# Pressure Drop Equations for a Partially  Penetrating Vertical Well in a Circular Cylinder  Drainage

Authors: Jalal Farhan Owayed; Jing Lu
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 unsteady-state pressure drop equations
for an off-center partially penetrating vertical well in a circular
cylinder drainage volume with constant pressure at outer boundary.
First, the point sink solution to the diffusivity equation is
derived, then using superposition principle, pressure drop equations
for a uniform line sink model are obtained. This paper also gives an
equation to calculate pseudoskin factor due to partial penetration.
The proposed equations provide fast analytical tools to evaluate the
performance of a vertical well which is located arbitrarily in a
circular cylinder drainage volume. It is concluded that the well off-center distance has significant
effect on well pressure drop behavior, but it does not have any effect on
pseudoskin factor due to partial penetration. Because the outer
boundary is at constant pressure, when producing time is
sufficiently long, steady-state is definitely reached. When well
producing length is equal to payzone thickness, the pressure drop
equations for a fully penetrating well are obtained.


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## Body

## 1. Introduction

For both fully and  partially penetrating vertical
wells, steady-state and unsteady-state pressure-transient testings are useful
tools for evaluating in situ reservoir and wellbore parameters that describe
the production characteristics of a well. The use of transient well testing for
determining reservoir parameters and well productivity has become common, in
the past years, analytic solutions have been presented for the pressure
behavior of partially penetrating vertical wells.The problem of fluid flow into wells with partial
penetration has received much attention in the past years in petroleum
engineering [1–7].In many oil and gas 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. The
exact solution of the partial penetration problem presents great analytical
problems because the boundary conditions that the solutions of the partial
differential equations must satisfy are mixed; that is, on one of the boundaries
the pressure is specified on one portion and the flux on the other. This
difficult occurs at the wellbore, for the flux over the nonproductive section of
the well is zero, the potential over the perforated interval must be constant.This problem may be overcome in the case of constant
rate production by making the assumption that the flux into the well is uniform
over the entire perforated interval, so that on the wellbore the flux is
specified over the total formation thickness. This approximation naturally
leads to an error in the solution since the potential (pressure) will not be
uniform over the perforated interval, but it has been shown that this occurrence
is not too significant.Many different techniques have been used for solving
the partial penetration problem, namely, finite difference method [2], Fourier, Hankel and
Laplace transforms [3–5], Green's functions  [6]. The analytical expressions
and the numerical results obtained for reservoir pressures by different methods
were essentially identical, however, there are some differences between the
values of wellbore pressures computed from numerical models and those obtained
from analytical solutions [7].The primary goal of this study is to present unsteady
state pressure drop equations for an off-center partially penetrating vertical
well in a circular cylinder drainage volume. Analytical solutions are derived
by making the assumption of uniform fluid withdrawal along the portion of the
wellbore open to flow. Taking the producing portion of a partially penetrating
well as a uniform line sink, using principle of potential superposition,
pressure drop equations for a partially penetrating well are obtained.


