Question

A ballasted flocculation process has two hydrocyclones in operation (two trains) with an influent flow of 6,000 gpm and a microsand slurry recirculation rate of 300 gpm. Six volumes of samples from the train produced the following volumes of settled microsand: 20 mL, 15 mL, 15 mL, 30 mL, 20 mL, and 20 mL. The six volumes of samples collected in the cone averaged 2,000 mL. Calculate Cm, the microsand concentration in the tanks, in grams of microsand per liter (g/L).

Answer 1

The  Cm of the Microsand concentration in the tanks when expressed in grams of Microsand per liter (g/L) is known to be 3.8 g/L.

What is Microsand?

This is known to be a kind of  aggregate, that is said to be exempted from clay and shale and it is one that can rightly fine to pass via a No. 100 (that is 150 µm) sieve.

Note that from the question:

The Average of six samples = 30mg/L.

Then one need to Multiply:

30 by 300gpm x 1 Train x 1700 ( this is the use of the bulk density conversion factor).

Then one need to also Divide by (4000 gpm x 2000 mL) =  1.8 g/L.

Lastly you then multiply by 2,

1.8 g/L x 2  = 3.8 g/L.

Therefore, looking at the solution above, the Cm of the Microsand concentration in the tanks  is seen to be 3.8 g/L.

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