Coupling Simulation Model of Transient Flow in Tubing and Unsteady Seepage in Reservoir During Hydraulic Fracturing with Unstable Injection

Abstract

The current research on hydraulic fracturing with unstable injection mainly focuses on laboratory experiments or field tests. However, due to the difference in scale, some conclusions derived from laboratory experiments are not suitable for field applications. Besides, there is little research on coupling analysis of transient flow in tubing and seepage in reservoir during fracturing. In this paper, the transient dynamic model of fluid in tubing and the model of unsteady seepage in reservoir were established, respectively. Furthermore, the simulation model was established by parameters coupling, which took into account the influence of reservoir breakdown on permeability. The variation of fluid pressure in the tubing of different scale was analyzed during unstable injection. Simultaneously, the amplitude frequency characteristics of fluid in the tubing were also compared between laboratory scale and field scale. The results show that the simulation model can be used to calculate the fluid pressure variation in the whole process of hydraulic fracturing. And the simulation results are basically consistent with the experimental results. The scale difference has a significant impact on the variation of fluid pressure in tubing. To improve the fluctuation amplitude of outlet pressure, it is necessary to design the frequency according to the amplitude frequency characteristics of the corresponding scale model. It can also be achieved by reducing the resistance coefficient. The research can provide guidance for parameters design in field applications. It is conducive to understand the difference between laboratory experiments and field applications.

Appendices

Appendix A: Derivation of Mass Conservation Equation

As shown in Fig. 1, based on the principle of mass conservation, the difference between the inflow and outflow mass of the control unit is equal to the mass variation in it within the same unit time.

$$ \frac{{\partial \left( {\rho_{s} A} \right)}}{\partial t} + \frac{{\partial \left( {\rho_{s} v_{s} A} \right)}}{\partial z} = 0 $$

(26)

where A is the cross-sectional area of the tubing, m2.

According to the definition of substantial (material) derivative [24], the following formula can be obtained:

$$ \frac{dA}{{dt}} = \frac{\partial A}{{\partial t}} + v_{s} \frac{\partial A}{{\partial z}} $$

(27)

$$ \frac{{d\rho_{s} }}{dt} = \frac{{\partial \rho_{s} }}{\partial t} + v_{s} \frac{{\partial \rho_{s} }}{\partial z} $$

(28)

$$ \frac{{dp_{s} }}{dt} = \frac{{\partial p_{s} }}{\partial t} + v_{s} \frac{{\partial p_{s} }}{\partial z} $$

(29)

Substitute Eqs. (27) and (28) into Eq. (26)

$$ \frac{{\partial v_{s} }}{\partial z} + \frac{1}{{\rho_{s} }}\frac{{d\rho_{s} }}{dt} + \frac{1}{A}\frac{dA}{{dt}} = 0 $$

(30)

The state equation of the fluid is as follows

$$ \frac{{dp_{s} }}{{d\rho_{s} }} = \frac{K}{{\rho_{s} }} $$

(31)

The following formula is obtained by transforming Eq. (31):

$$ \frac{1}{{\rho_{s} }}\frac{{d\rho_{s} }}{dt} = \frac{1}{K}\frac{dp}{{dt}} $$

(32)

The equilibrium relation between the pressure force and the circumferential stress force per unit pipe length can be expressed as [24]

$$ Ddp = 2e_{d} d\sigma_{\theta } $$

(33)

where σ θ is the circumferential stress in the tubing, MPa.

The elastic stress–strain relation

$$ d\varepsilon = \frac{{d\sigma_{\theta } }}{{E_{0} }} $$

(34)

where

$$ {\text{d}}\varepsilon = \frac{dD}{D} $$

(35)

$$ \frac{dA}{A} = \frac{2dD}{D} $$

(36)

Substitute Eqs. (34), (35) and (36) into Eq. (33):

$$ \frac{1}{A}\frac{dA}{{dt}} = \frac{D}{{eE_{0} }}\frac{dp}{{dt}} $$

(37)

Substituting Eqs. (32) and (37) into Eq. (30), the following formula can be obtained:

$$ \frac{1}{K}\frac{{dp_{s} }}{dt}\left( {1 + \frac{KD}{{E_{0} e_{d} }}} \right) + \frac{{\partial v_{s} }}{\partial z} = 0 $$

