Question 4 (Sedimentation) A drinking water treatment plant recently conducted a

Question

Question 4 (Sedimentation) A drinking water treatment plant recently conducted a test to characterize the size distribution and settling behavior of particles in the influent of their sedimentation basin. Based on the test result:s (see the table below), the plant operators are considering adjust the basin hydraulic residence time to improve the particle removal efficiency. What would be the hydraulic residence time required to achieve 84% particle removal if the overflow rate falls between 0.91 and 1.07 m/h (i.e., 0.91 m/h

Explanation / Answer

Let us consider that the Overflow rate be Vo , such that Vo falls between 0.91 and 1.07m/h.

Now for 100% removal of a particle of settling velocity Vs

Vs>Vo

And those particles having settling velocity less than overflow rate will not remove by 100%,but some percentage of that particle will also get removed,given by

P=Vs/Vo.

Now,As particles having settling velocity greater than 0.91m/h will settle by 100 percent.

Thus overall percentage removal is given by

P= (6*0.21/Vo)+ (10*0.39/Vo)+ (14*0.68/Vo) +22+21+16+11

As we have to achieve 84% removal that is P= 84%

84=(6*0.21/Vo)+ (10*0.39/Vo)+ (14*0.68/Vo) +22+21+16+11

Vo = 1.0485 m/h

As, Detention time Dt =Depth / Overflow rate

Dt = 2.88/1.0485 hr = 2.75 hours.

Please provide feedback. :)

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