Model Study of Mechanicochemical Degradation in a Planetary Ball Mill

Abstract

The process of ball milling and the materials that compose planetary ball mills are highly complex, and the existing research on the change in ball-milling energy is not mature. The theoretical model of a ball mill was established for the first time to simulate the motion, collision process, energy transfer, and temperature change of small balls during the ball-milling process. Furthermore, by comparing the information with the experimental data for a ball mill, the motion trajectory of the grinding ball, and the energy transfer between the balls and materials were studied, and the micro process during milling was discussed. This study provides a certain theoretical basis for the follow-up engineering application.

1. Introduction

In recent decades, although ball milling has been described using numerous types of theoretical models, a uniformly recognized model has not been established. These models are mainly divided into four categories, namely, kinematic, collision, energy transfer, and temperature rise models. The kinematic model mainly studies the law of motion of the ball-milling medium, including the distribution in the spherical tank, movement speed, and collision frequency, which form the bases for the complete theoretical model of ball milling. The collision model uses a computer program to simulate the dynamic change law for the size and shape of the material particles during collision. The energy transfer model is aimed at identifying how ball-milling energy is transferred from the ball-milling medium to the material particles during collision, and the impact of ball-milling energy on the ball-milling effect. Finally, the temperature rise model discusses the changes in the temperature of the material particles during ball milling [1]. Burn et al. [2] compared the trajectory obtained using a ball-milling model with the actual observed motion and found poor agreement between theoretical and practical results because of the relative sliding between the ball-milling media and the spherical tank. They also reported that the process mainly involves friction and wear rather than impact. Maurice et al. [3] studied ball collision, energy transfer, and temperature rise models and initially ascertained the motion law of the ball-milling medium through model-based estimation. Although Chattopadhyay et al. [4] have achieved certain results with their ball movement model, only a brief description of the collision model of the balls has been reported. The key controlling factors that affect the use of planetary ball mills for treating organic waste are the ball-milling parameters (including milling time, ball size, ball-to-material ratio, speed, and filling degree) and additives. Different parameters exert different effects on the degradation of organic chemicals. Furthermore, the energy change and material degradation mechanism vary significantly during the ball-milling process.

Establishing a ball-milling model with stronger applicability and higher accuracy than existing models is expected to provide important theoretical support for studying the ball-milling process. By establishing the theoretical model of ball milling, this paper investigated the trajectory of the ball as well as the energy transfer between the ball and material. In addition, the micro process of ball milling is intuitively calculated and discussed.

2. Ball Milling Model and Verification

2.1. Ball Milling Kinematic Model

According to the reported planetary kinematic model [4,5,6], this study simplified the model analysis process based on the following assumptions: (1) the ball is identified as a particle; (2) the milling balls do not interfere with each other during the movement; and (3) there is no relative slippage between the balls.

O₁ is the center of the ball mill, O₂ is the center of the ball tank, ${\omega }_{re}$ indicates the angular velocity of the ball-mill revolution, and ${\omega }_{ro}$ corresponds to the angular velocity of rotation of the spherical tank. P₀ is the position of the ball at t = 0, F_re is the centrifugal force of the revolution, and F_ro is the self-rotating centrifugal force.

In the milling process, the revolution and rotation of the milling ball are cyclical. Knowing how a single ball rotates in a spherical tank helps to understand the overall movement of the ball. This study referred to the motion model established by Chattopadhyay et al. [4] and established the motion model under experimental conditions, as shown in Figure 1.

2.1.1. The Derivation Process

A motion model was derived for the situation of a single milling ball rotating in a circle inside the spherical tank, as shown in Figure 2. The relationship between the position of the milling ball and the movement time t was established, and its motion trajectory was determined by analyzing the force of the milling ball. The milling ball speed v was obtained accordingly.

The milling ball moves from point P₀ to point P₁ from t = 0 to t = t₁. At this instant, the abscissa of the position is x₁-x₂, and the ordinate is y₁-y₂. When the angle is rotated under the common action, the position equations of the ball at time t are as follows:

$$x=R_{re}\cos {\omega }_{re}t+R_{ro}\cos \left[\left({\omega }_{ro}-{\omega }_{re}\right)t\right]$$

(1)

$$y=R_{re}\sin {\omega }_{re}t-R_{ro}\sin \left[\left({\omega }_{ro}-{\omega }_{re}\right)t\right]$$

(2)

where $R_{re}$ represents the distance from the spherical tank center O₂ to the ball mill axis O₁, and $R_{ro}$ represents the distance from the spherical tank center O₂ to the inner wall of the spherical tank. The derivative of t is taken at both ends of Equations (1) and (2) to obtain the velocity equation of point P at time t, as shown in Equations (3) and (4).

$$v_x=\frac{dx}{dt}=-{\omega }_{re}{R}_{re}\sin \left({\omega }_{re}t\right)-\left({\omega }_{ro}-{\omega }_{re}\right)R_{ro}\sin \left[\left({\omega }_{ro}-{\omega }_{re}\right)t\right]$$

(3)

$$v_y=\frac{dy}{dt}=-{\omega }_{re}{R}_{re}\cos \left({\omega }_{re}t\right)+\left({\omega }_{ro}-{\omega }_{re}\right)R_{ro}\cos \left[\left({\omega }_{ro}-{\omega }_{re}\right)t\right]$$

(4)

After obtaining the milling-ball speed equation, the running speed of the ball can be calculated at any time. However, when the milling ball moves away from the inner wall of the spherical tank, its running speed will not continue to increase. Therefore, it is necessary to analyze the force of ball. The running track of the milling ball was determined. As shown in Figure 1, at t = t₁, the ball is subjected to the resultant force F_co in the direction of the spherical tank axis (O₂P₁), which is composed of the component force of the revolution centrifugal force F_re and the self-rotation centrifugal force F_ro. The expression is shown in the Equations (5)–(7), and it can be further substituted into the Equation (9).