(38)

According to Eq. (38), the propagation velocity of pressure wave can be expressed as

$$ c^{2} = \frac{{\frac{K}{{\rho_{s} }}}}{{1 + \frac{KD}{{E_{0} e_{d} }}}} $$

(39)

Substituting Eqs. (29) and (39) into Eq. (38)

$$ \rho_{s} c^{2} \frac{{\partial v_{s} }}{\partial z} + \frac{{\partial p_{s} }}{\partial t} + v_{s} \frac{{\partial p_{s} }}{\partial z} = 0 $$

(40)

Appendix B: Numerical Simulation Method

In order to facilitate programming, the discretized Eqs. (22), (23), and (24) are written as follows:

$$ dy\left( 1 \right) = - \rho_{0} e^{{C_{L} (p_{s}^{i} - p_{0} )}} c^{2} \frac{{v_{s}^{i + 1} - v_{s}^{i - 1} }}{2\Delta z} - v_{s}^{i} \frac{{p_{s}^{i + 1} - p_{s}^{i - 1} }}{2\Delta z} $$

(41)

$$ \begin{gathered} dy\left( 2 \right) = \frac{1}{{1 + k_{3} }}\left\{ {g\sin \beta - k_{3} c \times sign(v_{s}^{i} )\left| {\frac{{v_{s}^{i + 1} - v_{s}^{i - 1} }}{2\Delta z}} \right|} \right. \hfill \\ - \frac{{f_{s} v_{s}^{i} \left| {v_{s}^{i} } \right|}}{2D} - \frac{1}{{\rho_{0} e^{{C_{L} (p_{s}^{i} - p_{0} )}} }}\frac{{p_{s}^{i + 1} - p_{s}^{i - 1} }}{2\Delta z} - \left. {v_{s}^{i} \frac{{v_{s}^{i + 1} - v_{s}^{i - 1} }}{2\Delta z}} \right\} \hfill \\ \end{gathered} $$

(42)

$$ dy\left( 3 \right) = \eta (\frac{{p_{r}^{j + 1} - 2p_{r}^{j} + p_{r}^{j - 1} }}{{\Delta r^{2} }} + \frac{1}{{0.5d_{w} + j\Delta r}}\frac{{p_{r}^{j + 1} - p_{r}^{j - 1} }}{2\Delta r}) $$

(43)

where y(1) represents ps. y(2) represents vs. y(3) represents pr.

The fourth-order-accurate Runge–Kutta method is then used to integrate Eqs. (41), (42) and (43) in time. This method has good stability and convergence. The calculation formula is as follows [31]:

$$ {\text{y}}\left( {m + 1} \right) = {\text{y}}\left( m \right) + \frac{{\Delta t\left( {k_{1} + 2k_{2} + 2k_{3} + k_{4} } \right)}}{6} $$

(44)

where

$$ \left\{ {\begin{array}{*{20}l} {{\text{k}}_{{1}} = f\left( {t(m),y(m)} \right)} \hfill \\ {k_{2} = f\left( {t(m) + \frac{\Delta t}{2},y(m) + \frac{\Delta t}{2}k_{1} } \right)} \hfill \\ {k_{3} = f\left( {t(m) + \frac{\Delta t}{2},y(m) + \frac{\Delta t}{2}k_{2} } \right)} \hfill \\ {k_{4} = f\left( {t(m) + \Delta t,y(m) + \Delta tk_{3} } \right)} \hfill \\ \end{array} } \right. $$

(45)

where m is the time node. △t is the time step. f(t,y) represents dy.

The initial conditions are as follows

$$ y = \left[ {p_{s0} + \rho_{s} gz\sin \beta , \, 0, \, p_{e} } \right] $$

(46)

At the end of each time step, the seepage velocity of fluid in the reservoir is calculated according to the following formula:

$$ v_{r}^{j} = - \frac{{K_{d} }}{\mu }\frac{{p_{r}^{j + 1} - p_{r}^{j - 1} }}{2\Delta r} $$

(47)

The numerical solution follows the flowchart shown in Fig. 3. The calculation formulas in each time step are shown above.