where
(5.11)dn={1,ifn=0,12,ifn>0.
Let
(5.12)P(x,y,z)=∑n=0∞‍φn(x,y)cos(λnz),
and substituting (5.12) into (5.1) and compare the coefficients of cos(λnz), we obtain
(5.13)Δφn-λn2φn=-cos(λnz′)δ(x-x′)δ(y)(Hdn),
in circular area Ω1={(x,y)∣x2+y2<Re2}, and
(5.14)φn=0,
on circumference Γ1={(x,y)∣x2+y2=Re2}, and Δ is two-dimensional Laplace operator,
(5.15)Δφn=∂2φn∂x2+∂2φn∂y2.Case 1.
Ifn=0, then
(5.16)Δφ0=-δ(x-x′)δ(y)H,inΩ1,φ0=0,onΓ1.
Using Green's function of Laplace problem in a circular domain, we obtain [9–11]
(5.17)φ0(x,y;x′,0)=(12πH){ln[|ρ-ρ*||ρ-ρ0||ρ0|Re]}.Case 2.
Ifn>0, then φn satisfies (5.13). Since [-1/(2π)]K0(λnR′) satisfies the equations:
(5.18)Δu-λn2u=δ(x-x′)δ(y),u=0,onΓ1.
So[-αn/(2π)]K0(λnR′) is a basic solution of (5.13), where
(5.19)αn=(-1Hdn)cos(nπz′H)=(-2H)cos(λnz′),(5.20)βn=αn(2π)=(-22πH)cos(nπz′H)=(-1πH)cos(λnz′).
Define(5.21)ψn=φn+βnK0(λnR′),
thus
(5.22)φn=ψn-βnK0(λnR′),
then ψn satisfies homogeneous equation
(5.23)Δψn-λn2ψn=0,inΩ1,ψn=βnK0(λnR′),onΓ1,
and R′ has the same meaning as in (5.6).
Under polar coordinates representation of Laplace operator and by using methods of separation of variables, we obtain a general solution [9–11]:
(5.24)ψn=[A0nI0(λnρ)+B0nK0(λnρ)][a0nθ+b0n]+∑m=1∞[AmnIm(λnρ)+BmnKm(λnρ)]×[amncos(mθ)+bmnsin(mθ)],
where Ain,Bin,ain,bin,i=0,1,… are undetermined coefficients.
Becauseψn is continuously bounded within Ω1, but Ki(0)=∞, so there holds
(5.25)Bin=0,i=0,1,….
There hold [7, 12]
(5.26)Kυ(z)=(πi2)eυπi/2Hυ(1)(zi),Iυ(z)=e-υπi/2Jυ(zi),
where Kυ(z) is modified Bessel function of second kind and order υ,Iυ(z) is modified Bessel function of first kind and order υ,Jυ(z) is Bessel function of first kind and order υ,Hυ(1)(z) is Hankel function of first kind and order υ, and i=-1.Also there hold [13, page 979]
(5.27)H0(1)(σR′)=J0(σρ0)H0(1)(σRe)+2∑m=1∞‍Jm(σρ0)Hm(1)(σRe)cos(mθ),(5.28)K0(λnR′)=(πi2)H0(1)(iλnR′).Letσ=iλn, (note that i2=-1) putting (5.26) into (5.27), and using (5.28), we have the following Cosine Fourier expansions of K0(λnR′) [13, page 952]:
(5.29)K0(λnR′)=(πi2)[J0(iλnρ0)H0(1)(iλnRe)+2∑m=1∞‍Jm(iλnρ0)Hm(1)(iλnRe)cos(mθ)]=J0(iλnρ0)K0(λnRe)+2∑m=1∞‍e-mπi/2Jm(iλnρ0)Km(λnRe)cos(mθ)=I0(λnρ0)K0(λnRe)+2∑m=1∞‍Im(λnρ0)Km(λnRe)cos(mθ).Note thatψn=βnK0(λnR′) on Γ1, comparing coefficients of Cosine Fourier expansions of K0(λnR′) in (5.29) and (5.24), we obtain
(5.30)a0n=0,b0n=1,bin=0,i=1,2,….Define(5.31)Ymn=amnAmn,n=0,1,2,…,
and recall (5.24), then we have
(5.32)ψn=∑m=0∞‍YmnIm(λnρ)cos(mθ),n=0,1,2,…,
where
(5.33)Y0n=βnK0(λnRe)I0(λnρ0)I0(λnRe),Ymn=2βnKm(λnRe)Im(λnρ0)Im(λnRe).There hold [13, page 919]
(5.34)Im(x)≃exp(x)(2πx)1/2,Km(x)≃[π/(2x)]1/2exp(x),x≫1,∀m≥0.Note thatH in Formula (5.8) is in dimensionless form, recall Formulas (3.9), (3.12) and (3.18), for dimensionless H,Re,ρ0,ρ, there hold
(5.35)λnRe≫1,λnρ0≫1,λnρ≫1,
thus
(5.36)Km(λnRe)Im(λnRe)≈πexp(-2λnRe),(5.37)Im(λnρ0)Im(λnρ)≈[exp(λnρ0)(2πλnρ0)1/2,exp(λnρ)(2πλnρ)1/2]=exp[λn(ρ+ρ0)](2πλn)(ρρ0)1/2,(5.38)YmnIm(λnρ)=2βnKm(λnRe)Im(λnρ0)Im(λnρ)Im(λnRe)≈(2βn)[πexp(-2λnRe)]{exp[λn(ρ+ρ0)](2πλn)(ρρ0)1/2}=(2βn)[π(2πλn)(ρρ0)1/2]exp[-λn(2Re-ρ0-ρ)]=[βnλn(ρρ0)1/2]exp[-λn(2Re-ρ0-ρ)].
There holds
(5.39)|ψn|=∑m=0∞|YmnIm(λnρ)cos(mθ)|<∑m=0∞|YmnIm(λnρ)|=∑m=0∞|2βnKm(λnRe)Im(λnρ0)Im(λnρ)Im(λnRe)|.
