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Mathematical Problems in Engineering
Volume 2009, Article ID 267964, 33 pages
http://dx.doi.org/10.1155/2009/267964
Research Article

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

1College of Engineering and Petroleum, Kuwait University, P.O. Box 5969, Safat 13060, Kuwait
2Department of Petroleum Engineering, The Petroleum Institute, P.O. Box 2533, Abu Dhabi, United Arab Emirates

Received 8 October 2008; Accepted 2 January 2009

Academic Editor: Saad A. Ragab

Copyright © 2009 Jalal Farhan Owayed and Jing Lu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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.

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 [17].

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 [35], 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

Figure 1 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.

267964.fig.001
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 constant 𝐾𝑥,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 pressure 𝑃𝑖 at the lateral surface.(3)The production occurs through a partially penetrating vertical well of radius 𝑅𝑤, 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 compressibility 𝐶𝑓, 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 isΩ=(𝑥,𝑦,𝑧)𝑥2+𝑦2<𝑅2𝑒,0<𝑧<𝐻,(2.1)where 𝑅𝑒 is cylinder radius, Ω is the cylindrical body.

Located at 𝑅0 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 holds𝐿𝑝𝑟=𝐿2𝐿1,𝐿𝑝𝑟𝐿𝐻.(2.2)

We assume𝐾𝑥=𝐾𝑦=𝐾,𝐾𝑧=𝐾𝑣(2.3)and define average permeability 𝐾𝑎=(𝐾𝑥𝐾𝑦𝐾𝑧)1/3=𝐾2/3𝐾𝑣1/3.(2.4)

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]:𝐾𝜕2𝑃𝜕𝑥2+𝐾𝜕2𝑃𝜕𝑦2+𝐾𝑣𝜕2𝑃𝜕𝑧2=𝜙𝜇𝐶𝑡𝜕𝑃𝜕𝑡+𝜇𝑞𝐵𝛿(𝑥𝑅0)𝛿(𝑦)𝛿(𝑧𝑧),inΩ,(2.5)where 𝐶𝑡 is total compressibility coefficient of porous media, 𝛿(𝑥𝑅0), 𝛿(𝑦), 𝛿(𝑧𝑧) are Dirac functions.

The initial condition is𝑃(𝑡,𝑥,𝑦,𝑧)|𝑡=0=𝑃𝑖,inΩ.(2.6)

The lateral boundary condition is𝑃(𝑡,𝑥,𝑦,𝑧)=𝑃𝑖,onΓ,(2.7)where Γ is the cylindrical lateral surface: Γ=(𝑥,𝑦,𝑧)𝑥2+𝑦2=𝑅2𝑒,0<𝑧<𝐻.(2.8)

The porous media domain is bounded by top and bottom impermeable boundaries, so𝜕𝑃|||𝜕𝑧𝑧=0=0;𝜕𝑃|||𝜕𝑧𝑧=𝐻=0.(2.9)

In order to simplify the above equations, we take the following dimensionless transforms:𝑥𝐷=2𝑥𝐿,𝑦𝐷=2𝑦𝐿,𝑧𝐷=2𝑧𝐿𝐾𝐾𝑣1/2,𝐿(2.10)𝐷𝐾=2𝐾𝑣1/2,𝐻𝐷=2𝐻𝐿𝐾𝐾𝑣1/2,𝐿(2.11)𝑝𝑟𝐷=𝐿2𝐷𝐿1𝐷=2(𝐿2𝐿1)𝐿𝐾𝐾𝑣1/2,𝑅(2.12)0𝐷=2𝑅0𝐿,𝑅𝑒𝐷=2𝑅𝑒𝐿,𝑅𝑤𝐷=2𝑅𝑤𝐿,𝑡(2.13)𝐷=4𝐾𝑡𝜙𝜇𝐶𝑡𝐿2.(2.14)

Assuming 𝑞 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 holds𝑄𝑞=𝐿𝑝𝑟𝐷.(2.15)

Define dimensionless pressures 𝑃𝐷=4𝜋𝐿(𝐾𝐾𝑣)1/2(𝑃𝑖𝑃),𝑃(𝜇𝑞𝐵)(2.16)𝑤𝐷=4𝜋𝐿(𝐾𝐾𝑣)1/2(𝑃𝑖𝑃𝑤).(𝜇𝑞𝐵)(2.17)

Note that if 𝑐 is a positive constant, there holds [9]𝛿(𝑐𝑥)=𝛿(𝑥)𝑐,(2.18)consequently, (2.5) becomes [8, 9]𝜕2𝑃𝐷𝜕𝑥𝐷2+𝜕2𝑃𝐷𝜕𝑦𝐷2+𝜕2𝑃𝐷𝜕𝑧𝐷2=𝜕𝑃𝐷𝜕𝑡𝐷8𝜋𝛿(𝑥𝐷𝑅0𝐷)𝛿(𝑦𝐷)𝛿(𝑧𝐷𝑧𝐷),inΩ𝐷,(2.19)whereΩ𝐷=(𝑥𝐷,𝑦𝐷,𝑧𝐷)𝑥2𝐷+𝑦2𝐷<𝑅2𝑒𝐷,0<𝑧𝐷<𝐻𝐷.(2.20)

If point 𝐫0 and point 𝐫 are with distances 𝜌0 and 𝜌, respectively, from the circular center, then the dimensionless off-center distances are𝜌0𝐷=2𝜌0𝐿,𝜌𝐷=2𝜌𝐿,(2.21)

There holds𝜋𝐻𝐷2𝑅𝑒𝐷𝜌0𝐷𝜌𝐷𝜌0𝐷𝜌𝐷=𝐾𝑣𝐾1/2𝜋𝐿2𝐻4𝑅𝑒𝐿2𝜌0𝐿2𝜌𝐿2𝜌0𝜌𝐿=𝐾𝑣𝐾1/2𝜋𝑅𝑒𝐻𝜌20𝑅𝑒𝜌𝑅𝑒𝜌0𝜌𝑅𝑒=𝐾𝑣𝐾1/2𝜋𝑅𝑒𝐻2𝜗0𝜗𝜗0𝜗,(2.22)where𝜗0=𝜌0𝑅𝑒𝜌,𝜗=𝑅𝑒.(2.23)

