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

Answer

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

$C_{down}=\dfrac{0.9\times 50}{8.7+0.9}$

$C_{down}=\dfrac{0.9\times 50}{9.6}$

               = 4.69 mg/L

b) discharge = 9.6 m³/s

cross sectional area = 10 m²

velocity steam = $\dfrac{8.7+0.9}{10}$

                        = 0.96 m/s

time taken to move 50 km down stream =$\dfrac{50 \times 1000}{0.96}$

                                                                  = 52083.3 s

                                                                  = $\dfrac{52083.3}{3600\times 24}$

                                                                   = 0.6 days

now,

$C_t=C_0e^{-kt}$

$C_t=4.6875\ e^{-0.2\times 0.6}$

$C_t = 4.16 mg/L$