Novel, Integrated and Revolutionary Well Test Interpretation and Analysis

Abstract

Well test interpretation is an important tool for reservoir characterization. There exist four methods to achieve this goal, which are as follows: type‐curve matching, conventional straight‐line method, non‐linear regression analysis, and TDS technique. The first method is basically a trial‐and‐error procedure; a deviation of a millimeter involves differences up to 200 psi and the difficulty of having so many matching charts. The second one, although very important, requires a plot for every flow regime, and there is no way for verification of the calculated parameters, and the third one has a problem of diversity of solutions but is the most used by engineers since it is automatically made by a computer program. This book focuses on the fourth method that uses a single plot of the pressure and pressure derivative plot for identifying different lines and feature for parameter estimation. It can be used alone and is applied practically to all the existing flow regime cases. In several cases, the same parameter can be estimated from different sources making a good way for verification. Combination of this method along with the second and third is recommended and widely used by the author.

Introduction

Well testing is a valuable and economical formation evaluation tool used in the hydrocarbon industry. It has been supported by mathematical modeling, computing, and the precision of measurement devices. The data acquired during a well test are used for reservoir characterization and description. However, the biggest drawback is that the system dealt with is neither designed nor seen by well test interpreters, and the only way to make contact with the reservoir is through the well by making indirect measurements.

Four methods are used for well test interpretation: (1) The oldest one is the conventional straight‐line method which consists of plotting pressure or the reciprocal rate—if dealing with transient rate analysis—in the y‐axis against a function of time in the x‐axis. This time function depends upon the governing equation for a given flow. For instance, radial flow uses the logarithm of time and linear flow uses the square root of time. The slope and intercept of such plot are used to find reservoir parameters. The main disadvantage of this method is the lack of confirmation and the difficulty to define a given flow regime. The method is widely used nowadays. (2) Type‐curve matching uses predefined dimensionless pressure and dimensionless time curves (some also use dimensionless pressure derivative), which are used as master guides to be matched with well pressure data to obtain a reference point for reservoir parameter determination. This method is basically a trial‐and‐error procedure which becomes into its biggest disadvantage. The method is practically unused. (3) Simulation of reservoir conditions and automatic adjustment to well test data by non‐linear regression analysis is the method widely used by petroleum engineers. This method is also being widely disused since engineers trust the whole task to the computer. They even perform inverse modeling trying to fit the data to any reservoir model without taking care of the actual conditions. However, the biggest weakness of this method lies on the none uniqueness of the solution. Depending on the input starting values, the results may be different. (4) The newest method known as Tiab’s direct synthesis (TDS) [1, 2] is the most powerful and practical one as will be demonstrated throughout the book. It employs characteristic points and features found on the pressure and pressure derivative versus time log‐log plot to be used into direct analytic equations for reservoir parameters’ calculation. It is even used, without using the original name, by all the commercial software. One of them calls it “Specialized lines.” Because of its practicality, accuracy and application is the main object of this book. Conventional analysis method will be also included for comparison purposes.

The TDS technique can be easily implemented for all kinds of conventional or unconventional systems. It can be easily applied on cases for which the other methods fail or are difficult to be applied. It is strongly based on the pressure derivative curve. The method works by sector or regions found on the test. This means once a given flow regime is identified, a straight line is drawn throughout it, and then, any arbitrary point on this line and the intersection with other lines as well are used into the appropriate equations for the calculation of reservoir parameters.

The book contains the application and detailed examples of the TDS technique to the most common or fundamental reservoir/fluid scenarios. It is divided into seven chapters that are recommended to be read in the other they appear, especially for academic purposes in senior undergraduate level or master degree level. Chapter 1 contains the governing equation and the superposition principle. Chapter 2 is the longest one since it includes drawdown for infinite and finite cases, elongated system, multi‐rate testing, and spherical/hemispherical flow. All the interpretation methods are studied in this chapter which covers about 45% of the book. Chapter 3 deals with pressure buildup testing and average reservoir pressure determination. Distance to barriers and interference testing are, respectively, treated in Chapters 4 and 5. Since the author is convinced that all reservoirs are naturally fractured, Chapter 6 covers this part which is also extended in hydraulically fractured wells in Chapter 7. In this last chapter, the most common flow regime shown in fractured wells: bilinear, linear, and elliptical are discussed with detailed for parameter characterization. The idea is to present a book on TDS technique as practical and short as possible; then, horizontal well testing is excluded here because of its complexity and extension, but the most outstanding and practical publications are named here.

My book entitled “Recent Advances in Practical Applied Well Test Analysis,” published in 2015, was written for people having some familiarity with the TDS technique, so that, it can be read in any order. This is not the case of the present textbook. It is recommended to be read in order from Chapter 1 and take especial care in Chapter 2 since many equations and concepts will be applied in the remaining chapters. TDS technique applies indifferently to both pressure drawdown and pressure buildup tests.

Finally, this book is an upgraded and updated version of a former one published in Spanish. Most of the type curves have been removed since they have never been used by the author on actual well test interpretations. However, the first motivation to publish this book is the author’s belief that TDS technique is the panacea for well test interpretation. TDS technique is such an easy and practical methodology that his creator, Dr. Djebbar Tiab, when day said to me “I still don’t believe TDS works!” But, it really does. Well, once things have been created, they look easy.

Fundamentals

1.1. Basic concepts

Pressure test fundamentals come from the application of Newton’s law, especially the third one: Principle of action‐reaction, since it comes from a perturbation on a well, as illustrated in Figure 1.1.

Figure 1.1.

Diagram of the mathematical representation of a pressure test.

A well can be produced under any of two given scenarios: (a) by keeping a constant flow rate and recording the well‐flowing pressure or (b) by keeping a constant well‐flowing pressure and measuring the flow rate. The first case is known as pressure transient analysis, PTA, and the second one is better known as rate transient analysis, RTA, which both are commonly run in very low permeable formations such as shales.

Basically, the objectives of the analysis of the pressure tests are:

• Reservoir evaluation and description: well delivery, properties, reservoir size, permeability by thickness (useful for spacing and stimulation), initial pressure (energy and forecast), and determination of aquifer existence.

• Reservoir management.

There are several types of tests with their particular applications. DST and pressure buildup tests are mainly used in primary production and exploration. Multiple tests are most often used during secondary recovery projects, and multilayer and vertical permeability tests are used in producing/injectors wells. Drawdown, interference, and pulse tests are used at all stages of production. Multi‐rate, injection, interference, and pulse tests are used in primary and secondary stages [3, 4, 5, 6, 7].

Pressure test analysis has a variety of applications over the life of a reservoir. DST and pressure buildup tests run in single wells are mainly used during primary production and exploration, while multiple tests are used more often during secondary recovery projects. Multilayer and vertical permeability tests are also run in producing/injectors wells. Drawdown, buildup, interference, and pulse tests are used at all stages of production. Multi‐rate, injection, interference, and pulse testing are used in the primary and secondary stages. Petroleum engineers should take into account the state of the art of interpreting pressure tests, data acquisition tools, interpretation methods, and other factors that affect the quality of the results obtained from pressure test analysis.

Once the data have been obtained from the well and reviewed, the pressure test analysis comprises two steps: (1) To establish the reservoir model and the identification of the different flow regimes encountered during the test and (2) the parameter estimation. To achieve this goal, several plots are employed; among them, we have log‐log plot of pressure and pressure derivative versus testing time (diagnostic tool), semilog graph of pressure versus time, Cartesian graph of the same parameters, etc. Pressure derivative will be dealt later in this chapter.

The interpretation of pressure tests is the primary method for determining average permeability, skin factor, average reservoir pressure, fracture length and fracture conductivity, and reservoir heterogeneity. In addition, it is the only fastest and cheapest method to estimate time‐dependent variables such as skin factor and permeability in stress‐sensitive reservoirs.

In general, pressure test analysis is an excellent tool to describe and define the model of a reservoir. Flow regimes are a direct function of the characteristics of the well/reservoir system, that is, a simple fracture that intercepts the well can be identified by detection of a linear flow. However, whenever there is linear flow, it does not necessarily imply the presence of a fracture. The infinite‐acing behavior occurs after the end of wellbore storage and before the influence of the limits of the deposit. Since the boundaries do not affect the data during this period, the pressure behavior is identical to the behavior of an infinite reservoir. The radial flow can be recognized by an apparent stabilization of the value of the derivative.

1.2. Type of well tests

Well tests can be classified in several ways depending upon the view point. Some classifications consider whether or not the well produces or is shut‐in. Other engineers focus on the number of flow rates. The two main pressure tests are (a) pressure drawdown and (b) buildup. While the first one involves only one flow rate, the second one involves two flow rates, one of which is zero. Then, a pressure buildup test can be considered as a multi‐rate test.

1.2.1 Pressure tests run in producer wells

Drawdown pressure test (see Figure 1.2): It is also referred as a flow test. After the well has been shut‐in for a long enough time to achieve stabilization, the well is placed in production, at a constant rate, while recording the bottom pressure against time. Its main disadvantage is that it is difficult to maintain the constant flow rate.

Figure 1.2.

Schematic representation of pressure drawdown and pressure buildup tests.

Pressure buildup test (see Figure 1.2): In this test, the well is shut‐in while recording the static bottom‐hole pressure as a function of time. This test allows obtaining the average pressure of the reservoir. Although since 2010, average reservoir pressures can be determined from drawdown tests. Its main disadvantage is economic since the shut‐in entails the loss of production.

1.2.2 Pressure tests run in injector wells

Injection test (see Figure 1.3): Since it considers fluid flow, it is a test similar to the pressure drawdown test, but instead of producing fluids, fluids, usually water, are injected.

Figure 1.3.

Injection pressure test (left) and falloff test (right).

Falloff test (see Figure 1.3): This test considers a pressure drawdown immediately after the injection period finishes. Since the well is shut‐in, falloff tests are identical to pressure buildup tests.

1.2.3 Other tests

Interference and/or multiple tests: They involve more than one well and its purpose is to define connectivity and find directional permeabilities. A well perturbation is observed in another well.

Drill stem test (DST): This test is used during or immediately after well drilling and consists of short and continuous shut‐off or flow tests. Its purpose is to establish the potential of the well, although the estimated skin factor is not very representative because well cleaning can occur during the first productive stage of the well (Figure 1.4).

Figure 1.4.

Well test classification based on the number of flow rates.

Short tests: There are some very short tests mainly run in offshore wells. They are not treated in this book. Some of them are slug tests, general close chamber tests (CCTs), surge tests, shoot and pool tests, FasTest, and impulse tests.

As stated before, in a pressure drawdown test, the well is set to a constant flow rate. This condition is, sometimes, difficult to be fulfilled; then, multi‐rate tests have to be employed. According to [8], multi‐rate tests fit into four categories: (a) uncontrolled variable rate [9, 10], series of constant rates [11, 12], pressure buildup testing, and constant bottom‐hole pressure with a continuous changing flow rate [13]. This last technique has been recently named as rate transient analysis (RTA) which is included in PTA, but its study is not treated in this book.

1.3 Diffusivity equation

At the beginning of production, the pressure in the vicinity of the well falls abruptly and the fluids near the well expand and move toward the area of lower pressure. Such movement is retarded by friction against the walls of the well and the inertia and viscosity of the fluid itself. As the fluid moves, an imbalance of pressure is created, which induces the surrounding fluids to move toward the well. The process continues until the pressure drop created by the production dissipates throughout the reservoir. The physical process that takes place in the reservoir can be described by the diffusivity equation whose deduction is shown below [5]:

According to the volume element given in Figure 1.5,

$${\displaystyle \left[\begin{array}{l}\text{Mass entering }\\ \text{the element}\end{array}\right]}-{\displaystyle \left[\begin{array}{l}\text{Mass coming out }\\ \text{from the element}\end{array}\right]}={\displaystyle \left[\begin{array}{l}\text{System }\\ \text{accumulation rate}\end{array}\right]}$$E1.1

Figure 1.5.

Radial volume element.

The right‐hand side part of Eq. (1.1) corresponds to the mass accumulated in the volume element. Darcy’s law for radial flow:

$$q=-\frac{kA}{\mu }\frac{dP}{dr}$$E1.2

The cross‐sectional area available for flow is provided by cylindrical geometry, 2πrh. Additionally, flow rate must be multiplied by density, ρ, to obtain mass flow. With these premises, Eq. (1.2) becomes:

$$q=-\frac{k}{\mu }2\pi \hspace{0.17em}rh\frac{\partial P}{\partial \hspace{0.17em}r}$$E1.3

Replacing Eq. (1.3) into (1.1) yields:

$$-\frac{k\rho }{\mu }{\displaystyle \left(\text{2}\pi \hspace{0.17em}rh\right)}\frac{\partial P}{\partial \hspace{0.17em}r}{|}_r+{\frac{k\rho }{\mu }{\displaystyle \left(\text{2}\pi \hspace{0.17em}rh\right)}\frac{\partial P\hspace{0.17em}}{\partial \hspace{0.17em}r}|}_{r+dr}=\frac{\partial }{\partial t}{\displaystyle \left(\text{[2}\pi rhdr\varphi ]\rho \right)}$$E1.4

If the control volume remains constant with time, then, Eq. (1.4) can be rearranged as:

$${-2\pi \hspace{0.17em}h\frac{k\rho }{\mu }r\frac{\partial P}{\partial \hspace{0.17em}r}|}_r+{2\pi \hspace{0.17em}h\frac{k\rho }{\mu }r\frac{\partial P}{\partial \hspace{0.17em}r}|}_{r+dr}=2\pi rhdr\frac{\partial }{\partial t}{\displaystyle \left(\varphi \rho \right)}$$E1.5

Rearranging further the above expression:

$$\frac{\frac{1}{r}{\displaystyle \left[\frac{k\rho }{\mu }r\frac{\partial P}{\partial \hspace{0.17em}r}|{}_{r+dr}-\frac{k\rho }{\mu }r\frac{\partial P}{\partial \hspace{0.17em}r}|{}_r\right]}}{dr}=\frac{\partial }{\partial t}{\displaystyle \left(\varphi \rho \right)}$$E1.6

The left‐hand side of Eq (1.6) corresponds to the definition of the derivative; then, it can be rewritten as:

$$\frac{1}{r}\frac{\partial }{\partial \hspace{0.17em}r}{\displaystyle \left(\frac{k\rho }{\mu }r\frac{\partial P}{\partial \hspace{0.17em}r}\right)}=\frac{\partial }{\partial \hspace{0.17em}t}{\displaystyle \left(\varphi \rho \right)}$$E1.7

The definition of compressibility has been widely used;

$$c=-\frac{1}{V}\frac{\partial \hspace{0.17em}V}{\partial P}=\frac{1}{\rho }\frac{\partial \rho }{\partial P}$$E1.8

By the same token, the pore volume compressibility is given by:

$$c_f=\frac{1}{\varphi }\frac{\partial \hspace{0.17em}\varphi }{\partial P}$$E1.9

The integration of Eq. (1.8) will lead to obtain:

$$\rho ={\rho }_o{e}^{c(P-P_o)}$$E1.10

The right‐hand side part of Eq. (1.7) can be expanded as:

$$\frac{\partial }{\partial \hspace{0.17em}t}{\displaystyle \left(\varphi \rho \right)}=\varphi \frac{\partial }{\partial \hspace{0.17em}t}\rho +\rho \frac{\partial }{\partial \hspace{0.17em}t}\varphi =\varphi \frac{\partial \rho }{\partial t}+\rho \frac{\partial \varphi }{\partial P}\frac{\partial P}{\partial \rho }\frac{\partial \rho }{\partial t}$$E1.11

Using the definitions given by Eqs. (1.9) and (1.10) into Eq. (1.11) leads to:

$$\frac{\partial }{\partial \hspace{0.17em}t}{\displaystyle \left(\varphi \rho \right)}=\varphi \frac{\partial \rho }{\partial t}+\frac{\rho \hspace{0.17em}\varphi \hspace{0.17em}{c}_f}{c\rho }\frac{\partial \rho }{\partial t}=\varphi \frac{\partial \rho }{\partial t}{\displaystyle \left[1+\frac{c_f}{c}\right]}=\frac{\varphi }{c}{\displaystyle \left[c_f+c\right]}\frac{\partial \rho }{\partial t}$$E1.12

Considering that the total compressibility, c_t, is the result of the fluid compressibility, c, plus the pore volume compressibility, c_f, it yields:

$$\frac{1}{r}\frac{\partial }{\partial \hspace{0.17em}r}{\displaystyle \left(\frac{k\rho }{\mu }r\frac{\partial P}{\partial \hspace{0.17em}r}\right)}=\frac{\varphi c_t}{c}\frac{\partial \rho }{\partial t}$$E1.13

The gradient term can be expanded as:

$$\frac{\partial P}{\partial r}=\frac{\partial P}{\partial \rho }\frac{\partial \rho }{\partial r}=\frac{1}{c\rho }\frac{\partial \rho }{\partial r}$$E1.14

Combination of Eqs. (1.14) and (1.13) results in:

$$\frac{1}{r}\frac{\partial }{\partial \hspace{0.17em}r}{\displaystyle \left(\frac{kr}{\mu c}\frac{\partial \rho }{\partial \hspace{0.17em}r}\right)}=\frac{\varphi }{c}{c}_t\frac{\partial \rho }{\partial t}$$E1.15

Taking derivative to Eq. (1.10) with respect to both time and radial distance and replacing these results into Eq. (1.15) yield:

$$\frac{1}{r}\frac{\partial }{\partial \hspace{0.17em}r}{\displaystyle \left(\frac{kr}{\mu c}{\rho }_o{e}^{c(P-P_o)}c\frac{\partial P}{\partial r}\right)}=\frac{\varphi }{c}{c}_t{\rho }_o{e}^{c(P-P_o)}c\frac{\partial P}{\partial t}$$E1.16

After simplification and considering permeability and viscosity to be constant, we obtain:

$$\frac{1}{r}\frac{k}{\mu }\frac{\partial }{\partial r}{\displaystyle \left(r\frac{\partial P}{\partial r}\right)}=\varphi \hspace{0.17em}{c}_t\frac{\partial P}{\partial t}$$E1.17

The hydraulic diffusivity constant is well known as

$$\frac{1}{\eta }=\frac{\varphi \mu c_t}{k}$$E1.18

Then, the final form of the diffusivity equation in oilfield units is obtained by combination of Eqs. (1.17) and (1.18):

$$\frac{1}{r}\frac{\partial }{\partial r}{\displaystyle \left(r\frac{\partial P}{\partial r}\right)}=\frac{\varphi \mu c_t}{0.0002637k}\frac{\partial P}{\partial t}=\frac{1}{\eta }\frac{\partial P}{\partial t}$$E1.19

In expanded form:

$$\frac{{\partial }^{2}P}{\partial r^2}+\frac{1}{r}\frac{\partial P}{\partial r}=\frac{1}{0.0002637\eta }\frac{\partial P}{\partial t}$$E1.20

The final form of the diffusivity equation strongly depends upon the flow geometry. For cylindrical, [11, 14], spherical [14], and elliptical coordinates [15], the diffusivity equation is given, respectively,

$$\frac{\partial {\hspace{0.17em}}^{2}P}{\partial \hspace{0.17em}{r}^2}+\frac{1}{r}\frac{\partial P}{\partial r}+\frac{k_{\theta }}{k_r}\frac{1}{r^2}\frac{{\partial }^{2}P}{\partial \hspace{0.17em}{\theta }^2}+\frac{k_z}{k_r}\frac{{\partial }^{2}P}{\partial z^2}=\frac{\varphi \mu c_t}{k_r}\frac{\partial P}{\partial t}$$E1.21

$$\frac{1}{r}{\displaystyle \left[\frac{\partial }{\partial r}{\displaystyle \left(r^2\frac{\partial P}{\partial r}\right)}+\frac{1}{\sin \mathrm{\theta }}\frac{\partial }{\partial \mathrm{\theta }}{\displaystyle \left(\sin \mathrm{\theta }\frac{\partial P}{\partial \mathrm{\theta }}\right)}+\frac{1}{{\sin }^2\mathrm{\theta }}\frac{{\partial }^{2}P}{\partial {\mathrm{\varphi }}^2}\right]}=\frac{\mathrm{\varphi }c\mu }{k}\frac{\partial P}{\partial t}$$E1.22

$$\frac{{\partial }^{2}P}{\partial {\xi }^2}+\frac{{\partial }^{2}P}{\partial {\mathrm{\eta }}^2}=\frac{1}{2}{a}^2{\displaystyle \left(\cosh 2\xi -\cos 2\mathrm{\eta }\right)}\frac{\mathrm{\varphi }c\mathrm{\mu }}{k}\frac{\partial P}{\partial t}$$E1.23

Here, ξ is a space coordinate and represents a family of confocal ellipses. The focal length of these ellipses is 2a. The space coordinate, η, represents a family of confocal hyperbolas that represent the streamlines for elliptical flow. These two coordinates are normal to each other.

1.4. Limitations of the diffusivity equation

1. Isotropic, horizontal, homogeneous porous medium, permeability, and constant porosity

2. A single fluid saturates the porous medium

3. Constant viscosity, incompressible, or slightly compressible fluid

4. The well completely penetrates the formation. Negligible gravitational forces

The density of the fluid is governed by an equation of state (EOS). For the case of slightly compressible fluid, Eq. (1.8) is used as the EOS.

1.5. Multiphase flow

Similar to the analysis of gas well tests as will be seen later, multiphase tests can be interpreted using the method of pressure approximation (Perrine method), [6, 7, 16], which is based on phase mobility:

$${\lambda }_t=\frac{k_o}{{\mu }_o}+\frac{k_g}{{\mu }_g}+\frac{k_w}{{\mu }_w}=\frac{k_{ro}}{{\mu }_o}+\frac{k_{rg}}{{\mu }_g}+\frac{k_{rw}}{{\mu }_w}$$E1.24

The total compressibility is defined by [17, 18]:

$$c_t=c_o{S}_o+c_g{S}_g+c_w{S}_w+c_f+\frac{S_o{B}_g}{5.615B_o}\frac{\partial R_s}{\partial P}+\frac{S_w{B}_g}{5.615B_w}\frac{\partial R_{sw}}{\partial P}$$E1.25

For practical purposes, Eq. (1.25) can be expressed as:

$$c_t\approx c_o{S}_o+c_g{S}_g+c_w{S}_w+c_f$$E1.26

As commented before Eq. (1.19) is limited to a single fluid. However, it can be extended to multiphase flow using the concept expressed by Eq. (1.24):

$$\frac{1}{r}\frac{\partial }{\partial r}{\displaystyle \left(r\frac{\partial P}{\partial r}\right)}=\frac{\varphi c_t}{0.0002637}\frac{1}{{\lambda }_t}\frac{\partial P}{\partial t}$$E1.27

Perrine method assumes negligible pressure and saturation gradients. Martin [19] showed that (a) the method loses accuracy as the gas saturation increases, (b) the estimation of the mobility is good, and (c) the mobility calculations are sensitive to the saturation gradients. Better estimates are obtained when the saturation distribution is uniform and (d) underestimates the effective permeability of the phase and overestimates the damage factor.

1.6. Gas flow

It is well known that gas compressibility, gas viscosity, and gas density are highly dependent pressure parameters; then, the liquid diffusivity equation may fail to observe pressure gas behavior. Therefore, there exist three forms for a better linearization of the diffusivity equation to better represent gas flow: (a) the pseudopressure approximation [20], (b) the P2 approximation, and (c) linear approximation. The first one is valid for any pressure range; the second one is valid for reservoir pressures between 2000 and 4000 psia, and the third one is for pressures above 4000 psia [20].

Starting from the equation of continuity and the equation of Darcy:

$$\frac{1}{r}\frac{\partial }{\partial r}{\displaystyle \left(r\rho u_r\right)}=-\frac{\partial }{\partial t}{\displaystyle \left(\mathrm{\varphi }\rho \right)}$$E1.28

$$u_r=-\frac{k}{\mathrm{\mu }}\frac{\partial P}{\partial r}$$E1.29

The state equation for slightly compressible liquids does not model gas flow; therefore, the law of real gases is used [21, 22]:

$$\mathrm{\rho }=-\frac{PM}{zRT}$$E1.30

Combining the above three equations:

$$\frac{1}{r}\frac{\partial }{\partial r}{\displaystyle \left(r\frac{kPM}{\mu zRT}\frac{\partial p}{\partial t}\right)}=\frac{\partial }{\partial t}{\displaystyle \left(\mathrm{\varphi }\frac{PM}{zRT}\right)}$$E1.31

Since M, R, and T are constants and assuming that the permeability is constant, the above equation reduces to:

$$\frac{1}{r}\frac{\partial }{\partial r}{\displaystyle \left(r\frac{P}{\mu z}\frac{\partial P}{\partial r}\right)}=\frac{1}{k}\frac{\partial }{\partial t}{\displaystyle \left(\mathrm{\varphi }\frac{P}{z}\right)}$$E1.32

Applying the differentiation chain rule to the right‐hand side part of Eq. (1.32) leads to:

$$\frac{1}{r}\frac{\partial }{\partial r}{\displaystyle \left(r\frac{P}{\mu z}\frac{\partial P}{\partial r}\right)}=\frac{1}{k}{\displaystyle \left[\frac{P}{z}\frac{\partial \mathrm{\varphi }}{\partial t}+\frac{\partial \mathrm{\varphi }}{\partial t}{\displaystyle \left(\frac{P}{z}\right)}\right]}$$E1.33

Expanding and rearranging,

$$\frac{1}{r}\frac{\partial }{\partial r}{\displaystyle \left(r\frac{P}{\mathrm{\mu }z}\frac{\partial P}{\partial r}\right)}=\frac{P\mathrm{\varphi }}{zk}\frac{\partial P}{\partial t}{\displaystyle \left[\frac{1}{\mathrm{\varphi }}\frac{\partial \mathrm{\varphi }}{\partial P}+\frac{z}{P}\frac{\partial }{\partial P}{\displaystyle \left(\frac{P}{z}\right)}\right]}$$E1.34

Using the definition of compressibility for gas flow:

$$c_g=\frac{1}{\rho }\frac{\partial \rho }{\partial P}=\frac{zRT}{PM}\frac{\partial }{\partial P}{\displaystyle \left(\frac{PM}{zRT}\right)}=\frac{z}{P}\frac{\partial }{\partial P}{\displaystyle \left(\frac{P}{z}\right)}$$E1.35

Using Eqs. (1.9) and (1.35) into Eq. (1.34),

$$\frac{1}{r}\frac{\partial }{\partial r}{\displaystyle \left(r\frac{P}{\mu z}\frac{\partial P}{\partial r}\right)}=\frac{P\mathrm{\varphi }}{zk}\frac{\partial P}{\partial t}{\displaystyle \left(c_f+c_g\right)}$$E1.36

If $c_t=c_g+c_f$ then,

$$\frac{1}{r}\frac{\partial }{\partial r}{\displaystyle \left(r\frac{P}{\mathrm{\mu }z}\frac{\partial P}{\partial r}\right)}=\frac{P\mathrm{\varphi }{c}_t}{zk}\frac{\partial P}{\partial t}$$E1.37

The above is a nonlinear partial differential equation and cannot be solved directly. In general, three limiting assumptions are considered for its solution, namely: (a) P/μz is constant; (b) μc_t is constant; and (c) the pseudopressure transformation, [20], for an actual gas.

