Mathematical models of steady-state temperature fields produced by multi-piped freezing

Abstract

The multi-piped freezing method is usually applied in artificial ground freezing (AGF) projects to fulfill special construction requirements, such as two-, three-, or four-piped freezing. Based on potential superposition theory, this paper gives analytical solutions to steady-state frozen temperature for two, three, and four freezing pipes with different temperatures and arranged at random. Specific solutions are derived for some particular arrangements, such as three freezing pipes in a linear arrangement with equal or unequal spacing, right and isosceles triangle arrangements, four freezing pipes in a linear arrangement with equal spacing, and rhombus and rectangle arrangements. A comparison between the analytical solutions and numerical thermal analysis shows that the analytical solutions are sufficiently precise. As a part of the theory of AGF, the analytical solutions of temperature fields for multi-piped freezing with arbitrary layouts and different temperatures of freezing pipes are approached for the first time.

中文概要

目 的

少量任意布置冻结管冻结的稳态温度场无解析解。建立任意布置少量管冻结稳态温度场模型, 获得解析解, 解决人工冻结温度场理论问题。

创新点

1. 基于势函数叠加原理, 确立人工地层冻结中少量管冻结稳态温度场的通用求解方法; 2. 建立任意布置的3 根和4 根非等温冻结管下冻结稳态温度场数学模型, 获得其解析解通解及特解。

方 法

1. 通过理论分析, 将冻结管简化为热汇点源, 确定人工地层冻结热势的拉普拉斯方程表述; 2. 应用势函数叠加原理建立少量管冻结稳态温度场的通用求解方法; 3. 建立少量管冻结稳态温度场的数学模型, 通过理论推导获得温度场解析解; 4. 通过数值模拟, 验证所提方法、数学模型和解析解的正确性和准确性。

结 论

1. 将冻结管简化为点源(热汇), 其冻结形成的热势场服从拉普拉斯方程, 其解即为热势函数; 2. 多根冻结管冻结时, 将单根冻结管的热势函数叠加, 由冻结管的位置决定每根冻结管的热流, 再根据边界条件定解。这一方法(即势函数叠加法)可以用于任意布置冻结管冻结稳态温度场解析解的求解; 3. 将冻结管简化为点源导致获得的解析解存在一定的误差, 但误差仅发生在冻结管附近极小的范围内, 并且误差微小, 完全满足工程上的精度要求。

1 Introduction

Artificial ground freezing (AGF) is a construction method that converts the water in the ground into ice by means of artificial refrigeration technology. This creates a strong, watertight frozen soil wall which serves as a temporary support structure during excavation. Due to its strong water-sealing ability and the mechanical strength of a frozen soil wall, and its superiority for safety and environmental conservation, AGF has been widely employed in many types of construction projects, including shield launching and receiving, retaining and protecting foundation pits, mine shaft sinking, tunnel boring machine (TBM) maintenance (Li et al., 2004; Itoh et al., 2005; Primentel et al., 2007; Schmall and Maishman, 2007; Viggiani and de Sanctis, 2009; Hu and Long, 2010; Russo et al., 2012; Viggiani and Casini, 2015; Casini et al., 2016), and recovery of shield tunnels (Ju et al., 1998; Wang, 2006; Xiao et al., 2006). As the properties of a frozen soil wall, such as its mechanical properties and thickness, are functions of temperature, the calculation of the temperature field is a pre-requisite for studying the temperature distribution, i.e., it is a significant part of the basis for design and construction using AGF.

Among the main methods for calculating AGF temperature fields, the analytical method is more reliable than simulation method and numerical analysis method, and is an important part of the theory of AGF. In AGF engineering applications, the rate of heat conduction is actually very slow. Consequently, we can adopt steady-state rather than transient temperature fields to calculate the temperature field at a certain point in the freezing process. This viewpoint has been accepted in academic and engineering circles (Trupak, 1954; Bakholdin, 1963; Frederick and Sanger, 1968; Tobe and Akimoto, 1979; Kato et al., 2007).

Based on steady-state analytical heat conduction theory, some solutions of steady-state temperature fields in AGF have been derived. Trupak (1954) studied the temperature field formed by single-piped freezing and obtained the analytical solution of a single-piped frozen temperature field. He also provided the analytical solution of a single-row-piped frozen temperature field according to the geometrical relationship between two adjacent frozen soil columns. The results, however, differed from experimental data because he had ignored the interaction between two adjacent pipes. Bakholdin (1963) considered that once two adjacent frozen soil columns were merging, the wave-shaped boundary of the frozen soil wall would soon become flat. Based on the theory of analogy between thermal and hydraulic problems, he obtained the analytical solutions of single-row-piped and double-row-piped steady-state frozen temperature fields. These solutions have been shown to be sufficiently accurate (Hu and Zhao, 2010; Hu C.P. et al., 2011). Frederick and Sanger (1968) presented a simplified formula for a single-row-piped frozen temperature field. Tobe and Akimoto (1979) and Kato et al. (2007) derived an analytical solution of a temperature field by multi-piped freezing with the pipes arranged in a straight line with equal spacing.

In China, there have been many studies of temperature fields of AGF (Chen and Tang, 1982; Tang et al., 1995; Cui, 1997; Wang and Cao, 2002; Xu, 2005; Dong et al., 2007; Yuan et al., 2011; Zhou and Zhou, 2011). Considering that the actual freezing temperature of soil is below 0 °C, Hu et al. (2008b) improved some existing analytical solutions. These analytical solutions have been applied in several other studies (Hu et al., 2008a; Hu and He, 2009; Hu, 2010a; 2010b; Hu and Zhao, 2010; Hu C.P. et al., 2011; Hu and Wang, 2012; Hu et al., 2012; 2013a).

In AGF engineering applications, a freezing method employing a small number of freezing pipes, called multi-piped freezing, is usually applied to meet the requirements of special construction practices, such as two-, three-, or four-piped freezing. Analytical solutions of multi-piped frozen temperature fields, such as for two freezing pipe arrangements, plus three and four freezing pipes arranged in a straight line with equal spacing, were derived by Tobe and Akimoto (1979) and Kato et al. (2007). However, there are no analytical solutions for three or four freezing pipes arranged in different forms. Based on potential superposition theory, this paper gives analytical solutions of steady-state frozen temperature for three and four freezing pipes arranged at random. Specific solutions are derived for some particular arrangements, including three freezing pipes in a linear arrangement with equal or unequal spacing, right and isosceles triangle arrangements, four freezing pipes in a linear arrangement with equal spacing, and rhombus and rectangle arrangements.

2 Heat potential and potential superposition

Heat conduction, convection, and radiation are the three main forms of heat transfer in soils. In AGF engineering, convection and radiation barely affect the distribution of the temperature field and therefore will be ignored in this paper.

According to Fourier’s law, in unit time, the heat flux passing the infinitesimal layer whose length is dx forms a direct ratio to the temperature rate and the area of the infinitesimal layer, which can be written as

$$q = - k{{\partial T} \over {\partial x}},$$

(1)

where k is the thermal conductivity of the soil and T is the soil temperature. We introduce a parameter Φ, defined as the heat potential, which is given by Φ=kT. Substituting Φ into Eq. (1) yields

$$q = - k{{\partial \Phi } \over {\partial x}}{.}$$

(2)

According to Fourier’s law and the first law of thermodynamics, the governing equation of heat conduction in the plane can be described by

$${{\partial {q_x}} \over {\partial x}} + {{\partial {q_y}} \over {\partial y}} - Q + \rho c{{\partial T} \over {\partial t}} = 0,$$

(3)

where q _x and q _y are the heat fluxes in unit time along the x and y axes, respectively. Q is the heat in unit time and unit volume exchanged between the system and the environment. ρ is the soil density and c the specific heat of soil.

In this paper, soil is as assumed to be isotropic, and the temperature field of frozen soil is approximated to be steady state. These assumptions have previously been shown to be sufficiently accurate (Trupak, 1954; Bakholdin, 1963; Frederick and Sanger, 1968; Tobe and Akimoto, 1979; Kato et al., 2007). Heat exchange between the system and the environment is ignored because the calculation region is infinite. Substituting Φ and the steady-state condition into Eq. (3) yields

$${{{\partial ^2}\Phi } \over {\partial {x^2}}} + {{{\partial ^2}\Phi } \over {\partial {y^2}}} = 0{.}$$

(4)

The polar form of Eq. (4) can be written as ${{\rm{d}} \over {{\rm{d}}r}}\left({r{{{\rm{d}}\Phi} \over {{\rm{d}}r}}} \right) = 0{.}$.

Eq. (4) is Laplace’s equation, the solution of which is the heat potential Φ.

For simplicity, the freezing pipe can be regarded as a point source, i.e., a cold source or heat sink, existing at the center of the pipe. Let q_c be the heat flow absorbed in unit time at steady state by the cold source. The heat potential at each point in the plane would be lowered due to q_c.

According to the above hypothesis, q_c can be written as

$${q_{\rm{c}}} = - 2{{\pi }}r{{d\Phi } \over {dr}},$$

(5)

where r is the distance from the cold source. Integrating Eq. (5), the heat potential Φ at the circle with radius r can be expressed as

$$\Phi = - {{{q_{\rm{c}}}} \over {2{{\pi }}}}\ln r + C,$$

(6)

where C is the integral constant to be determined by the boundary conditions.

According to potential superposition theory, if multiple point sources exist in the plane, the reduction of heat potential can be superposed. Therefore, when multiple freezing pipes are arranged in the plane, the heat potential at an arbitrary point in this plane equals the superposition of the heat potentials caused by the cold sources of each freezing pipe at this point. If there are n freezing pipes arranged in the plane, the heat potential at an arbitrary point can be expressed as

$$\Phi = - \left({{{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln {r_1} + {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}}\ln {r_2} + \cdots + {{{q_{{\rm{c}}n}}} \over {2{{\pi }}}}\ln {r_n}} \right) + C{.}$$

(7)

Eq. (7) can be written as $\Phi = - \sum\limits_{i = 1}^n {{{{q_{{\rm{c}}i}}} \over {2\pi }}\ln {r_i} + C}$, where r _i is the distance from this point to the ith cold source and q _ci is the heat flow of the ith cold source.

The potential superposition method, described above, is viable for solving plane problems of a steady-state temperature field frozen by any freezing pipes in any arrangement, including the most used engineering classes, such as multi-piped (which will be given in this paper), straight-row-piped (Hu et al., 2013c), and circle-piped arrangements.

The potential superposition method can also be used for space problems (Hu et al., 2013b).

3 Temperature field of a single freezing pipe

The freezing pipe is placed at the origin of the coordinates (Fig. 1). r₀ is the radius of the freezing pipe, ξ is the radius of the frozen soil zone, T_f is the surface temperature of the freezing pipe, T₀ is the freezing temperature of frozen soil, and q_c is the heat flow of the freezing pipe.

Fig. 1

One freezing pipe in the infinite plane

The heat potential expression at arbitrary point M(x, y) in the temperature field is the same as in Eq. (6). r is given by $r = \sqrt {{x^2} + {y^2}}$.

The heat potential at an arbitrary point at the boundary of frozen soil can be expressed as

$${\Phi _0} = - {{{q_{\rm{c}}}} \over {2{{\pi }}}}\ln \xi + C{.}$$

(8)

The heat potential at an arbitrary point on the freezing pipe surface can be expressed as

$${\Phi _{\rm{f}}} = - {{{q_{\rm{c}}}} \over {2{{\pi }}}}\ln {r_0} + C{.}$$

(9)

Solving Eqs. (8) and (9) simultaneously, the expression of q_c can be obtained as $- {{{q_{\rm{c}}}} \over {2\pi }} = {{{\Phi _{\rm{f}}} - {\Phi _0}} \over {\ln \;({r_0}/\xi)}}$. Substituting q_c, Φ=kT, Φ₀=kT₀), and Φ_f=kT_f into Eq. (6), we arrive at

$$T = {T_0} + {{\ln (r/\xi)} \over {\ln ({r_0}{{/}}\xi)}}({T_{\rm{f}}} - {T_0}){.}$$

(10)

Eq. (10) is the analytical solution of a steady-state temperature field of one freezing pipe in an infinite plane. Eq. (10) is Trupak’s formula (Trupak, 1954).

4 Temperature field of two freezing pipes

Φ_f1 and Φ_f2 are the heat potentials of two freezing pipes, respectively. The surface temperatures, T_f1 and T_f2, of the two freezing pipes are not equal. The radii of the two freezing pipes are equal (r₀). T₀ is the freezing temperature of frozen soil. q_c1 and q_c2 are the heat flows of two freezing pipes, respectively. The distance between two freezing pipes is 2d. We define a conditional point at the boundary of the frozen soil with the coordinates (0, ξ) (Fig. 2).

Fig. 2

Two freezing pipes in the infinite plane

According to potential superposition theory, the heat potential at an arbitrary point M(x, y) in the temperature field is superposed by heat potentials acquired from two freezing pipes acting alone, which can be written as

$$\Phi = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln {r_1} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}}\ln {r_2} + C,$$

(11)

where r₁ and r₂ are the distances of the arbitrary point M to the freezing pipes P₁ and P₂, respectively, which can be expressed as

$${r_1} = \sqrt {{{(x + d)}^2} + {y^2}}, \quad {r_2} = \sqrt {{{(x - d)}^2} + {y^2}} {.}$$

.

