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Solutions to Final Clarifier Math

 

The flow through an activated sludge plant is 15 mgd.  The BOD of the primary effluent is 110 mg/L and the BOD of the final effluent is 8 mg/L.  The MLSS is the aeration tank is 2,500 mg/L.  The 30-minute settleable solids test is 225 mL/L.  The TSS of the primary effluent is 115 mg/L and the TSS of the final effluent is 12 mg/L.  The return sludge ratio is 20%.  60,000 gallons of 2% sludge are wasted each day.  The flow is split evenly between two equal-sized rectangular aeration tanks and final clarifiers.  Each aeration tank is 175 feet long, 75 feet wide, and 12 feet deep.  Each final clarifier has a 75 feet diameter, a SWD of 8 feet, and they are 12 feet deep in the center.

  1. The circumference of a final clarifier is ________ feet.

3.14 x 75 ft = 235.5 ft

 

  1. The surface area of a final clarifier is ________ square feet.

0.785 x 75 ft x 75 ft = 4,415.625 sf

 

  1. The volume of a final clarifier is ________ cubic feet.

Cylinder:  0.785 x 75 ft x 75 ft x 8 ft = 35,325 cf

Cone:   (0.785 x 75 ft x 75 ft x 4 ft) ÷ 3 = 5,888 cf

Tank:    35,325 cf + 5,888 cf = 41,213 cf

 

  1. A final clarifier holds  ______ gallons.

41,213 cf x 7.48 gal/cf = 308,273 gal

 

  1. The detention time of a final clarifier is _______ minutes.

(308,273 gal ÷ 9,000,000 gpd) x 1440 min/day = 49 min

 

  1. The detention time of a final clarifier is _______ hours.

(308,273 gal ÷ 9,000,000 gpd) x 24 hrs/day = 0.82 hrs

 

49 min ÷ 60 min/hr = 0.82 hrs

 

  1. The surface settling rate of a final clarifier is _______ gpd/square foot

9,000,000 gpd/4415.625 sf = 2,038 gpd/sf

 

  1. The weir overflow rate of a final clarifier is ________ gpd/ft

9,000,000 gpd/235.5 ft = 38,216 gpd/ft

 

  1. The solids loading rate of a final clarifier is ________ lbs/day/sq ft

2,500 mg/L x 8.34 x 9 mgd =  187,650 lbs/day

 

(187,650 lbs/day)/4415.625 sf = 42.5 lbs/day/sq ft

 

  1. The BOD removed is ______ mg/L.

110 mg/L – 8 mg/L = 102 mg/L

 

  1. The BOD removed is ______ lbs/day.

102 mg/L x 8.34 lbs/gal x 15 mgd = 12,760 lbs/day

 

  1. The BOD discharged by the plant is ______ lbs/day.

8 mg/L x 8.34 lbs/gal x 15 mgd = 1,000.8 lbs/day

 

  1. The TSS removed is ______ mg/L.

115 mg/L – 12 mg/L = 103 mg/L

 

  1. The TSS removed is ______ lbs/day.

103 mg/L x 8.34 lbs/gal x 15 mgd = 12,885 lbs/day

 

  1. The TSS discharged by the plant is ______ lbs/day.

12 mg/L x 8.34 lbs/gal x 15 mgd = 1,501.2 lbs/day

 

  1. The BOD removal efficiency is _____ %.

 [(110 mg/L – 8 mg/L)/110] x 100% = 0.927 x 100% = 93%

 

  1. The TSS removal efficiency is _____ %.

 [(115 mg/L – 12 mg/L)/115 mg/L] x 100% =0.8956 x 100% = 90%

 

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