1.6.1 The equation of diffusivity in terms of pressure

Assuming the term P/μz remains constant with respect to the pressure, Eq. (1.17) is obtained.

1.6.2 The equation of diffusivity in terms of pressure squared

Eq. (1.37) can be written in terms of squared pressure, P², starting from the fact that, [3, 4, 5, 6, 7, 9, 17, 21, 22]:

$$P\frac{\partial P}{\partial r}=\frac{1}{2}\frac{\partial P^2}{\partial r}$$E1.38

$$P\frac{\partial P}{\partial t}=\frac{1}{2}\frac{\partial P^2}{\partial t}$$E1.39

$$\frac{1}{r}\frac{\partial }{\partial r}{\displaystyle \left(\frac{r}{\mathrm{\mu }z}\frac{\partial P^2}{\partial r}\right)}=\frac{\mathrm{\varphi }{c}_t}{kz}\frac{\partial P^2}{\partial t}$$E1.40

Assuming the term μz remains constant with respect to the pressure, and of course, the radius, then the above equation can be written as:

$$\frac{1}{r}\frac{\partial }{\partial r}{\displaystyle \left(r\frac{\partial P^2}{\partial r}\right)}=\frac{\mathrm{\varphi }\mathrm{\mu }{c}_t}{k}\frac{\partial P^2}{\partial t}$$E1.41

This expression is similar to Eq. (1.37), but the dependent variable is P². Therefore, its solution is similar to Eq. (1.17), except that it is given in terms of P². This equation also requires that μc_t remain constant.

1.6.3 Gas diffusivity equation in terms of pseudopressure, m(P)

The diffusivity equation in terms of P² can be applied at low pressures, and Eq. (1.17) can be applied at high pressures without incurring errors. Therefore, a solution is required that applies to all ranges. Ref. [20] introduced a more rigorous linearization method called pseudopressure that allows the general diffusivity equation to be solved without limiting assumptions that restrict certain properties of gases to remain constant with pressure [3, 4, 5, 6, 7, 9, 17, 20, 21, 22]:

$$m(P)=2{\displaystyle \underset{P_0}{\overset{P}{\int }}\frac{P}{\mathrm{\mu }z}dP}$$E1.42

Taking the derivative with respect to both time and radius and replacing the respective results in Eq. (1.37), we obtain:

$$\frac{1}{r}\frac{\partial }{\partial r}{\displaystyle \left[r\frac{P}{\mathrm{\mu }z}{\displaystyle \left(\frac{\mathrm{\mu }z}{2P}\frac{\partial m(P)}{\partial r}\right)}\right]}=\frac{P\mathrm{\varphi }{c}_t}{zk}{\displaystyle \left(\frac{\mathrm{\mu }z}{2P}\frac{\partial m(P)}{\partial t}\right)}$$E1.43

After simplification,

$$\frac{1}{r}\frac{\partial }{\partial r}{\displaystyle \left(r\frac{\partial m(P)}{\partial r}\right)}=\frac{\varphi \mu c_t}{k}\frac{\partial m(P)}{\partial t}$$E1.44

Expanding the above equation and expressing it in oilfield units:

$$\frac{{\partial }^{2}m(P)}{\partial r^2}+\frac{1}{r}\frac{\partial m(P)}{\partial r}=\frac{\varphi {\mu }_{gi}{c}_t}{0.0002637k_{gi}}\frac{\partial m(P)}{\partial t}$$E1.45

The solution to the above expression is similar to the solution of Eq. (1.17), except that it is now given in terms of m(P) which can be determined by numerical integration if the PVT properties are known at each pressure level.

For a more effective linearization of Eq. (1.45), [23] introduced pseudotime, t_a, since the product μ_gc_t in Eq. (1.45) is not constant:

$$t_a=2{\displaystyle \underset{0}{\overset{t}{\int }}\frac{d\varsigma }{\mu c_t}}$$E1.46

With this criterion, the diffusivity equation for gases is:

$$\frac{1}{r}\frac{\partial }{\partial r}{\displaystyle \left(r\frac{\partial m(P)}{\partial r}\right)}=\frac{2\mathrm{\varphi }(c_f+c_g)}{k\hspace{0.17em}{c}_g}\frac{\partial m(P)}{\partial t_a}$$E1.47

The incomplete linearization of the above expression leads to somewhat longer semilog slopes compared to those obtained for liquids. Sometimes it is recommended to use normalized variables in order to retain the units of time and pressure, [6]. The normalized pseudovariables are:

$$m{(P)}_n=P_i+\frac{{\mu }_i}{{\rho }_i}{\displaystyle \underset{P_0}{\overset{P}{\int }}\frac{\rho (\varsigma )}{\mu (\varsigma )}d\varsigma }$$E1.48

$$t_{an}={\mu }_i{c}_{ti}+{\displaystyle \underset{0}{\overset{t}{\int }}\frac{d\varsigma }{\mu (\varsigma )Z(\varsigma )}}$$E1.49

1.7. Solution to the diffusivity equation

The line‐source solution: The line‐source solution assumes that the wellbore radius approaches zero. Furthermore, the solution considers a reservoir of infinite extent and the well produces as a constant flow rate. Ref. [4] presents the solution of the source line using the Boltzmann transform, the Laplace transform, and Bessel functions. The following is the combinations of independent variables method, which is based on the dimensional analysis of Buckingham’s theorem [24]. This takes a function f = f(x, y, z, t), it must be transformed into a group or function containing fewer variables, f = f(s₁,s₂…). A group of variables whose general form is proposed as [24]:

$$s=a{r}^b{t}^c$$E1.50

The diffusivity equation is:

$$\frac{1}{r}\frac{\partial }{\partial \hspace{0.17em}r}{\displaystyle \left(r\frac{\partial \hspace{0.17em}f}{\partial \hspace{0.17em}r}\right)}=\frac{\partial \hspace{0.17em}f}{\partial \hspace{0.17em}t}$$E1.51

where f is a dimensionless term given by:

$$f=\frac{P-P_{wf}}{P_i-P_{wf}}$$E1.52

Eq. (1.51) is subjected to the following initial and boundary conditions:

$$f=0,0\le r\le \infty ,t=0$$E1.53

$$r\frac{\partial \hspace{0.17em}f}{\partial \hspace{0.17em}r}=1,r=0,t>0$$E1.54

$$f=0,r\to \infty ,t>0$$E1.55

Multiplying the Eq. (1.51) by ∂s/∂s:

$$\frac{1}{r}\frac{\partial \hspace{0.17em}s}{\partial \hspace{0.17em}s}\frac{\partial }{\partial \hspace{0.17em}r}{\displaystyle \left(r\frac{\partial \hspace{0.17em}s}{\partial \hspace{0.17em}s}\frac{\partial \hspace{0.17em}f}{\partial \hspace{0.17em}r}\right)}=\frac{\partial \hspace{0.17em}s}{\partial \hspace{0.17em}s}\frac{\partial \hspace{0.17em}f}{\partial \hspace{0.17em}t}$$E1.56

Exchanging terms:

$$\frac{1}{r}\frac{\partial \hspace{0.17em}s}{\partial \hspace{0.17em}r}\frac{\partial }{\partial \hspace{0.17em}s}{\displaystyle \left(r\frac{\partial \hspace{0.17em}s}{\partial \hspace{0.17em}r}\frac{\partial \hspace{0.17em}f}{\partial \hspace{0.17em}s}\right)}=\frac{\partial \hspace{0.17em}s}{\partial \hspace{0.17em}t}\frac{\partial \hspace{0.17em}f}{\partial \hspace{0.17em}s}$$E1.57

The new derivatives are obtained from Eq. (1.50):

$$\frac{\partial \hspace{0.17em}s}{\partial \hspace{0.17em}r}=ab{r}^{b-1}{t}^c$$E1.58

$$\frac{\partial \hspace{0.17em}s}{\partial \hspace{0.17em}t}=ac{r}^b{t}^{c-1}$$E1.59

Replacing the above derivatives into Eq. (1.56) and rearranging:

$$\frac{1}{r}{a}^2{b}^2\frac{r^b}{r}{t}^{2c}\frac{\partial }{\partial \hspace{0.17em}s}{\displaystyle \left(r\cdot \frac{r^b}{r}\frac{\partial \hspace{0.17em}f}{\partial \hspace{0.17em}s}\right)}=ac{r}^b{t}^{c-1}\frac{\partial \hspace{0.17em}f}{\partial \hspace{0.17em}s}$$E1.60

Solving from r^b from Eq. (1.50) and replacing this result into Eq. (1.6). After rearranging, it yields:

$$\frac{\partial }{\partial s}{\displaystyle \left(s\frac{\partial f}{\partial s}\right)}=\frac{c}{b^2}{\displaystyle \left[r^2{t}^{-1}\right]}s\frac{\partial f}{\partial s}$$E1.61

Comparing the term enclosed in square brackets with Eq. (1.50) shows that b = 2, c = −1, then

$$s=\frac{a{r}^2}{t}\hspace{0.17em}$$E1.62

From Eq. (1.61) follows r²t^‒1 = s/a, then

$$\frac{\partial }{\partial s}{\displaystyle \left(s\frac{\partial f}{\partial s}\right)}={\displaystyle \left[\frac{c}{b^{2}a}\right]}s\frac{\partial f}{\partial s}$$E1.63

The term enclosed in square brackets is a constant that is assumed equal to 1 for convenience. Since c/(b²a) = 1, then a = −1/4. Therefore, the above expression leads to:

$$\frac{\partial }{\partial s}{\displaystyle \left(s\frac{\partial f}{\partial s}\right)}=s\frac{\partial f}{\partial s}$$E1.64

Writing as an ordinary differential equation:

$$\frac{d}{ds}{\displaystyle \left(s\frac{df}{ds}\right)}=s\frac{df}{ds}$$E1.65

The differential equation is now ordinary, and only two conditions are required to solve it. Applying a similar mathematical treatment to both the initial and boundary conditions to convert them into function of s. Regarding Eq. (1.62) and referring to the initial condition, Eq. (1.53), when the time is set to zero; then, then s function tends to infinite:

$$at\hspace{0.17em}t=0,\hspace{0.17em}f=0\hspace{0.17em}\text{when}\hspace{0.17em}s\to \infty$$E1.66

Darcy’s law is used to convert the internal boundary condition. Eq. (1.54) multiplied by ∂s/∂s gives:

$$r\frac{\partial \hspace{0.17em}f}{\partial \hspace{0.17em}s}\frac{\partial \hspace{0.17em}s}{\partial \hspace{0.17em}r}=1$$E1.67

Replacing Eqs. (1.57) in the above equation; then, replacing Eq. (1.62) into the result, and after simplification, we obtain

$$\frac{\partial \hspace{0.17em}f}{\partial \hspace{0.17em}s}ab\frac{s}{a{t}^c}{t}^c=1$$E1.68

Since b = 2, then,

$$s\frac{\partial \hspace{0.17em}f}{\partial \hspace{0.17em}s}=\frac{1}{2}$$E1.69

For the external boundary condition, Eq. (1.55), consider the case of Eq. (1.62) when r → ∞ then:

$$s=\frac{a{r}^2}{t}\hspace{0.17em}\to \infty ;\hspace{0.17em}f=0,\hspace{0.17em}s\to \infty$$E1.70

Then, the new differential equation, Eq. (1.65) is subject to new conditions given by Eqs. (1.66), (1.69), and (1.70). Define now,

$$g=s\frac{df}{ds}$$E1.71

Applying this definition into the ordinary differential expression given by Eq. (1.65), it results:

$$\frac{d}{ds}g=g$$E1.72

Integration of the above expression leads to:

$$ln\hspace{0.17em}g=s+c_1$$E1.73

Rearranging the result and comparing to Eq. (1.71) and applying the boundary condition given by Eq. (1.69):

$$g=c_1{e}^s=s\frac{df}{ds}=\frac{1}{2}$$E1.74

Solving for df and integrating,

$${\displaystyle \int df}=c_1{\displaystyle \int \frac{e^s}{s}ds}$$E1.75

Eq. (1.75) cannot be analytically integrated (solved by power series). Simplifying the solution:

$$f=c_1{\displaystyle \int \frac{e^s}{s}ds}+c_2$$E1.76

When s = 0, e^s = 0, then c₁ = ½ and Eq. (1.76) becomes:

$$f=\frac{1}{2}{\displaystyle \underset{0}{\overset{s}{\int }}\frac{e^s}{s}ds}+c_2$$E1.77

Applying the external boundary condition, Eq. (1.69), when s → ∞, f = 0, therefore, Eq. (1.77) leads,

$$c_2=-\frac{1}{2}{\displaystyle \underset{0}{\overset{-\infty }{\int }}\frac{e^s}{s}ds}$$E1.78

Replacing c₁ and c₂ into Eq. (1.76) yields:

$$f=\frac{1}{2}{\displaystyle \underset{0}{\overset{s}{\int }}\frac{e^s}{s}ds}-\frac{1}{2}{\displaystyle \underset{0}{\overset{-\infty }{\int }}\frac{e^s}{s}ds}$$E1.79

This can be further simplified to:

$$f=-\frac{1}{2}{\displaystyle \underset{s}{\overset{\infty }{\int }}\frac{e^{-s}}{s}ds}$$E1.80

The integral given in Eq. (1.80) is well known as the exponential integral, Ei(−s). If the f variable is changed by pressure terms:

$$P(r,t)=-\frac{1}{2}Ei{\displaystyle \left(\frac{-r^2}{4t}\right)}$$E1.81

In dimensionless form,

$$P_D(r_D,t_D)=-\frac{1}{2}Ei{\displaystyle \left(-\frac{r_D^2}{4t_D}\right)}=-\frac{1}{2}{E}_i{\displaystyle \left(-x\right)}$$E1.82

The above equation is a very good approximation of the analytical solution when it is satisfied (Mueller and Witherspoon [2, 9, 18, 19, 25, 26]) that r_D ≥ 20 or t_D/r_D² ≥ 0.5, see Figure 1.6. If t_D/r_D²≥ 5, an error is less than 2%, and if t_D/r_D² ≥ 25, the error is less than 5%. Figure 1.7 is represented by the following adjustment which has a correlation coefficient, R² of 0.999998. This plot can be easily rebuilt using the algorithm provided in Figure 1.8. The fitted equation was achieved with the data generated from simulation.

Figure 1.6.

Dimensionless pressure for different values of the dimensionless radius, taken from [9, 25].

Figure 1.7.

Dimensionless well pressure behavior for a well without skin and storage effects in an infinite reservoir, taken from [9, 25].

Figure 1.8.

BASIC code function to calculate Ei function, taken from [29].

$$P_D={10}^{\frac{-0.\text{282}0\text{668952451542}+0.\text{447276}00\text{48}0\text{82251}x+0.\text{2581584173632316}{x}^2+0.0\text{499833292759}0\text{892}{x}^3}{1+\text{1}.0\text{47}0\text{15}0\text{81287319}x+0.\text{3493329681392351}{x}^2+0.0\text{295595578818}0\text{784}{x}^3-0.000\text{1636}0\text{472943}0\text{738}{x}^4}}$$E1.83

being x = log(t_D/r_D²) > −1.13.

The exponential function can be evaluated by the following formula, [27], for x ≤ 25:

$$Ei(x)=0.57721557+ln\hspace{0.17em}x-x+\frac{x^2}{\text{2}\cdot 2!}-\frac{x^3}{3\cdot 3!}+\frac{x^4}{4\cdot 4!}\mathrm{\dots .}$$E1.84

Figure 1.8 shows a listing of a program code in Basic, which can be easily added as a function in Microsoft Excel to calculate the exponential function. Figure 1.9 and Table 1.1, 1.2, 1.3, and 1.4 present solutions of the exponential function.

Figure 1.9.

Values of the exponential integral for 1 ≤ x ≤ 10 (left) and 0.0001 ≤ x ≤ 1 (right).

a	b	c	d	e	f
—0.0906765673563653	0.5133959845491270	—0.0243644307428167	—0.0000014346860800	—0.4865489789766050	—
0.7480202919199570	1.3629598993866700	—0.5960091961168400	0.0275653486990893	—0.7768782064908800	−0.0010740336145794

Table 1.1.

Constants for Eqs. (1.85) and (1.86).

x	0	1	2	3	4	5	6	7	8	9
0.000		8.63322	7.94018	7.53481	7.24723	7.02419	6.84197	6.68791	6.55448	6.43680
0.001	6.33154	6.23633	6.14942	6.06948	5.99547	5.92657	5.86214	5.80161	5.74455	5.69058
0.002	5.63939	5.59070	5.54428	5.49993	5.45747	5.41675	5.37763	5.33999	5.30372	5.26873
0.003	5.23493	5.20224	5.17059	5.13991	5.11016	5.08127	5.05320	5.02590	4.99934	4.97346
0.004	4.94824	4.92365	4.89965	4.87622	4.85333	4.83096	4.80908	4.78767	4.76672	4.74620
0.005	4.72610	4.70639	4.68707	4.66813	4.64953	4.63128	4.61337	4.59577	4.57847	4.56148
0.006	4.54477	4.52834	4.51218	4.49628	4.48063	4.46523	4.45006	4.43512	4.42041	4.40591
0.007	4.39162	4.37753	4.36365	4.34995	4.33645	4.32312	4.30998	4.29700	4.28420	4.27156
0.008	4.25908	4.24676	4.23459	4.22257	4.21069	4.19896	4.18736	4.17590	4.16457	4.15337
0.009	4.14229	4.13134	4.12052	4.10980	4.09921	4.08873	4.07835	4.06809	4.05793	4.04788
0.01	4.03793	3.94361	3.85760	3.77855	3.70543	3.63743	3.57389	3.51425	3.45809	3.40501
0.02	3.35471	3.30691	3.26138	3.21791	3.17634	3.13651	3.09828	3.06152	3.02614	2.99203
0.03	2.95912	2.92731	2.89655	2.86676	2.83789	2.80989	2.78270	2.75628	2.73060	2.70560
0.04	2.68126	2.65755	2.63443	2.61188	2.58987	2.56838	2.54737	2.52685	2.50677	2.48713
0.05	2.46790	2.44907	2.43063	2.41255	2.39484	2.37746	2.36041	2.34369	2.32727	2.31114
0.06	2.29531	2.27975	2.26446	2.24943	2.23465	2.22011	2.20581	2.19174	2.17789	2.16426
0.07	2.15084	2.13762	2.12460	2.11177	2.09913	2.08667	2.07439	2.06228	2.05034	2.03856
0.08	2.02694	2.01548	2.00417	1.99301	1.98199	1.97112	1.96038	1.94978	1.93930	1.92896
0.09	1.91874	1.90865	1.89868	1.88882	1.87908	1.86945	1.85994	1.85053	1.84122	1.83202
0.10	1.82292	1.81393	1.80502	1.79622	1.78751	1.77889	1.77036	1.76192	1.75356	1.74529
0.11	1.73711	1.72900	1.72098	1.71304	1.70517	1.69738	1.68967	1.68203	1.67446	1.66697
0.12	1.65954	1.65219	1.64490	1.63767	1.63052	1.62343	1.61640	1.60943	1.60253	1.59568
0.13	1.58890	1.58217	1.57551	1.56890	1.56234	1.55584	1.54940	1.54301	1.53667	1.53038
0.14	1.52415	1.51796	1.51183	1.50574	1.49970	1.49371	1.48777	1.48188	1.47603	1.47022
0.15	1.46446	1.45875	1.45307	1.44744	1.44186	1.43631	1.43080	1.42534	1.41992	1.41453
0.16	1.40919	1.40388	1.39861	1.39338	1.38819	1.38303	1.37791	1.37282	1.36778	1.36276
0.17	1.35778	1.35284	1.34792	1.34304	1.33820	1.33339	1.32860	1.32386	1.31914	1.31445
0.18	1.30980	1.30517	1.30058	1.29601	1.29147	1.28697	1.28249	1.27804	1.27362	1.26922
0.19	1.26486	1.26052	1.25621	1.25192	1.24766	1.24343	1.23922	1.23504	1.23089	1.22676
0.2	1.22265	1.21857	1.21451	1.21048	1.20647	1.20248	1.19852	1.19458	1.19067	1.18677

Table 1.2.

Values of the exponential integral for 0.0001 ≤ x ≤ 0.209.

x	0	1	2	3	4	5	6	7	8	9
4	37.7940000	37.7927530	33.4888052	29.6876209	26.3291192	23.3601005	20.7340078	18.4100584	16.3524950	14.5299393
5	11.4839049	11.4829557	10.2130008	9.0862158	8.0860830	7.1980442	6.4092603	5.7084015	5.0854647	4.5316127
6	3.6017735	3.6008245	3.2108703	2.8637634	2.5547143	2.2794796	2.0342987	1.8158374	1.6211385	1.4475779
7	1.1557663	1.1548173	1.0317127	0.9218812	0.8238725	0.7363972	0.6583089	0.5885877	0.5263261	0.4707165
8	0.3776052	0.3766562	0.3369951	0.3015486	0.2698641	0.2415382	0.2162112	0.1935625	0.1733060	0.1551866
9	0.1254226	0.1244735	0.1114954	0.0998807	0.0894849	0.0801790	0.0718477	0.0643883	0.0577086	0.0517267
10	0.0425187	0.0415697	0.0372704	0.0334186	0.0299673	0.0268747	0.0241031	0.0216191	0.0193925	0.0173966
11	0.0149520	0.0140030	0.0125645	0.0112746	0.0101178	0.0090804	0.0081498	0.0073151	0.0065663	0.0058946
12	0.0057001	0.0047511	0.0042658	0.0038303	0.0034395	0.0030888	0.0027739	0.0024913	0.0022377	0.0020099
13	0.0025709	0.0016219	0.0014570	0.0013090	0.0011761	0.0010567	0.0009495	0.0008532	0.0007667	0.0006890
14	0.0015056	0.0005566	0.0005002	0.0004496	0.0004042	0.0003633	0.0003266	0.0002936	0.0002640	0.0002373
15	0.0011409	0.00019186	0.00017251	0.00015513	0.00013950	0.00012545	0.00011282	0.00010146	9.1257E−05	8.2079E−05
16	0.0010155	6.6405E−09	5.9732E−09	5.3732E−09	4.8336E−09	4.3483E−09	3.9119E−09	3.5194E−09	3.1664E−09	2.8489E−09
17	0.0009725	2.3064E−09	2.0754E−09	1.8675E−09	1.6805E−09	1.5123E−09	1.3609E−09	1.2248E−09	1.1022E−09	9.9202E−10
18	0.0009563	8.0361E−10	7.2331E−10	6.5105E−10	5.8603E−10	5.2752E−10	4.7486E−10	4.2747E−10	3.8482E−10	3.4643E−10
19	0.0009511	2.8078E−10	2.5279E−10	2.2760E−10	2.0492E−10	1.8451E−10	1.6613E−10	1.4959E−10	1.3470E−10	1.2129E−10
20	0.0009526	9.8355E−11	8.8572E−11	7.9764E−11	7.1833E−11	6.4692E−11	5.8263E−11	5.2473E−11	4.7260E−11	4.2566E−11
21	0.0009248	3.4532E−11	3.1104E−11	2.8017E−11	2.5237E−11	2.2733E−11	2.0478E−11	1.8447E−11	1.6617E−11	1.4970E−11
22	0.0009183	1.2149E−11	1.0945E−11	9.8610E−12	8.8842E−12	8.0043E−12	7.2117E−12	6.4976E−12	5.8544E−12	5.2750E−12
23	0.0009464	4.2827E−12	3.8590E−12	3.4773E−12	3.1334E−12	2.8236E−12	2.5444E−12	2.2929E−12	2.0663E−2	1.8621E−12
24	0.0009316	1.5123E−12	1.3629E−12	1.2283E−12	1.1070E−12	9.9772E−13	8.9922E−13	8.1046E−13	7.3048E−13	6.5839E−13
25	0.0000779	5.3489E−13	4.8213E−13	4.3458E−13	3.9172E−13	3.5310E−13	3.1829E−13	2.8692E−13	2.5864E−13	2.3315E−13

Table 1.3.