The heat potential Φ₀ at the conditional point (0, ξ) at the boundary of the frozen soil can be expressed as

$${\Phi _0} = - {{{q_{{\rm{c}}1}}} \over {{{2\pi }}}} \cdot \ln \sqrt {{\xi ^2} + {d^2}} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}} \cdot \ln \sqrt {{\xi ^2} + {d^2}} + C{.}$$

(12)

The heat potentials Φ_f1 at point (−d, r₀) and Φ_f2 at (d, r₀) on the surface of each freezing pipe can be expressed as

$${\Phi _{{\rm{f}}1}} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}} \cdot \ln {r_0} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}} \cdot \ln \sqrt {4{d^2}+r_0^2} +C,$$

(13)

$${\Phi _{{\rm{f}}2}} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}} \cdot \ln \sqrt {4{d^2}{\bf{ + }}r_0^2} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}} \cdot \ln {r_0} + C{.}$$

(14)

As r₀ is small in comparison to the scale of the frozen zone, it has little impact on the final calculation. Consequently, Eqs. (13) and (14) can be simplified as

$${\Phi _{{\rm{f}}1}} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}} \cdot \ln {r_0} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}} \cdot \ln (2d) + C,$$

(15)

$${\Phi _{{\rm{f}}2}} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}} \cdot \ln (2d) - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}} \cdot \ln {r_0} + C{.}$$

(16)

Using Φ₀, Φ_f1, and Φ_f2 to stand for q_c1 and q_c2, we arrive at

$$\begin{array}{*{20}c} { - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}} = {{{\Phi _{{\rm{f}}1}}\ln {{{r_0}} \over {\sqrt {{\xi ^2}{{ + }}{d^2}} }}} \over {\ln {{{r_0}} \over {2d}}\ln {{2d{r_0}} \over {{\xi ^2}{{ + }}{d^2}}}}} - {{{\Phi _{{\rm{f}}2}}\ln {{2d} \over {\sqrt {{\xi ^2}{{ + }}{d^2}} }}} \over {\ln {{{r_0}} \over {2d}}\ln {{2d{r_0}} \over {{\xi ^2}{{ + }}{d^2}}}}}}\\{ - {{{\Phi _{{0}}}\ln {{{r_0}} \over {2d}}} \over {\ln {{{r_0}} \over {2d}}\ln {{2d{r_0}} \over {{\xi ^2}{{ + }}{d^2}}}}},\quad \quad \quad } \end{array}$$

(17)

$$\begin{array}{*{20}c} { - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}} = - {{{\Phi _{{{{\rm {f}}1}}}}\ln {{2d} \over {\sqrt {{\xi ^2}{{ + }}{d^2}} }}} \over {\ln {{{r_0}} \over {2d}}\ln {{2d{r_0}} \over {{\xi ^2}{{ + }}{d^2}}}}} + {{{\Phi _{{{{\rm {f}}2}}}}\ln {{{r_0}} \over {\sqrt {{\xi ^2}{{ + }}{d^2}} }}} \over {\ln {{{r_0}} \over {2d}}\ln {{2d{r_0}} \over {{\xi ^2}{{ + }}{d^2}}}}}}\\ { - {{{\Phi _{{0}}}\ln {{{r_0}} \over {2d}}} \over {\ln {{{r_0}} \over {2d}}\ln {{2d{r_0}} \over {{\xi ^2}{{ + }}{d^2}}}}}{.}\quad \quad \quad \quad } \end{array}$$

(18)

Substituting Eqs. (12), (17), and (18) into Eq. (11), the heat potential at an arbitrary point M(x, y) in the temperature field can be expressed as

$$\begin{array}{*{20}c} {\Phi = {{\ln {{{r_1}} \over {\sqrt {{\xi ^2} + {d^2}} }}\ln {{{r_0}} \over {\sqrt {{\xi ^2} + {d^2}} }}} \over {\ln {{{r_0}} \over {2d}}\ln {{2d{r_0}} \over {{\xi ^2} + {d^2}}}}}{\Phi _{{{{\rm {f}}1}}}}\quad \quad \quad \quad \quad \quad }\\ { - {{\ln {{{r_2}} \over {\sqrt {{\xi ^2} + {d^2}} }}\ln {{2d} \over {\sqrt {{\xi ^2} + {d^2}} }}} \over {\ln {{{r_0}} \over {2d}}\ln {{2d{r_0}} \over {{\xi ^2} + {d^2}}}}}{\Phi _{{{{\rm {f}}1}}}}}\\ { + {{\ln {{{r_2}} \over {\sqrt {{\xi ^2} + {d^2}} }}\ln {{{r_0}} \over {\sqrt {{\xi ^2} + {d^2}} }}} \over {\ln {{{r_0}} \over {2d}}\ln {{2d{r_0}} \over {{\xi ^2} + {d^2}}}}}{\Phi _{{{{\rm {f}}2}}}}}\\ { - {{\ln {{{r_1}} \over {\sqrt {{\xi ^2} + {d^2}} }}\ln {{2d} \over {\sqrt {{\xi ^2} + {d^2}} }}} \over {\ln {{{r_0}} \over {2d}}\ln {{2d{r_0}} \over {{\xi ^2} + {d^2}}}}}{\Phi _{{{{\rm {f}}2}}}} - \left({{{\ln {{{r_1}{r_2}} \over {{\xi ^2} + {d^2}}}} \over {\ln {{2d{r_0}} \over {{\xi ^2} + {d^2}}}}} - 1} \right){\Phi _{{0}}}{.}} \end{array}$$

(19)

Substituting Φ=kT, Φ₀=kT₀), Φ_fl=kT_fl, and Φ_f2= kT_f2 into Eq. (19), we arrive at

$$\begin{array}{*{20}c} {T = {{\ln {{{r_1}} \over {\sqrt {{\xi ^2} + {d^2}} }}\ln {{{r_0}} \over {\sqrt {{\xi ^2} + {d^2}} }}} \over {\ln {{{r_0}} \over {2d}}\ln {{2d{r_0}} \over {{\xi ^2} + {d^2}}}}}{T_{{{{\rm {f}}1}}}}\quad \quad \quad \quad \quad \quad \quad \;\;\quad \quad }\\ { - \; {{\ln {{{r_2}} \over {\sqrt {{\xi ^2} + {d^2}} }}\ln {{2d} \over {\sqrt {{\xi ^2} + {d^2}} }}} \over {\ln {{{r_0}} \over {2d}}\ln {{2d{r_0}} \over {{\xi ^2} + {d^2}}}}}{T_{{{{\rm {f}}1}}}}\quad \quad \quad \quad \quad \quad \quad \quad }\\ { + {{\ln {{{r_2}} \over {\sqrt {{\xi ^2} + {d^2}} }}\ln {{{r_0}} \over {\sqrt {{\xi ^2} + {d^2}} }}} \over {\ln {{{r_0}} \over {2d}}\ln {{2d{r_0}} \over {{\xi ^2} + {d^2}}}}}{T_{{{{\rm {f}}2}}}}\quad \quad \quad \quad \quad \quad \quad \quad }\\ { - \; {{\ln {{{r_1}} \over {\sqrt {{\xi ^2} + {d^2}} }}\ln {{2d} \over {\sqrt {{\xi ^2} + {d^2}} }}} \over {\ln {{{r_0}} \over {2d}}\ln {{2d{r_0}} \over {{\xi ^2} + {d^2}}}}}{T_{{{{\rm {f}}2}}}} - \left({{{\ln {{{r_1} \cdot {r_2}} \over {{\xi ^2} + {d^2}}}} \over {\ln {{2d{r_0}} \over {{\xi ^2} + {d^2}}} }} - 1} \right){T_{{0}}}{.}} \end{array}$$

(20)

Eq. (20) is the analytical solution of a steady-state temperature field of two freezing pipes with different surface temperatures in an infinite plane.

Assuming the surface temperatures of two freezing pipes are equal (T_f1=T_f2), Eq. (20) can be simplified as

$$\begin{array}{*{20}c} {T = {T_0}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;}\\ { + {{\ln {{\sqrt {{{(x + d)}^2} + {y^2}} \cdot \sqrt {{{(x - d)}^2} + {y^2}} } \over {{\xi ^2} + {d^2}}}} \over {\ln {{2d{r_0}} \over {{\xi ^2} + {d^2}}}}}({T_{\rm{f}}} - {T_0}){.}} \end{array}$$

(21)

Eq. (21) is consistent with the analytical solution derived by Tobe and Akimoto (1979) and Kato et al. (2007), which is a particular solution of our solution when T₀=0 °C.

Another solution to the steady-state temperature field frozen by two freezing pipes with different temperatures was given by Hu and Zhang (2013a). In their study, the boundary conditions were set differently and as a result, the expressions of the solution had different forms. Both solutions produce the same temperature field.

Hu and Zhang (2013b) solved the problem of a steady-state temperature field of one and two freezing pipes near a linear adiabatic boundary, using potential superposition method and mirror reflection of source and sink.

5 Temperature field of three freezing pipes

5.1 Temperature field of three freezing pipes in an arbitrary arrangement

Three-piped freezing method is usually applied in AGF projects. To fulfill construction requirements, three freezing pipes may be arranged in a particular form, such as a linear, right triangle or isosceles triangle arrangements. In this section, we derive the analytical solution of a steady-state temperature field of three-piped freezing with an arbitrary arrangement.

Φ_fl, Φ_f2, and Φ_f3 are the heat potentials of three freezing pipes, respectively. The surface temperatures, T_f1, T_f2, and T_f3, of the three freezing pipes are not equal. The radii of the pipes are equal (r₀). T₀ is the freezing temperature of frozen soil and q_c1, q_c2, and q_c3 are the heat flows of the three pipes, respectively. The distances between two pipes are d₁, d₂, and d₃. The center coordinates of the three pipes are P₁(x₁, y₁), P₂(x₂, y₂), and P₃(x₃, y₃), respectively (Fig. 3). To simplify the derivation, we establish a coordinate system with the line connecting the centers of pipes P₁ and P₃ as the x axis, and the vertical line from the center of pipe P₂ to the above connecting line as the y axis. Therefore, the center coordinates of the three pipes are changed to P₁(x₁, 0), P₂(0, y₂), and P₃(x₃, 0). Similarly, we define a conditional point at the boundary of the frozen soil, with the coordinates (0, y₂+ξ).

Fig. 3

Three freezing pipes arranged at random

Considering x₂, y₁, and y₃ are all zero, d₁, d₂, and d₃ in Fig. 3 can be expressed as ${d_1} = \sqrt {x_1^2 + y_2^2}$, ${d_2} = \sqrt {x_3^2 + y_2^2}$, and ${d_3} = \vert {x_1} - {x_3}\vert$. According to potential superposition theory, the heat potential at an arbitrary point M(x, y) in the temperature field is superposed by heat potentials acquired from the three freezing pipes acting alone, and can be expressed as

$$\Phi = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln {r_1} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}}\ln {r_2} - {{{q_{{\rm{c}}3}}} \over {2{{\pi }}}}\ln {r_3} + C,$$

(22)

where r₁, r₂, and r₃ are the distances of the arbitrary point M(x, y) to pipes P₁, P₂, and P₃, respectively, which can be expressed as ${r_1} = \sqrt {{{(x - {x_1})}^2} + {y^2}}$, ${r_2} = \sqrt {{x^2} + {{(y - {y_2})}^2}}$, and ${r_3} = \sqrt {{{(x - {x_3})}^2} + {y^2}}$.

