Comprehension The sludge from the aeration tank of the activated sludge process (ASP) has solids content (by weight) of 2 %. This sludge is put in a s

Question

What is the density of the sludge removed from the aeration tank?
1. 990 kg/m³
2. 1000 kg/m3
3. 1011 kg/m3
4. 1022 kg/m3

Answer 1

Correct Answer - Option 3 : 1011 kg/m3

Concept:

2% of solids (by weight) + 98% of water (by weight) = 100 % of sludge (by weight)

 \({\rm Specific~gravity(G_s)} = \frac{{{\rho _{solid}}}}{{{\rho _{water}}}}\)

Equating the volume 

\(\frac{{100}}{x} = \frac{{98}}{{1 \times {\rho _w}}} + \frac{2}{{{G_s} \times {\rho _w}}}\)

Calculation:

Given:

Solids content = 2 % by weight, Specific gravity of Sludge solid = 2.2

2 % of solids means there is 98% of water in sludge.

2% of solids + 98% of water = 100 % of sludge

Let ‘x’ be the density of sludge

Equating total volume 

 \(\frac{{100}}{x} = \frac{{98}}{{1 \times {\rho _w}}} + \frac{2}{{{G_s} \times {\rho _w}}}\)

\(\frac{{100}}{x} = \frac{{98}}{{1 \times 1000}} + \frac{2}{{2.2 \times 1000}}\)

X = 1011 kg/m²

∴ Density of sludge = 1011 kg/m²