Values of the exponential integral, Ei(−x) × 10⁻⁴, for 4 ≤ x ≤ 25.9.

x	0	1	2	3	4	5	6	7	8	9
0.20	1.222651	1.182902	1.145380	1.109883	1.076236	1.044283	1.013889	0.984933	0.957308	0.930918
0.30	0.905677	0.881506	0.858335	0.836101	0.814746	0.794216	0.774462	0.755442	0.737112	0.719437
0.40	0.702380	0.685910	0.669997	0.654614	0.639733	0.625331	0.611387	0.597878	0.584784	0.572089
0.50	0.559774	0.547822	0.536220	0.524952	0.514004	0.503364	0.493020	0.482960	0.473174	0.463650
0.60	0.454380	0.445353	0.436562	0.427997	0.419652	0.411517	0.403586	0.395853	0.388309	0.380950
0.70	0.373769	0.366760	0.359918	0.353237	0.346713	0.340341	0.334115	0.328032	0.322088	0.316277
0.80	0.310597	0.305043	0.299611	0.294299	0.289103	0.284019	0.279045	0.274177	0.269413	0.264750
0.90	0.260184	0.255714	0.251337	0.247050	0.242851	0.238738	0.234708	0.230760	0.226891	0.223100
1.00	0.2193840	0.2157417	0.2121712	0.2086707	0.2052384	0.2018729	0.1985724	0.1953355	0.1921606	0.1890462
1.10	0.1859910	0.1829936	0.1800526	0.1771667	0.1743347	0.1715554	0.1688276	0.1661501	0.1635218	0.1609417
1.20	0.1584085	0.1559214	0.1534793	0.1510813	0.1487263	0.1464135	0.1441419	0.1419107	0.1397191	0.1375661
1.30	0.1354511	0.1333731	0.1313314	0.1293253	0.1273541	0.1254169	0.1235132	0.1216423	0.1198034	0.1179960
1.40	0.1162194	0.1144730	0.1127562	0.1110684	0.1094090	0.1077775	0.1061734	0.1045960	0.1030450	0.1015197
1.50	0.1000197	0.0985445	0.0970936	0.0956665	0.0942629	0.0928822	0.0915241	0.0901880	0.0888737	0.0875806
1.60	0.0863084	0.0850568	0.0838252	0.0826134	0.0814211	0.0802477	0.0790931	0.0779568	0.0768385	0.0757379
1.70	0.0746547	0.0735886	0.0725392	0.0715063	0.0704896	0.0694888	0.0685035	0.0675336	0.0665788	0.0656387
1.80	0.0647132	0.0638020	0.0629048	0.0620214	0.0611516	0.0602951	0.0594516	0.0586211	0.0578032	0.0569977
1.90	0.0562045	0.0554232	0.0546538	0.0538960	0.0531496	0.0524145	0.0516904	0.0509771	0.0502745	0.0495824
2.00	0.0489006	0.0482290	0.0475673	0.0469155	0.0462733	0.0456407	0.0450173	0.0444032	0.0437981	0.0432019
2.10	0.0426144	0.0420356	0.0414652	0.0409032	0.0403493	0.0398036	0.0392657	0.0387357	0.0382133	0.0376986
2.20	0.0371912	0.0366912	0.0361984	0.0357127	0.0352340	0.0347622	0.0342971	0.0338387	0.0333868	0.0329414
2.30	0.0325024	0.0320696	0.0316429	0.0312223	0.0308077	0.0303990	0.0299961	0.0295988	0.0292072	0.0288210
2.40	0.0284404	0.0280650	0.0276950	0.0273301	0.0269704	0.0266157	0.0262659	0.0259210	0.0255810	0.0252457
2.50	0.0249150	0.0245890	0.0242674	0.0239504	0.0236377	0.0233294	0.0230253	0.0227254	0.0224296	0.0221380
2.60	0.0218503	0.0215666	0.0212868	0.0210109	0.0207387	0.0204702	0.0202054	0.0199443	0.0196867	0.0194326
2.70	0.0191820	0.0189348	0.0186909	0.0184504	0.0182131	0.0179790	0.0177481	0.0175204	0.0172957	0.0170740
2.80	0.0168554	0.0166397	0.0164269	0.0162169	0.0160098	0.0158055	0.0156039	0.0154050	0.0152087	0.0150151
2.90	0.0148241	0.0146356	0.0144497	0.0142662	0.0140852	0.0139066	0.0137303	0.0135564	0.0133849	0.0132155
3.00	0.0130485	0.0128836	0.0127209	0.0125604	0.0124020	0.0122457	0.0120915	0.0119392	0.0117890	0.0116408
3.10	0.0114945	0.0113502	0.0112077	0.0110671	0.0109283	0.0107914	0.0106562	0.0105229	0.0103912	0.0102613
3.20	0.0101331	0.0100065	0.0098816	0.0097584	0.0096367	0.0095166	0.0093981	0.0092811	0.0091656	0.0090516
3.30	0.0089391	0.0088281	0.0087185	0.0086103	0.0085035	0.0083981	0.0082940	0.0081913	0.0080899	0.0079899
3.40	0.0078911	0.0077935	0.0076973	0.0076022	0.0075084	0.0074158	0.0073244	0.0072341	0.0071450	0.0070571
3.50	0.0069702	0.0068845	0.0067999	0.0067163	0.0066338	0.0065524	0.0064720	0.0063926	0.0063143	0.0062369
3.60	0.0061605	0.0060851	0.0060106	0.0059371	0.0058645	0.0057929	0.0057221	0.0056523	0.0055833	0.0055152
3.70	0.0054479	0.0053815	0.0053160	0.0052512	0.0051873	0.0051242	0.0050619	0.0050003	0.0049396	0.0048796
3.80	0.0048203	0.0047618	0.0047041	0.0046470	0.0045907	0.0045351	0.0044802	0.0044259	0.0043724	0.0043195
3.90	0.0042672	0.0042157	0.0041647	0.0041144	0.0040648	0.0040157	0.0039673	0.0039194	0.0038722	0.0038255
4.00	0.0037794	0.0037339	0.0036890	0.0036446	0.0036008	0.0035575	0.0035148	0.0034725	0.0034308	0.0033896

Table 1.4.

Values of the exponential integral for 0.1 ≤ x ≤ 4.09.

1.8. Dimensionless quantities

Dimensional parameters do not provide a physical view of the parameter being measured but rather a general or universal description of these parameters. For example, a real time of 24 hours corresponds to a dimensionless time of approximately 300 hours in very low permeability formations or more than 10⁷ in very permeable formations [3, 9, 21, 25, 28].

A set number of Ei values for 0.0001 ≤ x ≤ 25 with the aid of the algorithm given in Figure 1.8. Then, a fitting of these data was performed to obtain the polynomials given by Eqs. (1.85) and (1.90). The first one has a R² of 1, and the second one has a R² of 0.999999999 which implies accuracy up to the fifth digit can be obtained.

$$Ei(-x)=a+bx+c{x}^{2.5}+dlnx+e\cdot \text{exp}(-x);\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x\le 1$$E1.85

$$lnEi(-x)=\frac{a+cx+e{x}^2}{1+bx+d{x}^2+f{x}^3};\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x>1$$E1.86

Adapted from [29] and generated with the Ei function code given in Figure 1.8.

Define dimensionless radius, dimensionless time, and dimensionless pressure as:

$$r_D=r/r_w$$E1.87

$$t_D=\frac{t}{t_o}$$E1.88

$$P_D=\frac{kh(P_i-P)}{141.2q\mu B}$$E1.89

Adapted from [29] and generated with the Ei function code given in Figure 1.8.

For pressure drawdown tests, ΔP = P_i − P_wf. For pressure buildup tests, ΔP = P_ws − P_wf (Δt = 0).

This means that the steady‐state physical pressure drop for radial flow is equal to the dimensionless pressure multiplied by a scalable factor, which in this case depends on the flow and the properties of the reservoir, [3, 4, 5, 6, 7, 9, 21, 26, 30]. The same concept applies to transient flow and to more complex situations, but in this case, the dimensionless pressure is different. For example, for transient flow, the dimensionless pressure is always a function of dimensionless time.

Taking derivative to Eqs. (1.87) and (1.88),

$$\partial \hspace{0.17em}r=r_w\partial \hspace{0.17em}{r}_D$$E1.90

$$\partial \hspace{0.17em}t=t_o\partial \hspace{0.17em}{t}_D$$E1.91

Replacing the above derivatives into Eq. (1.20),

Adapted from [5] and generated with the Ei function code given in Figure 1.8.

$$\frac{\partial {\hspace{0.17em}}^{2}P}{\partial \hspace{0.17em}\hspace{0.17em}{r}_D^2}+\frac{1}{\hspace{0.17em}{r}_D^{}}\frac{\partial \hspace{0.17em}P}{\partial \hspace{0.17em}\hspace{0.17em}{r}_D^{}}=\frac{\varphi \hspace{0.17em}\mu \hspace{0.17em}{c}_t\hspace{0.17em}{r}_w^2}{k{t}_o}\frac{\partial \hspace{0.17em}P}{\partial \hspace{0.17em}\hspace{0.17em}{t}_D}$$E1.92

Definition of t_o requires assuming $\frac{\varphi \mu c_t{r}_w^2}{k{t}_o}$ = 1, [24], then;

$$t_o=\frac{\varphi \hspace{0.17em}\mu \hspace{0.17em}{c}_t\hspace{0.17em}{r}_w^2}{k}$$E1.93

Replacing this definition into Eq. (1.88) and solving for the dimensionless time (oilfield units),

$$t_D=\frac{0.0002637kt}{\varphi \mu c_t{r}_w^2}$$E1.94

Replacing Eq. (1.93) in Eq. (1.92) leads, after simplification, to:

$$\frac{\partial {\hspace{0.17em}}^{2}P}{\partial \hspace{0.17em}\hspace{0.17em}{r}_D^2}+\frac{1}{\hspace{0.17em}{r}_D^{}}\frac{\partial \hspace{0.17em}P}{\partial \hspace{0.17em}\hspace{0.17em}{r}_D^{}}=\frac{\partial \hspace{0.17em}P}{\partial \hspace{0.17em}\hspace{0.17em}{t}_D}$$E1.95

The dimensionless pressure is also affected by the system geometry, other well systems, storage coefficient, anisotropic characteristics of the reservoir, fractures, radial discontinuities, double porosity, among others. In general, the pressure at any point in a single well system that produces the constant rate, q, is given by [25]:

$$[P_i-P(r,t)]=\frac{qB\mu }{\hspace{0.17em}kh}{P}_D(t_D,\hspace{0.17em}{r}_D,\hspace{0.17em}{C}_D,\hspace{0.17em}\text{geometry},\mathrm{\dots .})$$E1.96

Taking twice derivative to Eq. (1.87), excluding the conversion factor, will provide:

$$\partial \hspace{0.17em}{P}_D=-\frac{kh}{\hspace{0.17em}qB\mu }\partial P$$E1.97

$$\partial {\hspace{0.17em}}^2{P}_D=-\frac{kh}{\hspace{0.17em}qB\mu }{\partial }^{2}P$$E1.98

Replacing Eqs. (1.97) and (1.98) in Eq. (1.95) and simplifying leads to:

$$\frac{{\partial }^2{P}_D}{\partial \hspace{0.17em}{r}_D^2}+\frac{1}{\hspace{0.17em}{r}_D^{}}\frac{\partial P_D}{\partial \hspace{0.17em}{r}_D^{}}=\frac{1}{r_D}\frac{\partial }{\partial r_D}{\displaystyle \left(r_D\frac{\partial P_D}{\partial r_D}\right)}=\frac{\partial P_D}{\partial t_D}$$E1.99

If the characteristic length is the area, instead of wellbore radius, Eq. (1.92) can be expressed as:

$$t_{DA}=\frac{0.0002637kt}{\varphi \hspace{0.17em}\mu \hspace{0.17em}{c}_t\hspace{0.17em}A}=t_D{\displaystyle \left(\frac{r_w^2}{A}\right)}$$E1.100

Example 1.1

A square shaped reservoir produces 300 BPD through a well located in the center of one of its quadrants. See Figure 1.10. Estimate the pressure in the well after 1 month of production. Other relevant data:

Figure 1.10.

Geometry of the reservoir for example 1.1.

P_i = 3225 psia,            h = 42 ft

k_o = 1 darcy,         ϕ = 25%

μ_o = 25 cp,           c_t = 6.1 × 10⁻⁶/psia

B_o = 1.32 bbl/BF, r_w = 6 in

A = 150 Acres,  q = 300 BPD

Solution

Assuming the system behaves infinitely, it means, during 1 month of production the transient wave has not yet reached the reservoir boundaries, the problem can be solved by estimating the Ei function. Replacing Eqs. (1.82) and (1.92) into the argument of Eq. (1.82), it results:

$$x=-\frac{r_D^2}{4t_D}=-\frac{948\varphi \mu c_t{r}^2}{kt}$$E1.101

Using Eq. (1.101) with the above given reservoir and well data:

$$x=-\frac{948(0.25)(25)(6.1\times {10}^{-6})({0.5}^2)}{(1000)(720)}=1.25\times {10}^{-8}$$

This x value allows finding Ei(−x) = 17.6163 using the function provided in Figure 1.8. From the application of Eq. (82), P_D = 8.808. This dimensionless pressure is meaningless for practical purposes. Converting to oilfield units by means of Eq. (1.87), the well‐flowing pressure value after 1 month of production is given as:

$$8.808=\frac{(1000)(42)}{(141.2)(300)(1.32)(25)}(3225-P_{wf})$$

P_wf = 2931.84 psia.

How it can be now if the example was correctly done? A good approximation consists of considering a small pressure drop; let us say ± 0.002 psia (smallest value that can be read from current pressure recorders) at the closest reservoir boundary. Use Eq. (1.87) to convert from psia to dimensionless pressure:

$$P_D=\frac{(1000)(42)}{(141.2)(300)(1.32)(25)}(0.002)=6.0091\times {10}^{-5}$$

Eq. (1.82) allows finding Ei(−x) = 0.00012. This value can be used to determine an x value from Table 1.2. However, a trial‐and‐error procedure with the function given in Figure 1.8 was performed to find an x value of 6.97. Then, the time at which this value takes place at the nearest reservoir boundary is found from Eq. (1.101). The nearest boundary is obtained from one‐fourth of the reservoir size area (3.7 Ac or 1663500 ft²). Then, for a square geometry system (the system may also be approached to a circle):

$$L=\sqrt{\text{16635}00}=1278.09\hspace{0.17em}\text{ft}$$

The radial distance from the well to the nearest boundary corresponds to one half of the square side, the r = 639.04 ft. Solving for time from Eq. (1.101);

$$t=\frac{948\varphi \mu c_t{r}^2}{kx}=\frac{948(0.25)(25)(6.1\times {10}^{-6})({639.04}^2)}{(1000)(6.97)}=2.118\hspace{0.17em}\hspace{0.17em}\text{h}$$

This means that after 2 h and 7 min of flow, the wave has reached the nearest reservoir boundary; therefore, the infinite‐acting period no longer exists for this reservoir, then, a pseudosteady‐state solution ought to be applied (Figures 1.11–1.14). To do so, Eq. (1.98) is employed for the whole reservoir area:

$$t_{DA}=\frac{(0.0002637)(1000)(720)}{(0.25)(25)(6.1\times {10}^{-6})(6534000)}=0.76$$

With this t_DA value of 0.76, the normal procedure is to estimate the dimensionless pressure for a given reservoir‐well position configuration, which can be found in Figures C.13 through C.16 in [25] for which data were originally presented in [31]. These plots provide the pressure behavior for a well inside a rectangular/square no-flow system, without storage wellbore and skin factor; A^0.5/r_w = 2000 can also be found in [3, 9, 26]. This procedure is avoided in this textbook. Instead new set of data was generated and adjusted to the following polynomial fitting in which constants are reported in Table 1.5:

$$P_D=a+b*t_{DA}+c*t_{DA}^2+d*t_{DA}^{0.5}ln{t}_{DA}^{}+\frac{e}{t_{DA}^{0.5}}$$E1.102

Using Eq. (1.102) will result:

$$P_D=\text{4}.\text{4765}+\text{9}.\text{3437(12)}-0.{\text{2798(12}}^2)-\text{2}.\text{7516}\sqrt{12}ln(12)-\frac{0.0\text{16}0\text{98}}{\sqrt{12}}$$

Table 1.5.

Constants for Eq. (1.102).

P_D = 12.05597.

The well‐flowing pressure is estimated with Eq. (1.87); thus,

$$12.056=\frac{(1000)(42)}{(141.2)(300)(1.32)(25)}(P_i-P_{wf})$$

P_wf = 2823.75 psia.

1.9. Application of the diffusivity equation solution

A straight‐line behavior can be observed in mostly the whole range on the right‐hand plot of Ei versus x plot given in Figure 1.9. Then, it was concluded, [3, 4, 5, 6, 7, 9, 11, 19, 21, 26, 30], when x < 0.0025, the more complex mathematical representation of Eq. (1.82) can be replaced by a straight line function, given by:

$$E_i(-x)=ln(1.781x)$$E1.103

this leads to,

$$E_i(-x)=ln\hspace{0.17em}x+0.5772$$E1.104

Replacing this new definition into Eq. (1.82) will result in:

$$P_D=-\frac{1}{2}{\displaystyle \left[ln{\displaystyle \left(\frac{r_D^2}{4t_D}\right)}+0.5772\right]}$$E1.105

At the well r_D = 1, after rearranging,

$$P_D=\frac{1}{2}{\displaystyle \left[ln{t}_D+0.80907\right]}$$E1.106

The above indicates that the well pressure behavior obeys a semi‐logarithmic behavior of pressure versus time.

Example 1.2

A well and infinite reservoir has the following characteristics:

q = 2000 STB/D,   μ = 0.72 cp,   c_t = 1.5 × 10⁻⁵ psia⁻¹

ϕ = 23%,          P_i = 3000 psia,         h = 150 ft

B = 1.475 bbl/STB,     k = 10 md,       r_w = 0.5 ft

Estimate the well‐flowing pressure at radii of 0.5, 1, 5, 10, 20, 50, 70, 100, 200, 500, 1000, 2000, 2500, 3000, and 4000 feet after 1 month of production. Plot the results.

Solution

For the wellbore radius, find x with Eq. (1.101);

$$x=\frac{948(0.23)(0.72)(1.5\times {10}^{-5})({0.5}^2)}{(10)(720)}=8.177\times {10}^{-8}$$

Using the function given in Figure 1.9 or Eq. (1.103), a value of Ei(−x) of 15.7421 is found. Then, Eq. (1.82) indicates that P_D = 7.871. Use of Eq. (1.87) allows estimating both pressure drop and well‐flowing pressure:

$$\Delta P=P_i-P_{wf}=\frac{141.2q\mu B}{kh}{P}_D=\frac{141.2(2000)(0.72)(1.475)}{(10)(150)}7.871=1573.74\text{psia}$$

The remaining results are summarized in Table 1.6 and plotted in Figure 1.11. From this, it can be inferred that the highest pressure drop takes place in the near‐wellbore region which mathematically agrees with the continuity equation stating that when the area is reduced, the velocity has to be increased so the flow rate can be constant. The higher the fluid velocity, the higher the pressure drops.

r, ft	x	Ei(−x)	P, psia	Pwf, psia
0.5	8.18E−08	15.7421	1537.74	1462.26
1	3.27E−07	14.3558	1435.15	1564.85
5	8.18E−06	11.137	1113.36	1886.64
10	3.27E−04	9.75	974.78	2025.22
20	1.31E−04	8.365	836.2	2163.8
50	8.18E−04	6.533	653.07	2346.93
70	1.60E−03	5.86	585.87	2414.13
100	3.27E−03	5.149	514.72	2485.28
200	1.31E−02	3.772	377.11	2622.89
500	8.17E−02	2.007	200.616	2799.384
1000	3.27E−01	0.8425	84.225	2915.775
2000	1.31E+00	0.1337	13.368	2986.632
2500	2.04E+00	0.046	4.6	2995.4
3000	2.94E+00	0.014	1.401	2998.599
4000	5.23E+00	0.0009	0.087	2999.913

Table 1.6.

Summarized results for example 1.2.

Figure 1.11.

Pressure versus distance plot for example 1.2.

Example 1.3

Re‐work example 1.2 to estimate the sand‐face pressure at time values starting from 0.01 to 1000 h. Show the results in both Cartesian and semilog plots. What does this suggest?

Solution

Find x with Eq. (1.101);

$$x=\frac{948(0.23)(0.72)(1.5\times {10}^{-5})({0.5}^2)}{(10)(0.01)}=0.000948$$

A value of Ei(−x) of 6.385 is found with Eq. (1.103). Then, Eq. (1.82) gives a P_D value of 3.192 and Eq. (1.87) leads to calculate a well‐flowing pressure of;

$$P_{wf}=P_i-\frac{141.2q\mu B}{kh}{P}_D=3000-\frac{141.2(2000)(0.72)(1.475)}{(10)(150)}3.192=2361.71\hspace{0.17em}\text{psia}$$

The remaining well‐flowing pressure values against time are given in Table 1.7 and plotted in Figure 1.12. The semilog behavior goes in the upper part of the plot (solid line), and the Cartesian plot corresponds to the lower dashed line. The semilog line behaves linearly while the Cartesian curve does not. This situation perfectly agrees with Eq. (1.106), which ensures that the behavior of pressure drop versus time obeys a semilog trend. In other word, in a transient radial system, pressure drops is a linear function of the logarithm of time.

t, h	x	Ei(−x)	PD	Pwf, Psia	t, h	x	Ei(−x)	PD	Pwf, psia
0.01	9.480E−04	6.385	3.192	2361.71	6	1.580E−06	12.781	6.390	1722.30
0.02	4.740E−04	7.078	3.539	2292.46	7	1.354E−06	12.935	6.468	1706.89
0.03	3.160E−04	7.483	3.741	2251.94	8	1.185E−06	13.069	6.534	1693.54
0.04	2.370E−04	7.770	3.885	2223.19	9	1.053E−06	13.186	6.593	1681.77
0.05	1.896E−04	7.994	3.997	2200.89	10	9.480E−07	13.292	6.646	1671.23
0.06	1.580E−04	8.176	4.088	2182.66	20	4.740E−07	13.985	6.992	1601.94
0.07	1.354E−04	8.330	4.165	2167.25	30	3.160E−07	14.390	7.195	1561.41
0.08	1.185E−04	8.464	4.232	2153.91	40	2.370E−07	14.678	7.339	1532.65
0.09	1.053E−04	8.581	4.291	2142.13	50	1.896E−07	14.901	7.451	1510.34
0.1	9.480E−05	8.687	4.343	2131.60	60	1.580E−07	15.083	7.542	1492.11
0.2	4.740E−05	9.380	4.690	2062.31	70	1.354E−07	15.238	7.619	1476.70
0.3	3.160E−05	9.785	4.893	2021.78	80	1.185E−07	15.371	7.686	1463.35
0.4	2.370E−05	10.073	5.036	1993.02	90	1.053E−07	15.489	7.744	1451.58
0.5	1.896E−05	10.296	5.148	1970.71	100	9.480E−08	15.594	7.797	1441.05
0.6	1.580E−05	10.478	5.239	1952.49	200	4.740E−08	16.287	8.144	1371.75
0.7	1.354E−05	10.632	5.316	1937.08	300	3.160E−08	16.693	8.346	1331.22
0.8	1.185E−05	10.766	5.383	1923.73	400	2.370E−08	16.981	8.490	1302.46
0.9	1.053E−05	10.884	5.442	1911.95	500	1.896E−08	17.204	8.602	1280.15
1	9.480E−06	10.989	5.495	1901.42	600	1.580E−08	17.386	8.693	1261.92
2	4.740E−06	11.682	5.841	1832.13	700	1.354E−08	17.540	8.770	1246.51
3	3.160E−06	12.088	6.044	1791.59	800	1.185E−08	17.674	8.837	1233.17
4	2.370E−06	12.375	6.188	1762.84	900	1.053E−08	17.792	8.896	1221.39
5	1.896E−06	12.599	6.299	1740.53	1000	9.480E−09	17.897	8.948	1210.86

Table 1.7.

Summarized results for example 1.3.

Figure 1.12.

Pressure versus time plot for example 1.3.

1.10. Pressure distribution and skin factor

Once the dimensionless parameters are plugged in Eq. (1.82), this yields:

$$P(r,t)=P_i-70.6\frac{qB\mu }{kh}{E}_i{\displaystyle \left\{-\frac{948\varphi \mu c_t{r}^2}{kt}\right\}}$$E1.107

At point N, Figure 1.13, the pressure can be calculated by Eq. (1.107). At the wellbore r_D = r/r_w = 1, then, r = r_w and P(r,t) = P_wf. Note that application of the line‐source solution requires the reservoir to possess an infinite extent, [3, 9, 18, 21, 25, 26].

Figure 1.13.

Pressure distribution in the reservoir.

There are several ways to quantify damage or stimulation in an operating well (producer or injector). These conditions are schematically represented in Figure  1.14. The most popular method is to represent a well condition by a steady‐state pressure drop occurring at the wellbore, in addition to the transient pressure drop normally occurring in the reservoir. This additional pressure drop is called “skin pressure drop” and takes place in an infinitesimally thin zone: “damage zone,” [4, 5, 9, 11, 19, 30]. It can be caused by several factors:

Figure 1.14.

Skin factor influence.

1. Invasion of drilling fluids

2. Partial well penetration

3. Partial completion

4. Blocking of perforations

5. Organic/inorganic precipitation

6. Inadequate drilling density or limited drilling

7. Bacterial growth

8. Dispersion of clays

9. Presence of cake and cement

10. Presence of high gas saturation around the well

Skin factor is a dimensionless parameter; then, it has to be added to the dimensionless pressure in Eq. (1.87), so that:

$$P_i-P_{wf}=141.2\frac{q\mu B}{kh}{\displaystyle \left(P_D+s\right)}$$E1.108

From the above expression can be easily obtained:

$$P_i-P_{wf}=141.2\frac{q\mu B}{kh}{P}_D+141.2\frac{q\mu B}{kh}s$$E1.109

Therefore, the skin factor pressure drop is given by:

$$\Delta P_s=141.2\frac{q\mu B}{kh}s$$E1.110

Assuming steady state near the wellbore and the damage area has a finite radius, r_s, with an altered permeability, k_s, the pressure drop due to the damage is expressed as the pressure difference between the virgin zone and the altered zone, that is to say:

$$\Delta P_s=141.2\frac{q\mu B}{k_{s}h}ln\frac{r_s}{r_w}-141.2\frac{q\mu B}{kh}ln\frac{r_s}{r_w}$$E1.111

Rearranging;

$$\Delta P_s=141.2\frac{q\mu B}{k_{}h}{\displaystyle \left(\frac{k}{k_s}-1\right)}ln\frac{r_s}{r_w}$$E1.112

Comparing Eqs (1.112) and (1.107), the following can be concluded:

$$s={\displaystyle \left(\frac{k}{k_s}-1\right)}ln\frac{r_s}{r_w}$$E1.113

r_s and k_s are not easy to be obtained.

Equation (1.82) and (1.106) can be respectively written as:

$$P_D+s=-\frac{1}{2}{E}_i{\displaystyle \left(-x\right)}$$E1.114

$$P_D+s=\frac{1}{2}{\displaystyle \left[ln{t}_D+0.80907\right]}$$E1.115

Replacing the dimensionless quantities given by Eqs. (1.87) and (1.95) in Eq. (1.115) will result:

$$P_i=P_{wf}+\frac{70.6q\mu B}{kh}{\displaystyle \left[ln{\displaystyle \left(\frac{0.0002637kt}{\varphi \mu c_t{r}_w^2}\right)}+0.80908+2s\right]}$$E1.116

Taking natural logarithm to 0.0002637 and adding its result to 0.80908 results in:

$$P_i=P_{wf}+\frac{70.6q\mu B}{kh}{\displaystyle \left[-7.4316+ln{\displaystyle \left(\frac{kt}{\varphi \mu c_t{r}_w^2}\right)}+2s\right]}$$E1.117

Multiplying and dividing by the natural logarithm of 10 and solving for the well‐flowing pressure:

$$P_{wf}=P_i-\frac{162.6q\mu B}{kh}{\displaystyle \left[\log {\displaystyle \left(\frac{kt}{\varphi \mu c_t{r}_w^2}\right)}-3.2275+0.8686s\right]}$$E1.118

Thus, a straight line is expected to develop from a semilog plot of pressure against the time, as seen on the upper curve of Figure 1.12.

