Basics of Oil and Gas Flow in Reservoirs

Abstract

This chapter deals with basic concepts of oil and gas flow in porous media. In the first section (oil flow), the basics of oil flow in porous media and different flow equations, including viscous, Darcy, and Brinkman flow are introduced. Then, based on a new work the boundary between these kinds of flow are reported. Then, the well-known diffusivity equation, combination of continuity equation and Darcy velocity, is introduced. The relative permeability as an important concept in two-phase flow through porous media is described as well. After that, the forces affecting oil flow in porous media are introduced as well as different dimensionless numbers which reflect the relative importance of forces during oil flow. The gas flow through porous reservoir can occur as single or multiphase. A second phase releases as a result of pressure decline in reservoir and starts flowing when its saturation exceeds a minimum value, known as critical saturation. At these conditions, the relative permeability concept applies to take into account the multiphase flow. Also, the classic Darcy’s law equation for flow in porous media may fail in certain flowing conditions, and the non-Darcy flow equations need to be studied and applied for proper modelling of gas flow in porous medium. In addition, the role of wettability alteration on production enhancement of gas condensate reservoirs is a novel approach in gas reservoir performance. Also, the active forces during CO₂ storage and the challenges of gas flow in unconventional gas reservoirs are two important subjects. These subjects will be covered in the second section (gas flow) of this chapter.

Appendices

Appendix 3.1: The Viscous Flow Equation in Cylindrical Coordinates

The viscous equation is obtained in cylindrical coordinates according to Eq. (3.60) where $B = \frac{1}{\mu }\frac{dP}{{dz}}$.

$$\frac{{d^{2} v}}{{dr^{2} }} + { }\frac{1}{r}{ }\frac{dv}{{dr^{ } }} = B$$

(3.60)

This equation is an inhomogeneous ordinary differential equation because of its non-zero right-hand side. To solve this equation, first, the homogeneous part (B = 0) is solved and the general solution (v_g) is obtained, then the inhomogeneous part (B ≠ 0) is calculated, and the particular solution (v_p) is achieved, and finally, the total answer is resulted as in Eq. (3.61):

$$v_{\left( r \right)} = v_{g} + { }v_{p}$$

(3.61)

To solve the homogeneous part, the right-hand side of Eq. (3.60) is considered equal to zero and the general solution (v_g) is calculated. Therefore, Eq. (3.62) is obtained as follows:

$$\frac{{d^{2} v_{g} }}{{dr^{2} }} + { }\frac{1}{r}{ }\frac{{dv_{g} }}{{dr^{ } }} = 0{ }$$

(3.62)

Using the variation of variables in Eq. (3.63), Eq. (3.62) rewritten as Eq. (3.64):

$$u = { }\frac{{dv_{g} }}{dr}{ }.{ }\frac{du}{{dr}} = { }\frac{{d^{2} v_{g} }}{{dr^{2} }}{ }$$

(3.63)

$$\frac{du}{{dr}} + { }\frac{1}{r}{ }u = 0$$

(3.64)

By arranging and integrating from Eq. (3.64), Eq. (3.65) is obtained as follows:

$$u = { }\frac{{C_{1} }}{r}$$

(3.65)

where C₁ is a constant. By combining Eqs. (3.63) and (3.65) and reintegration, Eq. (3.66) is achieved:

$$v_{g} = { }C_{1} \ln \left( r \right) + { }C_{2}$$

(3.66)

where C₂ is constant. Based on the point that the left-hand side of Eq. (3.60) is a second-order differential equation, the particular solution of equation (v_p) is a relationship in the order of two. Therefore, the particular solution of the equation and its derivatives are obtained as Eq. (3.67):

$$v_{p} = ar^{2} + br + c \cdot \frac{{dv_{p} }}{{dr^{ } }} = 2ar + b \cdot \frac{{d^{2} v_{p} }}{{dr^{2} }} = 2a$$

(3.67)

where a, b, and c are constant. Using Eq. (3.67), Eq. (3.60) can be rewritten as Eq. (3.68):

$$\left( {2a} \right) + { }\frac{1}{r}\left( {2ar + b} \right) - B = 2a + 2a + { }\frac{b}{r} - B = 0$$

(3.68)

Assuming b = 0 in Eq. (3.68), the relationship of constants B and a results as Eq. (3.69):

$$a = { }\frac{B}{4}$$

(3.69)

Therefore, assuming c = 0 in Eq. (3.67), the particular solution of the differential equation is obtained as Eq. (3.70):

$$v_{p} = { }\frac{B}{4}r^{2}$$

(3.70)

The overall solution of the differential equation is expressed as Eq. (3.71) using Eqs. (3.61), (3.66) and (3.70):

$$v = { }\frac{B}{4}r^{2} + { }C_{1} \ln \left( r \right) + { }C_{2}$$

(3.71)

Boundary conditions are employed to calculate the constants C₁ and C₂. Applying the boundary conditions of Eqs. (3.72) and (3.73) in Eq. (3.71), these constants are calculated according to Eq. (3.74):

$$r = \pm R,\quad v = 0$$

(3.72)

$$r = 0,\quad \frac{dv}{{dr}} = 0$$

(3.73)

$$C_{1} = 0{ }.C_{2} = { } - \frac{B}{4}R^{2}$$

(3.74)

Therefore, by substitution of Eq. (3.74) and B value in Eq. (3.71), the viscous flow velocity profile equation in a cylinder can be obtained as Eq. (3.2).

$$v = \frac{1}{4\mu }{ }\frac{dp}{{dz}}\left( {R^{2} - { }r^{2} } \right)$$

To calculate the average value of a function in cylindrical coordinates, Eq. (3.75) is used.

$$\overline{v} = \frac{{\int_{0}^{R} {vrdr} }}{{\int_{0}^{R} {rdr} }}$$

(3.75)

Therefore, by substituting Eq. (3.2) in Eq. (3.75), the value of the average velocity can be calculated according to Eq. (3.76).

$$\overline{v} = \frac{1}{4\mu }{ }\frac{dp}{{dz}}\left( {\frac{{\int_{0}^{R} {r\left( {R^{2} - { }r^{2} } \right)} dr}}{{\int_{0}^{R} {rdr} }}} \right) = \frac{1}{4\mu }{ }\frac{dp}{{dz}}\left( {\frac{{\frac{{{\text{R}}^{4} }}{2} - \frac{{R^{4} }}{4}}}{{\frac{{R^{2} }}{2}}}} \right)$$

(3.76)

By simplifying Eq. (3.76), the viscous flow average velocity equation in a cylinder is obtained as Eq. (3.3).

$$\overline{v} = { }0.125\frac{{R^{2} }}{\mu }\frac{dp}{{dz}}$$

Appendix 3.2: The Brinkman Equation in Cylindrical Coordinates

Brinkman equation in cylindrical coordinates is known as Eq. (3.77):

$$r^{2} \frac{{d^{2} v}}{{dr^{2} }} + r\frac{dv}{{dr}} - { }Ar^{2} v = r^{2} B$$

(3.77)

In Eq. (3.77), parameter B is equal to the constant value of $\frac{1}{\mu }\frac{dP}{{dz}}$, and parameter A is equal to K⁻¹. This equation is an ordinary inhomogeneous differential equation. To solve this equation, both the general (v_g) and the particular (v_p) solutions which respectively indicate the homogeneous (r²B = 0) and inhomogeneous part (r²B ≠ 0) are calculated. According to Eq. (3.61), the sum of the particular and general solutions is equal to the total solution of the equation. To calculate v_g, first Eq. (3.77) is homogenized as Eq. (3.78):

$$r^{2} \frac{{d^{2} v}}{{dr^{2} }} + r\frac{dv}{{dr}} - { }Ar^{2} v = 0$$

(3.78)

