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Design a low rate filter to treat 6.0 Mld of sewage of BOD of 210 mg/l. The final effluent should be 30 mg/l and organic loading rate is 320 g/m^3/d. Find Hydraulic loading rate?(a) 326.18 g/d/m^3(b) 926.18 g/d/m^3(c) 126.18 g/d/m^3(d) 526.18 g/d/m^3The question was asked during a job interview.This interesting question is from Principle for the Preparation of Water Supply in division Valuation, Reports Technical and Design Data of Civil Engineering Drawing

Answer»

Right answer is (a) 326.18 g/d/m^3

The explanation is: Assume 30% of BOD load removed in primary sedimentation i.e., = 210 X 0.30 = 63 mg/l. Remaining BOD = 210 – 63 = 147 mg/l.

Percent of BOD REMOVAL required = (147-30) x 100/147 = 80%

BOD load applied to the FILTER = flow x conc. of sewage (kg/d) = 6 x 106 x 147/106 = 882 kg/d

To find out filter volume, using NRC equation

Depth of filter = 1.5 m, Fiter area = 2704/1.5 = 1802.66 m^2, and Diameter = 48 m < 60 m

Hydraulic loading rate = 6 x 106/103 x 1/1802.66 = 3.33m3/d/m2 < 4 hence o.k.

Organic loading rate = 882 x 1000 / 2704 = 326.18 g/d/m^3 which is approx. equal to 320.


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