1. Flotation Kinetics 0-t1, R1 t1-t2, R1+R2 t2-t3, R1+R2+R3
t3-t4, R1+R2+R3+R4
t4-t5, R1+R2+R3+R4+R5 R1 R2 R3 R4 R5 R1 R3 R4 R1 R2

2. R=R(1-e-kt) R=kt/(1+kt) Flotation Kinetics 0-t1, R1 t1-t2, R1+R2
t2-t3, R1+R2+R3
t3-t4, R1+R2+R3+R4
t4-t5, R1+R2+R3+R4+R5 R1 R2 R3 R4 R5
R=R(1-e-kt)
R=kt/(1+kt)
R recovery
t time
k rate constant

3. Flotation Bank Design
Mineral
% solids in feed
Retention time (min.)
No. of cell/bank
Barite
30-40
8-10
6-8
Copper
32-42
13-16
8-12
Flourspar
25-32
Feldspar
25-35
Lead
Molybdenum
35-45
14-20
10-14
Nickel
28-32
8-14
Phosphate
30-35
4-6
4-5
Potash
Tungsten
7-10
Zinc
Silica (iron ore)
40-50
Silica (phosphate)
Sand
7-9
Coal
4-12
effluents
As received
6-12
For cleaning applications 60% of the rougher percent solids.
Required retention time for cleaning is approx. 65% of rougher retention time.

4. Example min. g %Pb Pb mass Cum Pb Mass Pb Rec. Cum. Pb Rec. Con 1 0-2
83.28
83.46
69.51
64.99
Con 2
2-4
44.33
43.41
19.24
88.75
17.99
82.99
Con 3
4-8
60.99
15.78
9.62
98.37
9.00
91.99
Con 4
8-16
106.20
4.02
4.27
102.64
3.99
95.98
Tail
934.80
0.46
4.30
106.94
100.00
Total

5. Flotation Cell Selection
Sizing of the cells depends on the retantion time in flotation.
Q = V / t
Q: volumetric flowrate (m3/h)
V: volume (m3)
t: hours
Q=m3/h
V=m3
t=?

6. Flotation Cell Selection
Q=m3/h
Vf: total volume required (m3)
Q: volumetric flowrate (m3/h)
Tr: retention time
S: Scale up factor
Tr specified by customer S=1
Tr taken from continious pilot plant test S=1
Tr taken from typical industrial data S=1
Tr taken from laboratory scale test work S=
Ca= Aeration factor to account for air in pulp. = 0.85
V=m3
t=?

7. Flotation Cell Selection
The selection of the size, number and type of flotation cells for a particular application depends on
two important factors:
- required flotation residence time; and
- dry solids recovery rate for a given:
- froth surface area (t/hr/m2), and
- concentrate lip length (t/hr/m).
Residence time is influenced by ore type - mineralogy, associations, liberation, kinetics, and reagent
additions (collector and frother). In terms of the amount of concentrate to be recovered, selection also depends on ore type and characteristics such as particle size, specific gravity, water recovery, and mineral grade. The stage of flotation is important.
The rougher stage recovers 10% to 20% of the solids in the feed while scavengers recover 5% to 15% while the cleaner mass recovery can approach 80%.
Lab-scale tests and scale-up factors determine the residence time required in the plant.

8. Flotation Cell Selection
However, the froth carry-over flow rate (dry concentrate per froth surface area per hour, t/m2/hr) and
lip loading rate (dry concentrate per froth lip length per hour, 1.5 t/m/hr minimum and maximum) are
difficult to scale-up from the lab because the froth is continuously scraped-off during a lab test and the
type of process is batch rather than continuous flow through.
So these values are determined via calculation to give ranges that should not be exceeded by cell size selection and layout. With cleaner cells where a high percentage of the solids are recovered into the froth phase, carry-over flow rate and lip loading rate are much more important than residence time.
Application Rougher Scavenger Cleaner
Carry-over rate (t/hr/m2)