Combining Formulas (3.18), (5.38), and (5.39), we obtain
(5.40)∑n=1∞|ψncos(nπzH)|≤∑n=1∞|ψn|=∑n=1∞‍∑m=0∞|2βnKm(λnRe)Im(λnρ0)Im(λnρ)Im(λnRe)|≤∑n=1∞[|2βnK0(λnRe)I0(λnρ0)I0(λnρ)I0(λnRe)|+∑m=1∞|2βnKm(λnRe)Im(λnρ0)Im(λnρ)Im(λnRe)|]=∑n=1∞|2βnK0(λnRe)I0(λnρ0)I0(λnρ)I0(λnRe)|+∑n=1∞∑m=1∞|2βnKm(λnRe)Im(λnρ0)Im(λnρ)Im(λnRe)|=Ξ1+Ξ2,
where
(5.41)Ξ1=∑n=1∞|2βnK0(λnRe)I0(λnρ0)I0(λnρ)I0(λnRe)|,(5.42)Ξ2=∑n=1∞‍∑m=1∞|2βnKm(λnRe)Im(λnρ0)Im(λnρ)Im(λnRe)|.There holds(5.43)Ξ1=∑n=1∞|2βnK0(λnRe)I0(λnρ0)I0(λnρ)I0(λnRe)|≈∑n=1∞[|βn|λn(ρρ0)1/2]exp[-λn(2Re-ρ0-ρ)]=∑n=1∞[1nπ2(ρρ0)1/2]exp[-(nπH)(2Re-ρ0-ρ)]<∑n=1∞[1π2(ρρ0)1/2]exp[-(nπH)(2Re-ρ0-ρ)]≈[1π2(ρρ0)1/2]{exp[-(π/H)(2Re-ρ0-ρ)]1-exp[-(π/H)(2Re-ρ0-ρ)]}≈0,
where we use Formula (3.26),
(5.44)exp[-(πH)(2Re-ρ0-ρ)]≈0,(5.45)x+x2+x3+x4+x5+⋯=x1-x,0<x<1.Ifm>-1/2, there holds [13, page 916]
(5.46)Im(z)=[(z/2)mΓ(m+1/2)Γ(1/2)]∫-11(1-t2)m-1/2cosh(zt)dt,
thus for m≥1,
(5.47)Im(λnρ)≤[(λnρ/2)mΓ(m+1/2)Γ(1/2)]∫-11cosh(λnρt)dt=[2(λnρ/2)m(λnρ)Γ(m+1/2)Γ(1/2)]sinh(λnρ)=[(λnρ/2)m-1Γ(m+1/2)Γ(1/2)]sinh(λnρ)<[(λnρ/2)m-12Γ(m+1/2)Γ(1/2)]exp(λnρ),
where we use
(5.48)∫-11cosh(λnρt)dt=2sinh(λnρ)λnρ,sinh(λnρ)<exp(λnρ)2,
and if -1<t<1,m≥1, then
(5.49)(1-t2)m-1/2≤1.Putting Formula (5.47) into Formula (5.42), we obtain
(5.50)Ξ2=∑n=1∞‍∑m=1∞|2βnKm(λnRe)Im(λnρ0)Im(λnρ)Im(λnRe)|<∑n=1∞‍∑m=1∞(2π|βn|)exp[-λn(2Re-ρ0-ρ)][(λnρ/2)m-12Γ(m+1/2)Γ(1/2)][(λnρ0/2)m-12Γ(m+1/2)Γ(1/2)]=∑n=1∞‍∑m=1∞(2H)(λn2ρρ0/4)m-1[2Γ(m+1/2)Γ(1/2)]2exp[-λn(2Re-ρ0-ρ)]=∑n=1∞(2H)exp[-λn(2Re-ρ0-ρ)]∑m=1∞(λn2ρρ0/4)m-1[2Γ(m+1/2)Γ(1/2)]2.Note that(5.51)Γ(m+1/2)=1×3×5×⋯×(2m-1)π2m>1×2×6×⋯×(2m-2)π2m=2m-1(m-1)!π2m=(m-1)!π2,
then we obtain
(5.52)Ξ2<∑n=1∞(2H)exp[-λn(2Re-ρ0-ρ)]∑m=1∞(λn2ρρ0/4)m-1[2Γ(m+1/2)Γ(1/2)]2<∑n=1∞(2H)exp[-λn(2Re-ρ0-ρ)]∑m=1∞(λn2ρρ0/4)m-1[(m-1)!π]2=∑n=1∞(2H)exp[-λn(2Re-ρ0-ρ)]∑k=0∞(λnρρ0/2)2k(k!π)2=∑n=1∞(2π2H)exp[-λn(2Re-ρ0-ρ)]I0(λnρρ0),
where we use [13, page 919]
(5.53)I0(z)=∑k=0∞(z/2)2k(k!)2,Γ(12)=π.Note thatλnρρ0≫1, and we have
(5.54)I0(λnρρ0)≈exp(λnρρ0)(2πλnρρ0)1/2,
thus Formula (5.52) can be simplified as follows:
(5.55)Ξ2<∑n=1∞(2π2H)exp[-λn(2Re-ρ0-ρ)]I0(λnρρ0)≈∑n=1∞(2π2H)exp[-λn(2Re-ρ0-ρ)][exp(λnρρ0)(2πλnρρ0)1/2]<∑n=1∞[2π2H(2π)1/2(ρρ0)1/4]exp[-λn(2Re-ρ0-ρ-ρρ0)]λn1/2=∑n=1∞[2π3(2nH)1/2(ρρ0)1/4]exp[-(nπH)(2Re-ρ0-ρ-ρρ0)]<∑n=1∞[2π3(2H)1/2(ρρ0)1/4]exp[-(nπH)(2Re-ρ0-ρ-ρρ0)]=[2π3(2H)1/2(ρρ0)1/4]{exp[-(π/H)(2Re-ρ0-ρ-ρρ0)]1-exp[-(π/H)(2Re-ρ0-ρ-ρρ0)]}≈0,
where we use Formulas (5.45) and (3.26)
(5.56)exp[-(πH)(2Re-ρ0-ρ-ρρ0)]≈0.Combining Formulas (5.40), (5.43), and (5.55), we prove
(5.57)∑n=1∞‍ψncos(nπzH)≈0.Combining Formulas (5.4), (5.12), (5.17), (5.20), (5.22), (5.29), and (5.57), the point convergence pressure of point (x′,0,z′) is
(5.58)P(x,y,z;x′,0,z′)=(12πH){ln[|Re2-ρρ0||ρ-ρ0|Re]}+∑n=1∞[βnK0(λnRe)I0(λnR′)I0(λnRe)-βnK0(λnR′)]cos(λnz)=(-1πH)∑n=1∞[K0(λnRe)I0(λnR′)I0(λnRe)-K0(λnR′)]cos(λnz)cos(λnz′)+(12πH){ln[|Re2-ρρ0||ρ-ρ0|Re]}.


