Question

Wastewater from a small town is treated in a trickling filter which is preceded by a screen, sand-removal and primary clarification. The maximum hydraulic capacity, or peak flow, of the treatment plant is 1,500 m³/h. The dry weather flow (Qdwf) amounts 7,500 m³/day, with a BOD concentration of 300 mg/l.

The maximum scouring velocity or slip velocity is 0.3 m/s and the maximum surface load or Hazen velocity is 40 m/h.


Assume 4 channels. Take a W:L ratio of 1:15

a. Calculate the vertical surface area Avertical in m²:
b.Calculate the horizontal surface area Ahorizontal in m²:

Answer 1

Qdesing= (1500m3/s)/(3600s/h)= 0.42m3/s.
Qchannel= (0.42m3/s)/4channels = 0.104m3/s

assumptions: Depth = 0.7 and settling velocity of smallest grit particle V= 0.012 (derived by stoke´s law).

-Detention time:
t= Depth/vel ; t=(0.7m)/(0.012m/s) = 60s

-theorical length:
L= Vh*t ; L=(0.3m/s)*(60s)= 18m

-channel volume:
Vol= Q*t ; Q=(0.104m3/s)*(60s)= 6.24m3

-cross section area:
A= Vol/Length ; A= (6.24m3)/(18m)= 0.35m2

-channel width:
W= A/D ; W= (0.35m2)/(0.7m)= 0.5m

-And now we can calculate the vertical and horizontal area
Vertical area = 0.5m*0.7 = 0.35m2
Horizontal area = 0.5m*18m = 9m2

:)

Answer 2

Its wrong!