Question

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^3

The 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 1

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.