The heat potential Φ₀ at the conditional point (0, y₂+ξ) at the boundary of the frozen soil can be expressed as

$$\begin{array}{*{20}c} {{\Phi _0} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln \sqrt {x_1^2 + {{(\xi + {y_2})}^2}} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}}\ln \xi }\\ { - {{{q_{{\rm{c}}3}}} \over {2{{\pi }}}}\ln \sqrt {x_3^2 + {{(\xi + {y_2})}^2}} + C{.}} \end{array}$$

(23)

The heat potentials Φ_f1 at points (x₁, r₀), Φ_f2 at (0, y₂+r₀) and Φ_f3 at (x₃, r₀) on the surface of each freezing pipe can be expressed as

$$\begin{array}{*{20}c} {{\Phi _{{\rm{f}}1}} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln {r_0} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}}\ln \sqrt {x_1^2 + {{({y_2} - {r_0})}^2}} }\\ { - {{{q_{{\rm{c}}3}}} \over {2{{\pi }}}}\ln \sqrt {d_3^2 + r_0^2} + C,\quad \quad \quad } \end{array}$$

(24)

$$\begin{array}{*{20}c} {{\Phi _{{\rm{f}}2}} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln \sqrt {x_1^2 + {{({y_2} + {r_0})}^2}} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}}\ln {r_0}}\\ { - {{{q_{{\rm{c}}3}}} \over {2{{\pi }}}}\ln \sqrt {x_3^2 + {{({y_2} + {r_0})}^2}} + C,} \end{array}$$

(25)

$$\begin{array}{*{20}c} {{\Phi _{{\rm{f}}3}} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln \sqrt {d_3^2 + r_0^2} - {{{q_{{\rm{c}}3}}} \over {2{{\pi }}}}\ln {r_0}\quad \quad \quad }\\ { - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}}\ln \sqrt {x_3^2 + {{({y_2} - {r_0})}^2}} + C{.}} \end{array}$$

(26)

As r₀ is small in comparison to the scale of the frozen zone, it barely has any impact on the final calculation results. Consequently, Eqs. (24) to (26) can be simplified as

$${\Phi _{{\rm{f}}1}} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln {r_0} - {{{q_{c2}}} \over {2{{\pi }}}}\ln {d_1} - {{{q_{c3}}} \over {2{{\pi }}}}\ln {d_3} + C,$$

(27)

$${\Phi _{{\rm{f}}2}} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln {d_1} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}}\ln {r_0} - {{{q_{{\rm{c}}3}}} \over {2{{\pi }}}}\ln {d_2} + C,$$

(28)

$${\Phi _{{\rm{f}}3}} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln {d_3} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}}\ln {d_2} - {{{q_{{\rm{c}}3}}} \over {2{{\pi }}}}\ln {r_0} + C{.}$$

(29)

Using Φ₀, Φ_f1, Φ_f2, and Φ_f3 to stand for q_c1, q_c2, and q_c3, we arrive at

$$\begin{array}{*{20}c} { - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}} = {{{A_1}} \over D}({\Phi _{{\rm{f}}1}} - {\Phi _{{0}}}) + {{{B_1}} \over D}({\Phi _{{{{\rm {f}}2}}}} - {\Phi _{{0}}})}\\ { + {{{C_1}} \over D}({\Phi _{{{{\rm {f}}3}}}} - {\Phi _{{0}}}),\quad \quad \quad } \end{array}$$

(30)

$$\begin{array}{*{20}c} { - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}} = {{{A_2}} \over D}({\Phi _{{\rm{f}}1}} - {\Phi _{{0}}}) + {{{B_2}} \over D}({\Phi _{{{{\rm {f}}2}}}} - {\Phi _{{0}}}) }\\ { + {{{C_2}} \over D}({\Phi _{{{{\rm {f}}3}}}} - {\Phi _{{0}}}),\quad \quad \quad } \end{array}$$

(31)

$$\begin{array}{*{20}c} { - {{{q_{{{c3}}}}} \over {2{{\pi }}}} = {{{A_3}} \over D}({\Phi _{{\rm{f}}1}} - {\Phi _{{0}}}) + {{{B_3}} \over D}({\Phi _{{{{\rm {f}}2}}}} - {\Phi _{{0}}})}\\ { + {{{C_3}} \over D}({\Phi _{{{{\rm {f}}3}}}} - {\Phi _{{0}}}),\quad \quad \quad } \end{array}$$

(32)

where D=D₁+D₂+D₃. A₁, A₂, A₃, B₁, B₂, B₃, C₁, C₂, C₃, D₁, D₂, and D₃ are shown in Appendix A.

Substituting Eqs. (23), (30), (31), and (32) into Eq. (22), the heat potential at arbitrary point M(x, y) in the temperature field can be expressed as

$$\begin{array}{*{20}c} {\Phi = {{{A_1}\ln {{{r_1}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }} + {A_1}\ln {{{r_2}} \over \xi }} \over D}{\Phi _{{{{\rm {f}}1}}}}\quad \quad \quad \quad }\\ {{{ + }}{{{A_3}\ln {{{r_3}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}} \over D}{\Phi _{{{{\rm {f}}1}}}}\quad \quad \quad }\\ { + {{{B_1}\ln {{{r_1}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }} + {B_1}\ln {{{r_2}} \over \xi }} \over D}{\Phi _{{{{\rm {f}}2}}}}\quad \quad \;\;\;}\\ { + {{{B_3}\ln {{{r_3}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}} \over D}{\Phi _{{{{\rm {f}}2}}}}\quad \quad \;\;\;\quad \quad \;\;\quad }\\ { + {{{C_1}\ln {{{r_1}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }} + {C_1}\ln {{{r_2}} \over \xi }} \over D}{\Phi _{{{{\rm {f}}3}}}}\quad \;\;\quad }\\ { + {{{C_3}\ln {{{r_3}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}} \over D}{\Phi _{{{{\rm {f}}3}}}}\quad \quad \quad \quad \;\;\quad \;\;}\\ { - \left[ {{{({A_1} + {B_1} + {C_1})\ln {{{r_1}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}} \over D}\;\;\;} \right.\quad \;\;\;}\\ { + {{({A_2} + {B_2} + {C_2})\ln {{{r_2}} \over \xi }} \over D}\quad \quad \quad \quad \quad \quad \quad \quad \;\;}\\ { + \left. {{{({A_3} + {B_3} + {C_3})\ln {{{r_3}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}} \over D} - 1} \right]{\Phi _{{0}}}{{.}}} \end{array}$$

(33)

Substituting Φ=kT, Φ₀=kT₀, Φ_f1=kT_f1, Φ_f2=kT_f2, and Φ_f3=kT_f3 into Eq. (33), we arrive at

$$\begin{array}{*{20}c} {T = {{{A_1}\ln {{{r_1}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }} + {A_1}\ln {{{r_2}} \over \xi }} \over D}{T_{{{{\rm {f}}1}}}}\quad \quad }\\ { + {{{A_3}\ln {{{r_3}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}} \over D}{T_{{{{\rm {f}}1}}}}\quad \quad \quad \quad }\\ { + {{{B_1}\ln {{{r_1}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }} + {B_1}\ln {{{r_2}} \over \xi }} \over D}{T_{{{{\rm {f}}2}}}}}\\ { + {{{B_3}\ln {{{r_3}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}} \over D}{T_{{{{\rm {f}}2}}}}\quad \quad \quad \;\;}\\ { + {{{C_1}\ln {{{r_1}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }} + {C_1}\ln {{{r_2}} \over \xi }} \over D}{T_{{{{\rm {f}}3}}}}}\\ { + {{{C_3}\ln {{{r_3}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}} \over D}{T_{{{{\rm {f}}3}}}}\quad \quad \quad \quad \quad \quad \;\;}\\ { - \left[ {{{({A_1} + {B_1} + {C_1})\ln {{{r_1}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}} \over D}} \right.\quad \quad \;\;\;}\\ { + {{({A_2} + {B_2} + {C_2})\ln {{{r_2}} \over \xi }} \over D}\quad \quad \quad \quad \quad \quad \quad \quad \;\;}\\ {\left. { + {{({A_3} + {B_3} + {C_3})\ln {{{r_3}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}} \over D} - 1} \right]{T_{{0}}}{.}} \end{array}$$

(34)

Eq. (34) is the analytical solution of a steady-state temperature field of three freezing pipes in an arbitrary arrangement in an infinite plane.

Assuming the surface temperatures of the three freezing pipes are equal (T_f1=T_f2=T_f3), Eq. (34) can be simplified as

$$\begin{array}{*{20}c} {T = {T_0} + ({T_{\rm{f}}} - {T_0})\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;}\\ { \times {{\ln {{{r_1}} \over {\sqrt {x_1^2 + {{({y_2} + \xi)}^2}} }} + {\eta _1}\ln {{{r_2}} \over \xi } + {\eta _2}\ln {{{r_3}} \over {\sqrt {x_3^2 + {{({y_2} + \xi)}^2}} }}} \over {\ln {{{d_1}} \over {\sqrt {x_1^2 + {{({y_2} + \xi)}^2}} }} + {\eta _1}\ln {{{r_0}} \over \xi } + {\eta _2}\ln {{{d_2}} \over {\sqrt {x_3^2 + {{({y_2} + \xi)}^2}} }}}},} \end{array}$$

(35)

where, assuming q_c2=η₁q_c1 and q_c3=η₂q_c1, η₁ and η₂ can be expressed as

$${\eta _{{1}}} = {{\ln {{{d_1}} \over {{d_3}}} \cdot \ln {{{d_3}} \over {{d_2}}} - \ln {{{r_0}} \over {{d_1}}} \cdot \ln {{{d_2}} \over {{r_0}}}} \over {\ln {{{d_1}} \over {{r_0}}} \cdot \ln {{{d_2}} \over {{r_0}}} - \ln {{{r_0}} \over {{d_2}}} \cdot \ln {{{d_3}} \over {{d_2}}}}},$$

(36)

$${\eta _{{2}}} = {{\ln {{{d_1}} \over {{d_3}}} \cdot \ln {{{d_1}} \over {{r_0}}} - \ln {{{r_0}} \over {{d_1}}} \cdot \ln {{{r_0}} \over {{d_2}}}} \over {\ln {{{d_3}} \over {{d_2}}} \cdot \ln {{{r_0}} \over {{d_2}}} - \ln {{{d_2}} \over {{r_0}}} \cdot \ln {{{d_1}} \over {{r_0}}}}}{.}$$

(37)

5.2 Temperature field of three freezing pipes in a linear arrangement with equal spacing

When three freezing pipes are arranged in a straight line with equal spacing, the center of freezing pipe P₂ becomes the origin of the coordinates, and y₂ is zero. We use d to stand for d₁, d₂, and d₃ because 2d₁=2d₂=d₃ (Fig. 4).

Fig. 4

Three freezing pipes laid in a linear arrangement with equal spacing

With three freezing pipes arranged in this form, the coordinates of the conditional point at the boundary of the frozen soil are changed to (0, ξ), and the coordinates of arbitrary points on the surface of each freezing pipe are changed to (−d+r₀, 0), (0, r₀), and (d−r₀, 0), respectively. Eqs. (36) and (37) can be simplified as ${\eta _1} = {{\ln [d/(2{r_0})]} \over {\ln (d/{r_0})}}$ and ${\eta _2} = 1$. Eq. (36) can also be simplified as

$$T = {T_0} + {{\ln {{{r_1}{r_3}} \over {{\xi ^2} + {d^2}}} + {\eta _1}\ln {{{r_2}} \over \xi }} \over {\ln {{{d^2}} \over {{\xi ^2} + {d^2}}}{{ + }}{\eta _1}\ln {{{r_0}} \over \xi }}}({T_{\rm{f}}} - {T_0}),$$

(38)

where r₁, r₂, and r₃ can be expressed as

$${r_1} = \sqrt {{{(x + d)}^2} + {y^2}}, \quad {r_2} = \sqrt {{x^2} + {y^2}}, \quad {r_3} = \sqrt {{{(x - d)}^2} + {y^2}} {.}$$

.

Eq. (38) is the analytical solution of a steady-state temperature field of three freezing pipes in a linear arrangement with equal spacing in an infinite plane. It is consistent with the existing analytical solution derived by Tobe and Akimoto (1979) and Kato et al. (2007), which is a particular solution of our solution when T₀=0 °C.

5.3 Temperature field of three freezing pipes in a linear arrangement with unequal spacing

When three freezing pipes are laid in a linear arrangement with unequal spacing, the center of pipe P₂ is also the origin of the coordinates. We can see that d₃=d₁+d₂ (Fig. 5).

Fig. 5

Three freezing pipes laid in a linear arrangement with unequal spacing

When the three freezing pipes are arranged in this form, the coordinates of the conditional points at the boundary of the frozen soil and on the surface of pipe P₂ are the same as in Section 5.2. The coordinates of the other two points on the surface of each pipe are changed to (−d₁+r₀, 0) and (d₃−r₀, 0). According to the relationships among d₁, d₂, and d₃, Eqs. (36) and (37) can be expressed as

$${\eta _{{1}}} = {{\ln {{{d_1} + {d_2}} \over {{d_1}}} \cdot \ln {{{d_1} + {d_2}} \over {{d_2}}} - \ln {{{r_0}} \over {{d_1}}} \cdot \ln {{{r_0}} \over {{d_2}}}} \over {\ln {{{d_1}} \over {{r_0}}} \cdot \ln {{{r_0}} \over {{d_2}}} - \ln {{{d_1} + {d_2}} \over {{d_2}}} \cdot \ln {{{d_2}} \over {{r_0}}}}},$$

(39)

$${\eta _{{2}}} = {{\ln {{{d_1} + {d_2}} \over {{d_1}}} \cdot \ln {{{d_1}} \over {{r_0}}} - \ln {{{r_0}} \over {{d_1}}} \cdot \ln {{{d_2}} \over {{r_0}}}} \over {\ln {{{d_1} + {d_2}} \over {{d_2}}} \cdot \ln {{{d_2}} \over {{r_0}}} - \ln {{{r_0}} \over {{d_2}}} \cdot \ln {{{d_1}} \over {{r_0}}}}}{.}$$

(40)

The analytical solution to a steady-state temperature field of three freezing pipes in a linear arrangement with unequal spacing in an infinite plane can be obtained from

$$\begin{array}{*{20}c} {T = {T_0}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;}\\ { + {{\ln {{{r_1}} \over {\sqrt {{\xi ^2} + d_1^2} }} + {\eta _1}\ln {{{r_2}} \over \xi } + {\eta _2}\ln {{{r_3}} \over {\sqrt {{\xi ^2} + d_2^2} }}} \over {\ln {{{d_1}} \over {\sqrt {{\xi ^2} + d_1^2} }} + {\eta _1}\ln {{{r_0}} \over \xi } + {\eta _2}\ln {{{d_2}} \over {\sqrt {{\xi ^2} + d_2^2} }}}}({T_{\rm{f}}} - {T_0}),} \end{array}$$

(41)

where r₁, r₂, and r₃ can be expressed as

$${r_1} = \sqrt {{{(x + {d_1})}^2} + {y^2}}, \quad {r_2} = \sqrt {{x^2} + {y^2}}, \quad {r_3} = \sqrt {{{(x - {d_2})}^2} + {y^2}} {.}$$

.