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 assume𝜗00.6,𝜗0.6.(2.24)

If𝜗0𝐾=𝜗=0.6,𝑣𝐾𝑅=0.25,𝑒𝐻=15(2.25)then𝐾𝑣𝐾1/2𝜋𝑅𝑒𝐻2𝜗0𝜗𝜗0𝜗=0.251/2×(𝜋×15)×(2.00.60.60.6×0.6),exp(4.7124)=8.983×103;(2.26) and if𝜗0𝐾=𝜗=0.5,𝑣𝐾𝑅=0.5,𝑒𝐻=10,(2.27)then𝐾𝑣𝐾1/2𝜋𝑅𝑒𝐻2𝜗0𝜗𝜗0𝜗=0.51/2×(𝜋×10)×(2.00.50.50.5×0.5)=11.107,exp(11.107)=1.501×105.(2.28)

Recall (2.22), according to the above calculations, without losing generality, there holds𝜋exp𝐻𝐷2𝑅𝑒𝐷𝜌0𝐷𝜌𝐷𝜌0𝐷𝜌𝐷0.(2.29)

In the same manner, we have𝜋exp𝐻𝐷2𝑅𝑒𝐷𝜌0𝐷𝜌𝐷0.(2.30)

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𝜕𝑃𝜕𝑡Δ𝑃=8𝜋𝛿(𝑥𝑥)𝛿(𝑦)𝛿(𝑧𝑧),inΩ,(3.1)whereΩ=(𝑥,𝑦,𝑧)𝑥2+𝑦2<𝑅2𝑒,𝜕,0<𝑧<𝐻Δ𝑃=2𝑃𝜕𝑥2+𝜕2𝑃𝜕𝑦2+𝜕2𝑃𝜕𝑧2.(3.2)

The equation of initial condition is changed to𝑃(𝑡,𝑥,𝑦,𝑧)|𝑡=0=0,inΩ.(3.3)

The equation of lateral boundary condition is changed to𝑃(𝑡,𝑥,𝑦,𝑧)=0,onΓ,(3.4)whereΓ=(𝑥,𝑦,𝑧)𝑥2+𝑦2=𝑅2𝑒,0<𝑧<𝐻.(3.5)

The problem under consideration is that of fluid flow toward a point sink from an off-center position within a circular of radius 𝑅𝑒. We want to determine the pressure change at an observation point with a distance 𝜌 from the center of circle.

Figure 2 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 that𝜌=𝑅2𝑒𝜌0.(3.6)

267964.fig.002
Figure 2: Geometric representation of a circular system.

Assume 𝑅 is the distance between point 𝐫 and point 𝐫0, then [9, 10]𝑅=𝜌2+𝜌202𝜌𝜌0cos𝜃.(3.7)

If the observation point 𝐫 is on the drainage circle, 𝜌=𝑅𝑒, then𝑅=𝑅2𝑒+𝜌202𝑅𝑒𝜌0cos𝜃,𝑅𝑒>𝜌0>0.(3.8)

If the observation point 𝐫 is on the wellbore, then𝑅=𝑅𝑤.(3.9)

Recall (2.9), obviously for impermeable upper and lower boundary conditions, there holds [9, 10]𝛿(𝑧𝑧)=𝑘=0cos𝑘𝜋𝑧𝐻cos𝑘𝜋𝑧𝐻/(𝐻𝑑𝑘),(3.10)where𝑑𝑘=11,if𝑘=0,2,if𝑘>0.(3.11)

Let𝑃(𝑡;𝑥,𝑦,𝑧;𝑥,𝑦,𝑧)=𝑘=0𝜑𝑘(𝑡,𝑥,𝑦)cos𝑘𝜋𝑧𝐻,(3.12)and substitute (3.12) into (3.1) and compare the coefficients of cos(𝑘𝜋𝑧/𝐻), we obtain𝜕𝜑𝑘𝜕𝑡+𝜆2𝑘𝜑𝑘𝜕2𝜑𝑘𝜕𝑥2+𝜕2𝜑𝑘𝜕𝑦2=8𝜋cos𝑘𝜋𝑧𝐻𝛿(𝑥𝑥)𝛿(𝑦)/(𝐻𝑑𝑘)(3.13)in circular Ω1={(𝑥,𝑦)𝑥2+𝑦2<𝑅2𝑒}, and𝜑𝑘=0,(3.14)on circumference Γ1={(𝑥,𝑦)𝑥2+𝑦2=𝑅2𝑒}, and𝜑𝑘|𝑡=0=0,(3.15)where𝜆𝑘=𝑘𝜋𝐻.(3.16)

Taking the Laplace transform at the both sides of (3.13), then𝜕2𝜑𝑘𝜕𝑥2+𝜕2𝜑𝑘𝜕𝑦2𝑠+𝜆2𝑘𝜑𝑘=𝛼𝑘𝛿(𝑥𝑥)𝛿(𝑦)𝑠,inΩ1,(3.17)𝜑𝑘=0,onΓ1,(3.18)where𝛼𝑘=8𝜋𝐻𝑑𝑘cos𝑘𝜋𝑧𝐻,(3.19)and 𝑠 is Laplace transform variable.

Define𝛽𝑘=1𝛼2𝜋𝑘=4𝐻𝑑𝑘cos𝑘𝜋𝑧𝐻.(3.20)

Case 1. If 𝑘=0, then𝜕2𝜓0𝜕𝑥2+𝜕2𝜓0𝜕𝑦2𝑠𝜓0=𝛼0𝛿(𝑥𝑥)𝛿(𝑦)𝑠,inΩ1,(3.21)where𝛼0=8𝜋𝐻,𝜓0=0,onΓ1.(3.22)

