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# Solving unsteady flow problems with Laplace transform and source functions

There are many advantages of developing transient flow solutions in the Laplace transform domain. For example, in the Laplace transform domain, Duhamel’s theorem[1] provides a convenient means of developing transient flow solutions for variable rate production problems using the solutions for the corresponding constant rate production problem.

## Transient flow solutions in the Laplace domain

Duhamel’s theorem states that if Δp and Δpc denote the pressure drawdown corresponding to the variable production rate, q(t), and the constant production rate, qc, respectively, then

....................(1)

Applying the Laplace transform converts the convolution integral in Eq. 1 to an algebraic expression, and Duhamel’s theorem is given in the Laplace transform domain as

....................(2)

The simplicity of the expression given in Eq. 2 explains our interest in obtaining transient-flow solutions in the Laplace transform domain.

Another example to explain the convenience of the Laplace domain solutions is for the naturally fractured reservoirs. Common transient flow models of naturally fractured reservoirs lead to the following differential equation in radial coordinates in the Laplace transform domain: [2]

....................(3)

where the subscript f indicates the fracture property, and tD and rD are the dimensionless time and distance (as defined in Eqs. 12 and 16).

The naturally fractured reservoir function, f (s), is a function of matrix and fracture properties and depends on the model chosen to represent the naturally fractured reservoir.[2] The corresponding differential equation for a homogeneous reservoir is obtained by setting f (s) = 1 and is given by

....................(4)

The general solutions for Eqs. 3 and 4 are given, respectively, by

....................(5) and

....................(6)

To obtain a solution for constant-rate production from an infinite reservoir, for example, the following boundary conditions are imposed:

....................(7)

and

....................(8)

Then, it may be shown that

....................(9)

where the right side of Eq. 9 indicates the substitution of sf (s) for s in sΔp(s). This discussion demonstrates that it is possible to derive transient flow solutions for naturally fractured reservoirs by following the same lines as those for the homogeneous reservoirs. Furthermore, if the solution for the corresponding homogeneous reservoir system is known in the Laplace transform domain, then the solution for the naturally fractured reservoir problem may be directly obtained from Eq. 9.

Obtaining the Laplace transforms of the Green’s and source function solutions developed in the time domain with the methods explained on the Source function solutions of the diffusion equation and Solving unsteady flow problems with Green's and source functions pages usually poses a difficult problem. The problems arise mainly because of the use of the product method solution. For a specific class of functions, Chen et al.[3] presented a technique that may be used to apply the Laplace transform to the product solution technique. For a more general procedure to develop source function solutions in the Laplace transform domain, however, the product solution technique should be avoided.[4]

Ozkan and Raghavan[5][6] have shown that it is more convenient to develop source-function solutions in the Laplace transform domain if the point-source solution is used as a building block. Then, other source geometries may be obtained by the superposition (integration) of the point sources along the length, surface, or volume of the source.

## Point-source solution in the Laplace domain

Consider the flow of a slightly compressible fluid in an infinite, naturally fractured reservoir. We can use the double-porosity model suggested by Barenblatt et al.[7] and Warren and Root[8] to develop the governing flow equations for naturally fractured reservoirs. The results, however, will be applicable to the model suggested by Kazemi[9] and de Swaan-O[10] with a simple modification.

Flow around a point source in an infinite porous medium may be expressed conveniently in spherical coordinates. The differential equations governing flow in a naturally fractured reservoir are given in spherical coordinates by

....................(10)

and

....................(11)

In Eqs. 10 and 11, subscripts f and m indicate the property of the fracture and matrix systems, respectively. Initial pressure, pi, is assumed to be uniform in the entire system; that is, pfi = pmi = pi. The dimensionless time, tD, is defined by

....................(12)

where is a characteristic length in the system, and

....................(13)

The definitions of the other variables used in Eqs. 10 and 11 are

....................(14)

....................(15)

and

....................(16)

where

....................(17)

The initial and outer-boundary conditions are given, respectively, by

....................(18)

and

....................(19)

The inner-boundary condition corresponding to the instantaneous withdrawal of an amount of fluid, , at t = 0 from a point source is obtained by considering the mass balance on a small sphere. If we require that at any time t = T > 0, the sum of the flux through the surface of a small sphere around the source location must equal the volume of the fluid, , instantaneously withdrawn from the sphere at t = 0, we can write[11]

....................(20)

Although the withdrawal of fluids from the sphere is instantaneous, the resulting flow in the porous medium, and the flux across the surface of the sphere, is continuous. Therefore, if q represents the total flux across the surface of the small sphere during the time interval 0 ≤ tT, then the mass balance requires that the cumulative production (flux across the surface of the small sphere) at time T be equal to the instantaneous withdrawal volume of fluid from the sphere. That is,

....................(21)

For the condition expressed in Eq. 21 to hold for every T ≥ 0, we must have

....................(22)

where δ(t) is the Dirac delta function satisfying the properties expressed by Eqs. 23 and 24.