$$F_{co}=F_{ro}-F_{re}\cos \theta$$

(5)

$$F_{re}=m\sqrt{x^2+y^2}{\omega }_{re}^2$$

(6)

$$F_{ro}=m{R}_{ro}{\omega }_{ro}^2$$

(7)

$$\theta =\pi -\left({\omega }_{ro}-{\omega }_{re}\right)t$$

(8)

$$F_{co}=m{R}_{ro}{\omega }_{ro}^2-m\sqrt{x^2+y^2}{\omega }_{re}^2\cos \left[\pi -\left({\omega }_{ro}-{\omega }_{re}\right)t\right]$$

(9)

When the force between the milling ball and the inner wall of the spherical tank is 0, F_co = 0, the ball starts to separate from the inner wall of the spherical tank and reaches the take-off condition, thereby satisfying the Equation (10).

$$R_{ro}{\omega }_{ro}^2=\sqrt{x^2+y^2}{\omega }_{re}^2\cos \left[\pi -\left({\omega }_{ro}-{\omega }_{re}\right)t\right]$$

(10)

Equations (1) and (2) are incorporated into Equation (10) to obtain the one-variable nonlinear Equation (11) with respect to t as given below:

$$R_{ro}{\omega }_{ro}{}^2+\sqrt{R_{re}{}^2+R_{ro}{}^2+2R_{re}{R}_{ro}\cos ({\omega }_{ro}t)}{\omega }_{re}{}^2\cos \left[\right({\omega }_{ro}-{\omega }_{re})t]=0$$

(11)

This equation defines the radius and angular velocity of the rotation, as well as the revolution of the milling ball and the equation for time t. According to different ball mills and balls, the radius and angular velocity of the rotation and revolution are known but different. Based on this, a one-variable equation with respect to t can be obtained. The parameter t corresponds to the moment when the ball moves. A positive solution to the equation for t indicates the existence of a moment when the force on the ball is zero, that is, the milling ball is no longer bound to the wall of the spherical tank. When the take-off condition is reached, the equation for t without a positive solution indicates that the milling ball cannot reach the take-off condition and will continue to grind the wall of the spherical tank. Considering that the equation is related to the trigonometric function of t, if the value of t has a solution, then it will be a series of cyclic values, and at least one take-off condition will be reached for each cycle of operation. Determining whether the milling ball can reach the take-off condition is crucial for studying the movement of ball because this will determine the working conditions of the milling-ball process to ensure that it can fully generate the collision friction and effectively treat the pollutants.

Regardless of whether the ball reaches the take-off condition or not, the speed of the ball during the milling-ball process is an important indicator of the motion state of the ball and a crucial parameter in the milling-ball collision model. Therefore, the speed of the ball mill needs to be determined. When t has a solution, the milling ball reaches the take-off condition, and the milling ball in the air is not affected by other forces and cannot be accelerated. The take-off speed of the ball is its maximum speed:

$$v=\sqrt{{\nu }_x{}^2+v_x{}^2}$$

$$v=\sqrt{{\{-{\omega }_{r\mathrm{e}}{R}_{re}\sin ({\omega }_{re}t)-({\omega }_{re}-{\omega }_{ro})R_{ro}\sin [({\omega }_{re}-{\omega }_{ro})t]\}}^2+{\{-{\omega }_{re}{R}_{re}\cos ({\omega }_{re}t)-({\omega }_{re}-{\omega }_{ro})R_{ro}\cos [({\omega }_{re}-{\omega }_{ro})t]\}}^2}$$

$$v=\sqrt{{\omega }_{r\mathrm{e}}{}^2{R}_{re}{}^2+{({\omega }_{re}-{\omega }_{ro})}^2{R}_{ro}{}^2+2{\omega }_{r\mathrm{e}}({\omega }_{re}-{\omega }_{ro})R_{re}{R}_{ro}\cos ({\omega }_{ro}t)}$$

(12)

Then, the speed equation when the ball moves is as shown in Equation (12). Based on this, the speed of the milling ball can be calculated at any time before reaching the take-off condition. Determining the milling-ball speed is significant for the subsequent calculation of the milling-ball energy.

No solution for t in Equation (3) indicates that the milling ball has been running along the inner wall in the spherical tank at a constantly changing speed. To facilitate the calculation, the two points on the inner wall of the milling tank that are closest to the revolution axis and the two points farthest from the O₁O₂ axis are selected. A total of four representative points are used to calculate the speed of milling ball. The speed range of the ball is shown in Figure 3.

At this instant, the milling-ball movement speed is examined at the four characteristic positions of the spherical tank ABCD, where B represents the position of the inner wall of the spherical tank farthest from the center of the revolution, D represents the position of the inner wall of the spherical tank closest to the center of the revolution, and A and C are the positions of the inner wall of the spherical tank in the vertical direction connecting the center of revolution and the center of rotation, respectively. The above four positions are on the same plane.

