Answer:
a) 35%
b) yes it can be improved by moving the tray near the top
Tray should be located ( 1 to 2 meters below surface )
max removal efficiency ≈ 70%
c) The maximum removal will drop as the particle settling velocity = 0.5 m/h
Step-by-step explanation:
Given data:
flow rate = 8000 m^3/d
Detention time = 1h
depth = 3m
Full length movable horizontal tray : 1m below surface
a) Determine percent removal of particles having a settling velocity of 1m/h
velocity of critical sized particle to be removed = Depth / Detention time
= 3 / 1 = 3m/h
The percent removal of particles having a settling velocity of 1m/h ≈ 35%
b) Determine if the removal efficiency of the clarifier can be improved by moving the tray, the location of the tray and the maximum removal efficiency
The tray should be located near the top of the tray ( i.e. 1 to 2 meters below surface ) because here the removal efficiency above the tray will be 100% but since the tank is quite small hence the
Total Maximum removal efficiency
= percent removal
+ percent removal
= ( d
,v
) .
+ ( d
,v
) .
= 100
hence max removal efficiency ≈ 70%
c) what is the effect of moving the tray would be if the particle settling velocity were equal to 0.5m/h?
The maximum removal will drop as the particle settling velocity = 0.5 m/h