Assignment 3 Fall 2013_SOLN_revised

15mgvssmgvss008d457mgl 506 mgvssl iii vss that

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Unformatted text preview: = [1 + 9d(0.15mgVSS/mgVSS)(0.08/d)](45.7mg/L) = 50.6 mgVSS/L iii.) VSS that would be measured in the process effluent Solution: VSS in effluent = growth related biomass + nbVSS = 50.6 mgVSS/l + 50 m/l = 100.6 mg/L iv.) Oxygen requirements Solution: O2 Required = bCODremoved - bCODbiomass = Q (So - S) - 1.42mgCOD/mgVSS Px Where Px = QXvss Therefore: O2 Required = Q (So - S - 1.42VSS) = 2000m3/d[ 200mg/l - 3.7mg/l - 1.42(50.6mg/L)](1000l/m3)(1kg/10^6mg) = 249 kg/d Assume 1.42 kg COD/kg VSS 4.) A wastewater with a flow of 3800 m3/day and a soluble bCOD of 200 mg/L is to be treated in an aerated lagoon. The summer air temperature is expected to be 30oC while the winter air temperature is anticipated to be 10oC. The wastewater has a temperat...
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This document was uploaded on 02/10/2014.

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