According to the influence of the revolution and rotation of the milling balls at the four positions of v_A~v_D, the respective combined speeds can be expressed as Equations (13)–(16). When the milling ball does not reach the speed range under the take-off condition, its value is expected to be in the range of four speeds. Conversely, when the milling ball reaches the take-off condition, the maximum value of the milling-ball speed range is the takeoff speed calculated using Equation (9), and the minimum speed value is the same as that of a ball that does not reach take-off.

$$\begin{array}{l}{v}_A=\sqrt{{\left(v_{roA}-v_{reA}\sin \theta \right)}^2+{\left(v_{reA}\cos \theta \right)}^2}=\sqrt{v_{reA}{}^2+v_{roA}{}^2-2v_{reA}{v}_{roA}\sin \theta }\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\sqrt{{\left({\omega }_{re}\cdot O_{1}A\right)}^2+{\left({\omega }_{ro}\cdot O_{2}A\right)}^2-2\left({\omega }_{re}\cdot O_{1}A\right)\cdot \left({\omega }_{ro}\cdot O_{2}A\right)\cdot \frac{O_{2}A}{O_{1}A}}\end{array}$$

(13)

$$v_B=v_{reB}-v_{roB}={\omega }_{re}\cdot O_{1}B-{\omega }_{ro}\cdot O_{2}B$$

(14)

$$\begin{array}{l}{v}_C=\sqrt{{\left(v_{roC}-v_{reC}\sin \theta \right)}^2+{\left(v_{reC}\cos \theta \right)}^2}=\sqrt{v_{reC}{}^2+v_{roC}{}^2-2v_{reC}{v}_{roC}\sin \theta }\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\sqrt{{\left({\omega }_{re}\cdot O_{1}C\right)}^2+{\left({\omega }_{ro}\cdot O_{2}C\right)}^2-2\left({\omega }_{re}\cdot O_{1}C\right)\cdot \left({\omega }_{ro}\cdot O_{2}C\right)\cdot \frac{O_{2}C}{O_{1}C}}\end{array}$$

(15)

$$v_D=v_{reD}+v_{roD}={\omega }_{re}\cdot O_{1}D+{\omega }_{ro}\cdot O_{2}D$$

(16)

In summary, this section establishes a model to determine the motion state of the milling ball in the spherical tank. First, the existence of a solution for t in Equation (11) is determined to see whether the milling ball will take off and leave the wall of the spherical tank. Then, the speed range of the milling-ball movement corresponding to the take-off situation is determined according to Equations (12)–(16). Compared with the model of Chattopadhyay et al. [4], the motion model of the milling ball established in this study is more concise, which is convenient for analysis, judgment, and calculation. Determining the parameters of the transfer model provides a rudimentary basis.

2.1.2. The Experimental Verification Process

The experimental parameters are substituted into the established motion model Equation (11), and the motion state of a single milling ball under the experimental conditions is inspected. Then, the maximum speed and acceleration of the milling-ball motion are simulated and theoretically calculated.

First, the size parameters required for the calculation are determined. According to the internal geometrical dimensions of the spherical tank shown in Figure 4, the diameter of the ball is 8 mm, the inner radius of the spherical tank is 26.5 mm, and the distance between the center of revolution and the center of rotation of the spherical tank is O₁O₂ = 86.5 mm = R_re. The inner wall of the spherical tank is the longest distance to the center of the disk, 122 mm. The distances between the four characteristic positions ABCD of the spherical tank and the center of revolution and rotation are O₂A = O₂B = O₂C = O₂D = 22.5 mm = R_co, O₁A = O₁C = (22.5² + 86.5²)^−1/2 = 89.38 mm, O₁B = O₁O₂ + O₂B = 109 mm, and O₁D = O₁O₂ − O₂D = 64 mm.

Second, the operating parameters required for the calculation are determined. The rotation speed during milling ball is 550 rpm, and the ratio of revolution to rotation is set to 1:2. Then, ${\omega }_{ro}$, the rotation angular velocity of the spherical tank can be determined as:

$${\omega }_{ro}=2{\omega }_{re}=550\times \frac{2\pi }{60}=57.60\hspace{0.17em}\hspace{0.17em}\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}$$

Finally, according to the derived motion model Equation (11), the size and operating parameters are substituted as follows:

$$26.5\times {57.6}^2+\sqrt{{86.5}^2+{26.5}^2+2\times 86.5\times 26.5\times \cos \left(57.6t\right)}\times {28.8}^2\cos \left[\left(57.6-28.8\right)t\right]=0$$

To solve this equation, if we suppose $\cos \left(28.8t\right)=x$; then,$\cos \left(57.6t\right)=2x$. According to the above formula, x is negative, and the above formula can be simplified to:

$$\sqrt{3600+9169x^2}\times 829.44x=-87920.64$$

$$x^2=\frac{-3600+\sqrt{{3600}^2-4\times 9169\times \left(-11236\right)}}{2\times 9169}=0.9279$$

$$x=\pm 0.9633$$

Because x is a negative number, then x = −0.9633 and $\cos \left(28.8t\right)=x$; subsequently, $28.8t=164.4+360n$, $n=1,2,3\dots$. Finally, $t=5.7+12.5n$, $n=1,2,3\dots$

The calculation results show that t has a positive solution, indicating that there is a moment when the ball is subjected to zero force. At this moment, the ball reaches the take-off condition. The calculation results show that the ball reaches the take-off condition for the first time at t = 5.7 s and can reach the take-off condition again every 12.5 s. This result indicates that the ball will leave the wall of the ball tank and collide with other balls to degrade the pollutants during milling in this experiment, that is, the ideal state of the milling ball.

When the ball reaches the take-off condition and assuming that the ball is not affected by other forces in its motion, the take-off velocity of the ball is its maximum velocity. Considering that t has a solution, the maximum velocity of a single ball can be calculated by substituting t = 5.7 s into the deduced Equation (12) for the take-off velocity of the ball, v = 3.169 m/s. The maximum velocity of the ball in the experimental ball mill is determined, providing important parameters for the collision and energy transfer models.

To verify the accuracy of the established model, the milling-ball parameters of different types of planetary ball mills in the reported literature are analyzed and compared [1]. The milling-ball velocity calculated by simulation is within a reasonable range, indicating that the motion model established in this experiment is in line with reality and that the accuracy of simulation is high.

In addition to the speed of the ball, the acceleration of the grinding ball can measure the high-energy effect of the milling process. When the planetary ball mill is started, the revolution is clockwise, and the rotation is counterclockwise. Then, revolution and rotation are reversed once every 15 min. The characteristic position ABCD in Figure 3 is analyzed, where the ball at position B has the farthest distance from the center of the revolution, and the ball is subjected to the maximum centrifugal force in revolution. For the same centrifugal force in rotation, the ball at position B has the maximum acceleration. Therefore, position B is selected as the investigation point of the acceleration index to calculate the acceleration of the ball:

$$a_B=a_{re}+a_{ro}={\omega }_{re}{}^2\cdot O_{2}B+{\omega }_{ro}{}^2\cdot O_{1}B=165.06 \mathrm{m}/\mathrm{s}$$

The maximum centrifugal acceleration is 16.84 times of the acceleration due to gravity g. At this instant, the milling ball is subjected to a massive force, and the force of impact, extrusion, and friction between each other is also increased by more than ten times. Although the modeling and simulation results are accurate, the actual ball-milling process is the result of the combined action of several balls, and its motion process is highly complex. For the milling balls to collide and rub each other continuously, they need to constantly reach the take-off state, which can only be achieved with sufficient speed. To prevent the ball mill from not colliding and flying up after losing energy in multiple collisions, the ball mill was set to reverse every 15 min.

To summarize, this section establishes the motion model of the ball milling and determines the motion of the milling ball according to the parameters of the ball mill in this experiment. The maximum motion velocity of the milling ball is obtained as $v=3.169 \mathrm{m}/\mathrm{s}$, and the maximum centrifugal acceleration of the milling ball reaches 16.84 times the gravitational acceleration g. By comparing the results with those of other studies, the accuracy of the simulation calculation is verified. The planetary ball mill is a kind of high-energy ball mill that provides theoretical support and data reference for subsequent ball mill energy research.

2.2. Energy Transfer Model

For the research on the energy transfer model, the main problems are the modes and efficiency of energy transfer and the influence of milling-ball energy on the entire ball-milling results. The model also includes the relationship between the milling-ball medium radius, mass, quantity, and milling-ball energy. Magini et al. [3] believed that the energy transfer in the ball-milling process mainly includes collision, and milling exists simultaneously in the entire ball-milling process. The free-fall experiment shows that the milling-ball medium collision is categorized as completely inelastic collision, and the medium kinetic energy can be transmitted to the material particles. Chattopadhyay et al. [4] found that the elastic properties of the milling-ball medium and spherical tank determine the rate of material structure refinement during milling ball and influence the effect of collision. Schilz [6] demonstrated that the impact energy of material particles is only related to the ratio of rotation and revolution radius and angular velocity. Maurice [3] and Love [7] conducted a series of studies on the milling-ball model and found that a few ball collisions in the process of milling ball correspond to a positive collision; most of these collisions are caused by friction and shear; moreover, dynamic simulation proved that only 4–7% of collisions can produce considerable energy conversion, while the rest can only produce a small amount of energy conversion. This provides a good understanding of the milling-ball model, which shows that the front ball collision is not normal and that energy conversion without energy loss only exists in the hypothesis. However, despite the undisputable fact that the milling-ball model is the best way to analyze the energy transfer of the milling ball, various factors should be considered to the maximum extent in the simulation process to truly reflect the energy transfer of the milling ball.

To better study the energy transfer process, the specific process that the milling ball material particles undergo needs to be understood. The ball-milling process mainly occurs between the milling balls and between the milling balls and the spherical tank wall. The different impact angles induce the formation of the front extrusion and shear friction, which exerts aggregation and crushing effects on materials, as shown in Figure 5. The two functions have repeated effects on the materials. The extrusion function compresses the material particles between the milling-ball surfaces into a flat shape, resulting in severe plastic deformation. The shear friction function can break and separate the material particles uniformly compacted on the milling-ball surface, and then a new material particle surface is generated by extrusion. Because the highest energy transfer efficiency is obtained when the ball collides with another ball at the maximum speed, this experiment simulated and calculated this situation and then modified the overall milling-ball power according to parameters such as ball collision efficiency.

2.2.1. The Derivation Process

First, a collision surface between milling balls without materials is considered. As shown in Figure 6, such collision can occur within an angle range, and this geometric characteristic may have an important effect on the collision between milling balls. For example, depending on the angle, several types of collisions can occur. When two milling balls collide head on at the maximum speed, the collision energy is the highest. Considering that the maximum movement speed of the milling ball is 3.169 m/s and the relative speed of the material and the milling ball is considerably lower than the speed of sound in the tank, the kinetic energy generated by the collision relative movement of the ball during milling is considerably less than the elastic potential energy of the milling ball [7,8,9]. This finding shows that the collision during milling meets the classical Hertzian impact theory [3]. The following assumptions are needed when calculating the collision energy of ball according to the Hertzian impact theory:

(1)

The collision process involves completely elastic collision, and no energy loss occurs in the process;

(2)

The Hertz radius is significantly less than the radius of the impact milling ball; therefore, the contact surface of the impact milling ball is treated as a plane.

Figure 6. Geometry of maximum flat compression by the ball in collision.

Figure 6. Geometry of maximum flat compression by the ball in collision.

Figure 6 and Figure 7 show the geometric model of the maximum compression of the milling ball during impact and the schematic diagram of the milling ball upon collision with the material. According to the Hertzian impact theory [3], the impact time $\tau$, impact radius $r_h$, milling-ball compression deformation ${\delta }_{\max }$, maximum vertical impact pressure $P_n$, and powder layer thickness $h_0$ are determined.

Single Elastic Collision Energy

Elastic collision time $\tau$ is an important influencing variable that represents the time taken for milling-ball collision and separation. According to the derivation of Maurice [3], $\tau$ can be expressed using the following Formula (17):

$$\tau =0.5\cdot g_{\tau }{v}_n{}^{-0.2}{\left({\rho }_b/E\right)}^{0.4}{r}_b$$

(17)

In Equation (17): $g_{\tau }$ is the geometric coefficient whose value is determined by the geometric dimensions of the milling ball and spherical tank, as shown in

Table 1. Parameters of different types of collision.

Impact Type	${\mathit{g}}_{\mathit{\tau }}$	${\mathit{g}}_{\mathit{r}}$	${\mathit{g}}_{\mathit{p}}$
Ball to another ball of the same size and material	5.5744	0.9731	0.4646

; $v_n$ is the vertical impact velocity of the milling ball, $E$ is the effective elastic modulus of the collision medium, $r_b$ is the radius of the milling ball, and ${\rho }_b$ is the milling-ball density.

For maximum compression in the milling-ball process, the impact contact radius $r_h$ between the impact surfaces is important. It directly affects the compression deformation ${\delta }_{\max }$ of the milling ball, thus affecting the material volume upon each impact. The formula for $r_h$ according to the derivation of Maurice [3] and the Hertzian impact theory is given below as Equation (18), where $g_r$ is the geometric coefficient whose value is determined by the geometric dimensions of the milling ball and spherical tank in Table 1.

$$r_h=g_r{v}_n{}^{0.4}{\left({\rho }_b/E\right)}^{0.2}{r}_b$$

(18)

The compression deformation ${\delta }_{\max }$ can be calculated according to Figure 6 as ${\delta }_b{\sqrt{r_b{}^2-r_h{}^2}}_{\max }$. The impact radius $r_h$ is much smaller than the milling-ball radius $r_b$, so ${\delta }_b{\sqrt{1-\frac{r_h{}^2}{r_b{}^2}}}_{\max }$. According to the Equivalent Infinitesimal Substitution formula, ${\left(1+\beta x\right)}^{\partial }-1~\alpha \beta x$, because $r_h{}^2$ tends toward 0, and therefore, ${\delta }_{\max }$ is approximately equal to $r_b\left(\frac{1}{2}\cdot \frac{r_h{}^2}{r_b{}^2}\right)$. Simplified sorting can be obtained as ${\delta }_{\max }$, which is shown as Equation (19):

$${\delta }_{\max }=r_h^2/2r_b$$

(19)

When maximum elastic collision occurs during milling, the average pressure on the impact surface of the ball is the maximum vertical impact pressure $P_n$, which is a significant parameter that can be used to determine the elastic strain energy involved in the collision. This energy can affect the process of the milling ball to the maximum extent during milling. $P_n$ can be expressed by Equation (16), where $g_P$ is the geometric coefficient, whose value is determined by the geometric dimensions of the milling ball and spherical tank in

Table 2. Parameters of energy $U_E$ in a single collision.

Parameter	Value	Calculation Process	Remark
$v_n$	3.169 m/s	-	Section 2.1.2
$g_{\tau }$	5.5744	-	Table 1
$g_r$	0.9731	-
$g_p$	0.4646	-
$\tau$	$8.782\times {10}^{-6} \mathrm{s}$	$\tau =0.5g_{\tau }{v}_n{}^{-0.2}{({\rho }_b/E)}^{0.4}{r}_b$ $=0.5\times 5.5744\times {3.169}^{-0.2}(5.89\times {10}^3/190$ $\times {10}^9{)}^{0.4}\times 0.004$ $=8.782\times {10}^{-6} \mathrm{s}$	Equation (17)
$r_h$	$1.944\times {10}^{-4} \mathrm{m}$	$r_h=g_r{v}_n{}^{0.4}{({\rho }_b/E)}^{0.2}{r}_b$ $=0.9731\times {3.169}^{0.4}(5.89\times {10}^3/190$ $\times {10}^9{)}^{0.2}\times 0.004$ $=1.944\times {10}^{-4} \mathrm{m}$	Equation (18)
${\delta }_{\max }$	$4.723\times {10}^{-6} \mathrm{m}$	δmax = rh2/2rb $=(1.944\times {10}^{-4}{)}^2/0.008$ $=4.723\times {10}^{-6} \mathrm{m}$	Equation (19)
$P_n$		$P_n=g_p{v}_n{}^{0.4}{({\rho }_b/E)}^{0.2}E$ $=0.4646\times {3.169}^{0.4}(5.89\times {10}^3/190$ $\times {10}^9{)}^{0.2}190\times {10}^9$ $=4.407\times {10}^9 \mathrm{Pa}$	Equation (20)

.

$$P_n=g_p{v}_n{}^{0.4}{\left({\rho }_b/E\right)}^{0.2}E$$

(20)

The impact energy $U_E$ of the milling ball is the product of the elastic energy $r_b$ of unit volume deformation and the impact deformation volume 2V_E of two milling balls. Assuming that two colliding balls have the same influence radius, the elastic energy of deformation per unit volume is defined as [10]:

$$C_E=P_n{}^2/2E$$

(21)

Given that ${\delta }_{\max }$ is considerably smaller than $r_b$, the volume deformation of each milling ball is an approximate cone, and the impact deformation volume of the milling ball can be expressed by Equation (22):

$$V_E=\pi r_b^2{\delta }_{\max }/3$$

(22)

Therefore, the impact energy $U_E$ of the milling ball can be obtained as:

$$U_E=2\times C_E\times V_E=\pi r_b^2{n}^2{\delta }_{\max }/3E$$

(23)

Equations (18)–(20) are substituted into Equation (23) to yield:

$$U_E=C_E\times V_E=\frac{\pi }{6}{r}_b{}^{3}g{r}^{2}g{p}^2{v}_n{}^{1.6}{\rho }_b{}^{0.8}{E}^{0.2}$$

(24)

Equation (24) yields the energy generated by the single elastic collision of the milling ball. Evidently, the single elastic collision energy of the milling ball is directly proportional to the radius and speed of the milling ball along with the density and elastic modulus of the milling ball material. Among these parameters, the radius of the milling ball has the greatest influence, followed by the vertical impact speed of the milling ball; the density and elastic modulus of the milling ball are related to the material and influence of the collision energy to a certain extent. These results are consistent with those obtained in the ball-milling condition selection reported in the literature [10,11,12,13,14,15,16,17,18,19].

The Average Free Path of Milling Ball

Considering that many balls are present in the spherical tank, the average free path of the milling ball needs to be combined with the milling ball in the entire spherical tank for collision analysis. However, the movement of numerous milling balls is highly complex. To facilitate the analysis of the motion trajectory of the milling ball, the path length of the milling ball between two collisions is defined as the free path, and the average value of all free paths that the milling ball may pass is the average free path.

The first assumption is that all milling balls are evenly distributed in the spherical tank space and arranged in cube form before the milling-ball collision. As shown in Figure 8, the side length $a$ of the milling ball cube in the spherical tank space V_g can be expressed as Equation (25), where $D_m$ is the diameter of the spherical tank, L is the effective height in the spherical tank, and $n_b$ is the number of milling balls.

$$a=\sqrt[3]{\frac{V_g}{n_b}}=\sqrt[3]{\frac{\pi D_m{}^{2}L}{4n_b}}$$

(25)

The second assumption is that the average distance between each milling ball and all adjacent milling balls is the calculated average free path. Because the free path is the distance between two collisions in the milling ball, the average distance between the milling ball and the surrounding milling balls is half of the average free path S. Then, the average free path is:

$$\frac{S}{2}=\frac{6a+12\sqrt{2}a+8\sqrt{3}a}{26}-2r_b$$

(26)

Finally, the Equation of the milling ball cube side length (25) is substituted into Equation (26), and the average free path S is calculated as:

$$S=2.832\sqrt[3]{\frac{\pi D_m{}^{2}L}{4n_b}}-4r_b$$

(27)

Therefore, the average free path S is related to the diameter of the spherical tank, as well as the number and radius of the milling balls. The larger the diameter of the spherical tank, the larger the average free path S; by contrast, the larger the radius and quantity of milling-ball particles, the smaller the average free path S. The smaller the average free path S, the higher the frequency of milling-ball collision. Therefore, the selection of appropriate milling-ball quantity and ball-filling degree substantially affects the action efficiency of the milling-ball collision.

Collision Material Mass

After determining the energy of a single collision, the mass of the material affected by the energy of a single collision needs to be clarified to calculate the impact efficiency of the milling-ball collision and the temperature change of the material. According to diagram 7 of the impact surface of the milling ball, the mass $Q$ of the material affected by each collision process can be obtained, where ${\rho }_p$ is the density of the material.

$$Q={\rho }_P\pi r_h{}^2{h}_0$$

(28)

To determine the material thickness h₀ of the collision, it is assumed that the material powder is evenly distributed in the entire ball tank before the collision of the two balls. During each collision, the milling ball can act on all material powders on the free path within the impact radius. After the collision, the materials between the milling balls are again evenly distributed in the entire ball tank, as shown in Figure 9.

The volume $V_c$ of space swept by the milling ball in each collision:

$$V_c=\pi r_h{}^{2}S$$

(29)

Assuming that the powder is evenly distributed in the entire spherical tank space, the space volume V_p that the powder can occupy is the sum of the effective volume V_g of the spherical tank minus the volume $V_{bTotal}$ of all milling balls. Therefore, the volume fraction ${\varphi }_p$ of powder distributed in the tank space is:

$${\varphi }_p=V_p/\left(V_g-V_{bTotal}\right)=m_p{\rho }_p/\left[\left(V_g{\rho }_b-m_b{n}_b\right){\rho }_p\right]$$

(30)

In Equation (30), V_p is the volume of material powder, V_g is the effective volume in the milling-ball tank, $V_{bTotal}$ is the total volume of all milling balls, ${\rho }_p$ is the density of the material powder, $m_p$ is the volume of the material powder, ${\rho }_b$ is the density of the milling balls, $m_b$ is the mass of the milling balls, and $n_b$ is the number of milling balls.

As shown in Figure 7, the volume of the material powder at the time of collision can be regarded as a cylinder with a radius $r_h$ and a height $h_0$. Then, the thickness $h_0$ of the powder layer is given by:

$$h_0=\frac{V_c{\varphi }_p}{\pi r_h{}^2}=S{\varphi }_p$$

(31)

Then, substituting Equations (30) and (31) yields into Equation (28) for material mass $Q$ under collision:

$$Q={\rho }_p\pi r_h{}^{2}S{\varphi }_p$$

(32)

The mass $Q$ of the material under collision is directly proportional to the density of the material, impact radius of the milling ball, average free path, and volume fraction of the material powder in the spherical tank. The energy of the milling-ball collision on the unit mass material can be determined according to the calculated mass $Q$ of the collision material and the energy U_E transferred to the material in a single collision process.

Energy Transfer Power of Milling Ball System

The collision power of a single milling ball can be obtained by clarifying the energy generated by one collision of a single milling ball and the number of collisions of a single milling ball per unit time. Then, the power of the whole milling-ball process can be deduced.