....................(23)

....................(24)

Using the results given by Eqs. 21 and 22 in Eq. 20, we obtain

....................(25)

The Laplace transform of Eqs. 10, 11, 19, and 25 yields

....................(26)

where

....................(27)

....................(28)

and

....................(29)

In deriving these results, we have used the initial condition given by Eq. 18 and noted that

....................(30)

In Eq. 29, the term represents the strength of the source for the naturally fractured porous medium.

The solution of Eqs. 26, 28, and 29 yields the following solution for the pressure distribution in the reservoir, except at the source location (the origin), because of an instantaneous point source of strength acting at t = 0:

....................(31)

If the source is located at x′D, y′D, z′D, then, by translation, we can write

....................(32)

where

....................(33)

and

....................(34)

The instantaneous point-source solution for the model suggested by Barenblatt et al.[7] and Warren and Root[8] can also be used for the model suggested by Kazemi[9] and de Swaan-O,[10] provided that the appropriate f(s) function is invoked. To obtain the solution for a homogeneous reservoir, f(s) should be set to unity, Vf = 1, and Vm = 0.

If we consider continuous withdrawal of fluids from the point source, then, by the principle of superposition, we should have

....................(35)

The Laplace transform of Eq. 35 yields the following continuous point-source solution in an infinite reservoir:

....................(36)

where we have substituted Eq. 33 for S, dropped the subscript f, and defined

....................(37)

## Line-, surface-, and volumetric-source solution in the laplace domain

The point-source solution in the Laplace domain may be used to obtain the source solutions for different source geometries. If we define

....................(38)

where Δpp represents the appropriate point-source solution, then, by the application of the superposition principle, the solution for the withdrawal of fluids from a line, surface, or volume, Γw, is given by

....................(39)

If we assume a uniform-flux distribution in time and over the length, surface, or volume of the source, then

....................(40)

The following presentation of the source function approach in the Laplace domain assumes that the flux distribution is uniform, and . Also, the constant production rate from the length, area, or the volume of the source, Γw, is denoted by q so that .

Only sources in infinite reservoirs have been considered so far. These solutions may be easily extended to bounded reservoirs. The following sections present some useful solutions for transient-flow problems in bounded porous media. The first group of solutions is for laterally infinite reservoirs bounded by parallel planes in the vertical direction (infinite-slab reservoirs). The second and third groups comprise the solutions for cylindrical and rectangular reservoirs, respectively.

## Solutions for infinite-slab reservoirs

In this section, we consider one of the most common reservoir geometries used in pressure transient analysis of wells in porous media. It is assumed that the lateral boundaries of the reservoir are far enough not to influence the pressure response during the time period of interest. The top and bottom boundaries of the reservoir at z = 0 and z = h are parallel planes and may be of impermeable, constant pressure, or mixed type. Table 1 presents the solutions for the most common well geometries (point-source, vertical, fractured, and horizontal wells) in infinite-slab reservoirs. Next, we briefly discuss the derivation of these solutions.

Consider a point source in an infinite-slab reservoir with impermeable boundaries at the bottom, z = 0, and the top, z = h. To obtain the point-source solution for this case, we use the point-source solution in an infinite reservoir given by Eq. 36 with the method of images. The result is given by

....................(41)

where

....................(42)

....................(43)

....................(44)

and

....................(45)

The solution given in Eq. 41 is not very convenient for computational purposes. To obtain a computationally convenient form of the solution, we use the summation formula given by[11][12]

....................(46)

and recast Eq. 41 as

....................(47)

The point-source solutions for infinite-slab reservoirs with constant pressure and mixed boundaries at the top and bottom are obtained in a similar manner[12] and are given in Table 1. The point-source solutions can be used with Eqs. 38 and 40 to generate the solutions for the other well geometries given in Table 1. For example, to generate the solution for a partially penetrating vertical line-source well of length hw in an infinite-slab reservoir with impermeable slab boundaries, we integrate the right side of Eq. 47 from zwhw / 2 to zw + hw / 2 with respect to z′, where zw is the vertical coordinate of the midpoint of the open interval. If hw = h (i.e., the well penetrates the entire thickness of the slab reservoir), then this procedure yields the solution for a fully penetrating vertical line-source well. The solution for a partially penetrating fracture of height hf and half-length xf is obtained if the point-source solution is integrated once with respect to z′ from zwhf / 2 to zw + hf / 2 and then with respect to x′ from xwxf to xw + xf, where xw and zw are the coordinates of the midpoint of the fracture. Similarly, the solution for a horizontal-line source well of length Lh is obtained by integrating the point-source solution with respect to x′ from xwLh / 2 to xw + Lh / 2, where xw is the x-coordinate of the midpoint of the horizontal well.