## 2. 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 in circular cylinder drainage volume.The following assumptions are made.(1)
The porous media volume is circular cylinder which
has constantKx,Ky,Kz permeabilities, thickness H,
porosity ϕ.
And the porous volume is bounded by top and bottom impermeable boundaries.(2)
The pressure is initially constant in the
cylindrical body, during production the pressure remains constant and equal to
the initial pressurePi at the lateral surface.(3)
The production occurs through a partially
penetrating vertical well of radiusRw,
represented in the model by a uniform line sink which is located at R0 away from the axis of symmetry of the
cylindrical body. The drilled well length is L,
the producing well length is Lpr.(4)
A single-phase fluid, of small and constant
compressibilityCf,
constant viscosity μ,
and formation volume factor B,
flows from the porous media to the well. Fluids properties are independent of
pressure. Gravity forces are neglected.The porous media domain is(2.1)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 holds(2.2)Lpr=L2−L1,Lpr≤L≤H.We assume(2.3)Kx=Ky=Kh,Kz=Kvand define average permeability
(2.4)Ka=(KxKyKz)1/3=Kh2/3Kv1/3.Suppose point(R0,0,z′) is on 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 [8]:(2.5)Kh∂2P∂x2+Kh∂2P∂y2+Kv∂2P∂z2=ϕμCt∂P∂t+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.The initial condition is(2.6)P(t,x,y,z)|t=0=Pi,inΩ.The lateral boundary condition is(2.7)P(t,x,y,z)=Pi,onΓ,where Γ is the cylindrical lateral surface:
(2.8)Γ={(x,y,z)∣x2+y2=Re2,0<z<H}.The porous media domain is bounded by top and bottom
impermeable boundaries, so(2.9)∂P∂z|z=0=0;∂P∂z|z=H=0.In order to simplify the above equations, we take the
following dimensionless transforms:(2.10)xD=2xL,yD=2yL,zD=(2zL)(KhKv)1/2,(2.11)LD=2(KhKv)1/2,HD=(2HL)(KhKv)1/2,(2.12)LprD=L2D−L1D=[2(L2−L1)L](KhKv)1/2,(2.13)R0D=2R0L,ReD=2ReL,RwD=2RwL,(2.14)tD=4KhtϕμCtL2.Assumingq is the point convergence intensity at the
point sink (R0,0,z'),
the partially penetrating   well is a uniform line sink, the total flow rate of
the well is Q,
and there holds(2.15)q=QLprD.Define dimensionless pressures(2.16)PD=4πL(KhKv)1/2(Pi−P)(μqB),(2.17)PwD=4πL(KhKv)1/2(Pi−Pw)(μqB).Note that ifc is a positive constant, there holds [9](2.18)δ(cx)=δ(x)c,consequently,  (2.5)
becomes  [8, 9](2.19)∂2PD∂xD2+∂2PD∂yD2+∂2PD∂zD2=∂PD∂tD−8πδ(xD−R0D)δ(yD)δ(zD−zD′),inΩD,where(2.20)ΩD={(xD,yD,zD)∣xD2+yD2<ReD2,0<zD<HD}.If pointr0 and point r are with distances ρ0 and ρ,
respectively, from the circular center, then the dimensionless off-center
distances are(2.21)ρ0D=2ρ0L,ρD=2ρL,There holds(2.22)(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(2.23)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(2.24)ϑ0≤0.6,ϑ≤0.6.If(2.25)ϑ0=ϑ=0.6,KvKh=0.25,ReH=15then(2.26)(KvKh)1/2(πReH)(2−ϑ0−ϑ−ϑ0ϑ)=0.251/2×(π×15)×(2.0−0.6−0.6−0.6×0.6),exp(−4.7124)=8.983×10−3; and if(2.27)ϑ0=ϑ=0.5,KvKh=0.5,ReH=10,then(2.28)(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  (2.22), according to the above
calculations, without losing generality, there holds(2.29)exp[−(πHD)(2ReD−ρ0D−ρD−ρ0DρD)]≈0.In the same manner, we have(2.30)exp[−(πHD)(2ReD−ρ0D−ρD)]≈0.