Milling frequency is the reciprocal of the time from the beginning of a single milling-ball collision to that of the next collision and is given as Equation (33). The time of a single impact of the milling ball is the time when the ball passes through the average free path and the duration of the impact. $\tau$ is the duration of the collision, calculated from Equation (17), and corresponds to the movement time of the milling ball in the milling-ball cycle. $T$ is the time taken for the milling ball to pass through the average free path and is given as Equation (34).

$$f=\frac{1}{\tau +T}$$

(33)

$$T=\frac{s}{v_n}$$

(34)

Substituting Equations (17) and (33) into Equation (34), the milling-ball frequency can be further defined as follows:

$$f=\frac{v_n}{g_r{v}_n{}^{1.4}{({\rho }_b/E)}^{0.2}{r}_b+2.832\sqrt[3]{\frac{\pi D_m{}^{2}L}{4n_b}}-4r_b}$$

(35)

Combined with the above series of derivation results, the theoretical power of a single milling-ball collision is the single elastic collision energy $U_E$ multiplied by the milling-ball collision frequency $f$ and then by the number $n_b$ of milling balls to obtain the power $W$ transmitted by the energy of the whole milling-ball system.

$$W=0.5\eta U_E{n}_{b}f$$

(36)

In Equation (36), $\eta$ indicates the collision efficiency required for two milling balls during each collision; thus, the term is multiplied by a factor of 0.5.

According to the formula for energy transfer power $W$, the energy transfer power of the milling-ball system can be improved by increasing the energy generated by a single collision or increasing the number of milling balls and the collision frequency of milling balls.

2.2.2. The Experimental Verification Process

The size of the spherical tank used here is the one reported in various studies [10,11,12,13,14,15,16,17,18,19], with the inner diameter $D_m$ = 53 mm and the inner height L = 68 mm. The number of milling balls used or $n_b$ = 30 and the radius $r_b$ = 4 mm. The spherical tank material is zirconia with an elastic modulus $E$ = 190 Gpa [20] and density ${\rho }_b$ = 5.89 kg/m³ [21]. The densities of the experimental milling-ball objects ${\rho }_p$ are as follows: TBBPA, 2.1 g/cm³, compound sulfamethoxazole tablets, 1.08 g/cm³; and soil contaminated with lindane and polychlorinated biphenyls, 1.6 g/cm³ (the dry density of soil is generally 1.4–1.7 g/cm³).

Single Elastic Collision Energy

According to the energy Equation (14) derived from the single elastic collision of the milling ball in the previous section, the energy $U_E$ of a single collision can be calculated, and the calculation of various parameters is shown in Table 2.

$$\begin{array}{l}{U}_E=\pi r_b{}^2\delta n_{\max }^2=3.14159\times {0.004}^2\times 4.723\times {10}^{-6}\times (4.407\times {10}^9{)}^2/(3\times 190\times {10}^9)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=8.09\times {10}^{-3}\mathrm{J}\end{array}$$

Compared with the energy parameters of the three planetary mills in Table 2, the single collision energy of the milling ball belongs to the normal range with a value of $8.09\times {10}^{-3} \mathrm{J}$, which is a low value. However, a good degradation effect was still achieved in this experiment on various target organics, demonstrating that the free radicals produced by activated persulfate enhance the mechanochemical degradation of organics.

Average Free Path of the Milling Ball

The average free path S of the milling ball can be calculated according to Equation (27):

$$S=2.832\sqrt[3]{\frac{\pi D_m{}^{2}L}{4n_b}}-4r_b$$

$$=2.832\times \sqrt[3]{\frac{3.14159\times {(53\times {10}^{-3})}^2\times 68\times {10}^{-3}}{4\times 30}}-4\times 0.004=0.0324 \mathrm{m}$$

The average free path of the milling ball from the end of the collision to the beginning of the next collision is 0.0324 m. In the milling-ball tank with an internal diameter $D_m$ of 53 mm and an internal height L of 68 mm, the average free path can reach 32.4 mm. This result shows that the distance between the milling balls is relatively large when they are evenly distributed; moreover, each collision can act on as many materials as possible.

Mass of the Collision Material

The volume fraction ${\varphi }_p$ in Equation (30) of the powder distributed in the tank space is substituted into $h_0$ in Formula (31), which corresponds to the layer thickness $h_0$ of the extruded material powder when the milling ball collides:

$$\begin{gathered} h_0=S{m}_p{\rho }_b/[(V_g{\rho }_b-m_b{n}_b){\rho }_p] \\ =\frac{0.0324\times 5\times {10}^{-3}}{(\frac{3.14159\times {(53\times {10}^{-3})}^2\times 68\times {10}^{-3}}{4}-\frac{4}{3}\times 3.14159\times {0.004}^3\times 30)\times {\rho }_p}=\frac{162}{142{\rho }_p}m \end{gathered}$$

The following values are obtained for different materials: TBBPA, 5.432 × 10⁻⁴ m; sulfamethoxazole, $1.056\times {10}^{-3} \mathrm{m}$; and soil contaminated with lindane and polychlorinated biphenyls contaminated soil, $7.130\times {10}^{-4} \mathrm{m}$. Then, the mass of materials in each collision is:

$$Q={\rho }_p\pi r_h{}^2{h}_0={\rho }_p\pi {(1.944\times {10}^{-4})}^2\frac{162}{142{\rho }_p}=1.354\times {10}^{-7} \mathrm{kg}$$

Therefore, the energy of each collision acting on the material per unit mass can be further calculated as:

$$\frac{U_E}{Q}=\frac{8.09\times {10}^{-3}}{1.354\times {10}^{-7}}=5.975\times {10}^4 {\mathrm{Jkg}}^{-1}$$

At the same time, the mass of materials used in each milling ball is 5 g; thus, the number of collisions required to transfer energy to all materials at least once is calculated as:

$$\frac{5\times {10}^{-3} \mathrm{kg}}{1.354\times {10}^{-7} \mathrm{kg}}=3.693\times {10}^4$$

Energy Transfer Power

As derived in Equation (33), the milling-ball frequency $f$ is the reciprocal of the time required from the beginning of a single-ball collision to the beginning of the next collision. The theoretical power of a single-ball collision is the single-elastic collision energy $U_E$ multiplied by the ball-collision frequency $f$, and then multiplied by the number of balls $n_b$ to obtain the energy transfer power $W$ of the entire milling-ball system. The frequency $f$ can be obtained by substituting the experimental data into Equation (35):

$$f=\frac{1}{g_r{v}_n{}^{0.4}{({\rho }_b/E)}^{0.2}{r}_b+\frac{s}{v_n}}=\frac{1}{8.782\times {10}^{-6}+\frac{0.0324}{3.169}}=97.7 \mathrm{Hz}$$

Compared with the collision frequencies of other types of planetary ball mills [1], the frequency of the experimental ball mill falls within the normal range and belongs to a high degree. According to the value of $f$, the time required to transfer energy to all materials at least once is: $\frac{3.693\times {10}^4}{97.7}=378$ s.

