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tigry1 [53]
3 years ago
13

1)

Engineering
1 answer:
Kryger [21]3 years ago
7 0

Answer:

b

Explanation:

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5 0
2 years ago
The total solids production rate in an activated sludge aeration tank is 7240 kg/d on a dry mass basis. It is necessary to maint
snow_lady [41]

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

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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}

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S sludge = 1.152

so that

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density of sludge = 1152 kg/m³

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volume of biological sludge = 28.566 m³ per day

6 0
3 years ago
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