Research on the occurrence mechanism of mine earthquake in Longwanggou Coal Mine and its damage assessment to surface buildings
Hard roof fracture mine earthquake model
Mine earthquake originates from the fracturing of thick, competent rock layers, whose failure mechanisms exhibit notable diversity and complexity. These mechanisms are primarily categorized into tensile failure, shear failure, and combined fracture modes incorporating both tensile and shear mechanisms27.
During the fracturing of rock layers, cracks gradually nucleate and propagate, with simple crack types primarily classified into three models: opening mode (Mode I), in-plane shear mode (Mode II), and anti-plane shear mode (Mode III)28. When the hard roof strata in coal mines fracture, the three corresponding fracture modes exhibit distinct behaviors during the failure process. For Mode I fracture, when the hard roof strata bend and subside due to mining-induced stress, tensile stresses develop within the strata as a result of deformation. Once these tensile stresses exceed the intrinsic tensile strength of the strata, tensile failure is triggered, as specifically illustrated in Fig. 2, which is drawn by Microsoft Visio Professional 2024(Visio Professional 2024 (Digital download version) | Microsoft Authorized Store).
Schematic diagram of Model I rock fracture.
During Mode II and Mode III fracture processes, the hard roof strata also undergo bending and subsidence. However, unlike Mode I, the tensile stresses developed within the strata do not exceed their intrinsic tensile strength. Under the applied loads from overlying strata, shear failure occurs in the rock layers, macroscopically manifesting as roof shearing failure—a phenomenon clearly observable in Figs. 3 and 4, respectively.
Schematic diagram of Model II rock fracture.
Schematic diagram of Model III rock fracture.
Continuous monitoring of the dynamic evolution of rock strata structure during mining, combined with comprehensive analysis of detailed observational data from numerous engineering sites, reveals that during the extraction of extra-thick coal seams, the fracture of competent rock layers within the overlying strata is dominantly characterized by Mode III shear.
In accordance with Griffith’s theory based on the universal energy concept, the fracture of competent rock layers is brittle, and the stress state of the rock layers can be decomposed using the superposition principle. As illustrated in Fig. 5, Mode III fracture of hard roof strata is decomposed into in-situ stress and shear stresses acting on both sides of the crack. The in-situ stress \(\sigma_{y}^{c}\) is a compressive stress, tending to close the crack; the shear stress \(\tau_{yz}^{c}\) induces sliding of the crack. Due to the frictional resistance in shear sliding, when the rock layer is sheared and undergoes sliding, the equivalent shear stress \(\tau_{{{\text{yz}}}}^{e}\) equals the shear stress subtracted by the frictional stress, i.e., \(\tau_{{{\text{yz}}}}^{e} = \tau_{{{\text{yz}}}}^{c} – \tau_{f(x)} = \tau_{{{\text{yz}}}}^{c} – f_{d} \sigma_{y}^{c}\). Therefore, the fracture of the rock layer can be equivalent to the crack tearing under the action of \(\tau_{yx}^{e}\).
Force decomposition of Model III fracture in hard rock strata. In the figure, \(\sigma_{x}^{c}\) is the boundary stress in the x direction of the selected object, and \(\sigma_{y}^{c}\) is the boundary stress in the direction of the selected object.
Stress and energy analysis of Mode III fracture in competent rock layers
The Mode III fracture of rock strata belongs to spatial shear, not a plane problem. The solution of plane problem in elastic mechanics cannot be directly applied. However, in this problem, the changes of physical quantities (stress, strain and displacement) are independent of z and only depend on x and y coordinates. Therefore, it can still be simplified into a two-dimensional problem. The characteristics of mode III crack are as follows:
$$u = v = 0,\omega = \omega (x,y)$$
(1)
In the equation: x, y is the displacement in the depth direction.