The answer of Eq. (3.78) with its similarity to the modified Basel differential equation (reference) is obtained as Eq. (3.79):

$$v_{g} = C_{1} I_{0} \left( {r\sqrt A } \right) + C_{2} K_{0} \left( {r\sqrt A } \right)$$

(3.79)

where I₁ and K₁ are the Modified Bessel functions of the first and second kind in the zero-order. According to the point that the left-hand side of Eq. (3.77) is a second-order differential equation, the particular solution of equation (v_p) is in the same order. In this case, the particular solution and its derivatives can be assumed as Eq. (3.80):

$$v_{p} = ar^{2} + br + c \cdot \frac{{dv_{p} }}{dr} = 2ar + b \cdot \frac{{d^{2} v_{p} }}{{dr^{2} }} = 2a$$

(3.80)

where a, b, and c are constant. Therefore, Eq. (3.77) can be rewritten as Eq. (3.81) using Eq. (3.80).

$$r^{2} \left( {2a} \right) + r\left( {2ar + b} \right) - { }Ar^{2} \left( {ar^{2} + br + c} \right) = Br^{2}$$

(3.81)

According to Eq. (3.79), it can be seen that for the validity of this equation, the values a and b must be equal to zero (a = b = 0) and the value of c must be equal to –B/A (c = −B/A). Therefore, by placing these constants in Eq. (3.80), the particular solution can be calculated as Eq. (3.82).

$$v_{g} = - \frac{B}{A}$$

(3.82)

Finally, providing particular and general solutions and also considering Eq. (3.61), the velocity function is calculated as Eq. (3.83).

$$v_{\left( r \right)} = C_{1} I_{0} \left( {r\sqrt A } \right) + C_{2} K_{0} \left( {r\sqrt A } \right) - \frac{B}{A}$$

(3.83)

Boundary conditions are employed to calculate the constants C₁ and C₂ in Eq. (3.81). By applying the boundary conditions of Eq. (3.73) in Eq. (3.83), Eq. (3.84) can be written as follows:

$$C_{1} \sqrt A I_{1} \left( {0*\sqrt A } \right) - C_{2} \sqrt A K_{1} \left( {0*\sqrt A } \right) = 0$$

(3.84)

According to the Bessel function properties, the values I₁(0) and K₁(0) are equal to zero and infinity, respectively. Therefore, for the validity of Eq. (3.82), the value of C₂ must be equal to zero (C₂ = 0). Also, by applying the boundary conditions of Eq. (3.72) and the value of C₂ in Eq. (3.83), Eq. (3.85) is achieved. By solving Eq. (3.85), the C₁ constant results in Eq. (3.86).

$$C_{1} I_{0} \left( {R\sqrt A } \right) - \frac{B}{A} = 0$$

(3.85)

$$C_{1} = \frac{B}{{AI_{0} \left( {R\sqrt A } \right)}}$$

(3.86)

Finally, by placing the constants C₁ and C₂ as well as the values A and B in Eq. (3.83), the velocity function is calculated as Eq. (3.87).

$$v_{\left( r \right)} = \frac{K}{\mu }\frac{dP}{{dz}}\left( {1 - \frac{{I_{0} \left( {\frac{r}{\sqrt K }} \right)}}{{I_{0} \left( {\frac{R}{\sqrt K }} \right)}}{ }} \right)$$

(3.87)

To calculate the average value of a function in a cylindrical system Eq. (3.75) is used. Therefore, by placing Eq. (3.140) in Eq. (3.75), Eq. (3.88) is obtained as follows:

$$\overline{v} = \frac{K}{\mu }\frac{dP}{{dz}}\left( {\frac{{\int_{0}^{R} {rdr} }}{{\int_{0}^{R} {rdr} }} - \frac{{\int_{0}^{R} {rI_{0} \left( {\frac{r}{\sqrt K }} \right)} }}{{\int_{0}^{R} {rdr} I_{0} \left( {\frac{R}{\sqrt K }} \right)}}} \right) = \frac{K}{\mu }\frac{dP}{{dz}}\left( {1 - \frac{{\left[ {r\sqrt K I_{1} \left( {\frac{r}{\sqrt K }} \right)} \right]_{0}^{R} }}{{\left[ {\frac{{r^{2} }}{2}} \right]_{0}^{R} I_{0} \left( {\frac{R}{\sqrt K }} \right)}}} \right)$$

(3.88)

where I₁ is the modified Bessel function of the first kind in the order of one. By solving and simplification of Eq. (3.88), the average velocity function for the Brinkman flow in a cylindrical system is calculated as Eq. (3.89).

$$\overline{v} = { }\frac{dP}{{dz}}\frac{K}{\mu }{ }\left( {1 - \frac{{2\sqrt K I_{1} \left( {\frac{R}{\sqrt K }} \right)}}{{RI_{0} \left( {\frac{R}{\sqrt K }} \right)}}} \right)$$

(3.89)

Appendix 3.3: Linear Diffusivity Equation Solution

This solution is provided by John Lee for radial flow solution in an infinite-acting homogenous reservoir [178]. The basic partial differential equation is given in dimensionless format

$$\frac{{\partial^{2} p_{D} }}{{\partial r_{D}^{2} }} + \frac{1}{{r_{D} }}\frac{{\partial p_{D} }}{{\partial r_{D} }} = \frac{{\partial p_{D} }}{{\partial t_{D} }}$$

(3.90)

where

$$r_{D} = \frac{r}{{r_{w} }}$$

(3.91)

$$p_{D} = p_{DC} \frac{kh}{{qB_{\mu } }}\left( {p_{i} - p_{r} } \right)$$

(3.92)

$$t_{D} = t_{DC} \frac{k}{{Q\mu C_{t} r_{w}^{2} }}t$$

(3.93)

where $t_{DC}$ and $p_{DC}$ are given by

The “initial” condition is given as

$$p_{D} \left( {r_{D} \cdot t_{D} = 0} \right) = 0\quad ({\text{uniform}}\,{\text{pressure}}\,{\text{distribution}})$$

(3.94)

The constant rate inner boundary condition is

$$\left[ {r_{D} \frac{{\partial p_{D} }}{{\partial r_{D} }}} \right]_{{r_{D} = 1}} = - 1\quad ({\text{constant}}\,{\text{flow}}\,{\text{rate}}\,{\text{at}}\,{\text{the}}\,{\text{well}})$$

(3.95)

The “infinite-acting” outer boundary condition is given by

$$p_{D} \left( {r_{D} \to \infty \cdot t_{D} } \right) = 0$$

(3.96)

Rewriting Eq. (3.90):

$$\frac{1}{{r_{D} }}\frac{\partial }{{\partial r_{D} }}\left[ {r_{D} \frac{{\partial p_{D} }}{{\partial r_{D} }}} \right] = \frac{{\partial p_{D} }}{{\partial t_{D} }}$$

(3.97)

The Boltzmann transform variable, $\varepsilon_{D}$, is defined as

$$t_{D} = ar_{D}^{b} t_{D}^{c}$$

(3.98)

where in this problem

a = $1/4$, b = 2, c = −1.

