7.3 Hydraulic fracture design: Single fracture completion

7.3.1 Improvement of reservoir access and wellbore-reservoir surface area

The creation of a hydraulic fracture improves dramatically the surface area of the wellbore in contact with the formation, and also the corresponding production rates (at the same bottomhole pressure). The ratio between the area of a fracture (constant height $h$ and half-length $x_f$) and the area of an openhole wellbore (radius $r$ and height $h$) is $(4 x_f h)/(2\pi r_w h) \sim x_f/r_w$. Fractures are usually much longer than the radius of the wellbore and so does the fracture area in contact with the reservoir rock compared to the area of the wellbore. The use of a skin factor $s$ permits calculating the flow rates in the presence of the fracture using the wellbore equation (Figure 7.11).

Figure 7.11: Reservoir drainage: A fractured wellbore dramatically increases flow rates for the same bottomhole pressure. The rate is proportional to $x_f/r_w$.

7.3.2 The coupled fluid-driven fracture propagation problem

Several physical processes interact during the propagation of fluid-driven fractures (Figure 7.13). The main processes include:

1. Opening of the fracture by deformation of adjacent rock. The calculation of fracture width can be achieved by solving fracture opening, for example by solving a linear elasticity problem.
2. Creation of new fracture surface. Cemented rocks require tensile stresses at the tip surpass a certain limit in order to create new fracture surface. This problem is most times solved with the theory of linear elastic fracture mechanics.
3. Leak-off of fracturing fluid into the formation. Fracturing of porous media involves fluid flow into the fractured formation. Leak-off can be solved as any fluid flow problem or through simplified equations. Leak-off also affects effective stresses around the fracture face and tip.
4. Fluid flow within the fracture. Fracture propagation requires constant inflow of fluid into the fracture. Fluid-driven fractures in stiff rocks tend to be narrow (few millimeters) and therefore non-negligible viscous losses occur when fluids flow through fractures. This problem can be solved as a lubrication problem of fluid flow within planar surfaces.

The following subsections describe each of these problems in more detail.

Figure 7.12: The four coupled processes in fluid-driven fracture propagation in porous media.

Figure 7.13: The four coupled processes in fluid-driven fracture propagation in porous media.

7.3.2.1 Mechanics: The pressurized linear crack problem

The first step consists on solving the deformation of a fracture subjected to external far-field stresses $\uuline{S}$ and pressure in the fracture $p$. The example in Figure 7.14 shows a 2D example of a fracture within a infinitely large and impervious medium. The fracture has a fluid pressure that varies along the length of the fracture. The center is pressurized at pressure $p$ and the ends have zero pressure. The medium has a far-field minimum principal stress $S_3$ acting perpendicular to the fracture. The net pressure $p_{net}$ is defined as the difference between the fluid pressure in the fracture $p$ and least principal total stress $S_3$.

$$p_{net} = p - S_3$$ (7.2)

For example, the fracture in Figure 7.14 has positive net pressure in the center and negative net pressure in the ends. A fluid-driven fracture will open if and only if $p_{net}>0$, i.e., $p > S_3$. Also, the net pressure concept facilitates the development of analytical solutions, by combining two actions into one.

Figure 7.14: Example of the net pressure concept.

The second step is to solve the displacements and strains that occur due to pressuring a fracture with a given net pressure. This problem was solved by A. A. Griffith (https://en.wikipedia.org/wiki/Fracture_mechanics), and is known as the Griffith crack problem. Griffith assumed a homogeneous and isotropic solid to solve stresses around a fracture.

The simplest solution is the one in which the net pressure is constant along the fracture $p_{net} = p_o$. Hence, for a fracture with half-length $c$ (Fig. 7.15), the boundary conditions of the problem are:

$$\left\lbrace \begin{array}{rcl c l} \sigma_{yy} & = & p(x)= p... ...c \\ u_y & = & 0 & \text{ for } & x > c \\ \end{array}\right.$$ (7.3)

Figure 7.15: Griffith's crack problem: Geometry and assumptions of pressurized line crack in 2D elastic, isotropic, homogeneous solid.