Case 2. If 𝑘>0, then 𝜑𝑘 satisfies (3.17).
Define𝜁𝑘=𝜆2𝑘+𝑠.(3.23)
Recall (3.8), and [𝛽𝑘/𝑠]𝐾0(𝜁𝑘𝑅) is a basic solution of (3.17), since 𝑘>0, we have𝛼𝑘=16𝜋𝐻cos𝑘𝜋𝑧𝐻,𝛽𝑘=8𝐻cos𝑘𝜋𝑧𝐻,(3.24)so let𝜓𝑘=𝜑𝑘+𝜇𝑘,(3.25)where𝜇𝑘=𝛽𝑘𝐾0[𝜁𝑘𝑅]𝑠,(3.26)thus𝜑𝑘=𝜓𝑘𝜇𝑘,(3.27)and 𝜓𝑘 satisfies homogeneous equation:𝜕2𝜓𝑘𝜕𝑥2+𝜕2𝜓𝑘𝜕𝑦2𝑠+𝜆2𝑘𝜓𝑘=0,inΩ1,𝜓𝑘=𝛽𝑘𝐾0𝑠+𝜆2𝑘𝑅𝑠,onΓ1,(3.28)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 [1113]:𝜓𝑘(𝑠,𝑥,𝑦;𝑠,𝑥𝐴,0)=0𝑘𝐼0(𝜁𝑘𝜌)+𝐵0𝑘𝐾0(𝜁𝑘𝑎𝜌)0𝑘𝜃+𝑏0𝑘+𝑚=1𝐴𝑚𝑘𝐼𝑚(𝜁𝑘𝜌)+𝐵𝑚𝑘𝐾𝑚(𝜁𝑘𝑎𝜌)𝑚𝑘cos(𝑚𝜃)+𝑏𝑚𝑘,sin(𝑚𝜃)(3.29)where 𝐴𝑖𝑘,𝐵𝑖𝑘,𝑎𝑖𝑘,𝑏𝑖𝑘,𝑖=0,1,2,, are undetermined coefficients.

Because 𝜓𝑘(𝑠,𝑥,𝑦;𝑠,𝑥,0) is continuously bounded within Ω1, but 𝐾𝑖(0)=, there holds𝐵𝑖𝑘=0,𝑖=0,1,2,.(3.30)

There hold [9, 10]𝐾𝜐(𝑧)=𝜋𝑖2𝑒𝜐𝜋𝑖/2𝐻𝜐(1)𝐼(𝑧𝑖),𝜐(𝑧)=𝑒𝜐𝜋𝑖/2𝐽𝜐(𝑧𝑖),(3.31)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])𝐻0(1)(𝜎𝑅)=𝐽0(𝜎𝜌0)𝐻0(1)(𝜎𝑅𝑒)+2𝑚=1𝐽𝑚(𝜎𝜌0)𝐻𝑚(1)(𝜎𝑅𝑒𝐾)cos(𝑚𝜃),(3.32)0(𝜁𝑘𝑅)=𝜋𝑖2𝐻0(1)𝑖𝜁𝑘𝑅.(3.33)

Let 𝜎=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]):𝐾0𝜁𝑘𝑅=𝜋𝑖2𝐽0(𝑖𝜁𝑘𝜌0)𝐻0(1)(𝑖𝜁𝑘𝑅𝑒)+2𝑚=1𝐽𝑚(𝑖𝜁𝑘𝜌0)𝐻𝑚(1)(𝑖𝜁𝑘𝑅𝑒)cos(𝑚𝜃)=𝐽0(𝑖𝜁𝑘𝜌0)𝐾0(𝜁𝑘𝑅𝑒)+2𝑚=1𝑒𝑚𝜋𝑖/2𝐽𝑚(𝑖𝜁𝑘𝜌0)𝐾𝑚(𝜁𝑘𝑅𝑒)cos(𝑚𝜃)=𝐼0(𝜁𝑘𝜌0)𝐾0(𝜁𝑘𝑅𝑒)+2𝑚=1𝐼𝑚(𝜁𝑘𝜌0)𝐾𝑚(𝜁𝑘𝑅𝑒)cos(𝑚𝜃).(3.34)

So, we obtain𝛽𝑘𝐾0(𝜁𝑘𝑅)𝑠=𝛽𝑘𝐼0(𝜁𝑘𝜌0)𝐾0(𝜁𝑘𝑅𝑒)+2𝑚=1𝐼𝑚(𝜁𝑘𝜌0)𝐾𝑚(𝜁𝑘𝑅𝑒)cos(𝑚𝜃)𝑠.(3.35)

Note that 𝜓𝑘=𝛽𝑘𝐾0(𝜁𝑘𝑅)/𝑠 on Γ1, and comparing coefficients of Cosine Fourier expansions of 𝐾0(𝜁𝑘𝑅)/𝑠 in (3.35) and (3.29), we obtain𝑎0𝑘=0,𝑏0𝑘=1,𝑏𝑖𝑘=0,𝑖=1,2,.(3.36)

Define𝑌𝑚𝑘=𝑎𝑚𝑘𝐴𝑚𝑘,𝑘=0,1,2,(3.37)and recall (3.29), then we have𝜓𝑘(𝑠,𝑥,𝑦;𝑠,𝑥,0)=𝑚=0𝑌𝑚𝑘𝐼𝑚(𝜁𝑘𝜌)cos(𝑚𝜃),𝑘=0,1,2,,(3.38)where𝑌0𝑘=𝛽𝑘𝐾0(𝜁𝑘𝑅𝑒)𝐼0(𝜁𝑘𝜌0)𝑠𝐼0(𝜁𝑘𝑅𝑒),𝑌(3.39)𝑚𝑘=2𝛽𝑘𝐾𝑚(𝜁𝑘𝑅𝑒)𝐼𝑚(𝜁𝑘𝜌0)𝑠𝐼𝑚(𝜁𝑘𝑅𝑒).(3.40)

In the appendix, we can prove𝑘=1|||𝜓𝑘cos𝑘𝜋𝑧𝐻|||0.(3.41)

Thus we only consider the case 𝑘=0, in (3.38) and (3.40), let 𝑘=0, we have𝜓0(𝑠,𝑥,𝑦;𝑠,𝑥,0)=𝑚=0𝑌𝑚0𝐼𝑚(𝜁0𝜌)cos(𝑚𝜃),(3.42)where𝜁0=𝑌𝑠,(3.43)𝑚0𝐼𝑚(𝜁0𝛽𝜌)=0𝐾𝑚𝑠𝑅𝑒𝐼𝑚𝑠𝜌0𝐼𝑚𝑠𝜌𝑠𝐼𝑚𝑠𝑅𝑒=𝑓1𝑚(𝑠)×𝑓2𝑚(𝑠),(3.44)where𝑓1𝑚𝛽(𝑠)=0𝑅𝑚𝑒21+𝑚21+𝑚𝑠𝑚/2𝐾𝑚𝑠𝑅𝑒𝑅𝑚𝑒𝑓,𝑚=0,1,2,,(3.45)2𝑚𝐼(𝑠)=𝑚𝑠𝜌0𝐼𝑚𝑠𝜌𝑠𝑚/2+1𝐼𝑚𝑠𝑅𝑒,𝑚=0,1,2,.(3.46)

And there holds1𝑓1𝑚=𝛽(𝑠)0𝑅𝑚𝑒21+𝑚exp(𝑅2𝑒/4𝑡)𝑡𝑚+1,𝑚=0,1,2,,(3.47)where 1 is Inverse Laplace transform operator.