## 3. Point Sink Solution

For convenience in the following reference, we use
dimensionless transforms given by (2.10) through (2.17), every variable,
domain, initial and boundary conditions below should be taken as dimensionless,
but we drop the subscript D.Thus, if the point sink is at(x′,0,z′),
(2.19) can be written as(3.1)∂P∂t−ΔP=8πδ(x−x′)δ(y)δ(z−z′),inΩ,where(3.2)Ω={(x,y,z)∣x2+y2<Re2,0<z<H},ΔP=∂2P∂x2+∂2P∂y2+∂2P∂z2.The equation of initial condition is changed
to(3.3)P(t,x,y,z)|t=0=0,inΩ.The equation of lateral boundary condition is changed
to(3.4)P(t,x,y,z)=0,onΓ,where(3.5)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* 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(3.6)ρ*=Re2ρ0.Figure 2
Geometric representation of a circular system.AssumeR′ is the distance between point r and point r0,
then [9, 10](3.7)R′=ρ2+ρ02−2ρρ0cosθ.If the observation pointr is on the drainage circle, ρ=Re,
then(3.8)R′=Re2+ρ02−2Reρ0cosθ,Re>ρ0>0.If the observation pointr is on the wellbore, then(3.9)R′=Rw.Recall  (2.9), obviously for impermeable upper
and lower boundary conditions, there holds [9, 10](3.10)δ(z−z′)=∑k=0∞cos(kπz′H)coskπzH/(Hdk),where(3.11)dk={1,ifk=0,12,ifk>0.Let(3.12)P(t;x,y,z;x′,y′,z′)=∑k=0∞φk(t,x,y)cos(kπzH),and substitute  (3.12)
into  (3.1) and compare the coefficients of cos(kπz/H),
we obtain(3.13)∂φk∂t+λk2φk−(∂2φk∂x2+∂2φk∂y2)=8πcos(kπz′H)δ(x−x′)δ(y)/(Hdk)in circular Ω1={(x,y)∣x2+y2<Re2},
and(3.14)φk=0,on circumference Γ1={(x,y)∣x2+y2=Re2}, and(3.15)φk|t=0=0,where(3.16)λk=kπH.Taking the Laplace transform at the both sides of
(3.13), then(3.17)(∂2φk^∂x2+∂2φk^∂y2)−(s+λk2)φk^=αkδ(x−x′)δ(y)s,inΩ1,(3.18)φk^=0,onΓ1,where(3.19)αk=(−8πHdk)cos(kπz′H),and s is Laplace transform variable.Define(3.20)βk=(12π)αk=(−4Hdk)cos(kπz′H).Case 1.
Ifk=0,
then(3.21)∂2ψ0^∂x2+∂2ψ0^∂y2−sψ0^=α0δ(x−x′)δ(y)s,inΩ1,where(3.22)α0=(−8πH),ψ0^=0,onΓ1.Case 2.
Ifk>0,
then φk^ satisfies  (3.17).
Define(3.23)ζk=λk2+s.
Recall  (3.8), and [−βk/s]K0(ζkR′) is a basic solution of (3.17), since k>0,
we have(3.24)αk=(−16πH)cos(kπz′H),βk=(−8H)cos(kπz′H),so let(3.25)ψk^=φk^+μk^,where(3.26)μk^=βkK0[ζkR′]s,thus(3.27)φk^=ψk^−μk^,and ψk^ satisfies homogeneous equation:(3.28)∂2ψk^∂x2+∂2ψk^∂y2−(s+λk2)ψk^=0,inΩ1,ψk^=βkK0(s+λk2R′)s,onΓ1,R′ has the same meaning as in  (3.8).Under polar coordinates representation of Laplace
operator and by using methods of separation of variables, we obtain a general
solution [11–13]:(3.29)ψk^(s,x,y;s,x′,0)=[A0kI0(ζkρ)+B0kK0(ζkρ)][a0kθ+b0k]+∑m=1∞[AmkIm(ζkρ)+BmkKm(ζkρ)][amkcos(mθ)+bmksin(mθ)],where Aik,Bik,aik,bik,i=0,1,2,…,
are undetermined coefficients.Becauseψk^(s,x,y;s,x′,0) is continuously bounded within Ω1,