A visualized figure of the temperature field drawn according to the results of the analytical solution, Eq. (41), is shown in Fig. 6. To confirm that the numerical solution can exactly express the temperature distribution in the temperature field, the precision of the analytical solution Eq. (41) was examined by a steady-state numerical solution. The calculation was under the following particular conditions: d₁=0.4 m, d₂=0.8 m, ξ=1.0 m, T_f=−70 °C (considering adopting liquid nitrogen freezing), T₀=0 °C, r₀=0.054 m. The isothermal diagram of the steady-state numerical solution when ξ=1.0 m is shown in Fig. 7. Comparison of the analytical formula with the numerical simulation of the main section 1 in Fig. 5 is shown in Fig. 8. A comparison of the results of the analytical formula and numerical simulation of intersection 1 in Fig. 5 is shown in Fig. 9.

Fig. 6

Temperature field calculated by analytical formula for three freezing pipes in a linear arrangement with unequal spacing

Fig. 7

Isothermal diagram calculated by numerical simulation for three freezing pipes in a linear arrangement with unequal spacing

Fig. 8

Comparison of results from applying the analytical formula and numerical simulation in the main section 1 in Fig. 5

Fig. 9

Comparison of results from applying the analytical formula and numerical simulation in intersection 1 in Fig. 5

From Figs. 8 and 9, we find that the results calculated by the two methods coincide, which shows the analytical solution, Eq. (41), is precise enough. Results from the two methods in relation to other main sections and intersections in Fig. 5 also coincide (these are not discussed further here due to space limitations).

5.4 Temperature field of three freezing pipes in a right triangle arrangement

When three freezing pipes are in a right triangle arrangement, the relationship among d₁, d₂, and d₃ can be expressed as d ²₃ =d ²₁ +d ²₂ . According to the geometrical properties of right triangles, the center coordinates of the three pipes are changed to P₁(−d ²₁ /d₃, 0), P₂(0, d₁d₂/d₃), and P₃(d ²₂ /d₃, 0), respectively (Fig. 10).

Fig. 10

Three freezing pipes laid in a right triangle arrangement

When the three freezing pipes are arranged in this form, the coordinates of the conditional point at the boundary of the frozen soil are changed to (0, d₁d₂/d₃+ξ), and the coordinates of arbitrary points on the surface of each pipe are changed to (−d ²₁ /d₃+r₀, 0), (0, d₁d₂/d₃−r₀), and (d ²₂ /d₃−r₀, 0), respectively. The expressions of η₁ and η₂ can be still written as Eqs. (36) and (37). Then, we arrive at

$$T = {T_0} + {{\ln {{{r_1}} \over {{A_{11}}}} + {\eta _1}\ln {{{r_2}} \over \xi } + {\eta _2}\ln {{{r_3}} \over {{B_{11}}}}} \over {\ln {{{d_1}} \over {{A_{11}}}} + {\eta _1}\ln {{{r_0}} \over \xi } + {\eta _2}\ln {{{d_2}} \over {{B_{11}}}}}}({T_{\rm{f}}} - {T_0}),$$

(42)

where A₁₁, B₁₁, r₁, r₂, and r₃ can be expressed as

$$\begin{array}{*{20}c} {{A_{11}} = \sqrt {{{\left({\xi + {{{d_1}\cdot{d_2}} \over {{d_3}}}} \right)}^2} + {{d_1^4} \over {d_3^2}}},}\\ {{B_{11}} = \sqrt {{{\left({\xi + {{{d_1}\cdot{d_2}} \over {{d_3}}}} \right)}^2} + {{d_2^4} \over {d_3^2}}},}\\ {{r_1} = \sqrt {{{(x + d_1^2/{d_3})}^2} + {y^2}}, \quad }\\ {{r_2} = \sqrt {{x^2} + (y - {d_1}\cdot{d_2}/{d_3}{)^2}},} \\ {{r_3} = \sqrt {{{(x - d_2^2/{d_3})}^2} + {y^2}} {.}\quad } \end{array}$$

.

Eq. (42) is the analytical solution to a steady-state temperature field of three freezing pipes laid in a right triangle arrangement in an infinite plane.

A visualized figure of the temperature field drawn according to the results of the analytical solution Eq. (42) is shown in Fig. 11. The steady-state numerical calculation method was again applied to verify the analytical solution in this section, under the following particular conditions: d₁=0.6 m, d₂=0.8 m, d₃=1.0 m. The other parameters were the same as in Section 5.3. An isothermal diagram of the steady-state numerical solution when ξ=1.0 m is shown in Fig. 12. A comparison of the results from the analytical formula and numerical simulation of the main section 2 in Fig. 10 is shown in Fig. 13. A comparison of the results from the analytical formula and numerical simulation of intersection 2 in Fig. 10 is shown in Fig. 14.

Fig. 11

Temperature field calculated by the analytical formula for three freezing pipes in a right triangle arrangement

Fig. 12

Isothermal diagram calculated by numerical simulation for three freezing pipes in a right triangle arrangement

Fig. 13

Comparison of results from applying the analytical formula and numerical simulation to the main section 2 in Fig. 10

From Figs. 13 and 14, we find again that the results calculated by the two methods coincide, which shows the analytical solution Eq. (42) is precise enough. Due to space limitations, comparisons of the two methods in relation to the other main sections and intersections in Fig. 10 are not covered here.

Fig. 14

Comparison of results from applying the analytical formula and numerical simulation to intersection 2 in Fig. 10

5.5 Temperature field of three freezing pipes in an isosceles triangle arrangement

When three freezing pipes are laid in an isosceles triangle arrangement, we can see d₁−d₂ (Fig. 15). According to the geometrical properties of isosceles triangles, using d to stand for d₁ and d₂, the center coordinates of the three pipes are changed to P₁(−d₃/2, 0), ${P_2}(0,\;\sqrt {{d^2} - d_3^2/4})$, and P₃(d₃/2, 0), respectively.

Fig. 15

Three freezing pipes laid in an isosceles triangle arrangement

When the three freezing pipes are arranged in this form, the coordinates of the conditional point at the boundary of the frozen soil are changed to (0, $\sqrt {{d^2} - d_3^3/4} + \xi$), and the coordinates of arbitrary points on the surface of each pipe are changed to (−d₃/2+r₀, 0), ($(0,\;\sqrt {{d^2} - d_3^2/4} - {r_0})$) and (d₃/2+r₀, 0), respectively. Eqs. (36) and (37) can be simplified as η₁ = ln(r₀d₃/d²)/ln(r₀/d) and η₂=1. Then, we can arrive at

$$T = {T_0} + {{\ln {{{r_1}{r_3}} \over {{A_{22}}}} + {\eta _1}\ln {{{r_2}} \over \xi }} \over {\ln {{{d^2}} \over {{A_{22}}}} + {\eta _1}\ln {{{r_0}} \over \xi }}}({T_{\rm{f}}} - {T_0}),$$

(43)

where A₂₂, r₁, r₂, and r₃ can be expressed as

$$\begin{array}{*{20}c} {{A_{22}} = {{\left({\xi + \sqrt {{d^2} - d_3^2/4} } \right)}^2} + d_3^2/4,\quad \quad \quad \quad \quad \quad \quad \quad \quad }\\ {{r_1} = \sqrt {{{(x + {d_3}/2)}^2} + {y^2}}, \quad {r_2} = \sqrt {{x^2} + {{(y - \sqrt {{d^2} - d_3^2/4})}^2}},}\\ {{r_3} = \sqrt {{{(x - d_2^2/{d_3})}^2} + {y^2}} {.}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;} \end{array}$$

.

Eq. (43) is the analytical solution to a steady-state temperature field of three freezing pipes laid in an isosceles triangle arrangement in an infinite plane.

The steady-state numerical calculation method was again applied to verify the analytical solution. The results of the verification were the same as in Sections 5.3 and 5.4, which shows that the analytical solution, Eq. (42), is precise enough. Due to space limitations, comparison charts of the two methods are not included here.

6 Temperature field of four freezing pipes

6.1 Temperature field of four freezing pipes in an arbitrary arrangement

A four-piped freezing method is usually applied in AGF projects to meet particular construction requirements. Four freezing pipes may be arranged in particular forms, such as a linear arrangement, rhombus arrangement or rectangle arrangement. In this section, we derive the analytical solution of a steady-state temperature field of four freezing pipes with an arbitrary arrangement.

Φ_f1, Φ_f2, Φ_f3, and Φ_f4 are the heat potentials of four freezing pipes, respectively. The surface temperatures of the four pipes are not equal. T_f1, T_f2, T_f3, and T_f4 are the surface temperatures of the four pipes, respectively. The radii (r₀) of the four freezing pipes are equal. T₀ is the freezing temperature of frozen soil. q_c1, q_c2, q_c3, and q_c4 are the heat flows of the four pipes, respectively. The distances between two pipes are d₁, d₂, d₃, d₄, d₅, and d₆. The center coordinates of the four pipes are P₁(x₁, y₁), P₂(x₂, y₂), P₃(x₃, y₃), and P₄(x₄, y₄) (Fig. 16). The definition of the coordinate system is the same as in Section 5.1. The coordinates of a conditional point defined at the boundary of the frozen soil is also (0, y₂+ξ).

Fig. 16

Four freezing pipes arranged at random

Considering that x₂, y₁, and y₃ are all zero, d₁, d₂, d₃, d₄, d₅, and d₆ in Fig. 16 can be expressed as

$$\begin{array}{*{20}c} {{d_1} = \sqrt {x_1^2 + y_2^2}, \quad {d_2} = \sqrt {x_3^2 + y_2^2}, \quad \quad \quad \quad \quad \;\;\;}\\ {{d_3} = \sqrt {{{({x_3} - {x_4})}^2} + y_4^2}, \quad {d_4} = \sqrt {{{({x_1} - {x_4})}^2} + y_4^2},} \\ {{d_5} = \left\vert {{x_3} - {x_1}} \right\vert, \quad {d_6} = \sqrt {{{({x_4} - {x_2})}^2} + {{({y_4} - {y_2})}^2}} {.}\;\;} \end{array}$$

.

According to potential superposition theory, the heat potential of an arbitrary point M(x, y) in the temperature field is superposed by heat potentials acquired from four freezing pipes acting alone. Its expression can be written as

$$\begin{array}{*{20}c} {\Phi = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln {r_1} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}}\ln {r_2}\;\;\quad \quad \quad }\\ { - {{{q_{{\rm{c}}3}}} \over {2{{\pi }}}}\ln {r_3} - {{{q_{{\rm{c}}4}}} \over {2{{\pi }}}}\ln {r_4} + C,} \end{array}$$

(44)

where r₁, r₂, r₃, and r₄ are the distances of the arbitrary point M(x, y) to pipes P₁, P₂, P₃, and P₄, respectively, which can be expressed as

$$\begin{array}{*{20}c} {{r_1} = \sqrt {{{(x - {x_1})}^2} + {y^2}}, \quad {r_2} = \sqrt {{x^2} + {{(y - {y_2})}^2}}, \quad \quad \;\;}\\ {{r_3} = \sqrt {{{(x - {x_3})}^2} + {y^2}}, \quad {r_4} = \sqrt {{{(x - {x_4})}^2} + {{(y - {y_4})}^2}} {.}} \end{array}$$

.

The heat potential at the conditional point (0, y₂+ξ) at the boundary of the frozen soil can be expressed as

$$\begin{array}{*{20}c} {{\Phi _0} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln {a_1} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}}\ln \xi \quad \quad \quad \quad }\\ { - {{{q_{{\rm{c}}3}}} \over {2{{\pi }}}}\ln {a_2} - {{{q_{{\rm{c}}4}}} \over {2{{\pi }}}}\ln {a_3} + C,} \end{array}$$

(45)

where a₁, a₂, and a₃ can be expressed as

$$\begin{array}{*{20}c} {{a_1} = \sqrt {x_1^2 + {{(\xi + {y_2})}^2}}, \quad {a_2} = \sqrt {x_3^2 + {{(\xi + {y_2})}^2}},}\\ {{a_3} = \sqrt {x_4^2 + {{({y_4} - {y_2} - \xi)}^2}} {.}\quad \quad \quad \quad \quad \quad \quad \;\;} \end{array}$$

.