Which yields

$$\varepsilon_{D} = \frac{{r_{D}^{2} }}{{4t_{D} }}$$

(3.99)

Expanding $\frac{{\partial^{2} p_{D} }}{{\partial r_{D}^{2} }}$

$$\frac{\partial }{{\partial r_{D} }}\left[ {\frac{{\partial p_{D} }}{{\partial r_{D} }}} \right] + \frac{1}{{r_{D} }}\frac{{\partial p_{D} }}{{\partial r_{D} }} = \frac{{\partial p_{D} }}{{\partial t_{D} }}$$

(3.100)

Applying the chain rule

$$\frac{{\partial p_{D} }}{\partial x} = \frac{{\partial t_{D} }}{\partial x}\frac{{\partial p_{D} }}{{\partial t_{D} }}$$

(3.101)

Which combined with Eq. (3.100) gives

$$\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }}\frac{\partial }{{\partial \varepsilon_{D} }}\left[ {\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }}\frac{{\partial p_{D} }}{{\partial \varepsilon_{D} }}} \right] + \frac{1}{{r_{D} }}\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }}\frac{{\partial p_{D} }}{{\partial \varepsilon_{D} }} = \frac{{\partial \varepsilon_{D} }}{{\partial t_{D} }}\frac{{\partial p_{D} }}{{\partial \varepsilon_{D} }}$$

(3.102)

Expanding

$$\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }}\left[ {\frac{\partial }{{\partial \varepsilon_{D} }}\left( {\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }}} \right)\frac{{\partial p_{D} }}{{\partial \varepsilon_{D} }} + \frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }}\frac{{\partial^{2} p_{D} }}{{\partial \varepsilon_{D}^{2} }}} \right] + \left[ {\frac{1}{{r_{D} }}\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }} - \frac{{\partial \varepsilon_{D} }}{{\partial t_{D} }}} \right]\frac{{\partial p_{D} }}{{\partial t_{D} }} = 0$$

(3.103)

Isolating terms

$$\left[ {\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }}} \right]^{2} \frac{{\partial^{2} p_{D} }}{{\partial \varepsilon_{D}^{2} }} + \left[ {\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }}\frac{\partial }{{\partial \varepsilon_{D} }}\left( {\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }}} \right) + \frac{1}{{r_{D} }}\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }} - \frac{{\partial \varepsilon_{D} }}{{\partial t_{D} }}} \right]\frac{{\partial p_{D} }}{{\partial t_{D} }} = 0$$

(3.104)

Dividing through by $\left( {\partial \varepsilon_{D} /\partial r_{D} } \right)^{2}$ gives

$$\frac{{\partial^{2} p_{D} }}{{\partial \varepsilon_{D}^{2} }} + \frac{1}{{\left( {\partial \varepsilon_{D} /\partial r_{D} } \right)^{2} }}\left[ {\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }}\frac{\partial }{{\partial \varepsilon_{D} }}\left( {\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }}} \right) + \frac{1}{{r_{D} }}\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }} - \frac{{\partial \varepsilon_{D} }}{{\partial t_{D} }}} \right]\frac{{\partial p_{D} }}{{\partial t_{D} }} = 0$$

(3.105)

Reducing the $\partial ()/\partial \varepsilon_{D}$ term we have

$$\frac{{\partial^{2} p_{D} }}{{\partial \varepsilon_{D}^{2} }} + \frac{1}{{\left( {\partial \varepsilon_{D} /\partial r_{D} } \right)^{2} }}\left[ {\frac{\partial }{{\partial r_{D} }}\left( {\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }}} \right) + \frac{1}{{r_{D} }}\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }} - \frac{{\partial \varepsilon_{D} }}{{\partial t_{D} }}} \right]\frac{{\partial p_{D} }}{{\partial t_{D} }} = 0$$

(3.106)

Which can be further reduced to

$$\frac{{\partial^{2} p_{D} }}{{\partial \varepsilon_{D}^{2} }} + \frac{1}{{\left( {\partial \varepsilon_{D} /\partial r_{D} } \right)^{2} }}\left[ {\frac{{\partial^{2} \varepsilon_{D} }}{{\partial r_{D}^{2} }} + \frac{1}{{r_{D} }}\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }} - \frac{{\partial \varepsilon_{D} }}{{\partial t_{D} }}} \right]\frac{{\partial p_{D} }}{{\partial t_{D} }} = 0$$

(3.107)

Completing the factorization of the $\left( {\partial \varepsilon_{D} /\partial r_{D} } \right)^{2}$ gives

$$\frac{{\partial^{2} p_{D} }}{{\partial \varepsilon_{D}^{2} }} + \left[ {\frac{1}{{\left( {\partial \varepsilon_{D} /\partial r_{D} } \right)^{2} }}\frac{{\partial^{2} \varepsilon_{D} }}{{\partial r_{D}^{2} }} + \frac{1}{{r_{D} }}\frac{1}{{\partial \varepsilon_{D} /\partial r_{D} }} - \frac{1}{{\left( {\partial \varepsilon_{D} /\partial r_{D} } \right)^{2} }}\frac{{\partial \varepsilon_{D} }}{{\partial t_{D} }}} \right]\frac{{\partial p_{D} }}{{\partial t_{D} }} = 0$$

(3.108)

Using Eq. (3.99) the following derivation is taken

$$\frac{{\partial \varepsilon_{D} }}{{\partial t_{D} }} = \frac{\partial }{{\partial t_{D} }}\left( {\frac{{r_{D}^{2} }}{{4t_{D} }}} \right) = \frac{{r_{D}^{2} }}{4}\frac{\partial }{{\partial t_{D} }}\left( {\frac{1}{{t_{D} }}} \right) = \frac{ - 1}{{t_{D} }}\frac{{r_{D}^{2} }}{{4t_{D} }} = \frac{ - 1}{{t_{D} }}\varepsilon_{D}$$

(3.109)

$$\frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }} = \frac{\partial }{{\partial r_{D} }}\left( {\frac{{r_{D}^{2} }}{{4t_{D} }}} \right) = \frac{2}{{4t_{D} }}r_{D} = \frac{2}{{r_{D} }}\frac{{r_{D}^{2} }}{{4t_{D} }} = \frac{2}{{r_{D} }}t_{D}$$

(3.110)

$$\frac{{\partial^{2} \varepsilon_{D} }}{{\partial r_{D}^{2} }} = \frac{\partial }{{\partial r_{D} }}\left[ {\frac{\partial }{{\partial r_{D} }}\left( {\frac{{r_{D}^{2} }}{{4t_{D} }}} \right)} \right] = \frac{\partial }{{\partial r_{D} }}\left( {\frac{2}{{4t_{D} }}r_{D} } \right) = \frac{2}{{4t_{D} }} = \frac{2}{{r_{D}^{2} }}\frac{{r_{D}^{2} }}{{4t_{D} }} = \frac{2}{{r_{D}^{2} }}t_{D}$$

(3.111)

Substituting Eqs. (3.109)–(3.111) into Eq. (3.108) gives

$$\frac{{\partial^{2} p_{D} }}{{\partial \varepsilon_{D}^{2} }} + \left[ {\frac{1}{{\left( {2\varepsilon_{D} /r_{D} } \right)^{2} }}\frac{2}{{r_{D}^{2} }}\varepsilon_{D} + \frac{1}{{r_{D} }}\frac{1}{{(2\varepsilon_{D} /r_{D} )}} - \frac{1}{{\left( {\partial \varepsilon_{D} /\partial r_{D} } \right)^{2} }}\left( {\frac{ - 1}{{t_{D} }}\varepsilon_{D} } \right)} \right]\frac{{\partial p_{D} }}{{\partial \varepsilon_{D} }} = 0$$

(3.112)

Reducing

$$\frac{{\partial^{2} p_{D} }}{{\partial \varepsilon_{D}^{2} }} + \left[ {\frac{1}{{2\varepsilon_{D} }} + \frac{1}{{2\varepsilon_{D} }} + 1} \right]\frac{{\partial p_{D} }}{{\partial \varepsilon_{D} }} = 0$$

(3.113)

Finally

$$\frac{{\partial^{2} p_{D} }}{{\partial \varepsilon_{D}^{2} }} + \left[ {1 + \frac{1}{{\varepsilon_{D} }}} \right]\frac{{\partial p_{D} }}{{\partial \varepsilon_{D} }} = 0$$

(3.114)

where Eq. (3.114) is the “Boltzmann” transformed differential equation.