but Ki(0)=∞, there holds(3.30)Bik=0,i=0,1,2,….There hold [9, 10](3.31)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.And there hold (see [14, page 979])(3.32)H0(1)(σR′)=J0(σρ0)H0(1)(σRe)+2∑m=1∞Jm(σρ0)Hm(1)(σRe)cos(mθ),(3.33)K0(ζkR′)=(πi2)H0(1)(iζkR′).Letσ=iζk, (note that i2=−1), substituting  (3.31) into  (3.32) and using  (3.33),
we have the following Cosine Fourier expansions of K0(ζkR') (see [14, page 952]):(3.34)K0(ζkR′)=(πi2)[J0(iζkρ0)H0(1)(iζkRe)+2∑m=1∞Jm(iζkρ0)Hm(1)(iζkRe)cos(mθ)]=J0(iζkρ0)K0(ζkRe)+2∑m=1∞e−mπi/2Jm(iζkρ0)Km(ζkRe)cos(mθ)=I0(ζkρ0)K0(ζkRe)+2∑m=1∞Im(ζkρ0)Km(ζkRe)cos(mθ).So, we obtain(3.35)βkK0(ζkR′)s=βk[I0(ζkρ0)K0(ζkRe)+2∑m=1∞Im(ζkρ0)Km(ζkRe)cos(mθ)]s.Note thatψk^=βkK0(ζkR′)/s on Γ1,
and comparing coefficients of Cosine Fourier expansions of K0(ζkR′)/s in  (3.35) and  (3.29), we
obtain(3.36)a0k=0,b0k=1,bik=0,i=1,2,….Define(3.37)Ymk=amkAmk,k=0,1,2,…and recall  (3.29), then
we have(3.38)ψk^(s,x,y;s,x′,0)=∑m=0∞YmkIm(ζkρ)cos(mθ),k=0,1,2,…,where(3.39)Y0k=βkK0(ζkRe)I0(ζkρ0)sI0(ζkRe),(3.40)Ymk=2βkKm(ζkRe)Im(ζkρ0)sIm(ζkRe).In the appendix, we can prove(3.41)∑k=1∞|ψk^cos(kπzH)|≈0.Thus we only consider the casek=0,
in  (3.38) and (3.40), let k=0,
we have(3.42)ψ0^(s,x,y;s,x′,0)=∑m=0∞Ym0Im(ζ0ρ)cos(mθ),where(3.43)ζ0=s,(3.44)Ym0Im(ζ0ρ)=β0Km(sRe)Im(sρ0)Im(sρ)sIm(sRe)=f1m(s)×f2m(s),where(3.45)f1m(s)=(β0Rem21+m)[21+msm/2Km(sRe)Rem],m=0,1,2,…,(3.46)f2m(s)=Im(sρ0)Im(sρ)sm/2+1Im(sRe),m=0,1,2,….And there holds(3.47)ℒ−1{f1m(s)}=(β0Rem21+m)[exp(−(Re2/4t))tm+1],m=0,1,2,…,where ℒ−1 is Inverse Laplace transform operator.Sinces=0,s=−γmn are simple poles of meromorphic function f2m(s),
if using partial fraction expansion of meromorphic function, there holds
[15](3.48)f2m(s)=Bm0s+∑n=1∞Bmns+γmn,where Bm0,Bmn are residues at poles s=0,s=−γmn, respectively, and(3.49)Bm0=(ρ0ρ)m2mm!Rem,γmn=εmn2Re2,εmn is the nth root of equation Jm(x)=0.From  (3.48), we have(3.50)ℒ−1{f2m(s)}=Bm0+∑n=1∞Bmnexp(−γmnt)=∑n=0∞Bmnexp(−γmnt),where γm0=0.According to  the convolution theorem [12], from  (3.44), there
holds(3.51)ℒ−1{f1m(s)×f2m(s)}=(β0Rem21+m){∑n=0∞Bmn∫0t[exp(−(Re2/4τ))τm+1]exp[−γmn(t−τ)]dτ}=Cm0+Dm,where(3.52)Cm0=(β0RemBm021+m)∫0t[exp(−(Re2/4τ))τm+1]dτ=[β0(ρ0ρ)mm!×21+2m]∫0t[exp(−(Re2/4τ))τm+1]dτ,Dm=(β0Rem21+m){∑n=1∞Bmn∫0t[exp(−(Re2/4τ))τm+1]exp[−γmn(t−τ)]dτ}.Recall  (3.38), (3.44), and (3.51), there holds(3.53)ψ0(t,x,y;x′,0)=ℒ−1{ψ0^(s,x,y;x′,0)}=∑m=0∞ℒ−1{f1m(s)×f2m(s)}cos(mθ)=∑m=0∞(Cm0+Dm)cos(mθ).Using Laplace asymptotic integration (see
[16, page 221]), when γmn is sufficiently large, then(3.54)∫0t[exp(−(Re2/4τ))τm+1]exp[−γmn(t−τ)]dτ≈exp(−(Re2/4τ))γmntm+1,therefore,(3.55)Dm=[β0Remexp(−(Re2/4t))21+mtm+1]∑n=1∞Bmnγmn.There holds [9](3.56)Im(x)=∑k=0∞1k!(m+k)!(x2)m+2k=[(1m!)(x2)m]{1+m!(m+1)!(x2)2+[m!2!(m+2)!](x2)4+⋯}.Using  (3.48) and  (3.56), and note
