Most flotation plants use a jaw thickener as the first stage of dewatering equipment . Its calculation method is mainly calculated based on the production capacity per unit area and the sedimentation speed of the largest particles in the overflow in water. An example of its calculation method will now be given.
Copper ore processing plant embodiment of a copper concentrate thickener solution to the ore concentration: Solid = 5.7: 1, is required to concentrate the liquid: solid = 1: 1, the amount of processing plant daily output of 100 tons of copper concentrate, Q. How thick is the thicker machine?
It can be calculated in two ways:
(1) Calculate the required area according to the production capacity per unit area. Its calculation formula is:
Q
F = ——, m 2
q
Where F is the area of ​​the thickener required, m 2 ;
Q——The amount of solids fed into the thickener, tons; this example is entitled 100 tons/day;
q——Production capacity per unit area, t/m 2 · day.
Generally selected based on the index test or the like with reference to the beneficiation plant. This example uses a similar copper selection factory indicator, taking q = 0.5 tons / m 2 · day.
Therefore, the area of ​​the thickener required for this example is:
Q 100
F = —— = —— = 200 (m 2 )
q 0.5
The formula for the diameter of the thickener is:
Where D is the diameter of the thickener required, m;
F——The area of ​​the thickener.
The diameter of the thickener required for this example is:
(2) Calculate the required area according to the sedimentation velocity of the largest particle in the overflow. Its calculation formula is:
Q(R 1 - R 2 )K 1
F = ————————
86.4×υ o ×K[next]
Where Q is the amount of solids fed into the thickener, t/d, in this case, 100 tons/day;
R 1 , R 2 - the weight ratio of the slurry liquid to the solid before and after concentration. In this example, R 1 = 5.7; R 2 = 1;
K——The effective area factor of the thickener is generally 0.85~0.95. For small dense (ф<12m), take a small value, and the large thickener should take a large value. This question is 0.95. ;
K 1 —— The fluctuation coefficient of mineral quantity is 1.05~1.20, that is, considering the fluctuation of ore feeding of 5~20%, this question takes 1.10;
υ o ——The free settling velocity (Home/sec) of the largest 0.031 large particles in the overflow is usually obtained by experiment. If there is no test data, it can be obtained according to the Stokes settlement final velocity formula, namely:
υ o = 545(δ T - 1)d 2 , mm / sec
Where d is the maximum diameter (mm) of the solid particles allowed to be carried away in the overflow. For non-ferrous metal ore, the general requirements d≯5 mm,
For rare metal concentrates, d≯2~3 microns is required. This question takes d to be 5 microns, ie 0.005 mm;
δ T - relative density of ore, this topic is 3.3.
So the title of this example is = o = 545 × (δ T - 1)d 2
= 545 × (3.3 - 1) (5 × 10 -3 ) 2
= 0.031 mm / sec The area required for the thickener is:
Q(R 1 - R 2 )K 1
F = ————————————
86.4Ï… o K
100 (5.7 - 1) × 0.95
= ———————————
86.4×0.031×1.1
= 151.5 m 2
After determining the area of ​​the thickener, it is also necessary to correct it as follows. υ should be less than υ o ; the correction formula is:
V
υ = —— ×1000
F
Where V is the overflow of the thickener (m 3 / day), ie Q (R 1 -R 2 );
υ——The speed of the water rising for the thickener, mm/s.
V Q(R 1 - R 2 ) 100(5.7 - 1)
So υ = —— = ———————— = ————————
F F 151.5
= 0.0359 (mm/s)
Since 0.0359>0.031, the solid particles carried out in the overflow will have a particle size greater than 5 microns. Description F = 151.5 meters 2 is small. Reselect V
F=200 m 2 , press υ = ——calculate, get υ = 0.027 mm / sec, then υ < υ o , that is, the rising flow rate of the thickener overflow water is less than the overflow F
The settling velocity Ï… o of the solid particles allowed to be carried away indicates that the solid particle size overflowing in the overflow meets the requirements and will not cause serious metal loss in the future.
Therefore, the calculation of this example is based on the Ñ„16 m thick machine.
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