1.11. Finite reservoirs

In closed systems, the radial flow is followed by a transition period. This in turn is followed by the pseudosteady, semi‐stable, or quasi‐stable state, which is a transient flow regime where the pressures change over time, dP/dt, is constant at all points of the reservoir:

$$\frac{dP}{dt}=\frac{-q}{c{V}_p}$$E1.119

Eq. (1.99) is now subjected to the following initial and boundary conditions:

$$P_D{\displaystyle \left(r_D,t_D=0\right)}=0$$E1.120

$${{\displaystyle \left(\frac{\partial P_D}{\partial r_D}\right)}}_{r_{eD}}=0$$E1.121

$${{\displaystyle \left(\frac{\partial P_D}{\partial r_D}\right)}}_{r_D=1}=-1$$E1.122

Which solution is [9, 30]:

$$\begin{array}{l}{P}_D{\displaystyle \left(r_D,t_D\right)}=\frac{2}{{\displaystyle \left(r_{eD}^2-1\right)}}{\displaystyle \left(\frac{r_D^2}{4}+t_D\right)}-\frac{r_{eD}^{2}ln{r}_D}{{\displaystyle \left(r^2{}_{eD}-1\right)}}-\text{}\frac{{\displaystyle \left(3r_{eD}^4-4r_{eD}^{4}ln{r}_{eD}-2r_{eD}^2-1\right)}}{4{{\displaystyle \left(r_{eD}^2-1\right)}}^2}\\ +\pi {\displaystyle \sum _{n=1}^{\infty }{\displaystyle \{\frac{e^{-a_n^2{t}_D}\hspace{0.17em}{J}_1^2{\displaystyle \left(a_n{r}_{eD}\right)}\text{[}{J}_1{\displaystyle \left(a_n\right)}{Y}_0{\displaystyle \left(a_n{r}_D\right)}-Y_1{\displaystyle \left(a_n\right)}{\displaystyle \left(J_0\right)}{\displaystyle \left(a_n{r}_D\right)}\text{]}}{a_n\text{[}{J}_1^2{\displaystyle \left(a_n{r}_{eD}\right)}-J_1^2{\displaystyle \left(a_n\right)}\text{]}}\}}}\end{array}$$E1.123

The pseudosteady‐state period takes place at late times (t > 948ϕμc_tr_e²/k), so that as time tends to infinity, summation tends to zero, then:

$$P_D{\displaystyle \left(r_D,t_D\right)}=\frac{2}{{\displaystyle \left(r_{eD}^2-1\right)}}{\displaystyle \left(\frac{r_D^2}{4}+t_D\right)}-\frac{r_{eD}^{2}ln{r}_D}{{\displaystyle \left(r_{eD}^2-1\right)}}-\text{}\frac{{\displaystyle \left(3r_{eD}^4-4r_{eD}^{4}ln{r}_{eD}-2r_{eD}^2-1\right)}}{4{{\displaystyle \left(r_{eD}^2-1\right)}}^2}$$E1.124

At the well, r_D = 1 and as r_eD >>>> 1, the above expression is reduced to:

$$P_D(t_D)=\frac{2}{r_{eD}^2}+\frac{2t_D}{r_{eD}^2}-\frac{3}{4}+ln{r}_{eD}+\frac{1}{2r_{eD}^2}-\frac{1}{4r_{eD}^4}$$E1.125

This can be approximated to:

$$P_D(t_D)\cong \frac{2t_D}{r_{eD}^2}+ln{r}_{eD}-\frac{3}{4}$$E1.126

Invoking Eq. (1.98) for a circular reservoir area,

$$t_{DA}=t_D\frac{r_w^2}{\pi r_e^2}=\frac{t_D}{\pi r_{eD}^2}$$E1.127

It follows that;

$$\pi t_{DA}=\hspace{0.17em}\hspace{0.17em}\frac{t_D}{r_{eD}^2}$$E1.128

The final solution to the pseudosteady‐state diffusivity equation is obtained from using the definition given by Eq. (1.128) in Eq. (1.129):

$$P_D(t_D)=2\pi t_{DA}^{}+ln{r}_{eD}-\frac{3}{4}$$E1.129

The derivative with respect to time of the above equation in dimensional form allows obtaining the pore volume:

$$\frac{dP(r,t)}{dt}=-\frac{1.79qB}{h\varphi c_t{r}_e^2}$$E1.130

An important feature of this period is that the rate of change of pressure with respect to time is a constant, that is, dP_D/dt_DA = 2π.

When the reservoir pressure does not change over time at any point, the flow is said to be stable. In other words, the right side of Eq. (1.99) is zero, [3]:

$$\frac{1}{r_D}\frac{\partial }{\partial r_D}{\displaystyle \left(r_D\frac{\partial P_D}{\partial r_D}\right)}=0$$E1.131

Similar to the pseudosteady‐state case, steady state takes place at late times. Now, its initial, external, and internal boundary conditions are given by:

$$P_D{\displaystyle \left(r_D,t_D=0\right)}=0$$E1.132

$$P_D{\displaystyle \left(r_{De},0\right)}=0$$E1.133

$${{\displaystyle \left(\frac{\partial P_D}{\partial r_D}\right)}}_{r_D=1}=-1$$E1.134

The solution to the steady‐state diffusivity equation is [3]:

$$P_D{\displaystyle \left(r_D,t_D\right)}=ln\hspace{0.17em}{r}_{eD}-2{\displaystyle \sum _{n=1}^{\infty }{\displaystyle \left\{\frac{e^{-{\mathrm{\beta }}_n^2{t}_{D_n}{J}_0^2({\mathrm{\beta }}_n^{}{r}_{eD})}}{{\mathrm{\beta }}_n^2{\displaystyle \left[J_1^2{\displaystyle \left({\mathrm{\beta }}_n^{}\right)}-J_0^2{\displaystyle \left({\mathrm{\beta }}_n^{}{r}_{eD}\right)}\right]}}\right\}}}$$E1.135

As time tends to infinity, the summation tends to infinity, then:

$${(P_D)}_{ssr}=ln{r}_{eD}=ln\frac{r_e}{r_w}$$E1.136

In dimensional terms, the above expression is reduced to Darcy’s equation. The dimensionless pressure function for linear flow is given by:

$${(P_D)}_{ssL}=2\pi \frac{Lh}{A}$$E1.137

Steady state can occur in reservoirs only when the reservoir is fully recharged by an aquifer or when injection and production are balanced. However, a reservoir with a very active aquifer will not always act under steady‐state conditions. First, there has to be a period of unsteady state, which will be followed by the steady state once the pressure drop has reached the reservoir boundaries. Extraction of fluids from a pressurized reservoir with compressible fluids causes a pressure disturbance which travels throughout the reservoir. Although such disturbance is expected to travel at the speed of sound, it is rapidly attenuated so that for a given duration of production time, there is a distance, the drainage radius, beyond which no substantial changes in pressure will be observed. As more fluid is withdrawn (or injected), the disturbance moves further into the reservoir with continuous pressure decline at all points that have experienced pressure decline. Once a reservoir boundary is found, the pressure on the boundary continues to decline but at a faster rate than when the boundary was not detected. On the other hand, if the pressure transient reaches an open boundary (water influx), the pressure remains constant at some point; the pressure closest to the well will decline more slowly than if a closed boundary were found. Flow changes or the addition of new wells cause additional pressure drops that affect both the pressure decline and the pressure distribution. Each well will establish its own drainage area that supplies fluid. When a flow boundary is found, the pressure gradient—not the pressure level—tends to stabilize after sufficiently long production time. For the closed boundary case, the pressure reaches the pseudosteady state with a constant pressure gradient and general pressure drop everywhere, which is linear over time. For constant‐pressure boundaries, steady state is obtained; both the pressure and its gradient remain constant over time.

1.12. The pressure derivative function

Pressure derivative has been one of the most valuable tools ever introduced to the pressure transient analysis field. In fact, [32] affirms that pressure derivative and deconvolution have been the best elements added for well test interpretation. However, here it is affirmed that besides these two “blessings,” TDS technique, [1, 2], is the best and practical well test interpretation method in which application will be very devoted along this textbook. Actually, in the following chapters, TDS is extended for long, homogeneous reservoirs, [33], interference testing [34], drainage area determination in constant‐pressure reservoirs, [35], and recent applications on fractured vertical wells, [36], among others. More complex scenarios, for instance finite‐conductivity faults, [37], are treated extensively in [38].

Attempts to introduce the pressure derivative are not really new. Some of them try to even apply the derivative concept to material balance. Just to name a few of them, [39] in 1961, tried to approach the rate of pressure change with time for detection of reservoir boundaries. Later, in 1965, [40] presented drawdown curves of well pressure change with time for wells near intersecting faults (36 and 90°). These applications, however, use numerical estimations of the pressure rate change on the field data regardless of two aspects: (1) an understanding of the theoretical situation behind a given system and (2) noise in the pressure data.

Between 1975 and 1976, Tiab’s contributions on the pressure derivative were remarkable. Actually, he is the father of the pressure derivative concept as used nowadays. Refs. [41, 42] include detailed derivation and application of the pressure derivative function. These results are further summarized on [41, 42, 43, 44, 45]. Ref. [46] applied Tiab’s finding to provide a type‐curve matching technique using the natural logarithm pressure derivative.

It was required to obtain the pressure derivative from a continuous function, instead of attempting to work on discrete data in order to understand the pressure derivative behavior in an infinite system. Then, Tiab decided to apply the Leibnitz’s rule of derivation of an integral to the Ei function.

$$\frac{\partial }{\partial x}{\displaystyle {\int }_{f(x)}^{h(x)}g(u)du}{\displaystyle \left\{g{\displaystyle \left[h(x)\right]}\frac{\partial {\displaystyle \left[h(x)\right]}}{\partial x}-g{\displaystyle \left[f(x)\right]}\frac{\partial {\displaystyle \left[f(x)\right]}}{\partial x}\right\}}$$E1.138

Applying Leibnitz’s rule to the Ei function in Eq. (1.81) to differentiate with respect to t_D (see Appendix B in [42]),

$$\frac{\Delta }{\Delta t_D}{\displaystyle \left[E_i{\displaystyle \left(\frac{-r_D^2}{4t_D^{}}\right)}\right]}=-{\displaystyle {\int }_{\frac{r_{D^2}}{4t_D}}^{\infty }\frac{e^{-u}}{u}\hspace{0.17em}\Delta u}=-\frac{e^{-u}}{u}\hspace{0.17em}\frac{\Delta u}{\Delta t_D}\hspace{0.17em}|{}_{\frac{r_D^2}{4t_D}}^{{}^{{}^{\infty }}}$$E1.139

Taking the derivative Δu/Δt_D and replacing u by r_D²/4t_D,

$$\frac{\Delta }{\Delta t_D}{\displaystyle \left[E_i{\displaystyle \left(-\frac{r_D^2}{4t_D}\right)}\right]}=\frac{e^{-{\displaystyle \left(r_D^2/4t_D\right)}}}{r_D^2/4t_D}{\displaystyle \left(-\frac{r_D^2}{4t_D^2}\right)}$$E1.140

After simplification,

$$\frac{\Delta }{\Delta t_D}{\displaystyle \left[E_i{\displaystyle \left(-\frac{r_D^2}{4t_D}\right)}\right]}=-\frac{1}{t_D}{e}^{-{\displaystyle \left(r_D^2/4t_D\right)}}$$E1.141

From inspection of Eq. (1.81) results:

$$\frac{\partial P_D}{\partial t_D}=-\frac{1}{2}\frac{1}{t_D}\hspace{0.17em}{e}^{-\frac{r_D^2}{4t_D}}$$E1.142

In oilfield units,

$$\frac{\Delta P_{wf}}{\Delta t}=\frac{70.6q\mu B}{kht}{e}^{{\displaystyle \left(-\frac{948\varphi \mu c_t}{kt}\right)}}$$E1.143

At the well, r_D = 1, then, Eq. (1.142) becomes:

$$P_D\prime =\frac{1}{2t_D}{e}^{-\frac{1}{4t_D}}$$E1.144

For t_D > 250, e^−1/4tD = 1; then, Eq. (1.144) reduces to

$$P_D\prime =\frac{1}{2t_D}$$E1.145

The derivative of equation (1.145) is better known as the Cartesian derivative. The natural logarithmic derivative is obtained from:

$$t_D*P_D\prime =t_D\frac{\partial P_D}{\partial t_D}=t_D\frac{\partial P_D}{(\partial ln{t}_D)/t_D}=\frac{\partial P_D}{\partial ln{t}_D}$$E1.146

Later on, [46] use the natural logarithmic derivative to develop a type‐curve matching technique.

Appendix C in [42] also provides the derivation of the second pressure derivative:

$$\partial P_D″=P_D\prime \frac{1}{t_D}{\displaystyle \left(\frac{r_D^2}{4t_D}-1\right)}\hspace{0.17em}$$E1.147

Conversion of Eq. (1.145) to natural logarithmic derivative requires multiplying both sides of it by t_D; then, it results:

$$t_D*P_D\prime =\frac{1}{2}$$E1.148

Eq. (1.148) suggests that a log‐log plot of dimensionless pressure derivative against dimensionless time provides a straight line with zero slope and intercept of ½. Taking logarithm to both sides of Eq. (1.145) leads to:

$$\log P_D\prime =-\log t_D-0.301$$E1.149

$$P_{wf}\prime =\frac{\partial P_{wf}}{\partial t}=\frac{1}{t}{\displaystyle \left(\frac{70.6q\mu B}{kh}\right)}$$E1.150

The above expression corresponds of a straight line with negative unit slope. In dimensional form:

Taking logarithm to both sides of the above expression:

$$\log P_{wf}\prime =-\log \hspace{0.17em}t+\log {\displaystyle \left(\frac{70.6q\mu B}{kh}\right)}$$E1.151

As shown in Figure 1.15, Eq. (1.151) corresponds to a straight line with negative unit slope and intercept of:

Figure 1.15.

Log‐log plot of P_wf′ against t.

$$P{\prime }_{1hr}=\frac{70.6q\mu B}{kh}$$E1.152

Eq. (1.152) is applied to find permeability from the intersect plot of the Cartesian pressure derivative versus time plot. This type of plot is also useful to detect the presence of a linear boundary (fault) since the negative unit slope line displaces when the fault is felt as depicted in Figure 1.16.

Figure 1.16.

Fault identification by means of a log‐log plot of P_D′ vs. t_D.

The noise that occurs in a pressure test is due to such factors as (1) turbulence, (2) tool movements, (3) temperature variations, (4) opening and closing wells in the field, and (5) gravitational effects of the sun and moon on the tides (near the great lakes the noise is about 0.15 psia and offshore up to 1 psia).

The estimation of the pressure derivative with respect to time to actual data, of course, must be performed numerically since data recorded from wells are always discrete. During the derivative calculation, the noise is increased by the rate of change that the derivative imposes, so it is necessary to soften the derivative or to use smoothing techniques. The low resolution of the tool and the log‐log paper also increase or exaggerate the noise. Therefore, calculating the derivative of pressure requires some care because the process of data differentiation can amplify any noise that may be present. Numerical differentiation using adjacent points will produce a very noisy derivative, [8, 47, 48].

Ref. [8] conducted a comparative study of several algorithms for estimation of the pressure derivative. They obtained synthetic pressure derivatives for seven different reservoir and well configuration scenarios and, then, estimated the pressure derivative using several comparative methods. They found that the Spline algorithm (not presented here) is the best procedure to derive pressure versus time data since it produces minimal average errors. It is the only algorithm of polynomial character that to be continuous can be smoothed during any derivation process and the form of the curve obtained is in agreement with the worked model. The Horne and Bourdet algorithms when the smoothing window is of either 0.2 or 0.4 are good options for derivation processes. Ref. [8] also found the best procedure for data analysis of pressure against time is to differentiate and then smooth the data.

By itself, the central finite difference formula fails to provide good derivative computation. Instead, some modifications are introduced by [18, 20, 46], respectively:

Horne equation [32]:

$$t{{\displaystyle \left(\frac{\partial P}{\partial t}\right)}}_i=t{{\displaystyle \left(\frac{\partial P}{\partial \hspace{0.17em}ln\hspace{0.17em}t}\right)}}_i={\displaystyle \left\{\begin{array}{l}\frac{ln{\displaystyle \left(t_i/t_{i-k}\right)}\Delta P_{i+j}}{ln{\displaystyle \left(t_{i+j}/t_i\right)}ln{\displaystyle \left(t_{i+j}/t_{i-k}\right)}}+\frac{ln{\displaystyle \left(t_{i+j}{t}_{i-k}/t_i^2\right)}\Delta P_i}{ln{\displaystyle \left(t_{i+1}/t_i\right)}ln{\displaystyle \left(t_i/t_{i-1}\right)}}-\\ \frac{ln{\displaystyle \left(t_{i+j}/t_i\right)}\Delta P_{i-1}}{ln{\displaystyle \left(t_i/t_{i-k}\right)}ln{\displaystyle \left(t_{i+j}/t_{i-k}\right)}}\end{array}\right\}}$$E1.153

$ln\hspace{0.17em}{t}_{i+j}-ln\hspace{0.17em}{t}_i\ge 0.2$ and $ln\hspace{0.17em}{t}_i-ln\hspace{0.17em}{t}_{i-k}\ge 0.2$

When the data are distributed in a geometrical progression (with the time difference from one point to the next much larger as the test passes), then the noise in the derivative can be reduced using a numerical differentiation with respect to the logarithm of time. The best method to reduce noise is to use data that is separated by at least 0.2 logarithmic cycles, rather than points that are immediately adjacent. This procedure is recognized as smoothing and is best explained in Figure 1.17.

Equation of Bourdet et al. [46]:

Figure 1.17.

Smoothing diagram.

$${{\displaystyle \left(\frac{dP}{dx}\right)}}_i=\frac{\frac{P_i-P_{i-1}}{X_i-X_{i-1}}{\displaystyle \left(X_{i+1}-X_i\right)}+\frac{P_{i+1}-P_i}{X_{i+1}-X_i}{\displaystyle \left(X_i-X_{-1i}\right)}}{X_{i+1}-X_{i-1}}$$E1.154

Let X is the natural logarithm of the time function.

This differentiation algorithm reproduces the test type curve over the entire time interval. It uses a point before and a point after the point of interest, i, to calculate the corresponding derivative and places its weighted mean for the objective point. Smoothing can also be applied.

1.13. The principle of superposition

This principle is not new. It was first introduced to the petroleum literature by van Everdingen and Hurst in 1949, [49]. However, its application is too important and many field engineers fail or neglect to use it. Superposition is too useful for systems having one well producing at variable rate or the case when more than one well produces at different flow rates.

As quoted from [25], the superposition principle is defined by:

“Adding solutions to the linear differential equation will result in a new solution of that differential equation but for different boundary conditions,” which mathematically translates to:

$$\psi ={\psi }_1{f}_1+{\psi }_2{f}_2+{\psi }_3{f}_3\mathrm{\dots }$$E1.155

where ψ is the general solution and ψ₁ f₁, ψ₂ f₂ and ψ₃ f₃… are the particular solutions.

1.13.1 Space superposition

If the wells produce at a constant flow rate, the pressure drop at point N, Figure 1.18, will be [3, 9, 19, 21, 25]:

Figure 1.18.

Pressure at the point N.

$$\Delta P_N=\Delta P_{N,1}+\Delta P_{N,2}+\Delta P_{N,3}$$E1.156

If reservoir and fluid properties are considered constant, then, Eq. (1.87) can be applied to the above expression, so that:

$$\Delta P_N=\frac{141.2\mu \hspace{0.17em}}{kh}{\displaystyle \left[{{\displaystyle \left(q{B}_o\right)}}_1{P}_D(r_{D1},t_D)+{{\displaystyle \left(q{B}_o\right)}}_2{P}_D(r_{D2},t_D)+{{\displaystyle \left(q{B}_o\right)}}_3{P}_D(r_{D3},t_D)\right]}$$E1.157

The dimensionless radii are defined by:

$$r_{Dn}=\frac{r_n}{r_w};\hspace{0.17em}n=1,2,3$$E1.158

Extended to n number of wells:

$$\Delta P_N={\displaystyle \sum _{i=1}^n\frac{141.2q\mu B}{kh}}{\displaystyle \left[P_D(r_{Di},t_D)\right]}$$E1.159

If point N is an active well, its contribution to the total pressure drop plus the skin factor pressure drop, Eq. (1.108), must be included in Eq. (1.159), then,

$$\Delta P_N={\displaystyle \sum _{i=1}^n\frac{141.2q\mu B}{kh}}{\displaystyle \left[P_D(r_{DNi},t_D)\right]}+\frac{141.2q\mu B}{kh}s|{}_N$$E1.160

Notice that in Eqs. (1.159) and (1.160), changes of pressures or dimensionless pressures are added. If the point of interest is a well in operation, the damage factor should be added to the dimensionless pressure of that well only.

1.13.2 Time superposition

Sometimes there are changes in flow rate when a well produces as referred in Figures 1.19 and 1.22. Then, the superposition concept must be applied. To do this, [25], a single well is visualized as if there were two wells at the same point, one with a production rate of q₁ during a time period from t = 0 to t and another imaginary well with a production rate of q₂ − q₁ for a time frame between t₁ and t − t₁. The total rate after time t₁ is q₁+ (q₂ − q₁) = q₂. The change in well pressure due to the rate change [19, 25] is,

Figure 1.19.

Time superposition.

$$\Delta P=\frac{141.2\mu B}{kh}{\displaystyle \left[q_1{P}_D(r_D,t_{D1})+(q_2-q_1)P_D(r_D,t_{D2}+s)\right]}$$E1.161

where t_D2 = (t−t₁)_D. If there are more variations in flow rate,

$$\Delta P=\frac{141.2\mu }{kh}{\displaystyle \sum _{i=1}^n{\displaystyle \left[{(qB)}_i-{(qB)}_{i-1}\right]}}{\displaystyle \left\{P_D(r_D,{(t-t_i)}_D+s\right\}}$$E1.162

Example 1.4

This example is taken [25]. The below data and the schematic given in Figure 1.20 correspond to two wells in production:

Figure 1.20.

Flow rate changes for example 1.4.

k = 76 md,    ϕ = 20 %,  B = 1.08 bbl/STB

P_i = 2200 psia, μ = 1 cp,  c_t = 10 × 10⁻⁶/psia

h = 20 ft

Calculate the pressure in (a) well 1 after 7 h of production and (b) in well 2 after 11 h of production. Assume infinite behavior.

Solution

Part (a):

ΔP_{(7 hr)}= ΔP caused by production from well 1 to well 1 + ΔP caused by production from well 2 to well 1. Mathematically,

$$\Delta P_{7\text{hr}\hspace{0.17em}\hspace{0.17em}@\hspace{0.17em}\text{well}1}=\frac{141.2\mu q_{1}B}{kh}{\displaystyle \left[P_D(r_{D1},t_D)+s\right]}+\frac{141.2\mu q_{2}B}{kh}{\displaystyle \left(P_D(r_{D2},t_D)\right)}$$

Using Eq. (1.101) for the well,

$$x=-\frac{948\varphi \mu c_t{r}^2}{kt}=\frac{948(0.2)(1)(1\times {10}^{-5})1^2}{(76)(7)}=3.56\times {10}^{-6}$$

Since x <<<< 0.0025, it implies the use of Eq. (1.82) with Eq. (1.103); then,

$$P_D(r_D,t_D)=-\frac{1}{2}ln{\displaystyle \left(1.781x\right)}$$E1.163

$$P_D=\frac{1}{2}\left|{\displaystyle \left[ln{\displaystyle \left(1.781*3.56\times {10}^{-6}\right)}\right]}\right|=5.98$$

In well 2, x = 0.03564 from Eq. (1.101). Interpolating this value in Table 1.2, Ei(−x) = 2.7924; then, P_D ≅ 1.4. Estimating ΔP in well 1 will result:

$$\Delta P_{7hr,r_D=1}=\frac{141.2(100)(1.08)(1)}{(76)(20)}{\displaystyle \left(5.98+5\right)}+\frac{141.2(100)(1.08)(1)}{(76)(20)}{\displaystyle \left(1.4\right)}=113.7$$

P_{wf @ well1} = 2200−113.7 = 2086.4 psia (notice that skin factor was only applied to well 1)

Part (b);

At 11 h, it is desired to estimate the pressure in well 2. Two flow rates should be considered for in each well. Then, the use of Eq. (1.162) will provide:

$$\begin{array}{l}\Delta P_{(11\hspace{0.17em}\hspace{0.17em}\text{hr,well}\hspace{0.17em}2)}=\Delta P_{\text{well}\hspace{0.17em}\text{1}\to \text{Well}\hspace{0.17em}\text{2,}t\text{=11}\hspace{0.17em}\text{hr,}q\text{=100}\hspace{0.17em}\hspace{0.17em}\text{BPD,}\hspace{0.17em}{r}_D\text{=100}}+\Delta P_{\text{well}\hspace{0.17em}\text{1}\to \text{Well}\hspace{0.17em}\text{2,}t\text{=(11}\hspace{0.17em}-10)\hspace{0.17em}\text{hr,}q\text{=(50‐100)}\hspace{0.17em}\text{BPD,}\hspace{0.17em}{r}_D\text{=100}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\Delta P_{\text{well}\hspace{0.17em}\text{1}\to \text{Well}\hspace{0.17em}\text{1,}t\text{=11}\hspace{0.17em}\text{hr,}q\text{=25}\hspace{0.17em}\hspace{0.17em}\text{BPD,}\hspace{0.17em}{r}_D\text{=1,}{s}_2}+\Delta P_{\text{well}\hspace{0.17em}\text{1}\to \text{Well}\hspace{0.17em}\text{1,}t\text{=(11}\hspace{0.17em}-8)\hspace{0.17em}\text{hr,}q\text{=(100‐25)}\hspace{0.17em}\text{BPD,}\hspace{0.17em}{r}_D\text{=1,}{s}_2}\end{array}$$

Using Eq. (1.101), the four respective values of x are: x =0.02268, 0.2494, 2.268 × 10⁻⁶, and 8.316 × 10⁻⁶. Estimation of Ei requires the use of Table 1.2 for the first two values and use of Eq. (1.103) for the last two values. The four values of Ei(−x) are: 0.0227, 0.811, 12.42, and 11.12. Therefore, the respective values of P_D are 1.605, 0.405, 6.209, and 5.56. The total pressure drop is found with Eq. (1.161) as follows:

$$\Delta P_{\text{well}\hspace{0.17em}\text{2,}\hspace{0.17em}\text{11}\hspace{0.17em}\hspace{0.17em}\text{hr})}=\frac{141.2(1)(1.08)}{(76)(20)}{\displaystyle \left\{\begin{array}{l}(100)(1.605)+(50-100)(0.405)+\\ (25)(6.209+1.7)+(100-25)(5.56+1.7)\end{array}\right\}}=87.75\hspace{0.17em}\text{psia}$$

P_{wf @ well2} = 2200 − 87.75 = 2112.25 psia

1.13.3 Space superposition—method of images

The method of images applies to deal with either no‐flow or constant‐pressure boundaries. If a well operates at a constant flow rate at a distance, d, from an impermeable barrier (fault), the systems acts as if there were two wells separated 2d from each other [3, 25]. For no‐flow boundaries, the image well corresponds to the same operating well. For constant‐pressure boundary, the resulting image corresponds to an opposite operating well. In other words, if the well is a producer near a fault, the image well corresponds to an injector well. These two situations are sketched in Figure 1.21. For the no‐flow boundary, upper system in Figure 1.21, the dimensionless pressure can be expressed as:

$$P_{D\hspace{0.17em}\hspace{0.17em}\text{at}\hspace{0.17em}\text{real}\hspace{0.17em}\text{well}}=P_{D\hspace{0.17em}\text{at}\hspace{0.17em}\text{real}\hspace{0.17em}\text{well,}\hspace{0.17em}{r}_D\text{=1,}s}+P_{D\hspace{0.17em}\text{at}\hspace{0.17em}\text{image}\hspace{0.17em}\text{well}\to \text{real}\hspace{0.17em}\text{well,}\hspace{0.17em}{r}_D\text{= 2}d\text{/}{r}_w}$$E1.164

Figure 1.21.

