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?

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DeniceSandidge DeniceSandidge · Answer 1 · 2019-11-04T10:31:34+01:00

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 = [tex]\frac{100}{100 - P} M[/tex] kg per day

put her value

mass of sludge produced = [tex]\frac{100}{100 - 78} 7240 [/tex] kg

mass of sludge produced = 32909.09 kg

so

specific gravity of sludge = [tex]\frac{\rho sludge}{\rho water }[/tex]

and as we know that

[tex]\frac{100}{S sludge} = \frac{solid percentage}{S solid} = \frac{water percentage}{S water}[/tex]

[tex]\frac{100}{S sludge} = \frac{22}{2.5} = \frac{78}{1}[/tex]

[tex]S sludge[/tex] = 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 = [tex]\frac{mass sludge produce}{\rho sludge}[/tex]

volume of biological sludge = [tex]\frac{32909.09}{1152}[/tex]

volume of biological sludge = 28.566 m³ per day