Since 𝑠=0,𝑠=𝛾𝑚𝑛 are simple poles of meromorphic function 𝑓2𝑚(𝑠), if using partial fraction expansion of meromorphic function, there holds [15]𝑓2𝑚𝐵(𝑠)=𝑚0𝑠+𝑛=1𝐵𝑚𝑛𝑠+𝛾𝑚𝑛,(3.48)where 𝐵𝑚0,𝐵𝑚𝑛 are residues at poles 𝑠=0,𝑠=𝛾𝑚𝑛, respectively, and𝐵𝑚0=(𝜌0𝜌)𝑚2𝑚𝑚!𝑅𝑚𝑒,𝛾𝑚𝑛=𝜀2𝑚𝑛𝑅2𝑒,(3.49)𝜀𝑚𝑛 is the 𝑛th root of equation 𝐽𝑚(𝑥)=0.

From (3.48), we have1𝑓2𝑚(𝑠)=𝐵𝑚0+𝑛=1𝐵𝑚𝑛exp(𝛾𝑚𝑛𝑡)=𝑛=0𝐵𝑚𝑛exp(𝛾𝑚𝑛𝑡),(3.50)where 𝛾𝑚0=0.

According to the convolution theorem [12], from (3.44), there holds1𝑓1𝑚(𝑠)×𝑓2𝑚=𝛽(𝑠)0𝑅𝑚𝑒21+𝑚𝑛=0𝐵𝑚𝑛𝑡0exp((𝑅2𝑒/4𝜏))𝜏𝑚+1exp𝛾𝑚𝑛(𝑡𝜏)𝑑𝜏=𝐶𝑚0+𝐷𝑚,(3.51)where𝐶𝑚0=𝛽0𝑅𝑚𝑒𝐵𝑚021+𝑚𝑡0𝑅exp2𝑒/4𝜏𝜏𝑚+1=𝛽𝑑𝜏0(𝜌0𝜌)𝑚𝑚!×21+2𝑚𝑡0𝑅exp2𝑒/4𝜏𝜏𝑚+1𝐷𝑑𝜏,𝑚=𝛽0𝑅𝑚𝑒21+𝑚𝑛=1𝐵𝑚𝑛𝑡0𝑅exp2𝑒/4𝜏𝜏𝑚+1exp𝛾𝑚𝑛.(𝑡𝜏)𝑑𝜏(3.52)

Recall (3.38), (3.44), and (3.51), there holds𝜓0(𝑡,𝑥,𝑦;𝑥,0)=1𝜓0(𝑠,𝑥,𝑦;𝑥=,0)𝑚=01𝑓1𝑚(𝑠)×𝑓2𝑚=(𝑠)cos(𝑚𝜃)𝑚=0𝐶𝑚0+𝐷𝑚cos(𝑚𝜃).(3.53)

Using Laplace asymptotic integration (see [16, page 221]), when 𝛾𝑚𝑛 is sufficiently large, then𝑡0𝑅exp2𝑒/4𝜏𝜏𝑚+1exp𝛾𝑚𝑛𝑅(𝑡𝜏)𝑑𝜏exp2𝑒/4𝜏𝛾𝑚𝑛𝑡𝑚+1,(3.54)therefore,𝐷𝑚=𝛽0𝑅𝑚𝑒𝑅exp2𝑒/4𝑡21+𝑚𝑡𝑚+1𝑛=1𝐵𝑚𝑛𝛾𝑚𝑛.(3.55)

There holds [9]𝐼𝑚(𝑥)=𝑘=01𝑥𝑘!(𝑚+𝑘)!2𝑚+2𝑘=1𝑥𝑚!2𝑚1+𝑚!𝑥(𝑚+1)!22+𝑚!𝑥2!(𝑚+2)!24.+(3.56)

Using (3.48) and (3.56), and note that11+𝑥=1𝑥+𝑥2𝑥3+𝑂(𝑥3),(3.57)so (3.46) can be written as𝑓2𝑚=(𝑠)(1/𝑚!)𝑠𝜌0/2𝑚1+𝒜𝑠𝜌0/22+(1/𝑚!)𝑠𝜌/2𝑚1+𝒜𝑠𝜌/22+𝑠1+𝑚/2{(1/𝑚!)𝑠𝑅𝑒/2𝑚[1+(1/(𝑚+1))𝑠𝑅𝑒/22=𝐵+]}𝑚0𝑠𝜌1+20𝑠𝜌4(𝑚+1)+1+2𝑠𝑅4(𝑚+1)+12𝑒𝑠=𝐵4(𝑚+1)+𝑚0𝑠+1𝜌𝑚!0𝜌2𝑅𝑒𝑚1𝜌4(𝑚+1)20+𝜌2𝑅2𝑒,+𝑂(𝑠)(3.58)where 𝒜 denotes (𝑚!/(𝑚+1)!), thus from (3.48), we obtain𝑛=1𝐵𝑚𝑛𝛾𝑚𝑛=lim𝑠0𝑓2𝑚𝐵(𝑠)𝑚0𝑠=1𝜌4(𝑚+1)!0𝜌2𝑅𝑒𝑚𝜌20+𝜌2𝑅2𝑒,(3.59)therefore,𝐷𝑚=𝛽0𝑅𝑚𝑒𝑅exp2𝑒/4𝑡21+𝑚𝑡𝑚+1𝜌20+𝜌2𝑅2𝑒𝜌4(𝑚+1)!0𝜌2𝑅𝑒𝑚=𝛽0𝑅exp2𝑒/4𝑡𝜌8𝑡(𝑚+1)!20+𝜌2𝑅2𝑒𝜌0𝜌4𝑡𝑚.(3.60)

𝑅𝑃 is defined as real part operator, for example, 𝑅𝑃(𝑒𝑖𝑚𝜃) means real part of 𝑒𝑖𝑚𝜃,𝑅𝑃(𝑒𝑖𝑚𝜃)=cos(𝑚𝜃).(3.61) There holds𝜌exp0𝜌𝑒𝑖𝜃=4𝜏𝑚=01𝜌𝑚!0𝜌𝑒𝑖𝜃4𝜏𝑚,(3.62)and define𝑅𝜂=2𝑒4𝜌0𝜌𝑒𝑖𝜃4,(3.63)𝜂 is a complex number.

