Question

A wastewater with a flow rate of 0.035 m3 /s is discharged to a small stream (flow rate = 0.120 m3 /s) and is instantaneously mixed. The wastewater has an initial BOD of 240 mg/L just before being discharged into the stream and a k value of 0.22/d. What is the BOD of the stream water 890 miles downstream of the wastewater discharge point if the stream flows at a velocity of 280 miles per day and the BOD of the stream water just upstream of the point of wastewater discharge is negligible? The value of k does not change in the stream

Answer 1

Distance down the stream = 890 miles

velocity of the stream = 280 miles/day

Time to reach 890 miles downstream = Distance / velocity = 890/280 = 3.178 days

Given that before the wastewater stream,BOD of water stream is negligible. Thus the total BOD concentration after the wastewater enters the stream is :

BOD = Total flow of solids / Total flow of water

Total flow of solids = 0.035 * 10³ L/s * 240 mg/L = 8400 mg/s

Total flow of water = 0.120*10³ L/s + 0.035 * 10³ L/s = 155 L/s

BOD = Total flow of solids / Total flow of water = 8400 mg/s / 155 L/s = 54.194 mg/L

Given, k = 0.22/day

D_3.178 = D0(1-e^-kt) =54.194* ( 1 - e^-0.22*3.178) = 27.26 mg/L

The BOD at 890 miles downstream = 27.26 mg/L