CVL 300 Lecture Notes - Lecture 6: Mass Balance, Ryerson University, Environmental Engineering
15 Dec 2015
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Wastewater flow (qw) = 5000 m3/d = 5000/(24*60) = 3. 47 m3/min. Downstream river bod after mixing (cm) = 1 mg/l. Downstream river flow (qm) = qr + qw = 137. 7 + 3. 47 = 141. 2 m3/min. Qm cm = qr cr + qw cw. Allowable bod from wastewater= qw cw = 5000 * 40. 7 * 1000/106 = 204 kg/day. Degree of treatment = (250 40. 7)/250 = 84: for the problem 1 above, the dissolved oxygen concentration (do) and temperature (t) of the treated wastewater and the river are listed below: Treated wastewater: dow = 2 mg/l; t = 20 c. River upstream: dor = 9 mg/l; t = 15 c. Determine the dom, tm, and initial oxygen deficit (do) after the complete mixing of the treated wastewater and the river. It is assumed that air pressure is 1 atm and it contains. 20% oxygen and the saturated dissolved oxygen concentration (dos) equals 9. 75 mg/l.