Note that 𝛽0=4/𝐻, recall (3.53), defineΛ1=𝑚=0𝐶𝑚0=𝛽cos(𝑚𝜃)02𝑡0exp((𝑅2𝑒/4𝜏))𝜏𝑚=0cos(𝑚𝜃)𝜌𝑚!0𝜌4𝜏𝑚=𝛽𝑑𝜏02×𝑅𝑃𝑡0exp((𝑅2𝑒/4𝜏))𝜏𝜌exp0𝜌𝑒𝑖𝜃=𝛽4𝜏𝑑𝜏02×𝑅𝑃𝑡0exp((𝜂/𝜏))𝜏𝛽𝑑𝜏=02𝜂×𝑅𝑃𝐸𝑖𝑡=2𝐻𝜂×𝑅𝑃𝐸𝑖𝑡.(3.64)

In (3.60), let𝜌𝜒=0𝜌4𝑡,(3.65)and defineΛ2=𝑚=0𝐷𝑚=𝛽cos(𝑚𝜃)0exp((𝑅2𝑒/4𝑡))𝜌8𝑡20+𝜌2𝑅2𝑒𝑚=0cos(𝑚𝜃)𝜌(𝑚+1)!0𝜌4𝑡𝑚=𝛽0exp((𝑅2𝑒/4𝑡))𝜌8𝑡20+𝜌2𝑅2𝑒𝑚=01𝑅𝑃𝜌(𝑚+1)!0𝜌𝑒𝑖𝜃4𝑡𝑚=exp((𝑅2𝑒/4𝑡))𝜌2𝑡𝐻20+𝜌2𝑅2𝑒1×𝑅𝑃𝜒𝑒𝑖𝜃exp𝜒𝑒𝑖𝜃,1(3.66)thus we obtain𝜓0=1𝜓0=Λ1+Λ2,(3.67)and there holds [9, 10]1𝐾0(𝑎𝑠)𝑠1=2𝑎𝐸𝑖24𝑡,(3.68)thus if we recall (3.26) and defineΛ3=𝜇0=1𝜇0,(3.69)thenΛ3=1𝛽0𝐾0(𝑠𝑅)𝑠2=𝐻𝑅𝐸𝑖24𝑡(3.70)and 𝑅 has the same meaning as in (3.7).

In the above equations, 𝐸𝑖(𝑥) is exponential integral function,𝐸𝑖(𝑥)=𝑥exp(𝑢)𝑢𝑑𝑢,(0<𝑥<).(3.71)

Recall (3.27), there holds𝜑0(𝑡,𝑥,𝑦;𝑥,0)=𝜓0𝜇0=Λ1+Λ2+Λ3.(3.72)

Combining (3.12), (3.26), (3.27), and (3.41), we obtain𝑃(𝑡;𝑥,𝑦,𝑧;𝑥,𝑦,𝑧)=𝜑0+1𝑘=1𝜑𝑘cos𝑘𝜋𝑧𝐻=𝜑0+1𝑘=1(𝜓𝑘𝜇𝑘)cos𝑘𝜋𝑧𝐻=𝜑0𝑘=1cos𝑘𝜋𝑧𝐻1𝛽𝑘𝐾0𝑠+𝜆2𝑘𝑅𝑠.(3.73)

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, 𝜃0, 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, then𝜌0𝑅=0,𝜂=2𝑒4,𝜑0=2𝐻𝑅𝐸𝑖2𝑒𝜌4𝑡𝐸𝑖2+4𝑡exp((𝑅2𝑒/4𝑡))𝑅2𝑡𝐻2𝑒𝜌2.(3.74)

In Figure 2, if the point sink 𝐫0 and the observation point 𝐫 are on a radius, then𝑅𝜃=0,𝜂=2𝑒4𝜌0𝜌4,𝜑0=2𝐻𝑅𝐸𝑖2𝑒𝜌𝜌04𝑡𝐸𝑖(𝜌𝜌0)2𝜌4𝑡20+𝜌2𝑅2𝑒𝜌𝜌0exp𝜌𝜌0𝑅2𝑒𝑅4𝑡exp2𝑒.4𝑡(3.75)

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 subscript 𝐷.

The well line sink is located along the line {(𝑥,0,𝑧)𝐿1𝑧𝐿2}. If the observation point 𝐫 is on the wellbore, 𝑅=𝑅𝑤, note that 𝑅0𝑅𝑤, and there hold𝜃=0,𝜌=𝜌0+𝑅𝑤=𝑅0+𝑅𝑤,𝜌0=𝑅0,𝜌𝜌0=𝑅𝑤,𝜌0𝜌𝑅20,𝜌+𝜌02𝜌0=2𝑅0,(4.1) then𝑅𝜂=2𝑒4𝑅204,𝜒=𝜌𝜌0𝑅4𝑡20,4𝑡(4.2)and recall (3.64), thenΛ1(𝑡;𝑅0𝛽,0)=02𝑅𝐸𝑖2𝑒𝑅204𝑡.(4.3)

Recall (3.66), thenΛ2(𝑡;𝑅0𝑅,0)=2exp2𝑒/4𝑡𝐻𝑅202𝑅20𝑅2𝑒𝑅exp204𝑡1,(4.4)and recall (3.70), thenΛ3(𝑡;𝑅0𝛽,0)=02𝐸𝑖(𝜌𝜌0)2=𝛽4𝑡02𝑅𝐸𝑖2𝑤4𝑡.(4.5)

