A wastewater treatment plant discharges 1 m3/s of effluent with an ultimate BOD of 40 mg/L into a stream flowing at 10 m3/s. Just upstream of the discharge point, the stream has an ultimate BOD of 3 mg/L. The deoxygenation constant (kd) is estimated to be 0.22 1/d. (a) Assuming complete and instantaneous mixing, find the ultimate BOD of the mixture of the waste and the river just downstream of the outfall. (b) Assuming a constant cross-sectional area for the stream equal to 55 m2, what ultimate BOD would you expect to find at a point 10,000 m downstream

(a) 6.36 mg/L

(b) 5.60 mg/L

Explanation:

(a)

Using the formula below to find the required ultimate BOD of the mixture.

where;

Q_w = volumetric flow rate wastewater

Q_r = volumetric flow rate of the river just upstream of the discharge point

L_w = ultimate BOD of wastewater

Replacing the given values:

(b)

The Ultimate BOD is estimated as follows:

Recall that:

replacing;

distance with 10000 m and speed with

time (t) = 0.578 days

Finally;

here;

k = rate of coefficient reaction

Thus, the ultimate BOD = 5.60 mg/L

this is true

explanation:

i looked it up

answer: 500 wph.

explanation: