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The maximum particle diameter d allowed in the overflow; d=0.005mm for concentrate or medium ore, d=0.01mm for gangue minerals;
Concentrator selection and calculation
Wet beneficiation usually requires dehydration of some products (products). Most of the fine-grain concentrate dewatering uses concentration-filtration two-stage or concentration-filtration-drying three-stage operation; tailings generally need to be concentrated and transported to the tailings pond; sometimes, the intermediate products (middle mines) in the sorting process are also required. Concentrate to meet the concentration requirements of the next stage of operation.
First, the choice of thickener
At present, the most widely used at home and abroad is still the ordinary mechanical discharge concentrator. In recent years, high-efficiency concentrators, inclined plate concentrators, inclined plate concentrators, inclined plate thick boxes, etc. have also been developed and applied.
The ordinary concentrator has two kinds of center drive and peripheral drive. The central drive concentrator can automatically raise the raft, which is produced and applied abroad (including large-diameter concentrators). The maximum diameter of the foreign central drive concentrator is 20m. The peripheral drive is divided into two modes: roller drive and rack drive. The domestic roller drive concentrator is 15~30m in diameter and is mainly used for processing materials with low density. The domestically produced 45m, 50m and 53m peripheral drive concentrators are available in light and heavy duty, light for tailings and heavy for processing concentrates.
The high-efficiency concentrator is equipped with special equipment for flocculant addition and mixing, and the sedimentation speed is high; some have no inclined plates to increase the settlement area. Efficient concentration per unit area is high, and can be 7~8t/(m 2 ·d). The equipment is small in size and should be used in small and medium-sized concentrators, especially in the case where the plant area is narrow and the concentrator must be installed indoors.
The sloping plate concentrator uses the shallow sedimentation principle to shorten the material settling time, increase the settlement area, increase the concentration efficiency, and the processing capacity per unit area is higher. However, the structure of the inclined plate is complicated, and maintenance and repair are inconvenient. This equipment is also used in some concentrators.
The inclined thick box also applies the principle of shallow settlement, which has high concentration efficiency and no moving parts. The disadvantage is that the discharge port is easy to block and the concentration of ore is low. The equipment can be used for dewatering, de-sludge and concentration of small ore processing plants.
The magnetic separation plant strong magnetic mineral concentrate is generally concentrated by a magnetic separator or a magnetic dehydration tank.
Concentrating device type, select the size and the number of work stations to meet the requirements on their lower concentration of discharge, but also to meet the requirements for reducing the metal overflow and overflow drain water quality. The sedimentation speed of the maximum solid particles allowed in the overflow of the concentrator is an important factor in determining the concentration area. It is related to the material density and particle size composition, the ore and discharge ratio, the slurry viscosity, the flotation agent and flocculant type, the slurry temperature, etc. Related factors.
Second, the calculation of the thickener
The concentrator calculation has the following method
A is calculated according to the unit area throughput indicator
It is preferred to calculate the total area of ​​the thickener required as follows, and then determine the size and number of the thickener.
Where A - the total area of ​​the thickener required, m 2 ;
Q——the amount of solids fed into the thickener, t/d;
q o - the throughput per unit area, t / (m 2 · d), selected according to industrial tests or simulation tests. If there is no test, please refer to the actual indicator of similar concentrator or Table 1 (according to this, you can only do rough calculation).
Table 1 Concentrated unit per unit area processing amount q o value
Concentrated product name
Qo/t·m -2 ·d -1
Concentrated product name
Qo/t·m -2 ·d -1
Mechanical classifier overflow (before flotation)
0.7~1.5
Flotation iron concentrate
0.5~0.7
Lead oxide and lead concentrates - Copper Concentrate
0.4~0.5
Magnetically selected iron concentrate
3.0~35
Lead sulfide concentrate and lead-copper concentrate
0.6~1.0
White tungsten ore flotation concentrate and medium mine
0.4~0.7
Copper concentrate and copper-bearing pyrite concentrate
0.5~0.8
Fluorite flotation concentrate
0.8~1.0
Pyrite concentrate
1.0~2.0
Manganese concentrate
0.4~0.7
Molybdenite concentrate
0.4~0.6
Barite flotation concentrate
1.0~2.0
Zinc concentrate
0.5~1.0
Flotation tailings and medium mines
1.0~2.0
Antimony concentrate
0.5~0.8
Note: 1. The qo value in the table refers to the value when the ore particle size is -0.074mm, which accounts for 80~95%. When the particle size is coarse, the value is large.
2, the concentration of the other row of ore aluminum ore, sulfide ore pyrite concentrate, copper and zinc is no greater than 60 to 70%; no more than 60% other concentrates.
