CVL 300 Lecture Notes - Lecture 7: Ryerson University, Mass Balance, Mississippi Highway 1
15 Dec 2015
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Five million gallons per day (mgd) of wastewater, with a concentration of 10. 0 mg/l of a conservative pollutant, is released into a stream having an upstream flow of 10 mgd and pollutant concentration of 3. 0 mg/l. How many pounds of substance per day pass a given spot downstream (you may. No other inputs or outputs to the mixing zone want the conversions 3. 785 l/gal and 2. 2 kg/lbm from appendix a) Assumptions: wastewater will be completely mixed with stream. Mass rate of pollutants from the wastewater = 10 mgd * 3. 0 mg/l. Mass rate of pollutants from upstream = 5 mgd * 10. 0 mg/l. Volumetric balance at the mixing zone implies the total flow after mixing = 10+5 = 15 mgd. Let the concentration after mixing be c: 10*3+5*10=(10+5)*c. C= 80/15 = 5. 33 mg/l: the concentration after mixing is 5. 33 mg/l and the flow is 15 mgd. = 5. 33 mg/l*15*106 gal/day*3. 785 l/gal*1kg/106 mg*2. 2 lb/kg = 666 lb/day.