According to the relationship between the strain component and the displacement (Eq. 2), it can be obtained that (Eq. 3), and the remaining four strain components are always \(\varepsilon_{xx} = \varepsilon_{yy} = \varepsilon_{xy} = \varepsilon_{zz} = 0\).
$$\varepsilon_{ij} = (u_{i,j} + u_{j,i} )/2$$
(2)
$$\varepsilon_{xz} = \frac{1}{2}\frac{\partial \omega }{{\partial x}},\varepsilon_{yz} = \frac{1}{2}\frac{\partial \omega }{{\partial y}}$$
(3)
From the generalized Hooke’s law, \(\sigma_{xx} = \sigma_{xx} = \tau_{xy} = \sigma_{zz} = 0\) can be obtained.
$$\left. \begin{gathered} \tau_{xz} = 2\mu \varepsilon_{xz} = \mu \frac{\partial \omega }{{\partial x}} \hfill \\ \tau_{yz} = 2\mu \varepsilon_{yz} = \mu \frac{\partial \omega }{{\partial y}} \hfill \\ \end{gathered} \right\}$$
(4)
Therefore, two of the three equilibrium equations are automatically satisfied, and only one is left:
$$\frac{{\partial \tau_{xz} }}{\partial x} + \frac{{\partial \tau_{yz} }}{\partial y} = 0$$
(5)
Substitute Eq. (4) into Eq. (5) to obtain:
$$\nabla^{2} \omega = 0$$
(6)
Therefore, the problem comes down to solving the Laplace equation under the given boundary conditions, and \(\omega\) must be a harmonic function. We can take
$$\omega = \frac{1}{\mu }{\text{Im}} \mathop Z\limits^{\sim }_{III}$$
(7)
In the equation: \(\mu\) is the shear modulus; \(\mathop Z\limits^{\sim }_{III}\) is called the Westgaard function of Mode III crack. Equation (7) is the full-field equation of Mode III crack displacement.
According to the properties of analytic functions, if \(\mathop Z\limits^{\sim }_{III}\) is an analytic function, then Eq. (7) must satisfy the harmonic equation. The full field equation of Model III crack stress is obtained by substituting Eq. (7) for Eq. (4).
$$\left. \begin{gathered} \tau_{xz} = {\text{Im}} Z_{III} \hfill \\ \tau_{yz} = {\text{Re}} Z_{III} \hfill \\ \end{gathered} \right\}$$
(8)
For the equivalent stress decomposition crack problem, the boundary conditions are:
$$\left. \begin{gathered} {\text{z}} \to \infty {,}\tau_{yz} = \tau^{\infty } = \tau_{yz}^{e} \hfill \\ {\text{y}} = 0,\left| x \right| < a,\tau_{yz} = 0 \hfill \\ \end{gathered} \right\}$$
(9)
According to the above boundary adjustment
$$Z_{III} = \frac{{\tau_{yz}^{e} *z}}{{\sqrt {{\text{z}}^{2} – a^{2} } }}$$
(10)
The crack coordinate system \(Oxy\) is established, the crack length is \(2a\), and \(O\) is the midpoint of the crack. Then the crack polar coordinate system is established. Let \(r_{0}\) be the distance between the point \(P(x,y)\) and the crack center, \(\theta_{0}\) be the polar angle, \(r_{1}\) and \(r_{2}\) are the distance between the point \(P(x,y)\) and the two ends of the crack, \(\theta_{1}\) and \(\theta_{2}\) are the polar angles of the vector \(r_{1}\) and \(r_{2}\), respectively. Thus obtained
$$z = r_{0} e^{{i\theta_{0} }} ,z – a = r_{1} e^{{i\theta_{1} }} ,z + a = r_{2} e^{{i\theta_{2} }}$$
(11)
Substituting the Eq. (11) into the Eq. (10), the radial angle equation of the Westgaard stress function of the Model III crack is obtained.
$$Z_{III} = \frac{{r_{0} \tau_{yz}^{e} }}{{\sqrt {r_{1} *r_{2} } }}e^{i} (\theta_{0} – \frac{{\theta_{1} + \theta_{2} }}{2})$$
(12)
Substituting Eq. (12) into Eq. (8), the stress angle around the Mode III crack is obtained as follows:
$$\left. \begin{gathered} \tau_{xz} = \frac{{r_{0} \tau_{yz}^{e} }}{{\sqrt {r_{1} *r_{2} } }}\sin (\theta_{0} – \frac{{\theta_{1} + \theta_{2} }}{2}) \hfill \\ \tau_{yz} = \frac{{r_{0} \tau_{yz}^{e} }}{{\sqrt {r_{1} *r_{2} } }}\cos (\theta_{0} – \frac{{\theta_{1} + \theta_{2} }}{2}) \hfill \\ \end{gathered} \right\}$$
(13)
The remaining stress components are \(\sigma_{xx} = \sigma_{yy} = \tau_{xy} = \sigma_{zz} = 0\).
