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