Well near a linear barrier.

For the constant‐pressure boundary, lower part in Figure 1.21, the dimensionless pressure can be expressed as:

$$P_{D\hspace{0.17em}\hspace{0.17em}\text{at}\hspace{0.17em}\text{real}\hspace{0.17em}\text{well}}=P_{D\hspace{0.17em}\text{at}\hspace{0.17em}\text{real}\hspace{0.17em}\text{well,}\hspace{0.17em}{r}_D\text{=1,}s}-P_{D\hspace{0.17em}\text{at}\hspace{0.17em}\text{image}\hspace{0.17em}\text{well}\to \text{real}\hspace{0.17em}\text{well,}\hspace{0.17em}{r}_D\text{= 2}d\text{/}{r}_w}$$E1.165

The negative sign in Eq. (1.165) is because of dealing with an imaginary injector well.

For the case of two intersecting faults, the total number of wells depends on the value of the angle formed by the two faults, thus:

$$n_{\text{wells}}=\frac{360}{\mathrm{\theta }}$$E1.166

The image method is limited to one well per quadrant. If this situation fails to be fulfilled, then, the method cannot be applied. In the system of Figure 1.22, an angle of 90° is formed from the intersecting faults. According to Eq. (1.166), n_wells = 360/90 = 4 wells, as shown there. The ratio of the distances from the well to each fault is given by:

$$y_D=b_y/b_x$$E1.167

Figure 1.22.

Well between two intersecting faults.

The practical way to apply space superposition for generating the well system resulting from two intersecting faults consist of extending the length of the faults and setting as many divisions as suggested by Equation (1.166); that is, for example, 1.5, Figure 1.23 left, six well spaces are obtained. Then, draw a circle with center at the fault intersection and radius at well position. This guarantees that the total length corresponds to the double length value from the well to the fault. Draw from the well a line to be perpendicular to the nearest fault and keep drawing the line until the circle line has been reached. See Figure 1.24 left. Set the well. A sealing fault provides the same type of well as the source well, that is, a producing well generates another producing well to the other side of the fault. A constant‐pressure boundary provides the opposite well type of the source well, that is, a producing well generates an injector well on the other side of the line. Draw a new line from the just drawn imaginary well normal to the fault and keep drawing the line until the line circle is reached. See Figure 1.24 right. Repeat the procedure until the complete well set system has been drawn.

Figure 1.23.

Location of well A and resulting well number system for example 1.5.

Figure 1.24.

Generating the well system for two intersecting faults.

For more than six well spaces generated, that is angles greater than 60°, as the case of example 1.5, when a fault intersects a constant‐pressure boundary injector and producer imaginary wells ought to be generated. What type of line should be drawn? A solid line representing a sealing fault, or a dash line, representing a constant‐pressure boundary? The answer is any of both. The lines should be drawn alternatively and as long as the system closes correctly, superposition works well.

Example 1.5

Well A in Figure 1.23 has produced a constant rate of 380 BPD. It is desired (a) to estimate the well‐flowing pressure after one week of production. The properties of the reservoir, well and fluid are given as follows:

P_i = 2500 psia,  B = 1.3 bbl/STB,   μ = 0.87 cp

h = 40 ft,      c_t = 15×10⁻⁶/psia,    ϕ = 18 %

r_w = 6 in,       k = 220 md, s = −5

(b) What would be the well‐flowing pressure after a week of production if the well were in an infinite reservoir?

Solution

Part (a)

The pressure drop in well A is affected by its own pressure drop and pressure drop caused by its well images. The distance from well A to its imaginary wells is shown in Figure 1.23 (right‐hand side). The total pressure drop for well A is:

$$\begin{array}{l}\Delta P_A=\Delta P_{A,r=rw}+\Delta P_{image\hspace{0.17em}1\to \text{well}\hspace{0.17em}A,r=500\hspace{0.17em}ft}+\Delta P_{image\hspace{0.17em}2\to \text{well}\hspace{0.17em}A,r=866\hspace{0.17em}ft}+\\ \Delta P_{image\hspace{0.17em}3\to \text{well}\hspace{0.17em}A,r=1000\hspace{0.17em}ft}+\Delta P_{image\hspace{0.17em}4\to \text{well}\hspace{0.17em}A,r=866\hspace{0.17em}ft}+\Delta P_{image\hspace{0.17em}5\to \text{well}\hspace{0.17em}A,r=500\hspace{0.17em}ft}\end{array}$$

By symmetry, the above expression becomes:

$$\Delta P_A=\Delta P_{A,r=rw}+2\Delta P_{image\hspace{0.17em}1\to \text{well}\hspace{0.17em}A,r=500\hspace{0.17em}ft}+2\Delta P_{image\hspace{0.17em}2\to \text{well}\hspace{0.17em}A,r=866\hspace{0.17em}ft}+\Delta P_{image\hspace{0.17em}3\to \text{well}\hspace{0.17em}A,r=1000\hspace{0.17em}ft}$$

Using Eq. (1.101) for the well:

$$\frac{948(0.18)(0.87)(1.5\times {10}^{-5}){(0.5)}^2}{(220)(168)}=1.5\times {10}^{-8}$$

Since x <<<< 0.0025, Eq. (1.163) applies:

$$P_D(r_D,t_D)=-\frac{1}{2}\left|ln{\displaystyle \left[1.781(1.5\times {10}^{-8})\right]}\right|=8.72$$

Estimation for the image wells are given below. In all cases, x > 0.0025, then, Table 1.2 is used to find Ei and the resulting below divided by 2 for the estimation of P_D,

$$x_{\text{image}\hspace{0.17em}\text{well}\hspace{0.17em}\text{1}\hspace{0.17em}\text{or}\hspace{0.17em}\text{5}}=\frac{948(0.18)(0.87)(1.5\times {10}^{-5}){(500)}^2}{(220)(168)}=0.015,P_D=1.816$$

$$x_{\text{image}\hspace{0.17em}\text{well}\hspace{0.17em}\text{2}\hspace{0.17em}\text{or}\hspace{0.17em}\text{4}}=\frac{948(0.18)(0.87)(1.5\times {10}^{-5})({866}^2)}{(220)(168)}=0.0452,P_D=1.282$$

$$x_{\text{image}\hspace{0.17em}\text{well}\hspace{0.17em}\text{3}}=\frac{948(0.18)(0.87)(1.5\times {10}^{-5})({1000}^2)}{(220)(168)}=0.06,P_D=1.145$$

Then, the pressure drop in A will be:

$$\Delta P_A=141.2\frac{q\mu \hspace{0.17em}B}{kh}{\displaystyle \left[(P_D{}_{A,r=rw}+s)+2P_D{}_{\hspace{0.17em}\text{image}\hspace{0.17em}1,r=500\hspace{0.17em}ft}+2P_D{\hspace{0.17em}}_{\text{image}\hspace{0.17em}2,r=866\hspace{0.17em}ft}+P_{D\hspace{0.17em}}{}_{\text{image}\hspace{0.17em}3,r=1000\hspace{0.17em}ft}\right]}$$

$$\Delta P_A=\frac{141.2(380)(0.87)(1.3)}{(220)(40)}{\displaystyle \left[(8.72-5)+2(1.816)+2(1.282)+1.145\right]}=76.3\hspace{0.17em}\text{psia}$$

P_{wf @ well A} = 2500 − 76.3 = 2423.7 psia

Part (b)

If the well were located inside an infinite reservoir, the pressure drop would not include imaginary wells, then:

$$\Delta P_A=141.2\frac{q\mu \hspace{0.17em}B}{kh}{\displaystyle \left[P_D{}_{,r=rw}+s\right]}$$

$$\Delta P_A=\frac{141.2(380)(0.87)(1.3)}{(220)(40)}{\displaystyle \left[8.72-5\right]}=25.63\hspace{0.17em}\text{psia}$$

The well‐flowing pressure would be (2500 − 25.3) = 2474.4 psia. It was observed that the no‐flow boundaries contribute with 66.4% of total pressure drop in well A.

Nomenclature

A	area, ft2 or Ac
Bg	gas volume factor, ft3/STB
Bo	oil volume factor, bbl/STB
Bw	oil volume factor, bbl/STB
bx	distance from closer lateral boundary to well along the x‐direction, ft
by	distance from closer lateral boundary to well along the y‐direction, ft
c	compressibility, 1/psia
cf	pore volume compressibility, 1/psia
ct	total or system compressibility, 1/psia
d	distance from a well to a fault, ft
f	a given function
h	formation thickness, ft
k	permeability, md
ks	permeability in the damage zone, md
krf	phase relative permeability, f = oil, water or gas
L	reservoir length, ft
m	slope
m(P)	pseudopressure function, psia2/cp
M	gas molecular weight, lb/lbmol
P	pressure
dP/dr	pressure gradient, psia/ft
PD′	dimensionless pressure derivative
PD″	dimensionless second pressure derivative
PD	dimensionless pressure
Pi	initial reservoir pressure, psia
Pwf	well flowing pressure, psia
q	flow rate, bbl/D. For gas reservoirs the units are Mscf/D
Rs	gas dissolved in crude oil, SCF/STB
Rsw	gas dissolved in crude water, SCF/STB
rD	dimensionless radius
rDe	dimensionless drainage radius = re/rw
r	radial distance, radius, ft
re	drainage radius, ft
rs	radius of the damage zone, ft
rw	well radius, ft
Sf	fluid saturation, f = oil, gas or water
s	skin factor
T	reservoir temperature, ºR
t	time, h
ta	pseudotime, psia h/cp
to	dummy time variable
ur	radial flow velocity, ft/h
tD	dimensionless time based on well radius
tDA	dimensionless time based on reservoir area
tD*PD′	logarithmic pressure derivative
V	volume, ft3
z	vertical direction of the cylindrical coordinate, real gas constant

Greek

Δ	change, drop
Δt	shut‐in time, h
ϕ	porosity, fraction. Spherical coordinate
λ	phase mobility, md/cp
η	hydraulic diffusivity constant, md‐cp/psia
ρ	density, lbm/ft3
θ	cylindrical coordinate
μ	viscosity, cp
ζ	time function

Suffices

1 hr	reading at time of 1 h
D	dimensionless
DA	dimensionless with respect to area
f	formation
g	gas
i	initial conditions
o	oil, based condition
w	well, water
p	pore

Pressure Drawdown Testing

As can be seen in Figure 1.4, well pressure test analysis (PTA) considers this as the most basic and simple test, which does not mean that it is not important. In these tests, bottom‐hole well‐flowing pressure, P_wf, is continuously recorded keeping the flow constant. These tests are also referred as flow tests. Similar to an injection test, these tests require either production/injection from/into the well.

These tests are performed with the objective of (a) obtaining pore volume of the reservoir and (b) determining heterogeneities (in the drainage area). In fact, what is obtained is (a) transmissibility and (b) porous volume by total compressibility. In fact, a recent study by Agarwal [1] allows using drawdown tests to estimate the average permeability in the well drainage area. To run a pressure decline test, the following steps are generally followed:

• The well is shut‐in for a long enough time to achieve stabilization throughout the reservoir, if this is not achieved, multirate testing is probably required;

• The recording pressure tool is lowered to a level immediately above the perforations. This is to reduce Joule‐Thompson effects. It is important to have at least two pressure sensors for data quality control purposes;

• The well opens in production at constant flow and in the meantime the well‐flowing pressure is continuously recorded.

Ideally, the well is closed until the static reservoir pressure. The duration of a drawdown test may last for a few hours or several days, depending upon the test objectives and reservoir characteristics. There are extensive pressure drawdown tests or reservoir limit tests (RLT) that run to delimit the reservoir or estimate the well drainage volume. Other objectives are the determination of: well‐drainage area permeability, skin factor, wellbore storage coefficient (WBS), porosity, reservoir geometry, and size of an adjacent aquifer.

2.1. Wellbore storage coefficient

It is the continuous flow of the formation to the well after the well has been shut‐in for stabilization. It is also called after‐flow, postproduction, postinjection, loading, or unloading (for flow tests). The flow occurs by the expansion of fluids in the wellbore. In pressure buildup tests, after‐flow occurs. Figure 2.1 illustrates the above [2].

Traditional pressure tests had to be long enough to cope with both wellbore storage and skin effects so that a straight line could be obtained indicating the radial flow behavior. Even this approach has disadvantages since more than one apparent line can appear and analysts have problems deciding which line to use. In addition, the scale of the graph may show certain pressure responses as straight lines when in fact they are curves. To overcome these issues, analysts developed the method the type‐curve matching method.

There is flow in the wellbore face after shutting‐in the well in surface. Wellbore storage affects the behavior of the pressure transient at early times. Mathematically, the storage coefficient is defined as the total volume of well fluids per unit change in bottom‐hole pressure, or as the capacity of the well to discharge or load fluids per unit change in background pressure:

$$C=\frac{\mathrm{\Delta }V}{\mathrm{\Delta }P}$$E2.1

As commented by Earlougher [2], wellbore storage causes the flow rate at the face of the well to change more slowly than the surface flow rate. Figure 2.2 schematizes the relation q_sf/q when the surface rate is changed from 0 to q, when C = 0, q_sf/q = 1, while for C > 0, the relation q_sf/q gradually changes from 0 to 1. The greater the value of C, the greater the transition is. As the storage effects become less severe, the formation begins to influence more and more the bottom‐hole pressure until the infinite behavior is fully developed. Pressure data that are influenced by wellbore storage can be used for interpretation purposes since fluids unload or load has certain dependence on reservoir transmissibility; however, this analysis is risky and tedious. TDS technique, presented later in this chapter, can provide a better solution to this problem.

Typically, the flow rate is surface‐controlled (unless there is a bottom shut‐in tool), the fluids in the well do not allow an immediate transmission of the disturbance from the subsurface to the surface, resulting in uneven surface and wellbore face flow [2, 3, 4, 5, 6, 7]. Wellbore storage can change during a pressure test in both injector and producer wells. Various circumstances cause changes in storage, such as phase redistribution and increase or decrease in storage associated with pressure tests in injector wells. In injector wells, once the well is closed, the surface pressure is high but could decrease to atmospheric pressure and go to vacuum if the static pressure is lower than the hydrostatic pressure. This causes an increase in storage (up to 100 times) of an incompressible system to one in a system where the liquid level drops [2]. The inverse situation occurs in injector wells with a high level of increase of liquid storage level and in producing wells with a high gas‐oil ratio or by redissolution of the free gas. Both for increase or decrease of storage, the second storage coefficient determines the beginning of the semilogarithmic straight line.

When the relationship between ΔV and ΔP does not change during the test, the wellbore storage coefficient is constant and can be estimated from completion data [2, 3, 4].

$$C={\displaystyle \left(\frac{144}{\rho }\right)}{V}_u$$E2.2

where V_u is the wellbore volume/unit length, bbl/ft, r is the density of the fluid in the wellbore, lbm/ft³, and C is the wellbore storage coefficient, bbl/psia.

Figure 2.1.

Effects of wellbore storage on buildup and drawdown tests, taken from [2].

Figure 2.2.

Effect of storage on the flow rate at the face of the well, C₃>C₂>C₁, taken from [2].

For injector wells or wells completely filled with fluids:

$$C=c_{wb}{V}_{wb}$$E2.3

where C_wb is the wellbore fluid compressibility = 1/P_wb, V_wb is the total wellbore volume, and V_u can be estimated with internal casing, ID_csg, and external tubing, OD_tbg, diameters.

$$V_u=0.0009714(I{D}_{csg}^2-O{D}_{tbg}^2)$$E2.4

When opening a well, see Figure 2.3, the oil production will be given by the fluid that is stored in the well, q_sf = 0. As time goes by, q_sf tends to q and storage is neglected and the amount of liquid in the wellbore will be constant. The net accumulation volume will be (assuming constant B) [3, 5]:

$$V_{wb}=A_{wb}(Z)$$E2.5

Figure 2.3.

Schematic representation of wellbore storage, taken from [3].

The flow rate is given by:

$$\frac{d{V}_{wb}}{dt}=A_{wb}\frac{dZ}{dt}$$E2.6

The rate of volume change depends upon the difference between the subsurface and surface rates:

$$\frac{d{V}_{wb}}{dt}=(q_{sf}-q)B=\frac{24}{5.615}{A}_{wb}\frac{dZ}{dt}$$E2.7

Since (assuming g/g_c = 1):

$$P_w-P_t=\frac{\rho Z}{144}$$E2.8

Taking the derivative to Eq. (2.8),

$$\frac{d}{dt}(P_w-P_t)=\frac{\rho }{144}\frac{dZ}{dt}$$E2.9

Combining Eqs. (2.7) and (2.9) will result:

$$(q_{sf}-q)B=\frac{24(144)}{5.615}\frac{A_{wb}}{\rho }\frac{d(P_w-P_t)}{dt}$$E2.10

Define

$$C=\frac{144}{5.615}\frac{A_{wb}}{\rho }$$E2.11

Assuming constant, P_t, replacing the definition given by Eq. (2.11) and solving for the wellbore face flow rate, q_sf, leads to:

$$q_{sf}=q+\frac{24C}{B}\frac{d{P}_w}{dt}$$E2.12

Taking derivative to Eqs. (1.89) and (1.94) with respect to time and taking the ratio of these will yield:

$$\frac{d{P}_w}{dt}={\displaystyle \left(-\frac{0.0373qB}{\varphi \hspace{0.17em}h\hspace{0.17em}{c}_t\hspace{0.17em}{r}_w^2}\right)}\frac{d{P}_D}{d{t}_D}$$E2.13

Combining Eqs. (2.12) and (2.13);

$$q_{sf}=q-\frac{0.894qC}{\varphi \hspace{0.17em}{c}_{t}h{r}_w^2}\frac{d{P}_{wD}}{d{t}_D}$$E2.14

Defining the dimensionless wellbore storage coefficient;

$$C_D=\frac{0.894C}{\varphi c_{t}h{r}_w^2}$$E2.15

Rewriting Eq. (2.14);

$$\frac{q_{sf}}{q}=1-C_D\frac{d{P}_{wD}}{d{t}_D}$$E2.16

The main advantage of using downhole shut‐in devices is the minimization of wellbore storage effects and after‐flow duration.

Rhagavan [5] presents the solution for the radial flow diffusivity equation considering wellbore storage and skin effects in both Laplace and real domains, respectively:

$${\overline{P}}_D=\frac{K_0{\displaystyle \left(\sqrt{u}\right)}+s\sqrt{u}{K}_1{\displaystyle \left(\sqrt{u}\right)}}{u{\displaystyle \left\{\sqrt{u}{K}_1{\displaystyle \left(\sqrt{u}\right)}+C_{D}u{\displaystyle \left[K_0{\displaystyle \left(\sqrt{u}\right)}+s\sqrt{u}{K}_1{\displaystyle \left(\sqrt{u}\right)}\right]}\right\}}}$$E2.17

$$P_D=\frac{4}{{\pi }^2}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1-e^{-x^2{t}_D}}{x^3{\displaystyle \left\{{{\displaystyle \left[v{C}_D{J}_0{\displaystyle \left(x\right)}-f{\displaystyle \left(x\right)}{J}_1{\displaystyle \left(x\right)}\right]}}^2+{{\displaystyle \left[x{C}_D{Y}_0{\displaystyle \left(x\right)}-f{\displaystyle \left(x\right)}{Y}_1{\displaystyle \left(x\right)}\right]}}^2\right\}}}}dv$$E2.18

where f(x) = 1−C_D(s) x², and K₀, K₁, J₀, J₁,Y₀, and Y₁ are Bessel functions.

2.2. Well test interpretation methods

There exist four methods for well test interpretation as follows: (a) conventional straight‐line, (b) type‐curve matching, (c) regression analysis, and (d) modern method: TDS technique. Although they were named chronologically, from oldest to most recent, they will be presented in another way:

2.2.1 Regression analysis

This is the most widely used method. It consists of automatically matching the pressure versus time data to a given analytical solution (normally) of a specific reservoir model. The automatic procedure uses nonlinear regression analysis by taking the difference between a given matching point and the objective point from the analytical solution.

This method has been also widely misused. Engineers try to match the data with any reservoir model without considering the reservoir physics. The natural problem arid=sing with this method is the none‐uniqueness of the solution. This means that for a given problem, the results are different if the starting simulation values change. This can be avoided if the starting values for the simulation values are obtained from other techniques, such as TDS technique or conventional analysis, and then, the range of variation for a given variable is reduced. This technique will not be longer discussed here since this book focused on analytical and handy interpretation techniques.

2.2.2 Type‐curve matching

As seen before, this technique was the second one to appear. Actually, it came as a solution to the difficulty of identity flow regimes in conventional straight‐line plots. However, as observed later, the technique is basically a trial‐and‐error procedure. This makes the technique tedious and risky to properly obtain reservoir parameters.

The oldest type‐curve method was introduced by Ramey [2, 8, 9]. If C_D = 0 in Eq. (2.16), then, q_sf = q. Therefore;

$$1-C_D\frac{d{P}_D}{d{t}_D}=0$$E2.19

By integration between 0 and a given P_wD and from dimensionless time zero to t_D, and taking logarithm to both terms, it yields:

$$log\hspace{0.17em}{P}_D=log\hspace{0.17em}{t}_D-log\hspace{0.17em}{C}_D$$E2.20

Suffix w is used to emphasize that the pressure drop takes place at the wellbore bottom‐hole. This will be dropped for practical purposes. It is clearly observed in Eq. (2.19) that the slope is one. Then in any opportunity that is plotted P_D vs. t_D and a straight line with a unitary slope is observed at early times, is a good indication that storage exists. Substituting the dimensionless quantities given by Eqs. (1.89), (1.94) and (2.15) in Eq. (2.20), we have:

$$C=\frac{qB}{24}\frac{t}{\mathrm{\Delta }P}=\frac{qB}{24}\frac{t_N}{{(P_i-P_{wf})}_N}$$E2.21

Eq. (2.21) serves to determine the storage coefficient from data from a pressure decline test using a log‐log plot of ΔP versus time. Any point N is taken from the unit‐slope straight line portion. The value of C obtained using Eq. (2.21) must match the value obtained from Eq. (2.5). Otherwise, there may be an indication that the liquid level is going down or rising inside the well. The reasons most commonly attributed to this phenomenon are high gas‐oil ratios, highly stimulated wells, exhaust gaskets or spaces in the well connections caused by formation collapse or poor cementation and wells used for viscous fluid injection. In conclusion, the properties of Ramey's type curves allow (a) a unitary slope to be identified which indicates wellbore storage and (b) the fading of wellbore storage effects.

It can also be seen that each curve deviates from the unitary slope and forms a transition period lasting approximately 1.5 logarithmic cycles. This applies only to constant wellbore storage, otherwise, refer to [10]. If every ½ cycle is equal to (10^0.5 = 3.1622), it means that three half cycles (3.16223 = 31.62) represent approximately a value of 30. That is to say that a line that deviates at 2 min requires 1 h forming the transient state or radial flow regime. In other words, the test is masked for 1 h by wellbore storage effects [2, 5, 11]. It is also observed that a group of curves that present damage are mixed at approximately a dimensionless time,

$$t_D\cong (60+3.5s)C_D$$E2.22

After which time, the test is free of wellbore storage effects [2, 5, 6]. Along with TDS technique [10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73] which will be discussed later in this chapter, type‐curve matching is the only manual procedure that can be applied in short tests where radial flow has not been developed (semilog line). However, type‐curve matching is risky because it is a trial‐and‐error technique, but can provide approximate results even when conventional methods fail. One millimeter shifting can cause pressure differences of up to 200 psia. The procedure is as follows [2, 9]:

Figure 2.4.

Type curve of dimensionless pressure against dimensionless time for a well in an infinite reservoir (wellbore storage and skin), taken from [2, 9].

1. Prepare a plot of DP vs. t on logarithmic paper using the same scale as the master curve given in Figure 2.4. This is recognized as the field data plot, fdp.

2. Place the fdp on the master curve so that the axes are parallel.

3. Find the best match with one of the curves in Figure 2.4.

4. Choose a suitable match point and read the corresponding coordinates DPM, tM, PDM, tDM, and CDM. The two first parameters are read from the fdp. The remaining from the type‐curve (Figure 2.4).

5. Estimate permeability, porosity, and wellbore storage coefficient, respectively:

   $$k=141.2\frac{q\mu B}{h}{\displaystyle \left(\frac{P_{DM}}{\mathrm{\Delta }{P}_M}\right)}$$E2.23
   $$\varphi =\frac{0.0002637k}{\mu \hspace{0.17em}{c}_t\hspace{0.17em}{r}_w^2}{\displaystyle \left(\frac{t_M}{t_{DM}}\right)}$$E2.24
   $$C=\frac{\varphi c_{t}h\hspace{0.17em}{r}_w^2}{\hspace{0.17em}0.8936}{C}_{DM}$$E2.25

The results from the Ramey’s type curve must be verified with some other type curve. For instance, Earlougher and Kersch [8], formulated another type curve, Figure 2.5, which result should agree with those using Ramey method. The procedure for this method [8] is outlined as follows:

1. Plot ΔP/t vs. t (fdp) on logarithmic paper using the same scale as the master curve given in Figure 2.5. Match the plotted curve, fdp, with the appropriate curve of Figure 2.5. Choose any convenient point and read from the master graph (CDe2s)_M, (ΔP/t 24C/qB)_M and (kh/µ t/C)_M. Read from the fdp: (ΔP/t)_M and t_M.

Figure 2.5.

Earlougher and Kersch type‐curve for a well in infinite reservoir with wellbore storage and skin, taken from [2, 8].

