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Analysis of the transient flow of non-Newtonian power-law fluids in homogeneous reservoirs with the elastic outer boundary

Abstract

Since some oil fields have entered the late stage of high water-cut development, the use of polymer flooding and other oil recovery technologies makes the fluid be of non-Newtonian characteristics, but the existing elastic outer boundary percolation model ignores this point. Firstly, this paper establishes a non-Newtonian power-law fluid percolation model with the elastic outer boundary condition. Secondly, the analytical solution of the percolation model is obtained in the Laplace space by the structural method. Thirdly, the double logarithmic characteristic curves are drawn. Finally, the sensitivity analysis of the parameters is carried out. The results show that the wellbore storage coefficient and skin factor play important roles in the early stage of the double logarithmic curves. For the radial flow stage and the later flow stage, the main influence factors are the power-law index, the elastic coefficient and outer boundary radius, respectively. It is concluded that the percolation model established in this paper provides a theoretical basis for reasonably solving the problem that the previous data deviation could not explain the reservoir parameters well. Meanwhile, the application of the structural method can provide a scientific calculation method for the well test analysis of non-Newtonian power-law fluids.

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The data that support the findings of this study are available from the corresponding author upon reasonable request.

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Acknowledgements

This paper is supported by the National Natural Science Foundation of China (No. 11601451), the International Cooperation Program of Chengdu City (No. 2020-GH02-00023-HZ) and the Key R&D Project of Science and Technology Programming of Sichuan Province (No. 2020YFG0145).

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Correspondence to Chao Min.

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The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

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Communicated by Michael Nones, Ph.D. (CO-EDITOR-IN-CHIEF).

Appendix

Appendix

Lemma

If the boundary value problem (17a17c) satisfies the conditions (18), then the boundary value problem has only zero solutions.

$$ x^{2} \frac{{\partial^{2} y}}{{\partial x^{2} }} + Ax\frac{\partial y}{{\partial x}} + \left( {B - Cx^{q} } \right)y = 0; $$
(17a)
$$ \left. {\left[ {Ey + \left( {1 + EF} \right)\frac{\partial y}{{\partial x}}} \right]} \right|_{x = a} = 0; $$
(17b)
$$ \left. {\left[ {\varepsilon_{b} y + x\frac{\partial y}{{\partial x}}} \right]} \right|_{x = b} = 0 $$
(17c)
$$ \left\{ \begin{gathered} x^{2} > 0;{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} B - Cx^{q} \le 0; \hfill \\ E\left( {1 + EF} \right) \le 0; \hfill \\ \varepsilon_{b} b \ge 0;{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \varepsilon_{b}^{2} + b^{2} \ne 0. \hfill \\ \end{gathered} \right. $$
(18)

Proof

We multiply both sides of Eq. (17a) by \(y\left( x \right)\) and integrate it on the interval [a, b] and then have:

$$ \left. {\left[ { - x^{2} y\frac{\partial y}{{\partial x}}} \right]} \right|_{x = b} + \left. {\left[ {x^{2} y\frac{\partial y}{{\partial x}}} \right]} \right|_{x = a} + \int_{a}^{b} {x^{2} \left( {\frac{\partial y}{{\partial x}}} \right)^{2} } dx + \int_{a}^{b} {\left( {B - Cx^{q} } \right)y^{2} } dx = 0. $$
(19)

We multiply Eq. (17b) by \(E\frac{\partial y}{{\partial x}}\), there is:

$$ \left. {\left[ {E^{2} y\frac{\partial y}{{\partial x}}} \right]} \right|_{x = a} = \left. {\left[ { - E\left( {1 + EF} \right)\left( {\frac{\partial y}{{\partial x}}} \right)^{2} } \right]} \right|_{x = a} \ge 0. $$

So,

$$ \left. {\left[ {y\frac{\partial y}{{\partial x}}} \right]} \right|_{x = a} \ge 0. $$
(20)

Similarly, we have

$$ \left. {\left[ { - y\frac{\partial y}{{\partial x}}} \right]} \right|_{x = b} \ge 0. $$
(21)

We substitute Eqs. (20 ~ 21) into Eq. (19) and get:

$$ \frac{\partial y}{{\partial x}} \equiv 0,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} y \equiv 0,{\kern 1pt} {\kern 1pt} {\kern 1pt} x \in \left[ {a,b} \right] $$
(22)

The lemma is proved.