that(3.57)11+x=1−x+x2−x3+O(x3),so  (3.46) can be  written
as(3.58)f2m(s)={(1/m!)(sρ0/2)m[1+𝒜(sρ0/2)2+⋯]}{(1/m!)(sρ/2)m[1+𝒜(sρ/2)2+⋯]}s1+m/2{(1/m!)(sRe/2)m[1+(1/(m+1))(sRe/2)2+⋯]}=Bm0s[1+ρ02s4(m+1)+⋯][1+ρ2s4(m+1)+⋯][1−Re2s4(m+1)+⋯]=Bm0s+(1m!)(ρ0ρ2Re)m[14(m+1)(ρ02+ρ2−Re2)+O(s)],where 𝒜 denotes (m!/(m+1)!), thus from  (3.48), we
obtain(3.59)∑n=1∞Bmnγmn=lims→0[f2m(s)−Bm0s]=[14(m+1)!](ρ0ρ2Re)m(ρ02+ρ2−Re2),therefore,(3.60)Dm=[β0Remexp(−(Re2/4t))21+mtm+1][(ρ02+ρ2−Re2)4(m+1)!](ρ0ρ2Re)m=[β0exp(−(Re2/4t))8t(m+1)!](ρ02+ρ2−Re2)(ρ0ρ4t)m.RP is defined as real part operator, for example, RP(eimθ) means real part of eimθ,(3.61)RP(eimθ)=cos(mθ). There holds(3.62)exp(ρ0ρeiθ4τ)=∑m=0∞(1m!)(ρ0ρeiθ4τ)m,and define(3.63)η=Re24−(ρ0ρeiθ4),η is a complex number.Note thatβ0=−4/H,
recall  (3.53), define(3.64)Λ1=∑m=0∞Cm0cos(mθ)=(β02){∫0t[exp(−(Re2/4τ))τ]∑m=0∞[cos(mθ)m!(ρ0ρ4τ)m]dτ}=(β02)×RP{∫0t[exp(−(Re2/4τ))τ]exp(ρ0ρeiθ4τ)dτ}=(β02)×RP{∫0t[exp(−(η/τ))τ]dτ}=−(β02)×RP{Ei(−ηt)}=(2H)×RP{Ei(−ηt)}.In  (3.60), let(3.65)χ=ρ0ρ4t,and define(3.66)Λ2=∑m=0∞Dmcos(mθ)=[β0exp(−(Re2/4t))8t](ρ02+ρ2−Re2)∑m=0∞[cos(mθ)(m+1)!(ρ0ρ4t)m]=[β0exp(−(Re2/4t))8t](ρ02+ρ2−Re2)∑m=0∞RP{1(m+1)!(ρ0ρeiθ4t)m}=−[exp(−(Re2/4t))2tH](ρ02+ρ2−Re2)×RP{(1χeiθ)[exp(χeiθ)−1]},thus we obtain(3.67)ψ0=ℒ−1{ψ0^}=Λ1+Λ2,and there holds [9, 10](3.68)ℒ−1{K0(as)s}=−12Ei(−a24t),thus if we recall  (3.26)
and define(3.69)Λ3=−μ0=−ℒ−1{μ0^},then(3.70)Λ3=−ℒ−1[β0K0(sR′)s]=−(2H)Ei(−R′24t)and R' has the same meaning as in   (3.7).In the above equations,Ei(−x) is exponential integral
function,(3.71)Ei(−x)=∫−∞−xexp(u)udu,(0<x<∞).Recall  (3.27), there holds(3.72)φ0(t,x,y;x′,0)=ψ0−μ0=Λ1+Λ2+Λ3.Combining  (3.12), (3.26), (3.27), and (3.41), we
obtain(3.73)P(t;x,y,z;x′,y′,z′)=φ0+ℒ−1{∑k=1∞φk^cos(kπzH)}=φ0+ℒ−1{∑k=1∞(ψk^−μk^)cos(kπzH)}=φ0−∑k=1∞cos(kπzH)ℒ−1{βkK0(s+λk2R′)s}.Equation (3.73) is the pressure distribution equation of
an off-center point sink in the cylindrical body. If the point sink r0 and the observation point r are not on a radius of the drainage circle, θ≠0,
recall  (3.7), R' cannot be simplified, we cannot obtain exact
inverse Laplace transform of (3.73), but if necessary, we may obtain
numerical inverse Laplace transform  results.If the point sink is at the center of the drainage
circle, then(3.74)ρ0=0,η=Re24,φ0=(2H)[Ei(−Re24t)−Ei(−ρ24t)]+[exp(−(Re2/4t))2tH](Re2−ρ2).In Figure2, if the point sink r0 and the observation point r are on a radius, then(3.75)θ=0,η=Re24−(ρ0ρ4),φ0=(2H){Ei[−(Re2−ρρ04t)]−Ei[−(ρ−ρ0)24t]−(ρ02+ρ2−Re2ρρ0)[exp(ρρ0−Re24t)−exp(−Re24t)]}.


