Calculate the surface dimensions of Sedimentation Tank
A treatment plant upgrade calls for design of sedimentation basins to handle a flow rate of 8 MGD. The basins are for a Type I suspension with an overflow rate (q0) of 650 gpd/ft2 a length to width ratio of 4:1, a weir overflow rate (qw) of 20,000 gpd/ft, The minimum settling zone depth is 10 ft and 1.5 ft is allowed for freeboard.
(a) If 3 tanks are chosen, what are the surface dimensions of each tank?
(b) What is the required weir length for each tank?
(c) What is the detention time
(a) As = settling zone surface area, ft2 Q = flow rate = 8,000,000 gal/d
q0 = overflow rate =650 gpd/ft2
Surface area of settling tank As = Q/q0 = 8*106 / 650= 12307 ft2
As = L w and Given L = 4 w
Since 3 tanks are used, As per tank = 12307 / 3 = 4102 ft2 = LW= (4W)W = 4W2
W=32f t L = 4W = 128 ft Thus, L = 128 ft and W = 32 ft
(b) Overflow weir length Lw = (Q/tan K)/qw
Lw = overflow weir length (ft)
qw = weir overflow rate (gpd//ft)
Q/tank = flow rate per tank (gpd) = 8MGD/3 = 2.67 MGD
Lw = (Q/tan K)/qw = 2.67 * 106 gpd/ 20000 gpd/ft = 133.4 ft
(c) detention time = V/Q = [128*32*10 ft3] /2.67MGD
= 40960 ft3 / [2.67*106 gal/day* ft3 /7.48gal]
= 0.11475 days = 2.75 hours