### 5.2. Constant Pressure Upper or Lower Boundaries

If the reservoir is with gas cap and impermeable bottom boundary, then(5.59)P|z=0=0,∂P∂z|z=H=0,
and assume the outer boundary is at constant pressure
(5.60)P=0,
on cylindrical surface Γ={(x,y)∣x2+y2=Re2,0<z<H}.Define(5.61)ωn=[(2n-1)π](2H),gn(z)=2Hsin(ωnz),(n=1,2,3,…),
then under the boundary condition of (5.59), we have
(5.62)δ(z-z′)=∑n=1∞‍gn(z)gn(z′).Let(5.63)P(x,y,z)=∑n=1∞‍φn(x,y)sin(ωnz),
where φn satisfies
(5.64)Δφn-ωn2φn=(-2H)δ(x-x′)δ(y)sin(ωnz′),
in Ω1, and
(5.65)φn=0,
on Γ1.Let(5.66)ψn=φn+ζnK0(ωnR′),
where
(5.67)ζn=(-1πH)sin(ωnz′),
and R′ has the same meaning as in Formula (5.6).Thusψn satisfies homogeneous equation:
(5.68)Δψn-ωn2ψn=0,
in Ω1, and
(5.69)ψn=ζnK0(ωnR′),
on Γ1.Using polar coordinates, we have(5.70)ψn=ζnK0(ωnRe)I0(ωnR′)I0(ωnRe),
then
(5.71)φn=ψn-ζnK0(ωnR′),
and point convergence pressure of point (x′,0,z′) is:
(5.72)P(x,y,z;x′,0,z′)=∑n=1∞[ζnK0(ωnRe)I0(ωnR′)I0(ωnRe)-ζnK0(ωnR′)]sin(ωnz)=(-1πH)∑n=1∞[K0(ωnRe)I0(ωnR′)I0(ωnRe)-K0(ωnR′)]sin(ωnz)sin(ωnz′).If the reservoir is with bottom water and impermeable top boundary, then(5.73)P|z=H=0,∂P∂z|z=0=0,
and recall Formula (5.60), the outer boundary is at constant pressure.Define(5.74)hn(z)=2Hcos(ωnz),(n=1,2,3,…),
then under the boundary condition of (5.73), we have
(5.75)δ(z-z′)=∑n=1∞‍hn(z)hn(z′).Let(5.76)P(x,y,z)=∑n=1∞‍φn(x,y)cos(ωnz),
where φn satisfies
(5.77)Δφn-ωn2φn=(-2H)δ(x-x′)δ(y)cos(ωnz′),
in Ω1, and
(5.78)φn=0,
on Γ1.Let(5.79)ψn=φn+ηnK0(ωnR′),
where
(5.80)ηn=(-1πH)cos(ωnz′),
thus ψn satisfies
(5.81)Δψn-ωn2ψn=0,
in Ω1, and
(5.82)ψn=ηnK0(ωnR′),
on Γ1.Using polar coordinates, we have(5.83)ψn=ηnK0(ωnRe)I0(ωnR′)I0(ωnRe),
then
(5.84)φn=ψn-ηnK0(ωnR′),
and point convergence pressure of point (x′,0,z′) is
(5.85)P(x,y,z;x′,0,z′)=∑n=1∞[ηnK0(ωnRe)I0(ωnR′)I0(ωnRe)-ηnK0(ωnR′)]cos(ωnz)=(-1πH)∑n=1∞[K0(ωnRe)I0(ωnR′)I0(ωnRe)-K0(ωnR′)]cos(ωnz)cos(ωnz′).If the reservoir is with gas cap and bottom water, then(5.86)P|z=0=0,P|z=H=0,
and recall Formula (5.60), the outer boundary is at constant pressure.Define(5.87)fn(z)=2Hsin(λnz),(n=1,2,3,…),
where λn has the same meaning as in Formula (5.8).Under the boundary condition of (5.86), we have
(5.88)δ(z-z′)=∑n=1∞‍fn(z)fn(z′).Let(5.89)P(x,y,z)=∑n=1∞‍φn(x,y)sin(λnz),
where φn satisfies
(5.90)Δφn-λn2φn=(-2H)δ(x-x′)δ(y)sin(λnz′),
in Ω1, and
(5.91)φn=0,
on Γ1.Let(5.92)ψn=φn+ιnK0(λnR′),
where
(5.93)ιn=(-1πH)sin(λnz′),
thus ψn satisfies
(5.94)Δψn-λn2ψn=0,
in Ω1, and
(5.95)ψn=ιnK0(λnR′),
on Γ1.Using polar coordinates, we have(5.96)ψn=ιnK0(λnRe)I0(λnR′)I0(λnRe),
then
(5.97)φn=ψn-ιnK0(λnR′),
and point convergence pressure of point (x′,0,z′) is
(5.98)P(x,y,z;x′,0,z′)=∑n=1∞[ιnK0(λnRe)I0(λnR′)I0(λnRe)-ιnK0(λnR′)]sin(λnz)=(-1πH)∑n=1∞[K0(λnRe)I0(λnR′)I0(λnRe)-K0(λnR′)]sin(λnz)sin(λnz′).


## 5.1. Impermeable Upper and Lower Boundaries

If upper and lower boundaries are impermeable,(5.9)∂P∂z|z=0=0,∂P∂z|z=H=0,
obviously for such impermeable boundary conditions, we have
(5.10)δ(z-z′)=∑n=0∞‍cos(λnz′)cos(λnz)(Hdn),
where
(5.11)dn={1,ifn=0,12,ifn>0.
Let
(5.12)P(x,y,z)=∑n=0∞‍φn(x,y)cos(λnz),
and substituting (5.12) into (5.1) and compare the coefficients of cos(λnz), we obtain
(5.13)Δφn-λn2φn=-cos(λnz′)δ(x-x′)δ(y)(Hdn),
in circular area Ω1={(x,y)∣x2+y2<Re2}, and
(5.14)φn=0,
on circumference Γ1={(x,y)∣x2+y2=Re2}, and Δ is two-dimensional Laplace operator,
(5.15)Δφn=∂2φn∂x2+∂2φn∂y2.Case 1.
Ifn=0, then
(5.16)Δφ0=-δ(x-x′)δ(y)H,inΩ1,φ0=0,onΓ1.
Using Green's function of Laplace problem in a circular domain, we obtain [9–11]
(5.17)φ0(x,y;x′,0)=(12πH){ln[|ρ-ρ*||ρ-ρ0||ρ0|Re]}.Case 2.