## Solutions for cylindrical reservoir

Solutions for cylindrical reservoirs may also be obtained by starting from the point-source solution in the Laplace transform domain. The Laplace domain solution for a point source located at r′D, θ′, z′D should satisfy the following diffusion equation in cylindrical coordinates.[6]

....................(48)

where

....................(49) The point-source solution is also required to satisfy the following flux condition at the source location (rD →0+, θ = θ′, zD = z′D):

....................(50)

Assuming that the reservoir is bounded by a cylindrical surface at rD = reD and by the parallel planes at zD = 0 and hD, we should also impose the appropriate physical conditions at these boundaries.

We seek a point-source solution for a cylindrical reservoir in the following form:

....................(51)

In Eq. 51, is a solution of Eq. 48 that satisfies Eq. 50 and the boundary conditions at zD = 0 and hD. may be chosen as one of the point-source solutions in an infinite-slab reservoir given in Table 1, depending on the conditions imposed at the boundaries at zD = 0 and hD. If is chosen such that it satisfies the boundary conditions at zD = 0 and hD, its contribution to the flux vanishes at the source location, and + satisfies the appropriate boundary condition at rD = reD, then Eq. 51 should yield the point-source solution for a cylindrical reservoir with appropriate boundary conditions.

Consider the example of a closed cylindrical reservoir in which the boundary conditions are given by

....................(52)

and

....................(53)

According to the boundary condition given by Eq. 52, we should choose as the point-source solution given in Table 1 (or by Eq. 47). Then, with the addition theorem for the Bessel function K0(aRD) given by[13]

....................(54)

where

....................(55)

we can write

....................(56)

for rD < r′D. If rD > r′D, we interchange rD and r′D in Eq. 56. If we choose in Eq. 51 as

....................(57)

where ak and bk are constants, then satisfies the boundary condition given by Eq. 52, and the contribution of to the flux at the source location vanishes. If we also choose the constants ak and bk in Eq. 57 as

....................(58)

and

....................(59)

then satisfies the impermeable boundary condition at rD = reD given by Eq. 53. Thus, the point-source solution for a closed cylindrical reservoir is given by

....................(60)

This solution procedure may be extended to the cases in which the boundaries are at constant pressure or of mixed type.[6] Table 2 presents the point-source solutions for cylindrical reservoirs for all possible combinations of boundary conditions. Solutions for other source geometries in cylindrical reservoirs may be obtained by using the point-source solutions in Table 2 in Eq. 39 (or Eq. 40).

### Example 1 - Partially penetrating, uniform-flux fracture in an isotropic and closed cylindrical reservoir

Consider a partially penetrating, uniform-flux fracture of height hf and half-length xf in an isotropic and closed cylindrical reservoir. The center of the fracture is at r′ = 0, θ′ =0, z′ = zw, and the fracture tips extend from (r′ = xf, θ = α + π) to (r′ = xf, θ = α).

Solution. Fig. 1 shows the geometry of the fracture/reservoir system considered in this example. The solution for this problem is obtained by first generating a partially penetrating line source and then using this line-source solution to generate the plane source. The solution for a partially penetrating line source at r′D, θ′, zw is obtained by integrating the corresponding point-source solution given in Table 2 with respect to z′ from zwhf / 2 to zw + hf / 2 and is given by

....................(61)

To generate the solution for a partially penetrating plane source that represents the fracture, the partially penetrating line-source solution given in Eq. 61 is integrated with respect to r′ from 0 to xf with θ′ = α + π in the third quadrant and with θ′ = α in the first quadrant. This procedure yields

....................(62)

It is possible to obtain an alternate representation of the solution given in Eq. 62. With the addition theorem of the Bessel function K0(x) given by Eq. 54, the solution in Eq. 61 may be written as

....................(63)

where

....................(64)

and

....................(65)

The integration of the partially penetrating vertical well solution given in Eq. 63 with respect to r′ from 0 to xf (with θ′ = α + π in the third quadrant and with θ′ = α in the first quadrant) yields the following alternative form of the partially penetrating fracture solution:

....................(66)

where

....................(67)

### Example 2 - Uniform-flux, horizontal well in an isotropic and closed cylindrical reservoir

Consider a uniform-flux, horizontal line-source well of length Lh in an isotropic and closed cylindrical reservoir. The well extends from (r′ = Lh/2, θ = α + π) to (r′ = Lh/2, θ = α), and the center of the well is at r′ = 0, θ′ = 0, z′ = zw.