3. For the fine mud oxidized ore with more mud, the listed indicators should be appropriately reduced.
B Calculated according to the maximum particle sedimentation velocity in the overflow
First calculate the total area of ​​the concentrator required as follows, and then determine the size and number of concentrators.
Î¥o =545(Ï-1)d 2 (2a)
Wherein R 1 , R 2 are the solid-to-weight ratio of the slurry before and after concentration;
K 1 - the coefficient of fluctuation of the slurry, K 1 = 1.05 ~ 1.20;
K——the effective area coefficient of the thickener, generally taking K=0.85~0.95;
υ o —— the free settling velocity of the largest particles in the overflow, mm/s, υo is generally obtained by experiment, and no test can be calculated according to formula (2a).
A,q——same form (1).[next]
C Calculated according to the results of the settlement test
The sedimentation test needs to take a representative slurry sample and carry it out in the measuring cylinder. The slurry sample is allowed to stand after being thoroughly mixed, and the relationship between the height of the sediment line and the time is plotted with the settlement of the solid material. Based on this, calculate the concentrated area required for the unit throughput, and then calculate the total area of ​​the thickener, and determine the size and number of the thickener. There are three methods for calculating the concentrated area using the results of the settlement test. The Coe-Clevinger method is relatively old, and the trial and calculation process is cumbersome and has been rarely used. Now introduce the two methods.
a TalmadgrFitch
The main steps of the law are as follows:
(1) Find the compression point based on the height-time curve of the sediment line obtained by the test (see Figure 1). The compression point is the point where the sedimentation velocity is slower and slower, and the slope of the sedimentation curve changes significantly at this point.
(2) Draw the tangent of the sedimentation curve at the compression point (Talmej-Fitch plot line).
(3) Calculate the height of the sediment line Hu corresponding to the required underflow concentration (solid weight per unit of slurry volume) according to the following formula, and draw a horizontal line on the graph.
Hu=HoCo/Cu (3)
Where Hu is the height of the sediment line corresponding to the underflow concentration, cm;
Ho——the initial pulp height in the measuring cylinder, cm;
Co——initial pulp concentration (solid mass per unit volume of pulp), g/L;
Cu - underflow concentration (solid mass per unit volume of pulp), g/L.
(4) Find the time on the corresponding abscissa of the Hu horizontal line and the Talmej-Fich plot line intersection point on the graph. Then use the following formula to calculate the concentrated area required for the unit throughput.
A=t u /(HoCo) (4)
Where A is the concentrated area required for unit throughput, m 2 /(t·d-1);
t u ——to reach the underflow concentration time, d;
Ho - initial pulp height, m;
Co - initial pulp concentration (solid volume per unit volume), t/m 3 .
If the position of the compression point in the sedimentation curve is not obvious, the tangent line can be drawn by several points, and the concentrated area required for the unit throughput is calculated separately, and the largest one is taken.
b Oltmann method (Oltmann)
This method is an improved method of the Talmej-Fich method. It also uses the sedimentation test to obtain the relationship between the height of the sediment line and the time. From the starting point of the curve Ho, draw a straight line through the compression point Cp (see Figure 1), and the required underflow. The height line Hu with the same concentration intersects, and the time ty on the abscissa corresponding to the intersection point is the settling time to reach the underflow concentration. Then, the concentrated area required for the unit throughput is calculated by the following formula.
A=1.2t y /(HoCo) (5)
Where A is the concentrated area required for unit throughput, m 2 /(t·d -1 );
t y —— the settling time to reach the underflow concentration, d; other symbols are the same as (4).
Example 1 It is known that the concentration of the concentrate machine (solid content per unit volume of pulp) is 50g/L, the underflow concentration is required to be 340g/L, the initial settlement height is 34cm, and the sedimentation curve is shown in Figure 1. Try the Talmeiji-Fitch method and The Altman method calculates the concentration area required for unit throughput.
Solution 1 with the Talmej-Fich method
The sediment height corresponding to the required underflow concentration:
Hu=HoCo/Cu; Hu/cm=34×50/340=5
A horizontal line of Hu = 5 cm is drawn in the figure.
Determine the compression point on the sedimentation curve as Cp, and draw a tangent line at this point. The tangent line corresponding to the intersection of Hu=5cm horizontal line tu=31.5min or 0.022d, then the concentrated area required for unit throughput is:
A=tu/(HoCo); A/m 2 ·t -1 ·d -1 =0.022/(0.34×0.05)=1.29
Solution 2 using the Altman method
From the initial settlement height Ho, the straight line intersects the Hu horizontal line through the compression point Cp, and the corresponding abscissa time ty=22min or 0.0153d, the concentrated area required for the unit throughput is:
A=1.2ty/(HoCo); A/m 2 ·t -1 ·d -1 =1.2×0.0153/(0.34×0.05)=1.08