Ifn>0, then φn satisfies (5.13). Since [-1/(2π)]K0(λnR′) satisfies the equations:
(5.18)Δu-λn2u=δ(x-x′)δ(y),u=0,onΓ1.
So[-αn/(2π)]K0(λnR′) is a basic solution of (5.13), where
(5.19)αn=(-1Hdn)cos(nπz′H)=(-2H)cos(λnz′),(5.20)βn=αn(2π)=(-22πH)cos(nπz′H)=(-1πH)cos(λnz′).
Define(5.21)ψn=φn+βnK0(λnR′),
thus
(5.22)φn=ψn-βnK0(λnR′),
then ψn satisfies homogeneous equation
(5.23)Δψn-λn2ψn=0,inΩ1,ψn=βnK0(λnR′),onΓ1,
and R′ has the same meaning as in (5.6).
Under polar coordinates representation of Laplace operator and by using methods of separation of variables, we obtain a general solution [9–11]:
(5.24)ψn=[A0nI0(λnρ)+B0nK0(λnρ)][a0nθ+b0n]+∑m=1∞[AmnIm(λnρ)+BmnKm(λnρ)]×[amncos(mθ)+bmnsin(mθ)],
where Ain,Bin,ain,bin,i=0,1,… are undetermined coefficients.
Becauseψn is continuously bounded within Ω1, but Ki(0)=∞, so there holds
(5.25)Bin=0,i=0,1,….
There hold [7, 12]
(5.26)Kυ(z)=(πi2)eυπi/2Hυ(1)(zi),Iυ(z)=e-υπi/2Jυ(zi),
where Kυ(z) is modified Bessel function of second kind and order υ,Iυ(z) is modified Bessel function of first kind and order υ,Jυ(z) is Bessel function of first kind and order υ,Hυ(1)(z) is Hankel function of first kind and order υ, and i=-1.Also there hold [13, page 979]
(5.27)H0(1)(σR′)=J0(σρ0)H0(1)(σRe)+2∑m=1∞‍Jm(σρ0)Hm(1)(σRe)cos(mθ),(5.28)K0(λnR′)=(πi2)H0(1)(iλnR′).Letσ=iλn, (note that i2=-1) putting (5.26) into (5.27), and using (5.28), we have the following Cosine Fourier expansions of K0(λnR′) [13, page 952]:
(5.29)K0(λnR′)=(πi2)[J0(iλnρ0)H0(1)(iλnRe)+2∑m=1∞‍Jm(iλnρ0)Hm(1)(iλnRe)cos(mθ)]=J0(iλnρ0)K0(λnRe)+2∑m=1∞‍e-mπi/2Jm(iλnρ0)Km(λnRe)cos(mθ)=I0(λnρ0)K0(λnRe)+2∑m=1∞‍Im(λnρ0)Km(λnRe)cos(mθ).Note thatψn=βnK0(λnR′) on Γ1, comparing coefficients of Cosine Fourier expansions of K0(λnR′) in (5.29) and (5.24), we obtain
(5.30)a0n=0,b0n=1,bin=0,i=1,2,….Define(5.31)Ymn=amnAmn,n=0,1,2,…,
and recall (5.24), then we have
(5.32)ψn=∑m=0∞‍YmnIm(λnρ)cos(mθ),n=0,1,2,…,
where
(5.33)Y0n=βnK0(λnRe)I0(λnρ0)I0(λnRe),Ymn=2βnKm(λnRe)Im(λnρ0)Im(λnRe).There hold [13, page 919]
(5.34)Im(x)≃exp(x)(2πx)1/2,Km(x)≃[π/(2x)]1/2exp(x),x≫1,∀m≥0.Note thatH in Formula (5.8) is in dimensionless form, recall Formulas (3.9), (3.12) and (3.18), for dimensionless H,Re,ρ0,ρ, there hold
(5.35)λnRe≫1,λnρ0≫1,λnρ≫1,
thus
(5.36)Km(λnRe)Im(λnRe)≈πexp(-2λnRe),(5.37)Im(λnρ0)Im(λnρ)≈[exp(λnρ0)(2πλnρ0)1/2,exp(λnρ)(2πλnρ)1/2]=exp[λn(ρ+ρ0)](2πλn)(ρρ0)1/2,(5.38)YmnIm(λnρ)=2βnKm(λnRe)Im(λnρ0)Im(λnρ)Im(λnRe)≈(2βn)[πexp(-2λnRe)]{exp[λn(ρ+ρ0)](2πλn)(ρρ0)1/2}=(2βn)[π(2πλn)(ρρ0)1/2]exp[-λn(2Re-ρ0-ρ)]=[βnλn(ρρ0)1/2]exp[-λn(2Re-ρ0-ρ)].
There holds
(5.39)|ψn|=∑m=0∞|YmnIm(λnρ)cos(mθ)|<∑m=0∞|YmnIm(λnρ)|=∑m=0∞|2βnKm(λnRe)Im(λnρ0)Im(λnρ)Im(λnRe)|.