9. Flotation Cell Selection
In most cases, a quick estimate of these carry-over constraints can be obtained by limiting the residence time in any single cell to approximately 1.0 minutes. If the calculated value lies between 0.75 to 1.25 minutes, the circuit should perform well. However, should the design fall outside these ranges, then one should perform a detailed calculation of carry-over rate per surface area and lip length.
If the residence time requirement is met with a certain cell size selection but froth carry-over rate
and/or froth lip loading rate are exceeded, one must increase the number of flotation cells (i.e., select a
smaller size cell) to increase the surface area or concentrate lip length. Unfortunately this solution can
add to capital costs as well as floor space and operating and maintenance (O&M) costs.
Launder Configurations
- Central donut launder
- Internal launders
- External launders (smaller cells < 50 m3)
- Radial launders

10. Flotation Cell Selection
Froth Depth / Pulp Level
Froth depth is also an important factor to achieve good quality/quantity trade-off.
Generally the roughers are controlled to a froth depth between 15 to 30 cm, while the scavengers will range between 5 to 10 cm. Cleaner cells will generally be operated with the highest froth depth possible - perhaps as much as 20 cm, and on some column (or vertically configured) cells, froth depths as much as 1.2 m have been observed. At these depths, wash water onto the froth phase is generally required to sustain the froth. If the flow rate of concentrate becomes very low (changes in mineralogy or ore tonnage), then the froth may require reagent changes (amounts and type) to be sustainable.

11. Flotation Cell Selection
Example of Sizing a Flotation Circuit:
Design for 100,000 tpd of S.G. 3.0 and 40% solids
Calculate the volumetric flow rate of pulp into the cell:
Solids volumetric flow rate = (100,000/3.0)/(24*60) = m3/min
Water volumetric flow rate = (100,000/1.0)*(0.6/0.4)/(24*60) = m3/min
So the total volumetric flow rate = Q = m3/min
From lab testwork, the scaled-up required flotation time = 12 minutes
So, the total required plant volume = V = m3/min* 12 min = 1, m3
Adjust for impellor, air hold-up, and froth phase = V * 1.2 = 1,833.4 m3

12. Flotation Cell Selection
Example of Sizing a Flotation Circuit:
Try a unit cell size of 25 m3
Total number of cells required = 1,833.4 / 25 = 73.3 cells
Number of parallel banks required to achieve 1 min residence time per cell = m3/min* 1 min * 1.2 / 25 = 6.11 banks
For 6 banks: residence time / cell = 6 * 25 / ( * 1.2) = 0.98 minutes (excellent)
Design: 6 banks of 12 cells = 72 * 25 = total volume of 1,800 m3 to give a total residence time of 11.8 minutes

13. Flotation Cell Selection
Example of Sizing a Flotation Circuit:
Try a unit cell size of 50 m3
Total number of cells required = 1,833.4 / 50 = 36.7 cells
Number of parallel banks required to achieve 1 min residence time per cell = * 1 min * 1.2 / 50 = 3.05 banks
For 3 banks: residence time / cell = 3 * 50 / ( * 1.2) = 0.98 minutes (excellent)
Design: 3 banks of 12 cells = 36 * 50 = total volume of 1,800 m3
to give a total residence time of 11.8 minutes

14. Flotation Cell Selection
Example of Sizing a Flotation Circuit:
Try a unit cell size of 75 m3
Total number of cells required = 1,833.4 / 75 = 24.4 cells
Number of parallel banks required to achieve 1 min residence time per cell = * 1 min * 1.2 / 75 = 2.04 banks
For 2 banks: residence time / cell = 2 * 75 / ( * 1.2) = 0.98 minutes (excellent)
Design: 2 banks of 12 cells = total volume of 1,800 m3
to give a total residence time of 11.8 minutes