Similar to three-piped freezing, because r₀ is much smaller than the scale of the frozen zone, it has little impact on the final calculation. Consequently, the heat potentials at arbitrary points (x₁, r₀), (0, y₂+r₀), (x₃, r₀), and (x₄, y₄+r₀) on the surface of each pipe can be expressed simply as

$${\Phi _{{\rm{f}}1}} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln {r_0} - {{{q_{c2}}} \over {2{{\pi }}}}\ln {d_1} - {{{q_{c3}}} \over {2{{\pi }}}}\ln {d_5} - {{{q_{c4}}} \over {2{{\pi }}}}\ln {d_4} + C,$$

(46)

$${\Phi _{{\rm{f}}2}} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln {d_1} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}}\ln {r_0} - {{{q_{{\rm{c}}3}}} \over {2{{\pi }}}}\ln {d_2} - {{{q_{{\rm{c}}4}}} \over {2{{\pi }}}}\ln {d_6} + C,$$

(47)

$${\Phi _{{\rm{f}}3}} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln {d_5} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}}\ln {d_2} - {{{q_{{\rm{c}}3}}} \over {2{{\pi }}}}\ln {r_0} - {{{q_{{\rm{c}}4}}} \over {2{{\pi }}}}\ln {d_3} + C,$$

(48)

$${\Phi _{{\rm{f}}4}} = - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}}\ln {d_4} - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}}\ln {d_6} - {{{q_{{\rm{c}}3}}} \over {2{{\pi }}}}\ln {d_3} - {{{q_{{\rm{c}}4}}} \over {2{{\pi }}}}\ln {r_0} + C{.}$$

(49)

Using Φ₀, Φ_f1, Φ_f2, Φ_f3, and Φ_f4 to stand for q_c1, q_c2, q_c3, and q_c4, we arrive at

$$\begin{array}{*{20}c} { - {{{q_{{\rm{c}}1}}} \over {2{{\pi }}}} = {{{{A\prime}_1}} \over F}({\Phi _{{\rm{f}}1}} - {\Phi _{{0}}}) + {{{{B\prime}_1}} \over F}({\Phi _{{\rm{f}}2}} - {\Phi _{{0}}})\quad \quad \quad }\\ { + {{{{C\prime}_1}} \over F}({\Phi _{{\rm{f}}3}} - {\Phi _{{0}}}) + {{{{D\prime}_1}} \over F}({\Phi _{{\rm{f}}4}} - {\Phi _{{0}}}),} \end{array}$$

(50)

$$\begin{array}{*{20}c} { - {{{q_{{\rm{c}}2}}} \over {2{{\pi }}}} = {{{{A\prime}_2}} \over F}({\Phi _{{\rm{f}}1}} - {\Phi _{{0}}}) + {{{{B\prime}_2}} \over F}({\Phi _{{\rm{f}}2}} - {\Phi _{{0}}})\quad \quad \quad }\\ { + {{{{C\prime}_2}} \over F}({\Phi _{{\rm{f}}3}} - {\Phi _{{0}}}) + {{{{D\prime}_2}} \over F}({\Phi _{{\rm{f}}4}} - {\Phi _{{0}}}),} \end{array}$$

(51)

$$\begin{array}{*{20}c} { - {{{q_{{\rm{c}}3}}} \over {2{{\pi }}}} = {{{{A\prime}_3}} \over F}({\Phi _{{\rm{f}}1}} - {\Phi _{{0}}}) + {{{{B\prime}_3}} \over F}({\Phi _{{\rm{f}}2}} - {\Phi _{{0}}})\quad \quad \quad }\\ { + {{{{C\prime}_3}} \over F}({\Phi _{{\rm{f}}3}} - {\Phi _{{0}}}) + {{{{D\prime}_3}} \over F}({\Phi _{{\rm{f}}4}} - {\Phi _{{0}}}),} \end{array}$$

(52)

$$\begin{array}{*{20}c} { - {{{q_{{\rm{c}}4}}} \over {2{{\pi }}}} = {{{{A\prime}_4}} \over F}({\Phi _{{\rm{f}}1}} - {\Phi _{{0}}}) + {{{{B\prime}_4}} \over F}({\Phi _{{\rm{f}}2}} - {\Phi _{{0}}})\quad \quad \quad }\\ { + {{{{C\prime}_4}} \over F}({\Phi _{{\rm{f}}3}} - {\Phi _{{0}}}) + {{{{D\prime}_4}} \over F}({\Phi _{{\rm{f}}4}} - {\Phi _{{0}}}),} \end{array}$$

(53)

where

$$\begin{array}{*{20}c} {{A_1}^{\prime} = {A_{11}}^{\prime} + {A_{12}}^{\prime} + {A_{13}}^{\prime},\;{B_1}^{\prime} = {B_{11}}^{\prime} + {B_{12}}^{\prime} + {B_{13}}^{\prime},}\\ {{C_1}^{\prime} = {C_{11}}^{\prime} + {C_{12}}^{\prime} + {C_{13}}^{\prime},\;{D_1}^{\prime} = {D_{11}}^{\prime} + {D_{12}}^{\prime} + {D_{13}}^{\prime},}\\ {{A_2}^{\prime} = {A_{21}}^{\prime} + {A_{22}}^{\prime} + {A_{23}}^{\prime},\;{B_2}^{\prime} = {B_{21}}^{\prime} + {B_{22}}^{\prime} + {B_{23}}^{\prime},}\\ {{C_2}^{\prime} = {C_{21}}^{\prime} + {C_{22}}^{\prime} + {C_{23}}^{\prime},\;{D_2}^{\prime} = {D_{21}}^{\prime} + {D_{22}}^{\prime} + {D_{23}}^{\prime},}\\ {{A_3}^{\prime} = {A_{31}}^{\prime} + {A_{32}}^{\prime} + {A_{33}}^{\prime},\;{B_3}^{\prime} = {B_{31}}^{\prime} + {B_{32}}^{\prime} + {B_{33}}^{\prime},}\\ {{C_3}^{\prime} = {C_{31}}^{\prime} + {C_{32}}^{\prime} + {C_{33}}^{\prime},\;{D_3}^{\prime} = {D_{31}}^{\prime} + {D_{32}}^{\prime} + {D_{33}}^{\prime},}\\ {{A_4}^{\prime} = {A_{41}}^{\prime} + {A_{42}}^{\prime} + {A_{43}}^{\prime},\;{B_4}^{\prime} = {B_{41}}^{\prime} + {B_{42}}^{\prime} + {B_{43}}^{\prime},}\\ {{C_4}^{\prime} = {C_{41}}^{\prime} + {C_{42}}^{\prime} + {C_{43}}^{\prime},\;{D_4}^{\prime} = {D_{41}}^{\prime} + {D_{42}}^{\prime} + {D_{43}}^{\prime},}\\ {F = {F_1} + {F_2} + {F_3} + {F_4},\;{F_1} = {F_{11}} + {F_{12}} + {F_{13}},\quad \;}\\ {{F_2} = {F_{21}} + {F_{22}} + {F_{23}},\;{F_3} = {F_{31}} + {F_{32}} + {F_{33}},\quad \quad }\\ {{F_4} = {F_{41}} + {F_{42}} + {F_{43}}{.}\quad \quad \quad \quad \quad \quad \quad \quad \;\;\;\quad } \end{array}$$

A₁₁′ A₁₂′, A₁₃′, B₁₁′, B₁₂′, B₁₃′, C₁₁′, C₁₂′, C₁₃′, D₁₁′, D₁₂′, D₁₃′, A₂₁′, A₂₂′, A₂₃′, A₂₁′, B₂₁′, B₂₂′, B₂₃′, C₂₁′, C₂₂′, C₂₃′, D₂₁′, D₂₂′, D₂₃′, A₃₁′, A₃₂′, A₃₃′, B₃₁′, B₃₂′, B₃₃′, C₃₁′, C₃₂′, C₃₃′, D₃₁′, D₃₂′, D₃₃′, A₄₁′, A₄₂′, A₄₃′, B₄₁′, B₄₂′, B₄₃′, C₄₁′, C₄₂′, C₄₃′, D₄₁′, D₄₂′, D₄₃′, F₁₁′, F₁₂′, F₁₃, F₂₁′, F₂₂′, F₂₃′ F₃₁′, F₃₂′, F₃₃′, F₄₁′, F₄₂′, and F₄₃ are shown in Appendix B.

Substituting Eqs. (45) and (50)–(53) into Eq. (44), the heat potential at arbitrary point M(x, y) in the temperature field can be expressed as

$$\begin{array}{*{20}c} {\Phi = \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;\;\;}\\ {{{ }}{{{{A\prime}_1}\ln {{{r_1}} \over {{a_1}}} + {{A\prime}_2}\ln {{{r_2}} \over \xi }} \over F}{\Phi _{{\rm {f}}1}} + {{{{A\prime}_3}\ln {{{r_3}} \over {{a_2}}} + {{A\prime}_4}\ln {{{r_4}} \over {{a_3}}}} \over F}{\Phi _{{\rm {f}}1}}}\\ { + {{{{B\prime}_1}\ln {{{r_1}} \over {{a_1}}} + {{B\prime}_2}\ln {{{r_2}} \over \xi }} \over F}{\Phi _{{\rm {f}}2}} + {{{{B\prime}_3}\ln {{{r_3}} \over {{a_2}}} + {{B\prime}_4}\ln {{{r_4}} \over {{a_3}}}} \over F}{\Phi _{{\rm {f}}2}}}\\ {{{ }} + {{{{C\prime}_1}\ln {{{r_1}} \over {{a_1}}} + {{C\prime}_2}\ln {{{r_2}} \over \xi }} \over F}{\Phi _{{\rm {f}}3}} + {{{{C\prime}_3}\ln {{{r_3}} \over {{a_2}}} + {{C\prime}_4}\ln {{{r_4}} \over {{a_3}}}} \over F}{\Phi _{{\rm {f}}3}}}\\ {{{ }} + {{{{D\prime}_1}\ln {{{r_1}} \over {{a_1}}} + {{D\prime}_2}\ln {{{r_2}} \over \xi }} \over F}{\Phi _{{\rm {f}}4}} + {{{{D\prime}_3}\ln {{{r_3}} \over {{a_2}}} + {{D\prime}_4}\ln {{{r_4}} \over {{a_3}}}} \over F}{\Phi _{{\rm {f}}4}}}\\ { - \left[ {{{({{A\prime}_1} + {{B\prime}_1} + {{C\prime}_1} + {{D\prime}_1})\ln {{{r_1}} \over {{a_1}}}} \over F}} \right.\quad \quad \quad \quad \quad \quad \quad \quad }\\ { + {{({{A\prime}_2} + {{B\prime}_2} + {{C\prime}_2} + {{D\prime}_2})\ln {{{r_2}} \over \xi }} \over F}\quad \quad \quad \quad \quad \quad \quad \quad }\\ { + {{({{A\prime}_3} + {{B\prime}_3} + {{C\prime}_3} + {{D\prime}_3})\ln {{{r_3}} \over {{a_2}}}} \over F}\quad \quad \quad \quad \quad \quad \quad \quad }\\ {\left. { + {{({{A\prime}_4} + {{B\prime}_4} + {{C\prime}_4} + {{D\prime}_4})\ln {{{r_4}} \over {{a_3}}}} \over F} - 1} \right]{\Phi _0}{.}\quad \quad \quad \quad \quad } \end{array}$$

(54)

Substituting Φ=kT, Φ₀=kT₀, Φ_fl=kT_fl, Φ_f2=kT_f2, Φ_f3=kT_f3, and Φ_f4=kT_f4 into Eq. (54), we arrive at

$$\begin{array}{*{20}c} {T = {{{{A\prime}_1}\ln {{{r_1}} \over {{a_1}}} + {{A\prime}_2}\ln {{{r_2}} \over \xi }} \over F}{T_{{\rm{f}}1}} + {{{{A\prime}_3}\ln {{{r_3}} \over {{a_2}}} + {{A\prime}_4}\ln {{{r_4}} \over {{a_3}}}} \over F}{T_{{\rm{f}}1}}\quad \quad }\\ { + {{{{B\prime}_1}\ln {{{r_1}} \over {{a_1}}} + {{B\prime}_2}\ln {{{r_2}} \over \xi }} \over F}{T_{{\rm{f}}2}} + {{{{B\prime}_3}\ln {{{r_3}} \over {{a_2}}} + {{B\prime}_4}\ln {{{r_4}} \over {{a_3}}}} \over F}{T_{{\rm{f}}2}}}\\ { + {{{{C\prime}_1}\ln {{{r_1}} \over {{a_1}}} + {{C\prime}_2}\ln {{{r_2}} \over \xi }} \over F}{T_{{\rm{f}}3}} + {{{{C\prime}_3}\ln {{{r_3}} \over {{a_2}}} + {{C\prime}_4}\ln {{{r_4}} \over {{a_3}}}} \over F}{T_{{\rm{f}}3}}}\\ { + {{{{D\prime}_1}\ln {{{r_1}} \over {{a_1}}} + {{D\prime}_2}\ln {{{r_2}} \over \xi }} \over F}{T_{{\rm{f}}4}} + {{{{D\prime}_3}\ln {{{r_3}} \over {{a_2}}} + {{D\prime}_4}\ln {{{r_4}} \over {{a_3}}}} \over F}{T_{{\rm{f}}4}}}\\ { - \left[ {{{({{A\prime}_1} + {{B\prime}_1} + {{C\prime}_1} + {{D\prime}_1})\ln {{{r_1}} \over {{a_1}}}} \over F}} \right.\quad \quad \quad \quad \quad \quad \quad \;\;}\\ { + {{({{A\prime}_2} + {{B\prime}_2} + {{C\prime}_2} + {{D\prime}_2})\ln {{{r_2}} \over \xi }} \over F}\quad \quad \quad \quad \quad \quad \quad \quad }\\ { + {{({{A\prime}_3} + {{B\prime}_3} + {{C\prime}_3} + {{D\prime}_3})\ln {{{r_3}} \over {{a_2}}}} \over F}\quad \quad \quad \quad \quad \quad \quad \;\;}\\ {\left. { + {{({{A\prime}_4} + {{B\prime}_4} + {{C\prime}_4} + {{D\prime}_4})\ln {{{r_4}} \over {{a_3}}}} \over F} - 1} \right]{T_{{0}}}{.}\quad \quad \quad \quad \;\;} \end{array}$$

(55)

Eq. (55) is the analytical solution of a steady-state temperature field of four freezing pipes in an arbitrary arrangement in an infinite plane.