The solution of the Navier's equation (Eq. 3.40) yields the following values at the line $(x,y=0)$:

$$\left\lbrace \begin{array}{rcl c l} u_y(x,0) & = & \cfrac{2 p... ...2-c^2}} -1 \right) & \text{ for } & x >c \\ \end{array}\right.$$ (7.4)

where $E'= E / (1 - \nu^2)$ is the plane strain modulus. Let us investigate this solution at $x=0$ and $x=c$. The displacement at $x=0$ is $u_y(0,0) = 2 p_0 c /E'$, hence, the maximum width of the fracture is

$$w_o = \frac{4 p_0 c}{E'}$$ (7.5)

It makes sense that the width of the fracture will be proportional to the net pressure and the half-length, and inversely proportional to the stiffness of the medium. The solution of $u_y$ is an elliptical equation, thus, Griffith predicts an elliptically shaped fracture when the net pressure is constant.

Let us now solve the stress at the tip of the fracture $x=c$: $\sigma_{yy}(c,0) \rightarrow - \infty$ which in practice is impossible. We will find a solution for this problem in the next subsection. The stress beyond the tip is tensile. For example at $x=1.15c$, $\sigma_{yy}(1.15c,0) \sim - p_o$.

7.3.2.2 Mechanics: stress intensity factor and fracture toughness

The previous subsection shows that the solution of stresses at the the fracture tip yields an infinitely large tensile stress. Thus, we would not be able to use the tensile strength criterion because the stress would be always larger than the theoretical stress. In order to circumvent this problem, Griffith defined the stress intensity factor as

$$K_I = \lim_{r \rightarrow 0^+} \left[ (2 \pi r)^{1/2} \sigma_{yy} (x=c+r,y=0) \right]$$ (7.6)

where $r=x-c$ is the distance from a point in the solid to the fracture tip. The solution of the stress intensity factor equation for a line crack pressurized with constant pressure $p_o$ can be obtained plugging the solution for $\sigma_{yy}$ (Eq. 7.4) and is equal to

$$K_I = p_o ( \pi c)^{1/2}$$ (7.7)

The Griffith criterion establishes that a fracture may propagate if the stress intensity $K_I$ is larger than the fracture toughness $K_{IC}$, where $K_{IC}$ is a property of the material.

$$\left\lbrace \begin{array}{rcl} K_I \geq K_{IC} & \rightarrow... ...ow & \text{ Fracture does not propagate} \\ \end{array}\right.$$ (7.8)

The superscript $I$ means that this is “Mode I” fracture, also known as opening-mode fracture. There are two other modes of fracture intensity and propagation related to in-plane shear (Mode II) and out-of-plane shear (Mode III) (Figure 4.22).

The fracture toughness $K_{IC}$ is the property of a solid and can be measured with the semicircular bending test (Fig. 7.16). Linear elasticity permits calculating the stress intensity factor, such that, the fracture toughness is

$$K_{IC} = \frac{ P_{max} ( \pi a)^{1/2} }{2rL} Y_I$$ (7.9)

where $P_{max}$ [N] is the maximum load at failure and $a$ [m] is the length of the notch pre-carved in the sample (other geometrical values in figure). The parameter $Y_I$ [-] is geometrical factor and depends on $s/r$ and it is $Y_I \sim 4.5$ for $s/r = 0.8$ and $0.1 < a/r <0.2$. Typical rock fracture toughness values vary between $K_{IC} \sim 0.05$ to 1.5 MPa m$^{1/2}$.

Figure 7.16: Geometry and variables for the semicircular bending test. The notch simulates a pre-existing fracture and intensifies stresses at the tip.