Initial and boundary condition in terms of Boltzmann transform are

$$p_{D} \left( {r_{D} \cdot t_{D} = 0} \right) = 0$$

(3.115)

where for $t_{D} \to 0; \varepsilon_{D} \to \infty$, which gives

$$p_{D} \left( { \varepsilon_{D} \to \infty } \right) = 0$$

(3.116)

The outer boundary condition, Eq. (3.94),

$$p_{D} \left( { r_{D} \to \infty .t_{D} } \right) = 0$$

(3.117)

Or as $r_{D} \to \infty$; $\varepsilon_{D} \to \infty$ which yields

$$p_{D} \left( { \varepsilon_{D} \to \infty } \right) = 0$$

(3.117)

where Eqs. (3.116) and (3.118) are the same which illustrates that the Boltzmann transform “collapses” 2 conditions into 1. Combining this observation with the inner boundary condition, we have 2 “boundary” conditions. Coupling this observation with the fact that Eq. (3.114) is only a function of the Boltzmann variable, $\varepsilon_{D}$, we can solve Eq. (3.100) uniquely. Note that the “collapsing” of the initial and outer boundary conditions must occur or the Boltzmann transform is technically invalid. Recalling the constant rate inner boundary condition, Eq. (3.95)

$$\left[ { r_{D} + \frac{{\partial p_{D} }}{{\partial r_{D} }}} \right]_{{r_{D} = 1}} = - 1\,or\,\left[ { r_{D} + \frac{{\partial p_{D} }}{{\partial r_{D} }}} \right]_{{r_{D} \to 0}} = - 1\quad ({\text{line}}\,{\text{source}}\,{\text{condition}})$$

(3.119)

Or

$$\left[ { r_{D} \frac{{\partial \varepsilon_{D} }}{{\partial r_{D} }}\frac{{\partial p_{D} }}{{\partial \varepsilon_{D} }}} \right]_{{r_{D} \to 0}} = \left[ { r_{D} \left( {\frac{2}{{r_{D} }}\varepsilon_{D} } \right)\frac{{\partial p_{D} }}{{\partial \varepsilon_{D} }}} \right]_{{r_{D} \to 0,\varepsilon_{D} \to 0}} = 2\left[ { \varepsilon_{D} + \frac{{\partial p_{D} }}{{\partial \varepsilon_{D} }}} \right]_{{\varepsilon_{D} \to 0}} = - 1$$

(3.120)

Which can be rearranged to yield

$$\left[ {\varepsilon_{D} \frac{{\partial p_{D} }}{{\partial \varepsilon_{D} }}} \right]_{{\varepsilon_{D} \to 0}} = \frac{ - 1}{2}$$

(3.121)

Making the following variable of substitution

$$v = \frac{{dp_{D} }}{{d\varepsilon_{D} }}$$

(3.122)

Substituting Eq. (3.122) into Eq. (3.124) and noting that use of ordinary derivatives

$$\frac{dv}{{d\varepsilon_{D} }} + \left[ {1 + \frac{1}{{\varepsilon_{D} }}} \right]v = 0$$

(3.123)

$$\frac{1}{v}dv = - \left[ {1 + \frac{1}{{\varepsilon_{D} }}} \right]d\varepsilon_{D} = - d\varepsilon_{D} - \frac{1}{{\varepsilon_{D} }}d\varepsilon_{D}$$

(3.124)

Integrating

$$\ln \left( v \right) = - \varepsilon_{D} - \ln \left( {\varepsilon_{D} } \right) + \beta ;\,\beta = {\text{constant}}\,{\text{of}}\,{\text{integration}}$$

(3.125)

Exponentiation

$$v = {\text{exp}}\left[ { - \varepsilon_{D} - \ln \left( {\varepsilon_{D} } \right) + \beta } \right]$$

(3.126)

Or

$$v = {\text{exp}}\left[ { - \varepsilon_{D} \left] {{\text{exp}}} \right[ - \ln \left( {\varepsilon_{D} } \right)} \right]{\text{exp}}\left[ \beta \right]$$

(3.127)

Which reduces to

$$v = \frac{\alpha }{{\varepsilon_{D} }}{\text{exp}}\left[ { - \varepsilon_{D} } \right]$$

(3.128)

where $\alpha = {\text{exp}}\left[ \beta \right]$, i.e., the constant of integration. Recalling Eq. (3.120) and combining gives

$$\frac{{dp_{D} }}{{d\varepsilon_{D} }} = \frac{\alpha }{{\varepsilon_{D} }}{\text{exp}}\left[ { - \varepsilon_{D} } \right]$$

(3.129)

Multiplying through by $\varepsilon_{D}$ gives

$$\varepsilon_{D} \frac{{dp_{D} }}{{d\varepsilon_{D} }} = \alpha {\text{exp}}\left[ { - \varepsilon_{D} } \right]$$

(3.130)

Substitution of Eq. (3.128) into Eq. (3.119) gives

$$\alpha \mathop {\lim }\limits_{{\varepsilon_{D} \to 0}} \left[ {{\text{exp}}\left[ { - \varepsilon_{D} } \right]} \right] = \frac{ - 1}{2}$$

(3.131)

Or

$$\alpha = - 1/2$$

(3.132)

Substitution of Eq. (3.132) into Eq. (3.129) gives

$$\frac{{dp_{D} }}{{d\varepsilon_{D} }} = \frac{ - 1}{{2\varepsilon_{D} }}{\text{exp}}\left[ { - \varepsilon_{D} } \right]$$

(3.133)

Separating and integrating Eq. (3.133) gives

$$\int\limits_{{p_{D} = 0}}^{{p_{D} }} {dp_{D} } = \frac{ - 1}{2}\int\limits_{{\varepsilon_{D} = 0}}^{{\varepsilon_{D} }} {\frac{1}{{\varepsilon_{D} }}e^{{ - \varepsilon_{D} }} } d\varepsilon_{D}$$

(3.134)

where we note that $p_{D} = 0$ at $\varepsilon_{D} = \infty$ is the initial outer boundary condition. Completing the integration and reversing the limits we have

$$p_{D} = \frac{1}{2}\int\limits_{{\varepsilon_{D} = \frac{{r_{D}^{2} }}{{4t_{D} }}}}^{\infty } {\frac{1}{y}e^{{ - {\text{y}}}} } dy$$

(3.135)

We note that the integral in Eq. (3.135) is the exponential integral, E₁(x), which is given by

$${\text{E}}_{1} \left( {\text{x}} \right) = \int\limits_{X}^{\infty } {\frac{1}{y}e^{{ - {\text{y}}}} } dy$$

(3.136)

Combining Eqs. (3.135) and (3.136) gives our final result

$$p_{D} \left( {r_{D} ,t_{D} } \right) = \frac{1}{2}{\text{E}}_{1} \left( {\frac{{r_{D}^{2} }}{{4t_{D} }}} \right)$$

(3.137)

Appendix 3.4: Solution of Non-linear Diffusivity Equation

The analytical solution of non-linear diffusivity equation needs changing variables in order to perform linear equation to be solved analytically.

$$\frac{{\partial^{2} p}}{{\partial r^{2} }} + \frac{1}{r}\frac{\partial p}{{\partial r}} + c\left( {\frac{\partial p}{{\partial r}}} \right)^{2} = \frac{\emptyset \mu c}{{kt}}\frac{\partial p}{{\partial t}}$$

(3.138)

$$\Delta p = p - p_{i} \cdot T = \frac{kt}{{\emptyset \mu c}}\,and\,\Delta p = \frac{1}{c}\ln p^{*}$$

(3.139)

The linearized form is presented as [179]:

$$\frac{{\partial^{2} p^{*} }}{{\partial r^{2} }} + \frac{1}{r}\frac{{\partial p^{*} }}{\partial r} = \frac{{\partial p^{*} }}{\partial T}$$

(3.140)

The linear equation can be solved using Laplace transform and concludes:

$$\Delta p = \left( {\frac{ - 1}{c}} \right)\left( {{\text{ln}}\left( {1 + \lambda \left( {\ln \left( {\frac{T}{{r_{w}^{2} }}} \right) + 0.2319} \right)} \right) + \frac{0.5772\lambda }{{1 + \lambda \left( {\frac{{{\text{ln}}T}}{{r_{w}^{2} }} + 0.2319} \right)}}} \right)$$

(3.141)

$$\lambda = \frac{q\mu c}{{4\pi kh}}$$

(3.142)

where, h is formation thickness.