Solution. The solution for a horizontal line-source well in a closed cylindrical reservoir is obtained by integrating the corresponding point-source solution in Table 2 with respect to r′ from 0 to Lh / 2 with θ′ = α + π in the third quadrant and with θ′ = α in the first quadrant. The final form of the solution is given by

....................(68)

## Solutions for rectangular parallelepiped reservoir

Solutions for rectangular parallelepiped reservoirs may also be obtained by starting from the point-source solution in the Laplace transform domain in an infinite reservoir and using the method of images to generate the effects of the planar boundaries. Although the formal procedure to obtain the solution is fairly easy, the use of the method of images in three directions (x, y, z) yields triple infinite Fourier series, which may pose computational inconveniences. As an example, the solution for a continuous point source located at x′, y′, z′ in a rectangular porous medium occupying the region 0 < x < xe, 0 < y < ye, and 0 < z < h is obtained by applying the method of images to the point-source solution given by Eq. 36: [6][11]

....................(69)

where

....................(70)

and

....................(71)

....................(72)

....................(73)

Ozkan[11] shows that the triple infinite sums in Eq. 69 may be reduced to double infinite sums with

....................(74)

where

....................(75)

The resulting continuous point-source solution for a closed rectangular reservoir is given by

....................(76)

where

....................(77)

....................(78)

....................(79)

....................(80)

and

....................(81)

Following a procedure similar to the one explained here, it is possible to derive the point-source solutions in rectangular parallelepiped reservoirs for different combinations of boundary conditions.[11][12] Table 3 gives these solutions, which may be used to derive the solutions for the other source geometries with Eq. 39 (or Eq. 40). Examples 3.10 and 3.11 demonstrate the derivation of the solutions for the other source geometries in rectangular reservoirs.

### Example 3 - Fully penetrating vertical fracture in a closed rectangular reservoir

Consider a vertical fracture of half-length xf located at x′ = xw and y′ = yw in a closed rectangular reservoir.

Solution. Assuming uniform-flux distribution along the fracture surface, the solution for this problem is obtained by integrating the corresponding point-source solution in Table 3, first with respect to z′ from 0 to h and then with respect to x′ from xwxf to xw + xf. The result is

....................(82)

where , , and εk are given respectively by Eqs. 77, 78, and 80.

### Example 4 - Horizontal well in a closed rectangular reservoir

Consider a horizontal well of length Lh in the x-direction located at x′ = xw, y′ = yw, and z′ = zw in a closed rectangular reservoir.

Solution. The solution for a horizontal line-source well is obtained by integrating the corresponding point-source solution in Table 3, with respect to x′ from xwLh /2 to xw+Lh /2, and is given by

....................(83)

where

....................(84)

and

....................(85)

In Eq. 85, , , εn, εk, and εk, n are given by Eqs. 77 through 81.

## Conversion from 3D to 2D anisotropy

The solutions previously presented assume that the reservoir is anisotropic in all three principal directions, x, y, and z with kx, ky, and kz denoting the corresponding permeabilities. In these solutions, an equivalent isotropic permeability, k, has been defined by

....................(86)

For some applications, it may be more convenient to define an equivalent horizontal permeability by

....................(87)

and replace k in the solutions by kh. Note that k takes place in the definition of the dimensionless time tD (Eq. 12). Then, if we define a dimensionless time based on kh, the relation between and tD is given by

....................(88)

Because in the solutions given in this section the Laplace transformation is with respect to tD, conversion from 3D to 2D anisotropy requires the use of the following property of the Laplace transforms:

....................(89)

As an example, consider the solution for a horizontal well in an infinite-slab reservoir. Assuming that the midpoint of the well is the origin (xwD = 0, ywD = 0) and choosing the half-length of the horizontal well as the characteristic length (i.e., ℓ = Lh / 2), the horizontal-well solution given in Table 1 may be written as

....................(90)

In Eq. 90, s is the Laplace transform variable with respect to dimensionless time, tD, based on k and

....................(91)

....................(92)

....................(93)

and

....................(94)

If we define the following variables based on kh,

....................(95)

....................(96)

....................(97)

and also note that

....................(98)

then, we may rearrange Eq. 90 in terms of the dimensionless variables based on kh as

....................(99)

where

....................(100)

and

....................(101)

If we compare Eqs. 90 and 99, we can show that

....................(102)

where we have used the relation given by Eq. 90. If we now define as the Laplace transform variable with respect to , we may write

....................(103)

With the relation given by Eq. 103 and Eq. 90, we obtain the following horizontal-well solution in terms of dimensionless variables based on kh:

....................(104)

## Computational considerations and applications

The numerical evaluation of the solutions given previously may be sometimes difficult, inefficient, or even impossible. Alternative computational forms of some of these solutions have been presented in a few sources.[5][6][11] Here, we present a summary of the alternative formulas to be used in the computation of the source functions in the Laplace transform domain. Some of these formulas are for computations at early or late times and may be useful to derive asymptotic approximations of the solutions during the corresponding time periods.