Assuming the surface temperatures of the four pipes are equal (T_f1=T_f2=T_f3=T_f4), Eq. (55) can be simplified as

$$\begin{array}{*{20}c} {T = {T_0} + ({T_{\rm{f}}} - {T_0})\quad \quad \quad \quad \quad \quad \quad \quad }\\ { \times {{\ln {{{r_1}} \over {{a_1}}} + {\lambda _1}\ln {{{r_2}} \over \xi } + {\lambda _2}\ln {{{r_3}} \over {{a_2}}} + {\lambda _3}\ln {{{r_4}} \over {{a_3}}}} \over {\ln {{{d_1}} \over {{a_1}}} + {\lambda _1}\ln {{{r_0}} \over \xi } + {\lambda _2}\ln {{{d_2}} \over {{a_2}}} + {\lambda _3}\ln {{{d_6}} \over {{a_3}}}}},} \end{array}$$

(56)

where, assuming q_c2=λ₁q_c1, q_c3=λ₂q_c1, and q_c4=λ₃q_c1, λ₁, λ₂, and λ₃ can be written as

$${\lambda _{{1}}} = {{{b_1} + {b_2} + {b_3}} \over {{f_1} + {f_2} + {f_3}}},$$

(57)

$${\lambda _{{2}}} = - {{{c_1} + {c_2} + {c_3}} \over {{f_1} + {f_2} + {f_3}}},$$

(58)

$${\lambda _{{3}}} = {{{e_1} + {e_2} + {e_3}} \over {{f_1} + {f_2} + {f_3}}},$$

(59)

where

$$\begin{array}{*{20}c} {{b_1} = \ln {{{d_4}} \over {{r_0}}}\ln {{{d_5}} \over {{d_2}}}\ln {{{d_4}{r_0}} \over {{d_3}{d_5}}},\quad {b_2} = \ln {{{d_4}} \over {{d_6}}}\ln {{{d_5}} \over {{r_0}}}\ln {{{d_5}{r_0}} \over {{d_3}{d_4}}},}\\ {{b_3} = \ln {{{d_1}} \over {{r_0}}}\ln {{{d_3}} \over {{r_0}}}\ln {{{d_4}{d_5}} \over {{d_3}{r_0}}},\quad {c_1} = \ln {{{d_1}} \over {{d_2}}}\ln {{{d_4}} \over {{r_0}}}\ln {{{d_1}{d_6}} \over {{d_4}{r_0}}},}\\ {{c_2} = \ln {{{d_1}} \over {{r_0}}}\ln {{{d_4}} \over {{d_3}}}\ln {{{d_4}{d_6}} \over {{d_1}{r_0}}},\quad {c_3} = \ln {{{d_5}} \over {{r_0}}}\ln {{{d_6}} \over {{r_0}}}\ln {{{d_6}{r_0}} \over {{d_1}{d_4}}},}\\ {{e_1} = \ln {{{d_1}} \over {{r_0}}}\ln {{{d_5}} \over {{d_3}}}\ln {{{d_1}{r_0}} \over {{d_2}{d_5}}},\quad {e_2} = \ln {{{d_5}} \over {{r_0}}}\ln {{{d_1}} \over {{d_6}}}\ln {{{d_5}{r_0}} \over {{d_1}{d_2}}},}\\ {{e_3} = \ln {{{d_4}} \over {{r_0}}}\ln {{{d_2}} \over {{r_0}}}\ln {{{d_1}{d_5}} \over {{d_2}{r_0}}},\quad {f_1} = \ln {{{d_1}} \over {{r_0}}}\ln {{{d_3}} \over {{r_0}}}\ln {{{d_4}{d_5}} \over {{d_3}{r_0}}},}\\ {{f_2} = \ln {{{d_1}} \over {{d_6}}}\left({\ln {{{d_5}} \over {{d_2}}}\ln {{{d_4}} \over {{d_3}}} - \ln {{{d_4}} \over {{d_6}}}\ln {{{d_5}} \over {{r_0}}}} \right),\quad \quad \quad \quad \quad }\\ {{f_3} = \ln {{{d_1}} \over {{d_2}}}\left({\ln {{{d_4}} \over {{d_6}}}\ln {{{d_5}} \over {{d_3}}} - \ln {{{d_4}} \over {{r_0}}}\ln {{{d_5}} \over {{d_2}}}} \right){.}\quad \quad \quad \quad \quad } \end{array}$$

.

6.2 Temperature field of four freezing pipes in a linear arrangement with equal spacing

When four freezing pipes are laid in a linear arrangement with equal spacing, the centers of pipes P₂ and P₄ are moved to the x axis. Therefore, y₂ and y₄ are zero. We use d to stand for d₁, d₂, d₃, d₄, d₅, and d₆ due to d₁=d₂=d₃=d₄/4=d₅/2=d₆/2 (Fig. 17).

Fig. 17

Four freezing pipes laid in a linear arrangement with equal spacing

When four freezing pipes are arranged in this form, the coordinates of the conditional point at the boundary of the frozen soil are changed to (0, ξ), and the coordinates of arbitrary points on the surface of each pipe are changed to (−3d/2+r₀, 0), (−d/2+r₀, 0), (d/2−r₀, 0), and (3d/2−r₀, 0), respectively. Eqs. (57)–(59) can be simplified as ${\lambda _1} = {\lambda _2} = {{\ln [3{r_0}/(2d)]} \over {\ln [{r_0}/(2d)]}}$ and λ₃=1. Eq. (56) can also be simplified as

$$T = {T_0} + {{\ln {{{r_1}{r_4}} \over {{\xi ^2} + {{9{d^2}} \over 4}}} + {\lambda _1}\ln {{{r_2}{r_3}} \over {{\xi ^2} + {{{d^2}} \over 4}}}} \over {\ln {{3{r_0}d} \over {{\xi ^2} + {{9{d^2}} \over 4}}} + {\lambda _1}\ln {{2{d^2}} \over {{\xi ^2} + {{{d^2}} \over 4}}}}}({T_{\rm{f}}} - {T_0}),$$

(60)

where r₁, r₂, r₃, and r₄ can be expressed as

$$\begin{array}{*{20}c} {{r_1} = \sqrt {{{(x + 3d/2)}^2} + {y^2}}, \quad {r_2} = \sqrt {{{(x + d/2)}^2} + {y^2}},} \\ {{r_3} = \sqrt {{{(x - d/2)}^2} + {y^2}}, \quad {r_4} = \sqrt {{{(x - 3d/2)}^2} + {y^2}} {.}} \end{array}$$

Eq. (60) is the analytical solution to a steady-state temperature field of four freezing pipes in a linear arrangement with equal spacing in an infinite plane. It is also consistent with the existing analytical solution derived by Tobe and Akimoto (1979) and Kato et al. (2007), which is a particular solution of ours when T₀=0 °C.

6.3 Temperature field of four freezing pipes in a rhombus arrangement

When four freezing pipes are laid in a rhombus arrangement, we see that d₁=d₂=d₃=d₄ (Fig. 18), and we use d to stand for d₁, d₂, d₃, and d₄. According to the geometrical properties of a rhombus, d₅ and d₆ are satisfied by the expression 4d²=d ²₅ +d ²₆ .

Fig. 18

Four freezing pipes laid in a rhombus arrangement

With four freezing pipes arranged in this form, the coordinates of the conditional point at the boundary of the frozen soil are changed to (0, d₆/2+ξ), and the coordinates of arbitrary points on the surface of each pipe are changed to (−d₅/2+r₀, 0), (0, d₆/2−r₀), (d₅/2−r₀, 0), and (0, −d₆/2+r₀), respectively. Eqs. (49)–(51) can be simplified as ${\lambda _1} = {\lambda _3} = {{\ln [{d^2}/({d_5}{r_0})]} \over {\ln [{d^2}/({d_6}{r_0})]}}$ and λ₂=1. We arrive at

$$\begin{array}{*{20}c} {T = {T_0}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad }\\ { + {{\ln {{{r_1}{r_3}} \over {{{(\xi + {d_6}/2)}^2} + d_5^2/4}} + {\lambda _1}\ln {{{r_2}{r_4}} \over {\xi (\xi + {d_{{6}}})}}} \over {\ln {{{d_5}{r_0}} \over {{{(\xi + {d_6}/2)}^2} + d_5^2/4}} + {\lambda _1}\ln {{{d^2}} \over {\xi (\xi + {d_6})}}}}({T_{\rm{f}}} - {T_0}),} \end{array}$$

(61)

where r₁, r₂, r₃, and r₄ can be expressed as

$$\begin{array}{*{20}c} {{r_1} = \sqrt {{{(x + {d_5}/2)}^2} + {y^2}}, \quad {r_2} = \sqrt {{x^2} + {{(y - {d_6}/2)}^2}},} \\ {{r_3} = \sqrt {{{(x - {d_5}/2)}^2} + {y^2}}, \quad {r_4} = \sqrt {{x^2} + {{(y + {d_6}/2)}^2}} {.}} \end{array}$$

Eq. (61) is the analytical solution to a steady-state temperature field of four freezing pipes in a rhombus arrangement in an infinite plane.

A visualized figure of the temperature field drawn according to the results of the analytical solution, Eq. (61), is shown in Fig. 19. The steady-state numerical calculation method was again applied to verify the analytical solution, under the following particular conditions: d₅=1.0 m, d₆=0.8 m. The other parameters were the same as in Sections 5.3 to 5.5. According to the relationship of heat flow among four freezing pipes, the temperature fields of main sections 1 and 3, and of intersections 1 and 2 in Fig. 18 are the same. An isothermal diagram of the steady-state numerical solution when ξ=1.0 m is shown in Fig. 20. A comparison of results from the analytical formula and numerical simulation of the main section 1 is shown in Fig. 21. A comparison of results from the analytical formula and numerical simulation of intersection 1 is shown in Fig. 22.

Fig. 19

Temperature field calculated by the analytical formula for four freezing pipes in a rhombus arrangement

Fig. 20

Isothermal diagram calculated by numerical simulation for four freezing pipes in a rhombus arrangement

Fig. 21

Comparison of results from applying the analytical formula and numerical simulation to the main section 1 in Fig. 18

Fig. 22

Comparison of results from applying the analytical formula and numerical simulation to intersection 1 in Fig. 18

From Figs. 21 and 22, we find that the results calculated by the two methods coincide, which shows the analytical solution, Eq. (61), is precise enough. Comparisons of the two methods applied to the other main sections and intersections in Fig. 18 are not covered here.

6.4 Temperature field of four freezing pipes in a rectangle arrangement

When four freezing pipes are laid in a rectangle arrangement, for convenience of derivation, we put one pipe into each of the four quadrants (Fig. 23). According to the geometrical properties of rectangles, we see that d₁=d₃, d₂=d₄, and d₅=d₆, and d₁, d₂, and d₅ are satisfied by the expression d ²₅ =d ²₁ +d ²₂ . In addition, we use d₁ to stand for d₃, d₂ to stand for d₄, and d₅ to stand for d₆.

