A 75,000 ft3 clarifier is to be used to treat wastewater. The recycle ratio is 5

Question

A 75,000 ft3 clarifier is to be used to treat wastewater. The recycle ratio is 50%, the sludge volume index (SVI) is 125, and the return activated sludge concentration is 8000 mg/L. The biomass concentration is 3500 mg/L. The combined design flow rate of the primary and secondary clarifiers is 2.5 MGD. After primary treatment, the wastewater has an influent BOD concentration of 200 mg/L and an influent suspended solids flow rate of 200 mg/L. Two secondary clarifiers, each 28 ft in diameter, are then used. After secondary treatment, the effluent BOD concentration is 15 mg/L, and the effluent suspended solids flow rate is 20 mg/L. What is the solids loading rate to the secondary clarifier?

Hint – focus on the parameter you have been asked to calculate (i.e., the solids loading rate). You will not need to use most of the values they have provided you in the calculation

Explanation / Answer

As we have been provided with % Return sludge and it is 50%.

We can calculate mixed liquor suspended solids concentration in the given volume of primary clarifier by,

% Return sludge = (MLSS*100)/((1000000/SVI) -MLSS)

Where, MLSS (mg/l) is mixed liquor suspended solids concentration.

SVI is Sludge volume index.

=> 50 = (MLSS * 100)/((1000000/125) - MLSS)

=> 2*MLSS = 8000 - MLSS

=> MLSS = 8000/3

=> MLSS = 2667 mg/l.

Again we can calculate SV (Total settled solids per liter) by,

=> SV = MLSS * SVI

=> SV = 2667 * 125

=> SV= 333375 mg/l.

Since, % Return sludge rate is 50.

Solids loading rate to the secondary clarifier will be (1/2)*333375 mg/l

i.e 166687.5 mg/l.

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