## 4. Uniform Line Sink Solution

Although the off-center partially penetrating vertical
well is represented in the model by a line sink, we only concern in the
pressures at the wellbore face.For convenience, in the following reference, every
variable below is dimensionless but we drop the subscriptD.The well line sink is located along the line{(x′,0,z):L1≤z≤L2}.
If the observation point r is on the wellbore, R'=Rw,
note that R0≫Rw,
and there hold(4.1)θ=0,ρ=ρ0+Rw=R0+Rw,ρ0=R0,ρ−ρ0=Rw,ρ0ρ≈R02,ρ+ρ0≈2ρ0=2R0, then(4.2)η=Re24−R024,χ=ρρ04t≈R024t,and recall  (3.64),
then(4.3)Λ1(t;R0,0)=−(β02)Ei[−(Re2−R024t)].Recall  (3.66),  then(4.4)Λ2(t;R0,0)=−[2exp(−(Re2/4t))HR02](2R02−Re2)[exp(R024t)−1],and recall  (3.70),
then(4.5)Λ3(t;R0,0)=(β02)Ei[−(ρ−ρ0)24t]=(β02)Ei(−Rw24t).Define(4.6)Γ1=∫L1L2Λ1(t;R0,0)dz′=∫L1L2[−(β02)]Ei[−(Re2−R024t)]dz′=−(β02)(L2−L1)Ei[−(Re2−R024t)]=(2LprH)Ei[−(Re2−R024t)],Γ2=∫L1L2Λ2(t;R0,0)dz′=∫L1L2−[2exp(−(Re2/4t))HR02](2R02−Re2)[exp(R024t)−1]dz′=−[2Lpexp(−(Re2/4t))HR02](2R02−Re2)[exp(R024t)−1]=−(2LprHR02)(2R02−Re2)[exp(R02−Re24t)−exp(−Re24t)],Γ3=∫L1L2Λ3(t;R0,0)dz′=(β02)∫L1L2Ei(−Rw24t)dz′=−(2LprH)Ei(−Rw24t).In order to calculate the pressure at the wellbore,
using principle of potential superposition, integratingz′ at both sides of  (3.72) from L1 to L2,
then(4.7)Ψ0(t)=∫L1L2φ0(t;R0,0)dz′=Γ1+Γ2+Γ3=(2LprH)Ei[−(Re2−R024t)]−(2LprHR02)(2R02−Re2)[exp(R02−Re24t)−exp(−Re24t)]−(2LprH)Ei(−Rw24t)=(2LprH){Ei[−(Re2−R024t)]−Ei(−Rw24t)−(1R02)(2R02−Re2)[exp(R02−Re24t)−exp(−Re24t)]}.Recall  (3.26), and note that ρ−ρ0=Rw, we have(4.8)μ^k=βkK0(Rws+λk2)s,and define(4.9)σ^k=∫L1L2μ^kdz′=βk(1s)∫L1L2K0(Rws+λk2)dz′=−(8Hs)K0(Rws+λk2)∫L1L2cos(kπz′H)dz′=−(8kπs)K0(Rws+λk2)[sin(kπL2H)−sin(kπL1H)],because when s is very small, (time t is sufficiently long), there
holds(4.10)K0(Rws+λk2)=K0(Rwλk1+s/λk2)≈K0(Rwλk),so when time is sufficiently long,(4.11)σk=ℒ−1{σ^k}=−(8kπ)K0(Rwλk)[sin(kπL2H)−sin(kπL1H)].Recall  (3.12), (3.24),  (3.27), and (3.41), when time t is sufficiently long, define(4.12)U=∑k=1∞σkcos(kπzH)=∑k=1∞−(8kπ)K0(Rwλk)[sin(kπL2H)−sin(kπL1H)]cos(kπzH)=−(8π)∑n=1∞K0(nπRw/H)n[sin(nπL2H)−sin(nπL1H)]cos(nπzH).Therefore, the wellbore pressure at point(R0+Rw,z) is(4.13)P(Rw,z)=∫L1L2P(R0+Rw,0,z,t;R0,0,z′)dz′≈Ψ0(t)−U.Considering the bottom point of the well line sink,
thenz=Lpr,L1=0,
thus L2=Lpr,
in this case,  (4.12) reduces to(4.14)U=−(8π)∑n=1∞K0(nπRw/H)nsin(nπLprH)cos(nπzH)=−(4π)∑n=1∞K0(nπRw/H)nsin(2nπLprH)=I1+I2,where(4.15)I1=−(4π)∑n=1NK0(nπRw/H)nsin(2nπLprH),I2=−(4π)∑n=N+1∞K0(nπRw/H)nsin(2nπLprH),N=4[HπRw],where [H/πRw] is the integer part of H/πRw.ForI2 it holds the following
estimate:(4.16)|I2|=|∑n=N+1∞(4π)K0(nπRw/H)nsin(2nπLprH)|≤|∑n=N+1∞(4π)K0(nπRw/H)n|≤∫N∞(4π)K0(4x/N)xdx=(4π)∫4∞K0(y)ydy=2.7×10−3≈0.So,  (4.14) reduces to(4.17)U≈I1=−(4π)∑n=1NK0(nπRw/H)nsin(2nπLprH).Combining  (4.7), (4.13), and (4.17), pressure at
the bottom point of the producing portion is(4.18)P(Rw,Lpr)=Ψ0(t)+(4π)∑n=1NK0(nπRw/H)nsin(2nπLprH).In order to obtain average wellbore pressure, recall
(4.12) and (4.17), integrate both sides of  (4.13) with respect to z from L1 to L2,
then divided by Lpr,
average wellbore pressure is obtained:(4.19)Pa,w=1Lpr∫L1L2P(Rw,z)dz≈Ψ0(t)+(8π)∑n=1NK0(nπRw/H)n[sin(nπL2H)−sin(nπL1H)][1Lpr∫L1L2cos(nπzH)dz]=Ψ0(t)+(8Hπ2Lpr)∑n=1NK0(nπRw/H)n2[sin(nπL2H)−sin(nπL1H)]2=Ψ0(t)+(32Hπ2Lpr)∑n=1NK0(nπRw/H)n2sin2(nπLpr2H)cos2[nπ(L2+L1)2H],where we use(4.20)1Lpr∫L1L2cos(nπzH)dz=(HnπLpr)[sin(nπL2H)−sin(nπL1H)].