As Laplace transformation for solving transient flow problems notes, the short- and long-time approximations of the solutions correspond to the limiting forms of the solution in the Laplace transform domain as s→∞ and s→0, respectively. In the solutions given in this section, we have defined u = sf(s). From elementary considerations, it is possible to show that the definition of f(s) given in Eq. 27 yields the following limiting forms:

....................(105)

and

....................(106)

These limiting forms are used in the derivation of the short- and long-time asymptotic approximations. In the following expressions, homogeneous reservoir solutions are obtained by substituting ω = 1.

### The integral I

....................(107)

This integral arises in the computation of many practical transient-pressure solutions and may not be numerically evaluated, especially as yD→0; however, the following alternate forms of the integral are numerically computable.[6]

....................(108)

....................(109)

and

....................(110)

The integrals in Eqs. 108 through 110 may be evaluated with the standard numerical integration algorithms for yD ≠ 0. For yD = 0, the polynomial approximations given by Luke[14] or the following power series expansion given by Abramowitz and Stegun[15] may be used in the computation of the integrals in Eqs. 108 through 110:

....................(111)

For numerical computations and asymptotic evaluations, it may also be useful to note the following relations: [6]

....................(112)

and

....................(113)

It can be shown from Eqs. 112 and 113 that, for practical purposes, when z ≥ 20, the right sides of Eqs. 111 and 112 may be approximated by π/2 and π exp (−|c|)/2, respectively.[6][9]

As a few sources[5][6][11] show, it is possible to derive the following short- and long-time approximations (i.e., the limiting forms as s→∞ and s→0, respectively) for the integral given, respectively, by

....................(114)

where

....................(115)

and

....................(116)

where γ=0.5772… and

....................(117)

It is also useful to note the real inversions of Eqs. 114 and 116 given, respectively, by

....................(118)

and

....................(119)

### The series S1

....................(120)

Two alternative expressions for the series S1 may be convenient for the large and small values of u (i.e., for short and long times).[11] When u is large,

....................(121)

and when u + a2 << n2π2/h2D,

....................(122)

### The series S2

....................(123)

Alternative computational forms for the series S2 are given next.[11] When u is large,

....................(124)

and when u + a2 << n2π2/h2D,

....................(125)

### The series S3

....................(126)

The following alternative forms for the series may be convenient for the large and small values of u (i.e., for short and long times).[11] When u is large,

....................(127)

and when u + a2 << (2n − 1)2 π2/(4h2D),

....................(128)

### The series F

....................(129)

where

....................(130)

The series may be written in the following forms with the use of Eqs. 108 through 110.

....................(131)

....................(132)

and

....................(133)

The computation of the series in Eqs. 131 and 132 should not pose numerical difficulties; however, the series in Eq. 133 converges slowly. With the relation given in Eq. 112, we may write Eq. 133 as[11]

....................(134)

where

....................(135)

Before discussing the computation of the series given in Eq. 135, we first discuss the derivation of the asymptotic approximations for the series . When s is large (small times), may be approximated by[11]

....................(136)

where β is given by Eq. 115. If s is sufficiently large, then Eq. 136 may be further approximated by

....................(137)

The inverse Laplace transform of Eq. 137 yields

....................(138)

For small s (large times), depending on the value of xD, may be approximated by one of the following equations: [11]

....................(139)

....................(140)

....................(141)

where is given by Eq. 148.

### The series F1

....................(142)

where

....................(143)

With the relations given in Eqs. 121 and 122, the following alternative forms for the series may be obtained, respectively, for the large and small values of s (i.e., for short and long times).[11] When u is large,

....................(144)

and when u << n2π2/h2D,

....................(145)

It is also possible to derive asymptotic approximations for the series . When s is large (small times), may be approximated by[11]

....................(146)

If s is sufficiently large, then Eq. 146 may be further approximated by

....................(146)

The inverse Laplace transform of Eq. 146 yields

....................(147)

For small s (large times), may be approximated by[11]

....................(148)

### The ratio R1

....................(149)

By elementary considerations, the ratio may be written as[11]

....................(150)

The expression given in Eq. 150 provides computational advantages when s is small (time is large).

### Example 5 - Fully penetrating uniform flux fracture in an infinite-slab reservoir with closed top and bottom boundaries

Consider a fully penetrating, uniform-flux fracture of half-length xf located at x′=0, y′=0 in an infinite-slab reservoir with closed top and bottom boundaries.

Solution. Table 1 gives the solution for this problem. For simplicity, assuming an isotropic reservoir, choosing the characteristic length as ℓ = xf and noting that , the solution becomes

....................(151)

First consider the numerical evaluation of Eq. 151. We note from Eqs. 108 through 110 that Eq. 151 may be written in one of the following forms, depending on the value of xD.