PROBLEM 7.1: Calculate the maximum width and stress intensity of a fracture with half-length $x_f = 10$ m and pressurized to $p_f = 20.5$ MPa, immersed within a elastic medium with $E = 1$ GPa and $\nu =$ 0.25, and subjected to far field stress $S_3 = 20$ MPa.

SOLUTION

Let us first compute the plane strain modulus

$$E' = \frac{E}{1-\nu^2} = \frac{1 \text{ GPa}}{1-0.25^2} = 1.07 \text{ GPa} = 1070 \text{ MPa}$$

Using Eqs. 7.7 and 7.5, and noting that the net pressure is $p_{net} = p_f - S_3$ the results are:

$$w_o = \frac{4 p_0 c}{E'} = \frac{4 \times 0.5 \text{ MPa} \times 10 \text{ m}} {1070 \text{ MPa}} = 0.019 \text{ m} = 19 \text{ mm}$$

$$K_I = p_o ( \pi c)^{1/2} = 0.5 ( \pi \times 10 )^{1/2} = 2.8 ^{1/2}$$

Extra: The assumption of constant net pressure along the fracture is not accurate for a propagating fracture. The width and stress intensity for net pressure distribution starting at $p_o$ in the center and decreasing linearly to zero at the tips are:

$$w_o = \frac{2 p_0 c}{E'} = \frac{2 \times 0.5 \text{ MPa} \times 10 \text{ m}} {1070 \text{ MPa}} = 0.0095 \text{ m} = 9.5 \text{ mm}$$

$$K_I = \left( 1-\frac{2}{\pi} \right) p_o ( \pi c)^{1/2} = \left( 1-\frac{2}{\pi} \right) 0.5 ( \pi \times 10 )^{1/2} = 1.0 ^{1/2} \: \: \blacksquare$$

7.3.2.3 Transport: leak-off

The total injected fluid $V_i$ during fracturing goes into filling the open fracture $V$ and some leaks into the formation $V_L$. Hence, material balance dictates

$$V_i = V + V_L$$ (7.10)

The leak-off volume can be calculated as a function of time $t$ as a problem of fluid flow in porous media and approximated with the Carter leak-off equation:

$$V_L = A_L (2 c_L \sqrt{t} + S_p)$$ (7.11)

where $A_L$ is the area of leak-off, $c_L$ is the Carter leak off coefficient, and $S_p$ is the instantaneous fluid spurt loss. The efficiency of injection is defined through coefficient

$$\eta = \frac{V}{V_i}$$ (7.12)

Thus, $\eta \rightarrow 1$ indicates high efficiency of the use of fracturing fluid to propagate a fracture and $\eta \rightarrow 0$ indicates significant leak-off.

The amount of injected fluid for a constant injection rate $i$ is

$$V_i = i \: t$$ (7.13)

Both $V_i$ and $i$ are defined for one-wing of the fracture. Hence, the total injected fluid through the wellbore in a bi-wing fracture is $2 V_i$.

7.3.2.4 Transport: fluid flow through fracture

The calculation of fluid rate and pressure drop along the fracture can be simplified to a problem of fluid flow between two parallel plates. For a fixed width between two plates, the flow rate for laminar flow of a Newtonian fluid is

$$q = \frac {w^3 h_f}{12 \mu} \frac{\Delta P}{L}$$ (7.14)

where $w$ is the spacing between the plates (width of the fractures), $h_f$ is the height of the plates (fracture height), $\Delta P/L$ is the fluid pressure gradient, and $\mu$ is the viscosity. For a fracture of variable width $w(x)$ and constant flow rate, the pressure gradient is also a function of distance from the wellbore $\Delta P(x)/dx$. Hence, pressure gradient along the fracture depends on the fracture shape.

7.3.3 Fracture design: Single planar models

The design of a single (bi-wing) fracture consists in determining:

1. Optimum fracture size and orientation: $x_f$, $w$, and $h_f$, and
2. Injection schedule, including fracturing fluid selection and proppants.