DefineΓ1=𝐿2𝐿1Λ1(𝑡;𝑅0,0)𝑑𝑧=𝐿2𝐿1𝛽02𝑅𝐸𝑖2𝑒𝑅204𝑡𝑑𝑧𝛽=02(𝐿2𝐿1𝑅)𝐸𝑖2𝑒𝑅20=4𝑡2𝐿𝑝𝑟𝐻𝑅𝐸𝑖2𝑒𝑅20,Γ4𝑡2=𝐿2𝐿1Λ2(𝑡;𝑅0,0)𝑑𝑧=𝐿2𝐿12exp((𝑅2𝑒/4𝑡))𝐻𝑅202𝑅20𝑅2𝑒𝑅exp204𝑡1𝑑𝑧=2𝐿𝑝𝑅exp2𝑒/4𝑡𝐻𝑅202𝑅20𝑅2𝑒𝑅exp204𝑡1=2𝐿𝑝𝑟𝐻𝑅202𝑅20𝑅2𝑒𝑅exp20𝑅2𝑒𝑅4𝑡exp2𝑒,Γ4𝑡3=𝐿2𝐿1Λ3(𝑡;𝑅0,0)𝑑𝑧=𝛽02𝐿2𝐿1𝑅𝐸𝑖2𝑤4𝑡𝑑𝑧=2𝐿𝑝𝑟𝐻𝑅𝐸𝑖2𝑤.4𝑡(4.6)

In order to calculate the pressure at the wellbore, using principle of potential superposition, integrating 𝑧 at both sides of (3.72) from 𝐿1 to 𝐿2, thenΨ0(𝑡)=𝐿2𝐿1𝜑0(𝑡;𝑅0,0)𝑑𝑧=Γ1+Γ2+Γ3=2𝐿𝑝𝑟𝐻𝑅𝐸𝑖2𝑒𝑅204𝑡2𝐿𝑝𝑟𝐻𝑅202𝑅20𝑅2𝑒𝑅exp20𝑅2𝑒𝑅4𝑡exp2𝑒4𝑡2𝐿𝑝𝑟𝐻𝑅𝐸𝑖2𝑤=4𝑡2𝐿𝑝𝑟𝐻𝑅𝐸𝑖2𝑒𝑅20𝑅4𝑡𝐸𝑖2𝑤14𝑡𝑅202𝑅20𝑅2𝑒𝑅exp20𝑅2𝑒𝑅4𝑡exp2𝑒.4𝑡(4.7)

Recall (3.26), and note that 𝜌𝜌0=𝑅𝑤, we have𝜇𝑘=𝛽𝑘𝐾0𝑅𝑤𝑠+𝜆2𝑘𝑠,(4.8)and define𝜎𝑘=𝐿2𝐿1𝜇𝑘𝑑𝑧=𝛽𝑘1𝑠𝐿2𝐿1𝐾0𝑅𝑤𝑠+𝜆2𝑘𝑑𝑧8=𝐾𝐻𝑠0𝑅𝑤𝑠+𝜆2𝑘𝐿2𝐿1cos𝑘𝜋𝑧𝐻𝑑𝑧8=𝐾𝑘𝜋𝑠0𝑅𝑤𝑠+𝜆2𝑘sin𝑘𝜋𝐿2𝐻sin𝑘𝜋𝐿1𝐻,(4.9)because when 𝑠 is very small, (time 𝑡 is sufficiently long), there holds𝐾0𝑅𝑤𝑠+𝜆2𝑘=𝐾0𝑅𝑤𝜆𝑘1+𝑠/𝜆2𝑘𝐾0(𝑅𝑤𝜆𝑘),(4.10)so when time is sufficiently long,𝜎𝑘=1𝜎𝑘8=𝐾𝑘𝜋0(𝑅𝑤𝜆𝑘)sin𝑘𝜋𝐿2𝐻sin𝑘𝜋𝐿1𝐻.(4.11)

Recall (3.12), (3.24), (3.27), and (3.41), when time 𝑡 is sufficiently long, define𝑈=𝑘=1𝜎𝑘cos𝑘𝜋𝑧𝐻=𝑘=18𝐾𝑘𝜋0(𝑅𝑤𝜆𝑘)sin𝑘𝜋𝐿2𝐻sin𝑘𝜋𝐿1𝐻cos𝑘𝜋𝑧𝐻8=𝜋𝑛=1𝐾0(𝑛𝜋𝑅𝑤/𝐻)𝑛sin𝑛𝜋𝐿2𝐻sin𝑛𝜋𝐿1𝐻cos𝑛𝜋𝑧𝐻.(4.12)

Therefore, the wellbore pressure at point (𝑅0+𝑅𝑤,𝑧) is𝑃(𝑅𝑤,𝑧)=𝐿2𝐿1𝑃(𝑅0+𝑅𝑤,0,𝑧,𝑡;𝑅0,0,𝑧)𝑑𝑧Ψ0(𝑡)𝑈.(4.13)

Considering the bottom point of the well line sink, then 𝑧=𝐿𝑝𝑟,𝐿1=0, thus 𝐿2=𝐿𝑝𝑟, in this case, (4.12) reduces to8𝑈=𝜋𝑛=1𝐾0(𝑛𝜋𝑅𝑤/𝐻)𝑛sin𝑛𝜋𝐿𝑝𝑟𝐻cos𝑛𝜋𝑧𝐻4=𝜋𝑛=1𝐾0(𝑛𝜋𝑅𝑤/𝐻)𝑛sin2𝑛𝜋𝐿𝑝𝑟𝐻=𝐼1+𝐼2,(4.14)where𝐼14=𝜋𝑁𝑛=1𝐾0(𝑛𝜋𝑅𝑤/𝐻)𝑛sin2𝑛𝜋𝐿𝑝𝑟𝐻,𝐼24=𝜋𝑛=𝑁+1𝐾0(𝑛𝜋𝑅𝑤/𝐻)𝑛sin2𝑛𝜋𝐿𝑝𝑟𝐻,𝐻𝑁=4𝜋𝑅𝑤,(4.15)where [𝐻/𝜋𝑅𝑤] is the integer part of 𝐻/𝜋𝑅𝑤.