## 5. Dimensionless Wellbore Pressure
Equations

Combining  (4.7) and (4.19), the dimensionless
average wellbore pressure of an off-center partially penetrating vertical well
in a circular cylinder drainage volume is(5.1)PwD=(2LprDHD){Ei[−(ReD2−R0D24tD)]−Ei(−RwD24tD)−(1R0D2)(2R0D2−ReD2)[exp(R0D2−ReD24tD)−exp(−ReD24tD)]}+Sp,where(5.2)Sp=(32HDπ2LprD)∑n=1NK0(nπRwD/HD)n2sin2(nπLprD2HD)cos2[nπ(L2D+L1D)2HD],(5.3)N=4[HDπRwD],[HD/πRwD] is the integer part of HD/πRwD.Equation (5.1) is applicable to impermeable upper and
lower boundaries and long after the time when pressure transient reaches the
upper and lower boundaries. And Sp denotes pseudo-skin factor due to partial
penetration.IfLpr=L=H,
the drilled well length is equal to formation thickness, for a fully
penetrating well, Sp=0,  (5.1) reduces to(5.4)PwD=2{Ei[−(ReD2−R0D24tD)]−Ei(−RwD24tD)−(1R0D2)(2R0D2−ReD2)[exp(R0D2−ReD24tD)−exp(−ReD24tD)]}.If the well is located   at the  center of the
cylindrical body, thenx′=R0=0, there holds(5.5)limR0D→0[(2R0D2−ReD2)R0D2][exp(R0D2−ReD24tD)−exp(−ReD24tD)]=−(ReD24tD)exp(−ReD24tD).Thus,  (5.1) reduces to(5.6)PwD=(2LprDHD){Ei[−(ReD24tD)]−Ei(−RwD24tD)+(ReD24tD)exp[−(ReD24tD)]}+Sp,where Sp has the same meaning as in  (5.2).If the well is a fully penetrating well in an infinite
reservoir,Re=∞,
there holds(5.7)Ei(−ReD24tD)=0,(ReD24tD)exp(−ReD24tD)=0.Thus,  (5.6) reduces to(5.8)PwD=−2Ei(−RwD24tD).Substitute  (2.12) and (2.15) into  (2.17),
then simplify and rearrange the resulting equation, we obtain(5.9)Pi−Pw=(μQB8πKhLpr)PwD,where Q is total flow rate of the well, and PwD can be calculated by  (5.1), (5.4),
(5.6), and  (5.8) for different cases.During production, the unsteady state pressure drop of
an off-center partially penetrating vertical well in a circular cylinder
drainage volume can be calculated by (5.9).