Fig. 23

Four freezing pipes laid in a rectangle arrangement

With four freezing pipes arranged in this form, the coordinates of the conditional point at the boundary of the frozen soil are changed to (0, d₂/2+ξ), and the coordinates of arbitrary points on the surface of each pipe are changed to (−d₁/2+r₀, d₂/2), (d₁/2−r₀, d₂/2), (d₁/2−r₀, −d₂/2), and (−d₁/2+r₀, −d₂/2), respectively. Eqs. (49)–(51) can be simplified as λ₁=λ₂=λ₃=1. We arrive at

$$\begin{array}{*{20}c} {T = {T_0}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;}\\ { + {{\ln {{{r_1} \cdot {r_2} \cdot {r_3} \cdot {r_4}} \over {\left[ {{{(\xi + {d_2})}^2} + d_1^2/4} \right] \cdot ({\xi ^{{2}}} + d_1^2/4)}}} \over {\ln {{{d_1} \cdot {d_2} \cdot \sqrt {d_1^2 + d_2^2} \cdot {r_0}} \over {\left[ {{{(\xi + {d_2})}^2} + d_1^2/4} \right] \cdot ({\xi ^{{2}}} + d_1^2/4)}}}}({T_{\rm {f}}} - {T_0}),} \end{array}$$

(62)

where r₁, r₂, r₃, and r₄ can be expressed as

$$\begin{array}{*{20}c} {{r_1} = \sqrt {{{(x + {d_{{1}}}/2)}^2} + {{(y - {d_2}/2)}^2}},} \\ {{r_2} = \sqrt {{{(x - {d_{{1}}}/2)}^2} + {{(y - {d_{{2}}}/2)}^2}},} \\ {{r_3} = \sqrt {{{(x - {d_{{1}}}/2)}^2} + {{(y + {d_{{2}}}/2)}^2}},} \\ {{r_4} = \sqrt {{{(x + {d_{{1}}}/2)}^2} + {{(y + {d_{{2}}}/2)}^2}} {.}} \end{array}$$

Eq. (62) is the analytical solution of a steady-state temperature field of four freezing pipes in a rectangle arrangement in an infinite plane.

A visualized figure of the temperature field drawn according to the results from the analytical solution, Eq. (62), is shown in Fig. 24. The steady-state numerical calculation method was again applied to verify the analytical solution, under the following particular conditions: d₁=0.6 m, d₂=0.8 m. The other parameters were the same as in Sections 5.3–5.5 and Section 6.3. According to the relationship of heat flow among four freezing pipes, the temperature fields of main sections 1 and 2 in Fig. 23 are the same. An isothermal diagram of the steady-state numerical solution when ξ=1.0 m is shown in Fig. 25. A comparison of results from the analytical formula and numerical simulation of the main section 1 is shown in Fig. 26. A comparison of results from the analytical formula and numerical simulation of the intersection is shown in Fig. 27.

Fig. 24

Temperature field calculated by the analytical formula for four freezing pipes laid in a rectangle arrangement

Fig. 25

Isothermal diagram calculated by numerical simulation for four freezing pipes laid in a rectangle arrangement

Fig. 26

Comparison of results from applying the analytical formula and numerical simulation to the main section 1 in Fig. 23

From Figs. 26 and 27, we find that the results calculated by the two methods coincide, which shows the analytical solution, Eq. (62), is precise enough.

Fig. 27

Comparison of results from applying the analytical formula and numerical simulation to the intersection in Fig. 23

7 Conclusions

A method for deriving an analytical solution to the steady-state temperature field frozen by multiple freezing pipes has been found. The method is based on potential superposition theory. Using this method, solutions to the steady-state temperature field produced by one-, two-, three-, and four-piped freezing are presented. The results calculated by the analytical solutions are precise enough in comparison with those obtained by numerical simulation. Concerning the method, some conclusions can be drawn as follows:

• 1. The freezing pipe can be simplified to a point source, i.e., a cold source or heat sink, located at the center of the pipe. The potential field formed by the point source is governed by the Laplace equation, whose solution is the potential function.

• 2. In the case of multiple-pipe freezing, potential superposition theory can be applied to derive an analytical solution to the steady-state temperature field. The essence of the method is that the heat potential at an arbitrary point is equal to superposition of the heat potentials which are caused by the cold sources of each freezing pipe separately at this point. The heat flux to each pipe depends on the arrangement of all the pipes and the final solution is determined according the boundary conditions.

• 3. Simplifying a freezing pipe to a point source leads to certain errors in the solutions. However, the errors occur only within tiny areas around the pipes and are small enough to meet engineering accuracy requirements.

Using the method developed in this paper, we obtained analytical solutions to steady-state temperature fields of frequently used layouts of freezing pipes, such as two, three or four pipes arranged at random. In particular, solutions were derived for some specific arrangements of freezing pipes that are more commonly applied in AGF projects, such as three pipes in a linear arrangement with equal or unequal spacing, right and isosceles triangle arrangements, four pipes in a linear arrangement with equal spacing, and rhombus and rectangle arrangements. Theoretically, this method can serve as a universal method to solve the steady-state temperature field for any complicated layouts of freezing pipes.

Appendices

Appendix A

$$\begin{array}{*{20}c} {{A_1} = \ln {{{r_0}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}\ln {{{r_0}} \over \xi } - \ln {{{d_2}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_2}} \over \xi },}\\ {{B_1} = \ln {{{d_3}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_2}} \over \xi } - \ln {{{r_0}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_1}} \over \xi },}\\ {{C_1} = \ln {{{d_2}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_1}} \over \xi } - \ln {{{d_3}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}\ln {{{r_0}} \over \xi },}\\ {{A_2} = \ln {{{d_2}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_3}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\quad \quad \quad \quad \quad }\\ { - \ln {{{d_1}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{r_0}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }},\quad \quad }\\ {{B_2} = \ln {{{r_0}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{r_0}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}\quad \quad \quad \quad \quad }\\ { - \ln {{{d_3}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_3}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }},\quad \quad }\\ {{C_2} = \ln {{{d_1}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_3}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}\quad \quad \quad \quad \quad }\\ { - \ln {{{r_0}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_2}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }},\quad \quad }\\ {{A_3} = \ln {{{d_1}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_2}} \over \xi } - \ln {{{d_3}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{r_0}} \over \xi },}\\ {{B_3} = \ln {{{d_3}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_1}} \over \xi } - \ln {{{r_0}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_2}} \over \xi },}\\ {{C_3} = \ln {{{r_0}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{r_0}} \over \xi } - \ln {{{d_1}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_1}} \over \xi }{.}}\\ {D = {D_{{1}}} + {D_{{2}}} + {D_{{3}}},\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;}\\ {{D_1} = \left[ {\ln {{{d_2}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_3}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\quad \quad \quad \quad } \right.}\\ { - \left. {\ln {{{d_1}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{r_0}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}} \right] \ln {{{d_1}} \over \xi },}\\ {{D_2} = \left[ {\ln {{{r_0}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{r_0}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}\quad \quad \quad \quad \;} \right.}\\ {\left. { - \ln {{{d_3}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_3}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}} \right] \ln {{{r_0}} \over \xi },} \\ {{D_3} = \left[ {\ln {{{d_1}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_3}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}} \right.\quad \quad \quad \quad }\\ {\left. { - \ln {{{r_0}} \over {\sqrt {x_1^2 + {{(\xi + {y_2})}^2}} }}\ln {{{d_2}} \over {\sqrt {x_3^2 + {{(\xi + {y_2})}^2}} }}} \right] \ln {{{r_2}} \over \xi }{.}}\end{array}$$