Appendix 3.5: The Solution of the Warren-Root Equation in the Different Boundary Condition

3.1.1 Warren-Root Equations

Reservoir pressure distribution is achieved according to the mathematical model. Different reservoir boundary conditions can be applied to this end. The Warren-Root model is used to calculate pressure distribution in fractured reservoirs. Equations (3.28) and (3.29) are used for fluid velocity through matrix and fractures media, respectively:

$${\vec{\mathbf{U}}}_{{\mathbf{m}}} = - \frac{{{\mathbf{K}}_{{\mathbf{m}}} }}{{{\varvec{\upmu}}}}{\mathbf{gra}}{\mathbf{d}}\left( {{\mathbf{P}}_{{\mathbf{m}}} } \right)$$

$${\vec{\mathbf{U}}}_{{\mathbf{f}}} = - \frac{{{\mathbf{K}}_{2} }}{{{\varvec{\upmu}}}}{\mathbf{grad}}\left( {{\mathbf{P}}_{{\mathbf{f}}} } \right)$$

Subscripts m and f denote to matrix and fracture media, respectively. K is absolute permeability in (m²), U is velocity in (ms⁻¹), P is pressure in (Pa), and μ is dynamic viscosity in (Pa s). To calculate matrix and fracture pressure distribution, the mass conservation equation is formulated as follows:

$$\frac{{\partial \left( {\emptyset_{{\text{m}}}\uprho } \right)}}{{\partial {\text{t}}}} + {\text{div}}\left( {\uprho {\overline{\text{U}}}_{{\text{m}}} } \right) + {\text{U}}^{*} = 0$$

(3.143)

$$\frac{{\partial \left( {\emptyset_{{\text{f}}}\uprho } \right)}}{{\partial {\text{t}}}} + {\text{div}}\left( {\uprho {\overline{\text{U}}}_{{\text{f}}} } \right) + {\text{U}}^{*} = 0$$

(3.144)

$${\text{U}}^{*} = \frac{{\uprho \,{\text{SK}}_{{\text{m}}} }}{\upmu }\left( {{\text{P}}_{{\text{m}}} - {\text{P}}_{{\text{f}}} } \right)$$

(3.145)

The rate of mass flow per unit volume is defined by U^* in (kg s⁻¹ m⁻³), which indicates fluid transfer between matrix and fracture in quasi-steady-state condition. U̅ is fluid velocity in (ms⁻¹), and φ is porosity, which represents fluid storage capacity in each media. S is a characteristic coefficient of fractured rock proportional to the specific surface of a block, and ρ is fluid density in (kg m⁻³). For slightly compressible fluids, density is calculated by Eq. (3.146).

$$\uprho =\uprho _{0} \left( {1 + {\text{CP}}} \right)$$

(3.146)

where C is fluid compressibility in (Pa⁻¹), and indicates dependency of fluid volume on the pressure. Equations (3.147) and (3.148) are derived by combining Eqs. (3.143) to (3.146).

$$\emptyset_{{\text{m}}} {\text{C}}_{{\text{m}}} \frac{{\partial {\text{P}}_{{\text{m}}} }}{{\partial {\text{t}}}} + \frac{{{\text{SK}}_{{\text{m}}} }}{\upmu }\left( {{\text{P}}_{{\text{m}}} - {\text{P}}_{{\text{f}}} } \right) = 0$$

(3.147)

$$\emptyset_{{\text{f}}} {\text{C}}_{{\text{f}}} \frac{{\partial {\text{P}}_{{\text{f}}} }}{{\partial {\text{t}}}} - \frac{{{\text{K}}_{{\text{f}}} }}{\upmu }\left( {\frac{1}{{\text{r}}}\frac{\partial }{{\partial {\text{r}}}}\left( {{\text{r}}\frac{{\partial {\text{P}}_{{\text{f}}} }}{{\partial {\text{r}}}}} \right)} \right) + \frac{{{\text{SK}}_{{\text{m}}} }}{\upmu }\left( {{\text{P}}_{{\text{m}}} - {\text{P}}_{{\text{f}}} } \right) = 0$$

(3.148)

Dimensionless variables in Eqs. (3.149)–(3.153) are used to reduce the number of variables.

$${\text{P}}_{{\text{D}}} = \frac{{2\uppi {\text{K}}_{{\text{f}}} {\text{h}}\left( {{\text{P}}_{{\text{i}}} - {\text{P}}\left( {{\text{r}},{\text{t}}} \right)} \right)}}{{{\text{q}}\upmu }}$$

(3.149)

$${\text{r}}_{{\text{D}}} = \frac{{\text{r}}}{{{\text{r}}_{{\text{w}}} }}$$

(3.150)

$${\text{t}}_{{\text{D}}} = \frac{{{\text{K}}_{{\text{f}}} {\text{t}}}}{{\left( {{\text{C}}_{{\text{m}}} \emptyset_{{\text{m}}} + {\text{C}}_{{\text{f}}} \emptyset_{{\text{f}}} } \right)\upmu {\text{r}}_{{\text{w}}}^{2} }}$$

(3.151)

$$\uplambda = \frac{{\upalpha {\text{K}}_{{\text{m}}} {\text{r}}_{{\text{w}}}^{2} }}{{{\text{K}}_{{\text{f}}} }}$$

(3.152)

$$\upomega = \frac{{\emptyset_{{\text{f}}} {\text{C}}_{{\text{f}}} }}{{\emptyset_{{\text{m}}} {\text{C}}_{{\text{m}}} + \emptyset_{{\text{f}}} {\text{C}}_{{\text{f}}} }}$$

(3.153)

Dimensionless pressure is defined by P_D, r_D is dimensionless radius, and t_D is dimensionless time. Equations (3.152) and (3.153) represent fracture reservoir parameters, which indicate interporosity (λ) and fracture storage capacity (ω).

At the initial time, there is a pressure equilibrium state in the reservoir. Overall reservoir pressure is equal to reservoir initial pressure in this condition. Therefore, at the initial times, Eq. (3.154) is used as follows:

$${\text{P}} = {\text{P}}_{{\text{i}}} \quad {\text{t}} = 0$$

(3.154)

Dimensionless pressure at initial time is resulted by a combination of Eqs. (3.149) and (3.154).

$${\text{P}}_{{{\text{Dm}}}} = 0\quad {\text{t}}_{{\text{D}}} = 0$$

(3.155)

Equations (3.147) and (3.148) are rewritten in dimensionless forms by combining Eqs. (3.149)–(3.153) as follows:

$$\left( {1 -\upomega } \right)\frac{{\partial {\text{P}}_{{{\text{Dm}}}} }}{{\partial {\text{t}}_{{\text{D}}} }} -\uplambda \left( {{\text{P}}_{{{\text{Df}}}} - {\text{P}}_{{{\text{Dm}}}} } \right) = 0$$

(3.156)

$$-\upomega \frac{{\partial {\text{P}}_{{{\text{Df}}}} }}{{\partial {\text{t}}_{{\text{D}}} }} + \frac{1}{{{\text{r}}_{{\text{D}}} }}\frac{\partial }{{\partial {\text{r}}_{{\text{D}}} }}\left( {{\text{r}}_{{\text{D}}} \frac{{\partial {\text{P}}_{{{\text{Df}}}} }}{{\partial {\text{r}}_{{\text{D}}} }}} \right) + \left( {1 -\upomega } \right)\frac{{\partial {\text{P}}_{{{\text{Dm}}}} }}{{\partial {\text{t}}_{{\text{D}}} }} = 0$$