Combining Formulas (3.18), (5.38), and (5.39), we obtain
(5.40)∑n=1∞|ψncos(nπzH)|≤∑n=1∞|ψn|=∑n=1∞‍∑m=0∞|2βnKm(λnRe)Im(λnρ0)Im(λnρ)Im(λnRe)|≤∑n=1∞[|2βnK0(λnRe)I0(λnρ0)I0(λnρ)I0(λnRe)|+∑m=1∞|2βnKm(λnRe)Im(λnρ0)Im(λnρ)Im(λnRe)|]=∑n=1∞|2βnK0(λnRe)I0(λnρ0)I0(λnρ)I0(λnRe)|+∑n=1∞∑m=1∞|2βnKm(λnRe)Im(λnρ0)Im(λnρ)Im(λnRe)|=Ξ1+Ξ2,
where
(5.41)Ξ1=∑n=1∞|2βnK0(λnRe)I0(λnρ0)I0(λnρ)I0(λnRe)|,(5.42)Ξ2=∑n=1∞‍∑m=1∞|2βnKm(λnRe)Im(λnρ0)Im(λnρ)Im(λnRe)|.There holds(5.43)Ξ1=∑n=1∞|2βnK0(λnRe)I0(λnρ0)I0(λnρ)I0(λnRe)|≈∑n=1∞[|βn|λn(ρρ0)1/2]exp[-λn(2Re-ρ0-ρ)]=∑n=1∞[1nπ2(ρρ0)1/2]exp[-(nπH)(2Re-ρ0-ρ)]<∑n=1∞[1π2(ρρ0)1/2]exp[-(nπH)(2Re-ρ0-ρ)]≈[1π2(ρρ0)1/2]{exp[-(π/H)(2Re-ρ0-ρ)]1-exp[-(π/H)(2Re-ρ0-ρ)]}≈0,
where we use Formula (3.26),
(5.44)exp[-(πH)(2Re-ρ0-ρ)]≈0,(5.45)x+x2+x3+x4+x5+⋯=x1-x,0<x<1.Ifm>-1/2, there holds [13, page 916]
(5.46)Im(z)=[(z/2)mΓ(m+1/2)Γ(1/2)]∫-11(1-t2)m-1/2cosh(zt)dt,
thus for m≥1,
(5.47)Im(λnρ)≤[(λnρ/2)mΓ(m+1/2)Γ(1/2)]∫-11cosh(λnρt)dt=[2(λnρ/2)m(λnρ)Γ(m+1/2)Γ(1/2)]sinh(λnρ)=[(λnρ/2)m-1Γ(m+1/2)Γ(1/2)]sinh(λnρ)<[(λnρ/2)m-12Γ(m+1/2)Γ(1/2)]exp(λnρ),
where we use
(5.48)∫-11cosh(λnρt)dt=2sinh(λnρ)λnρ,sinh(λnρ)<exp(λnρ)2,
and if -1<t<1,m≥1, then
(5.49)(1-t2)m-1/2≤1.Putting Formula (5.47) into Formula (5.42), we obtain
(5.50)Ξ2=∑n=1∞‍∑m=1∞|2βnKm(λnRe)Im(λnρ0)Im(λnρ)Im(λnRe)|<∑n=1∞‍∑m=1∞(2π|βn|)exp[-λn(2Re-ρ0-ρ)][(λnρ/2)m-12Γ(m+1/2)Γ(1/2)][(λnρ0/2)m-12Γ(m+1/2)Γ(1/2)]=∑n=1∞‍∑m=1∞(2H)(λn2ρρ0/4)m-1[2Γ(m+1/2)Γ(1/2)]2exp[-λn(2Re-ρ0-ρ)]=∑n=1∞(2H)exp[-λn(2Re-ρ0-ρ)]∑m=1∞(λn2ρρ0/4)m-1[2Γ(m+1/2)Γ(1/2)]2.Note that(5.51)Γ(m+1/2)=1×3×5×⋯×(2m-1)π2m>1×2×6×⋯×(2m-2)π2m=2m-1(m-1)!π2m=(m-1)!π2,
then we obtain
(5.52)Ξ2<∑n=1∞(2H)exp[-λn(2Re-ρ0-ρ)]∑m=1∞(λn2ρρ0/4)m-1[2Γ(m+1/2)Γ(1/2)]2<∑n=1∞(2H)exp[-λn(2Re-ρ0-ρ)]∑m=1∞(λn2ρρ0/4)m-1[(m-1)!π]2=∑n=1∞(2H)exp[-λn(2Re-ρ0-ρ)]∑k=0∞(λnρρ0/2)2k(k!π)2=∑n=1∞(2π2H)exp[-λn(2Re-ρ0-ρ)]I0(λnρρ0),
where we use [13, page 919]
(5.53)I0(z)=∑k=0∞(z/2)2k(k!)2,Γ(12)=π.Note thatλnρρ0≫1, and we have
(5.54)I0(λnρρ0)≈exp(λnρρ0)(2πλnρρ0)1/2,
thus Formula (5.52) can be simplified as follows:
(5.55)Ξ2<∑n=1∞(2π2H)exp[-λn(2Re-ρ0-ρ)]I0(λnρρ0)≈∑n=1∞(2π2H)exp[-λn(2Re-ρ0-ρ)][exp(λnρρ0)(2πλnρρ0)1/2]<∑n=1∞[2π2H(2π)1/2(ρρ0)1/4]exp[-λn(2Re-ρ0-ρ-ρρ0)]λn1/2=∑n=1∞[2π3(2nH)1/2(ρρ0)1/4]exp[-(nπH)(2Re-ρ0-ρ-ρρ0)]<∑n=1∞[2π3(2H)1/2(ρρ0)1/4]exp[-(nπH)(2Re-ρ0-ρ-ρρ0)]=[2π3(2H)1/2(ρρ0)1/4]{exp[-(π/H)(2Re-ρ0-ρ-ρρ0)]1-exp[-(π/H)(2Re-ρ0-ρ-ρρ0)]}≈0,
where we use Formulas (5.45) and (3.26)
(5.56)exp[-(πH)(2Re-ρ0-ρ-ρρ0)]≈0.Combining Formulas (5.40), (5.43), and (5.55), we prove
(5.57)∑n=1∞‍ψncos(nπzH)≈0.Combining Formulas (5.4), (5.12), (5.17), (5.20), (5.22), (5.29), and (5.57), the point convergence pressure of point (x′,0,z′) is
(5.58)P(x,y,z;x′,0,z′)=(12πH){ln[|Re2-ρρ0||ρ-ρ0|Re]}+∑n=1∞[βnK0(λnRe)I0(λnR′)I0(λnRe)-βnK0(λnR′)]cos(λnz)=(-1πH)∑n=1∞[K0(λnRe)I0(λnR′)I0(λnRe)-K0(λnR′)]cos(λnz)cos(λnz′)+(12πH){ln[|Re2-ρρ0||ρ-ρ0|Re]}.