Theorem

The boundary value problems of modified Bessel equation as follows:

$$ x^{2} \frac{{\partial^{2} y}}{{\partial x^{2} }} + Ax\frac{\partial y}{{\partial x}} + \left( {B - Cx^{q} } \right)y = 0; $$
(23a)
$$ \left. {\left[ {Ey + \left( {1 + EF} \right)\frac{\partial y}{{\partial x}}} \right]} \right|_{x = a} = D; $$
(23b)
$$ \left. {\left[ {\varepsilon_{b} y + x\frac{\partial y}{{\partial x}}} \right]} \right|_{x = b} = 0. $$
(23c)

Here, A, B, C, D, E, F q, a, b, \(\varepsilon_{b}\) are constants, a < b, D ≠ 0.

If the boundary value problem (23a23c) satisfies the conditions (18), then the solution of (23a23c) can be expressed as follows:

$$ y = D \cdot \frac{1}{{E + \frac{1}{F + \Phi \left( a \right)}}} \cdot \frac{1}{F + \Phi \left( a \right)} \cdot \Phi \left( x \right), $$
(24)

where

$$ \Phi \left( x \right) = \frac{{\varepsilon_{b} \varphi_{v,v} \left( {x,b} \right) + b\varphi_{v,v + 1} \left( {x,b} \right)}}{{\varepsilon_{b} \varphi_{v + 1,v} \left( {a,b} \right) + b\varphi_{v + 1,v + 1} \left( {a,b} \right)}}, $$

among them

$$ \varphi_{v,v} \left( {x,\xi } \right) = \left( {x\xi } \right)^{\alpha } \psi_{v,v} \left( {x^{\beta } ,\xi^{\beta } ,\kappa } \right), $$
$$ \begin{aligned} \varphi_{v,v + 1} \left( {x,\xi } \right) & = \frac{\partial }{\partial \xi }\varphi_{v,v} \left( {x,\xi } \right) \\ & = x^{\alpha } \xi^{\alpha - 1} \left[ {\left( {\alpha + \beta v} \right)\psi_{v,v} \left( {x^{\beta } ,\xi^{\beta } ,\kappa } \right) + \kappa \beta \xi^{\beta } \psi_{v,v + 1} \left( {x^{\beta } ,\xi^{\beta } ,\kappa } \right)} \right], \\ \end{aligned} $$
$$ \begin{aligned} \varphi_{v + 1,v} \left( {x,\xi } \right) & = \frac{\partial }{\partial x}\varphi_{v,v} \left( {x,\xi } \right) \\ & {\kern 1pt} = x^{\alpha - 1} \xi^{\alpha } \left[ {\left( {\alpha + \beta v} \right)\psi_{v,v} \left( {x^{\beta } ,\xi^{\beta } ,\kappa } \right) - \kappa \beta x^{\beta } \psi_{v + 1,v} \left( {x^{\beta } ,\xi^{\beta } ,\kappa } \right)} \right], \\ \end{aligned} $$

\(\begin{aligned} \varphi_{v + 1,v + 1} \left( {x,\xi } \right) & = \frac{{\partial^{2} }}{\partial x\partial \xi }\varphi_{v,v} \left( {x,\xi } \right) = \frac{{\partial^{2} }}{\partial \xi \partial x}\varphi_{v,v} \left( {x,\xi } \right) \\ & = \left( {x\xi } \right)^{\alpha - 1} \left[ {\left( {\alpha + \beta v} \right)^{2} \psi_{v,v} \left( {x^{\beta } ,\xi^{\beta } ,\kappa } \right) - \kappa \beta \left( {\alpha + \beta v} \right)x^{\beta } \psi_{v + 1,v} \left( {x^{\beta } ,\xi^{\beta } ,\kappa } \right)} \right. \\ & \;\left. { + \kappa \beta \left( {\alpha + \beta v} \right)\xi^{\beta } \psi_{v,v + 1} \left( {x^{\beta } ,\xi^{\beta } ,\kappa } \right) - \kappa^{2} \beta^{2} \left( {x\xi } \right)^{\beta } \psi_{v + 1,v + 1} \left( {x^{\beta } ,\xi^{\beta } ,\kappa } \right)} \right]. \\ \end{aligned}\) where

$$ \alpha = \frac{1 - A}{2},\quad \beta = \frac{q}{2},\quad \kappa = \frac{2\sqrt C }{q},\quad v = \frac{{\sqrt {\left( {1 - A} \right)^{2} - 4B} }}{q}. $$

Proof

We first prove the existence of the solution and its representation.

