Question

The rate of flow through an ideal clarifier is 8000m3 /d, the detention time is 1h and the depth is 3m. If a full-length movable horizontal tray is set 1m below the surface of the water, a) determine the percent removal of particles having a settling velocity of 1m/h. b) Could the removal efficiency of the clarifier to be improved by moving the tray? If so, where should the tray be located and what would be the maximum removal efficiency? c) What effect would moving the tray have if the particle settling velocity were equal to 0.5m/h?

Answer 1

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

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$_{above}$ + percent removal$_{below}$

= ( d$_{a}$,v$_{p}$ ) . $\frac{d_{a} }{depth}$  + ( d$_{a}$,v$_{p}$ ) . $\frac{depth - d_{a} }{depth}$  = 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