Question

A city of 200,000 people deposits 37 m3/s of sewage having a BOD of 28.0 mg/L and 1.8 mg/L of DO into a river that has a flow rate of 250 m3/s and a flow speed of 0.5 m/s. Just upstream of the release point, the river has a BOD of 3.6 mg/L and a DO of 7.6 mg/L. The saturation value of DO is 8.5 mg/L. The deoxygenation coefficient kdis 0.61/day and the reaeration coefficient kr is 0.76/day. Assuming complete and instantaneous mixing of the sewage and river find a. The initial oxygen deficit and ultimate BOD just downstream of the outfall b. The time and distance to reach the minimum DO c. The minimumDO d. The DO that could be expected 10 miles downstream

Answer 1

Step-by-step explanation:

a. The initial oxygen deficit can be calculated as:

ODI = (DOsat - DOupstream) + (BODinflow / kD)

ODI = (8.5 - 7.6) + (28.0 / 0.61)

ODI = 22.0 mg/L

The ultimate BOD just downstream of the outfall can be calculated as:

BODultimate = (ODI * kD) + BODinflow

BODultimate = (22.0 * 0.61) + 28.0

BODultimate = 41.42 mg/L

b. The time and distance to reach the minimum DO can be calculated using the Streeter-Phelps equation:

DO = DOsat - (DOsat - DOupstream) * e^(-krt) - (BOD / kD) * (1 - e^(-kdt))

Setting DO to the minimum value of 1.8 mg/L, we can solve for the time and distance:

1.8 = 8.5 - (8.5 - 7.6) * e^(-0.76t) - (41.42 / 0.61) * (1 - e^(-0.61t))

t = 7.05 days

d = v * t

d = 0.5 * 86400 * 7.05

d = 30,096 m

c. The minimum DO can be calculated by substituting the time and distance values into the Streeter-Phelps equation:

DO = 8.5 - (8.5 - 7.6) * e^(-0.767.05) - (41.42 / 0.61) * (1 - e^(-0.617.05))

DO = 1.26 mg/L

d. The DO that could be expected 10 miles downstream can be estimated by assuming complete mixing and using the Streeter-Phelps equation with a distance of 10 miles:

d = 16093.4 m

DO = 8.5 - (8.5 - 7.6) * e^(-0.76*(d/v)) - (BODinflow / kD) * (1 - e^(-kD*(d/v)))

DO = 6.29 mg/L