Figure 7.17 summarizes the three most common pseudo-2D geometric models used for fracture design: PKN, KGD, and radial. The following subsections describe the PKN and KGD models. Section 7.3.4 discusses the determination of fracture height.

Figure 7.17: Pseudo-2D geometric models used to model propagation of planar bi-wing fractures.

7.3.3.1 The PKN model

This model was devoloped by Perkins, Kern and Nordgren. The most important assumption is plane-strain fracture deformation in a vertical plane perpendicular to the direction of fracture propagation 7.18. In addition the model assumes linear isotropic elasticity, negligible fracture toughness, and laminar flow. The solution consists in combining the equations seen in all the processes mentioned in Section 7.3.2.

Figure 7.18: Geometry of the PKN model (adapted from Adachi et al. [2007]). The gray plane represents the plane-strain assumption. PKN model is good for long fractures $x_f > h_f$.

Simple analytical solutions can be obtained by further assuming constant injection rate and no leak-off.

The solution is

• Fracture half-length:

  $$x_f = \left( \frac{625}{512 \pi^3} \right)^{1/5} \left( \frac{i^3... .../5} t^{4/5} = 0.524 \: \left( \frac{i^3 E'}{\mu \: h_f^4} \right)^{1/5} t^{4/5}$$ (7.15)

• Maximum width at the wellbore ($x=0$):

  $$w_{w,0} = \left( \frac{2560}{\pi^2} \right)^{1/5} \left( \frac{i^... ...{1/5} t^{1/5} = 3.040 \: \left( \frac{i^2 \mu}{E' \: h_f} \right)^{1/5} t^{1/5}$$ (7.16)

• Net pressure at the wellbore ($x=0$):

  $$p_{net,w} = \left( \frac{80}{\pi^2} \right)^{1/5} \left( \frac{E'... ...1/5} t^{1/5} = 1.520 \: \left( \frac{E'^4 i^2 \mu}{h_f^6} \right)^{1/5} t^{1/5}$$ (7.17)

The PKN model is suitable for long fractures with $x_f > h_f$. Hence, the PKN model is a good choice for representing long planar fractures done for reservoir completion with well defined bottom and top fracture barriers. Notice that the PKN model predicts increasing $p_{net}$ with increasing time to the power of $t^{1/5}$.

The volume of the fracture (one-wing) can be computed by multiplying the fracture height, the fracture half-length, and the average width (notice that the width varies from the wellbore to the tip and from top to bottom). The average width for a PKN fracture is $\bar{w}= (\pi/5) w_{w,0}$. Hence, the volume of one wing of the fracture is:

$$V_{frac} = (\pi/5) w_{w,0} h_f x_f$$ (7.18)

PROBLEM 7.2: Using the PKN model, calculate fracture half-length $x_f$, fracture width at the wellbore $w_{w,0}$, fracture net pressure $p_{net,w}$, and fracture (two-wing) volume after 30 min in a single fracture job with hydraulic fracture height $h_f = 100$ ft, rock plane-strain modulus $E'= 8.9 \times 10^6$ psi, fluid viscosity $\mu =$ 1 cP, and injection rate $i = 25$ bbl/min (one-wing).

SOLUTION
Let us first convert all quantities to a consistent unit system. We choose the SI units: Meter (m), kilogram (kg), second (s). Then, the values are $h_f = 30.5$ m, $E'= 61.4 \times 10^{9}$ Pa, $\mu =$ 0.001 Pa$\cdot$s, and $i = 0.066$ m$^3$/s (one-wing).

