Question

The total solids production rate in an activated sludge aeration tank is 7240 kg/d on a dry mass basis. It is necessary to maintain a constant solids residence time in the reactor for proper waste treatment. The maximum water content acceptable for disposal of the sludge is 78% and the specific gravity of the solids is 2.5. The density of the water is 1000 kg/m3. What volume of biological sludge from the activated sludge process will require disposal?

Answer 1

Answer:

volume of biological sludge = 28.566 m³ per day

Explanation:

given data

mass of solid = 7240 kg/day

initial moisture content = 78%

solution

here percentage of solid will be

% of solid = 100 - initial moisture content

% of solid = 100 - 78 = 22 %

so that

mass of sludge produced = $\frac{100}{100 - P} M$ kg  per day

put her value

mass of sludge produced = $\frac{100}{100 - 78} 7240$ kg

mass of sludge produced = 32909.09 kg

so

specific gravity of sludge =  $\frac{\rho sludge}{\rho water }$

and as we know that

$\frac{100}{S sludge} = \frac{solid percentage}{S solid} = \frac{water percentage}{S water}$

$\frac{100}{S sludge} = \frac{22}{2.5} = \frac{78}{1}$

$S sludge$ = 1.152

so that

density of sludge = S sludge × density of water

density of sludge = 1.152 × 1000

density of sludge = 1152 kg/m³

so that

volume of biological sludge = $\frac{mass sludge produce}{\rho sludge}$

volume of biological sludge = $\frac{32909.09}{1152}$

volume of biological sludge = 28.566 m³ per day