Question

There are two different suspended particles with settling velocities of 0.1 and 0.2 mm/s. The particles are present at 60:40 ratio with an overflow rate of (15+N*0.2) m/d. What is the overall removal efficiency of the system?

N=2

Answer 1

Given: Settling velocity of particle 1 (Vs1) = 0.1mm/s

Concentration of the particle 1=60%

Settling velocity of particle 2 (Vs2) = 0.2mm/s

Concentration of the particle 2=40%

Overflow rate (Vo) = 15+N*0.2

where N is given =2

then Overflow rate (Vo) = 15+N*0.2

(Vo) = 15+2*0.2

Vo=15.4m/d

to find Vo in mm/s

Vo= 15.4 x 1000/(86400)

Vo=0.178mm/s

We know that Particles having settling velocities more than the overflow rate will be 100% removed and other particles will be rmoved in the ratio of (Vs/Vo)

percentage of the particles 1 removed

=0.1 x 60/0.178

=33.70%

As Vs2> Vo All the particle having settling velocity =0.2mm/s will get removed

percentage of the particles 2 removed=40%

Total percentage of the particles removed =33.70%+40%=73.70%

So overall removal efficiency of the system will be 73.70%