Find wellbore storage coefficient, formation permeability, and skin factor using, respectively, the below expressions:

$$C=\frac{qB}{24}{{\displaystyle \left(\frac{\mathrm{\Delta }P}{t}\frac{24C}{qB}\right)}}_M/{{\displaystyle \left(\frac{\mathrm{\Delta }P}{t}\right)}}_M$$E2.26

$$k=\frac{\mu C}{h}\hspace{0.17em}{{\displaystyle \left(\frac{kh}{\mu }\frac{t}{c}\right)}}_M/t_M$$E2.27

$$s=\frac{1}{2}ln{\displaystyle \left[\frac{\varphi \hspace{0.17em}\mu \hspace{0.17em}{c}_t\hspace{0.17em}h\hspace{0.17em}{r}_w^2}{0.89359C}{(C_D{e}^{2s})}_M\right]}$$E2.28

Another important type curve that is supposed to provide a better match was presented by Bourdet et al. [73], Figure 2.6. This includes both pressure and pressure derivative curves. The variables to be matched are ΔP_M, (t*ΔP′)_M, (P_D)_M, [(t_D/C_D)P_D′]_M, t_M, (t_D/C_D)_M, and (C_De^2s)_M. The equations use after the matching are [73]:

$$k=\frac{141.2q\mu \hspace{0.17em}B}{h}\frac{P_{DM}}{\mathrm{\Delta }{P}_M}$$E2.29

$$C={\displaystyle \left(0.000295\frac{kh}{\mu }\right)}\frac{t_M}{{(t_D/C_D)}_M}$$E2.30

$$k=\frac{\frac{1412q\mu B}{h}\hspace{0.17em}{[(t_D/C_D)P_D\prime ]}_M}{{(t*\mathrm{\Delta }P\prime )}_M}$$E2.31

$$s=\frac{1}{2}ln\frac{{(C_D{e}^{2s})}_M}{C}$$E2.32

Figure 2.6.

Bourdet et al. [73] pressure and pressure‐derivative versus time‐type curve.

2.2.3 Straight‐line conventional analysis

The conventional method implies plotting either pressure or pressure drop against a given time function. The intercept and slope of such plot is used for reservoir and well parameters estimation. When the fluid initiates its path from the farthest reservoir point until the well head, several states and flow regimes are observed depending on the system geometry. For instance, if the reservoir has an elongated shape, probably linear flow will be observed. Linear flow obeys a pressure dependency on the square‐root of time, or, if the fluid experiences radial flow regime, the relation between pressure and time observes a semilog behavior, or, either inside the well or the limitation of the reservoir boundaries imply a pseudosteady‐state condition, then, pressure is a linear function of time.

$$P\hspace{0.17em}\alpha \hspace{0.17em}\hspace{0.17em}f(t)$$E2.33

The time function depends on the system geometry and could be any of the kinds described by Eq. (2.34).

Normally, the pressure or pressure drop are plotted in Cartesian coordinates, except certain few cases as for the Muskat method, see Chapter 3, which requires a potential plot, meaning, logarithm scale of pressure drop in the y‐axis and Cartesian scale for time in the x‐direction.

2.2.3.1 Semilog analysis

It is commonly referred as the “semilog method” since the radial flow is the most important regime found on a pressure test. Then, a semilogarithm plot is customary used in well test analysis.

$$f(t)={\displaystyle \{\begin{array}{l}log t\hspace{0.17em}-\text{Radial}\hspace{0.17em}\text{flow}\\ log\frac{t_p+\mathrm{\Delta }t}{\mathrm{\Delta }t}\hspace{0.17em}-\text{Radial}\hspace{0.17em}\text{flow}\hspace{0.17em}(\text{Horner}\hspace{0.17em}\text{plot})\\ t\hspace{0.17em}-\text{Pseudosteady}\hspace{0.17em}\text{state}\\ t^{0.135}-\text{Ellipsoidal}\hspace{0.17em}\text{flow}\\ t^{0.25}-\text{Bilinear}\hspace{0.17em}\text{flow}\\ t^{0.36}-\text{Birradial}\hspace{0.17em}\text{flow}\\ t^{0.5}-\text{Linear}\hspace{0.17em}\text{flow}\\ \text{1/}{t}^{0.5}-\text{Spherical/Hemispherical}\hspace{0.17em}\text{flow}\\ {(t_p+\mathrm{\Delta }t)}^{\xi }-\mathrm{\Delta }{t}^{\xi };\xi =0.135-Ellipsoidal,\hspace{0.17em}0.25,\hspace{0.17em}0.36,\hspace{0.17em}0.5\\ 1/\sqrt{t_p}+1/\sqrt{\mathrm{\Delta }t}-1/\sqrt{t_p+\mathrm{\Delta }t}-\text{Spherical/Hemispherical}\hspace{0.17em}\text{flow}\\ 1/\sqrt{\mathrm{\Delta }t}-1/\sqrt{t_p+\mathrm{\Delta }t}-\text{Spherical/Hemispherical}\hspace{0.17em}\text{flow}\end{array}}$$E2.34

Starting by including the skin factor in Eq. (1.106);

$$P_D=\frac{1}{2}{\displaystyle \left[ln{t}_D+0.80907\right]}+2s$$E2.35

Replacing the dimensionless terms given by Eqs. (1.89) and (1.94) into Eq. (2.35) and dividing both terms by ln 10 will lead to:

$$\frac{kh(P_i-P_{wf})}{162.6q\mu B}={\displaystyle \left[log{\displaystyle \left(\frac{0.0002637kt}{\varphi \mu c_t{r}_w^2}\right)}+0.3514+0.8686s\right]}$$E2.36

Solving for the well‐flowing pressure;

$$P_{wf}=P_i-\frac{162.6\hspace{0.17em}q\mu B}{kh}{\displaystyle \left[log{\displaystyle \left(\frac{kt}{\varphi \mu c_t{r}_w^2}\right)}-3.2275+0.8686s\right]}$$E2.37

Figure 2.7.

Behavior of the well‐flowing pressure observed in a semilog graph, taken from [68].

Eq. (2.37) suggests a straight‐line behavior which is represented in the central region of Figure 2.7. The other two regions are affected by wellbore storage and skin effects, at early times and boundary effects at late times. Reservoir transmissivity, mobility, or permeability can be determined from the slope;

$$m=T=\frac{kh}{\mu }=\left|\frac{162.6\hspace{0.17em}qB}{m}\right|$$E2.38

The intercept of Eq. (2.34) is used for the determination of the mechanical skin factor. For practical purposes, the well‐flowing pressure at time of 1 h, P_1hr, is read from the straight‐line portion of the semilog behavior, normally extrapolated as sketched in Figure 2.7, so solving for skin factor, s, from Eq. (2.34) results:

$$s=1.1513{\displaystyle \left[\frac{P_{1hr}-P_i}{m}-log{\displaystyle \left(\frac{k}{\varphi \mu c_t{r}_w^2}\right)}+3.23\right]}$$E2.39

Since the slope possesses a negative signed, so does the P_1hr− P_i term. Therefore, the first fractional in the above equation is always positive unless the well is highly stimulated.

According to Eq. (2.39), the contribution to the pressure drop caused by the mechanical skin factor is included to the last term: 0.8686s multiplied by the slope. Then:

$$\mathrm{\Delta }{P}_s=\left|0.87(m)\right|s,{\displaystyle \{\begin{array}{l}\text{if}\hspace{0.17em}s>0\to \mathrm{\Delta }{P}_s>0\\ \text{if}\hspace{0.17em}s<0\to \mathrm{\Delta }{P}_s<0\end{array}}$$E2.40

Eq. (2.40) is similar to Eq. (1.110) and works for either pressure drawdown or pressure buildup tests.

$$\mathrm{\Delta }{P}_s=-0.87(m){\displaystyle \left\{\frac{k}{k_s}-1\right\}}ln\frac{r_s}{r_w}$$E2.41

Eq. (1.110) is useful to find either skin factor, s, formation damaged permeability, k_s, or the damaged or affected skin zone radius, r_s. However, since the skin zone covers an infinitely thin area and the pressure wave travels at high speed, it is difficult to detect transmissivity changes, then, r_s and k_s are difficult to be measured.

Eqs. (1.110) and (2.37) imply the skin factor along flow rate just increases or decreases the well pressure drop. However, this occurs because the well radius behaves as if its radius was modified by the value of the skin factor. Brons and Miller [74] defined the apparent or effective wellbore radius, r_wa, to be used in Eqs. (1.89), (1.94), and (1.100)

$$r_{wa}=r_w{e}^{-s}$$E2.42

Example 2.1

A well with a radius of 0.25 ft was detected to have a skin factor of 2. A skin factor of −2 was obtained after a stimulation procedure. Find the apparent radii and the percentage of change in the radius due to the stimulation. What conclusion can be drawn?

Solution

Application of Eq. (2.42) for the damaged‐well case gives:

$$r_{wa}=r_w{e}^{-s}=0.25e^{-2}=0.034 \hspace{0.17em}\text{ft=0}\text{.406} \hspace{0.17em}\text{in}$$

Application of Eq. (2.42) for the damaged‐well case gives:

$$r_{wa}=r_w{e}^{-s}=0.25e^{-(-2)}=1.848\hspace{0.17em}\hspace{0.17em}\text{ft=22}\text{.17}\hspace{0.17em}\hspace{0.17em}\text{in}$$

It can be observed that 1.847 × 100/0.034 ≅ 5460%, meaning that the stimulation helps the well to increase its radius 55 times. It can be concluded from the example that for positive skin factor values, the effective wellbore radius decreases (r_wa<r_w) and for negative skin factor values, the effective wellbore radius increases (r_wa>r_w).

The starting time of the semilog straight line defined by Ramey [9] in Eq. (2.22) allows determining mathematically where the radial flow starts, i.e., the moment wellbore storage effects no longer affect the test. Replacing into Eq. (2.22) the dimensionless parameters given by Eqs. (1.94) and (2.15) results [2]:

$$t_{SSL}=\frac{(200000+12000s)\mu C}{kh}$$E2.43

The application of Eq. (2.40) is twofolded. (1) It can be used for test design purposes. The duration of a pressure drawdown test should be last 10 times the value of t_SSL, so a significant portion of the radial flow regime can be observed and analyzed and (2) finding the semilog slope can be somehow confusing. Once the semilog line is drawn and permeability, skin factor, and wellbore storage are calculated, then, Eq. (2.40) can be used to find the starting point of the radial flow regime. Radial flow is correctly found if the t_SSL value agrees with the one chosen in the plot. This last situation is avoid if the pressure and pressure derivative plot is available since radial flow is observed once the pressure derivative curve gets flat as seen in Figure 2.6.

The declination stabilization time (time required to reach the boundaries and develop the pseudosteady‐state period) during the test can be from the maximum time at which the maximum pressure drops (not shown here) take place. This is:

$$t_{\max }=\frac{948\varphi \mu c_t{r}^2}{k}$$E2.44

From which;

$$t_{pss}=\frac{948\varphi \mu c_t{r}_e^2}{k}$$E2.45

For square or circular geometries, t_DA = 0.1 from Table 2.1. Replacing this value in Eq. (1.100) and solving for time leads to:

$$t_{pss}=\frac{1190\varphi \mu c_t{r}_e^2}{k}$$E2.46

from

$$r_{inv}=0.0325\sqrt{\frac{k{t}_p}{\mu \varphi c_t\hspace{0.17em}}}$$E2.47

For any producing time, t_p, the radius of investigation—not bigger than r_e—can be found.

The point reached by the disturbance does not imply fluid movement occurs there. The drainage radius is about 90% that value, then

$$r_d=0.029\sqrt{\frac{k{t}_p}{\mu \varphi c_t\hspace{0.17em}}}$$E2.48

Skin factor is a dimensionless quantity. This does not necessarily reflect the degree of either damage or stimulation of a well. Then, more practical measurement parameters ought to be used. One of this is the flow efficiency, FE, which implies what percentage of the total pressure drawdown is due to skin factor. The flow efficiency is defined as the ratio between the actual productivity index, J, and the ideal productivity index. The productivity index involves money since it is defined as the amount of pressure drop needed to produce a barrel of fluid per day. In other words, it is the energy required to produce one BPD. Mathematically;

$$J=\frac{q}{\overline{P}-P_{wf}}$$E2.49

$$J_{ideal}=\frac{q}{\overline{P}-P_{wf}-\mathrm{\Delta }{P}_s}$$E2.50

$$FE=\frac{J}{J_{ideal}}=1-\frac{\mathrm{\Delta }{P}_s}{\overline{P}-P_{wf}}$$E2.51

Table 2.1.

Shape factors for different drainage areas, taken from [8, 75].

FE < 1 is an indication that well damage exists, otherwise there is stimulation. The productivity index can be increased by:

• Increasing the permeability in the zone near the well—hydraulic fracturing;

• Reduce viscosity—steam injection, dissolvent, or in situ combustion;

• Damage removing—acidification;

• Increase well penetration;

• Reduce volumetric factor—choosing correct surface separators.

Other parameters to quantify well damage are [68]:

Damage ratio, DR

$$DR=\text{1}/FE$$E2.52

Damage ratios less than the unity indicate stimulation.

Damage factor, DF

$$DR=\text{1}-FE$$E2.53

Negative values of damage factors indicate stimulation. The damage factor can also be estimated from [68]:

$$DF=\frac{s}{s+ln(r_e/r_w)}=1-\frac{q(\text{actual})}{q(\text{ideal})}$$E2.54

Eq. (2.54) applies to circular‐shaped reservoir.

Productivity ratio, PR

$$PR=\frac{q}{q_a}=\frac{q(\text{ideal})}{q(\text{actual})}=\frac{ln(r_e/r_w)}{ln(r_e/r_w)+s}$$E2.55

Annual loss income, FD$L (USD$)

$$FD\$L=365q(OP)DF$$E2.56

where OP is oil price.

Example 2.2

What will be the annual loss of a well that produces 500 BFD, which has a damage factor of 8, drains an area of 120 acres and has a radius of 6 inches? Assume circular reservoir area and a price of oil crude of USD $ 55/barrel.

Figure 2.8.

Characteristics found in the Cartesian graph, taken from [68].

Solution

120 acres = 5,227,200 ft². If the area is circular, then: r = 1290 ft. Find the damage factor from Eq. (2.54);

$$DF=\frac{s}{s+ln(r_e/r_w)}=\frac{8}{8+ln(1290/0.5)}=0.5046$$

Find the yearly loss income using Eq. (2.56)

$$FD\$L=365q(OP)DF={\displaystyle \left(\text{365}\right)}{\displaystyle \left(\text{5}00\right)}{\displaystyle \left(\text{55}\right)}{\displaystyle \left(0.\text{5}0\text{46}\right)}=\text{USD}\$ \text{5}0\text{64922}$$

This indicates that the well requires immediate stimulation.

2.2.3.2 Reservoir limit test, RLT

It is a drawdown test run long enough to reach the reservoir boundaries. Normal pressure drawdown tests, during either radial flow or transient period test, are used to estimate formation permeability and artificial well conditions (C and s), while an RLT test—introduced by [76]—deals with boundaries and is employed to determine well drainage area or well drainage pore volume. In a Cartesian graph for a closed boundary system, Figure 2.8, three zones are distinguished [8, 68]: (i) skin and wellbore storage dominated zone, (ii) transient zone (radial flow), and (iii) pseudosteady‐state zone. As indicated by Eq. (1.129), the pressure drop is a linear function of time. Eq. (1.129) is given for circular reservoir geometry. For any geometry, the late time pseudosteady‐state solution involves the Dietz shape factor, [75], to extent the use of Eq. (1.129) for other reservoir geometries, as described in Table 2.1. Under this condition, Eq. (1.129) becomes [77]:

$$P_D=2\pi t_{DA}+\frac{1}{2}ln{\displaystyle \left(\frac{A}{r_w^2}\right)}+ln{\displaystyle \left(\frac{2.2458}{C_A}\right)}$$E2.57

Replacing in the above expression the dimensionless quantities given by Eqs. (1.89) and (1.94), it results:

$$P_{wf}=-{\displaystyle \left[\frac{0.23395qB}{\varphi c_{t}Ah}\right]}t+P_i-\frac{70.6q\mu B}{kh}{\displaystyle \left[ln\frac{A}{r_w^2}+ln{\displaystyle \left(\frac{2.2458}{C_A}\right)}+2s\right]}$$E2.58

From the slope, m*, and intercept, P_INT, of Eq. (2.58), the reservoir pore volume and Dietz shape factor [74] can be obtained from either:

$$V_p=-\frac{0.23395qB}{\hspace{0.17em}{c}_t{m}^{*}}$$E2.59

$$C_A=5.456\frac{m}{m^{*}}{e}^{2.303\frac{P_{1hr}-P_{INT}}{m}}$$E2.60

Once the value of C_A is obtained from Eq. (2.60), the reservoir geometry can be obtained from Table 2.1 by using the closest tabulated value (“exact for t_DA”) and confront with the time to develop pseudosteady‐state regime, (t_DA)_pss which is found from:

$${(t_{DA})}_{pss}=0.1833\frac{m^{*}}{m}{t}_{pss}$$E2.61

t_pss can be read from the Cartesian plot. However, this reading is inexact; therefore, it is recommended to plot the Cartesian pressure derivative and to find the exact point at which this becomes flat.

2.2.4 Tiab’s direct synthesis (TDS) technique

TDS technique is the latest methodology for well test interpretation. Its basis started in 1989 [70]. TDS’ creator was Tiab [71], who provided analytical and practical solutions for reservoir characterization using characteristic points or features—called by him “fingerprints”—read from a log‐log plot of pressure and pressure derivative [15], versus time. Since the introduction of TDS in 1995, several scenarios, reservoir geometries, fluid types, well configurations, and operation conditions. For instance, extension of TDS technique to elongated systems can be found in [13, 14, 16–19, 23, 24, 28, 30, 31]. Some applications of conventional analysis in long reservoirs are given in [20, 29, 38, 54]. For vertical and horizontal gas wells with and without use of pseudotime, refer to [22, 36, 39]. Special cases of horizontal wells are found in [12, 47]. For transient rate analysis, refer to [27, 35, 49]. Applications on heavy oil (non‐Newtonian fluids) can be found in [32, 34, 41, 42, 45, 52, 56, 62, 64]. For cases on shales reservoirs, refer to [49, 51, 56, 78]. Well test analysis by the TDS technique on secondary and tertiary oil recovery is presented by [25, 33, 60, 79]. For multirate testing in horizontal and vertical wells, refer, respectively, to [65, 67]. References [43, 46] are given for conductive faults. For deviated and partially penetrated wells, refer to [37, 64], respectively. TDS technique extended to multiphase flow was presented by [26]. Wedged and T‐shaped reservoirs can be found in [48] and coalbed‐methane reservoirs with bottom water drive are given in [53]. TDS technique is excellent for interpreting pressure test in hydraulically fractured vertical wells since unseen flow regimes can be generated [50, 69, 80]. The first publications on horizontal wells in naturally fractured and anisotropic media are given in [81, 82]. The threshold pressure gradient is dealt by [57, 72]. For vertical wells in double porosity and double permeability formations, refer, respectively, to [41, 83]. A book published by Escobar [56] presents the most recent topics covered by the TDS technique, and a more comprehensive state‐of‐the‐art on TDS technique is given by [58]. This book revolves around this methodology; therefore, practically, the whole content of [71]—pioneer paper of TDS technique—will be brought here:

The starting point is the definition of the dimensionless pressure derivative from Eq. (1.89);

$$t_D*P_D\prime =\frac{kh(t*\mathrm{\Delta }P\prime )}{141.2q\mu B}$$E2.62

By looking at Eqs. (2.17) and (2.18), we can conclude the difficulty of using hand mathematical operations with them. Instead of using these general solutions, Tiab [71] obtained partial solutions to the differential equation for each flow regime or time period. For instance, during early pseudosteady‐state, the governing equation reduces to:

$$P_D=\frac{t_D}{C_D}$$E2.63

Combination of Eqs. (1.94) and (2.15) results in:

$$\frac{t_D}{C_D}={\displaystyle \left(2.95\hspace{0.17em}\times \hspace{0.17em}{10}^{-4}\frac{h}{\mu }\right)}\frac{t}{C}$$E2.64

Replacing Eq. (1.89) in the above expression yields;

$${\displaystyle \left(\frac{kh}{141.2q\mu B}\right)}\mathrm{\Delta }P={\displaystyle \left(2.95\times {10}^{-4}\frac{kh}{\mu }\right)}\frac{t}{C}$$E2.65

Solving for C;

$$C={\displaystyle \left(\frac{qB}{24}\right)}\frac{t}{\Delta P}$$E2.66

The pressure derivative curve also has a straight line of unitary slope at early times. The equation of this line is obtained by taking the derivative of Eq. (2.63) with respect to the natural logarithm of t_D/C_D. So:

$${\displaystyle \left(\frac{t_D}{C_D}\right)}{P}_D\prime =\frac{t_D}{C_D}$$E2.67

Where the derivative of the dimensionless pressure is:

$$P_D\prime =d{P}_D/d{t}_D={\displaystyle \left(\frac{kh}{141.2q\mu B}\right)}dP/{\displaystyle \left(\frac{0.0002637k}{\varphi \mu c_t{r}_w^2}\right)}dt$$E2.68

Rearranging;

$$P_D\prime ={\displaystyle \left(\frac{26.856r_w^2\varphi c_{t}h}{qB}\right)}\Delta P\prime$$E2.69

Converting to dimensional form, the left‐hand side of Eq. (2.67) by using the definitions given by Eqs. (2.64) and (2.68):

$${\displaystyle \left(\frac{t_D}{C_D}\right)}{P}_D\prime =0.00792252{\displaystyle \left(\frac{kh}{q\mu B}\right)}{\displaystyle \left(\frac{\varphi c_{t}h{r}_w^2}{C}\right)}(t*\Delta P\prime )$$E2.70

Multiplying and dividing by 0.8935;

$${\displaystyle \left(\frac{t_D}{C_D}\right)}{P}_D\prime =0.007087{\displaystyle \left(\frac{kh}{q\mu B}\right)}{\displaystyle \left(\frac{\varphi c_{t}h{r}_w^2}{0.8935C}\right)}(t*\Delta P\prime )$$E2.71

Recalling Eq. (2.15), the above becomes:

$${\displaystyle \left(\frac{t_D}{C_D}\right)}{P}_D\prime =0.007087{\displaystyle \left(\frac{kh}{q\mu B}\right)}{\displaystyle \left(\frac{1}{C_D}\right)}(t*\Delta P\prime )$$E2.72

Since the unit slope is one, then C_D = 1, thus;

$${\displaystyle \left(\frac{t_D}{C_D}\right)}{P}_D\prime =\frac{kh(t*\mathrm{\Delta }P\text{'})}{141.2q\mu B}$$E2.73

From looking at Figure 2.6, both pressure and pressure derivative curves display a unitary slope at early times. Replacing Eqs. (2.64), (2.73) in (2.67) and solving for C will result:

$$C={\displaystyle \left(\frac{qB}{24}\right)}\frac{t}{t*\mathrm{\Delta }{P}^{\prime }}$$E2.74

As seen in Figure 2.6, the infinitely acting radial flow portion of the pressure derivative is a horizontal straight line with intercept of 1/2. The governing equation is:

$${{\displaystyle \left[{\displaystyle \left(\frac{t_D}{C_D}\right)}{P}_D\prime \right]}}_{\hspace{0.17em}r}=\frac{1}{2}$$E2.75

Combining the above equation with Eq. (2.73) results the best expression to estimate reservoir permeability:

$$k=\frac{70.6q\mu B}{h{(t*\Delta P\prime )}_r}$$E2.76

Subscript r stands for radial flow line. A customary use of TDS, as established by Tiab [71], is to provide suffices to identify the different flow regimes. For instance, pss stands for pseudosteady state, i stands for either initial or intercept, etc. In terms of pressure, the equation of this line is:

$$P_{Dr}=0.5{\displaystyle \left\{ln{{\displaystyle \left(\frac{t_D}{C_D}\right)}}_r+0.80907+ln{\displaystyle \left[C_D{e}^{2s}\right]}\right\}}$$E2.77

It is recommended to draw a horizontal line throughout the radial flow regime and choose one convenient value of (t*ΔP')_r falling on such line.

Tiab [71] also obtained the start time of the infinite line of action of the pressure derivative is:

$${{\displaystyle \left(\frac{t_D}{C_D}\right)}}_{sr}=10\hspace{0.17em}log{(C_D{e}^{2s})}^{10}$$E2.78

Replacing Eqs. (1.92) and (2.15) in the above equation will yield:

$$t_{sr}=\frac{\mu \hspace{0.17em}C}{6.9\hspace{0.17em}\times \hspace{0.17em}{10}^{-5}kh}{\displaystyle \left[ln{\displaystyle \left(\frac{0.8935C}{\varphi c_{t}h{r}_w^2}\right)}+2s\right]}$$E2.79

A better form of Eq. (2.78) was given by [84];

$${{\displaystyle \left(\frac{t_D}{C_D}\right)}}_{sr}=\frac{1}{\alpha }{\displaystyle \left[ln(C_D{e}^{2s})+ln{{\displaystyle \left(\frac{t_D}{C_D}\right)}}_{SR}\right]}$$E2.80

Setting a = 0.05 in the above equation and solving for C:

$$C=0.056\varphi c_{t}h{r}_w^2\hspace{0.17em}{\displaystyle \left(\frac{t_{Dsr}}{2s+ln{t}_{Dsr}}\right)}$$E2.81

t_Dsr is calculated with Eq. (1.94) letting t = t_sr.

The point of intersection, i, between the early time unit‐slope line defined by Eqs. (2.63) and (2.67) and the late‐time infinite‐acting line of the pressure derivative, defined by Eq. (2.75), is given by:

$${{\displaystyle \left(\frac{t_D}{C_D}{P}_D\prime \right)}}_i=0.5$$E2.82

$${{\displaystyle \left(\frac{t_D}{C_D}\right)}}_i=0.5$$E2.83

where i stands for intersection. After replacing the definitions given by Eqs. (1.94), (2.15), and (2.72) will, respectively, provide:

$${{\displaystyle \left(t*\Delta P\prime \right)}}_i=\frac{70.6q\mu B}{kh}$$E2.84

$$k=\frac{1695\mu C}{h{t}_i}$$E2.85

For the unit‐slope line, the pressure curve is the same as for the pressure derivative curve. Then, at the intersection point:

$${(\mathrm{\Delta }P)}_i={(t*\mathrm{\Delta }P)}_i=(t*\mathrm{\Delta }P\prime {)}_r$$E2.86

Tiab [71] correlated for C_De^2s > 100 permeability, wellbore storage coefficient, and skin factor with the coordinates of the maximum point—suffix x—displayed once the “hump” observed once wellbore storage effects start diminishing. These correlations are given as follows:

$${{\displaystyle \left(\frac{t_D}{C_D}{P}_D\prime \right)}}_x=0.35717{{\displaystyle \left(\frac{t_D}{C_D}\right)}}_x-0.50$$E2.87

$$log(C_D{e}^{2s})=0.35{{\displaystyle \left(\frac{t_D}{c_D}\right)}}^{1.24}$$E2.88

$$log(C_D{e}^{2s})=1.71{{\displaystyle \left(\frac{t_D}{C_D}{P}_D\prime \right)}}^{1.24}$$E2.89

Replacing Eqs. (2.64) and (2.73) into Eq. (2.87) leads to:

$${{\displaystyle \left(t*\Delta P\prime \right)}}_x={\displaystyle \left(0.015\frac{qB}{C}\right)}{t}_x-0.42{\displaystyle \left(\frac{141.2q\mu B}{kh}\right)}$$E2.90

Either formation permeability or wellbore storage coefficient can be determined using the coordinates of the peak, t_x and (t*ΔP′)_x. Solving for both of these parameters from Eq. (2.90) results:

$$k={\displaystyle \left(\frac{70.6q\mu B}{h}\right)}\frac{1}{{\displaystyle \left(0.014879qB/C\right)}\hspace{0.17em}{t}_x-{(t*\mathrm{\Delta }P\prime )}_x}$$E2.91

$$C=\frac{0.014879qB{t}_x}{{(t*\mathrm{\Delta }P\prime )}_x+{(t*\mathrm{\Delta }P\prime )}_r}$$E2.92

The constants in Eqs. (2.91) and (2.92) are slightly different as those in [58]. These new unpublished versions were performed by TDS’ creator.