....................(152)

....................(153)

and

....................(154)

The numerical evaluation of the integrals in Eqs. 152 through 154 for yD ≠ 0 should be straightforward with the use of the standard numerical integration algorithms. For yD = 0, the polynomial approximations given by Luke[14] or the power series expansion given by Eq. 111 should be useful.

The short- and long-time asymptotic approximations of the fracture solution are also obtained by applying the relations given by Eqs. 114 and 116, respectively, to the right side of Eq. 151. This procedure yields, for short times,

....................(155)

or, in real-time domain,

....................(156)

where β is given by Eq. 115 with a = -1 and b = +1. At long times, the following asymptotic approximation may be used:

....................(157)

or, in real-time domain,

....................(158)

where γ = 0.5772… and σ(xD, yD, -1, +1) is given by Eq. 117.

### Example 6 - Horizontal well in an infinite-slab reservoir with closed top and bottom boundaries

Consider a horizontal well of length Lh located at x′ = 0, y′ = 0, and z′ = zw in an infinite-slab reservoir with closed top and bottom boundaries.

Solution. Table 1 gives the horizontal-well solution for an infinite-slab reservoir with impermeable boundaries. Assuming an isotropic reservoir, choosing the characteristic length as ℓ = Lh / 2 and noting that , the solution may be written as

....................(159)

where is the fracture solution given by the right side of Eq. 151 and is given by

....................(160)

with

....................(143)

....................(161)

and

....................(162)

The computation of the first term in the right side of Eq. 159 is the same as the computation of the fracture solution given by Eq. 151 and has been discussed in Example 5. The computational form of the second term in the right side of Eq. 159 is given by Eqs. 131 through 134. Of particular interest is the case for −1 ≤ xD ≤ +1. In this case, from Eqs. 134 and 135, we have

....................(163)

where

....................(164)

The computational considerations for the series have been discussed previously.

Next, we consider the short- and long-time approximations of the horizontal-well solution given by Eq. 159. To obtain a short-time approximation, we substitute the asymptotic expressions for and as s→∞ given, respectively, by Eqs. 155 and 137. This yields

....................(165)

where β is given by Eq. 115. The inverse Laplace transform of Eq. 165 is given by

....................(166)

To obtain the long-time approximation of Eq. 159, we substitute the asymptotic expressions for and as s→∞ given, respectively, by Eq. 158 and Eqs. 139 through 141. Of particular interest is the case for −1 ≤ xD ≤ +1, where we have

....................(167)

where γ=0.5772… and σ(xD, yD, -1, +1) is given by Eq. 117. The inverse Laplace transform of Eq. 167 yields

....................(168)

### Example 7 - Fully penetrating, uniform-flux fracture in an isotropic and closed cylindrical reservoir

Consider a fully penetrating, uniform-flux fracture of half-length xf in an isotropic and closed cylindrical reservoir. The center of the fracture is at r′ = 0, θ′ = 0 and the fracture tips extend from (r′ = xf, θ = α + π) to (r′ = xf, θ = α).

Solution. The solution for this problem has been obtained in Eq. 62 in Example 1 with hw = h. Choosing the characteristic length as ℓ = xf and noting that , the solution is given by

....................(169)

For the computation of the pressure responses at the center of the fracture (rD = 0), Eq. 169 simplifies to

....................(170)

It is also possible to find a very good approximation for Eq. 169, especially when reD is large. If we assume[6]

....................(171)

and use the following relation[16]

....................(172)

then Eq. 169 may be written as

....................(173)

Because[6]

....................(174)

where

....................(175)

Eq. 173 may also be written as

....................(176)

Although the assumption given in Eq. 171 may not be justified by itself, the solution given in Eq. 176 is a very good approximation for Eq. 169, especially when reD is large. For a fracture at the center of the cylindrical drainage region, Eq. 176 simplifies to

....................(177)

It is also possible to obtain short- and long-time approximations for the solution given in Eq. 177. For short times, u→∞ and the second term in the argument of the integral in Eq. 177 becomes negligible compared with the first term. Then, Eq. 177 reduces to the solution for an infinite-slab reservoir given by Eq. 151, of which the short-time approximation has been discussed in Example 5.

To obtain a long-time approximation, we evaluate Eq. 177 at the limit as s→0 (us). As shown in modified bessel functions, for small arguments we may approximate the Bessel functions in Eq. 177 by

....................(178)

....................(179)

....................(180)

and

....................(181)

where γ = 0.5772…. With Eqs. 178 through 181 and by neglecting the terms of the order s3/2, we may write[11]

....................(3.398)

If we substitute the right side of Eq. 182 into Eq. 177, we obtain

....................(183)

where σ(xD, yD, −1, +1) is given by Eq. 117 and

....................(184)

The inverse Laplace transform of Eq. 183 yields the following long-time approximation for a uniform-flux fracture at the center of a closed square:

....................(185)

### Example 8 - Fully penetrating uniform-flux fracture in an isotropic and closed parallelepiped reservoir

Consider a fully penetrating, uniform-flux fracture of half-length xf in an isotropic and closed parallelepiped reservoir of dimensions xe × ye × h. The fracture is parallel to the x axis and centered at xw, yw, zw.