We assume that \(y_{1} \left( x \right)\) and \(y_{2} \left( x \right)\) are two linearly independent solutions of Eq. (17a), and then, the general solution of Eq. (17a) is expressed as follows:

$$ y\left( x \right) = d_{1} y_{1} \left( x \right) + d_{2} y_{2} \left( x \right). $$
(25)

We substitute Eq. (25) into Eq. (17b) and Eq. (17c), respectively, and get:

$$ d_{1} \left. {\left[ {E\frac{{\partial y_{1} }}{\partial x} + \left( {1 + EF} \right)y_{1} \left( x \right)} \right]} \right|_{x = a} + d_{2} \left. {\left[ {E\frac{{\partial y_{2} }}{\partial x} + \left( {1 + EF} \right)y_{2} } \right]} \right|_{x = a} { = }0, $$
(26)
$$ d_{1} \left. {\left[ {\varepsilon_{b} \frac{{\partial y_{1} }}{\partial x} + by_{1} \left( x \right)} \right]} \right|_{x = b} + d_{2} \left. {\left[ {\varepsilon_{b} \frac{{\partial y_{2} }}{\partial x} + by_{2} } \right]} \right|_{x = b} { = 0}{\text{.}} $$
(27)

Here, Eq. (26) and Eq. (27) are linear algebraic equations about \(d_{1} ,d_{2}\), and the determinant of its coefficient matrix is as follows:

$$ \Delta { = }\left| {\begin{array}{*{20}c} \begin{gathered} \left. {\left[ {E\frac{{\partial y_{1} }}{\partial x} + \left( {1 + EF} \right)y_{1} \left( x \right)} \right]} \right|_{x = a} \\ \left. {\left[ {\varepsilon_{b} \frac{{\partial y_{1} }}{\partial x} + by_{1} \left( x \right)} \right]} \right|_{x = b} \\ \end{gathered} & \begin{gathered} \left. {\left[ {E\frac{{\partial y_{2} }}{\partial x} + \left( {1 + EF} \right)y_{2} } \right]} \right|_{x = a} \\ \left. {\left[ {\varepsilon_{b} \frac{{\partial y_{2} }}{\partial x} + by_{2} } \right]} \right|_{x = b} \\ \end{gathered} \\ \end{array} } \right|. $$

According to the lemma and the literature (Li and Wu 2013; Zhang and Hao 2007), we have

$$ \Delta \ne 0. $$

That is, the solution of the boundary value problem (23a ~ 23c) exists and is unique. Below we determine the expression of its solution. We introduce the following functions:

$$ \psi_{c,d} \left( {x,y,z} \right) = K_{c} (xz)I_{d} \left( {yz} \right) + \left( { - 1} \right)^{c - d + 1} I_{c} \left( {xz} \right)K_{d} \left( {yz} \right). $$

Among them, c, d are real numbers, x, y, z are variables, and \(I_{l} ( \cdot )\), \(K_{l} ( \cdot )\) are the first type and second type of modified Bessel functions of l-order, respectively.

We replace the variables as follows (Leng et al. 2017):

$$ u = x^{ - \alpha } y,\quad \xi = \kappa x^{\beta } . $$

Then, Eq. (23a) becomes the following standardized Bessel equation:

$$ \xi^{2} \frac{{\partial^{2} u}}{{\partial \xi^{2} }} + \xi \frac{\partial u}{{\partial \xi }} - \left( {\xi^{2} + \upsilon^{2} } \right)u = 0. $$
(28)

We know that the two linearly independent solutions of Eq. (28) are \(I_{\upsilon } \left( \xi \right),K_{\upsilon } \left( \xi \right)\), and then, the general solution of Eq. (23a) can be obtained as follows (Liu and Liu 2002):

$$ y = x^{\alpha } u\left( \xi \right) = x^{\alpha } u\left( {\kappa x^{\beta } } \right) = x^{\alpha } \left[ {C_{1} I_{\upsilon } \left( {\kappa x^{\beta } } \right) + C_{2} K_{\upsilon } \left( {\kappa x^{\beta } } \right)} \right]. $$
(29)