## 5.2. Constant Pressure Upper or Lower Boundaries

If the reservoir is with gas cap and impermeable bottom boundary, then(5.59)P|z=0=0,∂P∂z|z=H=0,
and assume the outer boundary is at constant pressure
(5.60)P=0,
on cylindrical surface Γ={(x,y)∣x2+y2=Re2,0<z<H}.Define(5.61)ωn=[(2n-1)π](2H),gn(z)=2Hsin(ωnz),(n=1,2,3,…),
then under the boundary condition of (5.59), we have
(5.62)δ(z-z′)=∑n=1∞‍gn(z)gn(z′).Let(5.63)P(x,y,z)=∑n=1∞‍φn(x,y)sin(ωnz),
where φn satisfies
(5.64)Δφn-ωn2φn=(-2H)δ(x-x′)δ(y)sin(ωnz′),
in Ω1, and
(5.65)φn=0,
on Γ1.Let(5.66)ψn=φn+ζnK0(ωnR′),
where
(5.67)ζn=(-1πH)sin(ωnz′),
and R′ has the same meaning as in Formula (5.6).Thusψn satisfies homogeneous equation:
(5.68)Δψn-ωn2ψn=0,
in Ω1, and
(5.69)ψn=ζnK0(ωnR′),
on Γ1.Using polar coordinates, we have(5.70)ψn=ζnK0(ωnRe)I0(ωnR′)I0(ωnRe),
then
(5.71)φn=ψn-ζnK0(ωnR′),
and point convergence pressure of point (x′,0,z′) is:
(5.72)P(x,y,z;x′,0,z′)=∑n=1∞[ζnK0(ωnRe)I0(ωnR′)I0(ωnRe)-ζnK0(ωnR′)]sin(ωnz)=(-1πH)∑n=1∞[K0(ωnRe)I0(ωnR′)I0(ωnRe)-K0(ωnR′)]sin(ωnz)sin(ωnz′).If the reservoir is with bottom water and impermeable top boundary, then(5.73)P|z=H=0,∂P∂z|z=0=0,
and recall Formula (5.60), the outer boundary is at constant pressure.Define(5.74)hn(z)=2Hcos(ωnz),(n=1,2,3,…),
then under the boundary condition of (5.73), we have
(5.75)δ(z-z′)=∑n=1∞‍hn(z)hn(z′).Let(5.76)P(x,y,z)=∑n=1∞‍φn(x,y)cos(ωnz),
where φn satisfies
(5.77)Δφn-ωn2φn=(-2H)δ(x-x′)δ(y)cos(ωnz′),
in Ω1, and
(5.78)φn=0,
on Γ1.Let(5.79)ψn=φn+ηnK0(ωnR′),
where
(5.80)ηn=(-1πH)cos(ωnz′),
thus ψn satisfies
(5.81)Δψn-ωn2ψn=0,
in Ω1, and
(5.82)ψn=ηnK0(ωnR′),
on Γ1.Using polar coordinates, we have(5.83)ψn=ηnK0(ωnRe)I0(ωnR′)I0(ωnRe),
then
(5.84)φn=ψn-ηnK0(ωnR′),
and point convergence pressure of point (x′,0,z′) is
(5.85)P(x,y,z;x′,0,z′)=∑n=1∞[ηnK0(ωnRe)I0(ωnR′)I0(ωnRe)-ηnK0(ωnR′)]cos(ωnz)=(-1πH)∑n=1∞[K0(ωnRe)I0(ωnR′)I0(ωnRe)-K0(ωnR′)]cos(ωnz)cos(ωnz′).If the reservoir is with gas cap and bottom water, then(5.86)P|z=0=0,P|z=H=0,
and recall Formula (5.60), the outer boundary is at constant pressure.Define(5.87)fn(z)=2Hsin(λnz),(n=1,2,3,…),
where λn has the same meaning as in Formula (5.8).Under the boundary condition of (5.86), we have
(5.88)δ(z-z′)=∑n=1∞‍fn(z)fn(z′).Let(5.89)P(x,y,z)=∑n=1∞‍φn(x,y)sin(λnz),
where φn satisfies
(5.90)Δφn-λn2φn=(-2H)δ(x-x′)δ(y)sin(λnz′),
in Ω1, and
(5.91)φn=0,
on Γ1.Let(5.92)ψn=φn+ιnK0(λnR′),
where
(5.93)ιn=(-1πH)sin(λnz′),
thus ψn satisfies
(5.94)Δψn-λn2ψn=0,
in Ω1, and
(5.95)ψn=ιnK0(λnR′),
on Γ1.Using polar coordinates, we have(5.96)ψn=ιnK0(λnRe)I0(λnR′)I0(λnRe),
then
(5.97)φn=ψn-ιnK0(λnR′),
and point convergence pressure of point (x′,0,z′) is
(5.98)P(x,y,z;x′,0,z′)=∑n=1∞[ιnK0(λnRe)I0(λnR′)I0(λnRe)-ιnK0(λnR′)]sin(λnz)=(-1πH)∑n=1∞[K0(λnRe)I0(λnR′)I0(λnRe)-K0(λnR′)]sin(λnz)sin(λnz′).


## 6. Uniform Line Sink Solutions

For convenience, we drop the subscriptD for dimensionless variables. In order to calculate the pressure at wellbore, assume the observation point r is on the wellbore, and recall Formula (5.7), R′=Rw  and
(6.1)|ρ-ρ0|=Rw,ρρ0=(ρ0+Rw)ρ0≈ρ02=R02.The partially penetrating vertical well is taken as a uniform line sink. Recall Formula (5.9), integrate z′ at both sides of Formula (5.58) from L1 to L2, if the upper and lower boundaries are impermeable, pressure at wellbore point (Rw,z) is
(6.2)P(Rw,z)=∫L1L2P(Rw,z;Rw,z′)dz′=(Lp2πH)ln(Re2-R02ReRw)+(-1πH)∑n=1∞[K0(λnRe)I0(λnRw)I0(λnRe)-K0(λnRw)]×cos(nπzH)∫L1L2cos(nπz′H)dz′.Recall Formula (5.34), there hold
(6.3)K0(x)≈π2xe-x,K0(x)I0(x)≈πe-2x,
if x>1, and
(6.4)I0(x)≈1.0,
if 0<x<0.5.Combining to Formulas (3.9), (3.12), (3.13), and (5.8), we obtain
(6.5)λnRe>1,λnRw<0.5,
so
(6.6)K0(λnRe)I0(λnRe)≈πe-2λnRe,I0(λnRw)≈1.0.According to Formula (6.6), we have
(6.7)|(-1πH)∑n=1∞[K0(λnRe)I0(λnRw)I0(λnRe)]cos(nπzH)∫L1L2cos(nπz′H)dz′|=|(-1πH)∑n=1∞[K0(λnRe)I0(λnRw)I0(λnRe)](Hnπ)cos(nπzH)[sin(nπL2H)-sin(nπL1H)]|≤|(-1πH)∑n=1∞[K0(λnRe)I0(λnRw)I0(λnRe)](Hnπ)|=(1π2)∑n=1∞‍1n|K0(λnRe)I0(λnRw)I0(λnRe)|≤(1π2)∑n=1∞|K0(λnRe)I0(λnRw)I0(λnRe)|≈(1π2)∑n=1∞‍πe-2λnRe=e-2πRe/Hπ(1-e-2πRe/H)=1π(e2πRe/H-1)≈0,