## 6. Examples and Discussions

Recall  (5.2), pseudo-skin factor due to
partial penetration Sp is a function of L1,L2 and H
is not a function of well off-center distance R0 or drainage radius Re.For an isotropic reservoir,  (5.2) reduces
to(6.1)Sp=(32Hπ2Lpr)∑n=1NK0(nπRw/H)n2sin2(nπLpr2H)cos2[nπ(L1+L2)2H],and  (5.3) reduces
to(6.2)N=4[HπRw],[H/πRw] is the integer part of H/πRw.If we define(6.3)f1=L1H,f2=L2H,f3=RwH,then  (6.1) can be
written as(6.4)Sp=[32π2(f2−f1)]∑n=1NK0(nπf3)n2sin2[nπ2(f2−f1)]cos2[nπ2(f2+f1)].Example 6.1.
Equation (6.4) shows that pseudo-skin factor Sp is a function of the three parameters f1,f2,f3,
fix two parameters, and generate plots that show the trend of Sp with the third  parameter.Solution
Case 1.
Figure3 shows the trend of Sp with f3 when f1=0.2,f2=0.8,
it can be found that Sp is a weak decreasing function of f3.Figure 3
Pseudo-skin
factor versusRw/H plot.
Case 2.
Figure4 shows the trend of Sp with f1 when f2=0.9,f3=0.002,
it can be found that Sp is an increasing function of f1.
When f2 is a constant, we may assume H is a constant, then L2 is also a constant; when f1 increases, L1 also increases, thus the well producing length Lpr=L2−L1 decreases, and pseudo-skin factor due to
partial penetration increases.
Figure 4
Pseudo-skin
factor versusL1/H plot.
Case 3.
Figure5 shows the trend of Sp with f2 when f1=0.1,f3=0.002,
it can be found that Sp is a decreasing function of f2.
When f1 is a constant, we may assume H is a constant, then L1 is also a constant; when f2 increases, L2 also increases, thus the well producing length Lpr=L2−L1 increases, and pseudo-skin factor due to
partial penetration decreases.
Figure 5
Pseudo-skin
factor versusL2/H plot.Example 6.2.
A fully penetrating off-center vertical well,
if(6.5)ReD=20,RwD=0.01,compare the wellbore pressure
responses when RoD=5,10,15.Solution
Equation (5.4) is used to calculate PwD,
the results are shown in Figure 6.
Figure6 shows that at early times, the well is in
infinite acting period. When producing time is long, the influence from outer
boundary appears. Because the outer boundary is at constant pressure, when the
producing time is sufficiently long, steady state will be reached, PwD becomes a constant.
At a given timetD,
if drainage radius ReD is a constant, when well off-center distance RoD increases, PwD decreases, which indicates the effect from
constant pressure outer boundary is more pronounced.Figure 6
The effect of
well off-center distance on wellbore pressure.Example 6.3.
A fully penetrating off-center vertical
well, if(6.6)RoD=10,RwD=0.01,compare the wellbore pressure
responses when ReD=20,30,40.Solution
Equation (5.4) is used to calculate PwD,
the results are shown in Figure 7.
Figure7 shows that at a given time tD,
if well off-center distance RoD is a constant, when drainage radius ReD increases, PwD also increases, which indicates the effect from
constant pressure outer boundary is more pronounced.Figure 7
The effect of
drainage radius on wellbore pressure.


## 7. Conclusions

The following conclusions are reached.(1)
The proposed equations provide fast analytical
tools to evaluate the performance of a vertical well which is located
arbitrarily in a circular drainage volume with constant pressure outer
boundary.(2)
The well off-center distance has significant
effect on well pressure drop behavior, but it does not have any effect on
pseudo-skin factor due to partial penetration.(3)
Because the outer boundary is at constant
pressure, when producing time is sufficiently long, steady-state is definitely
reached.(4)
At a given time in a given drainage volume, if the
well off-center distance increases, the pressure drop at wellbore decreases.(5)
When well producing length is equal to payzone
thickness, the pressure drop equations for a fully penetrating well are
obtained.



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*Source: 267964-2009-03-19.xml*