15. Flotation Cell Selection
Example of Sizing a Flotation Circuit:
Try a unit cell size of 100 m3
Total number of cells required = 1,833.4 / 100 = 18.3 cells
Number of parallel banks required to achieve 1 min residence time per cell = * 1 min * 1.2 / 100 = 1.53 banks
For 1 bank: residence time / cell = 1 * 100 / ( * 1.2) = minutes (too fast)
Try 2 banks = 2 * 100 / ( * 1.2) = 1.31 minutes (too slow)
Design: 2 banks of 9 cells = total volume of 1,800 m3
to give a total residence time of 11.8 minutes

16. Flotation Cell Selection
Example of Sizing a Flotation Circuit:
Try a unit cell size of 150 m3
Total number of cells required = 1,833.4 / 150 = 12.2 cells
Number of parallel banks required to achieve 1 min residence time per cell = * 1 min * 1.2 / 150 = 1.02 banks
For 1 bank: residence time / cell = 1 * 150 / ( * 1.2) = 0.98 minutes (excellent)
Design: 1 banks of 12 cells = total volume of 1,800 m3 to give a total residence time of 11.8 minutes

17. Flotation Cell Selection
So there are several possible answers (there may be others):
The most flexible design is for a cell size of 25 m3 - 6 banks of 12 cells per bank
It gives the highest operating flexibility, but will use more energy and have increased maintenance costs.
If one bank must be shut down for maintenance, 83% of the plant capacity is still available.
The least expensive design is a cell size of 300 m3, however, the mill will have no flexibility and the cells will be operating very slowly which will lead to level control problems.

18. Flotation Cell Selection
Summary
Cell Residence Number Number Total Total
Size Time/Cell of of Cells Volume Residence
(m3) (min.) Banks per bank (m3) Time (min.) , , ,
XX , ,
XX ,
XX ,
XX ,
Probably the best trade-off between capital costs and operating flexibility would be the 75 m3 design.

19. Flotation Cell Selection

20. Flotation Cell Selection

21. Flotation Cell Selection

22. Problem: As an example of sizing flotation cells, consider the following copper ore:
Specific gravity of dry ore = 2.7
Optimum Percent Solids in Laboratory Flotation Machine = 30%
Optimum Laboratory Flotation Retention Time = 6.5 minutes
Assume that the flotation facility will need to process 14,500 metric tons of dry ore per day.

23. Solution:
First, from Table 2, we see that the average laboratory flotation time for copper is 7 minutes, and the average plant flotation time is 14.5 minutes.
The scale-up factor is 14.5/7 = 2.07, and so the time needed for flotation in the plant for this particular ore is (6.5)(2.07) = 13.5 minutes.
A feedrate of 14,500 metric tons dry ore/day will require a flow of (14,500/1440
min/day = metric tons/minute,
and so for a 13.5 minute residence time there will need to be
(13.5)(10.07) = 136 metric tons of ore retained in the flotation bank.

24. Since the slurry must be 30% solids, 136 dry tons of ore will equal (136/0.3) = 453 metric tons of slurry.
Since the ore has a specific gravity of 2.7, the specific gravity of the slurry can be
calculated from the relation
(100/slurry) = (X/solids) + ((100-X)/water),
where X is the percent solids, and slurry, solids, and water are the specific gravities of the slurry, solids, and water,respectively.
We therefore have (100/slurry) = (30/2.7) + (70/1.0), and solving for slurry we find
29 that the specific gravity of the slurry is The total volume of the 453 tons of slurry that the cells must hold will be (453 metric tons)/1.23 = 368 cubic meters. However, when the pulp is aerated it will consist of approximately 15% air and 85% slurry by volume, so the actual volume of pulp needed will be 368/(0.85) = 433 cubic meters.

25. Cells of cubic meters (500 cubic feet) are a standard size, and so if these are used we will need 433/14.15 = 30.6 cells,
or approximately 31 cells.
Referring to Table 3, we see that for copper we should use between 12 and 17 cells per bank, with 14 to 17 cells being optimum.
Two banks of 16 cells would give 32 total cells, which would provide the 31 cells needed with some extra capacity.
The cells would then be arranged in the individual banks as per the
manufacturer’s recommendations.