## 2 Theory of Slug Test on Determining Permeability Coefficient of Anisotropic Fractured Rock Mass
### The Calculation of the Fractured Rock Mass Permeability Coefficient Tensor
3 in the second fracture group are basically the same, and the fracture width, dip angle and permeability coefficient of fracture No. 7 and No. 8 in the fourth fracture group are basically the same, \(b_{2}=b_{3}\), \(\varphi_{2}=\varphi_{3}\), \(k_{e2}=k_{e3}\), \(b_{7}=b_{8}\), \(\varphi_{7}=\varphi_{8}\), \(k_{e7}=k_{e8}\). So, according to the equivalent permeability coefficient of fractures in the corresponding test section determined by the slug test, the permeability coefficient of each fracture is inversely calculated, and then the equivalent permeability coefficient of all four fracture groups in the whole borehole test section is respectively determined. Finally, the fractured rock mass permeability coefficient tensor in the whole borehole test section is calculated. Because, according to the Equation (15),  
\[k_{12}=\frac{b_{1}}{L_{2}\cdot\cos\varphi_{1}}\cdot k_{e1}+\frac{b_{2}}{L_{2} \cdot\cos\varphi_{2}}\cdot k_{e2} \tag{19}\]  
\[k_{3}=\frac{b_{3}}{L_{3}\cdot\cos\varphi_{3}}\cdot k_{e3}=\frac{b_{2}}{L_{3} \cdot\cos\varphi_{2}}\cdot k_{e2} \tag{20}\]  
\[k_{e2}=\frac{k_{3}\cdot L_{3}\cdot\cos\varphi_{2}}{b_{2}} \tag{21}\]  
If Equation (21) is placed into Equation (19), then  
\[\frac{b_{1}}{L_{2}\cdot\cos\varphi_{1}}\cdot k_{e1}=k_{12}-\frac{b_{2}}{L_{2} \cdot\cos\varphi_{2}}\cdot\frac{k_{3}\cdot L_{3}\cdot\cos\varphi_{2}}{b_{2}} =k_{12}-\frac{L_{3}}{L_{2}}\cdot k_{3} \tag{22}\]  
So,  
\[k_{e1}=\left(k_{12}-\frac{L_{3}}{L_{2}}\cdot k_{3}\right)\cdot\frac{L_{2} \cdot\cos\varphi_{1}}{b_{1}} \tag{23}\]  
\[k_{e2}=k_{e3}=\frac{k_{3}\cdot L_{3}\cdot\cos\varphi_{3}}{b_{3}} \tag{24}\]  
\[k_{e4}=\frac{k_{4}\cdot L_{4}\cdot\cos\varphi_{4}}{b_{4}} \tag{25}\]  
\[k_{e5}=\frac{k_{5}\cdot L_{5}\cdot\cos\varphi_{5}}{b_{5}} \tag{26}\] \begin{table}
\begin{tabular}{c c c c} \hline \hline
**Fracture No. 1** & **Fracture Width** & **Dip Angle Symbol** &
\begin{tabular}{c} **Fracture** \\ **Permeability** \\ **Coefficient Symbol** \\ \end{tabular} \\ \hline No. 1 & \(b_{1}\) & \(\varphi_{1}\) & \(k_{e1}\) \\ \hline No. 2 & \(b_{2}\) & \(\varphi_{2}\) & \(k_{e2}\) \\ \hline No. 3 & \(b_{3}\) & \(\varphi_{3}\) & \(k_{e3}\) \\ \hline No. 4 & \(b_{4}\) & \(\varphi_{4}\) & \(k_{e4}\) \\ \hline No. 5 & \(b_{5}\) & \(\varphi_{5}\) & \(k_{e5}\) \\ \hline No. 6 & \(b_{6}\) & \(\varphi_{6}\) & \(k_{e6}\) \\ \hline No. 7 & \(b_{7}\) & \(\varphi_{7}\) & \(k_{e7}\) \\ \hline No. 8 & \(b_{8}\) & \(\varphi_{8}\) & \(k_{e8}\) \\ \hline No. 9 & \(b_{9}\) & \(\varphi_{9}\) & \(k_{e9}\) \\ \hline No. 10 & \(b_{10}\) & \(\varphi_{10}\) & \(k_{e10}\) \\ \hline \hline \end{tabular}
\end{table} 
Table 8: Symbol of fracture information. \[k_{e6}=\frac{k_{6}\cdot L_{6}\cdot\cos\varphi_{6}}{b_{6}} \tag{27}\]  
\[k_{78}=\frac{b_{7}}{L_{7}\cdot\cos\varphi_{7}}\cdot k_{e7}+\frac{b_{8}}{L_{7} \cdot\cos\varphi_{8}}\cdot k_{e8}=\frac{b_{7}}{L_{7}\cdot\cos\varphi_{7}}\cdot k _{e7}+\frac{b_{7}}{L_{7}\cdot\cos\varphi_{7}}\cdot k_{e7} \tag{28}\]  
\[ \begin{array}{c}k_{e7}=k_{e8}=\frac{L_{7}\cdot\cos\varphi_{7}}{2b_{7}}k_{78} \\ k_{e9}=\frac{L_{8}\cdot\cos\varphi_{9}}{b_{9}}k_{9}\\ k_{e10}=\frac{L_{9}\cdot\cos\varphi_{10}}{b_{10}}k_{10}\end{array}  \tag{29}\]  
The depth of the whole borehole test section is \(L=9.62\) m and one can calculate the equivalent permeability coefficient of every fracture group in the whole borehole test section. The equivalent permeability coefficient of the first fracture group in the whole borehole test section is as follows:  
\[k_{e1}=\frac{b_{1}}{L\cdot\cos\varphi_{1}}\cdot k_{e1} \tag{32}\]  
If Equation (23) is placed into Equation (32), then  
\[ \begin{array}{c}k_{e1}=\frac{b_{1}}{L\cdot\cos\varphi_{1}}\cdot\Big{(}k_{12 }-\frac{L_{2}}{L_{2}}\cdot k_{3}\Big{)}\cdot\frac{L_{2}\cdot\cos\varphi_{1}} {b_{1}}=\frac{L_{2}}{L}\Big{(}k_{12}-\frac{L_{2}}{L_{2}}\cdot k_{3}\Big{)}\\ =\frac{0.39}{9.62}\Big{(}1.06\times 10^{-4}-\frac{0.41}{0.39}\times 10^{-4} \Big{)}=3.53\times 10^{-8}\text{cm/s}\end{array}