(3.157)

Equation (3.157) is rewritten as Eq. (3.158):

$$\frac{{\partial {\text{P}}_{{{\text{Df}}}}^{2} }}{{\partial^{2} {\text{r}}_{{\text{D}}} }} + { }\frac{1}{{{\text{r}}_{{\text{D}}} }}\frac{{\partial {\text{P}}_{{{\text{Df}}}} }}{{\partial {\text{r}}_{{\text{D}}} }} = \left( {1 -\upomega } \right)\frac{{\partial {\text{P}}_{{{\text{Dm}}}} }}{{\partial {\text{t}}_{{\text{D}}} }} +\upomega \frac{{\partial {\text{P}}_{{{\text{Df}}}} }}{{\partial {\text{t}}_{{\text{D}}} }}$$

(3.158)

There are several methods for solving partial differential equations. Laplace transform is one of these methods. Laplace transform is applied to solve differential Eqs. (3.156) and (3.158). Consequently Eq. (3.159) is resulted by Eq. (3.156) Laplace transform:

$$\left( {1 -\upomega } \right){\text{zP}}_{{{\text{Dm}}}} \left( {{\text{z}},{\text{r}}_{{\text{D}}} } \right) - {\text{P}}_{{{\text{Dm}}}} \left( {{\text{t}} = 0} \right) =\uplambda \left( {{\text{P}}_{{{\text{Df}}}} \left( {{\text{z}},{\text{r}}_{{\text{D}}} } \right) - {\text{P}}_{{{\text{Dm}}}} \left( {{\text{z}},{\text{r}}_{{\text{D}}} } \right)} \right)$$

(3.159)

The Laplace transform variable is known as z. By substituting Eq. (3.159) in the initial condition (Eq. (3.155)), matrix dimensionless pressure is resulted as the following equation:

$${\text{P}}_{{{\text{D}}m}} \left( {\text{z}} \right) = \frac{{\uplambda {\text{P}}_{{{\text{Df}}}} \left( {{\text{z}} \cdot {\text{r}}_{{\text{D}}} } \right)}}{{\left( {1 -\upomega } \right){\text{z}} +\uplambda }}$$

(3.160)

To obtain the fracture pressure equation, the Laplace transform of Eq. (3.158) is rewritten as Eq. (3.161).

$$\frac{{\partial^{2} {\text{P}}_{{{\text{Df}}}} \left( {{\text{z}} \cdot {\text{r}}_{{\text{D}}} } \right)}}{{\partial {\text{r}}_{{\text{D}}}^{2} }} + \frac{1}{{{\text{r}}_{{\text{D}}} }}\frac{{\partial {\text{P}}_{{{\text{Df}}}} \left( {{\text{z}}.{\text{r}}_{{\text{D}}} } \right)}}{{\partial {\text{r}}_{{\text{D}}} }} = { }\left( {1 -\upomega } \right){\text{zP}}_{{{\text{Dm}}}} \left( {{\text{z}}.{\text{r}}_{{\text{D}}} } \right) +\upomega {\text{zP}}_{{{\text{Df}}}} \left( {{\text{z}}.{\text{r}}_{{\text{D}}} } \right)$$

(3.161)

Equation (3.160) is replaced in Eq. (3.161) and Eq. (3.162) is resulted as follows:

$$\frac{{{\text{d}}^{2} {\text{P}}_{{{\text{Df}}}} \left( {{\text{z}}.{\text{r}}_{{\text{D}}} } \right)}}{{{\text{dr}}_{{\text{D}}}^{2} }} + \frac{1}{{{\text{r}}_{{\text{D}}} }}\frac{{{\text{dP}}_{{{\text{Df}}}} \left( {{\text{z}}.{\text{r}}_{{\text{D}}} } \right)}}{{{\text{dr}}_{{\text{D}}} }} = [{\text{zf}}\left( {\text{z}} \right)]{\text{ P}}_{{{\text{Df}}}} \left( {{\text{z}},{\text{r}}_{{\text{D}}} } \right)$$

(3.162)

In Eq. (3.162), f(z) is a function of ω and λ and is described as Eq. (3.31):

$${\mathbf{f}}\left( {\mathbf{z}} \right) = \frac{{{{\varvec{\upomega}}}\left( {1 - {{\varvec{\upomega}}}} \right){\mathbf{z}} + {{\varvec{\uplambda}}}}}{{\left( {1 - {{\varvec{\upomega}}}} \right){\mathbf{z}} + {{\varvec{\uplambda}}}}}$$

The solution of the partial differential Eq. (3.162) is necessary to calculate fracture dimensionless pressure. Therefore, r_D² is multiplied by Eq. (3.162) and Eq. (3.163) is obtained as a standard form of Bessel equation.

$${\text{r}}_{{\text{D}}}^{2} \frac{{{\text{d}}^{2} {\text{P}}_{{{\text{Df}}}} \left( {z.{\text{r}}_{{\text{D}}} } \right)}}{{{\text{dr}}_{{\text{D}}}^{2} }} + {\text{r}}_{{\text{D}}} \frac{{{\text{dP}}_{{{\text{Df}}}} \left( {{\text{z}}.{\text{r}}_{{\text{D}}} } \right)}}{{{\text{dr}}_{{\text{D}}} }} - \left( {{\text{r}}_{{\text{D}}}^{2} {\text{zf}}\left( {\text{z}} \right)} \right){\text{P}}_{{{\text{D}}2}} \left( {{\text{z}}.{\text{r}}_{{\text{D}}} } \right) = 0$$

(3.163)

Therefore, the solution of Eq. (3.163) is written as Eq. (3.30) by I₀ and K₀ parameters, which are the first and second type of modified Bessel functions in zero-order, respectively.

$${\mathbf{P}}_{{{\mathbf{Df}}}} \left( {{\mathbf{z}}.{\mathbf{r}}_{{\mathbf{D}}} } \right) = {\mathbf{C}}_{1} {\mathbf{I}}_{0} \left( {{\mathbf{M}} {\mathbf{r}}_{{\mathbf{D}}} } \right) + {\mathbf{C}}_{2} {\mathbf{K}}_{0} \left( {{\mathbf{M}} {\mathbf{r}}_{{\mathbf{D}}} } \right)$$

Bessel equation constants are introduced by C₁ and C₂, and boundary conditions are applied to obtain these values. To simplify calculations, $M = \sqrt {{\text{zf}}\left( {\text{z}} \right)}$ is considered. Constants of Eq. (3.30) are calculated by different boundary conditions substitution, and pressure distribution equations are obtained as a result.