Appendix B

$$\begin{array}{*{20}c} {{{A\prime}_{11}} = \ln {{{d_1}} \over {{a_1}}}\left({\ln {{{r_0}} \over {{a_2}}}\ln {{{d_6}} \over \xi } - \ln {{{d_2}} \over \xi }\ln {{{d_3}} \over {{a_2}}}} \right),}\\ {{{A\prime}_{12}} = \ln {{{r_0}} \over \xi }\left({\ln {{{d_5}} \over {{a_1}}}\ln {{{d_3}} \over {{a_2}}} - \ln {{{r_0}} \over {{a_2}}}\ln {{{d_4}} \over {{a_1}}}} \right),}\\ {{{A\prime}_{13}} = \ln {{{d_2}} \over {{a_2}}}\left({\ln {{{d_2}} \over \xi }\ln {{{d_4}} \over {{a_1}}} - \ln {{{d_5}} \over {{a_1}}}\ln {{{d_6}} \over \xi }} \right),}\\ {{{B\prime}_{11}} = \ln {{{r_0}} \over {{a_1}}}\left({\ln {{{d_2}} \over \xi } \ln {{{d_3}} \over {{a_2}}} - \ln {{{r_0}} \over {{a_2}}}\ln {{{d_6}} \over \xi }} \right),}\\ {{{B\prime}_{12}} = \ln {{{d_1}} \over \xi }\left({\ln {{{r_0}} \over {{a_2}}}\ln {{{d_4}} \over {{a_1}}} - \ln {{{d_5}} \over {{a_1}}}\ln {{{d_3}} \over {{a_2}}}} \right),} \\ {{{B\prime}_{13}} = \ln {{ {d_5}} \over {{a_2}}}\left({\ln {{{d_5}} \over {{a_1}}}\ln {{{d_6}} \over \xi } - \ln {{{d_2}} \over \xi }\ln {{{d_4}} \over {{a_1}}}} \right),}\\ {{{C\prime}_{11}} = \ln {{{r_0}} \over {{a_1}}}\left({\ln {{{d_2}} \over {{a_2}}} \ln {{{d_6}} \over \xi } - \ln {{{r_0}} \over \xi }\ln {{{d_3}} \over {{a_2}}}} \right),}\\ {{{C\prime}_{12}} = \ln {{{d_1}} \over \xi }\left({\ln {{{d_1}} \over {{a_1}}}\ln {{{d_3}} \over {{a_2}}} - \ln {{{d_2}} \over {{a_2}}}\ln {{{d_4}} \over {{a_1}}}} \right),}\\ {{{C\prime}_{13}} = \ln {{ {d_5}} \over {{a_2}}}\left({\ln {{{r_0}} \over \xi }\ln {{{d_4}} \over {{a_1}}} - \ln {{{d_1}} \over {{a_1}}}\ln {{{d_6}} \over \xi }} \right),}\\ {{{D\prime}_{11}} = \ln {{{r_0}} \over {{a_1}}}\left({\ln {{{r_0}} \over \xi }\ln {{{r_0}} \over {{a_2}}} - \ln {{{d_2}} \over {{a_2}}}\ln {{{d_2}} \over \xi }} \right),}\\ {{{D\prime}_{12}} = \ln {{{d_1}} \over \xi }\left({\ln {{{d_2}} \over {{a_2}}}\ln {{{d_5}} \over {{a_1}}} - \ln {{{d_1}} \over {{a_1}}}\ln {{{r_0}} \over {{a_2}}}} \right),}\\ {{{D\prime}_{13}} = \ln {{ {d_5}} \over {{a_2}}}\left({\ln {{{d_1}} \over {{a_1}}}\ln {{{d_2}} \over \xi } - \ln {{{r_0}} \over \xi }\ln {{{d_5}} \over {{a_1}}}} \right),}\\ {{{A\prime}_{21}} = \ln {{{r_0}} \over \xi }\left({\ln {{{r_0}} \over {{a_2}}}\ln {{{r_0}} \over {{a_3}}} - \ln {{{d_3}} \over {{a_3}}}\ln {{{d_3}} \over {{a_2}}}} \right),}\\ {{{A\prime}_{22}} = \ln {{{d_2}} \over {{a_2}}}\left({\ln {{{d_3}} \over {{a_3}}}\ln {{{d_6}} \over \xi } - \ln {{{d_2}} \over \xi } \ln {{{r_0}} \over {{a_3}}}} \right),} \\ {{{A\prime}_{23}} = \ln {{{d_6}} \over {{a_3}}}\left({\ln {{{d_2}} \over \xi }\ln {{{d_3}} \over {{a_2}}} - \ln {{{r_0}} \over {{a_2}}}\ln {{{d_6}} \over \xi }} \right),}\\ {{{B\prime}_{21}} = \ln {{{d_1}} \over \xi }\left({\ln {{{d_3}} \over {{a_3}}}\ln {{{d_3}} \over {{a_2}}} - \ln {{{r_0}} \over {{a_2}}}\ln {{{r_0}} \over {{a_3}}}} \right),}\\ {{{B\prime}_{22}} = \ln {{ {d_5}} \over {{a_2}}}\left({\ln {{{d_2}} \over \xi }\ln {{{r_0}} \over {{a_3}}} - \ln {{{d_3}} \over {{a_3}}}\ln {{{d_6}} \over \xi }} \right),}\\ {{{B\prime}_{23}} = \ln {{{d_4}} \over {{a_3}}}\left({\ln {{{r_0}} \over {{a_2}}}\ln {{{d_6}} \over \xi } - \ln {{{d_2}} \over \xi }\ln {{{d_3}} \over {{a_2}}}} \right),}\\ {{{C\prime}_{21}} = \ln {{{d_1}} \over \xi }\left({\ln {{{d_2}} \over {{a_2}}} \ln {{{r_0}} \over {{a_3}}} - \ln {{{d_6}} \over {{a_3}}}\ln {{{d_3}} \over {{a_2}}}} \right),}\\ {{{C\prime}_{22}} = \ln {{ {d_5}} \over {{a_2}}}\left({\ln {{{d_6}} \over {{a_3}}}\ln {{{d_6}} \over \xi } - \ln {{{r_0}} \over \xi }\ln {{{r_0}} \over {{a_3}}}} \right),}\\ {{{C\prime}_{23}} = \ln {{{d_4}} \over {{a_3}}}\left({\ln {{{r_0}} \over \xi }\ln {{{d_3}} \over {{a_2}}} - \ln {{{d_2}} \over {{a_2}}}\ln {{{d_6}} \over \xi }} \right),}\\ {{{D\prime}_{21}} = \ln {{{d_1}} \over \xi }\left({\ln {{{d_6}} \over {{a_3}}}\ln {{{r_0}} \over {{a_2}}} - \ln {{{d_2}} \over {{a_2}}}\ln {{{d_3}} \over {{a_3}}}} \right),}\\ {{{D\prime}_{22}} = \ln {{ {d_5}} \over {{a_2}}}\left({\ln {{{r_0}} \over \xi }\ln {{{d_3}} \over {{a_3}}} - \ln {{{d_6}} \over {{a_3}}}\ln {{{d_2}} \over \xi }} \right),}\\ {{{D\prime}_{23}} = \ln {{{d_4}} \over {{a_3}}}\left({\ln {{{d_2}} \over {{a_2}}}\ln {{{d_2}} \over \xi } - \ln {{{r_0}} \over \xi }\ln {{{r_0}} \over {{a_2}}}} \right),}\\ {{{A\prime}_{31}} = \ln {{{d_1}} \over {{a_1}}}\left({\ln {{{d_3}} \over {{a_3}}}\ln {{{d_3}} \over {{a_2}}} - \ln {{{r_0}} \over {{a_2}}}\ln {{{r_0}} \over {{a_3}}}} \right),}\\ {{{A\prime}_{32}} = \ln {{{d_2}} \over {{a_2}}}\left({\ln {{{d_5}} \over {{a_1}}}\ln {{{r_0}} \over {{a_3}}} - \ln {{{d_3}} \over {{a_3}}}\ln {{{d_4}} \over {{a_1}}}} \right),}\\ {{{A\prime}_{33}} = \ln {{{d_6}} \over {{a_3}}}\left({\ln {{{r_0}} \over {{a_2}}}\ln {{{d_4}} \over {{a_1}}} - \ln {{{d_5}} \over {{a_1}}}\ln {{{d_3}} \over {{a_2}}}} \right),}\\ {{{B\prime}_{31}} = \ln {{{r_0}} \over {{a_1}}}\left({\ln {{{r_0}} \over {{a_2}}}\ln {{{r_0}} \over {{a_3}}} - \ln {{{d_3}} \over {{a_3}}}\ln {{{d_3}} \over {{a_2}}}} \right),}\\ {{{B\prime}_{32}} = \ln {{ {d_5}} \over {{a_2}}}\left({\ln {{{d_3}} \over {{a_3}}}\ln {{{d_4}} \over {{a_1}}} - \ln {{{d_5}} \over {{a_1}}}\ln {{{r_0}} \over {{a_3}}}} \right),}\\ {{{B\prime}_{33}} = \ln {{{d_4}} \over {{a_3}}}\left({\ln {{{d_5}} \over {{a_1}}}\ln {{{d_3}} \over {{a_2}}} - \ln {{{r_0}} \over {{a_2}}} \ln {{{d_4}} \over {{a_1}}}} \right),}\\ {{{C\prime}_{31}} = \ln {{{r_0}} \over {{a_1}}}\left({\ln {{{d_6}} \over {{a_3}}}\ln {{{d_3}} \over {{a_2}}} - \ln {{{d_2}} \over {{a_2}}}\ln {{{r_0}} \over {{a_3}}}} \right),}\\ {{{C\prime}_{32}} = \ln {{ {d_5}} \over {{a_2}}}\left({\ln {{{d_1}} \over {{a_1}}}\ln {{{r_0}} \over {{a_3}}} - \ln {{{d_6}} \over {{a_3}}}\ln {{{d_4}} \over {{a_1}}}} \right),} \\ {{{C\prime}_{33}} = \ln {{{d_4}} \over {{a_3}}}\left({\ln {{{d_2}} \over {{a_2}}}\ln {{{d_4}} \over {{a_1}}} - \ln {{{d_1}} \over {{a_1}}}\ln {{{d_3}} \over {{a_2}}}} \right),\quad \;\;}\\ {{{D\prime}_{31}} = \ln {{{r_0}} \over {{a_1}}}\left({\ln {{{d_2}} \over {{a_2}}}\ln {{{d_3}} \over {{a_3}}} - \ln {{{d_6}} \over {{a_3}}}\ln {{{r_0}} \over {{a_2}}}} \right),\quad \;\;}\\ {{{D\prime}_{32}} = \ln {{ {d_5}} \over {{a_2}}}\left({\ln {{{d_6}} \over {{a_3}}}\ln {{{d_5}} \over {{a_1}}} - \ln {{{d_1}} \over {{a_1}}}\ln {{{d_3}} \over {{a_3}}}} \right),\quad \;\;}\\ {{{D\prime}_{33}} = \ln {{{d_4}} \over {{a_3}}}\left({\ln {{{d_1}} \over {{a_1}}}\ln {{{r_0}} \over {{a_2}}} - \ln {{{d_2}} \over {{a_2}}}\ln {{{d_5}} \over {{a_1}}}} \right),\quad \;\;}\\ {{{A\prime}_{41}} = \ln {{{d_1}} \over {{a_1}}}\left({\ln {{{d_2}} \over \xi }\ln {{{r_0}} \over {{a_3}}} - \ln {{{d_3}} \over {{a_3}}}\ln {{{d_6}} \over \xi }} \right),\quad \;\;}\\ {{{A\prime}_{42}} = \ln {{{r_0}} \over \xi }\left({\ln {{{d_3}} \over {{a_3}}}\ln {{{d_4}} \over {{a_1}}} - \ln {{{d_5}} \over {{a_1}}}\ln {{{r_0}} \over {{a_3}}}} \right),\quad \;\;}\\ {{{A\prime}_{43}} = \ln {{{d_6}} \over {{a_3}}}\left({\ln {{{d_5}} \over {{a_1}}}\ln {{{d_6}} \over \xi } - \ln {{{d_2}} \over \xi }\ln {{{d_4}} \over {{a_1}}}} \right),\quad \;\;}\\ {{{B\prime}_{41}} = \ln {{{r_0}} \over {{a_1}}}\left({\ln {{{d_3}} \over {{a_3}}}\ln {{{d_6}} \over \xi } - \ln {{{d_2}} \over \xi }\ln {{{r_0}} \over {{a_3}}}} \right),\quad \;\;}\\ {{{B\prime}_{42}} = \ln {{{d_1}} \over \xi }\left({\ln {{{d_5}} \over {{a_1}}}\ln {{{r_0}} \over {{a_3}}} - \ln {{{d_3}} \over {{a_3}}}\ln {{{d_4}} \over {{a_1}}}} \right),\quad \;\;}\\ {{{B\prime}_{43}} = \ln {{{d_4}} \over {{a_3}}}\left({\ln {{{d_2}} \over \xi }\ln {{{d_4}} \over {{a_1}}} - \ln {{{d_5}} \over {{a_1}}}\ln {{{d_6}} \over \xi }} \right),\quad \;\;}\\ {{{C\prime}_{41}} = \ln {{{r_0}} \over {{a_1}}}\left({\ln {{{r_0}} \over \xi }\ln {{{r_0}} \over {{a_3}}} - \ln {{{d_6}} \over {{a_3}}}\ln {{{d_6}} \over \xi }} \right),\quad \;\;}\\ {{{C\prime}_{42}} = \ln {{{d_1}} \over \xi }\left({\ln {{{d_6}} \over {{a_3}}}\ln {{{d_4}} \over {{a_1}}} - \ln {{{d_1}} \over {{a_1}}}\ln {{{r_0}} \over {{a_3}}}} \right),\quad \;\;}\\ {{{C\prime}_{43}} = \ln {{{d_4}} \over {{a_3}}}\left({\ln {{{d_1}} \over {{a_1}}}\ln {{{d_6}} \over \xi } - \ln {{{r_0}} \over \xi }\ln {{{d_4}} \over {{a_1}}}} \right),\quad \;\;}\\ {{{D\prime}_{41}} = \ln {{{r_0}} \over {{a_1}}}\left({\ln {{{d_6}} \over {{a_3}}}\ln {{{d_2}} \over \xi } - \ln {{{r_0}} \over \xi }\ln {{{d_3}} \over {{a_3}}}} \right),\quad \;\;}\\ {{{D\prime}_{42}} = \ln {{{d_1}} \over \xi }\left({\ln {{{d_1}} \over {{a_1}}}\ln {{{d_3}} \over {{a_3}}} - \ln {{{d_6}} \over {{a_3}}}\ln {{{d_5}} \over {{a_1}}}} \right),\quad \;\;}\\ {{{D\prime}_{43}} = \ln {{{d_4}} \over {{a_3}}}\left({\ln {{{r_0}} \over \xi }\ln {{{d_5}} \over {{a_1}}} - \ln {{{d_1}} \over {{a_1}}}\ln {{{d_2}} \over \xi }} \right), \quad \quad }\\ {{F_{11}} = \ln {{{d_4}} \over {{a_3}}}\ln {{{d_1}} \over {{a_1}}}\left({\ln {{{r_0}} \over {{a_2}}}\ln {{{d_6}} \over \xi } - \ln {{{d_2}} \over \xi }\ln {{{d_3}} \over {{a_2}}}} \right),}\\ {{F_{12}} = \ln {{{d_4}} \over {{a_3}}}\ln {{{r_0}} \over \xi } \left({\ln {{{d_5}} \over {{a_1}}}\ln {{{d_3}} \over {{a_2}}} - \ln {{{r_0}} \over {{a_2}}}\ln {{{d_4}} \over {{a_1}}}} \right),} \\ {{F_{13}} = \ln {{{d_4}} \over {{a_3}}}\ln {{{d_2}} \over {{a_2}}}\left({\ln {{{d_2}} \over \xi }\ln {{{d_4}} \over {{a_1}}} - \ln {{{d_5}} \over {{a_1}}}\ln {{{d_6}} \over \xi }} \right),}\\ {{F_{21}} = \ln {{{d_6}} \over {{a_3}}}\ln {{{r_0}} \over {{a_1}}} \left({\ln {{{d_2}} \over \xi }\ln {{{d_3}} \over {{a_2}}} - \ln {{{r_0}} \over {{a_2}}}\ln {{{d_6}} \over \xi }} \right),}\\ {{F_{22}} = \ln {{{d_6}} \over {{a_3}}}\ln {{{d_1}} \over \xi }\left({\ln {{{r_0}} \over {{a_2}}}\ln {{{d_4}} \over {{a_1}}} - \ln {{{d_5}} \over {{a_1}}}\ln {{{d_3}} \over {{a_2}}}} \right),}\\ {{F_{23}} = \ln {{{d_6}} \over {{a_3}}}\ln {{{d_5}} \over {{a_2}}}\left({\ln {{{d_5}} \over {{a_1}}}\ln {{{d_6}} \over \xi } - \ln {{{d_2}} \over \xi }\ln {{{d_4}} \over {{a_1}}}} \right),}\\ {{F_{31}} = \ln {{{d_3}} \over {{a_3}}}\ln {{{r_0}} \over {{a_1}}}\left({\ln {{{d_2}} \over {{a_2}}}\ln {{{d_6}} \over \xi } - \ln {{{r_0}} \over \xi }\ln {{{d_3}} \over {{a_2}}}} \right),}\\ {{F_{32}} = \ln {{{d_3}} \over {{a_3}}}\ln {{{d_1}} \over \xi }\left({\ln {{{d_1}} \over {{a_1}}}\ln {{{d_3}} \over {{a_2}}} - \ln {{{d_2}} \over {{a_2}}}\ln {{{d_4}} \over {{a_1}}}} \right),}\\ {{F_{33}} = \ln {{{d_3}} \over {{a_3}}}\ln {{{d_5}} \over {{a_2}}}\left({\ln {{{r_0}} \over \xi }\ln {{{d_4}} \over {{a_1}}} - \ln {{{d_1}} \over {{a_1}}}\ln {{{d_6}} \over \xi }} \right),}\\ {{F_{41}} = \ln {{{r_0}} \over {{a_3}}}\ln {{{r_0}} \over {{a_1}}} \left({\ln {{{r_0}} \over \xi }\ln {{{r_0}} \over {{a_2}}} - \ln {{{d_2}} \over {{a_2}}}\ln {{{d_2}} \over \xi }} \right),}\\ {{F_{42}} = \ln {{{r_0}} \over {{a_3}}}\ln {{{d_1}} \over \xi }\left({\ln {{{d_2}} \over {{a_2}}}\ln {{{d_5}} \over {{a_1}}} - \ln {{{d_1}} \over {{a_1}}}\ln {{{r_0}} \over {{a_2}}}} \right),}\\ {{F_{43}} = \ln {{{r_0}} \over {{a_3}}}\ln {{{d_5}} \over {{a_2}}} \left({\ln {{{d_1}} \over {{a_1}}}\ln {{{d_2}} \over \xi } - \ln {{{r_0}} \over \xi }\ln {{{d_5}} \over {{a_1}}}} \right){.}}\end{array}$$