Question

Please determine the dimensions and air flowrate required for a 3m deep aerated grit chamber designed to remove 70% of 0.02 cm diameter sand [specific gravity = 2.65] at 20ºC. Assume that only particles of 0.001 cm will be scoured away. The wastewater flow rate is 10000 m3/d. Use a b of 0.04 and f of 0.03 for calculations.

Answer 1

Ans 1) Calculate settling velocity ,Vs

Vs = (1 / 18 ) (G - 1) g

where, = Kinematic viscosity of water = 10^(-6) m2/s

G = Specific gravity of sand particles = 2.65

D = Diameter of sand particles = 0.02 cm or 0.0002 m

Putting values,

=> Vs = [1/(18 x )] (2.65 - 1) x 9.81 x (0.0002 x 0.0002)

=> Vs = 0.0359 m/s

Calculate Reynold number,Re

Re = Vs D / = 0.0359 x 0.0002 /

=> Re = 7.18 > 1

Since, Reynold number is more than 1, we cannot use the above equation to calculate settling velocity so solution is iterative. Hence, use ,

   Vs =  [(4 / 3 Cd) (G - 1) g D]^0.5

where Cd = Drag coefficient = 18.5 /

Iteration 1:

Cd = 18.5 / = 18.5 / = 5.69

=> Vs = [(4 / 3(5.69)) (2.65 - 1) x 9.81 x 0.0002]^0.5

=> Vs = 0.0275 m/s

Iteration 2

Again, Re = 0.0275 x 0.0002 /   

=> Re = 5.5   

Cd = 18.5 / = 18.5 / = 6.65

=> Vs =  [(4 / 3(6.65)) (2.65 - 1) x 9.81 x 0.0002]^0.5

=> Vs = 0.0254 m/s

Iteration 3 :

Again, Re = 0.025 x 0.0002 /   

=> Re = 5.08

=> Cd = 18.5 / = 18.5 / = 6.97

=> Vs =  [(4 / 3(6.97)) (2.65 - 1) x 9.81 x 0.0002]^0.5

=> Vs = 0.025 m/s

Since , Vs = 0.025 m/s in teration 2 and 3, we can conclude that Vs = 0.025 m/s

Now, Vc = [ (8 b / f ) (G -1) g D]^0.5

Putting given values,

  Vc = [(8 x 0.04 /0.03) x (2.65 - 1) x 9.81 x 0.0002]^0.5

=> Vc = 0.186 m/s

Given, Q = 10000 m3/d or 0.1157 m3/s

Therefore, A = Q / Vc = 0.1157 / 0.186 = 0.622 m2

Length = A Vc / Vs = 0.622 x 0.186 / 0.025 = 4.627 m 4.63 m

Hence, Width = Area / Depth = 0.622 / 3   = 0.207 m

Hence, volume of tank = Length x Width x Depth = 4.63 x 0.207 x 3 = 2.875 m3

Detention time = Volume / Flow rate = 2.875 / 0.1157 = 24.85 sec

Let air supply rate = 0.20 m3 / min per meter length

=> Air flow required = 0.20 m3 / min per meter x 4.63 m = 0.926 m3/min