This finding indicates that all materials can be subjected to a milling-ball collision for at least 378 s.

By substituting the above calculation results into Equation (36), the milling-ball power of the material can be calculated as follows:

$$W=0.5\eta U_E{n}_{b}f=0.5\times 8.09\times {10}^{-3}\times 30\times 97.7\eta =11.86\eta W$$

Collision is always considered as the main energy transfer mode in the process of the milling ball [22] and the complete positive elastic collision is assumed in the simulation calculation. In addition to the abovementioned collision during the milling ball, only 4–7% of the collision can produce considerable energy conversion [23]. The material adheres to the surface of the ball during the experiment, considerably reducing the efficiency of energy transfer. The particles and fine powder produced during the milling ball gather through electrostatic adsorption and mechanical impact and finally adhere to the surface of the milling ball and spherical tank. This phenomenon is known as “ball wrapping” or “ball pasting”, and considerably reduces the effect of the milling ball. Experiments show that “the material layer with a thickness of 1 mm weakens the impact force to 1/3 of the direct impact”, and “the cushioning effect of 200-μm-thick material on the ball and pipe wall can reduce the impact energy by 80% [24].” This shows that the cushioning effect of the adhesive layer formed by the material consumes a large amount of energy. At the same time, the filling degree of the milling ball substantially affects the milling-ball efficiency. At a small filling degree, the collision frequency between the milling ball and tank wall is greater than that between balls [25]. The research shows that the highest efficiency of the milling ball is achieved when the filling degree reaches 35% [26]. In this experiment, the filling degree is approximately 8%. Evidently, many collisions occur between the milling balls and tank wall, and the energy transfer efficiency of this collision is lower than that between milling balls.

Considering the above points, the milling-ball efficiency $\eta$ in this experiment is calculated as 3%. Therefore, the action power of the milling ball on the material is $0.356 \mathrm{W}$, and the power converted to each ball is 0.0119 W. The milling-ball power is lower than that reported. This power value considers the calculated value after the correction of milling-ball efficiency; thus, the simulation calculation result is relatively reliable.

3. System Temperature Change

The temperature change of materials during the milling ball has a far-reaching impact on the process and the formation of intermediate products. The local high temperature generated in the process of the milling ball is accumulated by a continuous milling ball, thereby increasing the temperature of the whole system and changing the treatment conditions of pollutants. In general, the change in temperature has a certain impact on the chemical reaction rate and reaction conditions; therefore, the temperature change during milling is simulated and calculated to investigate the effect of temperature on the degradation process of ball-milled pollutants.

According to the energy conservation and heat calculation formula, the impact energy $U_E$ of the milling ball can be obtained as follows:

$$U_E=C_{p}Q\Delta T$$

(37)

In Equation (36), C_P refers to the weighted average specific heat capacity of materials under the milling ball. The value fluctuates according to different materials and milling-ball additives and is approximately 1 kJ/(kg·K).

Then, the calculation shows that: $C_p\Delta T=\frac{U_E}{Q}=5.975\times {10}^4 {\mathrm{Jkg}}^{-1}$.

ΔT represents the change in temperature, which is calculated as 59.75 K. Maurice et al. [3] showed that a time of 10⁻² s is required for the 50 μm heated particles to cool down, whereas the time from the impact of the experimental material particles with the milling ball to the next action is T_C:

$$T_c=\frac{s}{v_n}$$

(38)

T_c is calculated as 0.01 s, which reaches a magnitude of the order of 10⁻² s, indicating that the material particles have sufficient time to cool after collision and heating up, and that the temperature of the milling-ball system will not continue to increase with the milling ball. The experimental process is inspected as follows: after each run is stopped, the spherical tank is immediately removed, and the temperature in the tank is measured. The measured value is equal to the room temperature. Although the measured temperature is not high, the influence of the instantaneous high temperature generated during the grinding process on the ball-milled materials cannot be neglected. Further research and verification are required to elucidate these aspects.

4. Conclusions

This paper studied the movement of milling balls, derived the energy transfer model in the milling-ball process through simplified modeling, combined the experimental conditions to theoretically analyze the energy transfer power in the process of persulfate-enhanced mechanochemical degradation of organic matter, and verified the accuracy of the established model. The main findings can be summarized as follows:

(1)

Under the experimental milling-ball conditions, considering that a single ball is moving to reach the take-off condition, the maximum speed of the ball is v = 3.169 m/s, and the maximum centrifugal acceleration is 16.84 times the acceleration of gravity.

(2)

Under the experimental milling-ball conditions, the single-elastic collision energy of the milling ball is 8.09 × 10⁻³ J, the collision frequency of the ball mill is 97.7 Hz, and the power of the ball mill on the material is 0.356 W.

(3)

Under the experimental milling-ball conditions, the theoretical calculated time from the impact of the material particles in the milling-ball process to the next action is T_c = 0.01 s, and the material particles have sufficient time to cool after heating up due to the collisions. Therefore, no local high temperature accumulation occurs, and the temperature change of the entire milling-ball system is small.