Solution. The solution for this problem has been obtained in Example 3 and, by choosing ℓ = xf, is given by

....................(186)

where

....................(187)

The computation of the ratios of the hyperbolic functions in Eq. 186 may be difficult, especially when their arguments approach zero or infinity. When s is small (long times), Eq. 150 should be useful to compute the ratios of the hyperbolic functions. When s is large (small times), with Eq. 150 the solution given in Eq. 186 may be written as[11]

....................(188)

where

....................(189)

....................(190)

and

....................(191)

The last equality in Eq. 189 follows from the relation given by Eq. 133. The expression given in Eq. 189 may also be written as

....................(192)

where

....................(193)

and

....................(194)

Therefore, the solution given by Eq. 186 may be written as follows for computation at early times (for large values of s):

....................(195)

where is given by Eq. 193 and corresponds to the solution for a fractured well in an infinite-slab reservoir (see Eq. 151 in Example 5) and represents the contribution of the lateral boundaries and is given by

....................(196)

In Eq. 196, , , and are given, respectively, by Eqs. 190, 191, and 194. The integrals appearing in Eqs. 193 and 194 may be evaluated by following the lines outlined by Eqs. 108 through 110.

It is also possible to derive short- and long-time approximations for the fracture solution in a closed rectangular parallelepiped. The short-time approximation corresponds to the limit of the solution as s→∞. It can be easily shown that the term in Eq. 195 becomes negligible compared with the term for which a short-time approximation has been obtained in Example 5 (see Eqs. 155 and 156).

To obtain a long-time approximation (small values of s), the solution given in Eq. 186 may be written as[9]

....................(197)

where

....................(198)

and

....................(199)

The second equality in Eq. 198 results from[17]

....................(200)

For small values of s, replacing u by s and s + α by α, and with[17]

....................(201)

the term H given by Eq. 198 may be approximated by

....................(202)

The long-time approximation of the second term in Eq. 197 is obtained by assuming u << k2π2/x2eD and taking the inverse Laplace transform of the resulting expressions; therefore, we can obtain the following long-time approximation

....................(203)

### Example 9 - Uniform-flux horizontal well in an isotropic and closed parallelepiped reservoir

Consider a uniform-flux horizontal well of length Lh in an isotropic and closed parallelepiped reservoir of dimensions xe × ye × h. The center of the well is at xw, yw, zw, and the well is parallel to the x axis.

Solution. The solution for this problem was obtained in Example 4 and, by choosing ℓ = Lh / 2, is given by

....................(204)

where is the solution discussed in Example 8, and is given by

....................(205)

In Eq. 205, and are given by Eqs. 161 and 162, respectively,

....................(130)

and

....................(206)

The computation and the asymptotic approximations of the term have been discussed in Example 8. To compute the term for long times (small s), the relation for the ratios of the hyperbolic functions given by Eq. 150 should be useful. For computations at short times (large values of s), following the lines similar to those in Example 8, the term in Eq. 205 may be written as

....................(207)

where

....................(208)

....................(209)

....................(210)

....................(211)

and

....................(212)

The computational form of the term in Eq. 208 is obtained by applying the relations given by Eqs. 131 through 134 and Eq. 112. Of particular interest is the case for −1 ≤ xD ≤ +1 and yD = ywD given by

....................(213)

where

....................(214)

which can be written as follows by using the relation given in Eq. 121:

....................(215)

Similarly, for −1 ≤ xD ≤ +1 and yD = ywD, the term given in Eq. 212 may be written as

....................(216)

where

....................(217)

### Dimensionless fracture pressure

Example 8 discussed the short- and long-time approximations of the term in Eq. 204. A small-time approximation for given by Eq. 207 is obtained with u = ωs and by noting that as s→∞, . Then, substituting the short-time approximations for and given, respectively, by Eqs. 155 and 137 into Eq. 204, the following short-time approximation is obtained: [9]

....................(218)

where β is given by Eq. 115. The inverse Laplace transform of Eq. 218 yields

....................(219)

The long-time approximation of Eq. 204 is obtained by substituting the long-time approximations of and . The long time-approximation of is obtained in Example 8 (see Eq. 197 through 203). The long-time approximation of is obtained by evaluating the right side of Eq. 205 as s→0 (u→0) and is given by

....................(220)

where

....................(221)

and

....................(222)

Thus, the long-time approximation Eq. 204 is given by

....................(223)

where pDf and F1 are given, respectively, by Eqs. 203 and 220. For computational purposes, however, F1 may be replaced by

....................(224)

In Eq. 224, F, Fb1, Fb2, and Fb3 are given, respectively, by

....................(225)

....................(226)

....................(227)

and

....................(228)

When computing the integrals and the trigonometric series, the relations given by 108 through 110 and 129 through 134 are useful.