3.1.2 Constant Production Rate in Closed Outer Boundary

Well production rate is kept constant in constant production rate conditions. A choke valve is usually used to stabilize the production rate in operating conditions. This case is more common in hydrocarbon reservoirs production. The wellbore boundary condition is written as Eq. (3.164) in this case:

$$\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{r}}}}} \right)_{{{\text{r}}_{{\text{w}}} }} = {\text{constant}}\quad {\text{r }} = {\text{r}}_{{\text{w}}}$$

(3.164)

Some hydrocarbon reservoirs are confined with faults or impermeable layers. No fluid flows through the outer boundary in these closed reservoirs. Also, for a reservoir with several production wells, the closed boundary can be assumed among the wells. The closed boundary conditions are expressed in Eq. (3.165):

$$\left( {\frac{{\partial {\text{P}}_{ } }}{{\partial {\text{r}}}}} \right)_{{{\text{r}}_{{\text{e}}} }} = 0\quad {\text{r}} = {\text{r}}_{{\text{e}}}$$

(3.165)

Darcy equation is applied to make a relationship between flow and pressure for an inner boundary condition (the well condition) as Eq. (3.166). In this case, C₁ and C₂ are calculated by replacing the boundary conditions (Eqs. (3.164) and (3.165)) in Eq. (3.30).

$${\text{q}} = - 2\uppi {\text{r}}_{{\text{w}}} {\text{h }}\frac{{{\text{K}}_{{\text{f}}} }}{\upmu }{ }\frac{{\partial {\text{P}}_{{\text{f}}} }}{{\partial {\text{r}}}}\quad {\text{r}} = {\text{r}}_{{\text{w}}}$$

(3.166)

Equation (3.165) is written for closed boundary condition and dimensionless form of Eqs. (3.165) and (3.166) is resulted as Eqs. (3.168) and (3.167), respectively.

$$\frac{{\partial {\text{P}}_{{{\text{Df}}}} }}{{\partial {\text{r}}_{{\text{D}}} }} = - 1\quad {\text{r}}_{{\text{D}}} = 1$$

(3.167)

$$\frac{{\partial {\text{P}}_{{{\text{Df}}}} }}{{\partial r_{D} }} = 0\quad {\text{r}}_{{\text{D}}} = {\text{r}}_{{{\text{De}}}}$$

(3.168)

By differentiating Eq. (3.30) and integrating with Eq. (3.167), results can be represented as the following equations:

$${\text{C}}_{1} {\text{M}}\,{\text{I}}_{1} \left( {{\text{M}}\,{\text{r}}_{{\text{D}}} } \right) - {\text{C}}_{2} {\text{M}}\,{\text{K}}_{1} \left( {{\text{M}}\,{\text{r}}_{{\text{D}}} } \right) = \frac{{\partial {\text{P}}_{{{\text{Df}}}} }}{{\partial {\text{r}}_{{\text{D}}} }}$$

(3.169)

$${\text{C}}_{1} {\text{MI}}_{1} \left( {\text{M }} \right) - {\text{C}}_{2} {\text{M}}\,{\text{K}}_{1} \left( {\text{M }} \right) = \frac{ - 1}{z}$$

(3.170)

The first and second types of modified Bessel functions in the first order are known as I₁ and K₁. Equation (3.171) is resulted by differentiating Eq. (3.30) and integrating with Eq. (3.168).

$${\text{C}}_{1} {\text{M}}\,{\text{I}}_{1} \left( {{\text{M}}\,{\text{r}}_{{{\text{De}}}} { }} \right) - {\text{C}}_{2} {\text{M}}\,{\text{K}}_{1} \left( {{\text{M}}\,{\text{r}}_{{{\text{De}}}} { }} \right) = 0$$

(3.171)

A system of two linear Eqs. (3.170) and (3.171) is formed to calculate C₁ and C₂:

$${\text{C}}_{1} = \frac{{ - {\text{K}}_{1} \left( {{\text{M}}\,{\text{r}}_{{{\text{De}}}} { }} \right)}}{{{\text{z}}\left[ {\left( {{\text{K}}_{1} \left( {{\text{M}}\,{\text{r}}_{{{\text{De}}}} } \right){ }} \right) \times {\text{M}}\,{\text{I}}_{1} \left( {\text{M }} \right) - {\text{I}}_{1} \left( {{\text{M}}\,{\text{r}}_{{{\text{De}}}} { }} \right) \times {\text{M}}\,{\text{K}}_{1} \left( {\text{M }} \right)} \right]}}$$

(3.172)

$${\text{C}}_{2} = \frac{{ - {\text{I}}_{1} \left( {{\text{Mr}}_{{{\text{De}}}} { }} \right)}}{{{\text{ z}}\left[ {{\text{K}}_{1} \left( {{\text{ M}}\,{\text{r}}_{{{\text{De}}}} } \right) \times {\text{M}}\,{\text{I}}_{1} \left( {\text{M }} \right) - {\text{I}}_{1} \left( {{\text{M}}\,{\text{r}}_{{{\text{De}}}} { }} \right) \times {\text{M}}\,{\text{K}}_{1} \left( {\text{M }} \right)} \right]}}$$

(3.173)

Equation (3.32) is resulted by substitution C₁ and C₂ in Eq. (3.30) as follows:

$${\mathbf{P}}_{{{\mathbf{Df}}}} \left( {{\mathbf{z}}.{\mathbf{r}}_{{\mathbf{D}}} } \right) = \frac{{{\mathbf{K}}_{1} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right) {\mathbf{I}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{\mathbf{D}}} } \right) + {\mathbf{I}}_{1} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right){\mathbf{K}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{\mathbf{D}}} } \right) }}{{{\mathbf{z}} {\mathbf{M}} \left[ {{\mathbf{K}}_{1} \left( {\mathbf{M}} \right) {\mathbf{I}}_{1} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right) - {\mathbf{K}}_{1} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right) {\mathbf{I}}_{1} \left( {\mathbf{M}} \right)} \right]}}$$

Therefore, the dimensionless pressure equation in a closed fractured reservoir and constant production rate is given by Eq. (3.32).

3.1.3 Constant Production Rate in Constant Pressure Outer Boundary

For infinite reservoirs with high-pressure supplier, there is no pressure drop at the reservoir outer boundary, and the outer boundary pressure equals to reservoir initial pressure, as shown by Eq. (3.174). The dimensionless form of this equation is presented as Eq. (3.175).

$${\text{P}}_{{\text{e}}} = {\text{P}}_{{\text{i}}} \quad {\text{r }} = {\text{r}}_{{\text{e}}}$$

(3.174)

$${\text{P}}_{{{\text{Dm}}}} = {\text{P}}_{{{\text{Df}}}} = 0\quad {\text{r}}_{{\text{D}}} = {\text{r}}_{{{\text{De}}}}$$

(3.175)

Equation (3.175) is replaced in Eq. (3.30) and Eq. (3.176) is resulted as follows:

$${\text{C}}_{1} {\text{I}}_{0} \left( {{\text{M}}\,{\text{r}}_{{{\text{De}}}} { }} \right) + {\text{C}}_{2} {\text{K}}_{0} \left( {{\text{M}}\,{\text{r}}_{{{\text{De}}}} { }} \right) = 0$$

(3.176)

A system of two linear Eqs. (3.170) and (3.176) is formed to calculate C₁ and C₂ in this case:

$${\text{C}}_{1} = - \frac{{{\text{K}}_{0} \left( {{\text{M}}\,{\text{r}}_{{{\text{De}}}} { }} \right)}}{{{\text{z M}}\left[ {{\text{K}}_{0} \left( {{\text{M}}\,{\text{r}}_{{{\text{De}}}} } \right){\text{I}}_{1} \left( {\text{M }} \right) + {\text{I}}_{0} \left( {{\text{M}}\,{\text{r}}_{{{\text{De}}}} { }} \right){\text{K}}_{1} \left( {\text{M }} \right)} \right]}}$$

(3.177)

$${\text{C}}_{2} = \frac{{{\text{I}}_{0} \left( {{\text{M}}\,{\text{r}}_{{{\text{De}}}} { }} \right)}}{{{\text{ z M}}\left[ {{\text{K}}_{0} \left( {{\text{M}}\,{\text{r}}_{{{\text{De}}}} } \right){\text{I}}_{1} \left( {\text{M }} \right) + {\text{I}}_{0} \left( {{\text{M}}\,{\text{r}}_{{{\text{De}}}} } \right){\text{K}}_{1} \left( {\text{M}} \right)} \right]}}$$

(3.178)

Equation (3.33) is resulted by replacing C₁ and C₂ in Eq. (3.30).

$${\mathbf{P}}_{{{\mathbf{Df}}}} \left( {{\mathbf{z}}.{\mathbf{r}}_{{\mathbf{D}}} } \right) = \frac{{{\mathbf{K}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{\mathbf{D}}} } \right){\mathbf{I}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right) - {\mathbf{K}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right){\mathbf{I}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{\mathbf{D}}} } \right)}}{{ {\mathbf{z}} {\mathbf{M}}\left[ {{\mathbf{K}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right){\mathbf{I}}_{1} \left( {{\mathbf{M}} } \right) + {\mathbf{K}}_{1} \left( {{\mathbf{M}} } \right){\mathbf{I}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right)} \right]}}$$

Equation (3.33) is applicable for dimensionless flow rate calculation at constant pressure boundary in case of constant pressure production condition.

