Where w is the angle of the cylinder and t 2 is the time the ball moves on the parabolic trajectory. The ball cross-sectional area is therefore [next] The following is an example of the use of the formula (11).
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The value is less than zero, so H at the time of a = 55° 44' is the maximum value.
This method is more reasonable than the former because all ball loads are considered. This method is used in the production of mills with higher rotational speeds.
Second, the number of cycles of the ball load
The steel ball is forced to produce motion, and during the analysis of this motion, the trajectory of the steel ball is described. The purpose of studying steel ball kinematics is to establish the basis for determining the speed, ball loading rate, useful power and productivity. From the point of view of the grinding effect produced by the steel ball, these are necessarily related to the number of impacts of the steel ball on the ore.
The mill has always been a circular motion, but the movement of the steel ball is partly a circular curve and the other part is a parabola. Therefore, the rotation of the mill, the movement of the steel ball is not necessarily a cycle. The steel ball is thrown from the point of departure A to the point B, which is faster than the circular motion, so the steel ball is always ahead. In other words, when the mill turns one turn, the steel ball does not only circulate once. [next]
Let t 1 be the time for the steel ball to make a circular motion. When the mill turns one turn, the center angle of the steel ball rotated by the same speed for circular motion is
Therefore, the time required for the steel ball to move one cycle is
Let φ be the ball loading rate, and the volume of the ball inside the mill is . If the mill turns one turn and all the balls are cycled J times, then
Figure 1 Turnover rate of all ball loads in the mill
Figure 2 Relationship between J, Φ , φ and K Therefore, the value of R 2 (or K calculated by it) depends on the rotational speed n and has a limit value. Beyond this minimum, the extra balls can't be parabolic. Therefore, J in the above formula is related to the loading rate φ and the K value in relation to the rotational speed of the mill. As shown in Figure 2. The curve in the figure shows that the larger the rotation rate, the larger the K value and the smaller the J value at the same loading rate. If the transfer rate is the same, the more the ball loading rate, the smaller the K value and the larger the value. This is the relationship between the number of rotations, the amount of ball loading and the number of cycles of the ball affecting the impact, which clarifies the importance of correctly determining the speed of the mill. [next]
Example: set the mill diameter to D, its revolutions per minute is Try formula (11) to calculate its K, Corresponding value.
Solution: The steps to calculate such a problem are as follows:
From the data given in the question, obtain: [next]
Figure 1
First, the method of determining the speed of the mill
In the grinding machine of the throwing work, the grinding action mainly comes from the kinetic energy of the steel ball reaching the falling point. The magnitude of this kinetic energy is related to the falling height (H) of the steel ball, and the falling height of the steel ball is determined by the rotational speed of the mill. According to the previous discussion, it is inevitable to determine the appropriate rotational speed of the mill from the maximum drop height of the steel ball. Let it be equal to zero, you can find the angle of departure when the steel ball has the maximum parabolic drop height, and then find the corresponding speed from it.
Substituting a=54°44′
There are two ways to use the value and u t =0 (ie u p =u n ). The first type uses the outermost ball to have a maximum drop height to determine the rotational speed. If the radius of the outer sphere is the inner radius of the mill,
This method only considers that the outermost ball is in a proper state, and the other layers are not necessarily in a suitable state. The more balls are loaded, the more unsuitable the ball layer is. Therefore, it is not a reasonable method to obtain a lower transfer rate. The ball is guaranteed to be thrown down, and the mill with lower speed is now used.
The second is calculated by the relationship between the radius of gyration of the ball and the angle of departure. Imagine that the mass of all ball loads is concentrated in a certain layer of ball. The upper ball can represent the total ball load. Its radius of the ball layer (R 0 ) is the radius of gyration of the center of the ball mill around the center of the mill ( O ). According to the method of the fan shape to the pole moment of inertia of the O point,
When the ball has the maximum drop height, a 0 =54°44', thus
Let t 2 be the time when the steel ball is used as a parabola, and take the point A c as the coordinate origin.
The number of cycles in which the mill turns a steel ball is
It can be seen that the number of cycles of the steel ball depends on the angle of departure a C. When the rotational speed of the mill is constant, different spherical layers have different exit angles, and the number of cycles is different. The higher the rotational speed of the mill, the smaller the a C and the fewer the number of cycles. When the steel ball is centrifuged, a C =0, J=1, and the steel ball is attached to the liner to rotate together with the mill. [next]
From the perspective of all ball loads, as shown in Figure 1, the volume of the ball passing through the section AB along a circular path is within one revolution of the mill.
Available formula
The relationship between the ball loading rate (φ) and the ball load-cutting area (Ω) is
Ball Heqie cut area comprises an area for movement of the cutting area of the circular curve (Ω 1) and for the parabolic movement (Ω 2) in two parts, i.e.,
Therefore, Ω 1 and Ω 2 must be obtained first.
As shown in the figure below, take a sphere layer whose radius is R. The center angle θ of the circular arc of the layer ball from the falling point B to the point of departure from the point A is
When the change in the radius of the sphere is dR, the change in the area of ​​the arc is
Φ C and K C values ​​of various filling ratios in Schedule 2
Φ%
Φ C %
K C %
Φ%
Φ C %
K C %
0
30
35
52.9
68.3
71.3
1.000
0.603
0.550
40
45
50
74.8
78.3
81.8
0.501
0.458
0.419