We substitute Eq. (29) into Eq. (23b) and Eq. (23c), respectively, and get:

$$ \begin{gathered} \left\{ {Ea^{\alpha } I_{\upsilon } \left( {\kappa a^{\beta } } \right) + \left( {1 + EF} \right)\left[ {\left( {\alpha + \beta \upsilon } \right)a^{\alpha - 1} I_{\upsilon } \left( {\kappa a^{\beta } } \right) + \kappa \beta a^{\alpha + \beta - 1} I_{\upsilon + 1} \left( {\kappa a^{\beta } } \right)} \right]} \right\}C_{1} \hfill \\ + \left\{ {Ea^{\alpha } K_{\upsilon } \left( {\kappa a^{\beta } } \right) + \left( {1 + EF} \right)\left[ {\left( {\alpha + \beta \upsilon } \right)a^{\alpha - 1} K_{\upsilon } \left( {\kappa a^{\beta } } \right) - \kappa \beta a^{\alpha + \beta - 1} K_{\upsilon + 1} \left( {\kappa a^{\beta } } \right)} \right]} \right\}C_{2} = D, \hfill \\ \end{gathered} $$
(30)
$$ \begin{gathered} \left\{ {\left[ {\varepsilon_{b} b^{\alpha } + \left( {\alpha + \beta \upsilon } \right)b^{\alpha } } \right]I_{\upsilon } \left( {\kappa b^{\beta } } \right) + \kappa \beta b^{\alpha + \beta } I_{\upsilon + 1} \left( {\kappa b^{\beta } } \right)} \right\}C_{1} \hfill \\ \quad \quad {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} + \left\{ {\left[ {\varepsilon_{b} b^{\alpha } + \left( {\alpha + \beta \upsilon } \right)b^{\alpha } } \right]K_{\upsilon } \left( {\kappa b^{\beta } } \right) - \kappa \beta b^{\alpha + \beta } K_{\upsilon + 1} \left( {\kappa b^{\beta } } \right)} \right\}C_{2} = 0. \hfill \\ \end{gathered} $$
(31)

We get \(C_{1} ,{\kern 1pt} {\kern 1pt} C_{2}\) according to Eqs. (30~31), as follows:

$$ C_{1} = \frac{{ - D\left\{ {\left[ {\varepsilon_{b} b^{\alpha } + \left( {\alpha + \beta \upsilon } \right)b^{\alpha } } \right]K_{\upsilon } \left( {\kappa b^{\beta } } \right) - \kappa \beta b^{\alpha + \beta } K_{\upsilon + 1} \left( {\kappa b^{\beta } } \right)} \right\}}}{{E\left[ {\varepsilon_{b} \varphi_{v,v} \left( {a,b} \right) + b\varphi_{v,v + 1} \left( {a,b} \right)} \right] + \left( {1 + EF} \right)\left[ {\varepsilon_{b} \varphi_{v + 1,v} \left( {a,b} \right) + b\varphi_{v + 1,v + 1} \left( {a,b} \right)} \right]}} $$
$$ C_{2} = \frac{{D\left\{ {\left[ {\varepsilon_{b} b^{\alpha } + \left( {\alpha + \beta \upsilon } \right)b^{\alpha } } \right]K_{\upsilon } \left( {\kappa b^{\beta } } \right) - \kappa \beta b^{\alpha + \beta } K_{\upsilon + 1} \left( {\kappa b^{\beta } } \right)} \right\}}}{{E\left[ {\varepsilon_{b} \varphi_{v,v} \left( {a,b} \right) + b\varphi_{v,v + 1} \left( {a,b} \right)} \right] + \left( {1 + EF} \right)\left[ {\varepsilon_{b} \varphi_{v + 1,v} \left( {a,b} \right) + b\varphi_{v + 1,v + 1} \left( {a,b} \right)} \right]}} $$

Then put \(C_{1} ,C_{2}\) into Eq. (29) to obtain the solution Eq. (24) of the boundary value problem (23a ~ 23c).

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Zhao, C., Min, C. Analysis of the transient flow of non-Newtonian power-law fluids in homogeneous reservoirs with the elastic outer boundary. Acta Geophys. (2021). https://doi.org/10.1007/s11600-021-00641-2

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Keywords

  • Elastic function
  • Homogeneous reservoir
  • Non-Newtonian power-law fluid
  • Analytical solution