Substituting Eq. (12) into Eq. (7), the displacement angle around the Mode III crack is obtained as follows:
$$\omega = \frac{{\tau_{yz}^{e} }}{\mu }\sqrt {r_{1*} r_{2} } \sin \frac{1}{2}(\theta_{1} + \theta_{2} )$$
(14)
On the crack surface, \(y = 0\), that is, \(\theta_{0} = 0\) or \(\theta_{0} = \pm \pi\), on the upper crack surface, \(\theta_{1} = \pi\),\(\theta_{2} = 0\), on the lower crack surface,\(\theta_{1} = – \pi\), \(\theta_{2} = 0\). Replacing Eq. (14) respectively, the displacements of the upper and lower crack surfaces are obtained as
$$\omega^{ + } = \frac{{\tau_{yz}^{e} }}{\mu }\sqrt {a^{2} – x^{2} } ,\omega^{ – } = – \frac{{\tau_{yz}^{e} }}{\mu }\sqrt {a^{2} – x^{2} }$$
(15)
Therefore, the displacement discontinuity above and below the crack is
$$\Delta \omega = \omega^{ + } – \omega^{ – } = 2\frac{{\tau_{yz}^{e} }}{\mu }\sqrt {a^{2} – x^{2} }$$
(16)
By using the method of calculating the eigenvalue of stress tensor matrix, we can obtain that the three principal stresses are
$$\sigma_{1} = \tau_{yz}^{e} \frac{{r_{0} }}{{\sqrt {r_{1} r_{2} } }},\sigma_{2} = 0,\sigma_{3} = – \tau_{yz}^{e} \frac{{r_{0} }}{{\sqrt {r_{1} r_{2} } }}$$
(17)
The maximum shear stress is
$$\tau_{\max } = \frac{{\sigma_{1} – \sigma_{3} }}{2} = \tau_{yz}^{e} \frac{{r_{0} }}{{\sqrt {r_{1} *r_{2} } }}$$
(18)
The crack front coordinate system \(z = a + re^{i\theta } = a + \zeta\) is introduced. In the region near the crack tip (\(r/a1\)), the near field expressions of stress and displacement are
$$\left. \begin{gathered} \tau_{xz} = – \frac{{K_{III} }}{{\sqrt {2\pi *r} }}\sin \frac{\theta }{2} + o(r^{ – 1/2} ) \hfill \\ \tau_{yz} = \frac{{K_{III} }}{{\sqrt {2\pi *r} }}\cos \frac{\theta }{2} + o(r^{ – 1/2} ) \hfill \\ \omega = \frac{{2K_{III} }}{\mu }\sqrt {\frac{r}{2\pi }} \sin \frac{\theta }{2} + o(r^{ – 1/2} ) \hfill \\ \end{gathered} \right\}$$
(19)
The rest components are \(\sigma_{xx} = \sigma_{yy} = \tau_{xy} = \sigma_{zz} = 0\), where
$$K_{III} = \tau_{yz}^{e} \sqrt {\pi a}$$
(20)
It is called the stress intensity factor of Model III crack.
The stress component of the near-field formula of Model III crack in the cylindrical coordinate is obtained by coordinate transformation as follows :
$$\left. \begin{gathered} \tau_{rz} = \tau_{xz} \cos \theta + \tau_{yz} \sin \theta = \frac{{K_{III} }}{{\sqrt {2\pi *r} }}\sin \frac{\theta }{2} + o(r^{ – 1/2} ) \hfill \\ \tau_{\theta z} = – \tau_{xz} \sin \theta + \tau_{yz} \cos \theta = \frac{{K_{III} }}{{\sqrt {2\pi *r} }}\cos \frac{\theta }{2} + o(r^{ – 1/2} ) \hfill \\ \end{gathered} \right\}$$
(21)
The rest of the component \(\sigma_{rr} = \sigma_{\theta \theta } = \tau_{r\theta } = \sigma_{zz} = 0\), the stress component can be written as a unified
$$\sigma_{ij} = \frac{{K_{III} }}{{\sqrt {2\pi r} }}f_{ij} (\theta ) + o(r^{ – 1/2} )$$
(22)
Griffith applied the universal energy method to study the stability of cracks and pointed out that the surface energy can be understood as the energy loss caused by the destruction of chemical bonds between molecules when manufacturing the surface of the material29. When the crack begins to expand, its surface area will continue to increase. In this case, in order to meet the increasing demand for surface energy, it is necessary to provide corresponding energy. Denote the energy consumed by the crack propagation \(\Delta S_{c}\) area as \(\Delta \Gamma\), then the energy required by the crack propagation per unit area \(R\) is
$$R = \frac{\Delta \Gamma }{{\Delta S_{c} }}$$
(23)
Usually, \(R\) is also called crack propagation resistance.
