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Giventhe schematic flow and data of activated sludge below.

Aeration Basin 4

Clarifier 1

Aeration Basin 3

Aeration Basin 2

Aeration Basin 1

Clarifier 2

Influent Effluent

Plant Inflow: 20 MGD Clarifier: 2 MG

BOD5: 250 mg/L Core SS: 600 mg/L

NH3-N: 30 mg/L WAS Conc.: 7,500 mg/L

Aeration Basin: 6 MG WAS Flow: 110 GPM

MLSS: 3,000 mg/L Effluent TSS is negligible

Where: MGD = million gallons per day GPM = gallons per minute

MG = million gallons Core SS = core suspended solids in clarifier

WAS = wasted activated sludge TSS = total suspended solids

MLSS = mixed-liquor suspended solids

Required:

1.

Calculate the hydraulic retention time of the aeration basin. Then calculate the hydraulic

retention time of the clarifier. Express answers in hours.

2.

Calculate the Space Loading to the Aeration Basins

3.

Calculate the F/M ratio

4.

Calculate the SRT and the MCRT for this system. How much do they differ?

5.

Calculate the total oxygen (not air) demand on the system in one day. Use a value of 1.2 lbs of

oxygen per pound of BOD and 4.33 lbs of oxygen per lb of ammonia.

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6.

If the target MCRT is 10 days, what should be the wasting rate be adjusted to? Up or down?

7.

Assume that a total of four aeration basins make up the 6 MG of volume. If one basin is taken

off-line and the MLSS concentration does not change in the other three basins, what is the new

food to microorganism ratio?

8.

If each of the aeration basins is 16 feet deep and 33 feet wide, how long is each basin? Round tothe nearest whole foot.

9.

If there are two clarifiers and they are each 75 feet in diameter, what is the surface overflow

rate? Express your result in GPM/sq. ft.

Solution:

1.

= or = For Aeration Basin:

= 6 20 =0.30 . For Clarifier:

= 2 20 =0.10 . Therefore, HRT for aeration is 7.2 hours and HRT of clarifiers is 2.4 hours

2. = , = .., = . ,

3.

= =

.,. = .

4.

= = ,., . = ,.,. . = . =

= [3,00068.34] [60028.34]7,5000.1584 8.34 = . Note: The bigger the difference between SRT and MCRT means bigger sludge blanket

5.

First Step: Calculate for lbs BOD5& lbs NH3-N coming into the plant

lbs BOD5= (250)(20)(8.34) = 41,700 lbs x 1.2 lbs O2/lb BOD5= 50,040 lbs

lbs NH3-N = (30)(20)(8.34) = 5,004 lbs x 4.33 lbs O2/lb NH3-N= 21,667 lbs

Total oxygen demand = 50,040 + 21,667 = 71,707 lbs O2

6. = +

10 = [3,00068.34] [60028.34] =16,012.8 / = .8.34

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= 16,012.87,5008.34 = .

7.

6 MG for four (4) basins which means 1.5 MG for each basin

=250208.34

3,0003 1.58.34 = .

8.

Basins Dimensions:

==3316If each basins holds 1.5 MG (converted to ft3, 200,535 ft3)

= =200,5353316 = .

9.

The Surface Overflow Rate (SOR) is defined as the amount of flow moving through the area of

the clarifier.

First Step: Find the total available surface area of both clarifier

= = 3.141675 2 = , .

There are two clarifiers, so the total available surface area is 8,835.5 ft2

Next Step: Convert influent flow from MGD to GPM

20 101 (1

1,440 .) = 13,889 = 13,8898,835.5 = . /