At 30 min (= 1800 s),

• Fracture half-length:

  $$x_f = 0.524 \: \left[ \frac{(0.066 \text{ m}^3/\text{s})^3 \times... ...mes (30.5 \text{ m})^4} \right]^{1/5} (1800 \text{ s})^{4/5} = 1536.2 \text{ m}$$

• Maximum width at the wellbore ($x=0$):

  $$w_{w,0} = 3.040 \: \left[ \frac{(0.066 \text{ m}^3/\text{s})^2 \t... ... (30.5 \text{ m})} \right]^{1/5} (1800 \text{ s})^{1/5} = 4.05 \times 10^{-3} m$$

• Net pressure at the wellbore ($x=0$):

  $$p_{net,w} = 1.520 \: \left[ \frac{(61.4\times 10^9 \text{ Pa})^4 ... ...text{ m})^6} \right]^{1/5} (1800 \text{ s})^{1/5} = 4.1 \times 10^{6} \text{Pa}$$ (7.19)

The fracture (two-wing) volume at 30 min is

$$2 V_{frac} = 2 \frac{\pi}{5} (4.05 \times 10^{-3} m) \times (30.5 ) \times (1536.2 ) = 239 ^3$$

For example, if the fracture were to be filled with 80% fracturing fluid and 20% proppant, that would result in 191 m$^3$ of fracturing fluid (about 2 medium-size swimming pools or 1,200 bbl) and 48 m$^3$ of sand (about 126 tons of sand - one truck carries up to 20 tons of sand). $\: \: \blacksquare$

7.3.3.2 The KGD model

This model was devoloped by Khristianovich, Geertsma and deKlerk. The most important assumption is plane-strain fracture deformation in a horizontal plane 7.19. In addition the model assumes linear isotropic elasticity, negligible fracture toughness, and laminar flow. The solution consists in combining the equations seen in all the processes mentioned in Section 7.3.2.

Figure 7.19: Geometry of the KGD model (adapted from Adachi et al. [2007]). The plane-strain assumption is a horizontal plane (i.e., fracture width is independent of the vertical coordinate). KGD model is good for short fractures $x_f < h_f$.

Simple analytical solutions can be obtained by further assuming constant injection rate and no leak-off.

The solution is

• Fracture half-length:

  $$x_f = \left( \frac{16}{21 \pi^3} \right)^{1/6} \left( \frac{i^3 E... .../5} t^{2/3} = 0.539 \: \left( \frac{i^3 E'}{\mu \: h_f^3} \right)^{1/5} t^{2/3}$$ (7.20)

• Maximum width at the wellbore ($x=0$):

  $$w_{w,0} = \left( \frac{5376}{\pi^3} \right)^{1/6} \left( \frac{i^... .../6} t^{1/3} = 2.360 \: \left( \frac{i^3 \mu}{E' \: h_f^3} \right)^{1/6} t^{1/3}$$ (7.21)

• Net pressure at the wellbore ($x=0$):

  $$p_{net,w} = \left( \frac{21}{16} \right)^{1/3} \left( E'^2 \mu \right)^{1/3} t^{-1/3} = 1.090 \: \left( E'^2 \mu \right)^{1/3} t^{-1/3}$$ (7.22)

The KGD model is suitable for modeling short fractures with $x_f < h_f$. Hence, the KGD model is a good choice for representing short planar fractures as the ones present in leak-off tests or frac-pack completions. Notice that the KGD model predicts decreasing $p_{net}$ with increasing time to the power of $t^{-1/3}$. Furthermore, the KGD model implies that $p_{net}$ is independent of injection rate. Field measurements show that $p_{net}$ does vary with changes of injection rate.

7.3.4 Stress logs and implications on hydraulic fracture height

The first principle of hydraulic fracturing is that fluid-driven opening-mode fractures tend to open in isotropic media as planes perpendicular to the minimum principal stress $S_3$. However, the magnitude of the minimum principal stress tends to vary with space. In cases where $S_3 = S_{hmin}$, the minimum principal varies with depth as a result of rock mechanical heterogeneity and tectonic strains. The variation of mechanical properties and stresses with depth is known as “mechanical stratigraphy”.

The variation of minimum principal stress with depth (when $S_3 = S_{hmin}$) results in preferential locations for fracture growth. Local minima of $S_3$ result in zones that favor fracture containment, while local maxima result in fracture barriers (Fig. 7.20). Hence, stress contrast controls vertical propagation of fractures. Fractures will preferentially grow towards regions of low stress and may stop at fractures barriers depending on the net pressure in the fracture. The fracture height $h_f$ is approximately the distance between the fracture barriers.