because the value of Re/H is very big.Combining Formulas (6.2) and (6.7) yields
(6.8)P(Rw,z)=(Lp2πH)ln(Re2-R02ReRw)+(1πH)∑n=1∞K0(λnRw)cos(nπzH)∫L1L2cos(nπz′H)dz′=(Lp2πH)ln(Re2-R02ReRw)+∑n=1∞(1nπ2)K0(λnRw)cos(nπzH)[sin(nπL2H)-sin(nπL1H)],
where we use
(6.9)∫L1L2cos(nπz′H)dz′=(Hnπ)[sin(nπL2H)-sin(nπL1H)].In order to obtain the average wellbore pressure, integrate both sides of Formula (6.2) with respect to z from L1 to L2, then divided by Lp, we obtain
(6.10)Pa(Rw)=1Lp∫L1L2P(Rw,z)dz=(Lp2πH)ln(Re2-R02ReRw)+∑n=1∞(1nπ2)K0(λnRw)[sin(nπL2H)-sin(nπL1H)][1Lp∫L1L2cos(nπzH)dz]=(Lp2πH)ln(Re2-R02ReRw)+(Hπ3Lp)∑n=1∞(1n2)K0(λnRw)[sin(nπL2H)-sin(nπL1H)]2=(Lp2πH)ln(Re2-R02ReRw)+(4Hπ3Lp)∑n=1∞(1n2)K0(λnRw)sin2(nπLp2H)cos2[nπ(L2+L1)2H].Recall Formula (5.59), let R′=Rw, integrate z′ at both sides of Formula (5.72) from L1 to L2, if the reservoir is with gas cap and impermeable bottom boundary, pressure at wellbore point (Rw,z) is
(6.11)P(Rw,z)=∫L1L2P(Rw,z;Rw,z′)dz′=(-1πH)∑n=1∞[K0(ωnRe)I0(ωnRw)I0(ωnRe)-K0(ωnRw)]Ξ1,
where
(6.12)Ξ1=sin(ωnz)∫L1L2sin(ωnz′)dz′=sin[(2n-1)πz2H][(-2H)(2n-1)π]{cos[(2n-1)πL22H]-cos[(2n-1)πL12H]}.With a similar procedure in Formula (6.7), there holds
(6.13)|(-1πH)∑n=1∞[K0(ωnRe)I0(ωnRw)I0(ωnRe)]sin[(2n-1)πz2H][(-2H)(2n-1)π]|×|{cos[(2n-1)πL22H]-cos[(2n-1)πL12H]}|≈0,
thus
(6.14)P(Rw,z)=∫L1L2P(Rw,z;Rw,z′)dz′=(1πH)∑n=1∞K0(ωnRw)sin[(2n-1)πz2H][-2H(2n-1)π]×{cos[(2n-1)πL22H]-cos[(2n-1)πL12H]}=∑n=1∞[2(2n-1)π2]K0(ωnRw)sin[(2n-1)πz2H]×{cos[(2n-1)πL12H]-cos[(2n-1)πL22H]}.In order to obtain the average wellbore pressure, integrate both sides of Formula (6.14) with respect to z from L1 to L2, then divided by Lp, we obtain
(6.15)Pa(Rw)=1Lp∫L1L2P(Rw,z)dz=1Lp∑n=1∞[2(2n-1)π2]K0(ωnRw){cos[(2n-1)πL12H]-cos[(2n-1)πL22H]}×∫L1L2sin[(2n-1)πz2H]dz=(4Hπ3Lp)∑n=1∞1(2n-1)2K0(ωnRw){cos[(2n-1)πL22H]-cos[(2n-1)πL12H]}2=(16Hπ3Lp)∑n=1∞1(2n-1)2K0(ωnRw)sin2[(2n-1)πLp4H]sin2[(2n-1)π(L2+L1)4H].Recall Formula (5.73), let R′=Rw, integrate z′ at both sides of Formula (5.85) from L1 to L2, if the reservoir is with bottom water and impermeable top boundary, pressure at wellbore point (Rw,z) is
(6.16)P(Rw,z)=∫L1L2P(Rw,z;Rw,z′)dz′=(-1πH)∑n=1∞[K0(ωnRe)I0(ωnRw)I0(ωnRe)-K0(ωnRw)]Ξ2,
where
(6.17)Ξ2=cos(ωnz)∫L1L2cos(ωnz′)dz′=cos[(2n-1)πz2H][(2H)(2n-1)π]{sin[(2n-1)πL22H]-sin[(2n-1)πL12H]}.With a similar procedure in Formula (6.7), there hold
(6.18)|(-1πH)∑n=1∞[K0(ωnRe)I0(ωnRw)I0(ωnRe)]cos[(2n-1)πz2H][2H(2n-1)π]|×|{sin[(2n-1)πL22H]-sin[(2n-1)πL12H]}|≈0,(6.19)P(Rw,z)=∫L1L2P(Rw,z;Rw,z′)dz′=(1πH)∑n=1∞K0(ωnRw)cos[(2n-1)πz2H][2H(2n-1)π]×{sin[(2n-1)πL22H]-sin[(2n-1)πL12H]}=∑n=1∞[2(2n-1)π2]K0(ωnRw)cos[(2n-1)πz2H]×{sin[(2n-1)πL22H]-sin[(2n-1)πL12H]}.In order to obtain the average wellbore pressure, integrate both sides of Formula (6.19) with respect to z from L1 to L2, then divided by Lp, we obtain
(6.20)Pa(Rw)=1Lp∫L1L2P(Rw,z)dz=1Lp∑n=1∞[2(2n-1)π2]K0(ωnRw){sin[(2n-1)πL22H]-sin[(2n-1)πL12H]}×∫L1L2cos[(2n-1)πz2H]dz=(4Hπ3Lp)∑n=1∞1(2n-1)2K0(ωnRw){sin[(2n-1)πL22H]-sin[(2n-1)πL12H]}2=(16Hπ3Lp)∑n=1∞1(2n-1)2K0(ωnRw)sin2[(2n-1)πLp4H]cos2[(2n-1)π(L2+L1)4H].Recall Formula (5.86), let R′=Rw, integrate z′ at both sides of Formula (5.98) from L1 to L2, if the reservoir is with bottom water and gas cap, pressure at wellbore point (Rw,z) is
(6.21)P(Rw,z)=∫L1L2P(Rw,z;Rw,z′)dz′=(-1πH)∑n=1∞[K0(λnRe)I0(λnRw)I0(λnRe)-K0(λnRw)]Ξ3,
where
(6.22)Ξ3=sin(λnz)∫L1L2sin(λnz′)dz′=sin(nπzH)(-Hnπ)[cos(nπL2H)-cos(nπL1H)].With a similar procedure in Formula (6.7), there holds
(6.23)|(-1πH)∑n=1∞[K0(λnRe)I0(λnRw)I0(λnRe)]sin(nπzH)(-Hnπ)|×|[cos(nπL2H)-cos(nπL1H)]|≈0,
thus
(6.24)P(Rw,z)=∫L1L2P(Rw,z;Rw,z′)dz′=(1πH)∑n=1∞K0(λnRw)sin(nπzH)(-Hnπ)[cos(nπL2H)-cos(nπL1H)]=∑n=1∞(-1nπ2)K0(λnRw)sin(nπzH)[cos(nπL2H)-cos(nπL1H)].In order to obtain the average pressure, integrate both sides of Formula (6.24) with respect to z from L1 to L2, then divided by Lp, we obtain
(6.25)Pa(Rw)=1Lp∫L1L2P(Rw,z)dz=1Lp∑n=1∞(-1nπ2)K0(λnRw)[cos(nπL2H)-cos(nπL1H)]∫L1L2sin(nπzH)dz=(Hπ3Lp)∑n=1∞(1n2)K0(λnRw)[cos(nπL2H)-cos(nπL1H)]2=(4Hπ3Lp)∑n=1∞(1n2)K0(λnRw)sin2(nπLp2H)sin2[nπ(L2+L1)2H].