Eq. (2.91) is so helpful to find reservoir permeability in short test when radial flow is absent which is very common in fall‐off tests. Once permeability is found from Eq. (2.91), solved for (t*ΔP′)_r from Eq. (2.76) and plot on a horizontal line throughout this value. Then, compare with the actual derivative plot and use engineering criterion to determine if the permeability value is acceptable. This means, if the straight line is either lower or higher than expected. Otherwise, new coordinates of the peak ought to be read for repeating the calculations since the hump should look some flat.

Substitution of Eqs. (2.64) and (2.73) in Eqs. (2.88) and (2.89) allows obtaining two new respective correlations for the determination of the mechanical skin factor:

$$s=0.171{{\displaystyle \left(\frac{t_x}{t_i}\right)}}^{1.24}-0.5\hspace{0.17em}\hspace{0.17em}ln{\displaystyle \left(\frac{0.8935C}{\varphi \hspace{0.17em}h{c}_t{r}_w^2}\right)}$$E2.93

$$s=0.921{{\displaystyle \left(\frac{{(t*\Delta P\prime )}_x}{{(t*\Delta P\prime )}_i}\right)}}^{1.1}-0.5\hspace{0.17em}\hspace{0.17em}ln{\displaystyle \left(\frac{0.8935C}{\varphi h{c}_t{r}_w^2}\right)}$$E2.94

Sometimes, the reading of the peak coordinates may be wrong due to the flat appearance of it. Then, it should be a good practice to estimate the skin factor using both Eqs. (2.93) and (2.94). These values should match each other.

Divide Eq. (2.87) by Eq. (2.75); then, in the result replace Eqs. (2.64) and (2.73) and solve for both permeability and wellbore storage:

$$k=4745.36\frac{\mu C}{h{t}_x}{\displaystyle \left\{\frac{{(t*\mathrm{\Delta }P\prime )}_x}{{(t*\mathrm{\Delta }P\prime )}_r}+1\right\}}$$E2.95

$$C=\frac{0.014879qB{t}_x}{{(t*\mathrm{\Delta }P\text{'})}_x+{(t*\mathrm{\Delta }P\prime )}_r}$$E2.96

This last expression is useful to find wellbore storage coefficient when the early unitary slope line is absent.

TDS technique has a great particularity: for a given flow regime, the skin factor equation can be easily derived from dividing the dimensionless pressure equation by the dimensionless derivative equation of such flow regime. Then, the division of Eq. (2.77) by Eq. (2.75) leads to the below expression once the dimensionless parameters given by Eqs. (1.89), (1.94), and (2.73) are replaced in the resulting quotient. Solving for s from the final replacement leads to:

$$s=0.5{\displaystyle \left(\frac{\mathrm{\Delta }{P}_r}{{(t*\mathrm{\Delta }P\prime )}_r}-ln{\displaystyle \left[\frac{k\hspace{0.17em}{t}_r}{\varphi \mu c_t{r}_w^2}\right]}+7.43\right)}$$E2.97

being t_r any convenient time during the infinite‐acting radial flow regime throughout which a horizontal straight line should have been drawn. Read the ΔP_r corresponding to t_r. Comparison between Eqs. (2.38) and (2.76) allows concluding:

$$m=2.303{(t*\mathrm{\Delta }P\prime )}_r=ln(10){(t*\mathrm{\Delta }P\prime )}_r$$E2.98

which avoids the need of using the semilog plot if the skin pressure drop is needed to be estimated by Eq. (2.40), otherwise, Eq. (2.40) becomes:

$$\mathrm{\Delta }{P}_s=\left|2{(t*\mathrm{\Delta }P\prime )}_r\right|s,{\displaystyle \{\begin{array}{l}\text{if}\hspace{0.17em}s>0\to \mathrm{\Delta }{P}_s>0\\ \text{if}\hspace{0.17em}s<0\to \mathrm{\Delta }{P}_s<0\end{array}}$$E2.99

For the determination of well‐drainage area, Tiab [69] expressed Eq. (2.75) as:

$${(t_D*P_D\prime )}_r=\frac{1}{2}$$E2.100

Also, Tiab [69] differentiated the dimensionless pressure with respect to dimensionless time in Eq. (2.57), so:

$$t_D*P_D\prime =2\pi t_{DA}$$E2.101

Then, Tiab [69] based on the fact that two given flow regime governing equations can be intersected each other, regardless the physical meaning of such intersection, and solving for any given parameter, intercepted Eqs. (2.100) with (2.101), then, replaced in the resulting expression the dimensionless quantities given by Eqs. (2.92), (2.97), and (2.62) and solved for the area given in ft²:

$$A=\frac{k{t}_{rpssi}}{301.77\varphi \mu c_t}$$E2.102

Furthermore, Chacon et al. [85] replaced the dimensionless time given by Eq. (1.100) and the dimensionless pressure derivative of Eq. (2.62) into Eq. (2.102) and also solved for the area in ft²:

$$A=\frac{0.234qB{t}_{pss}}{\varphi c_t^{}h{(t*\mathrm{\Delta }P\prime )}_{pss}}$$E2.103

The above expression uses any convenient point, t_pss and (t*ΔP’)_pss, during the late time pseudosteady‐state period. Because of noisy pressure derivative data, the readings of several arbitrary points may provide, even close, different area values. Therefore, it is convenient to use an average value. To do so, it is recommended to draw the best late‐time unit‐slope line passing through the higher number of pressure derivative points and extrapolate the line at the time of 1 h and read the pressure derivative value, (t*ΔP')_pss1. Under these circumstances, Eq. (2.103) becomes:

$$A=\frac{0.234qB}{\varphi c_t^{}h{(t*\mathrm{\Delta }P\prime )}_{pss1}}$$E2.104

Eqs. (2.102) through (2.104) apply only to closed‐boundary reservoirs of any geometrical shape. For constant‐pressure reservoirs, the works by Escobar et al. [28, 54] for TDS technique (summary given in Table 2.2) and for conventional analysis are used for well‐drainage area determination in circular, square, and elongated systems.

TDS technique has certain step‐by‐step procedures which not necessarily are to be followed since the interpreter is welcome to explore and use TDS as desired. Then, they are not provided here but can be checked in [69, 71].

Example 2.3

Taken from [68]. The pressure and pressure derivative data given in Table 2.3 corresponds to a drawdown test of a well. Well, fluid, and reservoir data are given below:

rw = 0.267 ft	q = 250 BPD	μ = 1.2 cp
ct = 26.4 × 10^−5psi^−1	h = 16 pies	ϕ = 18%	B = 1.229 bbl/BF

Find permeability, skin factor, drainage area, and flow efficiency by conventional analysis. Find permeability, skin factor, and three values of drainage area using TDS technique:

Solution

Conventional analysis. Figure 2.9 and 2.10 present the semilog and Cartesian plots, respectively, to be used in conventional analysis. From Figure 2.9, the semilog slope, m, is of 18 psia/cycle and P_1hr = 2308 psia. Permeability and skin factor are calculated using Eqs. 2.38 and 2.39, respectively, thus:

$$k=\left|\frac{162.6\hspace{0.17em}(250)(1.2)(1.229)}{(-18)(16)}\right|=208\hspace{0.17em}\text{md}$$

$$s=1.1513{\displaystyle \left[\frac{2308-2733}{-18}-log{\displaystyle \left(\frac{208}{0.18(1.2)(26.4\times {10}^{-5}){(0.267)}^2}\right)}+3.23\right]}=22.15$$

Find the pressure loss due to skin factor with Eq. (2.40);

$$\mathrm{\Delta }{P}_s=\left|0.87(-18)\right|22.15=346.7\hspace{0.17em}\text{psia}$$

Since the average reservoir pressure is not reported, then, the initial pressure value is taken instead. Eq. (2.51) allows estimating the flow efficiency.

Table 2.2.

Summary of equations, taken from [28].

t, h	Pwf, psia	DP, psia	t*DP′, psia/h	t, h	Pwf, psia	DP, psia	t*DP′, psia/h
0.00	2733	0		5	2312	421	65.42
0.10	2703	30	31.05	7	2293	440	35.32
0.20	2672	61	58.95	9.6	2291	442	5.86
0.30	2644	89	84.14	12	2290	443	5.85
0.40	2616	117	106.30	16.8	2287	446	7.63
0.65	2553	180	129.70	33.6	2282	451	7.99
1.00	2500	233	135.15	50	2279	454	7.94
1.50	2440	293	151.90	72	2276	457	10.50
2.00	2398	335	127.26	85	2274	459	12.18
3.00	2353	380	102.10	100	2272	461	13.36
4.00	2329	404	81.44

Table 2.3.

Pressure and pressure derivative versus time data for example 2.3.

Figure 2.9.

Semilog plot for example 2.3.

Figure 2.10.

Cartesian plot for example 2.3.

Figure 2.11.

Pressure and pressure derivative plot for example 2.3.

$$FE=1-\frac{346.9}{2733-2272}=24.75\hspace{0.17em}\%$$

From the Cartesian plot, Figure 2.10, is read the following data:

m* = −0.13 psia/h

PINT = 2285 psia

tpss ≈ 50 h

Use Eq. (2.59) to find well drainage area:

$$A=-\frac{0.234qB}{\hspace{0.17em}\varphi h{c}_t{m}^{*}}=-\frac{0.234(250)(1.229)}{\hspace{0.17em}(0.18)(16)(26.4\times {10}^{-5})(-0.13)}=727391.1\hspace{0.17em}{\text{ft}}^{\text{2}}=16.7\hspace{0.17em}\text{Ac}$$

Find the Dietz shape factor with Eq. (2.60);

$$C_A=5.456\frac{-18}{-0.13}{e}^{{\displaystyle \left[\frac{2.303{\displaystyle \left(2308-2285\right)}}{-18}\right]}}=39.82$$

As observed in Table 2.1, there exist three possible well drainage area geometry values (hexagon, circle, and square) close to the above value. To discriminate which one should be the appropriate system geometry find the dimensionless time in which pseudosteady‐state period starts by using Eq. (2.61):

$${{\displaystyle \left(t_{DA}\right)}}_{pss}=0.1833\frac{-0.13}{-18}50=0.066\approx 0.1$$

TDS technique. The following are the characteristic points read from Figure 2.11:

(t*ΔP')r = 7.7 psia	tr = 33.6 h	ΔPr = 451 psia
tpss = 85 h	(t*ΔP')pss = 12.18 psia	trpi = 58 h
(t*ΔP')pss1 = 0.14 psia

Find permeability and skin factor with Eqs. (2.76) and (2.97), respectively:

$$k=\frac{70.6q\mu B}{h{(t*\mathrm{\Delta }P\text{'})}_r}=\frac{70.6(250)(1.2)(1.229)}{(16)(7.7)}=211.3\hspace{0.17em}\text{md}$$

$$s=0.5{\displaystyle \left[\frac{451}{7.7}-ln{\displaystyle \left(\frac{211.3(33.6)}{0.18(1.2)(26.4\times {10}^{-5})({0.267}^2)}\right)}+7.43\right]}=22.4$$

Determine the well drainage area with Eqs. (2.102) and (2.103), thus;

$$A=\frac{211.3(58)}{301.77(0.18)(1.2)(26.4\times {10}^{-5})(43560)}=16.35\hspace{0.17em}\text{Ac}$$

$$A=\frac{0.234(250)(1.229)(85)}{(0.18)(26.4\times {10}^{-5})(16)(12.18)(43560)}=15.15\hspace{0.17em}\text{Ac}$$

$$A=\frac{0.234(250)(1.229)}{(0.18)(26.4\times {10}^{-5})(16)(0.14)(43560)}=15.5\hspace{0.17em}\text{Ac}$$

Even, more parameters can be reestimated with TDS technique for verification purposes but it will not be performed for saving‐space reasons. However, the reader is invited to read the coordinates of the peak and the intersection point of the wellbore storage and radial flow lines. Then, estimate formation permeability with Eqs. (2.84), (2.85), and (2.91). Also, find the wellbore storage coefficient using Eqs. (2.74), (2.81), (2.92), and (2.96) and skin factor with Eqs. (2.93) and (2.94).

Example 2.4

Taken from [68] with the data from the previous example, Example 2.3, determine t_SSL and find if the well fluid level is increasing or decreasing in the annulus if the well has a drill pipe with 2 in external diameter inside a liner with 5 in of inner diameter including joint gaskets. The density of the wellbore fluid is 42.5 lbm/ft³.

Solution

From Figure 2.11, a point is chosen on the early unit‐slope line. This point has coordinates: DP = 59 psia and t = 0.2 h. Wellbore storage coefficient is found with Eq. (2.21):

$$C=\frac{(250)(1.229)}{24}\frac{0.2}{59}=0.0434\hspace{0.17em}\text{bbl/psia}$$

Solving for annulus capacity from Eq. (2.5);

$$V_u={\displaystyle \left(\frac{\rho }{144}\right)}C={\displaystyle \left(\frac{42.5}{144}\right)}0.0434=0.0128\hspace{0.17em}\text{bbl/ft}$$

The theoretical capacity is found with Eq. (2.45), so:

$$V_u=0.0009714(5^2-2^2)=0.0204\hspace{0.17em}\text{bbl/ft}$$

This leads to the conclusion that the annular liquid is falling.

2.3. Multiphase flow

According to Perrine [86], the single fluid flow may be applied to the multiple fluid flow systems when the gas does not dominate the pressure tests, it means liquid production is much more relevant than gas flow. Under this condition, the diffusivity equation, Eq. (1.27), will result and the total fluid mobility is determined by Eq. (1.24). We also mentioned in Chapter 1 that Martin [63] provided some tips for a better use of Perrine method. Actually, Perrine method works very well in liquid systems.

The semilog equations for drawdown and build tests are, respectively, given below:

$$P_{wf}=P_i-\frac{162.6q_t}{{\lambda }_{t}h}{\displaystyle \left(log\frac{{\lambda }_{t}t}{1688\varphi c_t{r}_w^2}+0.869s\right)}$$E2.105

$$P_{ws}=P_i-\frac{162.6q_t}{{\lambda }_{t}h}log{\displaystyle \left(\frac{t_p+\mathrm{\Delta }t}{\mathrm{\Delta }t}\right)}$$E2.106

The flow rate is estimated by:

$$q_t={\displaystyle \left(q_o{B}_o+(q_g-q_o{R}_s/1000)B_g+q_w{B}_w\right)}/B_o$$E2.107

Eq. (2.107) is recommended when oil flow dominates the test. It is removed from the denominator, otherwise. It advised to use consistent units in Eq. (2.107) meaning that the gas flow rate must be in Mscf/D and the gas volume factor bbl/SCF.

Once the semilog slope has been estimated, the total mobility, the phase effective permeabilities, and the mechanical skin factor are found from:

$${\lambda }_t=-\frac{162.6q_t}{mh}$$E2.108

$$k_L=-\frac{162.6q_L{B}_L{\mu }_L}{mh};\hspace{0.17em}L=\text{water}\hspace{0.17em}\text{or}\hspace{0.17em}\text{oil}$$E2.109

$$k_g=-\frac{162.6{\displaystyle \left(q_g-q_o{R}_s/1000\right)}{B}_g{\mu }_g}{mh}$$E2.110

$$s=1.1513{\displaystyle \left(\frac{P_{wf}-P_{1hr}}{m}-log{\displaystyle \left(\frac{{\lambda }_t}{\varphi c_t{r}_w^2}\right)}+3.23\right)}$$E2.111

The best way of interpreting multiphasic flow tests in by using biphasic and/or triphasic pseudofunctions. Normally, well test software uses empirical relationships to estimate relative permeability data. The accuracy of the following expression is sensitive to the relative permeability data:

$$m(P)={\displaystyle \underset{P_0}{\overset{P}{\int }}\frac{k_{ro}}{{\mu }_o{B}_o}dP}$$E2.112

The expressions used along this textbook for reservoir characterization may apply for both single fluid and multiple fluid production tests. Single mobility has to be changed by total fluid mobility and individual flow rate ought to be replaced by the total fluid rate. Just to cite a few of them, Eqs. (2.66), (2.76), (2.85), (2.91), (2.92), and (2.97) become:

$$C={\displaystyle \left(\frac{q_t}{24}\right)}\frac{t}{\mathrm{\Delta }P}$$E2.113

$${\lambda }_t=\frac{k}{\mu }|{}_t=\frac{70.6q_t}{h{(t*\mathrm{\Delta }P\text{'})}_r}$$E2.114

$${{\displaystyle \left(\frac{k}{\mu }\right)}}_t=\frac{1695C}{h{t}_i}$$E2.115

$${\lambda }_t={\displaystyle \left(\frac{70.6q_t}{h}\right)}\frac{1}{(0.014879q_t/C)t_x-{(t*\mathrm{\Delta }P\prime )}_x}$$E2.116

$$C=\frac{0.014879q_t{t}_x}{{(t*\mathrm{\Delta }P\text{'})}_x+{(t*\mathrm{\Delta }P\prime )}_r}$$E2.117

$$s=0.5{\displaystyle \left[\frac{\mathrm{\Delta }{P}_r}{{(t*\mathrm{\Delta }P)}_r}-ln{\displaystyle \left(\frac{{\lambda }_t{t}_r}{\varphi c_t{r}_w^2}\right)}+7.43\right]}$$E2.118

Also, the effective liquid permeabilities are found using the individual viscosity, rate, and volume factor. Then, Eq. (2.76) applied to oil and water will yield:

$$k_o=\frac{70.6q_o{\mu }_o{B}_o}{h{(t*\mathrm{\Delta }P\prime )}_r}$$E2.119

$$k_w=\frac{70.6q_w{\mu }_w{B}_w}{h{(t*\mathrm{\Delta }P\text{'})}_r}$$E2.120

However, from a multiple fluid test, it is a challenge to find the reservoir absolute permeability. Several methods have been presented. For instance, Al‐Khalifah et al. [87] presented a sophisticated method applied to either drawdown or multiple rate tests. Their method even includes the estimating of the saturation change respect to pressure. However, we presented the method by Kamal and Pan [88] which applies well for liquid fluid. Relative permeabilities must be known for its application. Once effective permeabilities are found, let us say from Eqs. (2.119) and (2.120), estimate the permeability ratio k_o/k_w and find the water saturation from the relative permeability curves as schematically depicted in Figure 2.12 (left). Then, using the estimated water saturation value, enter Figure 2.12 (right) and read a value from a relative permeability curve. Use the most dominant flow curve. The dominant phase is assumed to be oil for the example in Figure 2.12. Since both phase effective permeability and phase relative permeability are known, the absolute permeability is found from the definition of relative permeability:

Figure 2.12.

Determination of absolute permeability as outlined by Kamal and Pan [88].

$$k=\frac{k_o}{k_{ro}}$$E2.121

Further recommendations for handling multiphase flow tests are presented by Al‐Khalifah et al. [89] and are also reported by Stanislav and Kabir [7].

2.4. Partial penetration and partial completion

When a well penetrates a small part of the formation thickness, hemispherical flow takes place. See Figure 2.13 top. When the well is cased above the producer range and only a small part of the casing is perforated, spherical flow occurs in the region near the face of the well. See Figure 2.13 bottom. As the transient moves further into depth of the formation, the flow becomes radial, but if the test is short, the flow will be spherical. Both types of flow are characterized by a slope of −1/2 in the log‐log plot of pressure derivative versus time [90, 91]. Theoretically, before either hemispherical or spherical flow takes place, there exists a radial flow regime occurring by fluids withdrawn from the formation thickness that is close in height to the completion interval. This represents the transmissibility of the perforated interval. Actually, this flow regime is unpractical to be seen mainly because of wellbore storage effects. We will see further in this chapter that there are especial conditions for hemispherical/spherical flow to be observed which occur later that the completion‐interval‐limited radial flow regime. Both hemispherical and spherical flow vanished when the top and bottom boundaries have been fully reached by the transient wave; the true radial flow is developed throughout the full reservoir thickness.

Figure 2.13.

Ideal flow regimes in partial penetration (top) and partial completion (bottom) systems, after [66].

The apparent skin factor, s_a, obtained from pressure transient analysis is a combination of several “pseudoskin” factors such as [91]:

$$s_a=s+s_p+s_{\theta }+s_{cp}+\mathrm{\dots \dots }$$E2.122

where s is the true damage factor caused by damage to the well portion, s_p is the pseudoskin factor due to the restricted flow entry, s_q is the pseudoskin factor resulting from a well deviation angle, and s_cp is the pseudoskin due to a change in permeability near the face of the well. s_p can be estimated from [92]:

$$s_p={\displaystyle \left(\frac{h}{h_p}-1\right)}ln\hspace{0.17em}{h}_D$$E2.123

h_p = length of perforated or open interval. The equations of dimensionless thickness, h_D, for hemispherical and spherical flow, respectively:

$$h_D=\frac{h}{r_w}\sqrt{\frac{k_h}{k_z}}$$E2.124

$$h_D=\frac{h}{2r_w}\sqrt{\frac{k_h}{k_z}}$$E2.125

where k_h is the horizontal permeability, k_z = k_v is the vertical permeability. The contribution of the pseudoskin of an inclined well is given by Cinco et al. [92]:

$${\psi }^{\prime }={\tan }^{-1}{\displaystyle \left(\sqrt{\frac{k_z}{k_h}}tan\psi \right)}$$E2.126

$$s_{\theta }=-{{\displaystyle \left(\frac{{\psi }^{\prime }}{41}\right)}}^{2.06}-{{\displaystyle \left(\frac{{\psi }^{\prime }}{56}\right)}}^{1.865}\times log{\displaystyle \left(\frac{h}{100r_w}\right)}$$E2.127

According to Cinco et al. [92], the above equation is valid for 0° ≤ q ≤ 75°, h/r_w > 40, and t_D > 100. Note that Eq. (2.127) could provide a negative value. This is because the deviation at the face of the well increases the flow area or presents reservoir pseudothickness. The pseudoskin responding for permeability changes near wellbore is given by [93]:

$$s_{cp}=\frac{h}{h_p}{\displaystyle \left[1-0.2{\displaystyle \left(\frac{r_s-r_w}{h_p}\right)}\right]}{\displaystyle \left(\frac{k-k_s}{k_s}\right)}ln\frac{r_s}{r_w}$$E2.128

Example 2.4

Taken from [91]. A directional well which has an angle to the vertical of 24.1° has a skin factor s = −0.8. The thickness of the formation is 100 ft, the radius of the wellbore is 0.3 ft, and the horizontal to vertical permeability ratio is 5. Which portion of the damaged corresponds to the deviation of the well?

Solution

The deviation angle affected by the anisotropy is estimated with Eq. (2.126);

$${\psi }^{\prime }={\tan }^{-1}{\displaystyle \left(\sqrt{5}tan(24.1)\right)}=45°$$

The pseudoskin factor caused by well deviation is found from Eq. (2.127):

$$s_{\theta }=-{{\displaystyle \left(\frac{45}{41}\right)}}^{2.06}-{{\displaystyle \left(\frac{45}{56}\right)}}^{1.865}log{\displaystyle \left(\frac{100}{100(0.3)}\right)}=-1.56$$

From Eq. (2.122);

$$s_a=s+s_{\theta }=-0.8-1.56=-2.36$$

Therefore, 66.1 % of the skin factor is due to the well deviation.

2.4.1 Conventional analysis for spherical flow

The diffusivity equation for spherical flow assuming constant porosity, compressibility, and mobility is given by Abbott et al. [90]:

$$\frac{1}{r^2}\frac{\partial }{\partial r}{\displaystyle \left(r^2\frac{\partial P}{\partial r}\right)}=\frac{\varphi \mu c{}_t}{k_{sp}}\frac{\partial P}{\partial t}$$E2.129

where k_sp is the spherical permeability which is defined as the geometrical mean of the vertical and horizontal permeabilities:

$$k_{sp}=\sqrt[3]{k_{\text{}v}{k}_h^2}=k_{hs}$$E2.130

The physical system is illustrated in Figure 2.14, right. This region is called a “spherical sink.” r_sw is given by:

$$r_{sw}=\frac{h_p}{2ln{\displaystyle \left(\frac{h_p}{r_w}\right)}}$$E2.131

The spherical flow equations for pressure drawdown and pressure buildup when the flow time is much longer than the shut‐in time were presented by [94]:

$$P_{wf}=P_i-\frac{70.6q\mu B}{k_{sp}{r}_{sw}}(1+s_{sp})+\frac{2453q\mu B}{k_{sp}^{3/2}}\sqrt{\varphi \mu c_t}\frac{1}{\sqrt{t}}$$E2.132

$$P_{ws}=P_{wf}+\frac{70.6q\mu B}{k_{sp}{r}_{sw}}(1+s_{sp})-\frac{2453q\mu B}{k_{sp}^{3/2}}\sqrt{\varphi \mu c_t}{\displaystyle \left[\frac{1}{\sqrt{t_p}}+\frac{1}{\sqrt{\mathrm{\Delta }t}}-\frac{1}{\sqrt{t_p+\mathrm{\Delta }t}}\right]}$$E2.133

Figure 2.14.

Cylindrical, hemispherical, and spherical sinks, after [66].

