Question

data:highrate trickling filter

pipe dia = 30m

deep = 1m

swage flow = 4.5mld

recirulation ratio = 1.4

bod of raw sewage = 250mg/l

bod removed in primary clarifier = 25%

efficiency?

(ans:74%)

Answer 1

The BOD of the raw sewage is 250 mg/l.

BOD removed in primary clarifier is 25%

Therefore, BOD removed in primary clarifier is = 250 x 0.25 = 62.5 mg/l

Thus the remaining BOD = (250 - 62.5) = 187.5 mg/l.

The BOD load applied to the filter = 4.5 x 10^6 x 187.5/ 10^6 = 843.75 kg/day. (since the sewage flow is 4.5 mld )

The efficiency equation of the trickling filter is given by,

E=              100                         
     1+0.44(F_BOD/V.Rf)^1/2

Where, E is the efficiency, F_BOD is the BOD load of filter , V is the volume of filter, Rf is the recirculation ratio

Therefore, E=              100                       = 71.1 % ( volume(V) = = 706.5 m^3) diameter of filter = 30m
1+0.44(843.75/706.5 x 1.4)^1/2

Thus the efficiency of the filter is 71.1%

(The equation of the efficiency of trickling filter is according to National Research Council)