For 𝐼2 it holds the following estimate:|𝐼2|||||=𝑛=𝑁+14𝜋𝐾0(𝑛𝜋𝑅𝑤/𝐻)𝑛sin2𝑛𝜋𝐿𝑝𝑟𝐻||||||||𝑛=𝑁+14𝜋𝐾0(𝑛𝜋𝑅𝑤/𝐻)𝑛||||𝑁4𝜋𝐾0(4𝑥/𝑁)𝑥=4𝑑𝑥𝜋4𝐾0(𝑦)𝑦𝑑𝑦=2.7×1030.(4.16)

So, (4.14) reduces to𝑈𝐼14=𝜋𝑁𝑛=1𝐾0(𝑛𝜋𝑅𝑤/𝐻)𝑛sin2𝑛𝜋𝐿𝑝𝑟𝐻.(4.17)

Combining (4.7), (4.13), and (4.17), pressure at the bottom point of the producing portion is𝑃(𝑅𝑤,𝐿𝑝𝑟)=Ψ04(𝑡)+𝜋𝑁𝑛=1𝐾0(𝑛𝜋𝑅𝑤/𝐻)𝑛sin2𝑛𝜋𝐿𝑝𝑟𝐻.(4.18)

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:𝑃𝑎,𝑤=1𝐿𝑝𝑟𝐿2𝐿1𝑃(𝑅𝑤,𝑧)𝑑𝑧Ψ08(𝑡)+𝜋𝑁𝑛=1𝐾0(𝑛𝜋𝑅𝑤/𝐻)𝑛sin𝑛𝜋𝐿2𝐻sin𝑛𝜋𝐿1𝐻1𝐿𝑝𝑟𝐿2𝐿1cos𝑛𝜋𝑧𝐻𝑑𝑧=Ψ0(𝑡)+8𝐻𝜋2𝐿𝑝𝑟𝑁𝑛=1𝐾0(𝑛𝜋𝑅𝑤/𝐻)𝑛2sin𝑛𝜋𝐿2𝐻sin𝑛𝜋𝐿1𝐻2=Ψ0(𝑡)+32𝐻𝜋2𝐿𝑝𝑟𝑁𝑛=1𝐾0(𝑛𝜋𝑅𝑤/𝐻)𝑛2sin2𝑛𝜋𝐿𝑝𝑟2𝐻cos2𝑛𝜋(𝐿2+𝐿1),2𝐻(4.19)where we use1𝐿𝑝𝑟𝐿2𝐿1cos𝑛𝜋𝑧𝐻𝐻𝑑𝑧=𝑛𝜋𝐿𝑝𝑟sin𝑛𝜋𝐿2𝐻sin𝑛𝜋𝐿1𝐻.(4.20)

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 is𝑃𝑤𝐷=2𝐿𝑝𝑟𝐷𝐻𝐷𝑅𝐸𝑖2𝑒𝐷𝑅20𝐷4𝑡𝐷𝑅𝐸𝑖2𝑤𝐷4𝑡𝐷1𝑅20𝐷2𝑅20𝐷𝑅2𝑒𝐷𝑅exp20𝐷𝑅2𝑒𝐷4𝑡𝐷𝑅exp2𝑒𝐷4𝑡𝐷+𝑆𝑝,(5.1)where𝑆𝑝=32𝐻𝐷𝜋2𝐿𝑝𝑟𝐷𝑁𝑛=1𝐾0(𝑛𝜋𝑅𝑤𝐷/𝐻𝐷)𝑛2sin2𝑛𝜋𝐿𝑝𝑟𝐷2𝐻𝐷cos2𝑛𝜋(𝐿2𝐷+𝐿1𝐷)2𝐻𝐷,𝐻(5.2)𝑁=4𝐷𝜋𝑅𝑤𝐷,(5.3)[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.

If 𝐿𝑝𝑟=𝐿=𝐻, the drilled well length is equal to formation thickness, for a fully penetrating well, 𝑆𝑝=0, (5.1) reduces to𝑃𝑤𝐷𝑅=2𝐸𝑖2𝑒𝐷𝑅20𝐷4𝑡𝐷𝑅𝐸𝑖2𝑤𝐷4𝑡𝐷1𝑅20𝐷2𝑅20𝐷𝑅2𝑒𝐷𝑅exp20𝐷𝑅2𝑒𝐷4𝑡𝐷𝑅exp2𝑒𝐷4𝑡𝐷.(5.4)

If the well is located at the center of the cylindrical body, then 𝑥=𝑅0=0, there holdslim𝑅0𝐷02𝑅20𝐷𝑅2𝑒𝐷𝑅20𝐷𝑅exp20𝐷𝑅2𝑒𝐷4𝑡𝐷𝑅exp2𝑒𝐷4𝑡𝐷𝑅=2𝑒𝐷4𝑡𝐷exp𝑅2𝑒𝐷4𝑡𝐷.(5.5)

Thus, (5.1) reduces to𝑃𝑤𝐷=2𝐿𝑝𝑟𝐷𝐻𝐷𝑅𝐸𝑖2𝑒𝐷4𝑡𝐷𝑅𝐸𝑖2𝑤𝐷4𝑡𝐷+𝑅2𝑒𝐷4𝑡𝐷𝑅exp2𝑒𝐷4𝑡𝐷+𝑆𝑝,(5.6)where 𝑆𝑝 has the same meaning as in (5.2).

If the well is a fully penetrating well in an infinite reservoir, 𝑅𝑒=, there holds𝑅𝐸𝑖2𝑒𝐷4𝑡𝐷𝑅=0,2𝑒𝐷4𝑡𝐷𝑅exp2𝑒𝐷4𝑡𝐷=0.(5.7)

Thus, (5.6) reduces to𝑃𝑤𝐷𝑅=2𝐸𝑖2𝑤𝐷4𝑡𝐷.(5.8)

Substitute (2.12) and (2.15) into (2.17), then simplify and rearrange the resulting equation, we obtain𝑃𝑖𝑃𝑤=𝜇𝑄𝐵8𝜋𝐾𝐿𝑝𝑟𝑃𝑤𝐷,(5.9)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 to𝑆𝑝=32𝐻𝜋2𝐿𝑝𝑟𝑁𝑛=1𝐾0(𝑛𝜋𝑅𝑤/𝐻)𝑛2sin2𝑛𝜋𝐿𝑝𝑟2𝐻cos2𝑛𝜋(𝐿1+𝐿2)2𝐻,(6.1)and (5.3) reduces to𝐻𝑁=4𝜋𝑅𝑤,(6.2)[𝐻/𝜋𝑅𝑤] is the integer part of 𝐻/𝜋𝑅𝑤.