## 7. Partially Penetrating Vertical Well Productivity

Note that Formula (6.10) is in dimensionless form, substitute Formulas (3.14) and (3.15) into Formula (6.10), then rearrange and simplify the resulting formula, the productivity formula for a partially penetrating vertical well is obtained:
(7.1)Qw=FD2πKhH(Pe-Pw)/(μB)ln(ReD/RwD)+Sps,
where Pw is average wellbore pressure, Sps is pseudo-skin factor due to partial penetration:
(7.2)Sps=(8HD2π2LpD2)∑n=1N(1n2)K0[(nπHD)RwD]sin2(nπLpD2HD)cos2[nπ(L2D+L1D)2HD].However, (7.1) is only applicable to a circular cylinder drainage reservoir has both impermeable top and bottom boundaries, and has constant pressure lateral boundary.If the well is fully penetrating, (Lp=L=H,) then Sps=0, and if the reservoir is an isotropic permeability reservoir, that is, Ka=Kh=Kv=K, then Formula (7.1) reduces to Formula (2.1).Substituting Formulas (3.14) and (3.15) into Formula (6.15), then rearranging and simplifing the resulting formula, we obtain the productivity formula for a partially penetrating vertical well in a reservoir with impermeable bottom boundary and constant pressure top boundary (gas cap):
(7.3)Qw=FDπ3KhLp2(Pe-Pw)/(μB)16HΘ1,
where
(7.4)Θ1=∑n=1N1(2n-1)2K0{[(2n-1)π2HD]RwD}×sin2[(2n-1)πLpD4HD]sin2[(2n-1)π(L2D+L1D)4HD].Substituting Formulas (3.14) and (3.15) into Formula (6.20), then rearranging and simplifing the resulting formula, we obtain the productivity formula for a partially penetrating vertical well in a reservoir with impermeable top boundary and constant pressure bottom boundary (bottom water):
(7.5)Qw=FDπ3KhLp2(Pe-Pw)/(μB)16HΘ2,
where
(7.6)Θ2=∑n=1N1(2n-1)2K0{[(2n-1)π2HD]RwD}×sin2[(2n-1)πLpD4HD]cos2[(2n-1)π(L2D+L1D)4HD].Substituting Formulas (3.14) and (3.15) into Formula (6.25), then rearranging and simplifing the resulting formula, we obtain the productivity formula for a partially penetrating vertical well in a reservoir with both bottom water and gas cap:
(7.7)Qw=FDπ3KhLp2(Pe-Pw)/(μB)4HΘ3,
where
(7.8)Θ3=∑n=1N‍1n2K0[(nπHD)RwD]sin2(nπLpD2HD)sin2[nπ(L2D+L1D)2HD].It must be pointed out thatN=500 in the summation of the series in the aforementioned  formulas is sufficient to reach engineering accuracy.


## 8. Conclusions and Discussions

Comparing Formulas (7.1), (7.3), (7.5), and (7.7), we can reach the the following conclusions.If top and bottom reservoir boundaries are impermeable, the radius of the cylindrical system and off-center distance appears in the productivity formulas; if the reservoir has a gas cap or bottom water, the effects of the radius and off-center distance on productivity can be ignored.Because the pay zone thicknessH is very small compared with the circular cylinder drainage radius Re, when the reservoir has gas cap or bottom water which provides the main drive mechanism, the lateral boundary has little influence on productivity, the well is producing as if the reservoir is infinite. The performance is the same for a centered well and an off-center well, even the reservoir is with an edge water boundary (constant pressure outer boundary). The effects of drainage radius and off-center distance on productivity are negligible, thus Re and R0 do not show up in Formulas (7.3), (7.5), and (7.7). However, (7.3), (7.5), and (7.7) are applicable to both centered and off-center wells.If both top and bottom boundaries are impermeable, thenRe and R0 play important roles in well productivity, as indicated by Formula (7.1).Formula (7.2) for pseudo-skin factor is obtained by solving three-dimensional Laplace equation, thus it is more reliable than semianalytical and semiempirical expressions of pseudo-skin factor, that is, Formulas (2.4) and (2.7).Field data will be provided in forthcoming papers in the near future, which can show that the proposed formulas are reliable and accurate, they are fast analytical tools to evaluate well performance.In this paper, we always assume the lateral boundary is at constant pressure, so the proposed formulas are only applicable for steady-state, this study can be improved by considering:(1)
pseudo-steady-state productivity formulas for an oil reservoir with impermeable upper, lower, and lateral boundaries;(2)
steady-state productivity formulas for an oil reservoir with impermeable lateral boundary, but constant pressure upper or lower boundary;(3)
productivity formulas for a partially penetrating vertical well in a box-shaped reservoir, and the corresponding equation of pseudo-skin factor due to partial penetration;(4)
non-Darcy flow effect on an oil well productivity.



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*Source: 626154-2009-08-30.xml*