A wastewater plant is operating at 10MGD with a suspended solids concentration of 200mg/L. It has 6 clarifiers operating in parallel. Each clarifier is 40 foot in diameter and 10 feet deep.  

(a)  How many gallons flow to each clarifier?

     10 MGD/6= 1.67 MGD  

(b)  How many pounds of solids flow to each clarifier?

      Total Solids (lbs) = (MG)(8.34)(mg/L)  = (10)(8.34)(200)= 16,680 lbs

Solids per each clarifier = 16,680 lb/6 clar = 2,780 lb/clar

Or can calculate directly for one clarifier by substituting 1.67 MGD instead of 10  

(c) If the clarifier are 40% efficient in removing the solids. How many pounds of solids are produced per clarifier?

(0.4)(2,780) = 1,112lbs

(d)  How many total pounds of waste are removed by all 6 clarifiers?

 (6)(1,112) = 6,672 lbs  

(e) What is the suspended solids concentration of the wastewater leaving the clarifier?

40% of 200 mg/l solids are removed in the clarifier so concentration of solids left = 0.6*200 = 120 mg/l

 (f) What is the area of each clarifier?

Each Clarifier Area (A) = π (D²)/4  = (0.785)(40²)   = 1,256 ft²

(g) What is the surface loading rate (gpd/ft²) for each clarifier?

     (1,670,000 gpd)/ (1,256 ft²) = 1,329 gpd/ ft²

(h) What is the clarifier perimeter?

P = πD = (3.14) (40) = 126 ft  

(i)  What is the weir loading rate?

1,670,000 gpd) / (126 ft) = 13,254 gpd/ft  

(j)  What is the detention time?

Volume of clarifier = (0.785)(40²)(10) = 12,560 ft³   

(12,560 ft³* 7.4805 gal/ft³   = 93,955 gal

 Detention time = V/Q = 93955 gal/1,670,000 gal/day * 24 hr/day = 1.35 hrs