Figure 7.20: Example of fracture containment as a result of mechanical stratigraphy: (top) without stress contrast and (bottom) with stress contrast and determination of fracture heigth $h_f$. Inset figures from Oilfield review - Brady et. al 1992.

Horizontal stresses depend on rock properties, tectonic stress, and limits imposed by neighboring faults. Variations of horizontal stress with depth can be approximated by constructing a one-dimensional heterogeneous Mechanical Earth Model (MEM). This model consists in calculating stresses by applying a tectonic strain to a sequence of rock layers with data obtained from well logging and the laboratory. The workflow is the following (See Fig. 7.21):

1. Obtain rock mass density $\rho_{bulk}(z)$, elastic P-wave velocity $V_P(z)$, and elastic S-wave velocity $V_S(z)$ as a function of depth for a given location. You will need acoustic logging measurements which usually report travel times or slowness (units of inverse of velocity: [$\mu$s/ft]) $\Delta t_P = V_P^{-1}$ and $\Delta t_S = V_S^{-1}$.

2. Calculate the Young's modulus and Poisson's ratio (both as a function of depth) using elastic wave velocities according to the following equations:

   $$E_{dyn} = \rho_{bulk} V_S^2 \left( \frac{3V_P^2 - 4 V_S^2} {V_P^2 - V_S^2} \right)$$ (7.23)

   $$\nu_{dyn} = \frac{V_P^2 - 2 V_S^2}{2 \left( V_P^2 - V_S^2 \right)}$$ (7.24)

3. Calculate (quasi-)static Young's modulus $E_{sta}$ of the rock as a function of depth. The Young modulus calculated with elastic wave velocity $E_{dyn}$ is usually larger then the static Young's modulus $E_{sta}$ of the rock relevant to tectonic strains and strains imparted by hydraulic fracturing. The static Young's modulus $E_{sta}$ can be better approximated by laboratory measurements imparting large strains. Laboratory measurements can also measure the dynamic Young's modulus and permit establishing a relationship that links the two quantities

   $$E_{sta} = F_{ds} E_{dyn}$$ (7.25)

   where $F_{ds}<1$ is a parameter in the laboratory and usually varies from 0.3 to 0.8. The dynamic to static correction is usually done only for the Young modulus, thus, the Poisson's ratio remains the same $\nu_{sta}=\nu_{dyn}$.

4. Calculate total vertical stress $S_v$ and pore pressure$P_p$.

5. Utilize Eqs. 3.34 to calculate $S_{Hmax}$ and $S_{hmin}$ using a known tectonic strains $\varepsilon _{Hmax}$ and $\varepsilon_{hmin}$, and the static plane- strain modulus as a function of depth $E'_{sta}$. The result is what is known as a “stress log”.

Figure 7.21: Simplified workflow to calculate a stress log from wellbore logging data, field tests, and laboratory tests. The stress log permits identifying variation of horizontal stresses with depth and potential hydraulic fracture barriers.

The values of tectonic strain are usually unknown. Other measurements such as fracture tests and breakout widths along the wellbore are used to calibrate the values of tectonic strains, so that the calculated stresses match the $S_3$ values and $w_{BO}$ measured in the well.

Fig. 7.22-a shows an example of a stress log following the proposed procedure. Fig. 7.22-b-c show the results of numerical simulation of hydraulic fracture propagation started at the perforation interval shown in the stress log. Notice that the fracture avoids high stress regions, grows towards regions of low stress, and preferentially grows in regions of local stress minima.

Figure 7.22: Stress log example and fracture width from Adachi et al. [2007]. The fracture is not a simple 3D surface. Fracture width and length vary with depth. The fracture grows preferentially in at depths were minimum principal stress presents local minima.