Question

Wastewater is being treated in two well-mixed ponds in series. The flow rate of the wastewater is 4000 m3/d, and the ponds each have volumes of 20,000 m3 and 12,000 m3, respectively. The influent to the first pond has an ultimate BOD (i.e., L0) of 200 mg/L, and the system has been operating long enough to reach steady-state. If kd is 0.35 d-1 in each pond, how much BOD remains in the effluent as the water exits the second pond?

Answer 1

The BOD in the effluent is 12.16 mg/L.

Step-by-step explanation:

Form the given data

Flow rate=4000 m^3/day

The volume of pond 1 is 20000 m^3

The volume of pond 2 is 12000 m^3

So the Detention time is given as

`DT=(V_(P1)+V_(P2))/(FR)\\DT=(20000+12000)/(4000)\\DT=8 days`

So for the given value i.e. Lo=200 mg/L and K=0.35 day^-1

The value for 8 days is given as

`L=L_oe^(-kt)\\L=200e^(-0.35* 8)\\L=12.16 mg/L`

So the BOD in the effluent is 12.16 mg/L.