A mineral tailings slurry containing 25 wt % solids (S.G. = 1.7) in water is con
ID: 106836 • Letter: A
Question
A mineral tailings slurry containing 25 wt % solids (S.G. = 1.7) in water is concentrated by gravity sedimentation to remove as much water as possible before being pumped to disposal. The slurry is fed to a continuous thickener at a rate of 360 m^3/h, and the thickened underflow is discharged at 200m^3/h. The thickener has an inside diameter of 16 m in the vertical section. Laboratory batch solids flux data for the tailings slurry is attached. a. By means of the continuous solids flux method, assess the performance of the thickener by determining the solids concentrations in the underflow, the overflow, and in the thickening zone below the feed line of the thickener. b. Although the current operation may be satisfactory, the underflow slurry is not concentrated enough. Your boss has asked you to try the following two options to increase the underflow concentration further: Option 1: increasing the feed rate by 35%, keeping the underflow rate the same; Option 2: decreasing the underflow rate by 35%, keeping the feed rate constant. Which of the above options is feasible and how much increase in the underflow concentration can be achieved? Present your case with detailed calculations of the solids concentrations of all streams leaving the thickener.Explanation / Answer
a. Continuous solids flux method is given as,
G=(u+v)
where u is the normal slurry flow and v is the slurry flow in downward direction.
Rate of feeding slurry=360m3/h
Thickened underflow discharge=200m3/h
Performance of the thickener=(rate of slurry feeding+underflow discharge)×diameter of thickener
=(360+200)×16=8960m4/h
b.increasing feed rate by 35% and keeping the underflow rate the same would be better because the process if feeding slurry cannot be stopped and increase in the underflow rate causes deterioration in the process as the amount fed at upward direction decreases and the rate of feeding would not be constant.
Increasing the feed rate by 35%=360/(360+35)=0.91*100=91%
Hence the performance of the feeder can also be increased by 91%
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