HW2 Solutions - L day-= Suspended solids removed as sludge...

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CEE 3510 Answer Key Problem Set No. 2 Feb. 17, 2010 1. Rate of mass flow in = Rate of mass flow out + Rate of mass storage. 100 kg/day = (1 kg/day + 2 kg/day) + Rate of mass storage Rate of mass storage = (100 – 3) kg/day = 97 kg/day. 2. Suspended solids loading to clarifier = 3 6 3 3 x L mg kg kg 3780 10 220 10 832 day L mg day (1m 10 L) - = = Suspended solids in clarified effluent = 3 6 x x L kg kg 3780 10 5 10 18.9 day
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Unformatted text preview: L day-= Suspended solids removed as sludge = 831.6 18.9 = 812.9 kg/day. 3. Mass balance on contaminant: in = out + decay i) Q in = Q out = 10 3 m 3 /day V = 10 6 m 3 C = concentration in pond in in out in Q C Q C R V mg R 0.001 L day Q C R V mg C 2.4 Q L = + =--= = ii) in in out in Q C Q C k C V 1 k 0.01 day Q C mg C 0.309 Q k V L = + = = = +...
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This note was uploaded on 03/01/2010 for the course CEE 3510 taught by Professor Lion during the Spring '10 term at Cornell University (Engineering School).

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HW2 Solutions - L day-= Suspended solids removed as sludge...

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