$$R = \Gamma + \Gamma_{p}$$
(24)
In the equation : \(\Gamma\) is the surface energy per unit area of the material ; \(\Gamma_{p}\) is the energy consumed by plastic deformation during crack propagation.
The power provided by the system is \(G\) when the unit area of crack propagation is recorded, then \(G\) needs to overcome the energy consumption caused by crack propagation.
It is assumed that the fracture of hard rock is an ideal brittle fracture. In this ideal state, there is no plastic deformation when the crack propagates, and the corresponding plastic deformation energy \(\Gamma_{p}\) is 0, \(R = \Gamma\). Due to the lack of plastic energy, the Eq. (25) is simplified as
Under the critical condition, for the energy involved in the area of crack propagation \(\Delta S_{c}\), the energy consumed is equal to the energy provided by the system power, and the mathematical expression is
$$R\Delta S_{c} = G\Delta S_{c}$$
(27)
The potential energy of the whole system is \(\Pi\), and the variation of the potential energy of the system is \(\Delta \Pi\). When the potential energy of the system increases, \(\Delta \Pi\) takes a positive value, and when the potential energy of the system decreases, \(\Delta \Pi\) takes a negative value. According to the basic law of energy conservation, the decrease of the system potential energy \(\Delta \Pi\) is the energy supply on which the crack propagation can be carried out.
$$- \Delta \Pi = R\Delta S_{c} = G\Delta S_{c}$$
(28)
Therefore, the driving force \(G\) provided by the system when the crack extends per unit area (also known as the energy release rate of the system per unit area of crack extension) can be determined under the limiting condition where the crack propagation area \(\Delta S_{c}\) approaches 0 infinitely.
$$G = – \mathop {\lim }\limits_{{\Delta S_{c} \to 0}} \frac{\Delta \prod }{{\Delta S_{c} }}$$
(29)
The Griffith energy criterion can be further expressed as : in the category of material mechanics and fracture mechanics, when the elastic energy release rate at the crack tip is greater than or equal to the increase rate of surface energy, the crack will expand. Beckner proposed a relatively simple method to calculate the potential energy change of the system.
$$\Delta \Pi = \frac{1}{2}\iint\limits_{{\Delta S_{c} }} {X_{i} u_{i}^{*} dS}$$
(30)
In the equation : \(X_{i}\) is the surface force ; \(u_{i}^{*}\) is the displacement field of each point after crack propagation.
Substituting the above equation into Eq. (29) , we can obtain
$$G = – \mathop {\lim }\limits_{{\Delta S_{c} \to 0}} \frac{1}{{2\Delta S_{c} }}\iint\limits_{{\Delta S_{c} }} {X_{i} u_{i}^{*} dS} = – \mathop {\lim }\limits_{\Delta a \to 0} \frac{1}{2\Delta a}\int_{0}^{\Delta a} {X_{i} u_{i}^{*} dl}$$
(31)
In the equation : \(\Delta a\) is the length of crack propagation on the calculation plane ; \(l\) is the length variable in the direction of crack propagation on the calculation plane.
Equation (30) and Eq. (31) are called Beckner’s equation. According to Beckner’s equation, generalized Hooke’s law and Westgaard function, the relationship between energy release rate \(G_{III}\) of Model III crack system and stress intensity factor \(K_{III}\) of Model III crack can be obtained.
$$G_{III} = \frac{1 + \zeta }{E}K_{III}^{2}$$
(32)
In this equation : \(\zeta\) is Poisson ‘s ratio.