## Nomenclature

 a = radius of the spherical source, L B = formation volume factor, res cm3/std cm3 c = fluid compressibility, atm−1 cf = formation compressibility, atm−1 ct = total compressibility, atm−1 C = wellbore-storage coefficient, cm3/atm d = distance to a linear boundary, cm D = domain Ei(x) = exponential integral function f(s) = naturally fractured reservoir function = naturally fractured reservoir function based on = Laplace transform of a function f (t) G = Green’s function h = formation thickness, cm hf = fracture height (vertical penetration), cm hp = slab thickness, cm hw = well length (penetration), cm H(x - x′) = Heaviside’s unit step function = unit normal vector in the ξ direction, ξ = x, y, z, r, θ Iv(x) = modified Bessel function of the first kind of order v I′v(x) = derivative of Iv(x) Jv(x) = Bessel function of the first kind of order v k = isotropic permeability, md kf = fracture permeability, md kh = equivalent horizontal permeability, md ki j = permeability in i-direction as a result of pressure gradient in j-direction, md kξ = permeability in ξ-direction, ξ = x, y, z, md kξf = fracture permeability in ξ-direction, ξ = x, y, z, md Ki1(x) = first integral of K0(z) Kn(x) = modified Bessel function of the second kind of order n K′n(x) = derivative of Kn(x) ℓ = characteristic length of the system, cm L = Laplace transform operator L-1 = inverse Laplace transform operator Lh = horizontal-well length, cm m = pseudopressure, atm2/cp Mg = mass, g M = point in space M′ = source point in space Mw = point in Γw M′w = source point in Γw n = outward normal direction of the boundary surface = normal vector N = even integer in Stehfest’s algorithm p = pressure, atm pc = pressure for constant production rate, qc, atm = dimensionless fracture pressure pe = external boundary pressure, atm p f = fracture pressure, atm pf i = initial pressure in fracture system, atm pi = initial pressure, atm pj = pressure in medium j, j=m, f, atm pm = matrix pressure, atm pmi = initial pressure in matrix system, atm pw f = flowing wellbore pressure, atm = Laplace transform of p(t) p(t) = inverse of the Laplace domain function pa(T) = approximate inverse of at t=T, atm q = production rate, cm3/s = instantaneous production rate for a point source, cm3/s qc = constant production rate, cm3/s qs f = sandface production rate, cm3/s qwb = wellbore production rate as a result of storage, cm3/s r = radial coordinate and distance, cm r′ = source coordinate in r-direction, cm re = external radius of the reservoir, cm rw = wellbore radius, cm R = distance in 3D coordinates, cm RD = dimensionless radial distance in cylindrical coordinates s = Laplace transform parameter = Laplace transform paraeter based on sm = skin factor S = source function t = time, s = dimensionless time based on kh tAD = dimensionless time based on area tp = producing time, s T = Temperature, °C u = s f(s) = velocity vector vξ = velocity component in the ξ direction, ξ = x, y, z, r, θ, cm/s V = volume, cm3 Vi = constant in Stehfest’s algorithm Vf = fraction of the volume occupied by fractures Vm = fraction of the volume occupied by matrix x = distance in x-direction, cm x′ = source coordinate in x-direction, cm xe = distance to the external boundary in x-direction, cm xp = half slab thickness, cm xf = fracture half-length, cm = dimensionless fracture half-length xw = well coordinate in x-direction, cm y = distance in y-direction, cm y′ = source coordinate in y-direction, cm ye = distance to the external boundary in y-direction, cm yw = well coordinate in y-direction, cm Yn(x) = Bessel function of the second kind of order n z = distance in z-direction, cm z′ = source coordinate in z-direction, cm = dimensionless distance in z-direction, Eq. 161 zw = well coordinate in z-direction, cm = dimensionless well coordinate in z-direction, Eq. 162 Z = compressibility factor Γ = boundary surface, cm2 Γe = external boundary surface Γw = length, surface, or volume of the source Γ(x) = Gamma function γ = Euler’s constant (γ = 0.5772...) γ f = fundamental solution of diffusion equation Δ = difference operator δ(x) = Dirac delta function η = diffusivity constant ηi = diffusivity constant in i direction, i = x, y, z, or r θ = angle from positive x-direction, degrees radian θ′ = source coordinate in θ-direction, degrees radian λ = transfer coefficient for a naturally fractured reservoir = λ based on kh μ = viscosity, cp ρ = density, g/cm3 τ = time, s Φ = porosity, fraction φ(M) = any continuous function ω = storativity ratio for a naturally fractured reservoir

## References

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