3.1.4 Constant Pressure Production in Closed Outer Boundary

The wellbore pressure is kept constant, and flow rate changes i.e., it decreases with time in constant pressure production condition. In operating conditions, well pressure is maintained constant by using some gauges that alter the production rate. This production condition is mostly used in reservoirs with a high risk of gas or water coning. The wellbore boundary condition, in this case, is written as Eq. (3.179):

$${\text{P}}_{ } = {\text{constant}}\quad {\text{r }} = {\text{r}}_{{\text{w}}}$$

(3.179)

Production rate equation is obtained using the wellbore pressure equation suggested by Van Everdingen and Hurst [180]:

$${\text{q}}_{{\text{D}}} \left( {{\text{z}}.{\text{r}}_{{\text{D}}} } \right) = \frac{1}{{{\text{z}}^{2} {\text{ P}}_{{{\text{Dwf}}}} \left( {{\text{z}},{\text{r}}_{{\text{D}}} } \right)}}$$

(3.180)

The dimensionless production equation (Eq. (3.34)) is obtained by combining Eqs. (3.32) and (3.180).

$${\mathbf{q}}_{{\mathbf{D}}} \left( {{\mathbf{z}}.{\mathbf{r}}_{{\mathbf{D}}} } \right) = \frac{{ {\mathbf{M}}\left[ { {\mathbf{K}}_{1} \left( {\mathbf{M}} \right) {\mathbf{I}}_{1} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right) - {\mathbf{K}}_{1} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right) {\mathbf{I}}_{1} \left( {\mathbf{M}} \right)} \right] }}{{{\mathbf{z}}[{\mathbf{K}}_{1} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right) {\mathbf{I}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{\mathbf{D}}} } \right) + {\mathbf{I}}_{1} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right){\mathbf{K}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{\mathbf{D}}} } \right)]}}$$

The dimensionless flow rate at a closed boundary fractured reservoir with constant pressure production is calculated by Eq. (3.34).

3.1.5 Constant Pressure Production in Constant Pressure Outer Boundary

In this case, the inner and outer boundary conditions are described by Eqs. (3.179) and (3.174). The Van Everdingen and Hurst relations are applied as the same as the previous section. Equation (3.35) is concluded by combining Eqs. (3.33) and (3.180):

$${\mathbf{q}}_{{\mathbf{D}}} \left( {{\mathbf{z}}.{\mathbf{r}}_{{\mathbf{D}}} } \right) = \frac{{ {\mathbf{M}}\left[ {{\mathbf{K}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right){\mathbf{I}}_{1} \left( {{\mathbf{M}} } \right) + {\mathbf{K}}_{1} \left( {{\mathbf{M}} } \right){\mathbf{I}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right)} \right]}}{{{\mathbf{z}} \left[ {{\mathbf{K}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{\mathbf{D}}} } \right){\mathbf{I}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right) - {\mathbf{K}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{{\mathbf{De}}}} } \right){\mathbf{I}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{\mathbf{D}}} } \right)} \right]}}$$

Equation (3.35) is applicable for dimensionless flow rate calculations at constant pressure reservoir boundary in the case of constant pressure production.

3.1.6 Warren-Root Analytical Solution

The original solution provided by Warren and Root considers infinite outer boundary conditions and also the constant production rate for an inner boundary condition. The inner boundary condition is presented as Eq. (3.164) and the outer boundary condition is defined as Eq. (3.181):

$${\text{P}} = {\text{P}}_{{\text{i}}} \quad {\text{r }} = \infty$$

(3.181)

This assumption is applied to constant production rate conditions [65]. This simplification is applied to Eq. (3.32). The infinite values of the first and second-order Bessel function tend to infinite and zero, respectively. Thus, Eq. (3.36) is a simplified form of Eq. (3.32) which is resulted by using the Warren-Root assumptions:

$${\mathbf{P}}_{{\mathbf{D}}} \left( {{\mathbf{z}}.{\mathbf{r}}_{{\mathbf{D}}} } \right) = \frac{{{\mathbf{K}}_{0} \left( {{\mathbf{M}}\,{\mathbf{r}}_{{\mathbf{D}}} } \right) }}{{{\mathbf{z}} {\mathbf{M}}\,{\mathbf{K}}_{1} \left( {\mathbf{M}} \right) }}$$

The Laplace inverse calculation of Eq. (3.36) is not possible by conventional analytical methods. An approximate method for calculating K₀ and K₁ is possible by considering primary terms of Bessel functions:

$${\mathbf{K}}_{0} \left( {\mathbf{M}} \right) = - {{\varvec{\upgamma}}} - {\mathbf{Ln}}\left( {\frac{{{\mathbf{r}}_{{\mathbf{D}}} }}{2}{\mathbf{M}}^{2} } \right)$$

(3.182)

$${\mathbf{K}}_{1} \left( {\mathbf{M}} \right) = \frac{1}{{{\mathbf{M}}^{2} }}$$

(3.183)

In Eq. (3.182), ɤ is Euler number and equals 0.5772. This approximation is applicable only for M < 0.01 values. Also, r_D = 1 is assumed in the above equations. Equation (3.184) is obtained by substituting Eqs. (3.182) and (3.183) in Eq. (3.36) and dimensionless pressure is calculated by numerical inverse Laplace method, as follows:

$${\text{P}}_{{\text{D}}} \left( {{\text{t}}_{{\text{D}}} .1} \right) = \frac{1}{2}[0.80908 + {\text{Ln }}\left( {{\text{t}}_{{\text{D}}} } \right) + {\text{Ei}}\left( {\frac{{ -\uplambda {\text{t}}_{{\text{D}}} }}{{\upomega \left( {1 -\upomega } \right)}}} \right) - {\text{ Ei}}\left( {\frac{{ -\uplambda {\text{t}}_{{\text{D}}} }}{{\left( {1 -\upomega } \right)}}} \right)$$

(3.184)

Exponential integral, which is called Ei function, is defined as Eq. (3.185):

$${\text{Ei}}\left( { - {\text{X}}} \right) = - \int\limits_{{\text{X}}}^{\infty } {\frac{{{\text{e}}^{{ - {\text{u}}}} }}{{\text{u}}}} {\text{du}}$$

(3.185)

The Ei function values limit to zero at a very long time. Therefore, Eq. (3.184) is converted to Eq. (3.37):

$${\mathbf{P}}_{{\mathbf{D}}} \left( {{\mathbf{t}}_{{\mathbf{D}}} .1} \right) = \frac{1}{2}\left[ {0.80908 + {\mathbf{Ln}} \left( {{\mathbf{t}}_{{\mathbf{D}}} } \right)} \right] = \frac{1}{2}{\mathbf{Ln}} \left( {2.25{\mathbf{t}}_{{\mathbf{D}}} } \right) = 1.15\log \left( {2.25{\mathbf{t}}_{{\mathbf{D}}} } \right)$$

Therefore, Warren and Root solved the inverse Laplace of dimensionless pressure equation in a fractured reservoir with simple assumptions in specific conditions (M < 0.01). The solution provided by Warren and Root is a simplified form of the general solution shown in Eq. (3.32) for M < 0.01 at wellbore (r_D = 1) [63].