The total elastic strain energy \(U_{cIII}\) released when the Model III crack length extends from 0 to \(2a\) is
$$U_{cIII} = 2B\int_{0}^{a} {G_{III} da} = 2B\int_{0}^{a} {\frac{1 + \zeta }{E}K_{III}^{2} } da$$
(33)
And then get
$$U_{cIII} = 2B\int_{0}^{a} {\frac{1 + \zeta }{E}(\tau_{yz}^{e} \sqrt {\pi a} )^{2} } da = \frac{1 + \zeta }{E}B\pi (\tau_{yz}^{e} a)^{2}$$
(34)
In the equation: \(B\) is the thickness of hard rock, m; \(\tau_{yz}^{e}\) is the equivalent shear stress, Pa; \(a\) is half of the length of the working face, m.
Calculation of elastic strain energy of rock fracture
There must be some original damage such as joints, cracks and micro-holes in the rock mass30 In order to measure the influence of these damages on the rock mass, a section is taken in the rock mass, and the original damage variable \(D\) is the proportion of the area of the original damage in the section to the total area of the section. The value range of \(D\) is 0 ~ 1, and different \(D\) values correspond to different degrees of damage of rock mass.
The system energy release rate per unit area of crack propagation in rock mass with original damage is recorded as \(G_{D}\), and the system energy release rate per unit area of crack propagation in rock without original damage is recorded as \(G\), then \(G_{D}\) and \(G\) have the following relationship.
$$G_{D} = \left( {1 – D} \right)G$$
(35)
Therefore, in the rock mass with the original damage variable \(D\), the total elastic strain energy \(U_{D}\) released when the crack length extends from 0 to \(2a\) is
$${ }U_{D} = 2B\mathop \smallint \nolimits_{0}^{a} G_{D} da = 2B\mathop \smallint \nolimits_{0}^{a} \left( {1 – D} \right)Gda = \left( {1 – D} \right) \cdot 2B\mathop \smallint \nolimits_{0}^{a} Gda = \left( {1 – D} \right)U_{c}$$
(36)
Therefore, combined with Eq. (34), it can be obtained that
$$U_{DIII} = \frac{1 + \zeta }{E}\left( {1 – D} \right)B\pi \left( {\tau_{yz}^{e} a} \right)^{2}$$
(37)
The equivalent mechanical model calculation of hard rock fracture often assumes that the fracture surface is a standard plane. However, in reality, the internal rock strata are complex, and the fracture surface is not a plane, and its projection is the fracture plane we calculate. When the main crack expands, many branch cracks are derived, forming a large crack group. The actual fracture surface area is far more than the hypothesis, and the energy consumption and the influence on the stability of the rock mass are also greater31. Friedman et al.32 found that the surface area of microcracks is 100 times the fracture area. Aki et al.33 found that the new surface area actually generated by the macroscopic fracture is much larger than the apparent cross-sectional area. Therefore, the coefficient \(k\) is introduced to indicate that the new surface area formed by the actual fracture surface of the rock mass is the multiple of the projected plane area of the fracture surface, and the correction formula of the total elastic energy released by the rock mass fracture is obtained by Eq. (37):
$$U_{DkIII} = \frac{1 + \zeta }{E}\left( {1 – D} \right)kB\pi \left( {\tau_{yz}^{e} a} \right)^{2}$$
(38)
Vibration energy distribution and mine earthquake magnitude estimation
When the rock mass breaks, the total elastic energy released will be converted into a variety of energy forms. It includes the surface energy of the new surface, the frictional heat energy of the fracture surface sliding, the plastic energy consumption of the rock mass oscillation and the seismic wave vibration energy. Among them, seismic wave vibration energy is a part of the total elastic energy release, accounting for between (0, 1).
In view of the similarity between mine earthquake and earthquake in rock mass mechanical response, the mine earthquake efficiency \(\eta\) is temporarily defined as consistent with the concept of seismic efficiency, that is, in the process of mine earthquake, the proportion of mine earthquake wave vibration energy \(E_{S}\) in the total elastic energy released by rock mass fracture is expressed by mathematical formula as follows :
$$E_{S} = \eta U_{Dk}$$
(39)
Substituting Eq. (38) into the above equation, the mine earthquake vibration energy of Model III fracture in hard rock strata can be obtained.
