Question

Poorly treated municipal wastewater is discharged to a stream. The river flow rate upstream of the discharge point is Qu/s = 8.7 m3/s. The discharge occurs at a flow of Qd= 0.9 m3/s and has a BOD concentration of 50.0 mg/L. Assuming that the upstream BOD concentration is negligible. (a)What is the BOD concentration just downstream of the discharge point? (b) If the stream has a cross-sectional area of 10 m2, what would the BOD concentration be 50 km downstream? (BOD is removed with a first-order decay rate constant equal to 0.20/da

Answer 1

given,

Q u = 8.7 m³/s

Q d= 0.9 m³/s

BOD concentration = 50 mg/L

a) BOD concentration at the down stream

`C_(down)=(0.9* 50)/(8.7+0.9)`

`C_(down)=(0.9* 50)/(9.6)`

= 4.69 mg/L

b) discharge = 9.6 m³/s

cross sectional area = 10 m²

velocity steam =
`(8.7+0.9)/(10)`

= 0.96 m/s

time taken to move 50 km down stream =
`(50 * 1000)/(0.96)`

= 52083.3 s

=
`(52083.3)/(3600* 24)`

= 0.6 days

now,

`C_t=C_0e^(-kt)`

`C_t=4.6875\ e^(-0.2* 0.6)`

`C_t = 4.16 mg/L`