If we define𝑓1=𝐿1𝐻,𝑓2=𝐿2𝐻,𝑓3=𝑅𝑤𝐻,(6.3)then (6.1) can be written as𝑆𝑝=32𝜋2(𝑓2𝑓1)𝑁𝑛=1𝐾0(𝑛𝜋𝑓3)𝑛2sin2𝑛𝜋2(𝑓2𝑓1)cos2𝑛𝜋2(𝑓2+𝑓1).(6.4)

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. Figure 3 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.

267964.fig.003
Figure 3: Pseudo-skin factor versus 𝑅𝑤/𝐻 plot.

Case 2. Figure 4 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=𝐿2𝐿1 decreases, and pseudo-skin factor due to partial penetration increases.
267964.fig.004
Figure 4: Pseudo-skin factor versus 𝐿1/𝐻 plot.

Case 3. Figure 5 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=𝐿2𝐿1 increases, and pseudo-skin factor due to partial penetration decreases.
267964.fig.005
Figure 5: Pseudo-skin factor versus 𝐿2/𝐻 plot.

Example 6.2. A fully penetrating off-center vertical well, if𝑅𝑒𝐷=20,𝑅𝑤𝐷=0.01,(6.5)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.
Figure 6 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 time 𝑡𝐷, 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.

267964.fig.006
Figure 6: The effect of well off-center distance on wellbore pressure.

Example 6.3. A fully penetrating off-center vertical well, if𝑅𝑜𝐷=10,𝑅𝑤𝐷=0.01,(6.6)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.
Figure 7 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.

267964.fig.007
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.

Appendix

In this appendix, we want to prove (3.41).

For convenience, in the following reference, every variable below is dimensionless but we drop the subscript 𝐷.

There hold [14] 𝐼𝑚(𝑥)exp(𝑥)(2𝜋𝑥)1/2,𝐾𝑚(𝑥)[𝜋/(2𝑥)]1/2exp(𝑥),𝑥1,𝑚0.(A.1)

Since𝜁𝑘=𝜆2𝑘+𝑠>𝜆𝑘=𝑘𝜋𝐻>0,𝑘1,(A.2)and note that 𝐻 is in dimensionless form in the above equation, recall (2.11), (2.13) and (2.21), for dimensionless 𝐻,𝑅𝑒,𝜌0,𝜌, there hold𝜁𝑘𝑅𝑒1,𝜁𝑘𝜌01,𝜁𝑘𝜌1,(A.3)thus, we obtain𝐾𝑚(𝜁𝑘𝑅𝑒)𝐼𝑚(𝜁𝑘𝑅𝑒)𝜋exp(2𝜁𝑘𝑅𝑒𝐼),(A.4)𝑚(𝜁𝑘𝜌0)𝐼𝑚(𝜁𝑘𝜌)exp(𝜁𝑘𝜌0)(2𝜋𝜁𝑘𝜌0)1/2exp(𝜁𝑘𝜌)(2𝜋𝜁𝑘𝜌)1/2=exp[𝜁𝑘(𝜌+𝜌0)](2𝜋𝜁𝑘)(𝜌𝜌0)1/2,𝑌(A.5)𝑚𝑘𝐼𝑚(𝜁𝑘𝜌)=2𝛽𝑘𝐾𝑚(𝜁𝑘𝑅𝑒)𝐼𝑚(𝜁𝑘𝜌0)𝐼𝑚(𝜁𝑘𝜌)𝑠𝐼𝑚(𝜁𝑘𝑅𝑒)2𝛽𝑘𝑠𝜋exp(2𝜁𝑘𝑅𝑒)exp[𝜁𝑘(𝜌+𝜌0)](2𝜋𝜁𝑘)(𝜌𝜌0)1/2=2𝛽𝑘𝑠𝜋(2𝜋𝜁𝑘)(𝜌𝜌0)1/2exp𝜁𝑘(2𝑅𝑒𝜌0=𝛽𝜌)𝑘𝑠𝜁𝑘(𝜌𝜌0)1/2exp𝜁𝑘(2𝑅𝑒𝜌0.𝜌)(A.6)

There holds|𝜓𝑘|=𝑚=0||𝑌𝑚𝑘𝐼𝑚(𝜁𝑘||<𝜌)cos(𝑚𝜃)𝑚=0||𝑌𝑚𝑘𝐼𝑚(𝜁𝑘||=𝜌)𝑚=0|||2𝛽𝑘𝐾𝑚(𝜁𝑘𝑅𝑒)𝐼𝑚(𝜁𝑘𝜌0)𝐼𝑚(𝜁𝑘𝜌)𝑠𝐼𝑚(𝜁𝑘𝑅𝑒)|||.(A.7)

Combining (2.21), (3.20), (A.6), and (A.7), we obtain𝑘=1|||𝜓𝑘cos𝑘𝜋𝑧𝐻|||𝑘=1|𝜓𝑘|=𝑘=1𝑚=0|||2𝛽𝑘𝐾𝑚(𝜁𝑘𝑅𝑒)𝐼𝑚(𝜁𝑘𝜌0)𝐼𝑚(𝜁𝑘𝜌)𝑠𝐼𝑚(𝜁𝑘𝑅𝑒)|||=𝑘=1|||2𝛽𝑘𝐾0(𝜁𝑘𝑅𝑒)𝐼0(𝜁𝑘𝜌0)𝐼0(𝜁𝑘𝜌)𝑠𝐼0(𝜁𝑘𝑅𝑒)|||+𝑚=1|||2𝛽𝑘𝐾𝑚(𝜁𝑘𝑅𝑒)𝐼𝑚(𝜁𝑘𝜌0)𝐼𝑚(𝜁𝑘𝜌)