Since the 1960s, scholars have begun to study the value of seismic efficiency. So far, a unified method has not yet been formed, and the problem of value determination is still pending. Some scholars have determined a calculation method of seismic efficiency based on the relationship between apparent stress, vibration energy and seismic moment.
$$\eta = \frac{{E_{S} }}{{AD\overline{\sigma }}}{\text{ = G}}_{S} \frac{{E_{S} }}{{\overline{\sigma }M_{0} }}$$
(40)
In the equation : \(\overline{\sigma }\) is the average stress before and after the earthquake, Pa ; \(A\) is the focal fault area, m2 ; \(D\) is the average offset of fault plane, m ; \({\text{G}}_{S}\) is the shear modulus at the source, Pa ; \(M_{0}\) is scalar seismic moment.
Under the condition of ignoring friction and stress drop to 0, the relationship between seismic efficiency and rock fracture velocity \(V\) is :
$$\eta { = 1 – }g\left( V \right)$$
(41)
In the equation: \(g\left( V \right)\) is a function of rock fracture velocity.
Some scholars have found that there is a certain correlation between seismic efficiency and magnitude in the study of seismic efficiency. They believe that seismic efficiency will increase with the increase of magnitude. Based on this, a formula describing the relationship between the two is proposed.
$$\lg \eta = \left( { – 2.0 \pm 0.51} \right) + \left( {0.23 \pm 0.14} \right)M_{S}$$
(42)
In the equation: \(M_{S}\) is the magnitude of seismic surface wave.
Richter and Gutenberg finally proposed the magnitude calculation method after several modifications.
$$\lg E_{S} = 4.8 + 1.5M_{S}$$
(43)
Substituting Eq. (38) and Eq. (39) into the above equation, the calculation method of mine earthquake surface wave magnitude \(M_{SIII}\) of Model III fracture in hard rock strata can be obtained.
$$\lg (\eta U_{DkIII} ) = 4.8 + 1.5M_{SIII}$$
(44)
Most of the overlying strata in coal mines contain multi-layer hard and thick strata, forming the overlying strata spatial structure of multi-layer key strata. When the working face is mined, the key strata may break asynchronously or synchronously. When synchronous fracture occurs, the energy of each key stratum is released intensively, which is easy to cause large-scale mine earthquake and failure. Therefore, the total elastic energy \(U_{t}\) released by synchronous fracture needs to be considered.
$$U_{tIII} = \mathop \sum \nolimits_{i = 1}^{n} U_{DkIII}^{i}$$
(45)
In the equation : \(U_{tIII}\) denotes the total elastic energy released by the multi-critical layer Model III synchronous fracture, J ; n is the total number of key layers ; c represents the elastic energy released by Model III fracture of the key layer i, J.
Substituting the above equation into Eq. (43), the maximum surface wave magnitude of mine earthquake generated by synchronous fracture of different key strata can be obtained.
$$\lg \left( {\eta U_{tIII} } \right) = 4.8 + 1.5M_{S\max }$$
(46)
Taking 61,607 working face as an example, the relevant calculation parameters are selected, the thickness of hard rock stratum \(B\) = 64.55m, the buried depth of hard rock stratum \(H_{1}\) = 300m, the length of working face is 255m, so \(a\) is 127.5m. Usually, the mine earthquake is the most intense during the square period. Here, the calculation is based on the square area of the working face. The distance between the hard rock stratum and the working face is 25.2 m, the mining height of the working face is 21.6 m, and the lower part of the hard rock stratum is suspended before the fracture of the hard rock stratum. The hard rock stratum is calculated according to the simple support at both ends. The equivalent shear stress \(\tau_{yz}^{e}\) takes half of the vertical stress of the overlying rock stratum, and the rock stratum bulk density \(\gamma\) = 23 kN / m3.
$$\tau_{yz}^{e} = \frac{{0.5*\gamma *H_{1} *2a}}{B} = 13.6{\text{MPa}}$$
(47)
The elastic modulus of rock is 13.77 GPa and the Poisson’s ratio is 0.25. It is assumed that the original damage variable of rock mass is 0.3. Before the overall fracture of the hard rock stratum, many micro-cracks have been developed in different stages. It is assumed that the instantaneous fracture area of the hard rock stratum is 10% of the cross-section area of the rock stratum, and then combined with the law of rock surface energy, \(k\) is 2. It is calculated that the tearing-type fracture of hard rock triggers mine earthquake with a maximum magnitude of 2.4, which is close to the field—monitoring data, with a deviation of approximately 14%.
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