calculate the surface dimensions of sedimentation

Question

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) A_s = settling zone surface area, ft² Q = flow rate = 8,000,000 gal/d

q₀ = overflow rate =650 gpd/ft²

Surface area of settling tank A_s = Q/q₀ = 8*10⁶ / 650= 12307 ft²

A_s = L w and Given L = 4 w

Since 3 tanks are used, A_s 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 L_w = (Q/tan K)/q_w

L_w = overflow weir length (ft)           

q_w = weir overflow rate (gpd//ft)

Q/tank = flow rate per tank (gpd) = 8MGD/3 = 2.67 MGD

L_w = (Q/tan K)/q_w = 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