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

volume of biological sludge = 28.566 m³ per day

Step-by-step 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 =
`(100)/(100 - P) M` kg per day

put her value

mass of sludge produced =
`(100)/(100 - 78) 7240` kg

mass of sludge produced = 32909.09 kg

so

specific gravity of sludge =
`(\rho sludge)/(\rho water )`

and as we know that

`(100)/(S sludge) = (solid percentage)/(S solid) = (water percentage)/(S water)`

`(100)/(S sludge) = (22)/(2.5) = (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 =
`(mass sludge produce)/(\rho sludge)`

volume of biological sludge =
`(32909.09)/(1152)`

volume of biological sludge = 28.566 m³ per day