The spherical pressure buildup equation when the flow time is shorter than the shut‐in time:

$$P_{ws}=P_i-\frac{2453q\mu B}{k_{sp}^{3/2}}\sqrt{\varphi \mu c_t}{\displaystyle \left[\frac{1}{\sqrt{\mathrm{\Delta }t}}-\frac{1}{\sqrt{t_p+\mathrm{\Delta }t}}\right]}+\frac{70.6q\mu B}{k_{sp}{r}_{sw}}{s}_{sp}$$E2.134

Then, from a Cartesian plot of P_wf as a function of t ^−1/2, for drawdown, or P_ws as a function of either [t_p^−1/2 + Δt^−1/2−(t_p + Δt)^−1/2] or [Δt^−1/2−(t_p + Δt)^−1/2] for buildup, we obtain a line which slope, m, and intercept, I, can be used to estimate tridimensional permeability and geometrical (spherical) skin factor.

$$k_{sp}={{\displaystyle \left(-\frac{2453q\mu B}{m}\sqrt{\varphi \mu c_t}\right)}}^{2/3}$$E2.135

$$s_{sp}=\frac{(I-P_{wf})k_{sp}{r}_{sw}}{70.6q\mu B}-1$$E2.136

Once the spherical permeability is known, we solve for the vertical permeability from Eq. (2.130), and then, estimate the value of skin effects due to partial penetration [94]:

$$s_c={\displaystyle \left(\frac{1}{b}-1\right)}{\displaystyle \left[ln{h}_D-G\right]}$$E2.137

where b = h_p/h. h_D can be estimated from Eq. (2.125), and G is found from [94]:

$$G=2.948-7.363b+11.45b^2-4.576b^3$$E2.138

2.4.2 Conventional analysis for hemispherical flow

The model for hemispheric flow is very similar to that of spherical flow [94]. The difference is that a boundary condition considers half sphere. Figure 2.14 (left) outlines the geometry of such system. The drawdown and pressure equations are given below [94]:

$$P_{wf}=P_i-\frac{141.2q\mu B}{k_{sp}{r}_{sw}}(1+s_{sp})+\frac{4906q\mu B}{k_{sp}^{3/2}}\sqrt{\varphi \mu c_t}\frac{1}{\sqrt{t}}$$E2.139

$$P_{ws}=P_{wf}+\frac{141.2q\mu B}{k_{sp}{r}_{sw}}(1+s_{sp})-\frac{4906q\mu B}{k_{sp}^{3/2}}\sqrt{\varphi \mu c_t}{\displaystyle \left[\frac{1}{\sqrt{t_p}}+\frac{1}{\sqrt{\mathrm{\Delta }t}}-\frac{1}{\sqrt{t_p+\mathrm{\Delta }t}}\right]}$$E2.140

$$P_{ws}=P_i-\frac{4906q\mu B}{k_{sp}^{3/2}}\sqrt{\varphi \mu c_t}{\displaystyle \left[\frac{1}{\sqrt{\mathrm{\Delta }t}}-\frac{1}{\sqrt{t_p+\mathrm{\Delta }t}}\right]}+\frac{141.2q\mu B}{k_{sp}{r}_{sw}}{s}_{sp}$$E2.141

As for the spherical case, from a Cartesian plot of P_wf as a function of t^−1/2, for drawdown, or P_ws as a function of either [t_p^−1/2 + Δt^−1/2−(t_p + Δt)^−1/2] or [Δt^−1/2− (t_p + Δt)^−1/2] for buildup, we obtain a line which slope, m, and intercept, I, can be used to estimate spherical permeability and geometrical (spherical) skin factor.

$$k_{sp}={{\displaystyle \left(-\frac{4906q\mu B}{m}\sqrt{\varphi \mu c_t}\right)}}^{2/3}$$E2.142

$$s_{sp}=\frac{(I-P_{wf})k_{sp}{r}_{sw}}{141.2q\mu B}-1$$E2.143

2.4.3 TDS for spherical flow

Moncada et al. [66] presented the expressions for interpreting both pressure drawdown or buildup tests in either gas or oil reservoirs using the TDS methodology. Spherical permeability is estimated by reading the pressure derivative at any arbitrary time during which spherical flow can calculate spherical permeability and the spherical skin factor also uses the pressure reading at the same chosen time:

$$k_{sp}={{\displaystyle \left(\frac{1227qB\mu }{{(t*\mathrm{\Delta }P\text{'})}_{sp}}\sqrt{\frac{\varphi \mu c_t}{t_{sp}}}\right)}}^{2/3}$$E2.144

$$s_{sp}=34.74\sqrt{\frac{\varphi \mu c_t{r}_{sw}^2}{k_{sp}{t}_{sp}}}{\displaystyle \left[\frac{{{\displaystyle \left(\mathrm{\Delta }P\right)}}_{sp}}{2{{\displaystyle \left(t*\mathrm{\Delta }P\prime \right)}}_{sp}}+1\right]}-1$$E2.145

The total skin, s_t, is defined as the sum of all skin effects at the well surroundings:

$$s_t=\frac{s}{b}+s_c+s_{sp}$$E2.146

If the radial flow were seen, the horizontal permeability can be estimated from:

$$k_H=k=\frac{70.6qB\mu }{h_p{(t*\mathrm{\Delta }P\prime )}_{r1}}$$E2.147

The suffix r1 implies the first radial flow.

Moncada et al. [66] observed that the value of the derivative for the late radial flow in spherical geometry is equivalent to 0.0066 instead of 0.5 as of the radial system. In addition, the slope line −½ corresponding to the spherical flow and the late radial flow line of the curve of the dimensionless pressure derivative in spherical symmetry intersect in:

$${{\displaystyle \left(t_D*P_D\prime \right)}}_i=\frac{1}{2\sqrt{\pi }}{t}_{Dsp}^{-1/2}=0.0066$$E2.148

Replacing the dimensionless time results:

$$t_i=6927748.85\frac{\phi \mu c_t{r}_{sw}^2}{k_{sp}}$$E2.149

In the above equation, suffix i denotes the intersection between the spherical flow and the late radial flow. If the radial flow is not observed, this time can give an initial point to draw the horizontal line corresponding to the radial flow regime, from which horizontal permeability is determined. This point can also be used to verify spherical permeability, k_sp. Another equation defining the mentioned dimensionless time can be found from the intersection of the slope line −½ (spherical flow) with the radial line of late radial flow but in radial symmetry, knowing that:

$${{\displaystyle \left(t_D*P_D\text{'}\right)}}_i=\frac{k_{}^{3/2}h}{4k_{sp}^{3/2}\sqrt{\pi r_{sw}^2}}\frac{1}{\sqrt{t_D}}=0.5$$E2.150

Replacing the dimensionless time will give:

$$t_i=301.77\frac{k_{}^2{h}^2\varphi \mu c_t}{k_{sp}^3}$$E2.151

Combining Eqs. (2.149) and (2.151), an expression to find the spherical wellbore radius, r_sw:

$$r_{sw}=0.0066\frac{k_{}h}{k_{sp}}$$E2.152

2.4.4 TDS for hemispherical flow

Here the same considerations are presented in Section 2.4.3. Using a pressure and a pressure derivative value reading at any time during hemispherical flow allows finding hemispherical permeability and partial penetration skin [66],

$$k_{hs}={{\displaystyle \left(\frac{2453qB\mu }{{(t*\mathrm{\Delta }P\text{'})}_{hs}}\sqrt{\frac{\varphi \mu c_t}{t_{hs}}}\right)}}^{2/3}$$E2.153

$$s_{hs}=34.74\sqrt{\frac{\varphi \mu c_t{r}_{sw}^2}{k_{hs}{t}_{hs}}}{\displaystyle \left[\frac{{{\displaystyle \left(\mathrm{\Delta }P\right)}}_{hs}}{2{{\displaystyle \left(t*\mathrm{\Delta }P\text{'}\right)}}_{hs}}+1\right]}-1$$E2.154

Moncada et al. [66] also found that the derivative in spherical geometry of the late radial flow corresponds to 0.0033 instead of 0.5 as of the radial system. This time the line of radial flow and hemispheric flow, in hemispherical symmetry, intersect in:

$${{\displaystyle \left(t_D*P_D\text{'}\right)}}_i=\frac{1}{2\sqrt{\pi t_{Dsh}}}=0.0033$$E2.155

From where,

$$t_i=27710995.41\frac{\varphi \mu c_t{r}_{sw}^2}{k_{hs}}$$E2.156

As for the spherical case, there exists an expression to define the intersection time of the −½ slope line of the hemispherical flow regime pressure derivative and the late radial flow line pressure derivative but, now, in radial symmetry:

$${{\displaystyle \left(t_D*P_D\text{'}\right)}}_i=\frac{k_{}^{3/2}h}{2k_{hs}^{3/2}\sqrt{\pi r_{sw}^2}}\frac{1}{\sqrt{t_D}}=0.5$$E2.157

$$t_i=1207.09\frac{k_{}^2{h}^2\varphi \mu c_t}{k_{hs}^3}$$E2.158

This point of intersection in radial symmetry gives the following equation:

$$r_{sw}=0.0033\frac{k_{}h}{k_{hs}}$$E2.159

Skin factors are estimated in a manner similar to Section 2.4.3.

2.4.5 Wellbore storage and perforation length effects on hemispherical/spherical flow

It is important to identify the range of WBS values, which can influence the interpretation of the spherical and hemispheric flow regime. Figure 2.15 is a plot of P_D vs. t_D providing an idea of the storage effect. As can be seen, the pressure response for several C_D values can be distinguished when storage is low (<10), whereas for larger C_D values, the response is almost identical. For C_D < 10, the slope of −½ that characterizes both spherical and hemispherical flow is well distinguished. For values of 10 < C_D < 100, the slope of −½ is more difficult to identify. For values of C_D > 100, the spherical flow regime has been practically masked by storage, which makes it impossible to apply the technique presented above to estimate the vertical permeability. Then, to ensure there is no C_D masking, it should be less than 10 [66].

The length of the completed interval or the length of the partial penetration, h_p, plays an important role in defining the spherical/hemispherical flow. The presence of spherical or hemispheric flow is characterized by a slope of −½. This characteristic slope of −½ is absent when the penetration ratio, b = h_p/h, is greater than 20% [66], as shown in Figure 2.16.

Example 2.5

Abbott et al. [90] presented pressure‐time data for a pressure drawdown test. Well no. 20 is partially completed in a massive carbonate reservoir. The well was shut‐in for stabilization and then flowed to 5200 BOPD for 8.5 h. The pressure data are given in Table 2.4 and reservoir and fluid properties are given below:

h = 302 ft	rw = 0.246 f	Pi = 2298 psia
hp = 20 ft	q = 5200 BPD	B = 1.7 bbl/STB
φ = 0.2	μ = 0.21 cp	ct = 34.2 × 10^−6 psia^−1

Figure 2.15.

Pressure derivative spherical source solution for a single well in an infinite system including WBS and no skin, after [66].

Figure 2.16.

Pressure derivative behavior for a single well in an infinite reservoir with different partial penetration lengths (C_D = 0, s = 0), after [66].

Solution by conventional analysis

Using the slope value of −122 psia/cycle read from the semilog plot of Figure 2.17, the reservoir permeability is calculated with Eq. (2.38);

$$k=\frac{162.6qB\mu }{mh}=\left|\frac{162.6(5200)(1.7)(0.21)}{(-122)(302)}\right|=8.19\hspace{0.17em}\text{md}$$

The mechanical skin factor is determined with Eq. (2.39) once the intercept of 2252 psia is read from Figure 2.17.

$$s=1.1513{\displaystyle \left[\begin{array}{l}\frac{2252-2298}{-122}-\\ log{\displaystyle \left(\frac{8.19}{(0.2)(0.21)(34.2\times {10}^{-6}){(0.246)}^2}\right)}+3.2275\end{array}\right]}=-5.03$$

t, h	t^−0.5, h^−0.5	Pwf, psia	ΔP, psia	t*ΔP′, psia
0.0		2266	0
0.5	1.414	2255	11	11.5
1.0	1.000	2243	23	24.5
1.6	0.791	2228	38	40.0
2.0	0.707	2218	48	45.0
2.5	0.632	2208	58	52.5
3.0	0.577	2197	69	69.0
3.5	0.535	2185	81	66.5
4.0	0.500	2178	88	60.0
4.5	0.471	2170	96	56.3
5.5	0.426	2161	105	46.8
6.0	0.408	2157	109	48.0
6.5	0.392	2153	113	52.0
7.0	0.378	2149	117	49.0
7.5	0.365	2146	120	52.5
8.0	0.354	2142	124	48.0
8.5	0.343	2140	126

Table 2.4.

Pressure and pressure derivative versus time data for example 2.5.

Figure 2.18 contains a Cartesian graph of P_wf as a function of t^−1/2. From there, the observed slope is m = 250 psia (h^−1/2) and intercept, I = 2060 psia, spherical permeability, and spherical skin factors are calculated using Eqs. (2.239) and (2.240), respectively:

$$k_{sp}={{\displaystyle \left(\frac{2453q\mu B}{m}\sqrt{\varphi \mu c_t}\right)}}^{2/3}={{\displaystyle \left(\begin{array}{l}\frac{2453(5200)(0.21)(1.7)}{250}\\ \sqrt{(0.2)(0.21)(\text{3}4.\text{2}\times {\text{10}}^{\text{‐6}}}\end{array}\right)}}^{2/3}=7.81\hspace{0.17em}\text{md}$$

$$s_{sp}=\frac{{\displaystyle \left(P_i-I\right)}{k}_{sp}{r}_{sw}}{70.6q\mu B}-1=\frac{{\displaystyle \left(2298-2060\right)}(7.81)(9.69)}{70.6(5200)(0.21)(1.7)}-1=-0.86$$

Vertical permeability and spherical wellbore radius are found with Eq. (2.130) and (2.131), respectively,

$$k_v=\frac{k_{sp}^3}{k_h^2}=\frac{{7.81}^3}{{8.19}^2}=7.1\hspace{0.17em}\text{md}$$

$$r_{sw}=\frac{b}{2ln{\displaystyle \left(\frac{b}{r_w}\right)}}=\frac{120}{2ln{\displaystyle \left(\frac{120}{0.246}\right)}}=9.69\hspace{0.17em}\text{ft}$$

With the value of the vertical permeability, it is possible to estimate the skin factor caused by partial penetration with Eqs. (2.125), (2.138), and (2.137), thus:

Figure 2.17.

Semilog plot for well no. 20.

Figure 2.18.

Cartesian spherical plot for well no. 20.

Figure 2.19.

Pressure and pressure derivative versus time log‐log plot for well no. 20.

$$h_D=\sqrt{\frac{k_h}{k_v}}{\displaystyle \left(\frac{h}{r_w}\right)}=\sqrt{\frac{8.26}{7.1}}{\displaystyle \left(\frac{302}{0.246}\right)}=1324.1$$

$$\begin{array}{ll}G& =2.948-7.363b+11.45b^2-4.675b^3\\ & =2.948-7.363\left(\frac{120}{302}\right)+11.45{{\displaystyle \left(\frac{120}{302}\right)}}^2-4.675{{\displaystyle \left(\frac{120}{302}\right)}}^3=1.57\end{array}$$

$$s_c={\displaystyle \left(\frac{1}{h_p/h}-1\right)}[ln\hspace{0.17em}{h}_D-G]={\displaystyle \left(\frac{1}{120/302}-1\right)}[ln\hspace{0.17em}1318.5-1.57]=8.51$$

Solution by TDS technique

The following data points were read from Figure 2.19.

tN = 1 h	ΔP = 23 psia
(t*ΔP′)sp = 56.25 psia	ΔPs = 96 psia	tsp = 4.5 h
(t*ΔP′)r2 = 52.5 psia	ΔPr2 = 96 psia	tr2 = 7.5 h

Wellbore storage coefficient is found from Eq. (2.66)

$$C={\displaystyle \left(\frac{qB}{24}\right)}\frac{t_N}{{{\displaystyle \left(\mathrm{\Delta }P\right)}}_N}={\displaystyle \left(\frac{(5200)(1.7)}{24}\right)}\frac{1}{23}=\text{16}\text{.01 bbl/psi}$$

From the spherical flow pressure derivative line, m = −1/2, the spherical permeability and mechanical spherical skin factor are, respectively, estimated by Eqs. (2.144) and (2.145);

$$k_{sp}={{\displaystyle \left(\frac{1227qB\mu }{{(t*\mathrm{\Delta }P\text{'})}_{sp}}\sqrt{\frac{\varphi \mu c_t}{t_{sp}}}\right)}}^{2/3}={{\displaystyle \left(\begin{array}{l}1227\frac{(5200)(1.7)(0.21)}{56.25}\\ \sqrt{\frac{(0.2)(0.21)(\text{3}4.2\times {\text{10}}^{\text{‐6}}\text{)}}{4.5}}\end{array}\right)}}^{2/3}=8.05\hspace{0.17em}\text{md}$$

$$\begin{array}{l}{s}_{sp}=34.74\sqrt{\frac{\varphi \mu c_t{r}_{sw}^2}{k_{sp}{t}_{sp}}}{\displaystyle \left[\frac{{{\displaystyle \left(\mathrm{\Delta }{P}_w\right)}}_{sp}}{2{{\displaystyle \left(t*\mathrm{\Delta }P\text{'}\right)}}_{sp}}\right]}-1\\ s_{sp}=34.74\sqrt{\frac{(0.2)(0.21)(\text{3}4.2{\text{ x 10}}^{\text{‐6}}\text{)(9}{\text{.69}}^{\text{2}})}{(8.05)(4.5)}}{\displaystyle \left[\frac{{\displaystyle \left(96\right)}}{2{\displaystyle \left(56.25\right)}}\right]}-1=-0.93\end{array}$$

The horizontal permeability and mechanical skin are found during the late radial flow using Eqs. (2.76) and (2.97), respectively;

$$k_r=\frac{70.6qB\mu }{h{(t*\mathrm{\Delta }P\text{'})}_r}=\frac{70.6(5200)(1.7)(0.21)}{(302)(52.5)}=8.26\hspace{0.17em}\text{md}$$

$$s=0.5{\displaystyle \left[\begin{array}{l}\frac{\mathrm{\Delta }{P}_r}{{(t*\mathrm{\Delta }P\text{'})}_r}-\\ ln{\displaystyle \left(\frac{k_r{t}_r}{\varphi \mu c_t{r}_w^2}\right)}+7.43\end{array}\right]}=0.5{\displaystyle \left[\begin{array}{l}\frac{120}{52.5}-\\ ln{\displaystyle \left(\frac{(8.26)(7.5)}{(0.2)(0.21)(\text{3}4.2\times {10}^{\text{‐6}}\text{)(0}{\text{.246}}^{\text{2}}\text{)}}\right)}+7.43\end{array}\right]}=-5.33$$

Vertical permeability is determined from Eq. (2.130);

$$k_v=\frac{k_{sp}^3}{k_h^2}=\frac{{8.05}^3}{{8.26}^2}=7.65\hspace{0.17em}\text{md}$$

Table 2.5 presents the comparison of the results obtained by the conventional method and TDS technique.

Parameter	Conventional	TDS
ksp, md	7.01	8.05
ssp	−0.86	−0.93
$\overline{k}$, md	8.19	8.26
sr	−5.03	−5.53
kv, md	7.10	7.65

Table 2.5.

Comparison of results.

2.5. Multirate testing

So far, the considerations revolve around a single flow test, meaning the production rate is kept constant for the application of the solution of the diffusivity equation. However, there are cases in which the flow rate changes; in such cases, the use of the solution to the diffusivity equation requires the application of the time superposition principle already studied in Section 1.14.2. Some reasons for the use of multirate testing are outlined as follows:

• It is often impractical to keep a constant rate for a long time to perform a complete pressure drawdown test.

• When the well was not shut‐in long enough to reach the static pressure before the pressure drawdown test started. It implies superposition effects.

• When, it is not economically feasible shutting‐in a well to run a pressure buildup test.

Whether the production rates are constant or not during the flow periods, there are mainly four types of multirate tests:

1. Uncontrolled variable flow rate;

2. Series of constant flow rates;

3. Variable flow rate while keeping constant bottom‐hole pressure, Pwf. This test is common in gas wells producing very tight formations and more recently applied on testing of shale formations;

4. Pressure buildup (fall‐off) tests.

Actually, a holistic classification of transient well testing is given in Figure 1.4. It starts with PTA which is known in the oil argot as pressure transient analysis. As seen in the figure, it is divided in single well tests, normally known as drawdown (flow) tests for production cases or injection tests for injection fluid projects. Our field of interest focuses on more than one rate operation (multirate testing) which includes all the four types just above described. It is worth to mention types 3 and 4. Type three is also known as rate transient analysis (RTA) which has been dealt with in a full chapter by this book's author in reference [56]. As far as case 4 is concerned, pressure buildup testing is the most basic multirate test ever existed since it comprises two flow rates: (1) one time period at a given q value different than zero and (2) another time period with a zero flow rate. This is because when a well is shut‐in, the flow stops at surface by the formation keeps still providing fluid to the well due to inertia.

2.5.1 Conventional analysis

Considering the sketch of Figure 2.20, application of the superposition principle [2, 3, 4, 6, 7, 11, 27, 44, 56, 60, 65, 67, 95, 96, 97] leads to:

$$P_{wf}(t)=P_i-\frac{141.2\mu B}{kh}{\displaystyle \left\{\begin{array}{l}{q}_1[P_D(t_D)+s]+(q_2-q_1)[P_D({[t-t_1]}_D)+s]+\\ (q_3-q_2)[P_D({[t-t_2]}_D)+s]+(q_4-q_3)[P_D({[t-t_3]}_D)+s]\\ +\mathrm{\dots .}+(q_N-q_{N-1})[P_D({[t-t_N]}_D)+s]\end{array}\right\}}$$E2.160

Rearranging;

$$P_{wf}(t)=P_i-\frac{141.2\mu B}{kh}{\displaystyle \left\{\begin{array}{l}{q}_1\{P_D(t_D)-P_D({[t-t_1]}_D\}+q_2\{P_D{([t-t_1])}_D-\\ P_D{([t-t_2])}_D\}+\mathrm{\dots }+q_{N-1}\{P_D{([t-t_{N-2}])}_D-\\ P_D{([t-t_{N-1}])}_D\}++q_N\{P_D{([t-t_{N-1}])}_D\}+s\end{array}\right\}}$$E2.161

Next step is to replace P_D by an appropriate diffusivity equation solution which depends upon the flow regime dealt with. Figure 2.21 presents the most typical superposition functions applied to individual flow regimes. The normal case is to use radial flow, top function in Figure 2.21. However, Escobar et al. [44] presented the inconvenience of not applying the appropriate superposition function for a given flow regime. They found, for instance, that if the radial superposition is used, instead of the linear, for characterization of an infinite‐conductivity hydraulic fracture, the estimated half‐fracture length would be almost three times longer than the actual one.

Coming back to Eq. (2.161), the assumed superposition function to be used is the radial one; then, this equation becomes:

$$P_{wf}(t)=P_i-\frac{70.6\mu B}{kh}{\displaystyle \left\{\begin{array}{l}{q}_{1}ln{\displaystyle \left(\frac{t}{t-t_1}\right)}+q_{2}ln{\displaystyle \left(\frac{t-t_1}{t-t_2}\right)}+q_{3}ln{\displaystyle \left(\frac{t-t_2}{t-t_3}\right)}+\\ q_{N-1}ln{\displaystyle \left(\frac{t-t_{N-2}}{t-t_{N-1}}\right)}+q_N\{ln(t-t_{N-1})\}+\\ ln\frac{k}{\varphi \mu c_t{r}_w^2}-7.4316+2s\end{array}\right\}}$$E2.162

Figure 2.20.

Schematic representation of a multirate test (typ. 1).

Figure 2.21.

Flow regime superposition functions.

Since it is uneasy to find natural log paper in the stationary shops, then, dividing for the natural log of 10 is recommended to express Eq. (2.162) in decadic log; then,

$$P_{wf}(t)=P_i-\frac{162.6\mu B}{kh}{\displaystyle \left\{\begin{array}{l}{\displaystyle \sum _{j=1}^{N-1}{q}_{j}log{\displaystyle \left(\frac{t-t_{j-1}}{t-t_j}\right)}}+q_N\{log(t-t_{N-1})\}+\\ log\frac{k}{\varphi \mu c_t{r}_w^2}-3.2275+0.8686s\end{array}\right\}}$$E2.163

Simplifying;

$$\frac{P_i-P_{wf}(t)}{q_N}=\frac{162.6\mu B}{kh}{\displaystyle \left\{\begin{array}{l}{\displaystyle \sum _{j=1}^N{\displaystyle \left(\frac{q_j-q_{j-1}}{q_N}\right)}log(t-t_{j-1})}+log\frac{k}{\varphi \mu c_t{r}_w^2}\\ -3.2275+0.8686s\end{array}\right\}}$$E2.164

Let;

$$s\text{'}=log\frac{k}{\varphi \mu c_t{r}_w^2}-3.23+0.87\cdot s\hspace{0.17em}$$E2.165

$$m\text{'}=\frac{162.6\mu B}{kh}$$E2.166

Solving for skin factor from Eq. (2.165);

$$s=1.1513{\displaystyle \left[\frac{b\text{'}}{m\text{'}}-log\frac{k}{\varphi \mu c_t{r}_w^2}+3.23\right]}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}$$E2.167

$$X_n={\displaystyle \sum _{i=1}^n{\displaystyle \left(\frac{q_i-q_{i-1}}{q_n}\right)}log(t-t_{i-1})}$$E2.168

Plugging Eqs. (2.165), (2.166), and (2.168) into Eq. (2.164) will lead to:

$$\frac{P_i-P_{wf}{\displaystyle \left(t\right)}}{q_n}=m\text{'}{X}_n+m\text{'}s\text{'}$$E2.169

which indicates that a Cartesian plot of ΔP/q_n against the superposition time, X_n, provides a straight line which slope, m', and intercept, m'b’ allows finding reservoir permeability and skin factor using Eqs. (2.166) and (2.167), respectively. However, it is customary for radial flow well interpretation to employ a semilog plot instead of a Cartesian plot. This issue is easily solved by taking the antilogarithm to the superposition function resulting into the equivalent time, t_eq. Under this situation, Eq. (2.169) becomes:

$$\frac{P_i-P_{wf}{\displaystyle \left(t_n\right)}}{q_n}=m_{n}log{t}_{eq}+b_n$$E2.170

And the equivalent time is then defined by,

$$\frac{P_i-P_{wf}{\displaystyle \left(t_n\right)}}{q_n}=m_{n}log{t}_{eq}+b_n$$E2.171

$$t_{eq}={\displaystyle \prod _{i=1}^n{(t_n-t_{i-1})}^{(q_i-q_{i-1})/q_n}}={10}^{X_n}$$E2.172

For a two‐rate case, Russell [96] developed the governing well‐flowing pressure equation, as follows:

$$P_{wf}=m{\text{'}}_1{\displaystyle \left[log{\displaystyle \left(\frac{t_1+\mathrm{\Delta }t}{\mathrm{\Delta }t}\right)}+\frac{q_2}{q_1}log{\displaystyle \left(\mathrm{\Delta }t\right)}\right]}+P_{INT}$$E2.173

Therefore, the slope, m'₁, and intercept, P_INT, of a Cartesian plot of P_wf versus log[(t₁+Δt)/Δt] + (q₂/q₁)log(Δt) allows finding permeability and skin factor from the following relationships:

$$k=-\frac{162.6q_1\mu B}{m_1\text{'}h}\hspace{0.17em}$$E2.174

$$s=1.1513{\displaystyle [\frac{q_1}{q_1-q_2}{\displaystyle (\frac{P_{wf}{\displaystyle (\mathrm{\Delta }}}{}}}$$